BACKGROUNDThermodynamics is an exciting and fascinating subject that deals withenergy, which is essential for sustenance of life, and thermodynamics haslong been an essential part of engineering curricula all over the world. It hasa broad application area ranging from microscopic organisms to commonhousehold appliances, transportation vehicles, power generation systems,and even philosophy. This introductory book contains sufficient material fortwo sequential courses in thermodynamics. Students are assumed to have anadequate background in calculus and physics.

OBJECTIVESThis book is intended for use as a textbook by undergraduate engineeringstudents in their sophomore or junior year, and as a reference book for prac-ticing engineers. The objectives of this text are

• To cover the basic principles of thermodynamics.

• To present a wealth of real-world engineering examples to givestudents a feel for how thermodynamics is applied in engineeringpractice.

• To develop an intuitive understanding of thermodynamics by empha-sizing the physics and physical arguments.

It is our hope that this book, through its careful explanations of conceptsand its use of numerous practical examples and figures, helps studentsdevelop the necessary skills to bridge the gap between knowledge and theconfidence to properly apply knowledge.

PHILOSOPHY AND GOALThe philosophy that contributed to the overwhelming popularity of the prioreditions of this book has remained unchanged in this edition. Namely, ourgoal has been to offer an engineering textbook that

• Communicates directly to the minds of tomorrow’s engineers in asimple yet precise manner.

• Leads students toward a clear understanding and firm grasp of thebasic principles of thermodynamics.

• Encourages creative thinking and development of a deeper under-standing and intuitive feel for thermodynamics.

• Is read by students with interest and enthusiasm rather than beingused as an aid to solve problems.

PREFACE

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Special effort has been made to appeal to students’ natural curiosity andto help them explore the various facets of the exciting subject area of ther-modynamics. The enthusiastic responses we have received from users ofprior editions—from small colleges to large universities all over the world—indicate that our objectives have largely been achieved. It is our philosophythat the best way to learn is by practice. Therefore, special effort is madethroughout the book to reinforce material that was presented earlier.

Yesterday’s engineer spent a major portion of his or her time substitutingvalues into the formulas and obtaining numerical results. However, formulamanipulations and number crunching are now being left mainly to comput-ers. Tomorrow’s engineer will need a clear understanding and a firm grasp ofthe basic principles so that he or she can understand even the most complexproblems, formulate them, and interpret the results. A conscious effort ismade to emphasize these basic principles while also providing students witha perspective of how computational tools are used in engineering practice.

The traditional classical, or macroscopic, approach is used throughout thetext, with microscopic arguments serving in a supporting role as appropri-ate. This approach is more in line with students’ intuition and makes learn-ing the subject matter much easier.

NEW IN THIS EDITIONAll the popular features of the previous editions are retained while new onesare added. With the exception of reorganizing the first law coverage andupdating the steam and refrigerant properties, the main body of the textremains largely unchanged. The most significant changes in this fifth edi-tion are highlighted below.

EARLY INTRODUCTION OF THE FIRST LAW OF THERMODYNAMICSThe first law of thermodynamics is now introduced early in the new Chapter2, “Energy, Energy Transfer, and General Energy Analysis.” This introduc-tory chapter sets the framework of establishing a general understanding ofvarious forms of energy, mechanisms of energy transfer, the concept ofenergy balance, thermo-economics, energy conversion, and conversion effi-ciency using familiar settings that involve mostly electrical and mechanicalforms of energy. It also exposes students to some exciting real-world appli-cations of thermodynamics early in the course, and helps them establish asense of the monetary value of energy.

SEPARATE COVERAGE OF CLOSED SYSTEMS AND CONTROL VOLUME ENERGY ANALYSESThe energy analysis of closed systems is now presented in a separate chap-ter, Chapter 4, together with the boundary work and the discussion ofspecific heats for both ideal gases and incompressible substances. The con-servation of mass is now covered together with conservation of energy innew Chapter 5. A formal derivation of the general energy equation is alsogiven in this chapter as the Topic of Special Interest.

REVISED COVERAGE OF COMPRESSIBLE FLOWThe chapter on compressible flow that deals with compressibility effects(now Chapter 17) is greatly revised and expanded. This chapter now includes

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coverage of oblique shocks and flow with heat transfer (Rayleigh flow) withsome exciting photographs and extended discussions of shock waves.

UPDATED STEAM AND REFRIGERANT-134A TABLESThe steam and refrigerant-134a tables are updated using the most currentproperty data from EES. Tables A-4 through A-8 and A-11 through A-13, aswell as their counterparts in English units, have all been revised. All the exam-ples and homework problems in the text that involve steam or refrigerant-134a are also revised to reflect the small changes in steam and refrigerantproperties. An added advantage of this update is that students will get thesame result when solving problems whether they use steam or refrigerantproperties from EES or property tables in the appendices.

OVER 300 NEW COMPREHENSIVE PROBLEMSThis edition includes over 300 new comprehensive problems that comemostly from industrial applications. Problems whose solutions require para-metric investigations, and thus the use of a computer, are identified by acomputer-EES icon, as before.

CONTENT CHANGES AND REORGANIZATIONThe noteworthy changes in various chapters are summarized below forthose who are familiar with the previous edition.

• Chapter 1 is greatly revised, and its title is changed to “Introductionand Basic Concepts.” A new section Density and Specific Gravity anda new subsection The International Temperature Scale of 1990 areadded. The sections Forms of Energy and Energy and the Environmentare moved to new Chapter 2, and the Topic of Special Interest Ther-modynamic Aspects of Biological Systems is moved to new Chapter 4.

• The new Chapter 2 “Energy, Energy Transfer, and General EnergyAnalysis” mostly consists of the sections Forms of Energy and Energyand the Environment moved from Chapter 1, Energy Transfer by Heatand Energy Transfer by Work, and Mechanical Forms of Energy fromChapter 3, The First Law of Thermodynamics from Chapter 4, andEnergy Conversion Efficiencies from Chapter 5. The Topic of SpecialInterest in this chapter is Mechanisms of Heat Transfer moved fromChapter 3.

• Chapter 3 “Properties of Pure Substance” is essentially the previousedition Chapter 2, except that the last three sections on specific heatsare moved to new Chapter 4.

• Chapter 4 “Energy Analysis of Closed Systems” consists of MovingBoundary Work from Chapter 3, sections on Specific Heats fromChapter 2, and Energy Balance for Closed Systems from Chapter 4.Also, the Topic of Special Interest Thermodynamic Aspects of Biolog-ical Systems is moved here from Chapter 1.

• Chapter 5 “Mass and Energy Analysis of Control Volumes” consistsof Mass Balance for Control Volumes and Flow Work and the Energyof a Flowing Fluid from Chapter 3 and the sections on EnergyBalance for Steady- and Unsteady-Flow Systems from Chapter 4. The

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Topic of Special Interest Refrigeration and Freezing of Foods isdeleted and is replaced by a formal derivation of the General EnergyEquation.

• Chapter 6 “The Second Law of Thermodynamics” is identical to theprevious edition Chapter 5, except the section Energy ConversionEfficiencies is moved to Chapter 2.

• Chapters 7 through 15 are essentially identical to the previous editionChapters 6 through 14, respectively.

• Chapter 17 “Compressible Flow” is an updated version of the previ-ous edition Chapter 16. The entire chapter is greatly revised, the sec-tion Flow Through Actual Nozzles and Diffusers is deleted, and a newsection Duct Flow with Heat Transfer and Negligible Friction(Rayleigh Flow) is added.

• In Appendices 1 and 2, the steam and refrigerant-134a tables (Tables4 through 8 and 11 through 13) are entirely revised, but the tablenumbers are kept the same. The tables for isentropic compressibleflow functions and the normal shock functions (Tables A-32 and A-33) are updated and plots of functions are now included. Also,Rayleigh flow functions are added as Table A-34. Appendix 3 Intro-duction to EES is moved to the Student Resources DVD that comespackaged free with the text.

• The conversion factors on the inner cover pages and the physical con-stants are updated, and some nomenclature symbols are revised.

LEARNING TOOLSEMPHASIS ON PHYSICSA distinctive feature of this book is its emphasis on the physical aspects ofthe subject matter in addition to mathematical representations and manipula-tions. The authors believe that the emphasis in undergraduate educationshould remain on developing a sense of underlying physical mechanismsand a mastery of solving practical problems that an engineer is likely to facein the real world. Developing an intuitive understanding should also makethe course a more motivating and worthwhile experience for students.

EFFECTIVE USE OF ASSOCIATIONAn observant mind should have no difficulty understanding engineering sci-ences. After all, the principles of engineering sciences are based on oureveryday experiences and experimental observations. Therefore, a physical,intuitive approach is used throughout this text. Frequently, parallels aredrawn between the subject matter and students’ everyday experiences sothat they can relate the subject matter to what they already know. Theprocess of cooking, for example, serves as an excellent vehicle to demon-strate the basic principles of thermodynamics.

SELF-INSTRUCTINGThe material in the text is introduced at a level that an average student canfollow comfortably. It speaks to students, not over students. In fact, it isself-instructive. The order of coverage is from simple to general. That is, it

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starts with the simplest case and adds complexities gradually. In this way,the basic principles are repeatedly applied to different systems, and studentsmaster how to apply the principles instead of how to simplify a general for-mula. Noting that the principles of sciences are based on experimentalobservations, all the derivations in this text are based on physical arguments,and thus they are easy to follow and understand.

EXTENSIVE USE OF ARTWORKFigures are important learning tools that help students “get the picture,” andthe text makes very effective use of graphics. The fifth edition of Thermody-namics: An Engineering Approach contains more figures and illustrationsthan any other book in this category. This edition incorporates an expandedphoto program and updated art style. Figures attract attention and stimulatecuriosity and interest. Most of the figures in this text are intended to serveas a means of emphasizing some key concepts that would otherwise gounnoticed; some serve as page summaries. The popular cartoon feature“Blondie” is used to make some important points in a humorous way andalso to break the ice and ease the nerves. Who says studying thermodynam-ics can’t be fun?

LEARNING OBJECTIVES AND SUMMARIESEach chapter begins with an overview of the material to be covered andchapter-specific learning objectives. A summary is included at the end ofeach chapter, providing a quick review of basic concepts and important rela-tions, and pointing out the relevance of the material.

NUMEROUS WORKED-OUT EXAMPLES WITH A SYSTEMATIC SOLUTIONS PROCEDUREEach chapter contains several worked-out examples that clarify the materialand illustrate the use of the basic principles. An intuitive and systematicapproach is used in the solution of the example problems, while maintainingan informal conversational style. The problem is first stated, and the objec-tives are identified. The assumptions are then stated, together with their jus-tifications. The properties needed to solve the problem are listed separately,if appropriate. Numerical values are used together with their units to empha-size that numbers without units are meaningless, and that unit manipulationsare as important as manipulating the numerical values with a calculator. Thesignificance of the findings is discussed following the solutions. Thisapproach is also used consistently in the solutions presented in the instruc-tor’s solutions manual.

A WEALTH OF REAL-WORLD END-OF-CHAPTER PROBLEMSThe end-of-chapter problems are grouped under specific topics to makeproblem selection easier for both instructors and students. Within eachgroup of problems are Concept Questions, indicated by “C,” to check thestudents’ level of understanding of basic concepts. The problems underReview Problems are more comprehensive in nature and are not directly tiedto any specific section of a chapter—in some cases they require review ofmaterial learned in previous chapters. Problems designated as Design andEssay are intended to encourage students to make engineering judgments, toconduct independent exploration of topics of interest, and to communicate

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their findings in a professional manner. Problems designated by an “E” arein English units, and SI users can ignore them. Problems with the aresolved using EES, and complete solutions together with parametric studiesare included on the enclosed DVD. Problems with the are comprehen-sive in nature and are intended to be solved with a computer, preferablyusing the EES software that accompanies this text. Several economics- andsafety-related problems are incorporated throughout to enhance cost andsafety awareness among engineering students. Answers to selected problemsare listed immediately following the problem for convenience to students. Inaddition, to prepare students for the Fundamentals of Engineering Exam(that is becoming more important for the outcome-based ABET 2000 crite-ria) and to facilitate multiple-choice tests, over 200 multiple-choice prob-lems are included in the end-of-chapter problem sets. They are placed underthe title Fundamentals of Engineering (FE) Exam Problems for easy recog-nition. These problems are intended to check the understanding of funda-mentals and to help readers avoid common pitfalls.

RELAXED SIGN CONVENTIONThe use of a formal sign convention for heat and work is abandoned as itoften becomes counterproductive. A physically meaningful and engagingapproach is adopted for interactions instead of a mechanical approach. Sub-scripts “in” and “out,” rather than the plus and minus signs, are used to indi-cate the directions of interactions.

PHYSICALLY MEANINGFUL FORMULASThe physically meaningful forms of the balance equations rather than for-mulas are used to foster deeper understanding and to avoid a cookbookapproach. The mass, energy, entropy, and exergy balances for any systemundergoing any process are expressed as

Mass balance:

Energy balance:

Net energy transfer Change in internal, kinetic,by heat, work, and mass potential, etc., energies

Entropy balance:

Net entropy transfer Entropy Changeby heat and mass generation in entropy

Exergy balance:

Net exergy transfer Exergy Changeby heat, work, and mass destruction in exergy

These relations reinforce the fundamental principles that during an actualprocess mass and energy are conserved, entropy is generated, and exergy isdestroyed. Students are encouraged to use these forms of balances in earlychapters after they specify the system, and to simplify them for the particu-lar problem. A more relaxed approach is used in later chapters as studentsgain mastery.

X in � X out � X destroyed � ¢X system

Sin � Sout � Sgen � ¢Ssystem

E in � Eout � ¢E system

min � mout � ¢msystem

xxii | Preface

⎫⎪⎬⎪⎭ ⎫⎪⎬⎪⎭

⎫⎬⎭⎫⎪⎬⎪⎭ ⎫⎪⎬⎪⎭

⎫⎪⎬⎪⎭⎫⎪⎬⎪⎭ ⎫⎪⎬⎪⎭

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A CHOICE OF SI ALONE OR SI/ENGLISH UNITSIn recognition of the fact that English units are still widely used in someindustries, both SI and English units are used in this text, with an emphasison SI. The material in this text can be covered using combined SI/Englishunits or SI units alone, depending on the preference of the instructor. Theproperty tables and charts in the appendices are presented in both units,except the ones that involve dimensionless quantities. Problems, tables, andcharts in English units are designated by “E” after the number for easyrecognition, and they can be ignored by SI users.

TOPICS OF SPECIAL INTERESTMost chapters contain a section called “Topic of Special Interest” whereinteresting aspects of thermodynamics are discussed. Examples includeThermodynamic Aspects of Biological Systems in Chapter 4, HouseholdRefrigerators in Chapter 6, Second-Law Aspects of Daily Life in Chapter 8,and Saving Fuel and Money by Driving Sensibly in Chapter 9. The topicsselected for these sections provide intriguing extensions to thermodynamics,but they can be ignored if desired without a loss in continuity.

GLOSSARY OF THERMODYNAMIC TERMSThroughout the chapters, when an important key term or concept is intro-duced and defined, it appears in boldface type. Fundamental thermo-dynamic terms and concepts also appear in a glossary located on ouraccompanying website (www.mhhe.com/cengel). This unique glossary helpsto reinforce key terminology and is an excellent learning and review tool forstudents as they move forward in their study of thermodynamics. In addi-tion, students can test their knowledge of these fundamental terms by usingthe flash cards and other interactive resources.

CONVERSION FACTORSFrequently used conversion factors and physical constants are listed on theinner cover pages of the text for easy reference.

SUPPLEMENTSThe following supplements are available to the adopters of the book.

STUDENT RESOURCES DVDPackaged free with every new copy of the text, this DVD provides a wealthof resources for students including Physical Experiments in Thermodynam-ics, an Interactive Thermodynamics Tutorial, and EES Software.

Physical Experiments in Thermodynamics: A new feature of this book isthe addition of Physical Experiments in Thermodynamics created by RonaldMullisen of the Mechanical Engineering Department at California Polytech-nic State University (Cal Poly), San Luis Obispo. At appropriate places in themargins of Chapters 1, 3, and 4, photos with captions show physical experi-ments that directly relate to material covered on that page. The captions

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refer the reader to end-of-chapter problems that give a brief description ofthe experiments. These experiments cover thermodynamic properties, ther-modynamic processes, and thermodynamic laws. The Student ResourcesDVD contains complete coverage of the nine experiments. Each experimentcontains a video clip, a complete write-up including historical background,and actual data (usually in an Excel file). The results are also provided onthe website that accompanies the text, and they are password protected forinstructor use. After viewing the video and reading the write-up, the studentwill be ready to reduce the data and obtain results that directly connect withmaterial presented in the chapters. For all of the experiments the finalresults are compared against published information. Most of the experi-ments give final results that come within 10 percent or closer to these pub-lished values.

Interactive Thermodynamics Tutorial: Also included on the StudentResources DVD is the Interactive Thermodynamics Tutorial developed byEd Anderson of Texas Tech University. The revised tutorial is now tieddirectly to the text with an icon to indicate when students should refer tothe tutorial to further explore specific topics such as energy balance andisentropic processes.

Engineering Equation Solver (EES): Developed by Sanford Klein andWilliam Beckman from the University of Wisconsin–Madison, this softwarecombines equation-solving capability and engineering property data. EEScan do optimization, parametric analysis, and linear and nonlinear regres-sion, and provides publication-quality plotting capabilities. Thermodynam-ics and transport properties for air, water, and many other fluids are built in,and EES allows the user to enter property data or functional relationships.

ONLINE LEARNING CENTER (OLC)Web support is provided for the book on our Online Learning Center atwww.mhhe.com/cengel. Visit this robust site for book and supplement infor-mation, errata, author information, and further resources for instructors andstudents.

INSTRUCTOR’S RESOURCE CD-ROM (AVAILABLE TO INSTRUCTORS ONLY)This CD, available to instructors only, offers a wide range of classroompreparation and presentation resources including the solutions manual inPDF files by chapter, all text chapters and appendices as downloadable PDFfiles, and all text figures in JPEG format.

COSMOS CD-ROM (COMPLETE ONLINE SOLUTIONS MANUAL ORGANIZATION SYSTEM) (AVAILABLE TO INSTRUCTORS ONLY)This CD, available to instructors only, provides electronic solutions deliv-ered via our database management tool. McGraw-Hill’s COSMOS allowsinstructors to streamline the creation of assignments, quizzes, and tests byusing problems and solutions from the textbook—as well as their own cus-tom material.

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Chapter 9GAS POWER CYCLES

| 487

Two important areas of application for thermodynamicsare power generation and refrigeration. Both are usuallyaccomplished by systems that operate on a thermody-

namic cycle. Thermodynamic cycles can be divided into twogeneral categories: power cycles, which are discussed in thischapter and Chap. 10, and refrigeration cycles, which are dis-cussed in Chap. 11.

The devices or systems used to produce a net power outputare often called engines, and the thermodynamic cycles theyoperate on are called power cycles. The devices or systemsused to produce a refrigeration effect are called refrigerators,air conditioners, or heat pumps, and the cycles they operateon are called refrigeration cycles.

Thermodynamic cycles can also be categorized as gascycles and vapor cycles, depending on the phase of theworking fluid. In gas cycles, the working fluid remains in thegaseous phase throughout the entire cycle, whereas in vaporcycles the working fluid exists in the vapor phase during onepart of the cycle and in the liquid phase during another part.

Thermodynamic cycles can be categorized yet anotherway: closed and open cycles. In closed cycles, the workingfluid is returned to the initial state at the end of the cycle andis recirculated. In open cycles, the working fluid is renewed atthe end of each cycle instead of being recirculated. In auto-mobile engines, the combustion gases are exhausted andreplaced by fresh air–fuel mixture at the end of each cycle.The engine operates on a mechanical cycle, but the workingfluid does not go through a complete thermodynamic cycle.

Heat engines are categorized as internal combustionand external combustion engines, depending on how theheat is supplied to the working fluid. In external combustionengines (such as steam power plants), heat is supplied to theworking fluid from an external source such as a furnace, a

geothermal well, a nuclear reactor, or even the sun. In inter-nal combustion engines (such as automobile engines), this isdone by burning the fuel within the system boundaries. Inthis chapter, various gas power cycles are analyzed undersome simplifying assumptions.

ObjectivesThe objectives of Chapter 9 are to:

• Evaluate the performance of gas power cycles for which theworking fluid remains a gas throughout the entire cycle.

• Develop simplifying assumptions applicable to gas powercycles.

• Review the operation of reciprocating engines.

• Analyze both closed and open gas power cycles.

• Solve problems based on the Otto, Diesel, Stirling, andEricsson cycles.

• Solve problems based on the Brayton cycle; the Braytoncycle with regeneration; and the Brayton cycle withintercooling, reheating, and regeneration.

• Analyze jet-propulsion cycles.

• Identify simplifying assumptions for second-law analysis ofgas power cycles.

• Perform second-law analysis of gas power cycles.

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9–1 ■ BASIC CONSIDERATIONS IN THE ANALYSISOF POWER CYCLES

Most power-producing devices operate on cycles, and the study of powercycles is an exciting and important part of thermodynamics. The cyclesencountered in actual devices are difficult to analyze because of the pres-ence of complicating effects, such as friction, and the absence of sufficienttime for establishment of the equilibrium conditions during the cycle. Tomake an analytical study of a cycle feasible, we have to keep the complexi-ties at a manageable level and utilize some idealizations (Fig. 9–1). Whenthe actual cycle is stripped of all the internal irreversibilities and complexi-ties, we end up with a cycle that resembles the actual cycle closely but ismade up totally of internally reversible processes. Such a cycle is called anideal cycle (Fig. 9–2).

A simple idealized model enables engineers to study the effects of themajor parameters that dominate the cycle without getting bogged down in thedetails. The cycles discussed in this chapter are somewhat idealized, but theystill retain the general characteristics of the actual cycles they represent. Theconclusions reached from the analysis of ideal cycles are also applicable toactual cycles. The thermal efficiency of the Otto cycle, the ideal cycle forspark-ignition automobile engines, for example, increases with the compres-sion ratio. This is also the case for actual automobile engines. The numericalvalues obtained from the analysis of an ideal cycle, however, are not necessar-ily representative of the actual cycles, and care should be exercised in theirinterpretation (Fig. 9–3). The simplified analysis presented in this chapter forvarious power cycles of practical interest may also serve as the starting pointfor a more in-depth study.

Heat engines are designed for the purpose of converting thermal energy towork, and their performance is expressed in terms of the thermal efficiencyhth, which is the ratio of the net work produced by the engine to the totalheat input:

(9–1)

Recall that heat engines that operate on a totally reversible cycle, such asthe Carnot cycle, have the highest thermal efficiency of all heat enginesoperating between the same temperature levels. That is, nobody can developa cycle more efficient than the Carnot cycle. Then the following questionarises naturally: If the Carnot cycle is the best possible cycle, why do wenot use it as the model cycle for all the heat engines instead of botheringwith several so-called ideal cycles? The answer to this question is hardware-related. Most cycles encountered in practice differ significantly from theCarnot cycle, which makes it unsuitable as a realistic model. Each idealcycle discussed in this chapter is related to a specific work-producing deviceand is an idealized version of the actual cycle.

The ideal cycles are internally reversible, but, unlike the Carnot cycle,they are not necessarily externally reversible. That is, they may involve irre-versibilities external to the system such as heat transfer through a finite tem-perature difference. Therefore, the thermal efficiency of an ideal cycle, ingeneral, is less than that of a totally reversible cycle operating between the

hth �Wnet

Q inor hth �

wnet

qin

488 | Thermodynamics

OVEN

ACTUAL

IDEAL

175ºC

WATER

Potato

FIGURE 9–1Modeling is a powerful engineeringtool that provides great insight andsimplicity at the expense of some lossin accuracy.

P

Actual cycle

Ideal cycle

v

FIGURE 9–2The analysis of many complexprocesses can be reduced to amanageable level by utilizing someidealizations.

FIGURE 9–3Care should be exercised in the interpre-tation of the results from ideal cycles.

© Reprinted with special permission of KingFeatures Syndicate.

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same temperature limits. However, it is still considerably higher than thethermal efficiency of an actual cycle because of the idealizations utilized(Fig. 9–4).

The idealizations and simplifications commonly employed in the analysisof power cycles can be summarized as follows:

1. The cycle does not involve any friction. Therefore, the working fluiddoes not experience any pressure drop as it flows in pipes or devicessuch as heat exchangers.

2. All expansion and compression processes take place in a quasi-equilibrium manner.

3. The pipes connecting the various components of a system are well insu-lated, and heat transfer through them is negligible.

Neglecting the changes in kinetic and potential energies of the workingfluid is another commonly utilized simplification in the analysis of powercycles. This is a reasonable assumption since in devices that involve shaftwork, such as turbines, compressors, and pumps, the kinetic and potentialenergy terms are usually very small relative to the other terms in the energyequation. Fluid velocities encountered in devices such as condensers, boilers,and mixing chambers are typically low, and the fluid streams experience littlechange in their velocities, again making kinetic energy changes negligible.The only devices where the changes in kinetic energy are significant are thenozzles and diffusers, which are specifically designed to create large changesin velocity.

In the preceding chapters, property diagrams such as the P-v and T-s dia-grams have served as valuable aids in the analysis of thermodynamicprocesses. On both the P-v and T-s diagrams, the area enclosed by theprocess curves of a cycle represents the net work produced during the cycle(Fig. 9–5), which is also equivalent to the net heat transfer for that cycle.

Chapter 9 | 489

FIGURE 9–4An automotive engine with thecombustion chamber exposed.

Courtesy of General Motors

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The T-s diagram is particularly useful as a visual aid in the analysis of idealpower cycles. An ideal power cycle does not involve any internal irre-versibilities, and so the only effect that can change the entropy of the work-ing fluid during a process is heat transfer.

On a T-s diagram, a heat-addition process proceeds in the direction ofincreasing entropy, a heat-rejection process proceeds in the direction ofdecreasing entropy, and an isentropic (internally reversible, adiabatic)process proceeds at constant entropy. The area under the process curve on aT-s diagram represents the heat transfer for that process. The area under theheat addition process on a T-s diagram is a geometric measure of the totalheat supplied during the cycle qin, and the area under the heat rejectionprocess is a measure of the total heat rejected qout. The difference betweenthese two (the area enclosed by the cyclic curve) is the net heat transfer,which is also the net work produced during the cycle. Therefore, on a T-sdiagram, the ratio of the area enclosed by the cyclic curve to the area underthe heat-addition process curve represents the thermal efficiency of thecycle. Any modification that increases the ratio of these two areas will alsoincrease the thermal efficiency of the cycle.

Although the working fluid in an ideal power cycle operates on a closedloop, the type of individual processes that comprises the cycle depends onthe individual devices used to execute the cycle. In the Rankine cycle, whichis the ideal cycle for steam power plants, the working fluid flows through aseries of steady-flow devices such as the turbine and condenser, whereas inthe Otto cycle, which is the ideal cycle for the spark-ignition automobileengine, the working fluid is alternately expanded and compressed in a piston–cylinder device. Therefore, equations pertaining to steady-flow systemsshould be used in the analysis of the Rankine cycle, and equations pertainingto closed systems should be used in the analysis of the Otto cycle.

9–2 ■ THE CARNOT CYCLE AND ITS VALUE IN ENGINEERING

The Carnot cycle is composed of four totally reversible processes: isother-mal heat addition, isentropic expansion, isothermal heat rejection, and isen-tropic compression. The P-v and T-s diagrams of a Carnot cycle arereplotted in Fig. 9–6. The Carnot cycle can be executed in a closed system(a piston–cylinder device) or a steady-flow system (utilizing two turbinesand two compressors, as shown in Fig. 9–7), and either a gas or a vapor can

490 | Thermodynamics

P T

sv

1

2 3

4

1

2

3

4wnet wnet

FIGURE 9–5On both P-v and T-s diagrams, thearea enclosed by the process curverepresents the net work of the cycle.

P

T

s

v

1

2

3

4

1 2

34q out

q in

Isen

trop

ic

Isen

trop

ic

TH

TL

q in

Isentropic

q out

TH = const.

TL = const.

Isentropic

FIGURE 9–6P-v and T-s diagrams of a Carnotcycle.

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be utilized as the working fluid. The Carnot cycle is the most efficient cyclethat can be executed between a heat source at temperature TH and a sink attemperature TL, and its thermal efficiency is expressed as

(9–2)

Reversible isothermal heat transfer is very difficult to achieve in realitybecause it would require very large heat exchangers and it would take a verylong time (a power cycle in a typical engine is completed in a fraction of asecond). Therefore, it is not practical to build an engine that would operateon a cycle that closely approximates the Carnot cycle.

The real value of the Carnot cycle comes from its being a standardagainst which the actual or the ideal cycles can be compared. The thermalefficiency of the Carnot cycle is a function of the sink and source temper-atures only, and the thermal efficiency relation for the Carnot cycle(Eq. 9–2) conveys an important message that is equally applicable to bothideal and actual cycles: Thermal efficiency increases with an increasein the average temperature at which heat is supplied to the system or witha decrease in the average temperature at which heat is rejected fromthe system.

The source and sink temperatures that can be used in practice are notwithout limits, however. The highest temperature in the cycle is limited bythe maximum temperature that the components of the heat engine, such asthe piston or the turbine blades, can withstand. The lowest temperature islimited by the temperature of the cooling medium utilized in the cycle suchas a lake, a river, or the atmospheric air.

hth,Carnot � 1 �TL

TH

Chapter 9 | 491

qin

qout

Isothermalcompressor

Isentropiccompressor wnet

Isentropicturbine

Isothermalturbine

1

2

3

4

FIGURE 9–7A steady-flow Carnot engine.

EXAMPLE 9–1 Derivation of the Efficiency of the Carnot Cycle

Show that the thermal efficiency of a Carnot cycle operating between thetemperature limits of TH and TL is solely a function of these two tempera-tures and is given by Eq. 9–2.

Solution It is to be shown that the efficiency of a Carnot cycle depends onthe source and sink temperatures alone.

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9–3 ■ AIR-STANDARD ASSUMPTIONSIn gas power cycles, the working fluid remains a gas throughout the entirecycle. Spark-ignition engines, diesel engines, and conventional gas turbinesare familiar examples of devices that operate on gas cycles. In all theseengines, energy is provided by burning a fuel within the system boundaries.That is, they are internal combustion engines. Because of this combustionprocess, the composition of the working fluid changes from air and fuel tocombustion products during the course of the cycle. However, consideringthat air is predominantly nitrogen that undergoes hardly any chemical reac-tions in the combustion chamber, the working fluid closely resembles air atall times.

Even though internal combustion engines operate on a mechanical cycle(the piston returns to its starting position at the end of each revolution), theworking fluid does not undergo a complete thermodynamic cycle. It isthrown out of the engine at some point in the cycle (as exhaust gases)instead of being returned to the initial state. Working on an open cycle is thecharacteristic of all internal combustion engines.

The actual gas power cycles are rather complex. To reduce the analysis toa manageable level, we utilize the following approximations, commonlyknown as the air-standard assumptions:

1. The working fluid is air, which continuously circulates in a closed loopand always behaves as an ideal gas.

2. All the processes that make up the cycle are internally reversible.3. The combustion process is replaced by a heat-addition process from an

external source (Fig. 9–9).4. The exhaust process is replaced by a heat-rejection process that restores

the working fluid to its initial state.

Another assumption that is often utilized to simplify the analysis evenmore is that air has constant specific heats whose values are determined at

492 | Thermodynamics

Analysis The T-s diagram of a Carnot cycle is redrawn in Fig. 9–8. All fourprocesses that comprise the Carnot cycle are reversible, and thus the areaunder each process curve represents the heat transfer for that process. Heatis transferred to the system during process 1-2 and rejected during process3-4. Therefore, the amount of heat input and heat output for the cycle canbe expressed as

since processes 2-3 and 4-1 are isentropic, and thus s2 � s3 and s4 � s1.Substituting these into Eq. 9–1, we see that the thermal efficiency of aCarnot cycle is

Discussion Notice that the thermal efficiency of a Carnot cycle is indepen-dent of the type of the working fluid used (an ideal gas, steam, etc.) orwhether the cycle is executed in a closed or steady-flow system.

hth �wnet

qin� 1 �

qout

qin� 1 �

TL 1s2 � s1 2TH 1s2 � s1 2 � 1 �

TL

TH

qin � TH 1s2 � s1 2 and qout � TL 1s3 � s4 2 � TL 1s2 � s1 2

T

s

1 2

4 3

qin

qout

TH

TL

s1 = s4 s2 = s3

FIGURE 9–8T-s diagram for Example 9–1.

Combustionchamber

COMBUSTION

PRODUCTS

AIR

FUEL

AIR AIR

(a) Actual

(b) Ideal

Heatingsection

HEAT

FIGURE 9–9The combustion process is replaced bya heat-addition process in ideal cycles.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 492

room temperature (25°C, or 77°F). When this assumption is utilized, the air-standard assumptions are called the cold-air-standard assumptions.A cycle for which the air-standard assumptions are applicable is frequentlyreferred to as an air-standard cycle.

The air-standard assumptions previously stated provide considerable sim-plification in the analysis without significantly deviating from the actualcycles. This simplified model enables us to study qualitatively the influenceof major parameters on the performance of the actual engines.

9–4 ■ AN OVERVIEW OF RECIPROCATING ENGINESDespite its simplicity, the reciprocating engine (basically a piston–cylinderdevice) is one of the rare inventions that has proved to be very versatile andto have a wide range of applications. It is the powerhouse of the vast major-ity of automobiles, trucks, light aircraft, ships, and electric power genera-tors, as well as many other devices.

The basic components of a reciprocating engine are shown in Fig. 9–10.The piston reciprocates in the cylinder between two fixed positions calledthe top dead center (TDC)—the position of the piston when it forms thesmallest volume in the cylinder—and the bottom dead center (BDC)—theposition of the piston when it forms the largest volume in the cylinder.The distance between the TDC and the BDC is the largest distance that thepiston can travel in one direction, and it is called the stroke of the engine.The diameter of the piston is called the bore. The air or air–fuel mixture isdrawn into the cylinder through the intake valve, and the combustion prod-ucts are expelled from the cylinder through the exhaust valve.

The minimum volume formed in the cylinder when the piston is at TDCis called the clearance volume (Fig. 9–11). The volume displaced by thepiston as it moves between TDC and BDC is called the displacement vol-ume. The ratio of the maximum volume formed in the cylinder to the mini-mum (clearance) volume is called the compression ratio r of the engine:

(9–3)

Notice that the compression ratio is a volume ratio and should not be con-fused with the pressure ratio.

Another term frequently used in conjunction with reciprocating engines isthe mean effective pressure (MEP). It is a fictitious pressure that, if it actedon the piston during the entire power stroke, would produce the same amountof net work as that produced during the actual cycle (Fig. 9–12). That is,

or

(9–4)

The mean effective pressure can be used as a parameter to compare theperformances of reciprocating engines of equal size. The engine with a largervalue of MEP delivers more net work per cycle and thus performs better.

MEP �Wnet

Vmax � Vmin�

wnet

vmax � vmin1kPa 2

Wnet � MEP � Piston area � Stroke � MEP � Displacement volume

r �Vmax

Vmin�

VBDC

VTDC

Chapter 9 | 493

Intake valve

Exhaustvalve

BoreTDC

BDC

Stroke

FIGURE 9–10Nomenclature for reciprocatingengines.

TDC

BDC

Displacement volume

(a) Clearance volume

(b)

FIGURE 9–11Displacement and clearance volumesof a reciprocating engine.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 493

Reciprocating engines are classified as spark-ignition (SI) engines orcompression-ignition (CI) engines, depending on how the combustionprocess in the cylinder is initiated. In SI engines, the combustion of theair–fuel mixture is initiated by a spark plug. In CI engines, the air–fuelmixture is self-ignited as a result of compressing the mixture above its self-ignition temperature. In the next two sections, we discuss the Otto andDiesel cycles, which are the ideal cycles for the SI and CI reciprocatingengines, respectively.

9–5 ■ OTTO CYCLE: THE IDEAL CYCLE FOR SPARK-IGNITION ENGINES

The Otto cycle is the ideal cycle for spark-ignition reciprocating engines. Itis named after Nikolaus A. Otto, who built a successful four-stroke enginein 1876 in Germany using the cycle proposed by Frenchman Beau deRochas in 1862. In most spark-ignition engines, the piston executes fourcomplete strokes (two mechanical cycles) within the cylinder, and thecrankshaft completes two revolutions for each thermodynamic cycle. Theseengines are called four-stroke internal combustion engines. A schematic ofeach stroke as well as a P-v diagram for an actual four-stroke spark-ignitionengine is given in Fig. 9–13(a).

494 | Thermodynamics

Wnet = MEP(Vmax – Vmin)

Vmin Vmax V

MEP

P

TDC BDC

Wnet

FIGURE 9–12The net work output of a cycle isequivalent to the product of the meaneffective pressure and thedisplacement volume.

qin

qout

4

3

2

1

Patm

P

P

Compressionstroke

Power (expansion)stroke

Air–fuelmixture

(a) Actual four-stroke spark-ignition engine

(b) Ideal Otto cycle

Isentropiccompression

AIR(2)

(1)

End ofcombustion

Exhaust valveopens

Ignition

TDC BDC

Intake

Exhaust

Intakevalve opens

Expansion

Compression

Isentropic

Isentropic

AIR

(4)–(1)

Air–fuelmixture

AIR(2)–(3)

Exhauststroke

Intakestroke

AIR(3)

(4)

Exhaustgases

Isentropicexpansion

v = const.heat addition

v = const.heat rejection

qinqout

v

TDC BDC v

FIGURE 9–13Actual and ideal cycles in spark-ignition engines and their P-v diagrams.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 494

Initially, both the intake and the exhaust valves are closed, and the piston isat its lowest position (BDC). During the compression stroke, the piston movesupward, compressing the air–fuel mixture. Shortly before the piston reachesits highest position (TDC), the spark plug fires and the mixture ignites,increasing the pressure and temperature of the system. The high-pressuregases force the piston down, which in turn forces the crankshaft to rotate,producing a useful work output during the expansion or power stroke. At theend of this stroke, the piston is at its lowest position (the completion of thefirst mechanical cycle), and the cylinder is filled with combustion products.Now the piston moves upward one more time, purging the exhaust gasesthrough the exhaust valve (the exhaust stroke), and down a second time,drawing in fresh air–fuel mixture through the intake valve (the intakestroke). Notice that the pressure in the cylinder is slightly above the atmo-spheric value during the exhaust stroke and slightly below during the intakestroke.

In two-stroke engines, all four functions described above are executed injust two strokes: the power stroke and the compression stroke. In theseengines, the crankcase is sealed, and the outward motion of the piston isused to slightly pressurize the air–fuel mixture in the crankcase, as shown inFig. 9–14. Also, the intake and exhaust valves are replaced by openings inthe lower portion of the cylinder wall. During the latter part of the powerstroke, the piston uncovers first the exhaust port, allowing the exhaust gasesto be partially expelled, and then the intake port, allowing the fresh air–fuelmixture to rush in and drive most of the remaining exhaust gases out of thecylinder. This mixture is then compressed as the piston moves upward dur-ing the compression stroke and is subsequently ignited by a spark plug.

The two-stroke engines are generally less efficient than their four-strokecounterparts because of the incomplete expulsion of the exhaust gases andthe partial expulsion of the fresh air–fuel mixture with the exhaust gases.However, they are relatively simple and inexpensive, and they have highpower-to-weight and power-to-volume ratios, which make them suitable forapplications requiring small size and weight such as for motorcycles, chainsaws, and lawn mowers (Fig. 9–15).

Advances in several technologies—such as direct fuel injection, stratifiedcharge combustion, and electronic controls—brought about a renewed inter-est in two-stroke engines that can offer high performance and fuel economywhile satisfying the stringent emission requirements. For a given weight anddisplacement, a well-designed two-stroke engine can provide significantlymore power than its four-stroke counterpart because two-stroke engines pro-duce power on every engine revolution instead of every other one. In the newtwo-stroke engines, the highly atomized fuel spray that is injected into thecombustion chamber toward the end of the compression stroke burns muchmore completely. The fuel is sprayed after the exhaust valve is closed, whichprevents unburned fuel from being ejected into the atmosphere. With strati-fied combustion, the flame that is initiated by igniting a small amount of therich fuel–air mixture near the spark plug propagates through the combustionchamber filled with a much leaner mixture, and this results in much cleanercombustion. Also, the advances in electronics have made it possible to ensurethe optimum operation under varying engine load and speed conditions.

Chapter 9 | 495

Exhaustport Intake

port

Crankcase

Sparkplug

Fuel–airmixture

FIGURE 9–14Schematic of a two-strokereciprocating engine.

FIGURE 9–15Two-stroke engines are commonlyused in motorcycles and lawn mowers.

© Vol. 26/PhotoDisc

SEE TUTORIAL CH. 9, SEC. 2 ON THE DVD.

INTERACTIVETUTORIAL

cen84959_ch09.qxd 4/26/05 5:44 PM Page 495

Major car companies have research programs underway on two-strokeengines which are expected to make a comeback in the future.

The thermodynamic analysis of the actual four-stroke or two-stroke cyclesdescribed is not a simple task. However, the analysis can be simplified sig-nificantly if the air-standard assumptions are utilized. The resulting cycle,which closely resembles the actual operating conditions, is the ideal Ottocycle. It consists of four internally reversible processes:

1-2 Isentropic compression

2-3 Constant-volume heat addition

3-4 Isentropic expansion

4-1 Constant-volume heat rejection

The execution of the Otto cycle in a piston–cylinder device together witha P-v diagram is illustrated in Fig. 9–13b. The T-s diagram of the Otto cycleis given in Fig. 9–16.

The Otto cycle is executed in a closed system, and disregarding thechanges in kinetic and potential energies, the energy balance for any of theprocesses is expressed, on a unit-mass basis, as

(9–5)

No work is involved during the two heat transfer processes since both takeplace at constant volume. Therefore, heat transfer to and from the workingfluid can be expressed as

(9–6a)

and(9–6b)

Then the thermal efficiency of the ideal Otto cycle under the cold air stan-dard assumptions becomes

Processes 1-2 and 3-4 are isentropic, and v2 � v3 and v4 � v1. Thus,

(9–7)

Substituting these equations into the thermal efficiency relation and simpli-fying give

(9–8)

where

(9–9)

is the compression ratio and k is the specific heat ratio cp /cv.Equation 9–8 shows that under the cold-air-standard assumptions, the

thermal efficiency of an ideal Otto cycle depends on the compression ratioof the engine and the specific heat ratio of the working fluid. The thermalefficiency of the ideal Otto cycle increases with both the compression ratio

r �Vmax

Vmin�

V1

V2�

v1

v2

hth,Otto � 1 �1

r k�1

T1

T2� a v2

v1b k�1

� a v3

v4b k�1

�T4

T3

hth,Otto �wnet

qin� 1 �

qout

qin� 1 �

T4 � T1

T3 � T2� 1 �

T1 1T4>T1 � 1 2T2 1T3>T2 � 1 2

qout � u4 � u1 � cv 1T4 � T1 2qin � u3 � u2 � cv 1T3 � T2 2

1qin � qout 2 � 1win � wout 2 � ¢u 1kJ>kg 2

496 | Thermodynamics

T

s

1

2

3

4

v = co

nst.

v = const.

qout

qin

FIGURE 9–16T-s diagram of the ideal Otto cycle.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 496

and the specific heat ratio. This is also true for actual spark-ignition internalcombustion engines. A plot of thermal efficiency versus the compressionratio is given in Fig. 9–17 for k � 1.4, which is the specific heat ratio valueof air at room temperature. For a given compression ratio, the thermal effi-ciency of an actual spark-ignition engine is less than that of an ideal Ottocycle because of the irreversibilities, such as friction, and other factors suchas incomplete combustion.

We can observe from Fig. 9–17 that the thermal efficiency curve is rathersteep at low compression ratios but flattens out starting with a compressionratio value of about 8. Therefore, the increase in thermal efficiency with thecompression ratio is not as pronounced at high compression ratios. Also,when high compression ratios are used, the temperature of the air–fuel mix-ture rises above the autoignition temperature of the fuel (the temperature atwhich the fuel ignites without the help of a spark) during the combustionprocess, causing an early and rapid burn of the fuel at some point or pointsahead of the flame front, followed by almost instantaneous inflammation ofthe end gas. This premature ignition of the fuel, called autoignition, pro-duces an audible noise, which is called engine knock. Autoignition inspark-ignition engines cannot be tolerated because it hurts performance andcan cause engine damage. The requirement that autoignition not be allowedplaces an upper limit on the compression ratios that can be used in spark-ignition internal combustion engines.

Improvement of the thermal efficiency of gasoline engines by utilizinghigher compression ratios (up to about 12) without facing the autoignitionproblem has been made possible by using gasoline blends that have goodantiknock characteristics, such as gasoline mixed with tetraethyl lead.Tetraethyl lead had been added to gasoline since the 1920s because it is aninexpensive method of raising the octane rating, which is a measure of theengine knock resistance of a fuel. Leaded gasoline, however, has a veryundesirable side effect: it forms compounds during the combustion processthat are hazardous to health and pollute the environment. In an effort tocombat air pollution, the government adopted a policy in the mid-1970s thatresulted in the eventual phase-out of leaded gasoline. Unable to use lead, therefiners developed other techniques to improve the antiknock characteristicsof gasoline. Most cars made since 1975 have been designed to use unleadedgasoline, and the compression ratios had to be lowered to avoid engineknock. The ready availability of high octane fuels made it possible to raisethe compression ratios again in recent years. Also, owing to the improve-ments in other areas (reduction in overall automobile weight, improvedaerodynamic design, etc.), today’s cars have better fuel economy and conse-quently get more miles per gallon of fuel. This is an example of how engi-neering decisions involve compromises, and efficiency is only one of theconsiderations in final design.

The second parameter affecting the thermal efficiency of an ideal Ottocycle is the specific heat ratio k. For a given compression ratio, an idealOtto cycle using a monatomic gas (such as argon or helium, k � 1.667) asthe working fluid will have the highest thermal efficiency. The specific heatratio k, and thus the thermal efficiency of the ideal Otto cycle, decreases asthe molecules of the working fluid get larger (Fig. 9–18). At room tempera-ture it is 1.4 for air, 1.3 for carbon dioxide, and 1.2 for ethane. The working

Chapter 9 | 497

2 4 6 8 10 12 14Compression ratio, r

0.7

0.6

0.5

0.4

0.3

0.2

0.1

Typicalcompressionratios forgasolineenginesη

th,O

tto

FIGURE 9–17Thermal efficiency of the ideal Ottocycle as a function of compressionratio (k � 1.4).

0.8

0.6

0.4

0.2

2 4 6 8 10 12

k = 1.667

k = 1.4

k = 1.3

Compression ratio, r

η th

,Otto

FIGURE 9–18The thermal efficiency of the Ottocycle increases with the specific heatratio k of the working fluid.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 497

fluid in actual engines contains larger molecules such as carbon dioxide,and the specific heat ratio decreases with temperature, which is one of thereasons that the actual cycles have lower thermal efficiencies than the idealOtto cycle. The thermal efficiencies of actual spark-ignition engines rangefrom about 25 to 30 percent.

498 | Thermodynamics

EXAMPLE 9–2 The Ideal Otto Cycle

An ideal Otto cycle has a compression ratio of 8. At the beginning of thecompression process, air is at 100 kPa and 17°C, and 800 kJ/kg of heat istransferred to air during the constant-volume heat-addition process. Account-ing for the variation of specific heats of air with temperature, determine(a) the maximum temperature and pressure that occur during the cycle,(b) the net work output, (c) the thermal efficiency, and (d ) the mean effec-tive pressure for the cycle.

Solution An ideal Otto cycle is considered. The maximum temperature andpressure, the net work output, the thermal efficiency, and the mean effectivepressure are to be determined.Assumptions 1 The air-standard assumptions are applicable. 2 Kinetic andpotential energy changes are negligible. 3 The variation of specific heatswith temperature is to be accounted for.Analysis The P-v diagram of the ideal Otto cycle described is shown inFig. 9–19. We note that the air contained in the cylinder forms a closedsystem.

(a) The maximum temperature and pressure in an Otto cycle occur at theend of the constant-volume heat-addition process (state 3). But first we needto determine the temperature and pressure of air at the end of the isentropiccompression process (state 2), using data from Table A–17:

Process 1-2 (isentropic compression of an ideal gas):

Process 2-3 (constant-volume heat addition):

vr3 � 6.108

u3 � 1275.11 kJ>kg S T3 � 1575.1 K

800 kJ>kg � u3 � 475.11 kJ>kg

qin � u3 � u2

� 1100 kPa 2 a 652.4 K

290 Kb 18 2 � 1799.7 kPa

P2v2

T2�

P1v1

T1 S P2 � P1 a T2

T1b a v1

v2b

u2 � 475.11 kJ>kg

vr2

vr1�

v2

v1�

1r S vr2 �

vr1

r�

676.1

8� 84.51 S T2 � 652.4 K

vr1 � 676.1 T1 � 290 K S u1 � 206.91 kJ>kg

1

2

3

4

P, kPa

100

Isentropic

Isentropic

qin

qout

v2 = v1 v3 = v1 = v4v1–

8

FIGURE 9–19P-v diagram for the Otto cyclediscussed in Example 9–2.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 498

Chapter 9 | 499

(b) The net work output for the cycle is determined either by finding theboundary (P dV) work involved in each process by integration and addingthem or by finding the net heat transfer that is equivalent to the net workdone during the cycle. We take the latter approach. However, first we needto find the internal energy of the air at state 4:

Process 3-4 (isentropic expansion of an ideal gas):

Process 4-1 (constant-volume heat rejection):

Thus,

(c) The thermal efficiency of the cycle is determined from its definition:

Under the cold-air-standard assumptions (constant specific heat values atroom temperature), the thermal efficiency would be (Eq. 9–8)

which is considerably different from the value obtained above. Therefore,care should be exercised in utilizing the cold-air-standard assumptions.

(d ) The mean effective pressure is determined from its definition, Eq. 9–4:

where

Thus,

Discussion Note that a constant pressure of 574 kPa during the powerstroke would produce the same net work output as the entire cycle.

MEP �418.17 kJ>kg

10.832 m3>kg 2 11 � 182 a

1 kPa # m3

1 kJb � 574 kPa

v1 �RT1

P1�10.287 kPa # m3>kg # K 2 1290 K 2

100 kPa� 0.832 m3>kg

MEP �wnet

v1 � v2�

wnet

v1 � v1>r �wnet

v1 11 � 1>r 2

hth,Otto � 1 �1

r k�1 � 1 � r 1�k � 1 � 18 2 1�1.4 � 0.565 or 56.5%

hth �wnet

qin�

418.17 kJ>kg

800 kJ>kg� 0.523 or 52.3%

wnet � qnet � qin � qout � 800 � 381.83 � 418.17 kJ/kg

qout � 588.74 � 206.91 � 381.83 kJ>kg

�qout � u1 � u4 S qout � u4 � u1

u4 � 588.74 kJ>kg

vr4

vr3�

v4

v3� r S vr4 � rvr3 � 18 2 16.108 2 � 48.864 S T4 � 795.6 K

� 11.7997 MPa 2 a 1575.1 K

652.4 Kb 11 2 � 4.345 MPa

P3v3

T3�

P2v2

T2 S P3 � P2 a T3

T2b a v2

v3b

cen84959_ch09.qxd 4/26/05 5:44 PM Page 499

9–6 ■ DIESEL CYCLE: THE IDEAL CYCLE FOR COMPRESSION-IGNITION ENGINES

The Diesel cycle is the ideal cycle for CI reciprocating engines. The CIengine, first proposed by Rudolph Diesel in the 1890s, is very similar to theSI engine discussed in the last section, differing mainly in the method ofinitiating combustion. In spark-ignition engines (also known as gasolineengines), the air–fuel mixture is compressed to a temperature that is belowthe autoignition temperature of the fuel, and the combustion process is initi-ated by firing a spark plug. In CI engines (also known as diesel engines),the air is compressed to a temperature that is above the autoignition temper-ature of the fuel, and combustion starts on contact as the fuel is injected intothis hot air. Therefore, the spark plug and carburetor are replaced by a fuelinjector in diesel engines (Fig. 9–20).

In gasoline engines, a mixture of air and fuel is compressed during thecompression stroke, and the compression ratios are limited by the onset ofautoignition or engine knock. In diesel engines, only air is compressed dur-ing the compression stroke, eliminating the possibility of autoignition.Therefore, diesel engines can be designed to operate at much higher com-pression ratios, typically between 12 and 24. Not having to deal with theproblem of autoignition has another benefit: many of the stringent require-ments placed on the gasoline can now be removed, and fuels that are lessrefined (thus less expensive) can be used in diesel engines.

The fuel injection process in diesel engines starts when the pistonapproaches TDC and continues during the first part of the power stroke.Therefore, the combustion process in these engines takes place over alonger interval. Because of this longer duration, the combustion process inthe ideal Diesel cycle is approximated as a constant-pressure heat-additionprocess. In fact, this is the only process where the Otto and the Dieselcycles differ. The remaining three processes are the same for both idealcycles. That is, process 1-2 is isentropic compression, 3-4 is isentropicexpansion, and 4-1 is constant-volume heat rejection. The similaritybetween the two cycles is also apparent from the P-v and T-s diagrams ofthe Diesel cycle, shown in Fig. 9–21.

Noting that the Diesel cycle is executed in a piston–cylinder device,which forms a closed system, the amount of heat transferred to the workingfluid at constant pressure and rejected from it at constant volume can beexpressed as

(9–10a)

and

(9–10b)

Then the thermal efficiency of the ideal Diesel cycle under the cold-air-standard assumptions becomes

hth,Diesel �wnet

qin� 1 �

qout

qin� 1 �

T4 � T1

k 1T3 � T2 2 � 1 �T1 1T4>T1 � 1 2kT2 1T3>T2 � 1 2

�qout � u1 � u4 S qout � u4 � u1 � cv 1T4 � T1 2

� h3 � h2 � cp 1T3 � T2 2 qin � wb,out � u3 � u2 S qin � P2 1v3 � v2 2 � 1u3 � u2 2

500 | Thermodynamics

Gasoline engine Diesel engine

Sparkplug

Fuelinjector

AIR

Air–fuelmixture Fuel spray

Spark

FIGURE 9–20In diesel engines, the spark plug isreplaced by a fuel injector, and onlyair is compressed during thecompression process.

1

2 3

4

P

IsentropicIsentropic

s

v

1

2

3

4

T

P = constant

v = constant

(a) P- v diagram

v

(b) T-s diagram

qin

qout

qout

qin

FIGURE 9–21T-s and P-v diagrams for the idealDiesel cycle.

cen84959_ch09.qxd 4/26/05 5:44 PM Page 500

We now define a new quantity, the cutoff ratio rc, as the ratio of the cylin-der volumes after and before the combustion process:

(9–11)

Utilizing this definition and the isentropic ideal-gas relations for processes1-2 and 3-4, we see that the thermal efficiency relation reduces to

(9–12)

where r is the compression ratio defined by Eq. 9–9. Looking at Eq. 9–12carefully, one would notice that under the cold-air-standard assumptions, theefficiency of a Diesel cycle differs from the efficiency of an Otto cycle bythe quantity in the brackets. This quantity is always greater than 1. Therefore,

(9–13)

when both cycles operate on the same compression ratio. Also, as the cutoffratio decreases, the efficiency of the Diesel cycle increases (Fig. 9–22). For thelimiting case of rc � 1, the quantity in the brackets becomes unity (can youprove it?), and the efficiencies of the Otto and Diesel cycles become identical.Remember, though, that diesel engines operate at much higher compressionratios and thus are usually more efficient than the spark-ignition (gasoline)engines. The diesel engines also burn the fuel more completely since theyusually operate at lower revolutions per minute and the air–fuel mass ratio ismuch higher than spark-ignition engines. Thermal efficiencies of large dieselengines range from about 35 to 40 percent.

The higher efficiency and lower fuel costs of diesel engines make themattractive in applications requiring relatively large amounts of power, suchas in locomotive engines, emergency power generation units, large ships,and heavy trucks. As an example of how large a diesel engine can be, a 12-cylinder diesel engine built in 1964 by the Fiat Corporation of Italy had anormal power output of 25,200 hp (18.8 MW) at 122 rpm, a cylinder boreof 90 cm, and a stroke of 91 cm.

Approximating the combustion process in internal combustion engines as aconstant-volume or a constant-pressure heat-addition process is overly simplis-tic and not quite realistic. Probably a better (but slightly more complex)approach would be to model the combustion process in both gasoline anddiesel engines as a combination of two heat-transfer processes, one at constantvolume and the other at constant pressure. The ideal cycle based on this con-cept is called the dual cycle, and a P-v diagram for it is given in Fig. 9–23.The relative amounts of heat transferred during each process can be adjusted toapproximate the actual cycle more closely. Note that both the Otto and theDiesel cycles can be obtained as special cases of the dual cycle.

hth,Otto 7 hth,Diesel

hth,Diesel � 1 �1

r k�1 c r kc � 1

k 1rc � 1 2 d

rc �V3

V2�

v3

v2

Chapter 9 | 501

0.7

η th

,Die

sel

Compression ratio, r

0.6

0.5

0.4

0.3

0.2

0.1

2 4 6 8 10 12 14 16 18 20 22 24

Typicalcompression

ratios for dieselengines

rc = 1 (Otto)

23

4

FIGURE 9–22Thermal efficiency of the ideal Dieselcycle as a function of compression andcutoff ratios (k � 1.4).

1

2

3

4

P

IsentropicIsentropic

X

qin

qout

v

FIGURE 9–23P-v diagram of an ideal dual cycle.

EXAMPLE 9–3 The Ideal Diesel Cycle

An ideal Diesel cycle with air as the working fluid has a compression ratio of18 and a cutoff ratio of 2. At the beginning of the compression process, theworking fluid is at 14.7 psia, 80°F, and 117 in3. Utilizing the cold-air-standard assumptions, determine (a) the temperature and pressure of air at

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502 | Thermodynamics

the end of each process, (b) the net work output and the thermal efficiency,and (c) the mean effective pressure.

Solution An ideal Diesel cycle is considered. The temperature and pressureat the end of each process, the net work output, the thermal efficiency, andthe mean effective pressure are to be determined.Assumptions 1 The cold-air-standard assumptions are applicable and thusair can be assumed to have constant specific heats at room temperature.2 Kinetic and potential energy changes are negligible.Properties The gas constant of air is R � 0.3704 psia · ft3/lbm · R and itsother properties at room temperature are cp � 0.240 Btu/lbm · R, cv �0.171 Btu/lbm · R, and k � 1.4 (Table A–2Ea).Analysis The P-V diagram of the ideal Diesel cycle described is shown inFig. 9–24. We note that the air contained in the cylinder forms a closedsystem.

(a) The temperature and pressure values at the end of each process can bedetermined by utilizing the ideal-gas isentropic relations for processes 1-2and 3-4. But first we determine the volumes at the end of each process fromthe definitions of the compression ratio and the cutoff ratio:

Process 1-2 (isentropic compression of an ideal gas, constant specific heats):

Process 2-3 (constant-pressure heat addition to an ideal gas):

Process 3-4 (isentropic expansion of an ideal gas, constant specific heats):

(b) The net work for a cycle is equivalent to the net heat transfer. But firstwe find the mass of air:

m �P1V1

RT1�

114.7 psia 2 1117 in3 210.3704 psia # ft3>lbm # R 2 1540 R 2 a

1 ft3

1728 in3 b � 0.00498 lbm

P4 � P3 a V3

V4b k

� 1841 psia 2 a 13 in3

117 in3 b1.4

� 38.8 psia

T4 � T3 a V3

V4b k�1

� 13432 R 2 a 13 in3

117 in3 b1.4�1

� 1425 R

P2V2

T2�

P3V3

T3S T3 � T2 a V3

V2b � 11716 R 2 12 2 � 3432 R

P3 � P2 � 841 psia

P2 � P1 a V1

V2b k

� 114.7 psia 2 118 2 1.4 � 841 psia

T2 � T1 a V1

V2b k�1

� 1540 R 2 118 2 1.4�1 � 1716 R

V4 � V1 � 117 in3

V3 � rcV2 � 12 2 16.5 in3 2 � 13 in3

V2 �V1

r�

117 in3

18� 6.5 in3

1

2 3

4

P, psia

Isentropic

Isentropic

14.7

V2 = V1/18 V3 = 2V2 V1 = V4 V

qin

qout

FIGURE 9–24P-V diagram for the ideal Diesel cyclediscussed in Example 9–3.

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9–7 ■ STIRLING AND ERICSSON CYCLESThe ideal Otto and Diesel cycles discussed in the preceding sections arecomposed entirely of internally reversible processes and thus are internallyreversible cycles. These cycles are not totally reversible, however, since theyinvolve heat transfer through a finite temperature difference during the non-isothermal heat-addition and heat-rejection processes, which are irreversible.Therefore, the thermal efficiency of an Otto or Diesel engine will be lessthan that of a Carnot engine operating between the same temperature limits.

Consider a heat engine operating between a heat source at TH and a heatsink at TL. For the heat-engine cycle to be totally reversible, the temperaturedifference between the working fluid and the heat source (or sink) shouldnever exceed a differential amount dT during any heat-transfer process. Thatis, both the heat-addition and heat-rejection processes during the cycle musttake place isothermally, one at a temperature of TH and the other at a tem-perature of TL. This is precisely what happens in a Carnot cycle.

Chapter 9 | 503

Process 2-3 is a constant-pressure heat-addition process, for which theboundary work and �u terms can be combined into �h. Thus,

Process 4-1 is a constant-volume heat-rejection process (it involves no workinteractions), and the amount of heat rejected is

Thus,

Then the thermal efficiency becomes

The thermal efficiency of this Diesel cycle under the cold-air-standardassumptions could also be determined from Eq. 9–12.

(c) The mean effective pressure is determined from its definition, Eq. 9–4:

Discussion Note that a constant pressure of 110 psia during the powerstroke would produce the same net work output as the entire Diesel cycle.

� 110 psia

MEP �Wnet

Vmax � Vmin�

Wnet

V1 � V2�

1.297 Btu

1117 � 6.5 2 in3 a 778.17 lbf # ft

1 Btub a 12 in.

1 ftb

hth �Wnet

Q in�

1.297 Btu

2.051 Btu� 0.632 or 63.2%

Wnet � Q in � Q out � 2.051 � 0.754 � 1.297 Btu

� 0.754 Btu

� 10.00498 lbm 2 10.171 Btu>lbm # R 2 3 11425 � 540 2 R 4 Q out � m 1u4 � u1 2 � mcv 1T4 � T1 2

� 2.051 Btu

� 10.00498 lbm 2 10.240 Btu>lbm # R 2 3 13432 � 1716 2 R 4 Qin � m 1h3 � h2 2 � mcp 1T3 � T2 2

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There are two other cycles that involve an isothermal heat-addition processat TH and an isothermal heat-rejection process at TL: the Stirling cycle andthe Ericsson cycle. They differ from the Carnot cycle in that the two isen-tropic processes are replaced by two constant-volume regeneration processesin the Stirling cycle and by two constant-pressure regeneration processes inthe Ericsson cycle. Both cycles utilize regeneration, a process during whichheat is transferred to a thermal energy storage device (called a regenerator)during one part of the cycle and is transferred back to the working fluid dur-ing another part of the cycle (Fig. 9–25).

Figure 9–26(b) shows the T-s and P-v diagrams of the Stirling cycle,which is made up of four totally reversible processes:

1-2 T � constant expansion (heat addition from the external source)

2-3 v � constant regeneration (internal heat transfer from the workingfluid to the regenerator)

3-4 T � constant compression (heat rejection to the external sink)

4-1 v � constant regeneration (internal heat transfer from the regenerator back to the working fluid)

The execution of the Stirling cycle requires rather innovative hardware.The actual Stirling engines, including the original one patented by RobertStirling, are heavy and complicated. To spare the reader the complexities,the execution of the Stirling cycle in a closed system is explained with thehelp of the hypothetical engine shown in Fig. 9–27.

This system consists of a cylinder with two pistons on each side and aregenerator in the middle. The regenerator can be a wire or a ceramic mesh

504 | Thermodynamics

Energy

Energy

REGENERATOR

Working fluid

FIGURE 9–25A regenerator is a device that borrowsenergy from the working fluid duringone part of the cycle and pays it back(without interest) during another part.

s

1 2

34

T

s =

con

st.

s =

con

st.

TH

TL

1

2

3

4

P

TH = const.

TH = const. T

H = const.

TL = const.

TL = const.

TL = const.

1

2

3

4

P

Regeneration

Regeneration

1

23

4

P

s

1 2

34

T

v =

cons

t.

v =

cons

t.Regeneration

s

1 2

34

P =

cons

t.

P =

cons

t.Regeneration

(a) Carnot cycle (b) Stirling cycle (c) Ericsson cycle

qin

qout

TH

TL

T

TH

TL

qin

qout

qin

qout

qin

qout

qin

qin

qout qout

v v v

FIGURE 9–26T-s and P-v diagrams of Carnot,Stirling, and Ericsson cycles.

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or any kind of porous plug with a high thermal mass (mass times specificheat). It is used for the temporary storage of thermal energy. The mass ofthe working fluid contained within the regenerator at any instant is consid-ered negligible.

Initially, the left chamber houses the entire working fluid (a gas), which isat a high temperature and pressure. During process 1-2, heat is transferredto the gas at TH from a source at TH. As the gas expands isothermally, theleft piston moves outward, doing work, and the gas pressure drops. Duringprocess 2-3, both pistons are moved to the right at the same rate (to keep thevolume constant) until the entire gas is forced into the right chamber. As thegas passes through the regenerator, heat is transferred to the regenerator andthe gas temperature drops from TH to TL. For this heat transfer process to bereversible, the temperature difference between the gas and the regeneratorshould not exceed a differential amount dT at any point. Thus, the tempera-ture of the regenerator will be TH at the left end and TL at the right end ofthe regenerator when state 3 is reached. During process 3-4, the right pistonis moved inward, compressing the gas. Heat is transferred from the gas to asink at temperature TL so that the gas temperature remains constant at TLwhile the pressure rises. Finally, during process 4-1, both pistons are movedto the left at the same rate (to keep the volume constant), forcing the entiregas into the left chamber. The gas temperature rises from TL to TH as itpasses through the regenerator and picks up the thermal energy stored thereduring process 2-3. This completes the cycle.

Notice that the second constant-volume process takes place at a smallervolume than the first one, and the net heat transfer to the regenerator duringa cycle is zero. That is, the amount of energy stored in the regenerator duringprocess 2-3 is equal to the amount picked up by the gas during process 4-1.

The T-s and P-v diagrams of the Ericsson cycle are shown in Fig. 9–26c.The Ericsson cycle is very much like the Stirling cycle, except that the twoconstant-volume processes are replaced by two constant-pressure processes.

A steady-flow system operating on an Ericsson cycle is shown in Fig. 9–28.Here the isothermal expansion and compression processes are executed in acompressor and a turbine, respectively, and a counter-flow heat exchangerserves as a regenerator. Hot and cold fluid streams enter the heat exchangerfrom opposite ends, and heat transfer takes place between the two streams. Inthe ideal case, the temperature difference between the two fluid streams doesnot exceed a differential amount at any point, and the cold fluid stream leavesthe heat exchanger at the inlet temperature of the hot stream.

Chapter 9 | 505

State1

State2

State3

State4

Regenerator

TH

TH

TL

TL

qin

qout

FIGURE 9–27The execution of the Stirling cycle.

Regenerator

TL = const.Compressor

TH = const.Turbine

wnet

Heat

qinqout

FIGURE 9–28A steady-flow Ericsson engine.

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Both the Stirling and Ericsson cycles are totally reversible, as is the Carnotcycle, and thus according to the Carnot principle, all three cycles must havethe same thermal efficiency when operating between the same temperaturelimits:

(9–14)

This is proved for the Carnot cycle in Example 9–1 and can be proved in asimilar manner for both the Stirling and Ericsson cycles.

hth,Stirling � hth,Ericsson � hth,Carnot � 1 �TL

TH

506 | Thermodynamics

EXAMPLE 9–4 Thermal Efficiency of the Ericsson Cycle

Using an ideal gas as the working fluid, show that the thermal efficiency ofan Ericsson cycle is identical to the efficiency of a Carnot cycle operatingbetween the same temperature limits.

Solution It is to be shown that the thermal efficiencies of Carnot and Ericsson cycles are identical.Analysis Heat is transferred to the working fluid isothermally from an externalsource at temperature TH during process 1-2, and it is rejected again isother-mally to an external sink at temperature TL during process 3-4. For areversible isothermal process, heat transfer is related to the entropy change by

The entropy change of an ideal gas during an isothermal process is

The heat input and heat output can be expressed as

and

Then the thermal efficiency of the Ericsson cycle becomes

since P1 � P4 and P3 � P2. Notice that this result is independent ofwhether the cycle is executed in a closed or steady-flow system.

hth,Ericsson � 1 �qout

qin� 1 �

RTL ln 1P4>P3 2RTH ln 1P1>P2 2 � 1 �

TL

TH

qout � TL 1s4 � s3 2 � �TL a�R ln P4

P3b � RTL ln

P4

P3

qin � TH 1s2 � s1 2 � TH a�R ln P2

P1b � RTH ln

P1

P2

¢s � cp ln Te

Ti

� R ln Pe

Pi

� �R ln Pe

Pi

q � T ¢s

Stirling and Ericsson cycles are difficult to achieve in practice becausethey involve heat transfer through a differential temperature difference in allcomponents including the regenerator. This would require providing infi-nitely large surface areas for heat transfer or allowing an infinitely long timefor the process. Neither is practical. In reality, all heat transfer processes takeplace through a finite temperature difference, the regenerator does not havean efficiency of 100 percent, and the pressure losses in the regenerator areconsiderable. Because of these limitations, both Stirling and Ericsson cycles

0¡

cen84959_ch09.qxd 4/26/05 5:45 PM Page 506

have long been of only theoretical interest. However, there is renewed inter-est in engines that operate on these cycles because of their potential forhigher efficiency and better emission control. The Ford Motor Company,General Motors Corporation, and the Phillips Research Laboratories of theNetherlands have successfully developed Stirling engines suitable for trucks,buses, and even automobiles. More research and development are neededbefore these engines can compete with the gasoline or diesel engines.

Both the Stirling and the Ericsson engines are external combustion engines.That is, the fuel in these engines is burned outside the cylinder, as opposed togasoline or diesel engines, where the fuel is burned inside the cylinder.

External combustion offers several advantages. First, a variety of fuels canbe used as a source of thermal energy. Second, there is more time for com-bustion, and thus the combustion process is more complete, which meansless air pollution and more energy extraction from the fuel. Third, theseengines operate on closed cycles, and thus a working fluid that has the mostdesirable characteristics (stable, chemically inert, high thermal conductivity)can be utilized as the working fluid. Hydrogen and helium are two gasescommonly employed in these engines.

Despite the physical limitations and impracticalities associated with them,both the Stirling and Ericsson cycles give a strong message to design engi-neers: Regeneration can increase efficiency. It is no coincidence that moderngas-turbine and steam power plants make extensive use of regeneration. Infact, the Brayton cycle with intercooling, reheating, and regeneration, which isutilized in large gas-turbine power plants and discussed later in this chapter,closely resembles the Ericsson cycle.

9–8 ■ BRAYTON CYCLE: THE IDEAL CYCLE FOR GAS-TURBINE ENGINES

The Brayton cycle was first proposed by George Brayton for use in the recip-rocating oil-burning engine that he developed around 1870. Today, it is usedfor gas turbines only where both the compression and expansion processestake place in rotating machinery. Gas turbines usually operate on an opencycle, as shown in Fig. 9–29. Fresh air at ambient conditions is drawn intothe compressor, where its temperature and pressure are raised. The high-pressure air proceeds into the combustion chamber, where the fuel is burnedat constant pressure. The resulting high-temperature gases then enter the tur-bine, where they expand to the atmospheric pressure while producingpower. The exhaust gases leaving the turbine are thrown out (not recircu-lated), causing the cycle to be classified as an open cycle.

The open gas-turbine cycle described above can be modeled as a closedcycle, as shown in Fig. 9–30, by utilizing the air-standard assumptions. Herethe compression and expansion processes remain the same, but the combus-tion process is replaced by a constant-pressure heat-addition process froman external source, and the exhaust process is replaced by a constant-pressure heat-rejection process to the ambient air. The ideal cycle that theworking fluid undergoes in this closed loop is the Brayton cycle, which ismade up of four internally reversible processes:

1-2 Isentropic compression (in a compressor)

2-3 Constant-pressure heat addition

Chapter 9 | 507

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3-4 Isentropic expansion (in a turbine)

4-1 Constant-pressure heat rejection

The T-s and P-v diagrams of an ideal Brayton cycle are shown in Fig. 9–31.Notice that all four processes of the Brayton cycle are executed in steady-flow devices; thus, they should be analyzed as steady-flow processes. Whenthe changes in kinetic and potential energies are neglected, the energy bal-ance for a steady-flow process can be expressed, on a unit–mass basis, as

(9–15)

Therefore, heat transfers to and from the working fluid are

(9–16a)

and

(9–16b)

Then the thermal efficiency of the ideal Brayton cycle under the cold-air-standard assumptions becomes

Processes 1-2 and 3-4 are isentropic, and P2 � P3 and P4 � P1. Thus,

Substituting these equations into the thermal efficiency relation and simpli-fying give

(9–17)hth,Brayton � 1 �1

r 1k�12>kp

T2

T1� a P2

P1b 1k�12>k

� a P3

P4b 1k�12>k

�T3

T4

hth,Brayton �wnet

qin� 1 �

qout

qin� 1 �

cp 1T4 � T1 2cp 1T3 � T2 2 � 1 �

T1 1T4>T1 � 1 2T2 1T3>T2 � 1 2

qout � h4 � h1 � cp 1T4 � T1 2

qin � h3 � h2 � cp 1T3 � T2 2

1qin � qout 2 � 1win � wout 2 � hexit � hinlet

508 | Thermodynamics

Compressorwnet

Turbine

Combustionchamber

Freshair

Exhaustgases1

23

4

Fuel

FIGURE 9–29An open-cycle gas-turbine engine.

Compressor Turbine

1

23

4Heat

exchanger

Heatexchanger

wnet

qin

qout

FIGURE 9–30A closed-cycle gas-turbine engine.

P

s = const.

s = const.2

1 4

3

s

T

2

3

4

1

P = const.

P = const.

(a) T-s diagram

(b) P-v diagram

qout

qin

qout

qin

v

FIGURE 9–31T-s and P-v diagrams for the idealBrayton cycle.

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where

(9–18)

is the pressure ratio and k is the specific heat ratio. Equation 9–17 showsthat under the cold-air-standard assumptions, the thermal efficiency of anideal Brayton cycle depends on the pressure ratio of the gas turbine and thespecific heat ratio of the working fluid. The thermal efficiency increases withboth of these parameters, which is also the case for actual gas turbines.A plot of thermal efficiency versus the pressure ratio is given in Fig. 9–32 fork � 1.4, which is the specific-heat-ratio value of air at room temperature.

The highest temperature in the cycle occurs at the end of the combustionprocess (state 3), and it is limited by the maximum temperature that the tur-bine blades can withstand. This also limits the pressure ratios that can beused in the cycle. For a fixed turbine inlet temperature T3, the net work out-put per cycle increases with the pressure ratio, reaches a maximum, andthen starts to decrease, as shown in Fig. 9–33. Therefore, there should be acompromise between the pressure ratio (thus the thermal efficiency) and thenet work output. With less work output per cycle, a larger mass flow rate(thus a larger system) is needed to maintain the same power output, whichmay not be economical. In most common designs, the pressure ratio of gasturbines ranges from about 11 to 16.

The air in gas turbines performs two important functions: It supplies thenecessary oxidant for the combustion of the fuel, and it serves as a coolantto keep the temperature of various components within safe limits. The sec-ond function is accomplished by drawing in more air than is needed for thecomplete combustion of the fuel. In gas turbines, an air–fuel mass ratio of50 or above is not uncommon. Therefore, in a cycle analysis, treating thecombustion gases as air does not cause any appreciable error. Also, the massflow rate through the turbine is greater than that through the compressor, thedifference being equal to the mass flow rate of the fuel. Thus, assuming aconstant mass flow rate throughout the cycle yields conservative results foropen-loop gas-turbine engines.

The two major application areas of gas-turbine engines are aircraft propul-sion and electric power generation. When it is used for aircraft propulsion,the gas turbine produces just enough power to drive the compressor and asmall generator to power the auxiliary equipment. The high-velocity exhaustgases are responsible for producing the necessary thrust to propel the air-craft. Gas turbines are also used as stationary power plants to generate elec-tricity as stand-alone units or in conjunction with steam power plants on thehigh-temperature side. In these plants, the exhaust gases of the gas turbineserve as the heat source for the steam. The gas-turbine cycle can also be exe-cuted as a closed cycle for use in nuclear power plants. This time the work-ing fluid is not limited to air, and a gas with more desirable characteristics(such as helium) can be used.

The majority of the Western world’s naval fleets already use gas-turbineengines for propulsion and electric power generation. The General ElectricLM2500 gas turbines used to power ships have a simple-cycle thermal effi-ciency of 37 percent. The General Electric WR-21 gas turbines equipped withintercooling and regeneration have a thermal efficiency of 43 percent and

rp �P2

P1

Chapter 9 | 509

5Pressure ratio, rp

0.7

0.6

0.5

0.4

0.3

0.2

0.1

η th

,Bra

yton

Typical pressureratios for gas-

turbine engines

10 15 20 25

FIGURE 9–32Thermal efficiency of the idealBrayton cycle as a function of thepressure ratio.

s

T

2

3

wnet,max

Tmax1000 K

rp = 15

r p = 8.2

r p = 2

Tmin300 K 1

4

FIGURE 9–33For fixed values of Tmin and Tmax,the net work of the Brayton cyclefirst increases with the pressureratio, then reaches a maximum atrp � (Tmax/Tmin)

k/[2(k � 1)], andfinally decreases.

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produce 21.6 MW (29,040 hp). The regeneration also reduces the exhaust tem-perature from 600°C (1100°F) to 350°C (650°F). Air is compressed to 3 atmbefore it enters the intercooler. Compared to steam-turbine and diesel-propulsion systems, the gas turbine offers greater power for a given size andweight, high reliability, long life, and more convenient operation. The enginestart-up time has been reduced from 4 h required for a typical steam-propulsion system to less than 2 min for a gas turbine. Many modern marinepropulsion systems use gas turbines together with diesel engines because of thehigh fuel consumption of simple-cycle gas-turbine engines. In combined dieseland gas-turbine systems, diesel is used to provide for efficient low-power andcruise operation, and gas turbine is used when high speeds are needed.

In gas-turbine power plants, the ratio of the compressor work to the tur-bine work, called the back work ratio, is very high (Fig. 9–34). Usuallymore than one-half of the turbine work output is used to drive the compres-sor. The situation is even worse when the isentropic efficiencies of the com-pressor and the turbine are low. This is quite in contrast to steam powerplants, where the back work ratio is only a few percent. This is not surpris-ing, however, since a liquid is compressed in steam power plants instead ofa gas, and the steady-flow work is proportional to the specific volume of theworking fluid.

A power plant with a high back work ratio requires a larger turbine toprovide the additional power requirements of the compressor. Therefore, theturbines used in gas-turbine power plants are larger than those used in steampower plants of the same net power output.

Development of Gas TurbinesThe gas turbine has experienced phenomenal progress and growth since itsfirst successful development in the 1930s. The early gas turbines built in the1940s and even 1950s had simple-cycle efficiencies of about 17 percentbecause of the low compressor and turbine efficiencies and low turbine inlettemperatures due to metallurgical limitations of those times. Therefore, gasturbines found only limited use despite their versatility and their ability toburn a variety of fuels. The efforts to improve the cycle efficiency concen-trated in three areas:

1. Increasing the turbine inlet (or firing) temperatures This hasbeen the primary approach taken to improve gas-turbine efficiency. The tur-bine inlet temperatures have increased steadily from about 540°C (1000°F) inthe 1940s to 1425°C (2600°F) and even higher today. These increases weremade possible by the development of new materials and the innovative cool-ing techniques for the critical components such as coating the turbine bladeswith ceramic layers and cooling the blades with the discharge air from thecompressor. Maintaining high turbine inlet temperatures with an air-coolingtechnique requires the combustion temperature to be higher to compensate forthe cooling effect of the cooling air. However, higher combustion tempera-tures increase the amount of nitrogen oxides (NOx), which are responsible forthe formation of ozone at ground level and smog. Using steam as the coolantallowed an increase in the turbine inlet temperatures by 200°F without anincrease in the combustion temperature. Steam is also a much more effectiveheat transfer medium than air.

510 | Thermodynamics

wturbine

wnet

wcompressor

Back work

FIGURE 9–34The fraction of the turbine work usedto drive the compressor is called theback work ratio.

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2. Increasing the efficiencies of turbomachinery componentsThe performance of early turbines suffered greatly from the inefficiencies ofturbines and compressors. However, the advent of computers and advancedtechniques for computer-aided design made it possible to design these com-ponents aerodynamically with minimal losses. The increased efficiencies ofthe turbines and compressors resulted in a significant increase in the cycleefficiency.

3. Adding modifications to the basic cycle The simple-cycle efficien-cies of early gas turbines were practically doubled by incorporating intercool-ing, regeneration (or recuperation), and reheating, discussed in the next twosections. These improvements, of course, come at the expense of increasedinitial and operation costs, and they cannot be justified unless the decrease infuel costs offsets the increase in other costs. The relatively low fuel prices, thegeneral desire in the industry to minimize installation costs, and the tremen-dous increase in the simple-cycle efficiency to about 40 percent left little desirefor opting for these modifications.

The first gas turbine for an electric utility was installed in 1949 inOklahoma as part of a combined-cycle power plant. It was built by GeneralElectric and produced 3.5 MW of power. Gas turbines installed until themid-1970s suffered from low efficiency and poor reliability. In the past, thebase-load electric power generation was dominated by large coal andnuclear power plants. However, there has been an historic shift toward nat-ural gas–fired gas turbines because of their higher efficiencies, lower capitalcosts, shorter installation times, and better emission characteristics, and theabundance of natural gas supplies, and more and more electric utilities areusing gas turbines for base-load power production as well as for peaking.The construction costs for gas-turbine power plants are roughly half that ofcomparable conventional fossil-fuel steam power plants, which were the pri-mary base-load power plants until the early 1980s. More than half of allpower plants to be installed in the foreseeable future are forecast to be gas-turbine or combined gas–steam turbine types.

A gas turbine manufactured by General Electric in the early 1990s had apressure ratio of 13.5 and generated 135.7 MW of net power at a thermalefficiency of 33 percent in simple-cycle operation. A more recent gas turbinemanufactured by General Electric uses a turbine inlet temperature of 1425°C(2600°F) and produces up to 282 MW while achieving a thermal efficiencyof 39.5 percent in the simple-cycle mode. A 1.3-ton small-scale gas turbinelabeled OP-16, built by the Dutch firm Opra Optimal Radial Turbine, can runon gas or liquid fuel and can replace a 16-ton diesel engine. It has a pressureratio of 6.5 and produces up to 2 MW of power. Its efficiency is 26 percentin the simple-cycle operation, which rises to 37 percent when equipped witha regenerator.

Chapter 9 | 511

EXAMPLE 9–5 The Simple Ideal Brayton Cycle

A gas-turbine power plant operating on an ideal Brayton cycle has a pressureratio of 8. The gas temperature is 300 K at the compressor inlet and 1300 Kat the turbine inlet. Utilizing the air-standard assumptions, determine (a) the

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512 | Thermodynamics

gas temperature at the exits of the compressor and the turbine, (b) the backwork ratio, and (c) the thermal efficiency.

Solution A power plant operating on the ideal Brayton cycle is considered.The compressor and turbine exit temperatures, back work ratio, and the ther-mal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The air-standard assump-tions are applicable. 3 Kinetic and potential energy changes are negligible.4 The variation of specific heats with temperature is to be considered.Analysis The T-s diagram of the ideal Brayton cycle described is shown inFig. 9–35. We note that the components involved in the Brayton cycle aresteady-flow devices.(a) The air temperatures at the compressor and turbine exits are determinedfrom isentropic relations:

Process 1-2 (isentropic compression of an ideal gas):

Process 3-4 (isentropic expansion of an ideal gas):

(b) To find the back work ratio, we need to find the work input to the com-pressor and the work output of the turbine:

Thus,

That is, 40.3 percent of the turbine work output is used just to drive thecompressor.(c) The thermal efficiency of the cycle is the ratio of the net power output tothe total heat input:

Thus,

hth �wnet

qin�

362.4 kJ>kg

851.62 kJ>kg� 0.426 or 42.6%

wnet � wout � win � 606.60 � 244.16 � 362.4 kJ>kg

qin � h3 � h2 � 1395.97 � 544.35 � 851.62 kJ>kg

rbw �wcomp,in

wturb,out�

244.16 kJ>kg

606.60 kJ>kg� 0.403

wturb,out � h3 � h4 � 1395.97 � 789.37 � 606.60 kJ>kg

wcomp,in � h2 � h1 � 544.35 � 300.19 � 244.16 kJ>kg

h4 � 789.37 kJ>kg

Pr4 �P4

P3 Pr3 � a 1

8b 1330.9 2 � 41.36 S T4 � 770 K 1at turbine exit 2

Pr3 � 330.9

T3 � 1300 K S h3 � 1395.97 kJ>kg

h2 � 544.35 kJ>kg

Pr2 �P2

P1 Pr1 � 18 2 11.386 2 � 11.09 S T2 � 540 K 1at compressor exit 2

Pr1 � 1.386

T1 � 300 K S h1 � 300.19 kJ>kg

s

T, K

2

3

4

1

P = const.

P = const.

wturb

wcomp

rp = 8

1300

300

qout

qin

FIGURE 9–35T-s diagram for the Brayton cyclediscussed in Example 9–5.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 512

Deviation of Actual Gas-Turbine Cycles from Idealized OnesThe actual gas-turbine cycle differs from the ideal Brayton cycle on severalaccounts. For one thing, some pressure drop during the heat-addition and heat-rejection processes is inevitable. More importantly, the actual work input to thecompressor is more, and the actual work output from the turbine is lessbecause of irreversibilities. The deviation of actual compressor and turbinebehavior from the idealized isentropic behavior can be accurately accountedfor by utilizing the isentropic efficiencies of the turbine and compressor as

(9–19)

and

(9–20)

where states 2a and 4a are the actual exit states of the compressor and theturbine, respectively, and 2s and 4s are the corresponding states for the isen-tropic case, as illustrated in Fig. 9–36. The effect of the turbine and com-pressor efficiencies on the thermal efficiency of the gas-turbine engines isillustrated below with an example.

hT �wa

ws�

h3 � h4a

h3 � h4s

hC �ws

wa�

h2s � h1

h2a � h1

Chapter 9 | 513

The thermal efficiency could also be determined from

where

Discussion Under the cold-air-standard assumptions (constant specific heatvalues at room temperature), the thermal efficiency would be, from Eq. 9–17,

which is sufficiently close to the value obtained by accounting for the varia-tion of specific heats with temperature.

hth,Brayton � 1 �1

r 1k�12>kp

� 1 �1

8 11.4�12>1.4� 0.448

qout � h4 � h1 � 789.37 � 300.19 � 489.2 kJ>kg

hth � 1 �qout

qin

s

T

2s

3

1

2a

Pressure dropduring heataddition

Pressure dropduring heatrejection

4s

4a

FIGURE 9–36The deviation of an actual gas-turbinecycle from the ideal Brayton cycle as aresult of irreversibilities.

EXAMPLE 9–6 An Actual Gas-Turbine Cycle

Assuming a compressor efficiency of 80 percent and a turbine efficiency of85 percent, determine (a) the back work ratio, (b) the thermal efficiency,and (c) the turbine exit temperature of the gas-turbine cycle discussed inExample 9–5.

Solution The Brayton cycle discussed in Example 9–5 is reconsidered. Forspecified turbine and compressor efficiencies, the back work ratio, the ther-mal efficiency, and the turbine exit temperature are to be determined.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 513

514 | Thermodynamics

Analysis (a) The T-s diagram of the cycle is shown in Fig. 9–37. The actualcompressor work and turbine work are determined by using the definitions ofcompressor and turbine efficiencies, Eqs. 9–19 and 9–20:

Compressor:

Turbine:

Thus,

That is, the compressor is now consuming 59.2 percent of the work pro-duced by the turbine (up from 40.3 percent). This increase is due to theirreversibilities that occur within the compressor and the turbine.

(b) In this case, air leaves the compressor at a higher temperature andenthalpy, which are determined to be

Thus,

and

That is, the irreversibilities occurring within the turbine and compressorcaused the thermal efficiency of the gas turbine cycle to drop from 42.6 to26.6 percent. This example shows how sensitive the performance of agas-turbine power plant is to the efficiencies of the compressor and theturbine. In fact, gas-turbine efficiencies did not reach competitive valuesuntil significant improvements were made in the design of gas turbines andcompressors.

(c) The air temperature at the turbine exit is determined from an energy bal-ance on the turbine:

Then, from Table A–17,

Discussion The temperature at turbine exit is considerably higher than thatat the compressor exit (T2a � 598 K), which suggests the use of regenerationto reduce fuel cost.

T4a � 853 K

� 880.36 kJ>kg

� 1395.97 � 515.61

wturb,out � h3 � h4a S h4a � h3 � wturb,out

hth �wnet

qin�

210.41 kJ>kg

790.58 kJ>kg� 0.266 or 26.6%

wnet � wout � win � 515.61 � 305.20 � 210.41 kJ>kg

qin � h3 � h2a � 1395.97 � 605.39 � 790.58 kJ>kg

� 605.39 kJ>kg 1and T2a � 598 K 2 � 300.19 � 305.20

wcomp,in � h2a � h1 S h2a � h1 � wcomp,in

rbw �wcomp,in

wturb,out�

305.20 kJ>kg

515.61 kJ>kg� 0.592

wturb,out � hTws � 10.85 2 1606.60 kJ>kg 2 � 515.61 kJ>kg

wcomp,in �ws

hC�

244.16 kJ>kg

0.80� 305.20 kJ>kg

s

T, K

2s

3

4a

1

qin

qout

1300

300

2a

4s

FIGURE 9–37T-s diagram of the gas-turbine cyclediscussed in Example 9–6.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 514

9–9 ■ THE BRAYTON CYCLE WITH REGENERATIONIn gas-turbine engines, the temperature of the exhaust gas leaving the tur-bine is often considerably higher than the temperature of the air leaving thecompressor. Therefore, the high-pressure air leaving the compressor can beheated by transferring heat to it from the hot exhaust gases in a counter-flowheat exchanger, which is also known as a regenerator or a recuperator.A sketch of the gas-turbine engine utilizing a regenerator and the T-sdiagram of the new cycle are shown in Figs. 9–38 and 9–39, respectively.

The thermal efficiency of the Brayton cycle increases as a result of regener-ation since the portion of energy of the exhaust gases that is normally rejectedto the surroundings is now used to preheat the air entering the combustionchamber. This, in turn, decreases the heat input (thus fuel) requirements forthe same net work output. Note, however, that the use of a regenerator is rec-ommended only when the turbine exhaust temperature is higher than the com-pressor exit temperature. Otherwise, heat will flow in the reverse direction (tothe exhaust gases), decreasing the efficiency. This situation is encountered ingas-turbine engines operating at very high pressure ratios.

The highest temperature occurring within the regenerator is T4, the tem-perature of the exhaust gases leaving the turbine and entering the regenera-tor. Under no conditions can the air be preheated in the regenerator to atemperature above this value. Air normally leaves the regenerator at a lowertemperature, T5. In the limiting (ideal) case, the air exits the regenerator atthe inlet temperature of the exhaust gases T4. Assuming the regenerator tobe well insulated and any changes in kinetic and potential energies to benegligible, the actual and maximum heat transfers from the exhaust gases tothe air can be expressed as

(9–21)

and

(9–22)

The extent to which a regenerator approaches an ideal regenerator is calledthe effectiveness ` and is defined as

(9–23)P �qregen,act

qregen,max�

h5 � h2

h4 � h2

qregen,max � h5¿ � h2 � h4 � h2

qregen,act � h5 � h2

Chapter 9 | 515

Turbine

Regenerator

Compressorwnet

2 5

6

1

3

4Combustion

chamber

Heat

FIGURE 9–38A gas-turbine engine with regenerator.

Regeneration

s

T

4

qin

qout1

3

2

5'5

6

qsaved = qregen

qregen

FIGURE 9–39T-s diagram of a Brayton cycle withregeneration.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 515

When the cold-air-standard assumptions are utilized, it reduces to

(9–24)

A regenerator with a higher effectiveness obviously saves a greateramount of fuel since it preheats the air to a higher temperature prior to com-bustion. However, achieving a higher effectiveness requires the use of alarger regenerator, which carries a higher price tag and causes a larger pres-sure drop. Therefore, the use of a regenerator with a very high effectivenesscannot be justified economically unless the savings from the fuel costsexceed the additional expenses involved. The effectiveness of most regener-ators used in practice is below 0.85.

Under the cold-air-standard assumptions, the thermal efficiency of anideal Brayton cycle with regeneration is

(9–25)

Therefore, the thermal efficiency of an ideal Brayton cycle with regenera-tion depends on the ratio of the minimum to maximum temperatures as wellas the pressure ratio. The thermal efficiency is plotted in Fig. 9–40 for vari-ous pressure ratios and minimum-to-maximum temperature ratios. This fig-ure shows that regeneration is most effective at lower pressure ratios andlow minimum-to-maximum temperature ratios.

hth,regen � 1 � a T1

T3b 1rp 2 1k�12>k

P �T5 � T2

T4 � T2

516 | Thermodynamics

5 10 15 20 25

0.7

0.6

0.5

0.4

0.3

0.2

0.1

η th

,Bra

yton

Pressure ratio, rp

With regeneration

Without regeneration

T1/T3 = 0.2

T1/T3 = 0.25

T1/T3 = 0.33

FIGURE 9–40Thermal efficiency of the idealBrayton cycle with and withoutregeneration.

3

5

s

T, K

4a

1

qregen = qsaved

1300

300

2a

qin

FIGURE 9–41T-s diagram of the regenerativeBrayton cycle described in Example 9–7.

EXAMPLE 9–7 Actual Gas-Turbine Cycle with Regeneration

Determine the thermal efficiency of the gas-turbine described in Example9–6 if a regenerator having an effectiveness of 80 percent is installed.

Solution The gas-turbine discussed in Example 9–6 is equipped with aregenerator. For a specified effectiveness, the thermal efficiency is to bedetermined.Analysis The T-s diagram of the cycle is shown in Fig. 9–41. We first deter-mine the enthalpy of the air at the exit of the regenerator, using the defini-tion of effectiveness:

Thus,

This represents a savings of 220.0 kJ/kg from the heat input requirements.The addition of a regenerator (assumed to be frictionless) does not affect thenet work output. Thus,

hth �wnet

qin�

210.41 kJ>kg

570.60 kJ>kg� 0.369 or 36.9%

qin � h3 � h5 � 11395.97 � 825.37 2 kJ>kg � 570.60 kJ>kg

0.80 �1h5 � 605.39 2 kJ>kg

1880.36 � 605.39 2 kJ>kgS h5 � 825.37 kJ>kg

P �h5 � h2a

h4a � h2a

cen84959_ch09.qxd 4/26/05 5:45 PM Page 516

9–10 ■ THE BRAYTON CYCLE WITHINTERCOOLING, REHEATING, AND REGENERATION

The net work of a gas-turbine cycle is the difference between the turbinework output and the compressor work input, and it can be increased byeither decreasing the compressor work or increasing the turbine work, orboth. It was shown in Chap. 7 that the work required to compress a gasbetween two specified pressures can be decreased by carrying out the com-pression process in stages and cooling the gas in between (Fig. 9–42)—thatis, using multistage compression with intercooling. As the number of stagesis increased, the compression process becomes nearly isothermal at thecompressor inlet temperature, and the compression work decreases.

Likewise, the work output of a turbine operating between two pressurelevels can be increased by expanding the gas in stages and reheating it inbetween—that is, utilizing multistage expansion with reheating. This isaccomplished without raising the maximum temperature in the cycle. As thenumber of stages is increased, the expansion process becomes nearlyisothermal. The foregoing argument is based on a simple principle: Thesteady-flow compression or expansion work is proportional to the specificvolume of the fluid. Therefore, the specific volume of the working fluidshould be as low as possible during a compression process and as high aspossible during an expansion process. This is precisely what intercoolingand reheating accomplish.

Combustion in gas turbines typically occurs at four times the amount ofair needed for complete combustion to avoid excessive temperatures. There-fore, the exhaust gases are rich in oxygen, and reheating can be accom-plished by simply spraying additional fuel into the exhaust gases betweentwo expansion states.

The working fluid leaves the compressor at a lower temperature, and theturbine at a higher temperature, when intercooling and reheating are uti-lized. This makes regeneration more attractive since a greater potential forregeneration exists. Also, the gases leaving the compressor can be heated toa higher temperature before they enter the combustion chamber because ofthe higher temperature of the turbine exhaust.

A schematic of the physical arrangement and the T-s diagram of an idealtwo-stage gas-turbine cycle with intercooling, reheating, and regeneration areshown in Figs. 9–43 and 9–44. The gas enters the first stage of the compres-sor at state 1, is compressed isentropically to an intermediate pressure P2, iscooled at constant pressure to state 3 (T3 � T1), and is compressed in the sec-ond stage isentropically to the final pressure P4. At state 4 the gas enters theregenerator, where it is heated to T5 at constant pressure. In an ideal regenera-tor, the gas leaves the regenerator at the temperature of the turbine exhaust,that is, T5 � T9. The primary heat addition (or combustion) process takes

Chapter 9 | 517

Discussion Note that the thermal efficiency of the gas turbine has gone upfrom 26.6 to 36.9 percent as a result of installing a regenerator that helpsto recuperate some of the thermal energy of the exhaust gases.

P

P2

P1

D C

AB

Polytropicprocess paths

Work saved as a result ofintercooling

Isothermalprocess paths

Intercooling

1

v

FIGURE 9–42Comparison of work inputs to asingle-stage compressor (1AC) and atwo-stage compressor withintercooling (1ABD).

SEE TUTORIAL CH. 9, SEC. 5 ON THE DVD.

INTERACTIVETUTORIAL

cen84959_ch09.qxd 4/26/05 5:45 PM Page 517

place between states 5 and 6. The gas enters the first stage of the turbine atstate 6 and expands isentropically to state 7, where it enters the reheater. It isreheated at constant pressure to state 8 (T8 � T6), where it enters the secondstage of the turbine. The gas exits the turbine at state 9 and enters the regener-ator, where it is cooled to state 10 at constant pressure. The cycle is completedby cooling the gas to the initial state (or purging the exhaust gases).

It was shown in Chap. 7 that the work input to a two-stage compressor isminimized when equal pressure ratios are maintained across each stage. Itcan be shown that this procedure also maximizes the turbine work output.Thus, for best performance we have

(9–26)

In the analysis of the actual gas-turbine cycles, the irreversibilities that arepresent within the compressor, the turbine, and the regenerator as well as thepressure drops in the heat exchangers should be taken into consideration.

The back work ratio of a gas-turbine cycle improves as a result of inter-cooling and reheating. However, this does not mean that the thermal effi-ciency also improves. The fact is, intercooling and reheating alwaysdecreases the thermal efficiency unless they are accompanied by regenera-tion. This is because intercooling decreases the average temperature atwhich heat is added, and reheating increases the average temperature at whichheat is rejected. This is also apparent from Fig. 9–44. Therefore, in gas-turbine power plants, intercooling and reheating are always used in conjunc-tion with regeneration.

P2

P1�

P4

P3and

P6

P7�

P8

P9

518 | Thermodynamics

Regenerator

wnetCompressorII

CompressorI

Turbine I Turbine II

Combustionchamber

Reheater

Intercooler

10

1

23

4

5

6 7 8 9

FIGURE 9–43A gas-turbine engine with two-stage compression with intercooling, two-stage expansion withreheating, and regeneration.

s

T

3

4

6qin

qout 1

2

10

8

97

5

qregen = qsaved

qregen

FIGURE 9–44T-s diagram of an ideal gas-turbinecycle with intercooling, reheating, andregeneration.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 518

If the number of compression and expansion stages is increased, the idealgas-turbine cycle with intercooling, reheating, and regeneration approachesthe Ericsson cycle, as illustrated in Fig. 9–45, and the thermal efficiencyapproaches the theoretical limit (the Carnot efficiency). However, the contri-bution of each additional stage to the thermal efficiency is less and less, andthe use of more than two or three stages cannot be justified economically.

Chapter 9 | 519

s

T

TH,avg

TL,avg

P =

cons

t.

P = co

nst.

FIGURE 9–45As the number of compression andexpansion stages increases, the gas-turbine cycle with intercooling,reheating, and regenerationapproaches the Ericsson cycle.

EXAMPLE 9–8 A Gas Turbine with Reheating and Intercooling

An ideal gas-turbine cycle with two stages of compression and two stages ofexpansion has an overall pressure ratio of 8. Air enters each stage of thecompressor at 300 K and each stage of the turbine at 1300 K. Determinethe back work ratio and the thermal efficiency of this gas-turbine cycle,assuming (a) no regenerators and (b) an ideal regenerator with 100 percenteffectiveness. Compare the results with those obtained in Example 9–5.

Solution An ideal gas-turbine cycle with two stages of compression and twostages of expansion is considered. The back work ratio and the thermal effi-ciency of the cycle are to be determined for the cases of no regeneration andmaximum regeneration.Assumptions 1 Steady operating conditions exist. 2 The air-standard assump-tions are applicable. 3 Kinetic and potential energy changes are negligible.Analysis The T-s diagram of the ideal gas-turbine cycle described is shownin Fig. 9–46. We note that the cycle involves two stages of expansion, twostages of compression, and regeneration.

For two-stage compression and expansion, the work input is minimizedand the work output is maximized when both stages of the compressor andthe turbine have the same pressure ratio. Thus,

Air enters each stage of the compressor at the same temperature, and eachstage has the same isentropic efficiency (100 percent in this case). There-fore, the temperature (and enthalpy) of the air at the exit of each compres-sion stage will be the same. A similar argument can be given for the turbine.Thus,

At inlets:

At exits:

Under these conditions, the work input to each stage of the compressor willbe the same, and so will the work output from each stage of the turbine.

(a) In the absence of any regeneration, the back work ratio and the thermalefficiency are determined by using data from Table A–17 as follows:

h2 � 404.31 kJ>kg

Pr2 �P2

P1Pr1 � 28 11.386 2 � 3.92 S T2 � 403.3 K

Pr1 � 1.386

T1 � 300 K S h1 � 300.19 kJ>kg

T2 � T4, h2 � h4 and T7 � T9, h7 � h9

T1 � T3, h1 � h3 and T6 � T8, h6 � h8

P2

P1�

P4

P3� 28 � 2.83 and

P6

P7�

P8

P9� 28 � 2.83

s

T, K

qprimary

qout

6

1

10

8

97

5

4

3

2

qreheat

1300

300

FIGURE 9–46T-s diagram of the gas-turbine cyclediscussed in Example 9–8.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 519

520 | Thermodynamics

Then

Thus,

and

A comparison of these results with those obtained in Example 9–5 (single-stage compression and expansion) reveals that multistage compression withintercooling and multistage expansion with reheating improve the back workratio (it drops from 40.3 to 30.4 percent) but hurt the thermal efficiency (itdrops from 42.6 to 35.8 percent). Therefore, intercooling and reheating arenot recommended in gas-turbine power plants unless they are accompaniedby regeneration.

(b) The addition of an ideal regenerator (no pressure drops, 100 percent effec-tiveness) does not affect the compressor work and the turbine work. There-fore, the net work output and the back work ratio of an ideal gas-turbinecycle are identical whether there is a regenerator or not. A regenerator, how-ever, reduces the heat input requirements by preheating the air leaving thecompressor, using the hot exhaust gases. In an ideal regenerator, the com-pressed air is heated to the turbine exit temperature T9 before it enters thecombustion chamber. Thus, under the air-standard assumptions, h5 � h7 � h9.

The heat input and the thermal efficiency in this case are

and

Discussion Note that the thermal efficiency almost doubles as a result ofregeneration compared to the no-regeneration case. The overall effect of two-stage compression and expansion with intercooling, reheating, and regenera-

hth �wnet

qin�

477.04 kJ>kg

685.28 kJ>kg� 0.696 or 69.6%

� 11395.97 � 1053.33 2 � 11395.97 � 1053.33 2 � 685.28 kJ>kg

qin � qprimary � qreheat � 1h6 � h5 2 � 1h8 � h7 2

hth �wnet

qin�

477.04 kJ>kg

1334.30 kJ>kg� 0.358 or 35.8%

rbw �wcomp,in

wturb,out�

208.24 kJ>kg

685.28 kJ>kg� 0.304 or 30.4%

� 11395.97 � 404.31 2 � 11395.97 � 1053.33 2 � 1334.30 kJ>kg

qin � qprimary � qreheat � 1h6 � h4 2 � 1h8 � h7 2 wnet � wturb,out � wcomp,in � 685.28 � 208.24 � 477.04 kJ>kg

wturb,out � 2 1wturb,out,I 2 � 2 1h6 � h7 2 � 2 11395.97 � 1053.33 2 � 685.28 kJ>kg

wcomp,in � 2 1wcomp,in,I 2 � 2 1h2 � h1 2 � 2 1404.31 � 300.19 2 � 208.24 kJ>kg

h7 � 1053.33 kJ>kg

Pr7 �P7

P6Pr 6 �

1

281330.9 2 � 117.0 S T7 � 1006.4 K

Pr 6 � 330.9

T6 � 1300 K S h6 � 1395.97 kJ>kg

cen84959_ch09.qxd 4/26/05 5:45 PM Page 520

9–11 ■ IDEAL JET-PROPULSION CYCLESGas-turbine engines are widely used to power aircraft because they are lightand compact and have a high power-to-weight ratio. Aircraft gas turbinesoperate on an open cycle called a jet-propulsion cycle. The ideal jet-propulsion cycle differs from the simple ideal Brayton cycle in that thegases are not expanded to the ambient pressure in the turbine. Instead, theyare expanded to a pressure such that the power produced by the turbine isjust sufficient to drive the compressor and the auxiliary equipment, such asa small generator and hydraulic pumps. That is, the net work output of a jet-propulsion cycle is zero. The gases that exit the turbine at a relatively highpressure are subsequently accelerated in a nozzle to provide the thrust topropel the aircraft (Fig. 9–47). Also, aircraft gas turbines operate at higherpressure ratios (typically between 10 and 25), and the fluid passes through adiffuser first, where it is decelerated and its pressure is increased before itenters the compressor.

Aircraft are propelled by accelerating a fluid in the opposite direction tomotion. This is accomplished by either slightly accelerating a large mass offluid ( propeller-driven engine) or greatly accelerating a small mass of fluid( jet or turbojet engine) or both (turboprop engine).

A schematic of a turbojet engine and the T-s diagram of the ideal turbojetcycle are shown in Fig. 9–48. The pressure of air rises slightly as it is decel-erated in the diffuser. Air is compressed by the compressor. It is mixed withfuel in the combustion chamber, where the mixture is burned at constantpressure. The high-pressure and high-temperature combustion gases partiallyexpand in the turbine, producing enough power to drive the compressor andother equipment. Finally, the gases expand in a nozzle to the ambient pres-sure and leave the engine at a high velocity.

In the ideal case, the turbine work is assumed to equal the compressorwork. Also, the processes in the diffuser, the compressor, the turbine, andthe nozzle are assumed to be isentropic. In the analysis of actual cycles,however, the irreversibilities associated with these devices should be consid-ered. The effect of the irreversibilities is to reduce the thrust that can beobtained from a turbojet engine.

The thrust developed in a turbojet engine is the unbalanced force that iscaused by the difference in the momentum of the low-velocity air enteringthe engine and the high-velocity exhaust gases leaving the engine, and it is

Chapter 9 | 521

tion on the thermal efficiency is an increase of 63 percent. As the number ofcompression and expansion stages is increased, the cycle will approach theEricsson cycle, and the thermal efficiency will approach

Adding a second stage increases the thermal efficiency from 42.6 to 69.6percent, an increase of 27 percentage points. This is a significant increasein efficiency, and usually it is well worth the extra cost associated with thesecond stage. Adding more stages, however (no matter how many), canincrease the efficiency an additional 7.3 percentage points at most, andusually cannot be justified economically.

hth,Ericsson � hth,Carnot � 1 �TL

TH

� 1 �300 K

1300 K� 0.769

Turbine

Vexit

Nozzle

High T and P

FIGURE 9–47In jet engines, the high-temperatureand high-pressure gases leaving theturbine are accelerated in a nozzle toprovide thrust.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 521

determined from Newton’s second law. The pressures at the inlet and theexit of a turbojet engine are identical (the ambient pressure); thus, the netthrust developed by the engine is

(9–27)

where Vexit is the exit velocity of the exhaust gases and Vinlet is the inlet veloc-ity of the air, both relative to the aircraft. Thus, for an aircraft cruising in stillair, Vinlet is the aircraft velocity. In reality, the mass flow rates of the gases atthe engine exit and the inlet are different, the difference being equal to thecombustion rate of the fuel. However, the air–fuel mass ratio used in jet-propulsion engines is usually very high, making this difference very small.Thus, m

.in Eq. 9–27 is taken as the mass flow rate of air through the engine.

For an aircraft cruising at a constant speed, the thrust is used to overcome airdrag, and the net force acting on the body of the aircraft is zero. Commercialairplanes save fuel by flying at higher altitudes during long trips since air athigher altitudes is thinner and exerts a smaller drag force on aircraft.

The power developed from the thrust of the engine is called the propulsivepower W

.P, which is the propulsive force (thrust) times the distance this force

acts on the aircraft per unit time, that is, the thrust times the aircraft velocity(Fig. 9–49):

(9–28)

The net work developed by a turbojet engine is zero. Thus, we cannotdefine the efficiency of a turbojet engine in the same way as stationary gas-turbine engines. Instead, we should use the general definition of efficiency,which is the ratio of the desired output to the required input. The desiredoutput in a turbojet engine is the power produced to propel the aircraft W

.P,

and the required input is the heating value of the fuel Q.in. The ratio of these

two quantities is called the propulsive efficiency and is given by

(9–29)

Propulsive efficiency is a measure of how efficiently the thermal energyreleased during the combustion process is converted to propulsive energy. The

hP �Propulsive power

Energy input rate�

W#

P

Q#

in

W#

P � FVaircraft � m# 1Vexit � Vinlet 2Vaircraft 1kW 2

F � 1m# V 2 exit � 1m# V 2 inlet � m# 1Vexit � Vinlet 2 1N 2

522 | Thermodynamics

P = const.

P = const.

s

T

qin

qout

6

5

4

3

2

1Diffuser Compressor Burner section Turbine Nozzle

65

43

2

1

FIGURE 9–48Basic components of a turbojet engine and the T-s diagram for the ideal turbojet cycle.

Source: The Aircraft Gas Turbine Engine and Its Operation. © United Aircraft Corporation (now United Technologies Corp.), 1951, 1974.

·V, m/s

WP = F V

FF

FIGURE 9–49Propulsive power is the thrust actingon the aircraft through a distance perunit time.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 522

remaining part of the energy released shows up as the kinetic energy of theexhaust gases relative to a fixed point on the ground and as an increase inthe enthalpy of the gases leaving the engine.

Chapter 9 | 523

EXAMPLE 9–9 The Ideal Jet-Propulsion Cycle

A turbojet aircraft flies with a velocity of 850 ft/s at an altitude where the air isat 5 psia and �40°F. The compressor has a pressure ratio of 10, and the tem-perature of the gases at the turbine inlet is 2000°F. Air enters the compressorat a rate of 100 lbm/s. Utilizing the cold-air-standard assumptions, determine(a) the temperature and pressure of the gases at the turbine exit, (b) the veloc-ity of the gases at the nozzle exit, and (c) the propulsive efficiency of the cycle.

Solution The operating conditions of a turbojet aircraft are specified. Thetemperature and pressure at the turbine exit, the velocity of gases at thenozzle exit, and the propulsive efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 The cold-air-standardassumptions are applicable and thus air can be assumed to have constantspecific heats at room temperature (cp � 0.240 Btu/lbm · °F and k � 1.4).3 Kinetic and potential energies are negligible, except at the diffuser inletand the nozzle exit. 4 The turbine work output is equal to the compressorwork input.Analysis The T-s diagram of the ideal jet propulsion cycle described is shownin Fig. 9–50. We note that the components involved in the jet-propulsioncycle are steady-flow devices.(a) Before we can determine the temperature and pressure at the turbineexit, we need to find the temperatures and pressures at other states:

Process 1-2 (isentropic compression of an ideal gas in a diffuser): For con-venience, we can assume that the aircraft is stationary and the air is movingtoward the aircraft at a velocity of V1 � 850 ft/s. Ideally, the air exits thediffuser with a negligible velocity (V2 � 0):

Process 2-3 (isentropic compression of an ideal gas in a compressor):

T3 � T2 a P3

P2b 1k�12>k

� 1480 R 2 110 2 11.4�12>1.4 � 927 R

P3 � 1rp 2 1P2 2 � 110 2 18.0 psia 2 � 80 psia 1� P4 2

P2 � P1 a T2

T1b k>1k�12

� 15 psia 2 a 480 R

420 Rb 1.4>11.4�12

� 8.0 psia

� 480 R

� 420 R �1850 ft>s 2 2

2 10.240 Btu>lbm # R 2 a1 Btu>lbm

25,037 ft2>s2 b

T2 � T1 �V 2

1

2cp

0 � cp 1T2 � T1 2 �V 2

1

2

h2 �V 2

2

2� h1 �

V 21

2

P = const.

P = const.

s

T, °F

qin

qout

6

5

4

3

2

2000

–401

FIGURE 9–50T-s diagram for the turbojet cycledescribed in Example 9–9.

0¡

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524 | Thermodynamics

Process 4-5 (isentropic expansion of an ideal gas in a turbine): Neglectingthe kinetic energy changes across the compressor and the turbine andassuming the turbine work to be equal to the compressor work, we find thetemperature and pressure at the turbine exit to be

(b) To find the air velocity at the nozzle exit, we need to first determine thenozzle exit temperature and then apply the steady-flow energy equation.

Process 5-6 (isentropic expansion of an ideal gas in a nozzle):

(c) The propulsive efficiency of a turbojet engine is the ratio of the propul-sive power developed W

.P to the total heat transfer rate to the working fluid:

That is, 22.5 percent of the energy input is used to propel the aircraft andto overcome the drag force exerted by the atmospheric air.

hP �W#P

Q#

in

�8276 Btu>s

36,794 Btu>s � 22.5%

� 36,794 Btu>s � 1100 lbm>s 2 10.240 Btu>lbm # R 2 3 12460 � 927 2 R 4

Q#

in � m# 1h4 � h3 2 � m

#cp 1T4 � T3 2

� 8276 Btu>s 1or 11,707 hp 2 � 1100 lbm>s 2 3 13288 � 850 2 ft>s 4 1850 ft>s 2 a 1 Btu>lbm

25,037 ft2>s2 b W#

P � m# 1Vexit � Vinlet 2Vaircraft

� 3288 ft/s

� B2 10.240 Btu>lbm # R 2 3 12013 � 1114 2 R 4 a 25,037 ft2>s2

1 Btu>lbmb

V6 � 22cp 1T5 � T6 2 0 � cp 1T6 � T5 2 �

V 26

2

h6 �V 2

6

2� h5 �

V 25

2

T6 � T5 a P6

P5b 1k�12>k

� 12013 R 2 a 5 psia

39.7 psiab 11.4�12>1.4

� 1114 R

P5 � P4 a T5

T4b k>1k�12

� 180 psia 2 a 2013 R

2460 Rb 1.4>11.4�12

� 39.7 psia

T5 � T4 � T3 � T2 � 2460 � 927 � 480 � 2013 R

cp 1T3 � T2 2 � cp 1T4 � T5 2 h3 � h2 � h4 � h5

wcomp,in � wturb,out

0¡

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Modifications to Turbojet EnginesThe first airplanes built were all propeller-driven, with propellers poweredby engines essentially identical to automobile engines. The major break-through in commercial aviation occurred with the introduction of the turbo-jet engine in 1952. Both propeller-driven engines and jet-propulsion-drivenengines have their own strengths and limitations, and several attempts havebeen made to combine the desirable characteristics of both in one engine.Two such modifications are the propjet engine and the turbofan engine.

The most widely used engine in aircraft propulsion is the turbofan (orfanjet) engine wherein a large fan driven by the turbine forces a consider-able amount of air through a duct (cowl) surrounding the engine, as shownin Figs. 9–52 and 9–53. The fan exhaust leaves the duct at a higher veloc-ity, enhancing the total thrust of the engine significantly. A turbofan engineis based on the principle that for the same power, a large volume of slower-moving air produces more thrust than a small volume of fast-moving air.The first commercial turbofan engine was successfully tested in 1955.

Chapter 9 | 525

Discussion For those who are wondering what happened to the rest of theenergy, here is a brief account:

Thus, 32.2 percent of the energy shows up as excess kinetic energy (kineticenergy of the gases relative to a fixed point on the ground). Notice that forthe highest propulsion efficiency, the velocity of the exhaust gases relative tothe ground Vg should be zero. That is, the exhaust gases should leave thenozzle at the velocity of the aircraft. The remaining 45.3 percent of theenergy shows up as an increase in enthalpy of the gases leaving the engine.These last two forms of energy eventually become part of the internal energyof the atmospheric air (Fig. 9–51).

� 16,651 Btu>s 145.3% 2 � 1100 lbm>s 2 10.24 Btu>lbm # R 2 3 11114 � 420 2 R 4

Q#out � m

# 1h6 � h1 2 � m#cp 1T6 � T1 2

� 11,867 Btu>s 132.2% 2 KE

#out � m

# V 2g

2� 1100 lbm>s 2 e 3 13288 � 850 2 ft>s 4 2

2f a 1 Btu>lbm

25,037 ft2>s2 b

AIRCRAFT

(propulsive power)W·P

Qin

(excess thermal energy)

KEout

(excess kinetic energy)

·

·

Q·

out

FIGURE 9–51Energy supplied to an aircraft (fromthe burning of a fuel) manifests itselfin various forms.

Low-pressurecompressor

Fan Duct Burners

FanHigh-pressurecompressor

Low-pressure turbine

High-pressure turbine

Fan exhaust

Turbine exhaust

FIGURE 9–52A turbofan engine.

Source: The Aircraft Gas Turbine and ItsOperation. © United Aircraft Corporation (nowUnited Technologies Corp.), 1951, 1974.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 525

The turbofan engine on an airplane can be distinguished from the less-efficient turbojet engine by its fat cowling covering the large fan. All thethrust of a turbojet engine is due to the exhaust gases leaving the engine atabout twice the speed of sound. In a turbofan engine, the high-speed exhaustgases are mixed with the lower-speed air, which results in a considerablereduction in noise.

New cooling techniques have resulted in considerable increases in effi-ciencies by allowing gas temperatures at the burner exit to reach over1500°C, which is more than 100°C above the melting point of the turbineblade materials. Turbofan engines deserve most of the credit for the successof jumbo jets that weigh almost 400,000 kg and are capable of carrying over400 passengers for up to a distance of 10,000 km at speeds over 950 km/hwith less fuel per passenger mile.

The ratio of the mass flow rate of air bypassing the combustion chamber tothat of air flowing through it is called the bypass ratio. The first commercialhigh-bypass-ratio engines had a bypass ratio of 5. Increasing the bypass ratioof a turbofan engine increases thrust. Thus, it makes sense to remove thecowl from the fan. The result is a propjet engine, as shown in Fig. 9–54.Turbofan and propjet engines differ primarily in their bypass ratios: 5 or 6for turbofans and as high as 100 for propjets. As a general rule, propellers

526 | Thermodynamics

Fan

Airinlet

Low pressurecompressor

Fan air bypassingthe jet engine

High pressurecompressor

Low pressure turbineto turn inner shaft

2-stage high pressureturbine to turn outer shaft

Combustors

Thrust

Thrust

Twin spoolshaft to turn the fanand the compressors

FIGURE 9–53A modern jet engine used to powerBoeing 777 aircraft. This is a Pratt &Whitney PW4084 turbofan capable ofproducing 84,000 pounds of thrust. Itis 4.87 m (192 in.) long, has a 2.84 m(112 in.) diameter fan, and it weighs6800 kg (15,000 lbm).

Courtesy of Pratt & Whitney Corp.

Propeller

Compressor Burners Turbine

Gear reduction

FIGURE 9–54A turboprop engine.

Source: The Aircraft Gas Turbine Engine and ItsOperation. © United Aircraft Corporation (nowUnited Technologies Corp.), 1951, 1974.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 526

are more efficient than jet engines, but they are limited to low-speed andlow-altitude operation since their efficiency decreases at high speeds and alti-tudes. The old propjet engines (turboprops) were limited to speeds of aboutMach 0.62 and to altitudes of around 9100 m. The new propjet engines( propfans) are expected to achieve speeds of about Mach 0.82 and altitudesof about 12,200 m. Commercial airplanes of medium size and range pro-pelled by propfans are expected to fly as high and as fast as the planes pro-pelled by turbofans, and to do so on less fuel.

Another modification that is popular in military aircraft is the addition ofan afterburner section between the turbine and the nozzle. Whenever aneed for extra thrust arises, such as for short takeoffs or combat conditions,additional fuel is injected into the oxygen-rich combustion gases leaving theturbine. As a result of this added energy, the exhaust gases leave at a highervelocity, providing a greater thrust.

A ramjet engine is a properly shaped duct with no compressor or turbine,as shown in Fig. 9–55, and is sometimes used for high-speed propulsion ofmissiles and aircraft. The pressure rise in the engine is provided by the rameffect of the incoming high-speed air being rammed against a barrier. There-fore, a ramjet engine needs to be brought to a sufficiently high speed by anexternal source before it can be fired.

The ramjet performs best in aircraft flying above Mach 2 or 3 (two orthree times the speed of sound). In a ramjet, the air is slowed down to aboutMach 0.2, fuel is added to the air and burned at this low velocity, and thecombustion gases are expended and accelerated in a nozzle.

A scramjet engine is essentially a ramjet in which air flows through atsupersonic speeds (above the speed of sound). Ramjets that convert toscramjet configurations at speeds above Mach 6 are successfully tested atspeeds of about Mach 8.

Finally, a rocket is a device where a solid or liquid fuel and an oxidizerreact in the combustion chamber. The high-pressure combustion gases arethen expanded in a nozzle. The gases leave the rocket at very high veloci-ties, producing the thrust to propel the rocket.

9–12 ■ SECOND-LAW ANALYSIS OF GAS POWER CYCLES

The ideal Carnot, Ericsson, and Stirling cycles are totally reversible; thus theydo not involve any irreversibilities. The ideal Otto, Diesel, and Brayton cycles,however, are only internally reversible, and they may involve irreversibilities

Chapter 9 | 527

Air inlet

Fuel nozzles or spray bars

Jet nozzle

Flame holders

FIGURE 9–55A ramjet engine.

Source: The Aircraft Gas Turbine Engine and ItsOperation. © United Aircraft Corporation (nowUnited Technologies Corp.), 1951, 1974.

cen84959_ch09.qxd 4/26/05 5:45 PM Page 527

external to the system. A second-law analysis of these cycles reveals wherethe largest irreversibilities occur and where to start improvements.

Relations for exergy and exergy destruction for both closed and steady-flow systems are developed in Chap. 8. The exergy destruction for a closedsystem can be expressed as

(9–30)

where Tb,in and Tb,out are the temperatures of the system boundary whereheat is transferred into and out of the system, respectively. A similar relationfor steady-flow systems can be expressed, in rate form, as

(9–31)

or, on a unit–mass basis for a one-inlet, one-exit steady-flow device, as

(9–32)

where subscripts i and e denote the inlet and exit states, respectively.The exergy destruction of a cycle is the sum of the exergy destructions of

the processes that compose that cycle. The exergy destruction of a cycle canalso be determined without tracing the individual processes by consideringthe entire cycle as a single process and using one of the relations above.Entropy is a property, and its value depends on the state only. For a cycle,reversible or actual, the initial and the final states are identical; thus se � si.Therefore, the exergy destruction of a cycle depends on the magnitude ofthe heat transfer with the high- and low-temperature reservoirs involved andon their temperatures. It can be expressed on a unit–mass basis as

(9–33)

For a cycle that involves heat transfer only with a source at TH and a sink atTL, the exergy destruction becomes

(9–34)

The exergies of a closed system f and a fluid stream c at any state can bedetermined from

(9–35)

and

(9–36)

where subscript “0” denotes the state of the surroundings.

c � 1h � h0 2 � T0 1s � s0 2 �V2

2� gz 1kJ>kg 2

f � 1u � u0 2 � T0 1s � s0 2 � P0 1v � v0 2 �V2

2� gz 1kJ>kg 2

xdest � T0 a qout

TL

�qin

TH

b 1kJ>kg 2

xdest � T0 aa qout

Tb,out�a

qin

Tb,inb 1kJ>kg 2

X dest � T0sgen � T0 a se � si �qin

Tb,in�

qout

Tb,outb 1kJ>kg 2

X#

dest � T0S#gen � T0 1S# out � S

#in 2 � T0 aa

outm#s �a

inm#s �

Q#

in

Tb,in�

Q#

out

Tb,outb 1kW 2

� T0 c 1S2 � S1 2 sys �Q in

Tb,in�

Q out

Tb,outd 1kJ 2

X dest � T0Sgen � T0 1¢Ssys � Sin � Sout 2

528 | Thermodynamics

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Chapter 9 | 529

EXAMPLE 9–10 Second-Law Analysis of an Otto Cycle

Determine the exergy destruction associated with the Otto cycle (all fourprocesses as well as the cycle) discussed in Example 9–2, assuming thatheat is transferred to the working fluid from a source at 1700 K and heat isrejected to the surroundings at 290 K. Also, determine the exergy of theexhaust gases when they are purged.

Solution The Otto cycle analyzed in Example 9–2 is reconsidered. For spec-ified source and sink temperatures, the exergy destruction associated with thecycle and the exergy purged with the exhaust gases are to be determined.Analysis In Example 9–2, various quantities of interest were given or deter-mined to be

Processes 1-2 and 3-4 are isentropic (s1 � s2, s3 � s4) and therefore donot involve any internal or external irreversibilities; that is, Xdest,12 � 0 andXdest,34 � 0.

Processes 2-3 and 4-1 are constant-volume heat-addition and heat-rejectionprocesses, respectively, and are internally reversible. However, the heat trans-fer between the working fluid and the source or the sink takes place through afinite temperature difference, rendering both processes irreversible. Theexergy destruction associated with each process is determined from Eq. 9–32.However, first we need to determine the entropy change of air during theseprocesses:

Also,

Thus,

For process 4-1, s1 � s4 � s2 � s3 � �0.7540 kJ/kg · K, qR,41 � qout �381.83 kJ/kg, and Tsink � 290 K. Thus,

xdest,41 � T0 c 1s1 � s4 2 sys �qout

Tsinkd

� 82.2 kJ>kg

� 1290 K 2 c0.7540 kJ>kg # K �800 kJ>kg

1700 Kd

xdest,23 � T0 c 1s3 � s2 2 sys �qin

Tsourced

qin � 800 kJ>kg and Tsource � 1700 K

� 0.7540 kJ>kg # K

� 13.5045 � 2.4975 2 kJ>kg # K � 10.287 kJ>kg # K 2 ln 4.345 MPa

1.7997 MPa

s3 � s2 � s°3 � s°2 � R ln P3

P2

T3 � 1575.1 K wnet � 418.17 kJ>kg

T2 � 652.4 K qout � 381.83 kJ>kg

T1 � 290 K qin � 800 kJ>kg

T0 � 290 K P3 � 4.345 MPa

r � 8 P2 � 1.7997 MPa

cen84959_ch09.qxd 4/26/05 5:45 PM Page 529

530 | Thermodynamics

Therefore, the irreversibility of the cycle is

The exergy destruction of the cycle could also be determined from Eq. 9–34.Notice that the largest exergy destruction in the cycle occurs during theheat-rejection process. Therefore, any attempt to reduce the exergy destruc-tion should start with this process.

Disregarding any kinetic and potential energies, the exergy (work potential)of the working fluid before it is purged (state 4) is determined from Eq. 9–35:

where

Thus,

which is equivalent to the exergy destruction for process 4-1. (Why?)Discussion Note that 163.2 kJ/kg of work could be obtained from theexhaust gases if they were brought to the state of the surroundings in areversible manner.

f4 � 381.83 kJ>kg � 1290 K 2 10.7540 kJ>kg # K 2 � 0 � 163.2 kJ/kg

v4 � v0 � v4 � v1 � 0

u4 � u0 � u4 � u1 � qout � 381.83 kJ>kg

s4 � s0 � s4 � s1 � 0.7540 kJ>kg # K

f4 � 1u4 � u0 2 � T0 1s4 � s0 2 � P0 1v4 � v0 2

� 245.4 kJ/kg

� 0 � 82.2 kJ>kg � 0 � 163.2 kJ>kg

xdest,cycle � xdest,12 � xdest,23 � xdest,34 � xdest,41

� 163.2 kJ>kg

� 1290 K 2 c�0.7540 kJ>kg # K �381.83 kJ>kg

290 Kd

Two-thirds of the oil used in the United States is used for transportation. Halfof this oil is consumed by passenger cars and light trucks that are used to com-mute to and from work (38 percent), run a family business (35 percent), andfor recreational, social, and religious activities (27 percent). The overall fuelefficiency of the vehicles has increased considerably over the years due toimprovements primarily in aerodynamics, materials, and electronic controls.However, the average fuel consumption of new vehicles has not changed muchfrom about 20 miles per gallon (mpg) because of the increasing consumertrend toward purchasing larger and less fuel-efficient cars, trucks, and sportutility vehicles. Motorists also continue to drive more each year: 11,725 milesin 1999 compared to 10,277 miles in 1990. Consequently, the annual gasoline

*This section can be skipped without a loss in continuity. Information in this section

is based largely on the publications of the U.S. Department of Energy, Environmental

Protection Agency, and the American Automotive Association.

TOPIC OF SPECIAL INTEREST* Saving Fuel and Money by Driving Sensibly

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Chapter 9 | 531

use per vehicle in the United States has increased to 603 gallons in 1999(worth $1206 at $2.00/gal) from 506 gallons in 1990 (Fig. 9–56).

Saving fuel is not limited to good driving habits. It also involves purchas-ing the right car, using it responsibly, and maintaining it properly. A car doesnot burn any fuel when it is not running, and thus a sure way to save fuel isnot to drive the car at all—but this is not the reason we buy a car. We canreduce driving and thus fuel consumption by considering viable alternativessuch as living close to work and shopping areas, working at home, workinglonger hours in fewer days, joining a car pool or starting one, using publictransportation, combining errands into a single trip and planning ahead,avoiding rush hours and roads with heavy traffic and many traffic lights, andsimply walking or bicycling instead of driving to nearby places, with theadded benefit of good health and physical fitness. Driving only when neces-sary is the best way to save fuel, money, and the environment too.

Driving efficiently starts before buying a car, just like raising good chil-dren starts before getting married. The buying decision made now willaffect the fuel consumption for many years. Under average driving condi-tions, the owner of a 30-mpg vehicle will spend $400 less each year on fuelthan the owner of a 20-mpg vehicle (assuming a fuel cost of $2.00 per gal-lon and 12,000 miles of driving per year). If the vehicle is owned for5 years, the 30-mpg vehicle will save $2000 during this period (Fig. 9–57).The fuel consumption of a car depends on many factors such as the type ofthe vehicle, the weight, the transmission type, the size and efficiency of theengine, and the accessories and the options installed. The most fuel-efficient cars are aerodynamically designed compact cars with a smallengine, manual transmission, low frontal area (the height times the width ofthe car), and bare essentials.

At highway speeds, most fuel is used to overcome aerodynamic drag or airresistance to motion, which is the force needed to move the vehicle throughthe air. This resistance force is proportional to the drag coefficient and thefrontal area. Therefore, for a given frontal area, a sleek-looking aerodynami-cally designed vehicle with contoured lines that coincide with the stream-lines of air flow has a smaller drag coefficient and thus better fuel economythan a boxlike vehicle with sharp corners (Fig. 9–58). For the same overallshape, a compact car has a smaller frontal area and thus better fuel economycompared to a large car.

Moving around the extra weight requires more fuel, and thus it hurts fueleconomy. Therefore, the lighter the vehicle, the more fuel-efficient it is. Alsoas a general rule, the larger the engine is, the greater its rate of fuel con-sumption is. So you can expect a car with a 1.8 L engine to be more fuelefficient than one with a 3.0 L engine. For a given engine size, diesel enginesoperate on much higher compression ratios than the gasoline engines, andthus they are inherently more fuel-efficient. Manual transmissions are usu-ally more efficient than the automatic ones, but this is not always the case. Acar with automatic transmission generally uses 10 percent more fuel than acar with manual transmission because of the losses associated with thehydraulic connection between the engine and the transmission, and the addedweight. Transmissions with an overdrive gear (found in four-speed automatictransmissions and five-speed manual transmissions) save fuel and reduce

FIGURE 9–56The average car in the United States isdriven about 12,000 miles a year, usesabout 600 gallons of gasoline, worth$1200 at $2.00/gal.

30 MPG

20 MPG

$800/yr

$1200/yr

FIGURE 9–57Under average driving conditions, theowner of a 30-mpg vehicle spends$400 less each year on gasoline thanthe owner of a 20-mpg vehicle(assuming $2.00/gal and12,000 miles/yr).

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532 | Thermodynamics

noise and engine wear during highway driving by decreasing the engine rpmwhile maintaining the same vehicle speed.

Front wheel drive offers better traction (because of the engine weight ontop of the front wheels), reduced vehicle weight and thus better fuel econ-omy, with an added benefit of increased space in the passenger compart-ment. Four-wheel drive mechanisms provide better traction and braking thussafer driving on slippery roads and loose gravel by transmitting torque to allfour wheels. However, the added safety comes with increased weight, noise,and cost, and decreased fuel economy. Radial tires usually reduce the fuelconsumption by 5 to 10 percent by reducing the rolling resistance, but theirpressure should be checked regularly since they can look normal and still beunderinflated. Cruise control saves fuel during long trips on open roads bymaintaining steady speed. Tinted windows and light interior and exteriorcolors reduce solar heat gain, and thus the need for air-conditioning.

BEFORE DRIVING

Certain things done before driving can make a significant difference on thefuel cost of the vehicle while driving. Below we discuss some measures suchas using the right kind of fuel, minimizing idling, removing extra weight,and keeping the tires properly inflated.

Use Fuel with the Minimum Octane Number Recommended by the Vehicle ManufacturerMany motorists buy higher-priced premium fuel, thinking that it is better forthe engine. Most of today’s cars are designed to operate on regular unleadedfuel. If the owner’s manual does not call for premium fuel, using anythingother than regular gas is simply a waste of money. Octane number is not ameasure of the “power” or “quality” of the fuel, it is simply a measure offuel’s resistance to engine knock caused by premature ignition. Despite theimplications of flashy names like “premium,” “super,” or “power plus,” a fuelwith a higher octane number is not a better fuel; it is simply more expensivebecause of the extra processing involved to raise the octane number(Fig. 9–59). Older cars may need to go up one grade level from the recom-mended new car octane number if they start knocking.

Do Not Overfill the Gas TankTopping off the gas tank may cause the fuel to backflow during pumping.In hot weather, an overfilled tank may also cause the fuel to overflow dueto thermal expansion. This wastes fuel, pollutes the environment, and maydamage the car’s paint. Also, fuel tank caps that do not close tightly allowsome gasoline to be lost by evaporation. Buying fuel in cool weather suchas early in the mornings minimizes evaporative losses. Each gallon ofspilled or evaporated fuel emits as much hydrocarbon to the air as 7500miles of driving.

FIGURE 9–58Aerodynamically designed vehicleshave a smaller drag coefficient andthus better fuel economy than boxlikevehicles with sharp corners.

FIGURE 9–59Despite the implications of flashynames, a fuel with a higher octanenumber is not a better fuel; it is simplymore expensive.

© Vol. 21/PhotoDisc

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Chapter 9 | 533

Park in the GarageThe engine of a car parked in a garage overnight is warmer the next morn-ing. This reduces the problems associated with the warming-up period suchas starting, excessive fuel consumption, and environmental pollution. In hotweather, a garage blocks the direct sunlight and reduces the need for air-conditioning.

Start the Car Properly and Avoid Extended IdlingWith today’s cars, it is not necessary to prime the engine first by pumping theaccelerator pedal repeatedly before starting. This only wastes fuel. Warmingup the engine isn’t necessary either. Keep in mind that an idling enginewastes fuel and pollutes the environment. Don’t race a cold engine to warm itup. An engine warms up faster on the road under a light load, and the cat-alytic converter begins to function sooner. Start driving as soon as the engineis started, but avoid rapid acceleration and highway driving before the engineand thus the oil fully warms up to prevent engine wear.

In cold weather, the warm-up period is much longer, the fuel consumptionduring warm-up is much higher, and the exhaust emissions are much larger.At �20°C, for example, a car needs to be driven at least 3 miles to warm upfully. A gasoline engine uses up to 50 percent more fuel during warm-upthan it does after it is warmed up. Exhaust emissions from a cold engine dur-ing warm-up are much higher since the catalytic converters do not functionproperly before reaching their normal operating temperature of about 390°C.

Don’t Carry Unnecessary Weight in or on the VehicleRemove any snow or ice from the vehicle, and avoid carrying unneededitems, especially heavy ones (such as snow chains, old tires, books) in thepassenger compartment, trunk, or the cargo area of the vehicle (Fig. 9–60).This wastes fuel since it requires extra fuel to carry around the extra weight.An extra 100 lbm decreases fuel economy of a car by about 1–2 percent.

Some people find it convenient to use a roof rack or carrier for additionalcargo space. However, if you must carry some extra items, place theminside the vehicle rather than on roof racks to reduce drag. Any snow thataccumulates on a vehicle and distorts its shape must be removed for thesame reason. A loaded roof rack can increase fuel consumption by up to5 percent in highway driving. Even the most streamlined empty rackincreases aerodynamic drag and thus fuel consumption. Therefore, the roofrack should be removed when it is no longer needed.

Keep Tires Inflated to the Recommended Maximum PressureKeeping the tires inflated properly is one of the easiest and most importantthings one can do to improve fuel economy. If a range is recommended by themanufacturer, the higher pressure should be used to maximize fuel efficiency.Tire pressure should be checked when the tire is cold since tire pressurechanges with temperature (it increases by 1 psi for every 10°F rise in temper-ature due to a rise in ambient temperature or just road friction). Underinflatedtires run hot and jeopardize safety, cause the tires to wear prematurely, affect

FIGURE 9–60A loaded roof rack can increase fuelconsumption by up to 5 percent inhighway driving.

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534 | Thermodynamics

the vehicle’s handling adversely, and hurt the fuel economy by increasing therolling resistance. Overinflated tires cause unpleasant bumpy rides, and causethe tires to wear unevenly. Tires lose about 1 psi pressure per month due to airloss caused by the tire hitting holes, bumps, and curbs. Therefore, the tirepressure should be checked at least once a month. Just one tire underinflatedby 2 psi results in a 1 percent increase in fuel consumption (Fig. 9–61).Underinflated tires often cause fuel consumption of vehicles to increase by5 or 6 percent.

It is also important to keep the wheels aligned. Driving a vehicle with thefront wheels out of alignment increases rolling resistance and thus fuel con-sumption while causing handling problems and uneven tire wear. Therefore,the wheels should be aligned properly whenever necessary.

WHILE DRIVING

The driving habits can make a significant difference in the amount of fuelused. Driving sensibly and practicing some fuel-efficient driving techniquessuch as those discussed below can improve fuel economy easily by more than10 percent.

Avoid Quick Starts and Sudden StopsDespite the attention they may get, the abrupt, aggressive “jackrabbit” startswaste fuel, wear the tires, jeopardize safety, and are harder on vehicle com-ponents and connectors. The squealing stops wear the brake pads prema-turely, and may cause the driver to lose control of the vehicle. Easy startsand stops save fuel, reduce wear and tear, reduce pollution, and are safer andmore courteous to other drivers.

Drive at Moderate SpeedsAvoiding high speeds on open roads results in safer driving and better fueleconomy. In highway driving, over 50 percent of the power produced by theengine is used to overcome aerodynamic drag (i.e., to push air out of theway). Aerodynamic drag and thus fuel consumption increase rapidly atspeeds above 55 mph, as shown in Fig. 9–62. On average, a car uses about15 percent more fuel at 65 mph and 25 percent more fuel at 70 mph than itdoes at 55 mph. (A car uses about 10 percent more fuel at 100 km/h and 20percent more fuel at 110 km/h than it does at 90 km/h.)

The discussion above should not lead one to conclude that the lower thespeed, the better the fuel economy—because it is not. The number of milesthat can be driven per gallon of fuel drops sharply at speeds below 30 mph(or 50 km/h), as shown in the chart. Besides, speeds slower than the flow oftraffic can create a traffic hazard. Therefore, a car should be driven at moder-ate speeds for safety and best fuel economy.

Maintain a Constant SpeedThe fuel consumption remains at a minimum during steady driving at a mod-erate speed. Keep in mind that every time the accelerator is hard pressed,more fuel is pumped into the engine. The vehicle should be accelerated grad-ually and smoothly since extra fuel is squirted into the engine during quick

FIGURE 9–61Underinflated tires often cause fuelconsumption of vehicles to increase by5 or 6 percent.

© The McGraw-Hill Companies/Jill Braaten,photographer

15

15

20

25

MP

G

30

35

25 35 45

Speed (mph)

55 65 75

FIGURE 9–62Aerodynamic drag increases and thusfuel economy decreases rapidly atspeeds above 55 mph.

Source: EPA and U.S. Dept. of Energy.

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acceleration. Using cruise control on highway trips can help maintain a con-stant speed and reduce fuel consumption. Steady driving is also safer, easieron the nerves, and better for the heart.

Anticipate Traffic Ahead and Avoid TailgatingA driver can reduce fuel consumption by up to 10 percent by anticipatingtraffic conditions ahead and adjusting the speed accordingly, and avoidingtailgating and thus unnecessary braking and acceleration (Fig. 9–63). Accel-erations and decelerations waste fuel. Braking and abrupt stops can be mini-mized, for example, by not following too closely, and slowing down graduallyby releasing the gas pedal when approaching a red light, a stop sign, or slowtraffic. This relaxed driving style is safer, saves fuel and money, reduces pol-lution, reduces wear on the tires and brakes, and is appreciated by other driv-ers. Allowing sufficient time to reach the destination makes it easier to resistthe urge to tailgate.

Avoid Sudden Acceleration and Sudden Braking (Except in Emergencies)Accelerate gradually and smoothly when passing other vehicles or mergingwith faster traffic. Pumping or hard pressing the accelerator pedal while driv-ing causes the engine to switch to a “fuel enrichment mode” of operationthat wastes fuel. In city driving, nearly half of the engine power is used foracceleration. When accelerating with stick-shifts, the RPM of the engineshould be kept to a minimum. Braking wastes the mechanical energy pro-duced by the engine and wears the brake pads.

Avoid Resting Feet on the Clutch or Brake Pedal while DrivingResting the left foot on the brake pedal increases the temperature of the brakecomponents, and thus reduces their effectiveness and service life while wastingfuel. Similarly, resting the left foot on the clutch pedal lessens the pressure onthe clutch pads, causing them to slip and wear prematurely, wasting fuel.

Use Highest Gear (Overdrive) During Highway DrivingOverdrive improves fuel economy during highway driving by decreasing thevehicle’s engine speed (or RPM). The lower engine speed reduces fuel con-sumption per unit time as well as engine wear. Therefore, overdrive (the fifthgear in cars with overdrive manual transmission) should be used as soon asthe vehicle’s speed is high enough.

Turn the Engine Off Rather Than Letting It IdleUnnecessary idling during lengthy waits (such as waiting for someone or forservice at a drive-up window, being stuck in traffic, etc.) wastes fuel, pollutesthe air, and causes engine wear (more wear than driving) (Fig. 9–64). There-fore, the engine should be turned off rather than letting it idle. Idling formore than a minute consumes much more fuel than restarting the engine.Fuel consumption in the lines of drive-up windows and the pollution emittedcan be avoided altogether by simply parking the car and going inside.

FIGURE 9–63Fuel consumption can be decreased byup to 10 percent by anticipating trafficconditions ahead and adjustingaccordingly.

© Vol. 23/PhotoDisc

FIGURE 9–64Unnecessary idling during lengthywaits wastes fuel, costs money, andpollutes the air.

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536 | Thermodynamics

Use the Air Conditioner SparinglyAir-conditioning consumes considerable power and thus increases fuel con-sumption by 3 to 4 percent during highway driving, and by as much as 10 per-cent during city driving (Fig. 9–65). The best alternative to air-conditioning isto supply fresh outdoor air to the car through the vents by turning on the flow-through ventilation system (usually by running the air conditioner in the“economy” mode) while keeping the windows and the sunroof closed. Thismeasure is adequate to achieve comfort in pleasant weather, and it saves themost fuel since the compressor of the air conditioner is off. In warmer weather,however, ventilation cannot provide adequate cooling effect. In that case wecan attempt to achieve comfort by rolling down the windows or opening thesunroof. This is certainly a viable alternative for city driving, but not so onhighways since the aerodynamic drag caused by wide-open windows or sun-roof at highway speeds consumes more fuel than does the air conditioner.Therefore, at highway speeds, the windows or the sunroof should be closedand the air conditioner should be turned on instead to save fuel. This is espe-cially the case for the newer, aerodynamically designed cars.

Most air conditioners have a “maximum” or “recirculation” setting thatreduces the amount of hot outside air that must be cooled, and thus the fuelconsumption for air-conditioning. A passive measure to reduce the need forair conditioning is to park the vehicle in the shade, and to leave the windowsslightly open to allow for air circulation.

AFTER DRIVING

You cannot be an efficient person and accomplish much unless you take goodcare of yourself (eating right, maintaining physical fitness, having checkups,etc.), and the cars are no exception. Regular maintenance improves perfor-mance, increases gas mileage, reduces pollution, lowers repair costs, andextends engine life. A little time and money saved now may cost a lot later inincreased fuel, repair, and replacement costs.

Proper maintenance such as checking the levels of fluids (engine oil, coolant,transmission, brake, power steering, windshield washer, etc.), the tightness ofall belts, and formation of cracks or frays on hoses, belts, and wires, keepingtires properly inflated, lubricating the moving components, and replacingclogged air, fuel, or oil filters maximizes fuel efficiency (Fig. 9–66). Cloggedair filters increase fuel consumption (by up to 10 percent) and pollution byrestricting airflow to the engine, and thus they should be replaced. The carshould be tuned up regularly unless it has electronic controls and a fuel-injection system. High temperatures (which may be due to a malfunction ofthe cooling fan) should be avoided as they may cause the break down of theengine oil and thus excessive wear of the engine, and low temperatures(which may be due to a malfunction of the thermostat) may extend theengine’s warm-up period and may prevent the engine from reaching the opti-mum operating conditions. Both effects reduce fuel economy.

Clean oil extends engine life by reducing engine wear caused by friction,removes acids, sludge, and other harmful substances from the engine, improvesperformance, reduces fuel consumption, and decreases air pollution. Oil alsohelps to cool the engine, provides a seal between the cylinder walls and the

FIGURE 9–65Air conditioning increases fuelconsumption by 3 to 4 percent duringhighway driving, and by as much as10 percent during city driving.

FIGURE 9–66Proper maintenance maximizes fuelefficiency and extends engine life.

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pistons, and prevents the engine from rusting. Therefore, oil and oil filter shouldbe changed as recommended by the vehicle manufacturer. Fuel-efficient oils(indicated by “Energy Efficient API” label) contain certain additives that reducefriction and increase a vehicle’s fuel economy by 3 percent or more.

In summary, a person can save fuel, money, and the environment by pur-chasing an energy-efficient vehicle, minimizing the amount of driving, beingfuel-conscious while driving, and maintaining the car properly. These mea-sures have the added benefits of enhanced safety, reduced maintenance costs,and extended vehicle life.

SUMMARY

A cycle during which a net amount of work is produced iscalled a power cycle, and a power cycle during which theworking fluid remains a gas throughout is called a gas powercycle. The most efficient cycle operating between a heatsource at temperature TH and a sink at temperature TL is theCarnot cycle, and its thermal efficiency is given by

The actual gas cycles are rather complex. The approxi-mations used to simplify the analysis are known as the air-standard assumptions. Under these assumptions, all theprocesses are assumed to be internally reversible; the work-ing fluid is assumed to be air, which behaves as an ideal gas;and the combustion and exhaust processes are replaced byheat-addition and heat-rejection processes, respectively. Theair-standard assumptions are called cold-air-standard assump-tions if air is also assumed to have constant specific heats atroom temperature.

In reciprocating engines, the compression ratio r and themean effective pressure MEP are defined as

The Otto cycle is the ideal cycle for the spark-ignition recip-rocating engines, and it consists of four internally reversibleprocesses: isentropic compression, constant-volume heat addi-tion, isentropic expansion, and constant-volume heat rejection.Under cold-air-standard assumptions, the thermal efficiencyof the ideal Otto cycle is

where r is the compression ratio and k is the specific heatratio cp /cv.

hth,Otto � 1 �1

r k�1

MEP �wnet

vmax � vmin

r �Vmax

Vmin�

VBDC

VTDC

hth,Carnot � 1 �TL

TH

The Diesel cycle is the ideal cycle for the compression-ignition reciprocating engines. It is very similar to the Ottocycle, except that the constant-volume heat-addition processis replaced by a constant-pressure heat-addition process. Itsthermal efficiency under cold-air-standard assumptions is

where rc is the cutoff ratio, defined as the ratio of the cylindervolumes after and before the combustion process.

Stirling and Ericsson cycles are two totally reversiblecycles that involve an isothermal heat-addition process at THand an isothermal heat-rejection process at TL. They differfrom the Carnot cycle in that the two isentropic processes arereplaced by two constant-volume regeneration processes inthe Stirling cycle and by two constant-pressure regenerationprocesses in the Ericsson cycle. Both cycles utilize regenera-tion, a process during which heat is transferred to a thermalenergy storage device (called a regenerator) during one partof the cycle that is then transferred back to the working fluidduring another part of the cycle.

The ideal cycle for modern gas-turbine engines is the Bray-ton cycle, which is made up of four internally reversibleprocesses: isentropic compression, constant-pressure heat addi-tion, isentropic expansion, and constant-pressure heat rejection.Under cold-air-standard assumptions, its thermal efficiency is

where rp � Pmax/Pmin is the pressure ratio and k is the specificheat ratio. The thermal efficiency of the simple Brayton cycleincreases with the pressure ratio.

The deviation of the actual compressor and the turbine fromthe idealized isentropic ones can be accurately accounted forby utilizing their isentropic efficiencies, defined as

hC �ws

wa�

h2s � h1

h2a � h1

hth,Brayton � 1 �1

r 1k�12>kp

hth,Diesel � 1 �1

r k�1 c r kc � 1

k 1rc � 1 2 d

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and

where states 1 and 3 are the inlet states, 2a and 4a are theactual exit states, and 2s and 4s are the isentropic exit states.

In gas-turbine engines, the temperature of the exhaust gasleaving the turbine is often considerably higher than the tem-perature of the air leaving the compressor. Therefore, thehigh-pressure air leaving the compressor can be heated bytransferring heat to it from the hot exhaust gases in a counter-flow heat exchanger, which is also known as a regenerator.The extent to which a regenerator approaches an ideal regen-erator is called the effectiveness P and is defined as

Under cold-air-standard assumptions, the thermal effi-ciency of an ideal Brayton cycle with regeneration becomes

where T1 and T3 are the minimum and maximum tempera-tures, respectively, in the cycle.

The thermal efficiency of the Brayton cycle can also beincreased by utilizing multistage compression with intercool-ing, regeneration, and multistage expansion with reheating.The work input to the compressor is minimized when equalpressure ratios are maintained across each stage. This proce-dure also maximizes the turbine work output.

hth,regen � 1 � a T1

T3b 1rp 2 1k�12>k

P �qregen,act

qregen,max

hT �wa

ws�

h3 � h4a

h3 � h4s

538 | Thermodynamics

Gas-turbine engines are widely used to power aircraftbecause they are light and compact and have a high power-to-weight ratio. The ideal jet-propulsion cycle differs fromthe simple ideal Brayton cycle in that the gases arepartially expanded in the turbine. The gases that exit theturbine at a relatively high pressure are subsequently accel-erated in a nozzle to provide the thrust needed to propel theaircraft.

The net thrust developed by the engine is

where m. is the mass flow rate of gases, Vexit is the exit veloc-ity of the exhaust gases, and Vinlet is the inlet velocity of theair, both relative to the aircraft.

The power developed from the thrust of the engine iscalled the propulsive power W

.P, and it is given by

Propulsive efficiency is a measure of how efficiently theenergy released during the combustion process is convertedto propulsive energy, and it is defined as

For an ideal cycle that involves heat transfer only with asource at TH and a sink at TL, the exergy destruction is

xdest � T0 a qout

TL

�qin

TH

b

hP �Propulsive power

Energy input rate�

W#

P

Q#

in

W#

P � m# 1Vexit � Vinlet 2Vaircraft

F � m# 1Vexit � Vinlet 2

REFERENCES AND SUGGESTED READINGS

1. W. Z. Black and J. G. Hartley. Thermodynamics. NewYork: Harper & Row, 1985.

2. V. D. Chase. “Propfans: A New Twist for the Propeller.”Mechanical Engineering, November 1986, pp. 47–50.

3. C. R. Ferguson and A. T. Kirkpatrick, InternalCombustion Engines: Applied Thermosciences, 2nd ed.,New York: Wiley, 2000.

4. R. A. Harmon. “The Keys to Cogeneration and CombinedCycles.” Mechanical Engineering, February 1988,pp. 64–73.

5. J. Heywood, Internal Combustion Engine Fundamentals,New York: McGraw-Hill, 1988.

6. L. C. Lichty. Combustion Engine Processes. New York:McGraw-Hill, 1967.

7. H. McIntosh. “Jumbo Jet.” 10 Outstanding Achievements1964–1989. Washington, D.C.: National Academy ofEngineering, 1989, pp. 30–33.

8. W. Pulkrabek, Engineering Fundamentals of the InternalCombustion Engine, 2nd ed., Upper Saddle River, NJ:Prentice-Hall, 2004.

9. W. Siuru. “Two-stroke Engines: Cleaner and Meaner.”Mechanical Engineering. June 1990, pp. 66–69.

10. C. F. Taylor. The Internal Combustion Engine in Theoryand Practice. Cambridge, MA: M.I.T. Press, 1968.

11. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

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PROBLEMS*

Actual and Ideal Cycles, Carnot Cycle, Air-StandardAssumptions, Reciprocating Engines

9–1C Why is the Carnot cycle not suitable as an ideal cyclefor all power-producing cyclic devices?

9–2C How does the thermal efficiency of an ideal cycle, ingeneral, compare to that of a Carnot cycle operating betweenthe same temperature limits?

9–3C What does the area enclosed by the cycle representon a P-v diagram? How about on a T-s diagram?

9–4C What is the difference between air-standard assump-tions and the cold-air-standard assumptions?

9–5C How are the combustion and exhaust processes mod-eled under the air-standard assumptions?

9–6C What are the air-standard assumptions?

9–7C What is the difference between the clearance volumeand the displacement volume of reciprocating engines?

9–8C Define the compression ratio for reciprocating engines.

9–9C How is the mean effective pressure for reciprocatingengines defined?

9–10C Can the mean effective pressure of an automobileengine in operation be less than the atmospheric pressure?

9–11C As a car gets older, will its compression ratio change?How about the mean effective pressure?

9–12C What is the difference between spark-ignition andcompression-ignition engines?

9–13C Define the following terms related to reciprocatingengines: stroke, bore, top dead center, and clearance volume.

9–14 An air-standard cycle with variable specific heats isexecuted in a closed system and is composed of the followingfour processes:

1-2 Isentropic compression from 100 kPa and 27°C to800 kPa

2-3 v � constant heat addition to 1800 K3-4 Isentropic expansion to 100 kPa4-1 P � constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the net work output per unit mass.(c) Determine the thermal efficiency.

9–15 Reconsider Problem 9–14. Using EES (or other)software, study the effect of varying the temper-

ature after the constant-volume heat addition from 1500 K to2500 K. Plot the net work output and thermal efficiency as afunction of the maximum temperature of the cycle. Plot theT-s and P-v diagrams for the cycle when the maximum tem-perature of the cycle is 1800 K.

9–16 An air-standard cycle is executed in a closed systemand is composed of the following four processes:

1-2 Isentropic compression from 100 kPa and 27°C to1 MPa

2-3 P � constant heat addition in amount of 2800kJ/kg

3-4 v � constant heat rejection to 100 kPa4-1 P � constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the maximum temperature in the cycle.(c) Determine the thermal efficiency.

Assume constant specific heats at room temperature.Answers: (b) 3360 K, (c) 21.0 percent

9–17E An air-standard cycle with variable specific heats isexecuted in a closed system and is composed of the followingfour processes:

1-2 v � constant heat addition from 14.7 psia and80°F in the amount of 300 Btu/lbm

2-3 P � constant heat addition to 3200 R3-4 Isentropic expansion to 14.7 psia4-1 P � constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the total heat input per unit mass.(c) Determine the thermal efficiency.Answers: (b) 612.4 Btu/lbm, (c) 24.2 percent

9–18E Repeat Problem 9–17E using constant specific heatsat room temperature.

9–19 An air-standard cycle is executed in a closed systemwith 0.004 kg of air and consists of the following threeprocesses:

1-2 Isentropic compression from 100 kPa and 27°C to1 MPa

2-3 P � constant heat addition in the amount of 2.76 kJ3-1 P � c1v + c2 heat rejection to initial state (c1 and

c2 are constants)(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the heat rejected.(c) Determine the thermal efficiency.

Assume constant specific heats at room temperature.Answers: (b) 1.679 kJ, (c) 39.2 percent

* Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

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9–20 An air-standard cycle with variable specific heats isexecuted in a closed system with 0.003 kg of air and consistsof the following three processes:

1-2 v � constant heat addition from 95 kPa and 17°Cto 380 kPa

2-3 Isentropic expansion to 95 kPa3-1 P � constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the net work per cycle, in kJ.(c) Determine the thermal efficiency.

9–21 Repeat Problem 9–20 using constant specific heats atroom temperature.

9–22 Consider a Carnot cycle executed in a closed systemwith 0.003 kg of air. The temperature limits of the cycle are300 and 900 K, and the minimum and maximum pressures thatoccur during the cycle are 20 and 2000 kPa. Assuming con-stant specific heats, determine the net work output per cycle.

9–23 An air-standard Carnot cycle is executed in a closedsystem between the temperature limits of 350 and 1200 K. Thepressures before and after the isothermal compression are150 and 300 kPa, respectively. If the net work output per cycleis 0.5 kJ, determine (a) the maximum pressure in the cycle,(b) the heat transfer to air, and (c) the mass of air. Assumevariable specific heats for air. Answers: (a) 30,013 kPa,(b) 0.706 kJ, (c) 0.00296 kg

9–24 Repeat Problem 9–23 using helium as the working fluid.

9–25 Consider a Carnot cycle executed in a closed systemwith air as the working fluid. The maximum pressure in thecycle is 800 kPa while the maximum temperature is 750 K. Ifthe entropy increase during the isothermal heat rejectionprocess is 0.25 kJ/kg � K and the net work output is 100kJ/kg, determine (a) the minimum pressure in the cycle,(b) the heat rejection from the cycle, and (c) the thermal effi-ciency of the cycle. (d) If an actual heat engine cycle operatesbetween the same temperature limits and produces 5200 kWof power for an air flow rate of 90 kg/s, determine the secondlaw efficiency of this cycle.

Otto Cycle

9–26C What four processes make up the ideal Otto cycle?

9–27C How do the efficiencies of the ideal Otto cycle andthe Carnot cycle compare for the same temperature limits?Explain.

9–28C How is the rpm (revolutions per minute) of an actualfour-stroke gasoline engine related to the number of thermo-dynamic cycles? What would your answer be for a two-strokeengine?

9–29C Are the processes that make up the Otto cycle ana-lyzed as closed-system or steady-flow processes? Why?

9–30C How does the thermal efficiency of an ideal Ottocycle change with the compression ratio of the engine and thespecific heat ratio of the working fluid?

540 | Thermodynamics

9–31C Why are high compression ratios not used in spark-ignition engines?

9–32C An ideal Otto cycle with a specified compressionratio is executed using (a) air, (b) argon, and (c) ethane as theworking fluid. For which case will the thermal efficiency bethe highest? Why?

9–33C What is the difference between fuel-injected gaso-line engines and diesel engines?

9–34 An ideal Otto cycle has a compression ratio of 8. Atthe beginning of the compression process, air is at 95 kPaand 27°C, and 750 kJ/kg of heat is transferred to air duringthe constant-volume heat-addition process. Taking into accountthe variation of specific heats with temperature, determine(a) the pressure and temperature at the end of the heat-addition process, (b) the net work output, (c) the thermal effi-ciency, and (d ) the mean effective pressure for the cycle.Answers: (a) 3898 kPa, 1539 K, (b) 392.4 kJ/kg, (c) 52.3 per-cent, (d ) 495 kPa

9–35 Reconsider Problem 9–34. Using EES (or other)software, study the effect of varying the compres-

sion ratio from 5 to 10. Plot the net work output and thermalefficiency as a function of the compression ratio. Plot the T-sand P-v diagrams for the cycle when the compression ratio is 8.

9–36 Repeat Problem 9–34 using constant specific heats atroom temperature.

9–37 The compression ratio of an air-standard Otto cycle is9.5. Prior to the isentropic compression process, the air is at100 kPa, 35°C, and 600 cm3. The temperature at the end ofthe isentropic expansion process is 800 K. Using specificheat values at room temperature, determine (a) the highesttemperature and pressure in the cycle; (b) the amount of heattransferred in, in kJ; (c) the thermal efficiency; and (d ) themean effective pressure. Answers: (a) 1969 K, 6072 kPa,(b) 0.59 kJ, (c) 59.4 percent, (d) 652 kPa

9–38 Repeat Problem 9–37, but replace the isentropic expan-sion process by a polytropic expansion process with the poly-tropic exponent n � 1.35.

9–39E An ideal Otto cycle with air as the working fluid hasa compression ratio of 8. The minimum and maximum tem-peratures in the cycle are 540 and 2400 R. Accounting for thevariation of specific heats with temperature, determine (a) theamount of heat transferred to the air during the heat-additionprocess, (b) the thermal efficiency, and (c) the thermal effi-ciency of a Carnot cycle operating between the same temper-ature limits.

9–40E Repeat Problem 9–39E using argon as the workingfluid.

9–41 A four-cylinder, four-stroke, 2.2-L gasoline engineoperates on the Otto cycle with a compression ratio of 10. Theair is at 100 kPa and 60°C at the beginning of the compres-sion process, and the maximum pressure in the cycle is8 MPa. The compression and expansion processes may be

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modeled as polytropic with a polytropic constant of 1.3. Usingconstant specific heats at 850 K, determine (a) the tempera-ture at the end of the expansion process, (b) the net work out-put and the thermal efficiency, (c) the mean effective pressure,(d ) the engine speed for a net power output of 70 kW, and (e)the specific fuel consumption, in g/kWh, defined as the ratioof the mass of the fuel consumed to the net work produced.The air–fuel ratio, defined as the amount of air divided by theamount of fuel intake, is 16.

DIESEL CYCLE

9–42C How does a diesel engine differ from a gasolineengine?

9–43C How does the ideal Diesel cycle differ from theideal Otto cycle?

9–44C For a specified compression ratio, is a diesel orgasoline engine more efficient?

9–45C Do diesel or gasoline engines operate at higher com-pression ratios? Why?

9–46C What is the cutoff ratio? How does it affect the ther-mal efficiency of a Diesel cycle?

9–47 An air-standard Diesel cycle has a compression ratioof 16 and a cutoff ratio of 2. At the beginning of the com-pression process, air is at 95 kPa and 27°C. Accounting forthe variation of specific heats with temperature, determine(a) the temperature after the heat-addition process, (b) thethermal efficiency, and (c) the mean effective pressure.Answers: (a) 1724.8 K, (b) 56.3 percent, (c) 675.9 kPa

9–48 Repeat Problem 9–47 using constant specific heats atroom temperature.

9–49E An air-standard Diesel cycle has a compression ratioof 18.2. Air is at 80°F and 14.7 psia at the beginning of thecompression process and at 3000 R at the end of the heat-addition process. Accounting for the variation of specificheats with temperature, determine (a) the cutoff ratio, (b) theheat rejection per unit mass, and (c) the thermal efficiency.

9–50E Repeat Problem 9–49E using constant specific heatsat room temperature.

9–51 An ideal diesel engine has a compression ratio of 20and uses air as the working fluid. The state of air at thebeginning of the compression process is 95 kPa and 20°C. Ifthe maximum temperature in the cycle is not to exceed 2200K, determine (a) the thermal efficiency and (b) the meaneffective pressure. Assume constant specific heats for air atroom temperature. Answers: (a) 63.5 percent, (b) 933 kPa

9–52 Repeat Problem 9–51, but replace the isentropic expan-sion process by polytropic expansion process with the poly-tropic exponent n � 1.35.

9–53 Reconsider Problem 9–52. Using EES (or other)software, study the effect of varying the com-

pression ratio from 14 to 24. Plot the net work output, mean

effective pressure, and thermal efficiency as a function of thecompression ratio. Plot the T-s and P-v diagrams for thecycle when the compression ratio is 20.

9–54 A four-cylinder two-stroke 2.4-L diesel engine thatoperates on an ideal Diesel cycle has a compression ratio of17 and a cutoff ratio of 2.2. Air is at 55°C and 97 kPa at thebeginning of the compression process. Using the cold-air-standard assumptions, determine how much power the enginewill deliver at 1500 rpm.

9–55 Repeat Problem 9–54 using nitrogen as the workingfluid.

9–56 The compression ratio of an ideal dual cycle is14. Air is at 100 kPa and 300 K at the beginning

of the compression process and at 2200 K at the end of theheat-addition process. Heat transfer to air takes place partlyat constant volume and partly at constant pressure, and itamounts to 1520.4 kJ/kg. Assuming variable specific heatsfor air, determine (a) the fraction of heat transferred at con-stant volume and (b) the thermal efficiency of the cycle.

9–57 Reconsider Problem 9–56. Using EES (or other)software, study the effect of varying the com-

pression ratio from 10 to 18. For the compression ratio equalto 14, plot the T-s and P-v diagrams for the cycle.

9–58 Repeat Problem 9–56 using constant specific heats atroom temperature. Is the constant specific heat assumptionreasonable in this case?

9–59 A six-cylinder, four-stroke, 4.5-L compression-ignitionengine operates on the ideal diesel cycle with a compressionratio of 17. The air is at 95 kPa and 55°C at the beginning ofthe compression process and the engine speed is 2000 rpm.The engine uses light diesel fuel with a heating value of42,500 kJ/kg, an air–fuel ratio of 24, and a combustion effi-ciency of 98 percent. Using constant specific heats at 850 K,determine (a) the maximum temperature in the cycle and thecutoff ratio (b) the net work output per cycle and the thermalefficiency, (c) the mean effective pressure, (d ) the net poweroutput, and (e) the specific fuel consumption, in g/kWh,defined as the ratio of the mass of the fuel consumed to thenet work produced. Answers: (a) 2383 K, 2.7 (b) 4.36 kJ, 0.543,(c) 969 kPa, (d ) 72.7 kW, (e) 159 g/kWh

Stirling and Ericsson Cycles9–60C Consider the ideal Otto, Stirling, and Carnot cyclesoperating between the same temperature limits. How wouldyou compare the thermal efficiencies of these three cycles?

9–61C Consider the ideal Diesel, Ericsson, and Carnotcycles operating between the same temperature limits. Howwould you compare the thermal efficiencies of these threecycles?

9–62C What cycle is composed of two isothermal and twoconstant-volume processes?

9–63C How does the ideal Ericsson cycle differ from theCarnot cycle?

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9–64E An ideal Ericsson engine using helium as the work-ing fluid operates between temperature limits of 550 and3000 R and pressure limits of 25 and 200 psia. Assuming amass flow rate of 14 lbm/s, determine (a) the thermal effi-ciency of the cycle, (b) the heat transfer rate in the regenera-tor, and (c) the power delivered.

9–65 Consider an ideal Ericsson cycle with air as the work-ing fluid executed in a steady-flow system. Air is at 27°C and120 kPa at the beginning of the isothermal compressionprocess, during which 150 kJ/kg of heat is rejected. Heattransfer to air occurs at 1200 K. Determine (a) the maximumpressure in the cycle, (b) the net work output per unit mass ofair, and (c) the thermal efficiency of the cycle. Answers:(a) 685 kPa, (b) 450 kJ/kg, (c) 75 percent

9–66 An ideal Stirling engine using helium as the workingfluid operates between temperature limits of 300 and 2000 Kand pressure limits of 150 kPa and 3 MPa. Assuming the massof the helium used in the cycle is 0.12 kg, determine (a) thethermal efficiency of the cycle, (b) the amount of heat transferin the regenerator, and (c) the work output per cycle.

Ideal and Actual Gas-Turbine (Brayton) Cycles

9–67C Why are the back work ratios relatively high in gas-turbine engines?

9–68C What four processes make up the simple ideal Bray-ton cycle?

9–69C For fixed maximum and minimum temperatures, whatis the effect of the pressure ratio on (a) the thermal efficiencyand (b) the net work output of a simple ideal Brayton cycle?

9–70C What is the back work ratio? What are typical backwork ratio values for gas-turbine engines?

9–71C How do the inefficiencies of the turbine and thecompressor affect (a) the back work ratio and (b) the thermalefficiency of a gas-turbine engine?

9–72E A simple ideal Brayton cycle with air as the work-ing fluid has a pressure ratio of 10. The air enters the com-pressor at 520 R and the turbine at 2000 R. Accounting forthe variation of specific heats with temperature, determine(a) the air temperature at the compressor exit, (b) the backwork ratio, and (c) the thermal efficiency.

9–73 A simple Brayton cycle using air as the workingfluid has a pressure ratio of 8. The minimum

and maximum temperatures in the cycle are 310 and 1160 K.Assuming an isentropic efficiency of 75 percent for the com-pressor and 82 percent for the turbine, determine (a) the airtemperature at the turbine exit, (b) the net work output, and(c) the thermal efficiency.

9–74 Reconsider Problem 9–73. Using EES (or other)software, allow the mass flow rate, pressure ratio,

turbine inlet temperature, and the isentropic efficiencies of theturbine and compressor to vary. Assume the compressor inlet

542 | Thermodynamics

pressure is 100 kPa. Develop a general solution for the prob-lem by taking advantage of the diagram window method forsupplying data to EES software.

9–75 Repeat Problem 9–73 using constant specific heats atroom temperature.

9–76 Air is used as the working fluid in a simple idealBrayton cycle that has a pressure ratio of 12, a compressorinlet temperature of 300 K, and a turbine inlet temperature of1000 K. Determine the required mass flow rate of air for anet power output of 70 MW, assuming both the compressorand the turbine have an isentropic efficiency of (a) 100 per-cent and (b) 85 percent. Assume constant specific heats atroom temperature. Answers: (a) 352 kg/s, (b) 1037 kg/s

9–77 A stationary gas-turbine power plant operates on asimple ideal Brayton cycle with air as the working fluid. Theair enters the compressor at 95 kPa and 290 K and the tur-bine at 760 kPa and 1100 K. Heat is transferred to air at arate of 35,000 kJ/s. Determine the power delivered by thisplant (a) assuming constant specific heats at room tempera-ture and (b) accounting for the variation of specific heatswith temperature.

9–78 Air enters the compressor of a gas-turbine engine at300 K and 100 kPa, where it is compressed to 700 kPa and580 K. Heat is transferred to air in the amount of 950 kJ/kgbefore it enters the turbine. For a turbine efficiency of 86 per-cent, determine (a) the fraction of the turbine work outputused to drive the compressor and (b) the thermal efficiency.Assume variable specific heats for air.

9–79 Repeat Problem 9–78 using constant specific heats atroom temperature.

9–80E A gas-turbine power plant operates on a simpleBrayton cycle with air as the working fluid. The air enters theturbine at 120 psia and 2000 R and leaves at 15 psia and1200 R. Heat is rejected to the surroundings at a rate of 6400Btu/s, and air flows through the cycle at a rate of 40 lbm/s.Assuming the turbine to be isentropic and the compresssor tohave an isentropic efficiency of 80 percent, determine the netpower output of the plant. Account for the variation of spe-cific heats with temperature. Answer: 3373 kW

9–81E For what compressor efficiency will the gas-turbinepower plant in Problem 9–80E produce zero net work?

9–82 A gas-turbine power plant operates on the simple Bray-ton cycle with air as the working fluid and delivers 32 MW ofpower. The minimum and maximum temperatures in the cycleare 310 and 900 K, and the pressure of air at the compressorexit is 8 times the value at the compressor inlet. Assuming anisentropic efficiency of 80 percent for the compressor and86 percent for the turbine, determine the mass flow rate of airthrough the cycle. Account for the variation of specific heatswith temperature.

9–83 Repeat Problem 9–82 using constant specific heats atroom temperature.

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9–84 A gas-turbine power plant operates on the simpleBrayton cycle between the pressure limits of 100 and 1200kPa. The working fluid is air, which enters the compressor at30°C at a rate of 150 m3/min and leaves the turbine at 500°C.Using variable specific heats for air and assuming a compres-sor isentropic efficiency of 82 percent and a turbine isen-tropic efficiency of 88 percent, determine (a) the net poweroutput, (b) the back work ratio, and (c) the thermal efficiency.Answers: (a) 659 kW, (b) 0.625, (c) 0.319

Rover in the United Kingdom. The world’s first gas-turbine-powered automobile, the 200-hp Rover Jet 1, was built in1950 in the United Kingdom. This was followed by the pro-duction of the Plymouth Sport Coupe by Chrysler in 1954under the leadership of G. J. Huebner. Several hundred gas-turbine-powered Plymouth cars were built in the early 1960sfor demonstration purposes and were loaned to a select groupof people to gather field experience. The users had no com-plaints other than slow acceleration. But the cars were nevermass-produced because of the high production (especiallymaterial) costs and the failure to satisfy the provisions of the1966 Clean Air Act.

A gas-turbine-powered Plymouth car built in 1960 had aturbine inlet temperature of 1700°F, a pressure ratio of 4, anda regenerator effectiveness of 0.9. Using isentropic efficien-cies of 80 percent for both the compressor and the turbine,determine the thermal efficiency of this car. Also, determinethe mass flow rate of air for a net power output of 95 hp.Assume the ambient air to be at 540 R and 14.5 psia.

9–91 The 7FA gas turbine manufactured by GeneralElectric is reported to have an efficiency of 35.9

percent in the simple-cycle mode and to produce 159 MW ofnet power. The pressure ratio is 14.7 and the turbine inlettemperature is 1288°C. The mass flow rate through the tur-bine is 1,536,000 kg/h. Taking the ambient conditions to be20°C and 100 kPa, determine the isentropic efficiency of theturbine and the compressor. Also, determine the thermal effi-ciency of this gas turbine if a regenerator with an effective-ness of 80 percent is added.

9–92 Reconsider Problem 9–91. Using EES (or other)software, develop a solution that allows different

isentropic efficiencies for the compressor and turbine andstudy the effect of the isentropic efficiencies on net workdone and the heat supplied to the cycle. Plot the T-s diagramfor the cycle.

9–93 An ideal Brayton cycle with regeneration has a pres-sure ratio of 10. Air enters the compressor at 300 K and theturbine at 1200 K. If the effectiveness of the regenerator is100 percent, determine the net work output and the thermalefficiency of the cycle. Account for the variation of specificheats with temperature.

9–94 Reconsider Problem 9–93. Using EES (or other)software, study the effects of varying the isen-

tropic efficiencies for the compressor and turbine and regen-erator effectiveness on net work done and the heat supplied tothe cycle for the variable specific heat case. Plot the T-s dia-gram for the cycle.

9–95 Repeat Problem 9–93 using constant specific heats atroom temperature.

9–96 A Brayton cycle with regeneration using air as theworking fluid has a pressure ratio of 7. The minimum and max-imum temperatures in the cycle are 310 and 1150 K. Assumingan isentropic efficiency of 75 percent for the compressor and

Compressor Turbine

Combustionchamber

100 kPa30°C

1.2 MPa

500°C1

2 3

4

FIGURE P9–84

Brayton Cycle with Regeneration

9–85C How does regeneration affect the efficiency of aBrayton cycle, and how does it accomplish it?

9–86C Somebody claims that at very high pressure ratios,the use of regeneration actually decreases the thermal effi-ciency of a gas-turbine engine. Is there any truth in this claim?Explain.

9–87C Define the effectiveness of a regenerator used ingas-turbine cycles.

9–88C In an ideal regenerator, is the air leaving the com-pressor heated to the temperature at (a) turbine inlet, (b) tur-bine exit, (c) slightly above turbine exit?

9–89C In 1903, Aegidius Elling of Norway designed andbuilt an 11-hp gas turbine that used steam injection betweenthe combustion chamber and the turbine to cool the combus-tion gases to a safe temperature for the materials available atthe time. Currently there are several gas-turbine power plantsthat use steam injection to augment power and improve ther-mal efficiency. For example, the thermal efficiency of theGeneral Electric LM5000 gas turbine is reported to increasefrom 35.8 percent in simple-cycle operation to 43 percentwhen steam injection is used. Explain why steam injectionincreases the power output and the efficiency of gas turbines.Also, explain how you would obtain the steam.

9–90E The idea of using gas turbines to power automobileswas conceived in the 1930s, and considerable research wasdone in the 1940s and 1950s to develop automotive gas tur-bines by major automobile manufacturers such as theChrysler and Ford corporations in the United States and

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82 percent for the turbine and an effectiveness of 65 percent forthe regenerator, determine (a) the air temperature at the turbineexit, (b) the net work output, and (c) the thermal efficiency.Answers: (a) 783 K, (b) 108.1 kJ/kg, (c) 22.5 percent

9–97 A stationary gas-turbine power plant operates on anideal regenerative Brayton cycle (P � 100 percent) with airas the working fluid. Air enters the compressor at 95 kPaand 290 K and the turbine at 760 kPa and 1100 K. Heat is transferred to air from an external source at a rate of75,000 kJ/s. Determine the power delivered by this plant(a) assuming constant specific heats for air at room temper-ature and (b) accounting for the variation of specific heatswith temperature.

9–98 Air enters the compressor of a regenerative gas-turbineengine at 300 K and 100 kPa, where it is compressed to 800kPa and 580 K. The regenerator has an effectiveness of 72percent, and the air enters the turbine at 1200 K. For a tur-bine efficiency of 86 percent, determine (a) the amount ofheat transfer in the regenerator and (b) the thermal efficiency.Assume variable specific heats for air. Answers: (a) 152.5kJ/kg, (b) 36.0 percent

9–99 Repeat Problem 9–98 using constant specific heats atroom temperature.

9–100 Repeat Problem 9–98 for a regenerator effectivenessof 70 percent.

Brayton Cycle with Intercooling, Reheating, and Regeneration

9–101C Under what modifications will the ideal simplegas-turbine cycle approach the Ericsson cycle?

9–102C The single-stage compression process of an idealBrayton cycle without regeneration is replaced by a multi-stage compression process with intercooling between thesame pressure limits. As a result of this modification,(a) Does the compressor work increase, decrease, or remainthe same?(b) Does the back work ratio increase, decrease, or remainthe same?(c) Does the thermal efficiency increase, decrease, orremain the same?

9–103C The single-stage expansion process of an idealBrayton cycle without regeneration is replaced by a multi-stage expansion process with reheating between the samepressure limits. As a result of this modification,(a) Does the turbine work increase, decrease, or remain thesame?(b) Does the back work ratio increase, decrease, or remainthe same?(c) Does the thermal efficiency increase, decrease, or remainthe same?

9–104C A simple ideal Brayton cycle without regenerationis modified to incorporate multistage compression with inter-

544 | Thermodynamics

cooling and multistage expansion with reheating, withoutchanging the pressure or temperature limits of the cycle. As aresult of these two modifications,(a) Does the net work output increase, decrease, or remainthe same?(b) Does the back work ratio increase, decrease, or remainthe same?(c) Does the thermal efficiency increase, decrease, orremain the same?(d ) Does the heat rejected increase, decrease, or remain thesame?

9–105C A simple ideal Brayton cycle is modified to incor-porate multistage compression with intercooling, multistageexpansion with reheating, and regeneration without chang-ing the pressure limits of the cycle. As a result of thesemodifications,(a) Does the net work output increase, decrease, or remainthe same?(b) Does the back work ratio increase, decrease, or remainthe same?(c) Does the thermal efficiency increase, decrease, orremain the same?(d ) Does the heat rejected increase, decrease, or remain thesame?

9–106C For a specified pressure ratio, why does multistagecompression with intercooling decrease the compressor work,and multistage expansion with reheating increase the turbinework?

9–107C In an ideal gas-turbine cycle with intercooling,reheating, and regeneration, as the number of compressionand expansion stages is increased, the cycle thermal effi-ciency approaches (a) 100 percent, (b) the Otto cycle effi-ciency, or (c) the Carnot cycle efficiency.

9–108 Consider an ideal gas-turbine cycle with two stages ofcompression and two stages of expansion. The pressure ratioacross each stage of the compressor and turbine is 3. The airenters each stage of the compressor at 300 K and each stage ofthe turbine at 1200 K. Determine the back work ratio and thethermal efficiency of the cycle, assuming (a) no regenerator isused and (b) a regenerator with 75 percent effectiveness isused. Use variable specific heats.

9–109 Repeat Problem 9–108, assuming an efficiency of 80percent for each compressor stage and an efficiency of 85percent for each turbine stage.

9–110 Consider a regenerative gas-turbine power plant withtwo stages of compression and two stages of expansion. Theoverall pressure ratio of the cycle is 9. The air enters eachstage of the compressor at 300 K and each stage of the tur-bine at 1200 K. Accounting for the variation of specific heatswith temperature, determine the minimum mass flow rate ofair needed to develop a net power output of 110 MW.Answer: 250 kg/s

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9–111 Repeat Problem 9–110 using argon as the workingfluid.

Jet-Propulsion Cycles

9–112C What is propulsive power? How is it related tothrust?

9–113C What is propulsive efficiency? How is it deter-mined?

9–114C Is the effect of turbine and compressor irreversibil-ities of a turbojet engine to reduce (a) the net work, (b) thethrust, or (c) the fuel consumption rate?

9–115E A turbojet is flying with a velocity of 900 ft/s at analtitude of 20,000 ft, where the ambient conditions are 7 psiaand 10°F. The pressure ratio across the compressor is 13, andthe temperature at the turbine inlet is 2400 R. Assuming idealoperation for all components and constant specific heats forair at room temperature, determine (a) the pressure at the tur-bine exit, (b) the velocity of the exhaust gases, and (c) thepropulsive efficiency.

9–116E Repeat Problem 9–115E accounting for the varia-tion of specific heats with temperature.

9–117 A turbojet aircraft is flying with a velocity of320 m/s at an altitude of 9150 m, where the ambient condi-tions are 32 kPa and �32°C. The pressure ratio across thecompressor is 12, and the temperature at the turbine inlet is1400 K. Air enters the compressor at a rate of 60 kg/s, andthe jet fuel has a heating value of 42,700 kJ/kg. Assumingideal operation for all components and constant specific heatsfor air at room temperature, determine (a) the velocity of theexhaust gases, (b) the propulsive power developed, and(c) the rate of fuel consumption.

9–118 Repeat Problem 9–117 using a compressor effi-ciency of 80 percent and a turbine efficiency of 85 percent.

9–119 Consider an aircraft powered by a turbojet enginethat has a pressure ratio of 12. The aircraft is stationary onthe ground, held in position by its brakes. The ambient air isat 27°C and 95 kPa and enters the engine at a rate of 10 kg/s.The jet fuel has a heating value of 42,700 kJ/kg, and it isburned completely at a rate of 0.2 kg/s. Neglecting the effectof the diffuser and disregarding the slight increase in mass atthe engine exit as well as the inefficiencies of engine compo-nents, determine the force that must be applied on the brakesto hold the plane stationary. Answer: 9089 N

9–120 Reconsider Problem 9–119. In the problemstatement, replace the inlet mass flow rate by

an inlet volume flow rate of 9.063 m3/s. Using EES (or other)software, investigate the effect of compressor inlet tempera-ture in the range of –20 to 30°C on the force that must beapplied to the brakes to hold the plane stationary. Plot thisforce as a function in compressor inlet temperature.

9–121 Air at 7°C enters a turbojet engine at a rate of 16 kg/s and at a velocity of 300 m/s (relative to the engine).

Air is heated in the combustion chamber at a rate 15,000 kJ/sand it leaves the engine at 427°C. Determine the thrust produced by this turbojet engine. (Hint: Choose the entireengine as your control volume.)

Second-Law Analysis of Gas Power Cycles

9–122 Determine the total exergy destruction associatedwith the Otto cycle described in Problem 9–34, assuming asource temperature of 2000 K and a sink temperature of 300K. Also, determine the exergy at the end of the power stroke.Answers: 245.12 kJ/kg, 145.2 kJ/kg

9–123 Determine the total exergy destruction associatedwith the Diesel cycle described in Problem 9–47, assuming asource temperature of 2000 K and a sink temperature of 300K. Also, determine the exergy at the end of the isentropiccompression process. Answers: 292.7 kJ/kg, 348.6 kJ/kg

9–124E Determine the exergy destruction associated withthe heat rejection process of the Diesel cycle described inProblem 9–49E, assuming a source temperature of 3500 Rand a sink temperature of 540 R. Also, determine the exergyat the end of the isentropic expansion process.

9–125 Calculate the exergy destruction associated witheach of the processes of the Brayton cycle described in Prob-lem 9–73, assuming a source temperature of 1600 K and asink temperature of 290 K.

9–126 Determine the total exergy destruction associatedwith the Brayton cycle described in Problem 9–93, assuminga source temperature of 1800 K and a sink temperature of300 K. Also, determine the exergy of the exhaust gases at theexit of the regenerator.

9–127 Reconsider Problem 9–126. Using EES (orother) software, investigate the effect of vary-

ing the cycle pressure ratio from 6 to 14 on the total exergydestruction for the cycle and the exergy of the exhaust gasleaving the regenerator. Plot these results as functions ofpressure ratio. Discuss the results.

9–128 Determine the exergy destruction associated witheach of the processes of the Brayton cycle described inProblem 9–98, assuming a source temperature of 1260 Kand a sink temperature of 300 K. Also, determine theexergy of the exhaust gases at the exit of the regenerator.Take Pexhaust � P0 � 100 kPa.

9–129 A gas-turbine power plant operates on the simpleBrayton cycle between the pressure limits of 100 and 700kPa. Air enters the compressor at 30°C at a rate of 12.6 kg/sand leaves at 260°C. A diesel fuel with a heating value of42,000 kJ/kg is burned in the combustion chamber with anair–fuel ratio of 60 and a combustion efficiency of 97 per-cent. Combustion gases leave the combustion chamber andenter the turbine whose isentropic efficiency is 85 percent.Treating the combustion gases as air and using constant spe-cific heats at 500°C, determine (a) the isentropic efficiency

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of the compressor, (b) the net power output and the backwork ratio, (c) the thermal efficiency, and (d) the second-lawefficiency.

9–130 A four-cylinder, four-stroke, 2.8-liter modern, high-speed compression-ignition engine operates on the ideal dualcycle with a compression ratio of 14. The air is at 95 kPa and55°C at the beginning of the compression process and theengine speed is 3500 rpm. Equal amounts of fuel are burnedat constant volume and at constant pressure. The maximumallowable pressure in the cycle is 9 MPa due to materialstrength limitations. Using constant specific heats at 850 K,determine (a) the maximum temperature in the cycle, (b) thenet work output and the thermal efficiency, (c) the meaneffective pressure, and (d ) the net power output. Also, deter-mine (e) the second-law efficiency of the cycle and the rate ofexergy output with the exhaust gases when they are purged.Answers: (a) 3254 K, (b) 1349 kJ/kg, 0.587, (c) 1466 kPa, (d) 120 kW, (e) 0.646, 50.4 kW

9–131 A gas-turbine power plant operates on the regenera-tive Brayton cycle between the pressure limits of 100 and700 kPa. Air enters the compressor at 30°C at a rate of 12.6kg/s and leaves at 260°C. It is then heated in a regenerator to400°C by the hot combustion gases leaving the turbine. Adiesel fuel with a heating value of 42,000 kJ/kg is burned inthe combustion chamber with a combustion efficiency of 97percent. The combustion gases leave the combustion chamberat 871°C and enter the turbine whose isentropic efficiency is85 percent. Treating combustion gases as air and using con-stant specific heats at 500°C, determine (a) the isentropicefficiency of the compressor, (b) the effectiveness of theregenerator, (c) the air–fuel ratio in the combustion chamber,(d ) the net power output and the back work ratio, (e) the ther-mal efficiency, and ( f ) the second-law efficiency of the plant.Also determine (g) the second-law (exergetic) efficiencies ofthe compressor, the turbine, and the regenerator, and (h) therate of the exergy flow with the combustion gases at theregenerator exit. Answers: (a) 0.881, (b) 0.632, (c) 78.1,(d) 2267 kW, 0.583, (e) 0.345, (f ) 0.469, (g) 0.929, 0.932,0.890, (h) 1351 kW

546 | Thermodynamics

Review Problems

9–132 A four-stroke turbocharged V-16 diesel engine builtby GE Transportation Systems to power fast trains produces3500 hp at 1200 rpm. Determine the amount of power pro-duced per cylinder per (a) mechanical cycle and (b) thermo-dynamic cycle.

9–133 Consider a simple ideal Brayton cycle operatingbetween the temperature limits of 300 and 1500 K. Usingconstant specific heats at room temperature, determine thepressure ratio for which the compressor and the turbine exittemperatures of air are equal.

9–134 An air-standard cycle with variable coefficients isexecuted in a closed system and is composed of the followingfour processes:

1-2 v � constant heat addition from 100 kPa and27°C to 300 kPa

2-3 P � constant heat addition to 1027°C3-4 Isentropic expansion to 100 kPa4-1 P � constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the net work output per unit mass.(c) Determine the thermal efficiency.

9–135 Repeat Problem 9–134 using constant specific heatsat room temperature.

9–136 An air-standard cycle with variable specific heats isexecuted in a closed system with 0.003 kg of air, and it con-sists of the following three processes:

1-2 Isentropic compression from 100 kPa and 27°C to700 kPa

2-3 P � constant heat addition to initial specific volume3-1 v � constant heat rejection to initial state

(a) Show the cycle on P-v and T-s diagrams.(b) Calculate the maximum temperature in the cycle.(c) Determine the thermal efficiency.Answers: (b) 2100 K, (c) 15.8 percent

Compressor Turbine

Combustionchamber

100 kPa30°C

700 kPa260°C

Diesel fuel

1

2 3

4

FIGURE P9–129

Compressor Turbine

Combustionchamber

Regenerator

100 kPa30°C 700 kPa

260°C

400°C871°C1 2

6

5

3

4

FIGURE P9–131

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9–137 Repeat Problem 9–136 using constant specific heatsat room temperature.

9–138 A Carnot cycle is executed in a closed system anduses 0.0025 kg of air as the working fluid. The cycle effi-ciency is 60 percent, and the lowest temperature in the cycleis 300 K. The pressure at the beginning of the isentropicexpansion is 700 kPa, and at the end of the isentropic com-pression it is 1 MPa. Determine the net work output per cycle.

9–139 A four-cylinder spark-ignition engine has acompression ratio of 8, and each cylinder has a

maximum volume of 0.6 L. At the beginning of the compres-sion process, the air is at 98 kPa and 17°C, and the maximumtemperature in the cycle is 1800 K. Assuming the engine tooperate on the ideal Otto cycle, determine (a) the amount ofheat supplied per cylinder, (b) the thermal efficiency, and(c) the number of revolutions per minute required for a netpower output of 60 kW. Assume variable specific heats for air.

9–140 Reconsider Problem 9–139. Using EES (orother) software, study the effect of varying the

compression ratio from 5 to 11 on the net work done andthe efficiency of the cycle. Plot the P-v and T-s diagrams forthe cycle, and discuss the results.

9–141 An ideal Otto cycle has a compression ratio of 9.2and uses air as the working fluid. At the beginning of thecompression process, air is at 98 kPa and 27°C. The pressureis doubled during the constant-volume heat-addition process.Accounting for the variation of specific heats with tempera-ture, determine (a) the amount of heat transferred to the air,(b) the net work output, (c) the thermal efficiency, and (d ) themean effective pressure for the cycle.

9–142 Repeat Problem 9–141 using constant specific heatsat room temperature.

9–143 Consider an engine operating on the ideal Dieselcycle with air as the working fluid. The volume of the cylin-der is 1200 cm3 at the beginning of the compression process,75 cm3 at the end, and 150 cm3 after the heat-additionprocess. Air is at 17°C and 100 kPa at the beginning of thecompression process. Determine (a) the pressure at the begin-ning of the heat-rejection process, (b) the net work per cycle,in kJ, and (c) the mean effective pressure.

9–144 Repeat Problem 9–143 using argon as the workingfluid.

9–145E An ideal dual cycle has a compression ratio of 12and uses air as the working fluid. At the beginning of thecompression process, air is at 14.7 psia and 90°F, and occu-pies a volume of 75 in3. During the heat-addition process,0.3 Btu of heat is transferred to air at constant volume and1.1 Btu at constant pressure. Using constant specific heatsevaluated at room temperature, determine the thermal effi-ciency of the cycle.

9–146 Consider an ideal Stirling cycle using air as the work-ing fluid. Air is at 350 K and 200 kPa at the beginning of the

isothermal compression process, and heat is supplied to airfrom a source at 1800 K in the amount of 900 kJ/kg. Deter-mine (a) the maximum pressure in the cycle and (b) the network output per unit mass of air. Answers: (a) 5873 kPa,(b) 725 kJ/kg

9–147 Consider a simple ideal Brayton cycle with air as theworking fluid. The pressure ratio of the cycle is 6, and theminimum and maximum temperatures are 300 and 1300 K,respectively. Now the pressure ratio is doubled without chang-ing the minimum and maximum temperatures in the cycle.Determine the change in (a) the net work output per unitmass and (b) the thermal efficiency of the cycle as a result ofthis modification. Assume variable specific heats for air.Answers: (a) 41.5 kJ/kg, (b) 10.6 percent

9–148 Repeat Problem 9–147 using constant specific heatsat room temperature.

9–149 Helium is used as the working fluid in a Braytoncycle with regeneration. The pressure ratio of the cycle is 8,the compressor inlet temperature is 300 K, and the turbineinlet temperature is 1800 K. The effectiveness of the regener-ator is 75 percent. Determine the thermal efficiency and therequired mass flow rate of helium for a net power output of60 MW, assuming both the compressor and the turbine havean isentropic efficiency of (a) 100 percent and (b) 80 percent.

9–150 A gas-turbine engine with regeneration operates withtwo stages of compression and two stages of expansion. Thepressure ratio across each stage of the compressor and tur-bine is 3.5. The air enters each stage of the compressor at300 K and each stage of the turbine at 1200 K. The compres-sor and turbine efficiencies are 78 and 86 percent, respec-tively, and the effectiveness of the regenerator is 72 percent.Determine the back work ratio and the thermal efficiency ofthe cycle, assuming constant specific heats for air at roomtemperature. Answers: 53.2 percent, 39.2 percent

9–151 Reconsider Problem 9–150. Using EES (orother) software, study the effects of varying the

isentropic efficiencies for the compressor and turbine andregenerator effectiveness on net work done and the heat sup-plied to the cycle for the variable specific heat case. Let theisentropic efficiencies and the effectiveness vary from 70 per-cent to 90 percent. Plot the T-s diagram for the cycle.

9–152 Repeat Problem 9–150 using helium as the workingfluid.

9–153 Consider the ideal regenerative Brayton cycle. Deter-mine the pressure ratio that maximizes the thermal efficiencyof the cycle and compare this value with the pressure ratiothat maximizes the cycle net work. For the same maximum-to-minimum temperature ratios, explain why the pressureratio for maximum efficiency is less than the pressure ratiofor maximum work.

9–154 Consider an ideal gas-turbine cycle with one stage ofcompression and two stages of expansion and regeneration.

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The pressure ratio across each turbine stage is the same. Thehigh-pressure turbine exhaust gas enters the regenerator andthen enters the low-pressure turbine for expansion to thecompressor inlet pressure. Determine the thermal efficiencyof this cycle as a function of the compressor pressure ratioand the high-pressure turbine to compressor inlet temperatureratio. Compare your result with the efficiency of the standardregenerative cycle.

9–155 A four-cylinder, four-stroke spark-ignition engineoperates on the ideal Otto cycle with a compression ratio of11 and a total displacement volume of 1.8 liter. The air is at90 kPa and 50°C at the beginning of the compressionprocess. The heat input is 1.5 kJ per cycle per cylinder.Accounting for the variation of specific heats of air with tem-perature, determine (a) the maximum temperature and pres-sure that occur during the cycle, (b) the net work per cycleper cyclinder and the thermal efficiency of the cycle, (c) themean effective pressure, and (d ) the power output for anengine speed of 3000 rpm.

9–156 A gas-turbine plant operates on the regenerativeBrayton cycle with two stages of reheating and two-stages ofintercooling between the pressure limits of 100 and1200 kPa. The working fluid is air. The air enters the first andthe second stages of the compressor at 300 K and 350 K,respectively, and the first and the second stages of the turbineat 1400 K and 1300 K, respectively. Assuming both the com-pressor and the turbine have an isentropic efficiency of80 percent and the regenerator has an effectiveness of 75 per-cent and using variable specific heats, determine (a) the backwork ratio and the net work output, (b) the thermal efficiency,and (c) the second-law efficiency of the cycle. Also deter-mine (d ) the exergies at the exits of the combustion chamber(state 6) and the regenerator (state 10) (See Figure 9–43 inthe text). Answers: (a) 0.523, 317 kJ/kg, (b) 0.553, (c) 0.704,(d) 931 kJ/kg, 129 kJ/kg

9–157 Electricity and process heat requirements of a manu-facturing facility are to be met by a cogeneration plant consist-ing of a gas turbine and a heat exchanger for steam production.

548 | Thermodynamics

500°C

350°C 25°C

HeatexchangerCombustion

chamber

TurbineCompressor

100 kPa30°C

1

23

1.2 MPa Sat. vapor200°C

4

FIGURE P9–157

The plant operates on the simple Brayton cycle between thepressure limits of 100 and 1200 kPa with air as the workingfluid. Air enters the compressor at 30°C. Combustion gasesleave the turbine and enter the heat exchanger at 500°C, andleave the heat exchanger of 350°C, while the liquid waterenters the heat exchanger at 25°C and leaves at 200°C as a sat-urated vapor. The net power produced by the gas-turbine cycleis 800 kW. Assuming a compressor isentropic efficiency of82 percent and a turbine isentropic efficiency of 88 percent andusing variable specific heats, determine (a) the mass flow rateof air, (b) the back work ratio and the thermal efficiency, and(c) the rate at which steam is produced in the heat exchanger.Also determine (d) the utilization efficiency of the cogenera-tion plant, defined as the ratio of the total energy utilized to theenergy supplied to the plant.

9–158 A turbojet aircraft flies with a velocity of 900 km/hat an altitude where the air temperature and pressure are�35°C and 40 kPa. Air leaves the diffuser at 50 kPa with avelocity of 15 m/s, and combustion gases enter the turbine at450 kPa and 950°C. The turbine produces 500 kW of power,all of which is used to drive the compressor. Assuming anisentropic efficiency of 83 percent for the compressor, tur-bine, and nozzle, and using variable specific heats, determine(a) the pressure of combustion gases at the turbine exit,(b) the mass flow rate of air through the compressor, (c) thevelocity of the gases at the nozzle exit, and (d ) the propulsivepower and the propulsive efficiency for this engine. Answers:(a) 147 kPa, (b) 1.76 kg/s, (c) 719 m/s, (d) 206 kW, 0.156

9–159 Using EES (or other) software, study the effectof variable specific heats on the thermal effi-

ciency of the ideal Otto cycle using air as the working fluid.At the beginning of the compression process, air is at 100kPa and 300 K. Determine the percentage of error involved inusing constant specific heat values at room temperature forthe following combinations of compression ratios and maxi-mum cycle temperatures: r � 6, 8, 10, 12, and Tmax � 1000,1500, 2000, 2500 K.

9–160 Using EES (or other) software, determine theeffects of compression ratio on the net work

output and the thermal efficiency of the Otto cycle for a maxi-mum cycle temperature of 2000 K. Take the working fluid tobe air that is at 100 kPa and 300 K at the beginning of thecompression process, and assume variable specific heats. Varythe compression ratio from 6 to 15 with an increment of 1.Tabulate and plot your results against the compression ratio.

9–161 Using EES (or other) software, determine theeffects of pressure ratio on the net work output

and the thermal efficiency of a simple Brayton cycle for amaximum cycle temperature of 1800 K. Take the workingfluid to be air that is at 100 kPa and 300 K at the beginningof the compression process, and assume variable specificheats. Vary the pressure ratio from 5 to 24 with an incrementof 1. Tabulate and plot your results against the pressure ratio.At what pressure ratio does the net work output become a

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Chapter 9 | 549

maximum? At what pressure ratio does the thermal efficiencybecome a maximum?

9–162 Repeat Problem 9–161 assuming isentropicefficiencies of 85 percent for both the turbine

and the compressor.

9–163 Using EES (or other) software, determine theeffects of pressure ratio, maximum cycle tem-

perature, and compressor and turbine efficiencies on the network output per unit mass and the thermal efficiency of asimple Brayton cycle with air as the working fluid. Air is at100 kPa and 300 K at the compressor inlet. Also, assumeconstant specific heats for air at room temperature. Deter-mine the net work output and the thermal efficiency for allcombinations of the following parameters, and draw conclu-sions from the results.

Pressure ratio: 5, 8, 14Maximum cycle temperature: 800, 1200, 1600 KCompressor isentropic efficiency: 80, 100 percentTurbine isentropic efficiency: 80, 100 percent

9–164 Repeat Problem 9–163 by considering the vari-ation of specific heats of air with temperature.

9–165 Repeat Problem 9–163 using helium as theworking fluid.

9–166 Using EES (or other) software, determine theeffects of pressure ratio, maximum cycle tem-

perature, regenerator effectiveness, and compressor and tur-bine efficiencies on the net work output per unit mass and onthe thermal efficiency of a regenerative Brayton cycle withair as the working fluid. Air is at 100 kPa and 300 K at thecompressor inlet. Also, assume constant specific heats for airat room temperature. Determine the net work output and thethermal efficiency for all combinations of the followingparameters.

Pressure ratio: 6, 10Maximum cycle temperature: 1500, 2000 KCompressor isentropic efficiency: 80, 100 percentTurbine isentropic efficiency: 80, 100 percentRegenerator effectiveness: 70, 90 percent

9–167 Repeat Problem 9–166 by considering the vari-ation of specific heats of air with temperature.

9–168 Repeat Problem 9–166 using helium as theworking fluid.

9–169 Using EES (or other) software, determine theeffect of the number of compression and expan-

sion stages on the thermal efficiency of an ideal regenerativeBrayton cycle with multistage compression and expansion.Assume that the overall pressure ratio of the cycle is 12, andthe air enters each stage of the compressor at 300 K and eachstage of the turbine at 1200 K. Using constant specific heatsfor air at room temperature, determine the thermal efficiencyof the cycle by varying the number of stages from 1 to 22 inincrements of 3. Plot the thermal efficiency versus the number

of stages. Compare your results to the efficiency of an Erics-son cycle operating between the same temperature limits.

9–170 Repeat Problem 9–169 using helium as theworking fluid.

Fundamentals of Engineering (FE) Exam Problems9–171 An Otto cycle with air as the working fluid has acompression ratio of 8.2. Under cold-air-standard conditions,the thermal efficiency of this cycle is

(a) 24 percent (b) 43 percent (c) 52 percent(d) 57 percent (e) 75 percent

9–172 For specified limits for the maximum and minimumtemperatures, the ideal cycle with the lowest thermal effi-ciency is

(a) Carnot (b) Stirling (c) Ericsson(d ) Otto (e) All are the same

9–173 A Carnot cycle operates between the temperaturelimits of 300 and 2000 K, and produces 600 kW of netpower. The rate of entropy change of the working fluid dur-ing the heat addition process is

(a) 0 (b) 0.300 kW/K (c) 0.353 kW/K(d ) 0.261 kW/K (e) 2.0 kW/K

9–174 Air in an ideal Diesel cycle is compressed from 3 to0.15 L, and then it expands during the constant pressure heataddition process to 0.30 L. Under cold air standard condi-tions, the thermal efficiency of this cycle is

(a) 35 percent (b) 44 percent (c) 65 percent(d ) 70 percent (e) 82 percent

9–175 Helium gas in an ideal Otto cycle is compressedfrom 20°C and 2.5 to 0.25 L, and its temperature increases byan additional 700°C during the heat addition process. Thetemperature of helium before the expansion process is

(a) 1790°C (b) 2060°C (c) 1240°C(d ) 620°C (e) 820°C

9–176 In an ideal Otto cycle, air is compressed from 1.20kg/m3 and 2.2 to 0.26 L, and the net work output of the cycleis 440 kJ/kg. The mean effective pressure (MEP) for thiscycle is

(a) 612 kPa (b) 599 kPa (c) 528 kPa(d ) 416 kPa (e) 367 kPa

9–177 In an ideal Brayton cycle, air is compressed from 95kPa and 25°C to 800 kPa. Under cold-air-standard conditions,the thermal efficiency of this cycle is

(a) 46 percent (b) 54 percent (c) 57 percent(d ) 39 percent (e) 61 percent

9–178 Consider an ideal Brayton cycle executed betweenthe pressure limits of 1200 and 100 kPa and temperature lim-its of 20 and 1000°C with argon as the working fluid. The network output of the cycle is

(a) 68 kJ/kg (b) 93 kJ/kg (c) 158 kJ/kg(d ) 186 kJ/kg (e) 310 kJ/kg

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9–179 An ideal Brayton cycle has a net work output of 150kJ/kg and a back work ratio of 0.4. If both the turbine and thecompressor had an isentropic efficiency of 85 percent, the network output of the cycle would be

(a) 74 kJ/kg (b) 95 kJ/kg (c) 109 kJ/kg(d ) 128 kJ/kg (e) 177 kJ/kg

9–180 In an ideal Brayton cycle, air is compressed from100 kPa and 25°C to 1 MPa, and then heated to 1200°Cbefore entering the turbine. Under cold-air-standard condi-tions, the air temperature at the turbine exit is

(a) 490°C (b) 515°C (c) 622°C(d ) 763°C (e) 895°C

9–181 In an ideal Brayton cycle with regeneration, argongas is compressed from 100 kPa and 25°C to 400 kPa, andthen heated to 1200°C before entering the turbine. The high-est temperature that argon can be heated in the regenerator is

(a) 246°C (b) 846°C (c) 689°C(d ) 368°C (e) 573°C

9–182 In an ideal Brayton cycle with regeneration, air iscompressed from 80 kPa and 10°C to 400 kPa and 175°C, isheated to 450°C in the regenerator, and then further heated to1000°C before entering the turbine. Under cold-air-standardconditions, the effectiveness of the regenerator is

(a) 33 percent (b) 44 percent (c) 62 percent(d ) 77 percent (e) 89 percent

9–183 Consider a gas turbine that has a pressure ratio of 6and operates on the Brayton cycle with regeneration betweenthe temperature limits of 20 and 900°C. If the specific heatratio of the working fluid is 1.3, the highest thermal effi-ciency this gas turbine can have is

(a) 38 percent (b) 46 percent (c) 62 percent(d ) 58 percent (e) 97 percent

9–184 An ideal gas turbine cycle with many stages of com-pression and expansion and a regenerator of 100 percenteffectiveness has an overall pressure ratio of 10. Air entersevery stage of compressor at 290 K, and every stage of turbineat 1200 K. The thermal efficiency of this gas-turbine cycle is

(a) 36 percent (b) 40 percent (c) 52 percent(d ) 64 percent (e) 76 percent

9–185 Air enters a turbojet engine at 260 m/s at a rate of 30kg/s, and exits at 800 m/s relative to the aircraft. The thrustdeveloped by the engine is

(a) 8 kN (b) 16 kN (c) 24 kN(d ) 20 kN (e) 32 kN

Design and Essay Problems9–186 Design a closed-system air-standard gas power cyclecomposed of three processes and having a minimum thermalefficiency of 20 percent. The processes may be isothermal,isobaric, isochoric, isentropic, polytropic, or pressure as a lin-ear function of volume. Prepare an engineering report describ-

550 | Thermodynamics

ing your design, showing the system, P-v and T-s diagrams,and sample calculations.

9–187 Design a closed-system air-standard gas power cyclecomposed of three processes and having a minimum thermalefficiency of 20 percent. The processes may be isothermal,isobaric, isochoric, isentropic, polytropic, or pressure as a lin-ear function of volume; however, the Otto, Diesel, Ericsson,and Stirling cycles may not be used. Prepare an engineeringreport describing your design, showing the system, P-v andT-s diagrams, and sample calculations.

9–188 Write an essay on the most recent developments onthe two-stroke engines, and find out when we might be see-ing cars powered by two-stroke engines in the market. Whydo the major car manufacturers have a renewed interest intwo-stroke engines?

9–189 In response to concerns about the environment, somemajor car manufacturers are currently marketing electric cars.Write an essay on the advantages and disadvantages of elec-tric cars, and discuss when it is advisable to purchase an elec-tric car instead of a traditional internal combustion car.

9–190 Intense research is underway to develop adiabaticengines that require no cooling of the engine block. Suchengines are based on ceramic materials because of the abilityof such materials to withstand high temperatures. Write anessay on the current status of adiabatic engine development.Also determine the highest possible efficiencies with theseengines, and compare them to the highest possible efficien-cies of current engines.

9–191 Since its introduction in 1903 by Aegidius Elling ofNorway, steam injection between the combustion chamberand the turbine is used even in some modern gas turbinescurrently in operation to cool the combustion gases to ametallurgical-safe temperature while increasing the massflow rate through the turbine. Currently there are several gas-turbine power plants that use steam injection to augmentpower and improve thermal efficiency.

Consider a gas-turbine power plant whose pressure ratio is8. The isentropic efficiencies of the compressor and the turbineare 80 percent, and there is a regenerator with an effectivenessof 70 percent. When the mass flow rate of air through the com-pressor is 40 kg/s, the turbine inlet temperature becomes 1700K. But the turbine inlet temperature is limited to 1500 K, andthus steam injection into the combustion gases is being consid-ered. However, to avoid the complexities associated with steaminjection, it is proposed to use excess air (that is, to take inmuch more air than needed for complete combustion) to lowerthe combustion and thus turbine inlet temperature whileincreasing the mass flow rate and thus power output of the tur-bine. Evaluate this proposal, and compare the thermodynamicperformance of “high air flow” to that of a “steam-injection”gas-turbine power plant under the following design conditions:the ambient air is at 100 kPa and 25°C, adequate water supplyis available at 20°C, and the amount of fuel supplied to thecombustion chamber remains constant.

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Chapter 10VAPOR AND COMBINED POWER CYCLES

| 551

In Chap. 9 we discussed gas power cycles for which theworking fluid remains a gas throughout the entire cycle. Inthis chapter, we consider vapor power cycles in which the

working fluid is alternatively vaporized and condensed. Wealso consider power generation coupled with process heatingcalled cogeneration.

The continued quest for higher thermal efficiencies hasresulted in some innovative modifications to the basic vaporpower cycle. Among these, we discuss the reheat and regen-erative cycles, as well as combined gas–vapor power cycles.

Steam is the most common working fluid used in vaporpower cycles because of its many desirable characteristics,such as low cost, availability, and high enthalpy of vaporiza-tion. Therefore, this chapter is mostly devoted to the discus-sion of steam power plants. Steam power plants are commonlyreferred to as coal plants, nuclear plants, or natural gasplants, depending on the type of fuel used to supply heat tothe steam. However, the steam goes through the same basiccycle in all of them. Therefore, all can be analyzed in thesame manner.

ObjectivesThe objectives of Chapter 10 are to:

• Analyze vapor power cycles in which the working fluid isalternately vaporized and condensed.

• Analyze power generation coupled with process heatingcalled cogeneration.

• Investigate ways to modify the basic Rankine vapor powercycle to increase the cycle thermal efficiency.

• Analyze the reheat and regenerative vapor power cycles.

• Analyze power cycles that consist of two separate cyclesknown as combined cycles and binary cycles.

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10–1 ■ THE CARNOT VAPOR CYCLEWe have mentioned repeatedly that the Carnot cycle is the most efficientcycle operating between two specified temperature limits. Thus it is naturalto look at the Carnot cycle first as a prospective ideal cycle for vapor powerplants. If we could, we would certainly adopt it as the ideal cycle. Asexplained below, however, the Carnot cycle is not a suitable model forpower cycles. Throughout the discussions, we assume steam to be the work-ing fluid since it is the working fluid predominantly used in vapor powercycles.

Consider a steady-flow Carnot cycle executed within the saturation domeof a pure substance, as shown in Fig. 10-1a. The fluid is heated reversiblyand isothermally in a boiler (process 1-2), expanded isentropically in a tur-bine (process 2-3), condensed reversibly and isothermally in a condenser(process 3-4), and compressed isentropically by a compressor to the initialstate (process 4-1).

Several impracticalities are associated with this cycle:

1. Isothermal heat transfer to or from a two-phase system is not diffi-cult to achieve in practice since maintaining a constant pressure in thedevice automatically fixes the temperature at the saturation value. Therefore,processes 1-2 and 3-4 can be approached closely in actual boilers and con-densers. Limiting the heat transfer processes to two-phase systems, how-ever, severely limits the maximum temperature that can be used in the cycle(it has to remain under the critical-point value, which is 374°C for water).Limiting the maximum temperature in the cycle also limits the thermal effi-ciency. Any attempt to raise the maximum temperature in the cycle involvesheat transfer to the working fluid in a single phase, which is not easy toaccomplish isothermally.

2. The isentropic expansion process (process 2-3) can be approximatedclosely by a well-designed turbine. However, the quality of the steam decreasesduring this process, as shown on the T-s diagram in Fig. 10–1a. Thus theturbine has to handle steam with low quality, that is, steam with a highmoisture content. The impingement of liquid droplets on the turbine bladescauses erosion and is a major source of wear. Thus steam with qualities lessthan about 90 percent cannot be tolerated in the operation of power plants.

552 | Thermodynamics

s

T

34

1 2

s

T

34

1 2

(a) (b)

FIGURE 10–1T-s diagram of two Carnot vaporcycles.

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This problem could be eliminated by using a working fluid with a verysteep saturated vapor line.

3. The isentropic compression process (process 4-1) involves the com-pression of a liquid–vapor mixture to a saturated liquid. There are two difficul-ties associated with this process. First, it is not easy to control the condensationprocess so precisely as to end up with the desired quality at state 4. Second, itis not practical to design a compressor that handles two phases.

Some of these problems could be eliminated by executing the Carnotcycle in a different way, as shown in Fig. 10–1b. This cycle, however, pre-sents other problems such as isentropic compression to extremely high pres-sures and isothermal heat transfer at variable pressures. Thus we concludethat the Carnot cycle cannot be approximated in actual devices and is not arealistic model for vapor power cycles.

10–2 ■ RANKINE CYCLE: THE IDEAL CYCLEFOR VAPOR POWER CYCLES

Many of the impracticalities associated with the Carnot cycle can be elimi-nated by superheating the steam in the boiler and condensing it completelyin the condenser, as shown schematically on a T-s diagram in Fig. 10–2. Thecycle that results is the Rankine cycle, which is the ideal cycle for vaporpower plants. The ideal Rankine cycle does not involve any internal irre-versibilities and consists of the following four processes:

1-2 Isentropic compression in a pump

2-3 Constant pressure heat addition in a boiler

3-4 Isentropic expansion in a turbine

4-1 Constant pressure heat rejection in a condenser

Chapter 10 | 553

wturb,out

Pump

Turbine

3

4

Boiler

Condenser

wpump,in

2

1

qin

qout

s

T

3

2

41

wturb,out

wpump,in

qout

qin

FIGURE 10–2The simple ideal Rankine cycle.

SEE TUTORIAL CH. 10, SEC. 1 ON THE DVD.

INTERACTIVETUTORIAL

cen84959_ch10.qxd 4/25/05 3:56 PM Page 553

Water enters the pump at state 1 as saturated liquid and is compressedisentropically to the operating pressure of the boiler. The water temperatureincreases somewhat during this isentropic compression process due to aslight decrease in the specific volume of water. The vertical distancebetween states 1 and 2 on the T-s diagram is greatly exaggerated for clarity.(If water were truly incompressible, would there be a temperature change atall during this process?)

Water enters the boiler as a compressed liquid at state 2 and leaves as asuperheated vapor at state 3. The boiler is basically a large heat exchangerwhere the heat originating from combustion gases, nuclear reactors, or othersources is transferred to the water essentially at constant pressure. Theboiler, together with the section where the steam is superheated (the super-heater), is often called the steam generator.

The superheated vapor at state 3 enters the turbine, where it expands isen-tropically and produces work by rotating the shaft connected to an electricgenerator. The pressure and the temperature of steam drop during thisprocess to the values at state 4, where steam enters the condenser. At thisstate, steam is usually a saturated liquid–vapor mixture with a high quality.Steam is condensed at constant pressure in the condenser, which is basicallya large heat exchanger, by rejecting heat to a cooling medium such as alake, a river, or the atmosphere. Steam leaves the condenser as saturated liq-uid and enters the pump, completing the cycle. In areas where water is pre-cious, the power plants are cooled by air instead of water. This method ofcooling, which is also used in car engines, is called dry cooling. Severalpower plants in the world, including some in the United States, use drycooling to conserve water.

Remembering that the area under the process curve on a T-s diagramrepresents the heat transfer for internally reversible processes, we seethat the area under process curve 2-3 represents the heat transferred to thewater in the boiler and the area under the process curve 4-1 representsthe heat rejected in the condenser. The difference between these two(the area enclosed by the cycle curve) is the net work produced during thecycle.

Energy Analysis of the Ideal Rankine CycleAll four components associated with the Rankine cycle (the pump, boiler,turbine, and condenser) are steady-flow devices, and thus all four processesthat make up the Rankine cycle can be analyzed as steady-flow processes.The kinetic and potential energy changes of the steam are usually small rel-ative to the work and heat transfer terms and are therefore usuallyneglected. Then the steady-flow energy equation per unit mass of steamreduces to

(10–1)

The boiler and the condenser do not involve any work, and the pump andthe turbine are assumed to be isentropic. Then the conservation of energyrelation for each device can be expressed as follows:

Pump (q � 0): (10–2)wpump,in � h2 � h1

1qin � qout 2 � 1win � wout 2 � he � hi 1kJ>kg 2

554 | Thermodynamics

cen84959_ch10.qxd 4/19/05 2:17 PM Page 554

or,

(10–3)

where

(10–4)

Boiler (w � 0): (10–5)

Turbine (q � 0): (10–6)

Condenser (w � 0): (10–7)

The thermal efficiency of the Rankine cycle is determined from

(10–8)

where

The conversion efficiency of power plants in the United States is oftenexpressed in terms of heat rate, which is the amount of heat supplied, inBtu’s, to generate 1 kWh of electricity. The smaller the heat rate, the greaterthe efficiency. Considering that 1 kWh � 3412 Btu and disregarding thelosses associated with the conversion of shaft power to electric power, therelation between the heat rate and the thermal efficiency can be expressed as

(10–9)

For example, a heat rate of 11,363 Btu/kWh is equivalent to 30 percentefficiency.

The thermal efficiency can also be interpreted as the ratio of the areaenclosed by the cycle on a T-s diagram to the area under the heat-additionprocess. The use of these relations is illustrated in the following example.

EXAMPLE 10–1 The Simple Ideal Rankine Cycle

Consider a steam power plant operating on the simple ideal Rankine cycle.Steam enters the turbine at 3 MPa and 350°C and is condensed in the con-denser at a pressure of 75 kPa. Determine the thermal efficiency of thiscycle.

Solution A steam power plant operating on the simple ideal Rankine cycleis considered. The thermal efficiency of the cycle is to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.Analysis The schematic of the power plant and the T-s diagram of the cycleare shown in Fig. 10–3. We note that the power plant operates on the idealRankine cycle. Therefore, the pump and the turbine are isentropic, there areno pressure drops in the boiler and condenser, and steam leaves the con-denser and enters the pump as saturated liquid at the condenser pressure.

hth �3412 1Btu>kWh 2

Heat rate 1Btu>kWh 2

wnet � qin � qout � wturb,out � wpump,in

hth �wnet

qin� 1 �

qout

qin

qout � h4 � h1

wturb,out � h3 � h4

qin � h3 � h2

h1 � hf @ P1and v � v1 � vf @ P1

wpump,in � v 1P2 � P1 2

Chapter 10 | 555

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It is also interesting to note the thermal efficiency of a Carnot cycle oper-ating between the same temperature limits

The difference between the two efficiencies is due to the large external irre-versibility in Rankine cycle caused by the large temperature differencebetween steam and combustion gases in the furnace.

10–3 ■ DEVIATION OF ACTUAL VAPOR POWERCYCLES FROM IDEALIZED ONES

The actual vapor power cycle differs from the ideal Rankine cycle, as illus-trated in Fig. 10–4a, as a result of irreversibilities in various components.Fluid friction and heat loss to the surroundings are the two common sourcesof irreversibilities.

Fluid friction causes pressure drops in the boiler, the condenser, and thepiping between various components. As a result, steam leaves the boiler at asomewhat lower pressure. Also, the pressure at the turbine inlet is somewhatlower than that at the boiler exit due to the pressure drop in the connectingpipes. The pressure drop in the condenser is usually very small. To compen-sate for these pressure drops, the water must be pumped to a sufficientlyhigher pressure than the ideal cycle calls for. This requires a larger pumpand larger work input to the pump.

The other major source of irreversibility is the heat loss from the steam tothe surroundings as the steam flows through various components. To main-tain the same level of net work output, more heat needs to be transferred to

hth,Carnot � 1 �Tmin

Tmax� 1 �

191.76 � 273 2 K1350 � 273 2 K � 0.415

Chapter 10 | 557

wturb,out

Pump

Turbine

3

Boiler

Condenser

wpump,in

2

1

qout

qin

s

T, °C

2

41

3 MPa350°C

4

75 kPa

75 kPa

75 kPa

3 MPa

350

s1 = s2 s3 = s4

3

75 k

Pa

3 M

Pa

FIGURE 10–3Schematic and T-s diagram for Example 10–1.

SEE TUTORIAL CH. 10, SEC. 2 ON THE DVD.

INTERACTIVETUTORIAL

cen84959_ch10.qxd 4/25/05 3:56 PM Page 557

the steam in the boiler to compensate for these undesired heat losses. As aresult, cycle efficiency decreases.

Of particular importance are the irreversibilities occurring within thepump and the turbine. A pump requires a greater work input, and a turbineproduces a smaller work output as a result of irreversibilities. Under idealconditions, the flow through these devices is isentropic. The deviation ofactual pumps and turbines from the isentropic ones can be accounted for byutilizing isentropic efficiencies, defined as

(10–10)

and

(10–11)

where states 2a and 4a are the actual exit states of the pump and the turbine,respectively, and 2s and 4s are the corresponding states for the isentropiccase (Fig. 10–4b).

Other factors also need to be considered in the analysis of actual vaporpower cycles. In actual condensers, for example, the liquid is usually sub-cooled to prevent the onset of cavitation, the rapid vaporization and conden-sation of the fluid at the low-pressure side of the pump impeller, which maydamage it. Additional losses occur at the bearings between the moving partsas a result of friction. Steam that leaks out during the cycle and air thatleaks into the condenser represent two other sources of loss. Finally, thepower consumed by the auxiliary equipment such as fans that supply air tothe furnace should also be considered in evaluating the overall performanceof power plants.

The effect of irreversibilities on the thermal efficiency of a steam powercycle is illustrated below with an example.

hT �wa

ws�

h3 � h4a

h3 � h4s

hP �ws

wa�

h2s � h1

h2a � h1

558 | Thermodynamics

3

IDEAL CYCLE

ACTUAL CYCLE

Pressure dropin the condenser

Irreversibilityin the turbine

Irreversibilityin the pump Pressure drop

in the boiler

s

T

2

4

1

(a)

3

s

T

2s

4s1

4a

2a

(b)

FIGURE 10–4(a) Deviation of actual vapor power cycle from the ideal Rankine cycle. (b) The effect of pump andturbine irreversibilities on the ideal Rankine cycle.

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EXAMPLE 10–2 An Actual Steam Power Cycle

A steam power plant operates on the cycle shown in Fig. 10–5. If the isen-tropic efficiency of the turbine is 87 percent and the isentropic efficiency ofthe pump is 85 percent, determine (a) the thermal efficiency of the cycleand (b) the net power output of the plant for a mass flow rate of 15 kg/s.

Solution A steam power cycle with specified turbine and pump efficienciesis considered. The thermal efficiency and the net power output are to bedetermined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.Analysis The schematic of the power plant and the T-s diagram of the cycleare shown in Fig. 10–5. The temperatures and pressures of steam at variouspoints are also indicated on the figure. We note that the power plant involvessteady-flow components and operates on the Rankine cycle, but the imper-fections at various components are accounted for.

(a) The thermal efficiency of a cycle is the ratio of the net work output to theheat input, and it is determined as follows:

Pump work input:

� 19.0 kJ>kg

�10.001009 m3>kg 2 3 116,000 � 9 2 kPa 4

0.85 a 1 kJ

1 kPa # m3 b

wpump,in �ws,pump,in

hp�

v1 1P2 � P1 2hp

Chapter 10 | 559

wturb,out

wpump,in

PumpηP = 0.85

5

Boiler

Condenser

2

TurbineηT = 0.87

6

9 kPa38°C

1

15 MPa600°C

10 kPa

16 MPa

15.9 MPa35°C 15.2 MPa

625°C

4

3

s

T

4

5

66s13

2s2

FIGURE 10–5Schematic and T-s diagram for Example 10–2.

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Turbine work output:

Boiler heat input:

Thus,

(b) The power produced by this power plant is

Discussion Without the irreversibilities, the thermal efficiency of this cyclewould be 43.0 percent (see Example 10–3c).

10–4 ■ HOW CAN WE INCREASE THE EFFICIENCYOF THE RANKINE CYCLE?

Steam power plants are responsible for the production of most electricpower in the world, and even small increases in thermal efficiency can meanlarge savings from the fuel requirements. Therefore, every effort is made toimprove the efficiency of the cycle on which steam power plants operate.

The basic idea behind all the modifications to increase the thermal effi-ciency of a power cycle is the same: Increase the average temperature atwhich heat is transferred to the working fluid in the boiler, or decrease theaverage temperature at which heat is rejected from the working fluid in thecondenser. That is, the average fluid temperature should be as high as possi-ble during heat addition and as low as possible during heat rejection. Nextwe discuss three ways of accomplishing this for the simple ideal Rankinecycle.

Lowering the Condenser Pressure (Lowers Tlow,avg)Steam exists as a saturated mixture in the condenser at the saturationtemperature corresponding to the pressure inside the condenser. Therefore,lowering the operating pressure of the condenser automatically lowers thetemperature of the steam, and thus the temperature at which heat is rejected.

The effect of lowering the condenser pressure on the Rankine cycle effi-ciency is illustrated on a T-s diagram in Fig. 10–6. For comparison pur-poses, the turbine inlet state is maintained the same. The colored area onthis diagram represents the increase in net work output as a result of lower-ing the condenser pressure from P4 to . The heat input requirements alsoincrease (represented by the area under curve 2�-2), but this increase is verysmall. Thus the overall effect of lowering the condenser pressure is anincrease in the thermal efficiency of the cycle.

P4¿

W#

net � m# 1wnet 2 � 115 kg>s 2 11258.0 kJ>kg 2 � 18.9 MW

hth �wnet

qin�

1258.0 kJ>kg

3487.5 kJ>kg� 0.361 or 36.1%

wnet � wturb,out � wpump,in � 11277.0 � 19.0 2 kJ>kg � 1258.0 kJ>kg

qin � h4 � h3 � 13647.6 � 160.1 2 kJ>kg � 3487.5 kJ>kg

� 1277.0 kJ>kg

� hT 1h5 � h6s 2 � 0.87 13583.1 � 2115.3 2 kJ>kg

wturb,out � hTws,turb,out

560 | Thermodynamics

3

s

T

41

2

1'

2'

4'

P '4< P 4

Increase in wnet

FIGURE 10–6The effect of lowering the condenserpressure on the ideal Rankine cycle.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 560

To take advantage of the increased efficiencies at low pressures, the con-densers of steam power plants usually operate well below the atmosphericpressure. This does not present a major problem since the vapor powercycles operate in a closed loop. However, there is a lower limit on the con-denser pressure that can be used. It cannot be lower than the saturation pres-sure corresponding to the temperature of the cooling medium. Consider, forexample, a condenser that is to be cooled by a nearby river at 15°C. Allow-ing a temperature difference of 10°C for effective heat transfer, the steamtemperature in the condenser must be above 25°C; thus the condenser pres-sure must be above 3.2 kPa, which is the saturation pressure at 25°C.

Lowering the condenser pressure is not without any side effects, however.For one thing, it creates the possibility of air leakage into the condenser.More importantly, it increases the moisture content of the steam at the finalstages of the turbine, as can be seen from Fig. 10–6. The presence of largequantities of moisture is highly undesirable in turbines because it decreasesthe turbine efficiency and erodes the turbine blades. Fortunately, this prob-lem can be corrected, as discussed next.

Superheating the Steam to High Temperatures(Increases Thigh,avg)The average temperature at which heat is transferred to steam can beincreased without increasing the boiler pressure by superheating the steam tohigh temperatures. The effect of superheating on the performance of vaporpower cycles is illustrated on a T-s diagram in Fig. 10–7. The colored area onthis diagram represents the increase in the net work. The total area under theprocess curve 3-3� represents the increase in the heat input. Thus both the network and heat input increase as a result of superheating the steam to a highertemperature. The overall effect is an increase in thermal efficiency, however,since the average temperature at which heat is added increases.

Superheating the steam to higher temperatures has another very desirableeffect: It decreases the moisture content of the steam at the turbine exit, ascan be seen from the T-s diagram (the quality at state 4� is higher than thatat state 4).

The temperature to which steam can be superheated is limited, however, bymetallurgical considerations. Presently the highest steam temperature allowedat the turbine inlet is about 620°C (1150°F). Any increase in this valuedepends on improving the present materials or finding new ones that canwithstand higher temperatures. Ceramics are very promising in this regard.

Increasing the Boiler Pressure (Increases Thigh,avg)Another way of increasing the average temperature during the heat-additionprocess is to increase the operating pressure of the boiler, which automati-cally raises the temperature at which boiling takes place. This, in turn, raisesthe average temperature at which heat is transferred to the steam and thusraises the thermal efficiency of the cycle.

The effect of increasing the boiler pressure on the performance of vaporpower cycles is illustrated on a T-s diagram in Fig. 10–8. Notice that for afixed turbine inlet temperature, the cycle shifts to the left and the moisture con-tent of steam at the turbine exit increases. This undesirable side effect can becorrected, however, by reheating the steam, as discussed in the next section.

Chapter 10 | 561

3

s

T

41

2

Increase in wnet

3'

4'

FIGURE 10–7The effect of superheating the steam tohigher temperatures on the idealRankine cycle.

3

s

T

1

2

Increase in wnet

3'

4

2'

4'

Decrease in wnet

Tmax

FIGURE 10–8The effect of increasing the boilerpressure on the ideal Rankine cycle.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 561

Operating pressures of boilers have gradually increased over the yearsfrom about 2.7 MPa (400 psia) in 1922 to over 30 MPa (4500 psia) today,generating enough steam to produce a net power output of 1000 MW or morein a large power plant. Today many modern steam power plants operate atsupercritical pressures (P � 22.06 MPa) and have thermal efficiencies ofabout 40 percent for fossil-fuel plants and 34 percent for nuclear plants.There are over 150 supercritical-pressure steam power plants in operation inthe United States. The lower efficiencies of nuclear power plants are due tothe lower maximum temperatures used in those plants for safety reasons.The T-s diagram of a supercritical Rankine cycle is shown in Fig. 10–9.

The effects of lowering the condenser pressure, superheating to a highertemperature, and increasing the boiler pressure on the thermal efficiency ofthe Rankine cycle are illustrated below with an example.

EXAMPLE 10–3 Effect of Boiler Pressure and Temperature on Efficiency

Consider a steam power plant operating on the ideal Rankine cycle. Steamenters the turbine at 3 MPa and 350°C and is condensed in the condenser ata pressure of 10 kPa. Determine (a) the thermal efficiency of this powerplant, (b) the thermal efficiency if steam is superheated to 600°C instead of350°C, and (c) the thermal efficiency if the boiler pressure is raised to 15MPa while the turbine inlet temperature is maintained at 600°C.

Solution A steam power plant operating on the ideal Rankine cycle is con-sidered. The effects of superheating the steam to a higher temperature andraising the boiler pressure on thermal efficiency are to be investigated.Analysis The T-s diagrams of the cycle for all three cases are given inFig. 10–10.

562 | Thermodynamics

3

s

T

1

2

4

Criticalpoint

FIGURE 10–9A supercritical Rankine cycle.

3

s

T

1

2

4

T3 = 350°C3 MPa

10 kPa

(a)

3

s

T

1

2

4

T3 = 600°C

3 MPa

10 kPa

(b)

3

s

T

1

2

4

15 MPa

10 kPa

(c)

T3 = 600°C

FIGURE 10–10T-s diagrams of the three cycles discussed in Example 10–3.

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Chapter 10 | 563

(a) This is the steam power plant discussed in Example 10–1, except thatthe condenser pressure is lowered to 10 kPa. The thermal efficiency isdetermined in a similar manner:

State 1:

State 2:

State 3:

State 4:

Thus,

and

Therefore, the thermal efficiency increases from 26.0 to 33.4 percent as aresult of lowering the condenser pressure from 75 to 10 kPa. At the sametime, however, the quality of the steam decreases from 88.6 to 81.3 percent(in other words, the moisture content increases from 11.4 to 18.7 percent).

(b) States 1 and 2 remain the same in this case, and the enthalpies atstate 3 (3 MPa and 600°C) and state 4 (10 kPa and s4 � s3) are deter-mined to be

Thus,

and

hth � 1 �qout

qin� 1 �

2188.5 kJ>kg

3488.0 kJ>kg� 0.373 or 37.3%

qout � h4 � h1 � 2380.3 � 191.81 � 2188.5 kJ>kg

qin � h3 � h2 � 3682.8 � 194.83 � 3488.0 kJ>kg

h4 � 2380.3 kJ>kg 1x4 � 0.915 2 h3 � 3682.8 kJ>kg

hth � 1 �qout

qin� 1 �

1944.3 kJ>kg

2921.3 kJ>kg� 0.334 or 33.4%

qout � h4 � h1 � 12136.1 � 191.81 2 kJ>kg � 1944.3 kJ>kg

qin � h3 � h2 � 13116.1 � 194.83 2 kJ>kg � 2921.3 kJ>kg

h4 � hf � x4hfg � 191.81 � 0.8128 12392.1 2 � 2136.1 kJ>kg

x4 �s4 � sf

sfg�

6.7450 � 0.6492

7.4996� 0.8128

s4 � s3

P4 � 10 kPa 1sat. mixture 2 P3 � 3 MPa

T3 � 350°Cf h3 � 3116.1 kJ>kg

s3 � 6.7450 kJ>kg # K

h2 � h1 � wpump,in � 1191.81 � 3.02 2 kJ>kg � 194.83 kJ>kg

� 3.02 kJ>kg

wpump,in � v1 1P2 � P1 2 � 10.00101 m3>kg 2 3 13000 � 10 2 kPa 4 a 1 kJ

1 kPa # m3 b s2 � s1

P2 � 3 MPa

P1 � 10 kPa

Sat. liquidf h1 � hf @ 10 kPa � 191.81 kJ>kg

v1 � vf @ 10 kPa � 0.00101 m3>kg

cen84959_ch10.qxd 4/19/05 2:17 PM Page 563

564 | Thermodynamics

Therefore, the thermal efficiency increases from 33.4 to 37.3 percent as aresult of superheating the steam from 350 to 600°C. At the same time, thequality of the steam increases from 81.3 to 91.5 percent (in other words,the moisture content decreases from 18.7 to 8.5 percent).

(c) State 1 remains the same in this case, but the other states change. Theenthalpies at state 2 (15 MPa and s2 � s1), state 3 (15 MPa and 600°C),and state 4 (10 kPa and s4 � s3) are determined in a similar manner to be

Thus,

and

Discussion The thermal efficiency increases from 37.3 to 43.0 percent as aresult of raising the boiler pressure from 3 to 15 MPa while maintaining theturbine inlet temperature at 600°C. At the same time, however, the qualityof the steam decreases from 91.5 to 80.4 percent (in other words, the mois-ture content increases from 8.5 to 19.6 percent).

10–5 ■ THE IDEAL REHEAT RANKINE CYCLEWe noted in the last section that increasing the boiler pressure increases thethermal efficiency of the Rankine cycle, but it also increases the moisturecontent of the steam to unacceptable levels. Then it is natural to ask the fol-lowing question:

How can we take advantage of the increased efficiencies at higher boilerpressures without facing the problem of excessive moisture at the final stages of the turbine?

Two possibilities come to mind:

1. Superheat the steam to very high temperatures before it enters theturbine. This would be the desirable solution since the average temperatureat which heat is added would also increase, thus increasing the cycle effi-ciency. This is not a viable solution, however, since it requires raising thesteam temperature to metallurgically unsafe levels.

2. Expand the steam in the turbine in two stages, and reheat it inbetween. In other words, modify the simple ideal Rankine cycle with areheat process. Reheating is a practical solution to the excessive moistureproblem in turbines, and it is commonly used in modern steam power plants.

The T-s diagram of the ideal reheat Rankine cycle and the schematic ofthe power plant operating on this cycle are shown in Fig. 10–11. The idealreheat Rankine cycle differs from the simple ideal Rankine cycle in that the

hth � 1 �qout

qin� 1 �

1923.5 kJ>kg

3376.2 kJ>kg� 0.430 or 43.0%

qout � h4 � h1 � 2115.3 � 191.81 � 1923.5 kJ>kg

qin � h3 � h2 � 3583.1 � 206.95 � 3376.2 kJ>kg

h4 � 2115.3 kJ>kg 1x4 � 0.804 2 h3 � 3583.1 kJ>kg

h2 � 206.95 kJ>kg

SEE TUTORIAL CH. 10, SEC. 3 ON THE DVD.

INTERACTIVETUTORIAL

cen84959_ch10.qxd 4/25/05 3:56 PM Page 564

expansion process takes place in two stages. In the first stage (the high-pressure turbine), steam is expanded isentropically to an intermediate pres-sure and sent back to the boiler where it is reheated at constant pressure,usually to the inlet temperature of the first turbine stage. Steam then expandsisentropically in the second stage (low-pressure turbine) to the condenserpressure. Thus the total heat input and the total turbine work output for areheat cycle become

(10–12)

and

(10–13)

The incorporation of the single reheat in a modern power plant improvesthe cycle efficiency by 4 to 5 percent by increasing the average temperatureat which heat is transferred to the steam.

The average temperature during the reheat process can be increased byincreasing the number of expansion and reheat stages. As the number ofstages is increased, the expansion and reheat processes approach an isother-mal process at the maximum temperature, as shown in Fig. 10–12. The useof more than two reheat stages, however, is not practical. The theoreticalimprovement in efficiency from the second reheat is about half of thatwhich results from a single reheat. If the turbine inlet pressure is not highenough, double reheat would result in superheated exhaust. This is undesir-able as it would cause the average temperature for heat rejection to increaseand thus the cycle efficiency to decrease. Therefore, double reheat is usedonly on supercritical-pressure (P � 22.06 MPa) power plants. A third reheatstage would increase the cycle efficiency by about half of the improvementattained by the second reheat. This gain is too small to justify the added costand complexity.

wturb,out � wturb,I � wturb,II � 1h3 � h4 2 � 1h5 � h6 2

qin � qprimary � qreheat � 1h3 � h2 2 � 1h5 � h4 2

Chapter 10 | 565

Pump

Low-Pturbine

3

6

Boiler

Condenser

2

s

T

4

5

61

3

2

4

1

High-Pturbine

5

Reheater

High-pressureturbine

Low-pressureturbine

Reheating

P4 = P5 = Preheat

FIGURE 10–11The ideal reheat Rankine cycle.

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566 | Thermodynamics

Tavg,reheat

s

T

FIGURE 10–12The average temperature at which heatis transferred during reheatingincreases as the number of reheatstages is increased.

The reheat cycle was introduced in the mid-1920s, but it was abandonedin the 1930s because of the operational difficulties. The steady increase inboiler pressures over the years made it necessary to reintroduce singlereheat in the late 1940s and double reheat in the early 1950s.

The reheat temperatures are very close or equal to the turbine inlet tem-perature. The optimum reheat pressure is about one-fourth of the maximumcycle pressure. For example, the optimum reheat pressure for a cycle with aboiler pressure of 12 MPa is about 3 MPa.

Remember that the sole purpose of the reheat cycle is to reduce the mois-ture content of the steam at the final stages of the expansion process. If wehad materials that could withstand sufficiently high temperatures, therewould be no need for the reheat cycle.

EXAMPLE 10–4 The Ideal Reheat Rankine Cycle

Consider a steam power plant operating on the ideal reheat Rankine cycle.Steam enters the high-pressure turbine at 15 MPa and 600°C and is con-densed in the condenser at a pressure of 10 kPa. If the moisture content ofthe steam at the exit of the low-pressure turbine is not to exceed 10.4 per-cent, determine (a) the pressure at which the steam should be reheated and(b) the thermal efficiency of the cycle. Assume the steam is reheated to theinlet temperature of the high-pressure turbine.

Solution A steam power plant operating on the ideal reheat Rankine cycleis considered. For a specified moisture content at the turbine exit, the reheatpressure and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.Analysis The schematic of the power plant and the T-s diagram of the cycleare shown in Fig. 10–13. We note that the power plant operates on the idealreheat Rankine cycle. Therefore, the pump and the turbines are isentropic,there are no pressure drops in the boiler and condenser, and steam leavesthe condenser and enters the pump as saturated liquid at the condenserpressure.

(a) The reheat pressure is determined from the requirement that theentropies at states 5 and 6 be the same:

State 6:

Also,

Thus,

State 5:

Therefore, steam should be reheated at a pressure of 4 MPa or lower to pre-vent a moisture content above 10.4 percent.

T5 � 600°C

s5 � s6f P5 � 4.0 MPa

h5 � 3674.9 kJ>kg

h6 � hf � x6hfg � 191.81 � 0.896 12392.1 2 � 2335.1 kJ>kg

s6 � sf � x6sfg � 0.6492 � 0.896 17.4996 2 � 7.3688 kJ>kg # K

x6 � 0.896 1sat. mixture 2 P6 � 10 kPa

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Chapter 10 | 567

Pump

3

6

Boiler

Condenser

2 s

T, °C

4

1

5

Reheater

Reheating

P4 = P5 = Preheat

15 MPa

10 kPa

10 kPa

15 MPa

15 MPa 600

4

5

61

3

2

10 kPa

15 MPaLow-Pturbine

High-Pturbine

FIGURE 10–13Schematic and T-s diagram for Example 10–4.

(b) To determine the thermal efficiency, we need to know the enthalpies atall other states:

State 1:

State 2:

State 3:

State 4:

Thus

� 2143.3 kJ>kg

qout � h6 � h1 � 12335.1 � 191.81 2 kJ>kg

� 3896.1 kJ>kg

� 13583.1 � 206.95 2 kJ>kg � 13674.9 � 3155.0 2 kJ>kg

qin � 1h3 � h2 2 � 1h5 � h4 2

P4 � 4 MPa

s4 � s3f h4 � 3155.0 kJ>kg

1T4 � 375.5°C 2

P3 � 15 MPa

T3 � 600°Cf h3 � 3583.1 kJ>kg

s3 � 6.6796 kJ>kg # K

h2 � h1 � wpump,in � 1191.81 � 15.14 2 kJ>kg � 206.95 kJ>kg

� 15.14 kJ>kg

� 3 115,000 � 10 2kPa 4 a 1 kJ

1 kPa # m3 b wpump,in � v1 1P2 � P1 2 � 10.00101 m3>kg 2

s2 � s1

P2 � 15 MPa

P1 � 10 kPa

Sat. liquidf h1 � hf @ 10 kPa � 191.81 kJ>kg

v1 � vf @ 10 kPa � 0.00101 m3>kg

cen84959_ch10.qxd 4/19/05 2:17 PM Page 567

and

Discussion This problem was solved in Example 10–3c for the same pres-sure and temperature limits but without the reheat process. A comparison ofthe two results reveals that reheating reduces the moisture content from19.6 to 10.4 percent while increasing the thermal efficiency from 43.0 to45.0 percent.

10–6 ■ THE IDEAL REGENERATIVE RANKINE CYCLEA careful examination of the T-s diagram of the Rankine cycle redrawn inFig. 10–14 reveals that heat is transferred to the working fluid duringprocess 2-2� at a relatively low temperature. This lowers the average heat-addition temperature and thus the cycle efficiency.

To remedy this shortcoming, we look for ways to raise the temperature ofthe liquid leaving the pump (called the feedwater) before it enters the boiler.One such possibility is to transfer heat to the feedwater from the expandingsteam in a counterflow heat exchanger built into the turbine, that is, to useregeneration. This solution is also impractical because it is difficult todesign such a heat exchanger and because it would increase the moisturecontent of the steam at the final stages of the turbine.

A practical regeneration process in steam power plants is accomplished byextracting, or “bleeding,” steam from the turbine at various points. This steam,which could have produced more work by expanding further in the turbine, isused to heat the feedwater instead. The device where the feedwater is heatedby regeneration is called a regenerator, or a feedwater heater (FWH).

Regeneration not only improves cycle efficiency, but also provides a con-venient means of deaerating the feedwater (removing the air that leaks in atthe condenser) to prevent corrosion in the boiler. It also helps control thelarge volume flow rate of the steam at the final stages of the turbine (due tothe large specific volumes at low pressures). Therefore, regeneration hasbeen used in all modern steam power plants since its introduction in theearly 1920s.

A feedwater heater is basically a heat exchanger where heat is transferredfrom the steam to the feedwater either by mixing the two fluid streams(open feedwater heaters) or without mixing them (closed feedwater heaters).Regeneration with both types of feedwater heaters is discussed below.

Open Feedwater HeatersAn open (or direct-contact) feedwater heater is basically a mixing cham-ber, where the steam extracted from the turbine mixes with the feedwaterexiting the pump. Ideally, the mixture leaves the heater as a saturated liquidat the heater pressure. The schematic of a steam power plant with one openfeedwater heater (also called single-stage regenerative cycle) and the T-sdiagram of the cycle are shown in Fig. 10–15.

In an ideal regenerative Rankine cycle, steam enters the turbine at theboiler pressure (state 5) and expands isentropically to an intermediate pres-

hth � 1 �qout

qin� 1 �

2143.3 kJ>kg

3896.1 kJ>kg� 0.450 or 45.0%

568 | Thermodynamics

s

T

41

3

2'

2

Steam enteringboiler

Low-temperatureheat addition

Steam exitingboiler

FIGURE 10–14The first part of the heat-additionprocess in the boiler takes place atrelatively low temperatures.

SEE TUTORIAL CH. 10, SEC. 4 ON THE DVD.

INTERACTIVETUTORIAL

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sure (state 6). Some steam is extracted at this state and routed to the feed-water heater, while the remaining steam continues to expand isentropicallyto the condenser pressure (state 7). This steam leaves the condenser as a sat-urated liquid at the condenser pressure (state 1). The condensed water,which is also called the feedwater, then enters an isentropic pump, where itis compressed to the feedwater heater pressure (state 2) and is routed to thefeedwater heater, where it mixes with the steam extracted from the turbine.The fraction of the steam extracted is such that the mixture leaves the heateras a saturated liquid at the heater pressure (state 3). A second pump raisesthe pressure of the water to the boiler pressure (state 4). The cycle is com-pleted by heating the water in the boiler to the turbine inlet state (state 5).

In the analysis of steam power plants, it is more convenient to work withquantities expressed per unit mass of the steam flowing through the boiler.For each 1 kg of steam leaving the boiler, y kg expands partially in the turbineand is extracted at state 6. The remaining (1 � y) kg expands completely tothe condenser pressure. Therefore, the mass flow rates are different in dif-ferent components. If the mass flow rate through the boiler is m. , for exam-ple, it is (1 � y)m. through the condenser. This aspect of the regenerativeRankine cycle should be considered in the analysis of the cycle as well as inthe interpretation of the areas on the T-s diagram. In light of Fig. 10–15, theheat and work interactions of a regenerative Rankine cycle with one feed-water heater can be expressed per unit mass of steam flowing through theboiler as follows:

(10–14)

(10–15)

(10–16)

(10–17) wpump,in � 11 � y 2wpump I,in � wpump II,in

wturb,out � 1h5 � h6 2 � 11 � y 2 1h6 � h7 2 qout � 11 � y 2 1h7 � h1 2 qin � h5 � h4

Chapter 10 | 569

Pump I

Turbine Boiler

CondenserPump II

5

2

76

OpenFWH

4

1

3

y 1 – y

7

5

6

1

3

4

2

s

T

FIGURE 10–15The ideal regenerative Rankine cycle with an open feedwater heater.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 569

where

The thermal efficiency of the Rankine cycle increases as a result of regen-eration. This is because regeneration raises the average temperature atwhich heat is transferred to the steam in the boiler by raising the tempera-ture of the water before it enters the boiler. The cycle efficiency increasesfurther as the number of feedwater heaters is increased. Many large plantsin operation today use as many as eight feedwater heaters. The optimumnumber of feedwater heaters is determined from economical considerations.The use of an additional feedwater heater cannot be justified unless it savesmore from the fuel costs than its own cost.

Closed Feedwater HeatersAnother type of feedwater heater frequently used in steam power plants isthe closed feedwater heater, in which heat is transferred from the extractedsteam to the feedwater without any mixing taking place. The two streamsnow can be at different pressures, since they do not mix. The schematic of asteam power plant with one closed feedwater heater and the T-s diagram ofthe cycle are shown in Fig. 10–16. In an ideal closed feedwater heater, thefeedwater is heated to the exit temperature of the extracted steam, whichideally leaves the heater as a saturated liquid at the extraction pressure. Inactual power plants, the feedwater leaves the heater below the exit tempera-

wpump II,in � v3 1P4 � P3 2 wpump I,in � v1 1P2 � P1 2

y � m#

6>m# 5 1fraction of steam extracted 2

570 | Thermodynamics

Boiler

Condenser s

T

5

6

81

2

Pump II

Turbine

1

9

2

8Mixingchamber Closed

FWH

6

34

77

3

45

9

Pump I

FIGURE 10–16The ideal regenerative Rankine cycle with a closed feedwater heater.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 570

ture of the extracted steam because a temperature difference of at least afew degrees is required for any effective heat transfer to take place.

The condensed steam is then either pumped to the feedwater line or routedto another heater or to the condenser through a device called a trap. A trapallows the liquid to be throttled to a lower pressure region but traps thevapor. The enthalpy of steam remains constant during this throttling process.

The open and closed feedwater heaters can be compared as follows. Openfeedwater heaters are simple and inexpensive and have good heat transfercharacteristics. They also bring the feedwater to the saturation state. Foreach heater, however, a pump is required to handle the feedwater. Theclosed feedwater heaters are more complex because of the internal tubingnetwork, and thus they are more expensive. Heat transfer in closed feedwa-ter heaters is also less effective since the two streams are not allowed to bein direct contact. However, closed feedwater heaters do not require a sepa-rate pump for each heater since the extracted steam and the feedwater canbe at different pressures. Most steam power plants use a combination ofopen and closed feedwater heaters, as shown in Fig. 10–17.

EXAMPLE 10–5 The Ideal Regenerative Rankine Cycle

Consider a steam power plant operating on the ideal regenerative Rankinecycle with one open feedwater heater. Steam enters the turbine at 15 MPaand 600°C and is condensed in the condenser at a pressure of 10 kPa.

Chapter 10 | 571

Pump

Boiler

Condenser

Pump

Trap Trap Trap

ClosedFWH

ClosedFWH

ClosedFWH

OpenFWH

Turbine

Deaerating

FIGURE 10–17A steam power plant with one open and three closed feedwater heaters.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 571

572 | Thermodynamics

Boiler

Condenser s

T

5

71

2

Pump II

Turbine

1

4

7

63

4

32

1.2 MPa

1.2 MPa 6

10 kPa

Pump I10 kPa

wturb,out

qout

15 MPa600°C

5

1.2 MPa15 MPa

qin

OpenFWH

FIGURE 10–18Schematic and T-s diagram for Example 10–5.

Some steam leaves the turbine at a pressure of 1.2 MPa and enters the openfeedwater heater. Determine the fraction of steam extracted from the turbineand the thermal efficiency of the cycle.

Solution A steam power plant operates on the ideal regenerative Rankinecycle with one open feedwater heater. The fraction of steam extracted fromthe turbine and the thermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.Analysis The schematic of the power plant and the T-s diagram of the cycleare shown in Fig. 10–18. We note that the power plant operates on the idealregenerative Rankine cycle. Therefore, the pumps and the turbines are isen-tropic; there are no pressure drops in the boiler, condenser, and feedwaterheater; and steam leaves the condenser and the feedwater heater as satu-rated liquid. First, we determine the enthalpies at various states:

State 1:

State 2:

h2 � h1 � wpump I,in � 1191.81 � 1.20 2 kJ>kg � 193.01 kJ>kg

� 1.20 kJ>kg

wpump I,in � v1 1P2 � P1 2 � 10.00101 m3>kg 2 3 11200 � 10 2 kPa 4 a 1 kJ

1 kPa # m3 bs2 � s1

P2 � 1.2 MPa

P1 � 10 kPa

Sat. liquidf h1 � hf @ 10 kPa � 191.81 kJ>kg

v1 � vf @ 10 kPa � 0.00101 m3>kg

cen84959_ch10.qxd 4/19/05 2:17 PM Page 572

Chapter 10 | 573

State 3:

State 4:

State 5:

State 6:

State 7:

The energy analysis of open feedwater heaters is identical to the energyanalysis of mixing chambers. The feedwater heaters are generally well insu-lated (Q

.� 0), and they do not involve any work interactions (W

.� 0). By

neglecting the kinetic and potential energies of the streams, the energy bal-ance reduces for a feedwater heater to

or

where y is the fraction of steam extracted from the turbine (�m. 6/m. 5). Solv-ing for y and substituting the enthalpy values, we find

Thus,

and

hth � 1 �qout

qin� 1 �

1486.9 kJ>kg

2769.1 kJ>kg� 0.463 or 46.3%

� 1486.9 kJ>kg

qout � 11 � y 2 1h7 � h1 2 � 11 � 0.2270 2 12115.3 � 191.81 2 kJ>kg

qin � h5 � h4 � 13583.1 � 814.03 2 kJ>kg � 2769.1 kJ>kg

y �h3 � h2

h6 � h2�

798.33 � 193.01

2860.2 � 193.01� 0.2270

yh6 � 11 � y 2h2 � 1 1h3 2

E#in � E

#out S a

inm#h �a

outm#h

h7 � hf � x7hfg � 191.81 � 0.8041 12392.1 2 � 2115.3 kJ>kg

s7 � s5 x7 �s7 � sf

sfg�

6.6796 � 0.6492

7.4996� 0.8041

P7 � 10 kPa

P6 � 1.2 MPa

s6 � s5f h6 � 2860.2 kJ>kg

1T6 � 218.4°C 2

P5 � 15 MPa

T5 � 600°Cf

h5 � 3583.1 kJ>kg

s5 � 6.6796 kJ>kg # K

h4 � h3 � wpump II,in � 1798.33 � 15.70 2 kJ>kg � 814.03 kJ>kg

� 15.70 kJ>kg

� 10.001138 m3>kg 2 3 115,000 � 1200 2 kPa 4 a 1 kJ

1 kPa # m3 bwpump II,in � v3 1P4 � P3 2

s4 � s3

P4 � 15 MPa

P3 � 1.2 MPa

Sat. liquidf v3 � vf @ 1.2 MPa � 0.001138 m3>kg

h3 � hf @ 1.2 MPa � 798.33 kJ>kg

cen84959_ch10.qxd 4/19/05 2:17 PM Page 573

574 | Thermodynamics

Discussion This problem was worked out in Example 10–3c for the samepressure and temperature limits but without the regeneration process. Acomparison of the two results reveals that the thermal efficiency of the cyclehas increased from 43.0 to 46.3 percent as a result of regeneration. The network output decreased by 171 kJ/kg, but the heat input decreased by607 kJ/kg, which results in a net increase in the thermal efficiency.

EXAMPLE 10–6 The Ideal Reheat–Regenerative Rankine Cycle

Consider a steam power plant that operates on an ideal reheat–regenerativeRankine cycle with one open feedwater heater, one closed feedwater heater,and one reheater. Steam enters the turbine at 15 MPa and 600°C and iscondensed in the condenser at a pressure of 10 kPa. Some steam isextracted from the turbine at 4 MPa for the closed feedwater heater, and theremaining steam is reheated at the same pressure to 600°C. The extractedsteam is completely condensed in the heater and is pumped to 15 MPabefore it mixes with the feedwater at the same pressure. Steam for the openfeedwater heater is extracted from the low-pressure turbine at a pressure of0.5 MPa. Determine the fractions of steam extracted from the turbine aswell as the thermal efficiency of the cycle.

Solution A steam power plant operates on the ideal reheat–regenerativeRankine cycle with one open feedwater heater, one closed feedwater heater,and one reheater. The fractions of steam extracted from the turbine and thethermal efficiency are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible. 3 In both open and closed feedwater heaters,feedwater is heated to the saturation temperature at the feedwater heaterpressure. (Note that this is a conservative assumption since extracted steamenters the closed feedwater heater at 376°C and the saturation temperatureat the closed feedwater pressure of 4 MPa is 250°C).Analysis The schematic of the power plant and the T-s diagram of the cycleare shown in Fig. 10–19. The power plant operates on the ideal reheat–regenerative Rankine cycle and thus the pumps and the turbines are isen-tropic; there are no pressure drops in the boiler, reheater, condenser, and feed-water heaters; and steam leaves the condenser and the feedwater heaters assaturated liquid.

The enthalpies at the various states and the pump work per unit mass offluid flowing through them are

h8 � 1089.8 kJ>kg wpump III,in � 13.77 kJ>kg

h7 � 1101.2 kJ>kg wpump II,in � 3.83 kJ>kg

h6 � 1087.4 kJ>kg wpump I,in � 0.49 kJ>kg

h5 � 1087.4 kJ>kg h13 � 2335.7 kJ>kg

h4 � 643.92 kJ>kg h12 � 3014.8 kJ>kg

h3 � 640.09 kJ>kg h11 � 3674.9 kJ>kg

h2 � 192.30 kJ>kg h10 � 3155.0 kJ>kg

h1 � 191.81 kJ>kg h9 � 3155.0 kJ>kg

cen84959_ch10.qxd 4/27/05 5:49 PM Page 574

Chapter 10 | 575

Boiler

Condenser

s

T

5

9

1

2

Pump II

1

11

2

13

ClosedFWH

9

37

12

3

87

6

Low-Pturbine

High-Pturbine

OpenFWH

Pump III Pump I

4 MPa

8

4

6

0.5 MPa

Reheater

1 kg

Mixingchamber

10 kPa

15 MPa600°C

1 – y – z

10

y P10 = P11 = 4 MPa z600°C

11

10

12

13

5

4

15 MPa

4 MPa

0.5 MPa z

10 kPa

1 – y

1 –

y

1 kg

y

1 – y – z

1 – y

FIGURE 10–19Schematic and T-s diagram for Example 10–6.

The fractions of steam extracted are determined from the mass and energybalances of the feedwater heaters:Closed feedwater heater:

Open feedwater heater:

The enthalpy at state 8 is determined by applying the mass and energyequations to the mixing chamber, which is assumed to be insulated:

� 1089.8 kJ>kg

h8 � 11 � 0.1766 2 11087.4 2 kJ>kg � 0.1766 11101.2 2 kJ>kg

11 2h8 � 11 � y 2h5 � yh7

E#

in � E#

out

z �11 � y 2 1h3 � h2 2

h12 � h2�11 � 0.1766 2 1640.09 � 192.30 2

3014.8 � 192.30� 0.1306

zh12 � 11 � y � z 2h2 � 11 � y 2h3

E#

in � E#out

y �h5 � h4

1h10 � h6 2 � 1h5 � h4 2 �1087.4 � 643.92

13155.0 � 1087.4 2 � 11087.4 � 643.92 2 � 0.1766

yh10 � 11 � y 2h4 � 11 � y 2h5 � yh6

E#in � E

#out

cen84959_ch10.qxd 4/19/05 2:17 PM Page 575

576 | Thermodynamics

Thus,

and

Discussion This problem was worked out in Example 10–4 for the same pres-sure and temperature limits with reheat but without the regeneration process.A comparison of the two results reveals that the thermal efficiency of the cyclehas increased from 45.0 to 49.2 percent as a result of regeneration.

The thermal efficiency of this cycle could also be determined from

where

Also, if we assume that the feedwater leaves the closed FWH as a satu-rated liquid at 15 MPa (and thus at T5 � 342°C and h5 � 1610.3 kJ/kg), itcan be shown that the thermal efficiency would be 50.6.

10–7 ■ SECOND-LAW ANALYSIS OF VAPOR POWER CYCLES

The ideal Carnot cycle is a totally reversible cycle, and thus it does notinvolve any irreversibilities. The ideal Rankine cycles (simple, reheat, orregenerative), however, are only internally reversible, and they may involveirreversibilities external to the system, such as heat transfer through a finitetemperature difference. A second-law analysis of these cycles reveals wherethe largest irreversibilities occur and what their magnitudes are.

Relations for exergy and exergy destruction for steady-flow systems aredeveloped in Chap. 8. The exergy destruction for a steady-flow system canbe expressed, in the rate form, as

(10–18)

or on a unit mass basis for a one-inlet, one-exit, steady-flow device as

(10–19)xdest � T0sgen � T0 a se � si �qout

Tb,out�

qin

Tb,inb 1kJ>kg 2

X#

dest � T0 S#gen � T0 1S# out � S

#in 2 � T0 aa

outm# s �

Q#

out

Tb,out�a

inm#s �

Q#

in

Tb,inb 1kW 2

wpump,in � 11 � y � z 2wpump I,in � 11 � y 2wpump II,in � 1y 2wpump III,in

wturb,out � 1h9 � h10 2 � 11 � y 2 1h11 � h12 2 � 11 � y � z 2 1h12 � h13 2

hth �wnet

qin�

wturb,out � wpump,in

qin

hth � 1 �qout

qin� 1 �

1485.3 kJ>kg

2921.4 kJ>kg� 0.492 or 49.2%

� 1485.3 kJ>kg

� 11 � 0.1766 � 0.1306 2 12335.7 � 191.81 2 kJ>kg

qout � 11 � y � z 2 1h13 � h1 2 � 2921.4 kJ>kg

� 13583.1 � 1089.8 2 kJ>kg � 11 � 0.1766 2 13674.9 � 3155.0 2 kJ>kg

qin � 1h9 � h8 2 � 11 � y 2 1h11 � h10 2

cen84959_ch10.qxd 4/19/05 2:17 PM Page 576

where Tb,in and Tb,out are the temperatures of the system boundary whereheat is transferred into and out of the system, respectively.

The exergy destruction associated with a cycle depends on the magnitudeof the heat transfer with the high- and low-temperature reservoirs involved,and their temperatures. It can be expressed on a unit mass basis as

(10–20)

For a cycle that involves heat transfer only with a source at TH and a sink atTL , the exergy destruction becomes

(10–21)

The exergy of a fluid stream c at any state can be determined from

(10–22)

where the subscript “0” denotes the state of the surroundings.

EXAMPLE 10–7 Second-Law Analysis of an Ideal Rankine Cycle

Determine the exergy destruction associated with the Rankine cycle (all fourprocesses as well as the cycle) discussed in Example 10–1, assuming thatheat is transferred to the steam in a furnace at 1600 K and heat is rejectedto a cooling medium at 290 K and 100 kPa. Also, determine the exergy ofthe steam leaving the turbine.

Solution The Rankine cycle analyzed in Example 10–1 is reconsidered. Forspecified source and sink temperatures, the exergy destruction associatedwith the cycle and exergy of the steam at turbine exit are to be determined.Analysis In Example 10–1, the heat input was determined to be 2728.6 kJ/kg,and the heat rejected to be 2018.6 kJ/kg.

Processes 1-2 and 3-4 are isentropic (s1 � s2, s3 � s4) and therefore donot involve any internal or external irreversibilities, that is,

Processes 2-3 and 4-1 are constant-pressure heat-addition and heat-rejection processes, respectively, and they are internally reversible. But theheat transfer between the working fluid and the source or the sink takesplace through a finite temperature difference, rendering both processes irre-versible. The irreversibility associated with each process is determined fromEq. 10–19. The entropy of the steam at each state is determined from thesteam tables:

Thus,

� 1110 kJ/kg

� 1290 K 2 c 16.7450 � 1.2132 2 kJ>kg # K �2728.6 kJ>kg

1600 Kd

xdest,23 � T0 a s3 � s2 �qin,23

Tsourceb

s4 � s3 � 6.7450 kJ>kg # K 1at 3 MPa, 350°C 2 s2 � s1 � sf @ 75 kPa � 1.2132 kJ>kg # K

xdest,12 � 0 and xdest,34 � 0

c � 1h � h0 2 � T0 1s � s0 2 �V 2

2� gz 1kJ>kg 2

xdest � T0 a qout

TL

�qin

THb 1kJ>kg 2

xdest � T0 aa qout

Tb,out�a

qin

Tb,inb 1kJ>kg 2

Chapter 10 | 577

cen84959_ch10.qxd 4/19/05 2:17 PM Page 577

Therefore, the irreversibility of the cycle is

The total exergy destroyed during the cycle could also be determined fromEq. 10–21. Notice that the largest exergy destruction in the cycle occursduring the heat-addition process. Therefore, any attempt to reduce theexergy destruction should start with this process. Raising the turbine inlettemperature of the steam, for example, would reduce the temperature differ-ence and thus the exergy destruction.

The exergy (work potential) of the steam leaving the turbine is determinedfrom Eq. 10–22. Disregarding the kinetic and potential energies, it reduces to

where

Thus,

Discussion Note that 449 kJ/kg of work could be obtained from the steamleaving the turbine if it is brought to the state of the surroundings in areversible manner.

10–8 ■ COGENERATIONIn all the cycles discussed so far, the sole purpose was to convert a portionof the heat transferred to the working fluid to work, which is the most valu-able form of energy. The remaining portion of the heat is rejected to rivers,lakes, oceans, or the atmosphere as waste heat, because its quality (or grade)is too low to be of any practical use. Wasting a large amount of heat is aprice we have to pay to produce work, because electrical or mechanicalwork is the only form of energy on which many engineering devices (suchas a fan) can operate.

Many systems or devices, however, require energy input in the form ofheat, called process heat. Some industries that rely heavily on process heatare chemical, pulp and paper, oil production and refining, steel making,

� 449 kJ/kg

c4 � 12403.0 � 71.355 2 kJ>kg � 1290 K 2 3 16.7450 � 0.2533 2 kJ>kg # K 4

s0 � s@ 290 K,100 kPa � sf @ 290 K � 0.2533 kJ>kg # K

h0 � h@ 290 K,100 kPa � hf @ 290 K � 71.355 kJ>kg

� 1h4 � h0 2 � T0 1s4 � s0 2 c4 � 1h4 � h0 2 � T0 1s4 � s0 2 �

V 24

2� gz4

� 1524 kJ/kg

� 0 � 1110 kJ>kg � 0 � 414 kJ>kg

xdest,cycle � xdest,12 � xdest,23 � xdest,34 � xdest,41

� 414 kJ/kg

� 1290 K 2 c 11.2132 � 6.7450 2 kJ>kg # K �2018.6 kJ>kg

290 Kd

xdest,41 � T0 a s1 � s4 �qout,41

Tsinkb

578 | Thermodynamics

Q0

Q0

cen84959_ch10.qxd 4/19/05 2:17 PM Page 578

food processing, and textile industries. Process heat in these industries isusually supplied by steam at 5 to 7 atm and 150 to 200°C (300 to 400°F).Energy is usually transferred to the steam by burning coal, oil, natural gas,or another fuel in a furnace.

Now let us examine the operation of a process-heating plant closely. Disre-garding any heat losses in the piping, all the heat transferred to the steam in theboiler is used in the process-heating units, as shown in Fig. 10–20. Therefore,process heating seems like a perfect operation with practically no waste ofenergy. From the second-law point of view, however, things do not look so per-fect. The temperature in furnaces is typically very high (around 1400°C), andthus the energy in the furnace is of very high quality. This high-quality energyis transferred to water to produce steam at about 200°C or below (a highly irre-versible process). Associated with this irreversibility is, of course, a loss inexergy or work potential. It is simply not wise to use high-quality energy toaccomplish a task that could be accomplished with low-quality energy.

Industries that use large amounts of process heat also consume a largeamount of electric power. Therefore, it makes economical as well as engi-neering sense to use the already-existing work potential to produce powerinstead of letting it go to waste. The result is a plant that produces electric-ity while meeting the process-heat requirements of certain industrial pro-cesses. Such a plant is called a cogeneration plant. In general, cogenerationis the production of more than one useful form of energy (such as processheat and electric power) from the same energy source.

Either a steam-turbine (Rankine) cycle or a gas-turbine (Brayton) cycle oreven a combined cycle (discussed later) can be used as the power cycle in acogeneration plant. The schematic of an ideal steam-turbine cogenerationplant is shown in Fig. 10–21. Let us say this plant is to supply process heat Q

.p

at 500 kPa at a rate of 100 kW. To meet this demand, steam is expanded in theturbine to a pressure of 500 kPa, producing power at a rate of, say, 20 kW.The flow rate of the steam can be adjusted such that steam leaves the process-heating section as a saturated liquid at 500 kPa. Steam is then pumped to theboiler pressure and is heated in the boiler to state 3. The pump work is usuallyvery small and can be neglected. Disregarding any heat losses, the rate of heatinput in the boiler is determined from an energy balance to be 120 kW.

Probably the most striking feature of the ideal steam-turbine cogenerationplant shown in Fig. 10–21 is the absence of a condenser. Thus no heat isrejected from this plant as waste heat. In other words, all the energy trans-ferred to the steam in the boiler is utilized as either process heat or electricpower. Thus it is appropriate to define a utilization factor �u for a cogener-ation plant as

(10–23)

or

(10–24)

where Q.out represents the heat rejected in the condenser. Strictly speaking,

Q.out also includes all the undesirable heat losses from the piping and other

components, but they are usually small and thus neglected. It also includescombustion inefficiencies such as incomplete combustion and stack losses

�u � 1 �Q#

out

Q#

in

�u �Net work output � Process heat delivered

Total heat input�

W#

net � Q#

p

Q#

in

Chapter 10 | 579

Boiler

Qin

Pump

Processheater

Qp

FIGURE 10–20A simple process-heating plant.

Wpump = 0

Pump

Turbine

3

4

Boiler

100 kW

Processheater

2

1

120 kW

20 kW

~

FIGURE 10–21An ideal cogeneration plant.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 579

when the utilization factor is defined on the basis of the heating value of thefuel. The utilization factor of the ideal steam-turbine cogeneration plant isobviously 100 percent. Actual cogeneration plants have utilization factors ashigh as 80 percent. Some recent cogeneration plants have even higher uti-lization factors.

Notice that without the turbine, we would need to supply heat to thesteam in the boiler at a rate of only 100 kW instead of at 120 kW. The addi-tional 20 kW of heat supplied is converted to work. Therefore, a cogenera-tion power plant is equivalent to a process-heating plant combined with apower plant that has a thermal efficiency of 100 percent.

The ideal steam-turbine cogeneration plant described above is not practi-cal because it cannot adjust to the variations in power and process-heatloads. The schematic of a more practical (but more complex) cogenerationplant is shown in Fig. 10–22. Under normal operation, some steam isextracted from the turbine at some predetermined intermediate pressure P6.The rest of the steam expands to the condenser pressure P7 and is thencooled at constant pressure. The heat rejected from the condenser representsthe waste heat for the cycle.

At times of high demand for process heat, all the steam is routed to theprocess-heating units and none to the condenser (m

.7 � 0). The waste heat is

zero in this mode. If this is not sufficient, some steam leaving the boiler isthrottled by an expansion or pressure-reducing valve (PRV) to the extractionpressure P6 and is directed to the process-heating unit. Maximum processheating is realized when all the steam leaving the boiler passes through thePRV (m

.5 � m

.4). No power is produced in this mode. When there is no

demand for process heat, all the steam passes through the turbine and thecondenser (m

.5 � m

.6 � 0), and the cogeneration plant operates as an ordi-

nary steam power plant. The rates of heat input, heat rejected, and processheat supply as well as the power produced for this cogeneration plant can beexpressed as follows:

(10–25)

(10–26)

(10–27)

(10–28)

Under optimum conditions, a cogeneration plant simulates the idealcogeneration plant discussed earlier. That is, all the steam expands in theturbine to the extraction pressure and continues to the process-heating unit.No steam passes through the PRV or the condenser; thus, no waste heat isrejected (m

.4 � m

.6 and m

.5 � m

.7 � 0). This condition may be difficult to

achieve in practice because of the constant variations in the process-heatand power loads. But the plant should be designed so that the optimumoperating conditions are approximated most of the time.

The use of cogeneration dates to the beginning of this century whenpower plants were integrated to a community to provide district heating,that is, space, hot water, and process heating for residential and commercialbuildings. The district heating systems lost their popularity in the 1940sowing to low fuel prices. However, the rapid rise in fuel prices in the 1970sbrought about renewed interest in district heating.

W#

turb � 1m# 4 � m#

5 2 1h4 � h6 2 � m#

7 1h6 � h7 2 Q#

p � m#

5h5 � m#

6h6 � m#

8h8

Q#

out � m#

7 1h7 � h1 2 Q#

in � m#

3 1h4 � h3 2

580 | Thermodynamics

Pump I

Turbine

7

Boiler

Condenser

1

Pump II

Processheater

Expansionvalve

38

4

56

2

FIGURE 10–22A cogeneration plant with adjustableloads.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 580

Cogeneration plants have proved to be economically very attractive. Con-sequently, more and more such plants have been installed in recent years,and more are being installed.

EXAMPLE 10–8 An Ideal Cogeneration Plant

Consider the cogeneration plant shown in Fig. 10–23. Steam enters the turbineat 7 MPa and 500°C. Some steam is extracted from the turbine at 500 kPa forprocess heating. The remaining steam continues to expand to 5 kPa. Steam isthen condensed at constant pressure and pumped to the boiler pressure of7 MPa. At times of high demand for process heat, some steam leaving theboiler is throttled to 500 kPa and is routed to the process heater. The extrac-tion fractions are adjusted so that steam leaves the process heater as a satu-rated liquid at 500 kPa. It is subsequently pumped to 7 MPa. The mass flowrate of steam through the boiler is 15 kg/s. Disregarding any pressure dropsand heat losses in the piping and assuming the turbine and the pump to beisentropic, determine (a) the maximum rate at which process heat can besupplied, (b) the power produced and the utilization factor when no processheat is supplied, and (c) the rate of process heat supply when 10 percent ofthe steam is extracted before it enters the turbine and 70 percent of thesteam is extracted from the turbine at 500 kPa for process heating.

Solution A cogeneration plant is considered. The maximum rate of processheat supply, the power produced and the utilization factor when no processheat is supplied, and the rate of process heat supply when steam is extractedfrom the steam line and turbine at specified ratios are to be determined.Assumptions 1 Steady operating conditions exist. 2 Pressure drops and heatlosses in piping are negligible. 3 Kinetic and potential energy changes arenegligible.Analysis The schematic of the cogeneration plant and the T-s diagram ofthe cycle are shown in Fig. 10–23. The power plant operates on an ideal

Chapter 10 | 581

Boiler

Condenser

s

T

Pump I

Turbine

Pump II

5 kPa

7 MPa500°C

7 MPa

1, 2, 3

4

5

6

10

8

711

911

7

8

Processheater 5 kPa

54500 kPa500 kPa

Mixingchamber

9

10

1

6

32

Expansionvalve

FIGURE 10–23Schematic and T-s diagram for Example 10–8.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 581

cycle and thus the pumps and the turbines are isentropic; there are no pres-sure drops in the boiler, process heater, and condenser; and steam leavesthe condenser and the process heater as saturated liquid.

The work inputs to the pumps and the enthalpies at various states are asfollows:

(a) The maximum rate of process heat is achieved when all the steam leavingthe boiler is throttled and sent to the process heater and none is sent to theturbine (that is, m

.4 � m

.7 � m

.1 � 15 kg/s and m

.3 � m

.5 � m

.6 � 0). Thus,

The utilization factor is 100 percent in this case since no heat is rejected inthe condenser, heat losses from the piping and other components areassumed to be negligible, and combustion losses are not considered.

(b) When no process heat is supplied, all the steam leaving the boiler passesthrough the turbine and expands to the condenser pressure of 5 kPa (that is,m.

3 � m.

6 � m.

1 � 15 kg/s and m.

2 � m.

5 � 0). Maximum power is producedin this mode, which is determined to be

Thus,

That is, 40.8 percent of the energy is utilized for a useful purpose. Noticethat the utilization factor is equivalent to the thermal efficiency in this case.

(c) Neglecting any kinetic and potential energy changes, an energy balanceon the process heater yields

�u �W#net � Q

#p

Q#

in

�119,971 � 0 2 kW

48,999 kW� 0.408 or 40.8%

Q#

in � m#

1 1h1 � h11 2 � 115 kg>s 2 3 13411.4 � 144.78 2 kJ>kg 4 � 48,999 kW

W#

net,out � W#

turb,out � W#

pump,in � 120,076 � 105 2 kW � 19,971 kW � 20.0 MW

W#

pump,in � 115 kg>s 2 17.03 kJ>kg 2 � 105 kW

W#

turb,out � m# 1h3 � h6 2 � 115 kg>s 2 3 13411.4 � 2073.0 2 kJ>kg 4 � 20,076 kW

Q#

p,max � m#

1 1h4 � h7 2 � 115 kg>s 2 3 13411.4 � 640.09 2 kJ>kg 4 � 41,570 kW

h10 � h7 � wpump II,in � 1640.09 � 7.10 2 kJ>kg � 647.19 kJ>kg

h9 � h8 � wpump I,in � 1137.75 � 7.03 2 kJ>kg � 144.78 kJ>kg

h8 � hf @ 5 kPa � 137.75 kJ>kg

h7 � hf @ 500 kPa � 640.09 kJ>kg

h6 � 2073.0 kJ>kg

h5 � 2739.3 kJ>kg

h1 � h2 � h3 � h4 � 3411.4 kJ>kg

� 7.10 kJ>kg

wpump II,in � v7 1P10 � P7 2 � 10.001093 m3>kg2 3 17000 � 5002 kPa 4 a 1 kJ

1 kPa # m3b � 7.03 kJ>kg

wpump I,in � v8 1P9 � P8 2 � 10.001005 m3>kg 2 3 17000 � 5 2kPa 4 a 1 kJ

1 kPa # m3 b

582 | Thermodynamics

cen84959_ch10.qxd 4/19/05 2:17 PM Page 582

Chapter 10 | 583

or

where

Thus

Discussion Note that 26.2 MW of the heat transferred will be utilized in theprocess heater. We could also show that 11.0 MW of power is produced inthis case, and the rate of heat input in the boiler is 43.0 MW. Thus the uti-lization factor is 86.5 percent.

10–9 ■ COMBINED GAS–VAPOR POWER CYCLESThe continued quest for higher thermal efficiencies has resulted in ratherinnovative modifications to conventional power plants. The binary vaporcycle discussed later is one such modification. A more popular modificationinvolves a gas power cycle topping a vapor power cycle, which is called thecombined gas–vapor cycle, or just the combined cycle. The combinedcycle of greatest interest is the gas-turbine (Brayton) cycle topping a steam-turbine (Rankine) cycle, which has a higher thermal efficiency than either ofthe cycles executed individually.

Gas-turbine cycles typically operate at considerably higher temperaturesthan steam cycles. The maximum fluid temperature at the turbine inlet isabout 620°C (1150°F) for modern steam power plants, but over 1425°C(2600°F) for gas-turbine power plants. It is over 1500°C at the burner exitof turbojet engines. The use of higher temperatures in gas turbines is madepossible by recent developments in cooling the turbine blades and coatingthe blades with high-temperature-resistant materials such as ceramics. Becauseof the higher average temperature at which heat is supplied, gas-turbinecycles have a greater potential for higher thermal efficiencies. However, thegas-turbine cycles have one inherent disadvantage: The gas leaves the gasturbine at very high temperatures (usually above 500°C), which erases anypotential gains in the thermal efficiency. The situation can be improvedsomewhat by using regeneration, but the improvement is limited.

It makes engineering sense to take advantage of the very desirable charac-teristics of the gas-turbine cycle at high temperatures and to use the high-temperature exhaust gases as the energy source for the bottoming cycle suchas a steam power cycle. The result is a combined gas–steam cycle, as shown

� 26.2 MW

� 112 kg>s 2 1640.09 kJ>kg 2 Q#

p,out � 11.5 kg>s 2 13411.4 kJ>kg 2 � 110.5 kg>s 2 12739.3 kJ>kg 2

m#

7 � m#

4 � m#

5 � 1.5 � 10.5 � 12 kg>s m#

5 � 10.7 2 115 kg>s 2 � 10.5 kg>s m#

4 � 10.1 2 115 kg>s 2 � 1.5 kg>s

Q#

p,out � m#

4h4 � m#

5h5 � m#

7h7

m#

4h4 � m#

5h5 � Q#

p,out � m#

7h7

E#in � E

#out

cen84959_ch10.qxd 4/19/05 2:17 PM Page 583

in Fig. 10–24. In this cycle, energy is recovered from the exhaust gases bytransferring it to the steam in a heat exchanger that serves as the boiler. Ingeneral, more than one gas turbine is needed to supply sufficient heat to thesteam. Also, the steam cycle may involve regeneration as well as reheating.Energy for the reheating process can be supplied by burning some addi-tional fuel in the oxygen-rich exhaust gases.

Recent developments in gas-turbine technology have made the combinedgas–steam cycle economically very attractive. The combined cycle increasesthe efficiency without increasing the initial cost greatly. Consequently, manynew power plants operate on combined cycles, and many more existingsteam- or gas-turbine plants are being converted to combined-cycle powerplants. Thermal efficiencies well over 40 percent are reported as a result ofconversion.

A 1090-MW Tohoku combined plant that was put in commercial operationin 1985 in Niigata, Japan, is reported to operate at a thermal efficiency of44 percent. This plant has two 191-MW steam turbines and six 118-MW gasturbines. Hot combustion gases enter the gas turbines at 1154°C, and steamenters the steam turbines at 500°C. Steam is cooled in the condenser by cool-ing water at an average temperature of 15°C. The compressors have a pressureratio of 14, and the mass flow rate of air through the compressors is 443 kg/s.

584 | Thermodynamics

4

9

45

8

1

T

s

6

7

3

Qout

9

Qin

STEAMCYCLE

GAS CYCLE

GAS CYCLE

Heat exchanger

STEAMCYCLE

Exhaustgases

Airin

Condenser

Pump

Qout

Qin

2

1

6 7

Compressor

Combustionchamber

Gasturbine

5 8

3

Steamturbine 2

FIGURE 10–24Combined gas–steam power plant.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 584

A 1350-MW combined-cycle power plant built in Ambarli, Turkey, in1988 by Siemens of Germany is the first commercially operating thermalplant in the world to attain an efficiency level as high as 52.5 percent atdesign operating conditions. This plant has six 150-MW gas turbines andthree 173-MW steam turbines. Some recent combined-cycle power plantshave achieved efficiencies above 60 percent.

EXAMPLE 10–9 A Combined Gas–Steam Power Cycle

Consider the combined gas–steam power cycle shown in Fig. 10–25. The top-ping cycle is a gas-turbine cycle that has a pressure ratio of 8. Air enters thecompressor at 300 K and the turbine at 1300 K. The isentropic efficiency ofthe compressor is 80 percent, and that of the gas turbine is 85 percent. Thebottoming cycle is a simple ideal Rankine cycle operating between the pressurelimits of 7 MPa and 5 kPa. Steam is heated in a heat exchanger by the exhaustgases to a temperature of 500°C. The exhaust gases leave the heat exchangerat 450 K. Determine (a) the ratio of the mass flow rates of the steam and thecombustion gases and (b) the thermal efficiency of the combined cycle.

Solution A combined gas–steam cycle is considered. The ratio of the massflow rates of the steam and the combustion gases and the thermal efficiencyare to be determined.Analysis The T-s diagrams of both cycles are given in Fig. 10–25. The gas-turbine cycle alone was analyzed in Example 9–6, and the steam cycle inExample 10–8b, with the following results:

Gas cycle:

h5¿ � h@ 450 K � 451.80 kJ>kg

qin � 790.58 kJ>kg wnet � 210.41 kJ>kg hth � 26.6%

h4¿ � 880.36 kJ>kg 1T4¿ � 853 K 2

Chapter 10 | 585

41'

4'

1

T, K

s

2'3

5'

500°C7 MPa

5 kPa

7 MPa450

300

3'

2

1300

FIGURE 10–25T-s diagram of the gas–steamcombined cycle described inExample 10–9.

cen84959_ch10.qxd 4/19/05 2:17 PM Page 585

Steam cycle:

(a) The ratio of mass flow rates is determined from an energy balance on theheat exchanger:

Thus,

That is, 1 kg of exhaust gases can heat only 0.131 kg of steam from 33 to500°C as they are cooled from 853 to 450 K. Then the total net work out-put per kilogram of combustion gases becomes

Therefore, for each kg of combustion gases produced, the combined plantwill deliver 384.8 kJ of work. The net power output of the plant is deter-mined by multiplying this value by the mass flow rate of the working fluid inthe gas-turbine cycle.

(b) The thermal efficiency of the combined cycle is determined from

Discussion Note that this combined cycle converts to useful work 48.7 per-cent of the energy supplied to the gas in the combustion chamber. This valueis considerably higher than the thermal efficiency of the gas-turbine cycle(26.6 percent) or the steam-turbine cycle (40.8 percent) operating alone.

hth �wnet

qin�

384.8 kJ>kg gas

790.6 kJ>kg gas� 0.487 or 48.7%

� 384.8 kJ>kg gas

� 1210.41 kJ>kg gas 2 � 10.131 kg steam>kg gas 2 11331.4 kJ>kg steam 2 wnet � wnet,gas � ywnet,steam

m#

s

m#

g

� y � 0.131

m#

s 13411.4 � 144.78 2 � m#

g 1880.36 � 451.80 2 m#

s 1h3 � h2 2 � m#

g 1h¿4 � h¿5 2 m#

g h5¿ � m#

s h3 � m#

g h4¿ � m#

s h2

E#

in � E#

out

wnet � 1331.4 kJ>kg hth � 40.8%

h3 � 3411.4 kJ>kg 1T3 � 500°C 2 h2 � 144.78 kJ>kg 1T2 � 33°C 2

586 | Thermodynamics

With the exception of a few specialized applications, the working fluid pre-dominantly used in vapor power cycles is water. Water is the best workingfluid presently available, but it is far from being the ideal one. The binarycycle is an attempt to overcome some of the shortcomings of water and toapproach the ideal working fluid by using two fluids. Before we discuss thebinary cycle, let us list the characteristics of a working fluid most suitablefor vapor power cycles:

TOPIC OF SPECIAL INTEREST* Binary Vapor Cycles

*This section can be skipped without a loss in continuity.

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Chapter 10 | 587

1. A high critical temperature and a safe maximum pressure. A criti-cal temperature above the metallurgically allowed maximum temperature(about 620°C) makes it possible to transfer a considerable portion of theheat isothermally at the maximum temperature as the fluid changesphase. This makes the cycle approach the Carnot cycle. Very high pres-sures at the maximum temperature are undesirable because they creatematerial-strength problems.

2. Low triple-point temperature. A triple-point temperature below thetemperature of the cooling medium prevents any solidification problems.

3. A condenser pressure that is not too low. Condensers usuallyoperate below atmospheric pressure. Pressures well below the atmo-spheric pressure create air-leakage problems. Therefore, a substancewhose saturation pressure at the ambient temperature is too low is not agood candidate.

4. A high enthalpy of vaporization (hfg) so that heat transfer to theworking fluid is nearly isothermal and large mass flow rates are notneeded.

5. A saturation dome that resembles an inverted U. This eliminatesthe formation of excessive moisture in the turbine and the need forreheating.

6. Good heat transfer characteristics (high thermal conductivity).7. Other properties such as being inert, inexpensive, readily avail-

able, and nontoxic.

Not surprisingly, no fluid possesses all these characteristics. Water comesthe closest, although it does not fare well with respect to characteristics 1, 3,and 5. We can cope with its subatmospheric condenser pressure by carefulsealing, and with the inverted V-shaped saturation dome by reheating, butthere is not much we can do about item 1. Water has a low critical tempera-ture (374°C, well below the metallurgical limit) and very high saturationpressures at high temperatures (16.5 MPa at 350°C).

Well, we cannot change the way water behaves during the high-temperaturepart of the cycle, but we certainly can replace it with a more suitable fluid.The result is a power cycle that is actually a combination of two cycles, onein the high-temperature region and the other in the low-temperature region.Such a cycle is called a binary vapor cycle. In binary vapor cycles, thecondenser of the high-temperature cycle (also called the topping cycle) servesas the boiler of the low-temperature cycle (also called the bottoming cycle).That is, the heat output of the high-temperature cycle is used as the heat inputto the low-temperature one.

Some working fluids found suitable for the high-temperature cycle are mer-cury, sodium, potassium, and sodium–potassium mixtures. The schematic andT-s diagram for a mercury–water binary vapor cycle are shown in Fig. 10–26.The critical temperature of mercury is 898°C (well above the current metal-lurgical limit), and its critical pressure is only about 18 MPa. This makes mer-cury a very suitable working fluid for the topping cycle. Mercury is notsuitable as the sole working fluid for the entire cycle, however, since at a con-denser temperature of 32°C its saturation pressure is 0.07 Pa. A power plantcannot operate at this vacuum because of air-leakage problems. At an accept-able condenser pressure of 7 kPa, the saturation temperature of mercury is

cen84959_ch10.qxd 4/19/05 2:17 PM Page 587

237°C, which is too high as the minimum temperature in the cycle. Therefore,the use of mercury as a working fluid is limited to the high-temperaturecycles. Other disadvantages of mercury are its toxicity and high cost. Themass flow rate of mercury in binary vapor cycles is several times that of waterbecause of its low enthalpy of vaporization.

It is evident from the T-s diagram in Fig. 10–26 that the binary vapor cycleapproximates the Carnot cycle more closely than the steam cycle for thesame temperature limits. Therefore, the thermal efficiency of a power plantcan be increased by switching to binary cycles. The use of mercury–waterbinary cycles in the United States dates back to 1928. Several such plantshave been built since then in the New England area, where fuel costs are typ-ically higher. A small (40-MW) mercury–steam power plant that was in ser-vice in New Hampshire in 1950 had a higher thermal efficiency than most ofthe large modern power plants in use at that time.

Studies show that thermal efficiencies of 50 percent or higher are possiblewith binary vapor cycles. However, binary vapor cycles are not economicallyattractive because of their high initial cost and the competition offered by thecombined gas–steam power plants.

588 | Thermodynamics

4

5 8

1

T

s

6

3

STEAMCYCLE

MERCURY CYCLE

MERCURY CYCLE

Heat exchanger

STEAMCYCLE

Steamturbine

Condenser

Steampump

6

5

1

Mercuryturbine

4

Boiler

Mercurypump

7

Q2

7

Saturationdome(steam)

Saturation dome(mercury)

2 3

Superheater

8

FIGURE 10–26Mercury–water binary vapor cycle.

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Chapter 10 | 589

SUMMARY

The Carnot cycle is not a suitable model for vapor powercycles because it cannot be approximated in practice. Themodel cycle for vapor power cycles is the Rankine cycle,which is composed of four internally reversible processes:constant-pressure heat addition in a boiler, isentropic expan-sion in a turbine, constant-pressure heat rejection in a con-denser, and isentropic compression in a pump. Steam leavesthe condenser as a saturated liquid at the condenser pressure.

The thermal efficiency of the Rankine cycle can be increasedby increasing the average temperature at which heat is trans-ferred to the working fluid and/or by decreasing the averagetemperature at which heat is rejected to the cooling medium.The average temperature during heat rejection can bedecreased by lowering the turbine exit pressure. Conse-quently, the condenser pressure of most vapor power plants iswell below the atmospheric pressure. The average tempera-ture during heat addition can be increased by raising theboiler pressure or by superheating the fluid to high tempera-tures. There is a limit to the degree of superheating, however,since the fluid temperature is not allowed to exceed a metal-lurgically safe value.

Superheating has the added advantage of decreasing themoisture content of the steam at the turbine exit. Lowering theexhaust pressure or raising the boiler pressure, however, increasesthe moisture content. To take advantage of the improved effi-ciencies at higher boiler pressures and lower condenser pres-sures, steam is usually reheated after expanding partially in thehigh-pressure turbine. This is done by extracting the steam afterpartial expansion in the high-pressure turbine, sending it backto the boiler where it is reheated at constant pressure, andreturning it to the low-pressure turbine for complete expansionto the condenser pressure. The average temperature during the

reheat process, and thus the thermal efficiency of the cycle, canbe increased by increasing the number of expansion and reheatstages. As the number of stages is increased, the expansion andreheat processes approach an isothermal process at maximumtemperature. Reheating also decreases the moisture content atthe turbine exit.

Another way of increasing the thermal efficiency of theRankine cycle is regeneration. During a regeneration process,liquid water (feedwater) leaving the pump is heated by steambled off the turbine at some intermediate pressure in devicescalled feedwater heaters. The two streams are mixed in openfeedwater heaters, and the mixture leaves as a saturated liquidat the heater pressure. In closed feedwater heaters, heat istransferred from the steam to the feedwater without mixing.

The production of more than one useful form of energy(such as process heat and electric power) from the sameenergy source is called cogeneration. Cogeneration plants pro-duce electric power while meeting the process heat require-ments of certain industrial processes. This way, more of theenergy transferred to the fluid in the boiler is utilized for auseful purpose. The fraction of energy that is used for eitherprocess heat or power generation is called the utilization fac-tor of the cogeneration plant.

The overall thermal efficiency of a power plant can beincreased by using a combined cycle. The most commoncombined cycle is the gas–steam combined cycle where agas-turbine cycle operates at the high-temperature range anda steam-turbine cycle at the low-temperature range. Steam isheated by the high-temperature exhaust gases leaving the gasturbine. Combined cycles have a higher thermal efficiencythan the steam- or gas-turbine cycles operating alone.

REFERENCES AND SUGGESTED READINGS

1. R. L. Bannister and G. J. Silvestri. “The Evolution ofCentral Station Steam Turbines.” Mechanical Engineer-ing, February 1989, pp. 70–78.

2. R. L. Bannister, G. J. Silvestri, A. Hizume, and T. Fuji-kawa. “High Temperature Supercritical Steam Turbines.”Mechanical Engineering, February 1987, pp. 60–65.

3. M. M. El-Wakil. Powerplant Technology. New York:McGraw-Hill, 1984.

4. K. W. Li and A. P. Priddy. Power Plant System Design.New York: John Wiley & Sons, 1985.

5. H. Sorensen. Energy Conversion Systems. New York: JohnWiley & Sons, 1983.

6. Steam, Its Generation and Use. 39th ed. New York:Babcock and Wilcox Co., 1978.

7. Turbomachinery 28, no. 2 (March/April 1987). Norwalk,CT: Business Journals, Inc.

8. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

9. J. Weisman and R. Eckart. Modern Power Plant Engi-neering. Englewood Cliffs, NJ: Prentice-Hall, 1985.

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590 | Thermodynamics

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

PROBLEMS*

Carnot Vapor Cycle

10–1C Why is excessive moisture in steam undesirable insteam turbines? What is the highest moisture contentallowed?

10–2C Why is the Carnot cycle not a realistic model forsteam power plants?

10–3E Water enters the boiler of a steady-flow Carnotengine as a saturated liquid at 180 psia and leaves with aquality of 0.90. Steam leaves the turbine at a pressure of 14.7psia. Show the cycle on a T-s diagram relative to the satura-tion lines, and determine (a) the thermal efficiency, (b) thequality at the end of the isothermal heat-rejection process,and (c) the net work output. Answers: (a) 19.3 percent,(b) 0.153, (c) 148 Btu/lbm

10–4 A steady-flow Carnot cycle uses water as the workingfluid. Water changes from saturated liquid to saturated vaporas heat is transferred to it from a source at 250°C. Heat rejec-tion takes place at a pressure of 20 kPa. Show the cycle ona T-s diagram relative to the saturation lines, and determine(a) the thermal efficiency, (b) the amount of heat rejected, inkJ/kg, and (c) the net work output.

10–5 Repeat Prob. 10–4 for a heat rejection pressure of10 kPa.

10–6 Consider a steady-flow Carnot cycle with water as theworking fluid. The maximum and minimum temperatures inthe cycle are 350 and 60°C. The quality of water is 0.891 atthe beginning of the heat-rejection process and 0.1 at the end.Show the cycle on a T-s diagram relative to the saturationlines, and determine (a) the thermal efficiency, (b) the pres-sure at the turbine inlet, and (c) the net work output.Answers: (a) 0.465, (b) 1.40 MPa, (c) 1623 kJ/kg

The Simple Rankine Cycle

10–7C What four processes make up the simple ideal Rank-ine cycle?

10–8C Consider a simple ideal Rankine cycle with fixedturbine inlet conditions. What is the effect of lowering thecondenser pressure on

Pump work input: (a) increases, (b) decreases,(c) remains the same

Turbine work (a) increases, (b) decreases,output: (c) remains the same

Heat supplied: (a) increases, (b) decreases,(c) remains the same

Heat rejected: (a) increases, (b) decreases,(c) remains the same

Cycle efficiency: (a) increases, (b) decreases,(c) remains the same

Moisture content (a) increases, (b) decreases,at turbine exit: (c) remains the same

10–9C Consider a simple ideal Rankine cycle with fixedturbine inlet temperature and condenser pressure. What is theeffect of increasing the boiler pressure on

Pump work input: (a) increases, (b) decreases,(c) remains the same

Turbine work (a) increases, (b) decreases,output: (c) remains the same

Heat supplied: (a) increases, (b) decreases,(c) remains the same

Heat rejected: (a) increases, (b) decreases,(c) remains the same

Cycle efficiency: (a) increases, (b) decreases,(c) remains the same

Moisture content (a) increases, (b) decreases,at turbine exit: (c) remains the same

10–10C Consider a simple ideal Rankine cycle with fixedboiler and condenser pressures. What is the effect of super-heating the steam to a higher temperature on

Pump work input: (a) increases, (b) decreases,(c) remains the same

Turbine work (a) increases, (b) decreases,output: (c) remains the same

Heat supplied: (a) increases, (b) decreases,(c) remains the same

Heat rejected: (a) increases, (b) decreases,(c) remains the same

Cycle efficiency: (a) increases, (b) decreases,(c) remains the same

Moisture content (a) increases, (b) decreases,at turbine exit: (c) remains the same

10–11C How do actual vapor power cycles differ from ide-alized ones?

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Chapter 10 | 591

10–12C Compare the pressures at the inlet and the exit ofthe boiler for (a) actual and (b) ideal cycles.

10–13C The entropy of steam increases in actual steam tur-bines as a result of irreversibilities. In an effort to controlentropy increase, it is proposed to cool the steam in the tur-bine by running cooling water around the turbine casing. It isargued that this will reduce the entropy and the enthalpy ofthe steam at the turbine exit and thus increase the work out-put. How would you evaluate this proposal?

10–14C Is it possible to maintain a pressure of 10 kPa in acondenser that is being cooled by river water entering at20°C?

10–15 A steam power plant operates on a simple idealRankine cycle between the pressure limits of 3 MPa and50 kPa. The temperature of the steam at the turbine inlet is300°C, and the mass flow rate of steam through the cycle is35 kg/s. Show the cycle on a T-s diagram with respect to sat-uration lines, and determine (a) the thermal efficiency of thecycle and (b) the net power output of the power plant.

10–16 Consider a 210-MW steam power plant that operateson a simple ideal Rankine cycle. Steam enters the turbine at10 MPa and 500°C and is cooled in the condenser at a pres-sure of 10 kPa. Show the cycle on a T-s diagram with respectto saturation lines, and determine (a) the quality of the steamat the turbine exit, (b) the thermal efficiency of the cycle,and (c) the mass flow rate of the steam. Answers: (a) 0.793,(b) 40.2 percent, (c) 165 kg/s

10–17 Repeat Prob. 10–16 assuming an isentropic effi-ciency of 85 percent for both the turbine and the pump.Answers: (a) 0.874, (b) 34.1 percent, (c) 194 kg/s

10–18E A steam power plant operates on a simple idealRankine cycle between the pressure limits of 1250 and2 psia. The mass flow rate of steam through the cycle is75 lbm/s. The moisture content of the steam at the turbine exitis not to exceed 10 percent. Show the cycle on a T-s diagramwith respect to saturation lines, and determine (a) the mini-mum turbine inlet temperature, (b) the rate of heat input inthe boiler, and (c) the thermal efficiency of the cycle.

10–19E Repeat Prob. 10–18E assuming an isentropic effi-ciency of 85 percent for both the turbine and the pump.

10–20 Consider a coal-fired steam power plant that pro-duces 300 MW of electric power. The power plant operateson a simple ideal Rankine cycle with turbine inlet conditionsof 5 MPa and 450°C and a condenser pressure of 25 kPa. Thecoal has a heating value (energy released when the fuel isburned) of 29,300 kJ/kg. Assuming that 75 percent of thisenergy is transferred to the steam in the boiler and that theelectric generator has an efficiency of 96 percent, determine(a) the overall plant efficiency (the ratio of net electric poweroutput to the energy input as fuel) and (b) the required rate ofcoal supply. Answers: (a) 24.5 percent, (b) 150 t/h

10–21 Consider a solar-pond power plant that operates on asimple ideal Rankine cycle with refrigerant-134a as the work-ing fluid. The refrigerant enters the turbine as a saturatedvapor at 1.4 MPa and leaves at 0.7 MPa. The mass flow rateof the refrigerant is 3 kg/s. Show the cycle on a T-s diagramwith respect to saturation lines, and determine (a) the thermalefficiency of the cycle and (b) the power output of this plant.

10–22 Consider a steam power plant that operates on a sim-ple ideal Rankine cycle and has a net power output of45 MW. Steam enters the turbine at 7 MPa and 500°C and iscooled in the condenser at a pressure of 10 kPa by runningcooling water from a lake through the tubes of the condenserat a rate of 2000 kg/s. Show the cycle on a T-s diagram withrespect to saturation lines, and determine (a) the thermal effi-ciency of the cycle, (b) the mass flow rate of the steam, and(c) the temperature rise of the cooling water. Answers:(a) 38.9 percent, (b) 36 kg/s, (c) 8.4°C

10–23 Repeat Prob. 10–22 assuming an isentropic effi-ciency of 87 percent for both the turbine and the pump.Answers: (a) 33.8 percent, (b) 41.4 kg/s, (c) 10.5°C

10–24 The net work output and the thermal efficiency forthe Carnot and the simple ideal Rankine cycles with steam asthe working fluid are to be calculated and compared. Steamenters the turbine in both cases at 10 MPa as a saturatedvapor, and the condenser pressure is 20 kPa. In the Rankinecycle, the condenser exit state is saturated liquid and in theCarnot cycle, the boiler inlet state is saturated liquid. Drawthe T-s diagrams for both cycles.

10–25 A binary geothermal power plant uses geothermalwater at 160°C as the heat source. The cycle operates on thesimple Rankine cycle with isobutane as the working fluid.Heat is transferred to the cycle by a heat exchanger in whichgeothermal liquid water enters at 160°C at a rate of 555.9 kg/sand leaves at 90°C. Isobutane enters the turbine at 3.25 MPaand 147°C at a rate of 305.6 kg/s, and leaves at 79.5°C and

3

41

2

Air-cooledcondenser

Geothermalwater in Geothermal

water out

TurbinePump

Heat exchanger

FIGURE P10–25

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410 kPa. Isobutane is condensed in an air-cooled condenserand pumped to the heat exchanger pressure. Assuming thepump to have an isentropic efficiency of 90 percent, determine(a) the isentropic efficiency of the turbine, (b) the net poweroutput of the plant, and (c) the thermal efficiency of the cycle.

10–26 The schematic of a single-flash geothermal powerplant with state numbers is given in Fig. P10–26. Geothermalresource exists as saturated liquid at 230°C. The geothermalliquid is withdrawn from the production well at a rate of 230kg/s, and is flashed to a pressure of 500 kPa by an essentiallyisenthalpic flashing process where the resulting vapor is sepa-rated from the liquid in a separator and directed to the tur-bine. The steam leaves the turbine at 10 kPa with a moisturecontent of 10 percent and enters the condenser where it iscondensed and routed to a reinjection well along with the liq-uid coming off the separator. Determine (a) the mass flowrate of steam through the turbine, (b) the isentropic efficiencyof the turbine, (c) the power output of the turbine, and (d) thethermal efficiency of the plant (the ratio of the turbine workoutput to the energy of the geothermal fluid relative to stan-dard ambient conditions). Answers: (a) 38.2 kg/s, (b) 0.686,(c) 15.4 MW, (d) 7.6 percent

592 | Thermodynamics

10–28 Reconsider Prob. 10–26. Now, it is proposed that theliquid water coming out of the separator be used as the heatsource in a binary cycle with isobutane as the working fluid.Geothermal liquid water leaves the heat exchanger at 90°Cwhile isobutane enters the turbine at 3.25 MPa and 145°Cand leaves at 80°C and 400 kPa. Isobutane is condensed in anair-cooled condenser and then pumped to the heat exchangerpressure. Assuming an isentropic efficiency of 90 percent forthe pump, determine (a) the mass flow rate of isobutane inthe binary cycle, (b) the net power outputs of both the flash-ing and the binary sections of the plant, and (c) the thermalefficiencies of the binary cycle and the combined plant.Answers: (a) 105.5 kg/s, (b) 15.4 MW, 6.14 MW, (c) 12.2 percent,10.6 percent

Separator

Flashchamber

Productionwell

Reinjectionwell

2

1

6

3

Steamturbine

Condenser

4

5

FIGURE P10–26

Flashchamber

Flashchamber

SeparatorI

SeparatorII

Condenser

Steamturbine

Productionwell

Reinjectionwell

1

6

3

8

4

59

7

2

FIGURE P10–27

10–27 Reconsider Prob. 10–26. Now, it is proposed that theliquid water coming out of the separator be routed throughanother flash chamber maintained at 150 kPa, and the steamproduced be directed to a lower stage of the same turbine.Both streams of steam leave the turbine at the same state of10 kPa and 90 percent quality. Determine (a) the temperatureof steam at the outlet of the second flash chamber, (b) thepower produced by the lower stage of the turbine, and (c) thethermal efficiency of the plant.

Pump

Productionwell

Reinjectionwell

Flashchamber

Heat exchanger

Air-cooledcondenser

Condenser

Isobutaneturbine BINARY

CYCLE

Steamturbine

Separator

1

2

3

4

9

10

11

5

6

7

8

FIGURE P10–28

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Chapter 10 | 593

The Reheat Rankine Cycle

10–29C How do the following quantities change when asimple ideal Rankine cycle is modified with reheating?Assume the mass flow rate is maintained the same.

Pump work input: (a) increases, (b) decreases,(c) remains the same

Turbine work (a) increases, (b) decreases,output: (c) remains the same

Heat supplied: (a) increases, (b) decreases,(c) remains the same

Heat rejected: (a) increases, (b) decreases,(c) remains the same

Moisture content (a) increases, (b) decreases,at turbine exit: (c) remains the same

10–30C Show the ideal Rankine cycle with three stages ofreheating on a T-s diagram. Assume the turbine inlet tempera-ture is the same for all stages. How does the cycle efficiencyvary with the number of reheat stages?

10–31C Consider a simple Rankine cycle and an idealRankine cycle with three reheat stages. Both cycles operatebetween the same pressure limits. The maximum temperatureis 700°C in the simple cycle and 450°C in the reheat cycle.Which cycle do you think will have a higher thermalefficiency?

10–32 A steam power plant operates on the idealreheat Rankine cycle. Steam enters the high-

pressure turbine at 8 MPa and 500°C and leaves at 3 MPa.Steam is then reheated at constant pressure to 500°C before itexpands to 20 kPa in the low-pressure turbine. Determine theturbine work output, in kJ/kg, and the thermal efficiency ofthe cycle. Also, show the cycle on a T-s diagram with respectto saturation lines.

10–33 Reconsider Prob. 10–32. Using EES (or other)software, solve this problem by the diagram

window data entry feature of EES. Include the effects of theturbine and pump efficiencies and also show the effects ofreheat on the steam quality at the low-pressure turbine exit.Plot the cycle on a T-s diagram with respect to the saturationlines. Discuss the results of your parametric studies.

10–34 Consider a steam power plant that operates on areheat Rankine cycle and has a net power output of 80 MW.Steam enters the high-pressure turbine at 10 MPa and 500°Cand the low-pressure turbine at 1 MPa and 500°C. Steamleaves the condenser as a saturated liquid at a pressure of10 kPa. The isentropic efficiency of the turbine is 80 percent,and that of the pump is 95 percent. Show the cycle on a T-sdiagram with respect to saturation lines, and determine(a) the quality (or temperature, if superheated) of the steam atthe turbine exit, (b) the thermal efficiency of the cycle, and(c) the mass flow rate of the steam. Answers: (a) 88.1°C,(b) 34.1 percent, (c) 62.7 kg/s

10–35 Repeat Prob. 10–34 assuming both the pump and theturbine are isentropic. Answers: (a) 0.949, (b) 41.3 percent,(c) 50.0 kg/s

10–36E Steam enters the high-pressure turbine of a steampower plant that operates on the ideal reheat Rankine cycle at800 psia and 900°F and leaves as saturated vapor. Steam isthen reheated to 800°F before it expands to a pressure of1 psia. Heat is transferred to the steam in the boiler at a rate of6 � 104 Btu/s. Steam is cooled in the condenser by the coolingwater from a nearby river, which enters the condenser at 45°F.Show the cycle on a T-s diagram with respect to saturationlines, and determine (a) the pressure at which reheating takesplace, (b) the net power output and thermal efficiency, and(c) the minimum mass flow rate of the cooling water required.

10–37 A steam power plant operates on an ideal reheat Rank-ine cycle between the pressure limits of 15 MPa and 10 kPa.The mass flow rate of steam through the cycle is 12 kg/s. Steamenters both stages of the turbine at 500°C. If the moisture con-tent of the steam at the exit of the low-pressure turbine is not toexceed 10 percent, determine (a) the pressure at which reheat-ing takes place, (b) the total rate of heat input in the boiler, and(c) the thermal efficiency of the cycle. Also, show the cycle ona T-s diagram with respect to saturation lines.

10–38 A steam power plant operates on the reheat Rankinecycle. Steam enters the high-pressure turbine at 12.5 MPaand 550°C at a rate of 7.7 kg/s and leaves at 2 MPa. Steam isthen reheated at constant pressure to 450°C before it expandsin the low-pressure turbine. The isentropic efficiencies of theturbine and the pump are 85 percent and 90 percent, respec-tively. Steam leaves the condenser as a saturated liquid. If themoisture content of the steam at the exit of the turbine is notto exceed 5 percent, determine (a) the condenser pressure,(b) the net power output, and (c) the thermal efficiency.Answers: (a) 9.73 kPa, (b) 10.2 MW, (c) 36.9 percent

Boiler

3

4

5

1

Turbine

6

Condenser

Pump

2

FIGURE P10–38

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Regenerative Rankine Cycle

10–39C How do the following quantities change when thesimple ideal Rankine cycle is modified with regeneration?Assume the mass flow rate through the boiler is the same.

Turbine work (a) increases, (b) decreases,output: (c) remains the same

Heat supplied: (a) increases, (b) decreases,(c) remains the same

Heat rejected: (a) increases, (b) decreases,(c) remains the same

Moisture content (a) increases, (b) decreases,at turbine exit: (c) remains the same

10–40C During a regeneration process, some steam isextracted from the turbine and is used to heat the liquid waterleaving the pump. This does not seem like a smart thing to dosince the extracted steam could produce some more work inthe turbine. How do you justify this action?

10–41C How do open feedwater heaters differ from closedfeedwater heaters?

10–42C Consider a simple ideal Rankine cycle and an idealregenerative Rankine cycle with one open feedwater heater.The two cycles are very much alike, except the feedwater inthe regenerative cycle is heated by extracting some steam justbefore it enters the turbine. How would you compare the effi-ciencies of these two cycles?

10–43C Devise an ideal regenerative Rankine cycle that hasthe same thermal efficiency as the Carnot cycle. Show thecycle on a T-s diagram.

10–44 A steam power plant operates on an ideal regenera-tive Rankine cycle. Steam enters the turbine at 6 MPa and450°C and is condensed in the condenser at 20 kPa. Steam isextracted from the turbine at 0.4 MPa to heat the feedwater inan open feedwater heater. Water leaves the feedwater heateras a saturated liquid. Show the cycle on a T-s diagram, anddetermine (a) the net work output per kilogram of steamflowing through the boiler and (b) the thermal efficiency ofthe cycle. Answers: (a) 1017 kJ/kg, (b) 37.8 percent

10–45 Repeat Prob. 10–44 by replacing the open feedwaterheater with a closed feedwater heater. Assume that the feed-water leaves the heater at the condensation temperature of theextracted steam and that the extracted steam leaves the heateras a saturated liquid and is pumped to the line carrying thefeedwater.

10–46 A steam power plant operates on an ideal regenera-tive Rankine cycle with two open feedwater heaters. Steamenters the turbine at 10 MPa and 600°C and exhausts to thecondenser at 5 kPa. Steam is extracted from the turbine at 0.6and 0.2 MPa. Water leaves both feedwater heaters as asaturated liquid. The mass flow rate of steam through theboiler is 22 kg/s. Show the cycle on a T-s diagram, and deter-mine (a) the net power output of the power plant and (b) the

594 | Thermodynamics

thermal efficiency of the cycle. Answers: (a) 30.5 MW,(b) 47.1 percent

10–47 Consider an ideal steam regenerative Rankinecycle with two feedwater heaters, one closed

and one open. Steam enters the turbine at 12.5 MPa and550°C and exhausts to the condenser at 10 kPa. Steam isextracted from the turbine at 0.8 MPa for the closed feedwa-ter heater and at 0.3 MPa for the open one. The feedwater isheated to the condensation temperature of the extracted steamin the closed feedwater heater. The extracted steam leaves theclosed feedwater heater as a saturated liquid, which is subse-quently throttled to the open feedwater heater. Show the cycleon a T-s diagram with respect to saturation lines, and deter-mine (a) the mass flow rate of steam through the boiler for anet power output of 250 MW and (b) the thermal efficiencyof the cycle.

Condenser

1 – y – z

TurbineBoiler

ClosedFWH

P II

P I

5

4

9

8

10

271

z

y

11

OpenFWH3

6

FIGURE P10–47

10–48 Reconsider Prob. 10–47. Using EES (or other)software, investigate the effects of turbine and

pump efficiencies as they are varied from 70 percent to 100percent on the mass flow rate and thermal efficiency. Plot themass flow rate and the thermal efficiency as a function of tur-bine efficiency for pump efficiencies of 70, 85, and 100 per-cent, and discuss the results. Also plot the T-s diagram forturbine and pump efficiencies of 85 percent.

10–49 A steam power plant operates on an ideal reheat–regenerative Rankine cycle and has a net power output of 80MW. Steam enters the high-pressure turbine at 10 MPa and550°C and leaves at 0.8 MPa. Some steam is extracted at thispressure to heat the feedwater in an open feedwater heater.The rest of the steam is reheated to 500°C and is expanded inthe low-pressure turbine to the condenser pressure of 10 kPa.Show the cycle on a T-s diagram with respect to saturationlines, and determine (a) the mass flow rate of steam throughthe boiler and (b) the thermal efficiency of the cycle.Answers: (a) 54.5 kg/s, (b) 44.4 percent

10–50 Repeat Prob. 10–49, but replace the open feedwaterheater with a closed feedwater heater. Assume that the feed-

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Chapter 10 | 595

water leaves the heater at the condensation temperature of theextracted steam and that the extracted steam leaves the heateras a saturated liquid and is pumped to the line carrying thefeedwater.

10–52 A steam power plant operates on the reheat-regenerative Rankine cycle with a closed feedwater heater.Steam enters the turbine at 12.5 MPa and 550°C at a rate of24 kg/s and is condensed in the condenser at a pressure of20 kPa. Steam is reheated at 5 MPa to 550°C. Some steam isextracted from the low-pressure turbine at 1.0 MPa, is com-pletely condensed in the closed feedwater heater, and pumpedto 12.5 MPa before it mixes with the feedwater at the samepressure. Assuming an isentropic efficiency of 88 percent forboth the turbine and the pump, determine (a) the temperatureof the steam at the inlet of the closed feedwater heater,(b) the mass flow rate of the steam extracted from the turbinefor the closed feedwater heater, (c) the net power output, and(d ) the thermal efficiency. Answers: (a) 328°C, (b) 4.29 kg/s,(c) 28.6 MW, (d) 39.3 percent

10–51E A steam power plant operates on an idealreheat–regenerative Rankine cycle with one reheater and twoopen feedwater heaters. Steam enters the high-pressure tur-bine at 1500 psia and 1100°F and leaves the low-pressure tur-bine at 1 psia. Steam is extracted from the turbine at 250 and40 psia, and it is reheated to 1000°F at a pressure of 140 psia.Water leaves both feedwater heaters as a saturated liquid.Heat is transferred to the steam in the boiler at a rate of 4 �105 Btu/s. Show the cycle on a T-s diagram with respect tosaturation lines, and determine (a) the mass flow rate ofsteam through the boiler, (b) the net power output of theplant, and (c) the thermal efficiency of the cycle.

Condenser

1 – yBoiler

ClosedFWH

P II

4 9

5

10

2

7

1

y 8Mixingchamber

P I

High-Pturbine

Low-Pturbine

3

6

FIGURE P10–50

Condenser

1 – y

Boiler

OpenFWH

II

4

7

32

z

y

12

8

P III

1 – y – z

P II P I

9

y

z

10

15

6

Reheater

Low-Pturbine

OpenFWH

I

High-Pturbine

11

FIGURE P10–51E

Boiler

Mixingchamber

ClosedFWH

5

6

7

8

1

Low-Pturbine

High-Pturbine

91 – y

y

Condenser

P IP II

4

10 2

311

FIGURE P10–52

Second-Law Analysis of Vapor Power Cycles

10–53C How can the second-law efficiency of a simpleideal Rankine cycle be improved?

10–54 Determine the exergy destruction associated witheach of the processes of the Rankine cycle described in Prob.10–15, assuming a source temperature of 1500 K and a sinktemperature of 290 K.

10–55 Determine the exergy destruction associated witheach of the processes of the Rankine cycle described in Prob.10–16, assuming a source temperature of 1500 K and a sinktemperature of 290 K. Answers: 0, 1112 kJ/kg, 0, 172.3 kJ/kg

10–56 Determine the exergy destruction associated with theheat rejection process in Prob. 10–22. Assume a source tem-perature of 1500 K and a sink temperature of 290 K. Also,determine the exergy of the steam at the boiler exit. Take P0� 100 kPa.

10–57 Determine the exergy destruction associated witheach of the processes of the reheat Rankine cycle describedin Prob. 10–32. Assume a source temperature of 1800 K anda sink temperature of 300 K.

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10–58 Reconsider Prob. 10–57. Using EES (or other)software, solve this problem by the diagram

window data entry feature of EES. Include the effects of theturbine and pump efficiencies to evaluate the irreversibilitiesassociated with each of the processes. Plot the cycle on a T-sdiagram with respect to the saturation lines. Discuss theresults of your parametric studies.

10–59 Determine the exergy destruction associated with theheat addition process and the expansion process in Prob.10–34. Assume a source temperature of 1600 K and a sinktemperature of 285 K. Also, determine the exergy of thesteam at the boiler exit. Take P0 � 100 kPa. Answers: 1289kJ/kg, 247.9 kJ/kg, 1495 kJ/kg

10–60 Determine the exergy destruction associated with theregenerative cycle described in Prob. 10–44. Assume a sourcetemperature of 1500 K and a sink temperature of 290 K.Answer: 1155 kJ/kg

10–61 Determine the exergy destruction associated with thereheating and regeneration processes described in Prob.10–49. Assume a source temperature of 1800 K and a sinktemperature of 290 K.

10–62 The schematic of a single-flash geothermal powerplant with state numbers is given in Fig. P10–62. Geothermalresource exists as saturated liquid at 230°C. The geothermalliquid is withdrawn from the production well at a rate of 230kg/s and is flashed to a pressure of 500 kPa by an essentiallyisenthalpic flashing process where the resulting vapor is sepa-rated from the liquid in a separator and is directed to the tur-bine. The steam leaves the turbine at 10 kPa with a moisturecontent of 5 percent and enters the condenser where it iscondensed; it is routed to a reinjection well along with theliquid coming off the separator. Determine (a) the power out-put of the turbine and the thermal efficiency of the plant,(b) the exergy of the geothermal liquid at the exit of the flash

596 | Thermodynamics

chamber, and the exergy destructions and the second-law(exergetic) efficiencies for (c) the flash chamber, (d) the tur-bine, and (e) the entire plant. Answers: (a) 10.8 MW, 0.053,(b) 17.3 MW, (c) 5.1 MW, 0.898, (d) 10.9 MW, 0.500,(e) 39.0 MW, 0.218

Cogeneration

10–63C How is the utilization factor Pu for cogenerationplants defined? Could Pu be unity for a cogeneration plantthat does not produce any power?

10–64C Consider a cogeneration plant for which the uti-lization factor is 1. Is the irreversibility associated with thiscycle necessarily zero? Explain.

10–65C Consider a cogeneration plant for which the uti-lization factor is 0.5. Can the exergy destruction associatedwith this plant be zero? If yes, under what conditions?

10–66C What is the difference between cogeneration andregeneration?

10–67 Steam enters the turbine of a cogeneration plant at7 MPa and 500°C. One-fourth of the steam is extracted fromthe turbine at 600-kPa pressure for process heating. Theremaining steam continues to expand to 10 kPa. Theextracted steam is then condensed and mixed with feedwaterat constant pressure and the mixture is pumped to the boilerpressure of 7 MPa. The mass flow rate of steam through theboiler is 30 kg/s. Disregarding any pressure drops and heatlosses in the piping, and assuming the turbine and the pumpto be isentropic, determine the net power produced and theutilization factor of the plant.

Condenser

Steamturbine

Productionwell

Reinjectionwell

Flashchamber

Separator

2

1

6

4

5

3

FIGURE P10–62

P I

Turbine

8

Boiler

Condenser

1P II

Processheater

53

6

7Qprocess

4 2

·

FIGURE P10–67

10–68E A large food-processing plant requires 2 lbm/s ofsaturated or slightly superheated steam at 80 psia, which isextracted from the turbine of a cogeneration plant. The boilergenerates steam at 1000 psia and 1000°F at a rate of 5 lbm/s,

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Chapter 10 | 597

and the condenser pressure is 2 psia. Steam leaves theprocess heater as a saturated liquid. It is then mixed with thefeedwater at the same pressure and this mixture is pumped tothe boiler pressure. Assuming both the pumps and the turbinehave isentropic efficiencies of 86 percent, determine (a) therate of heat transfer to the boiler and (b) the power output ofthe cogeneration plant. Answers: (a) 6667 Btu/s, (b) 2026 kW

10–69 Steam is generated in the boiler of a cogenerationplant at 10 MPa and 450°C at a steady rate of 5 kg/s. Innormal operation, steam expands in a turbine to a pressure of0.5 MPa and is then routed to the process heater, where itsupplies the process heat. Steam leaves the process heater asa saturated liquid and is pumped to the boiler pressure. In thismode, no steam passes through the condenser, which operatesat 20 kPa.(a) Determine the power produced and the rate at whichprocess heat is supplied in this mode.(b) Determine the power produced and the rate of processheat supplied if only 60 percent of the steam is routed to theprocess heater and the remainder is expanded to the con-denser pressure.

10–70 Consider a cogeneration power plant modified withregeneration. Steam enters the turbine at 6 MPa and 450°Cand expands to a pressure of 0.4 MPa. At this pressure, 60percent of the steam is extracted from the turbine, and theremainder expands to 10 kPa. Part of the extracted steam isused to heat the feedwater in an open feedwater heater. Therest of the extracted steam is used for process heating andleaves the process heater as a saturated liquid at 0.4 MPa. Itis subsequently mixed with the feedwater leaving the feedwa-ter heater, and the mixture is pumped to the boiler pressure.

Assuming the turbines and the pumps to be isentropic, showthe cycle on a T-s diagram with respect to saturation lines,and determine the mass flow rate of steam through the boilerfor a net power output of 15 MW. Answer: 17.7 kg/s

10–71 Reconsider Prob. 10–70. Using EES (or other)software, investigate the effect of the extraction

pressure for removing steam from the turbine to be used forthe process heater and open feedwater heater on the requiredmass flow rate. Plot the mass flow rate through the boiler as afunction of the extraction pressure, and discuss the results.

10–72E Steam is generated in the boiler of a cogenerationplant at 600 psia and 800°F at a rate of 18 lbm/s. The plant isto produce power while meeting the process steam require-ments for a certain industrial application. One-third of thesteam leaving the boiler is throttled to a pressure of 120 psiaand is routed to the process heater. The rest of the steam isexpanded in an isentropic turbine to a pressure of 120 psiaand is also routed to the process heater. Steam leaves theprocess heater at 240°F. Neglecting the pump work, deter-mine (a) the net power produced, (b) the rate of process heatsupply, and (c) the utilization factor of this plant.

10–73 A cogeneration plant is to generate power and 8600kJ/s of process heat. Consider an ideal cogeneration steamplant. Steam enters the turbine from the boiler at 7 MPa and500°C. One-fourth of the steam is extracted from the turbineat 600-kPa pressure for process heating. The remainder of thesteam continues to expand and exhausts to the condenser at10 kPa. The steam extracted for the process heater is con-densed in the heater and mixed with the feedwater at 600kPa. The mixture is pumped to the boiler pressure of 7 MPa.Show the cycle on a T-s diagram with respect to saturationlines, and determine (a) the mass flow rate of steam that mustbe supplied by the boiler, (b) the net power produced by theplant, and (c) the utilization factor.

Condenser

TurbineBoiler

P I

P I

6

21

8

FWH

Processheater

34

95

7

FIGURE P10–70

Boiler

Processheater

Turbine

Condenser

P IP II 1

3

4

7

2

85

6

FIGURE P10–73

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Combined Gas–Vapor Power Cycles

10–74C In combined gas–steam cycles, what is the energysource for the steam?

10–75C Why is the combined gas–steam cycle more effi-cient than either of the cycles operated alone?

10–76 The gas-turbine portion of a combined gas–steampower plant has a pressure ratio of 16. Air enters the com-pressor at 300 K at a rate of 14 kg/s and is heated to 1500 Kin the combustion chamber. The combustion gases leavingthe gas turbine are used to heat the steam to 400°C at 10MPa in a heat exchanger. The combustion gases leave theheat exchanger at 420 K. The steam leaving the turbine iscondensed at 15 kPa. Assuming all the compression andexpansion processes to be isentropic, determine (a) the massflow rate of the steam, (b) the net power output, and (c) thethermal efficiency of the combined cycle. For air, assumeconstant specific heats at room temperature. Answers:(a) 1.275 kg/s, (b) 7819 kW, (c) 66.4 percent

10–77 Consider a combined gas–steam power plantthat has a net power output of 450 MW. The

pressure ratio of the gas-turbine cycle is 14. Air enters thecompressor at 300 K and the turbine at 1400 K. The combus-tion gases leaving the gas turbine are used to heat the steamat 8 MPa to 400°C in a heat exchanger. The combustiongases leave the heat exchanger at 460 K. An open feedwaterheater incorporated with the steam cycle operates at a pres-sure of 0.6 MPa. The condenser pressure is 20 kPa. Assumingall the compression and expansion processes to be isentropic,determine (a) the mass flow rate ratio of air to steam, (b) therequired rate of heat input in the combustion chamber, and(c) the thermal efficiency of the combined cycle.

10–78 Reconsider Prob. 10–77. Using EES (or other)software, study the effects of the gas cycle

pressure ratio as it is varied from 10 to 20 on the ratio of gasflow rate to steam flow rate and cycle thermal efficiency. Plotyour results as functions of gas cycle pressure ratio, and dis-cuss the results.

10–79 Repeat Prob. 10–77 assuming isentropic efficienciesof 100 percent for the pump, 82 percent for the compressor,and 86 percent for the gas and steam turbines.

10–80 Reconsider Prob. 10–79. Using EES (or other)software, study the effects of the gas cycle

pressure ratio as it is varied from 10 to 20 on the ratio of gasflow rate to steam flow rate and cycle thermal efficiency. Plotyour results as functions of gas cycle pressure ratio, and dis-cuss the results.

10–81 Consider a combined gas–steam power cycle. Thetopping cycle is a simple Brayton cycle that has a pressureratio of 7. Air enters the compressor at 15°C at a rate of 10kg/s and the gas turbine at 950°C. The bottoming cycle is a

598 | Thermodynamics

reheat Rankine cycle between the pressure limits of 6 MPaand 10 kPa. Steam is heated in a heat exchanger at a rate of1.15 kg/s by the exhaust gases leaving the gas turbine and theexhaust gases leave the heat exchanger at 200°C. Steamleaves the high-pressure turbine at 1.0 MPa and is reheated to400°C in the heat exchanger before it expands in the low-pressure turbine. Assuming 80 percent isentropic efficiencyfor all pumps and turbine, determine (a) the moisture contentat the exit of the low-pressure turbine, (b) the steam tempera-ture at the inlet of the high-pressure turbine, (c) the net poweroutput and the thermal efficiency of the combined plant.

Gasturbine

Steamturbine

Compressor

Condenser

Pump

Combustionchamber

Heatexchanger

9

1

6

2

3

4

5

7

8

1011

FIGURE P10–81

Special Topic: Binary Vapor Cycles

10–82C What is a binary power cycle? What is its purpose?

10–83C By writing an energy balance on the heat exchangerof a binary vapor power cycle, obtain a relation for the ratioof mass flow rates of two fluids in terms of their enthalpies.

10–84C Why is steam not an ideal working fluid for vaporpower cycles?

10–85C Why is mercury a suitable working fluid for thetopping portion of a binary vapor cycle but not for the bot-toming cycle?

10–86C What is the difference between the binary vaporpower cycle and the combined gas–steam power cycle?

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Chapter 10 | 599

Review Problems

10–87 Show that the thermal efficiency of a combinedgas–steam power plant hcc can be expressed as

where hg � Wg /Qin and hs � Ws /Qg,out are the thermal effi-ciencies of the gas and steam cycles, respectively. Using thisrelation, determine the thermal efficiency of a combinedpower cycle that consists of a topping gas-turbine cycle withan efficiency of 40 percent and a bottoming steam-turbinecycle with an efficiency of 30 percent.

10–88 It can be shown that the thermal efficiency of a com-bined gas–steam power plant hcc can be expressed in terms ofthe thermal efficiencies of the gas- and the steam-turbinecycles as

Prove that the value of hcc is greater than either of hg or hs.That is, the combined cycle is more efficient than either ofthe gas-turbine or steam-turbine cycles alone.

10–89 Consider a steam power plant operating on the idealRankine cycle with reheat between the pressure limits of 25MPa and 10 kPa with a maximum cycle temperature of600°C and a moisture content of 8 percent at the turbine exit.For a reheat temperature of 600°C, determine the reheat pres-sures of the cycle for the cases of (a) single and (b) doublereheat.

10–90E The Stillwater geothermal power plant in Nevada,which started full commercial operation in 1986, is designedto operate with seven identical units. Each of these sevenunits consists of a pair of power cycles, labeled Level I andLevel II, operating on the simple Rankine cycle using anorganic fluid as the working fluid.

The heat source for the plant is geothermal water (brine)entering the vaporizer (boiler) of Level I of each unit at325°F at a rate of 384,286 lbm/h and delivering 22.79MBtu/h (“M” stands for “million”). The organic fluid thatenters the vaporizer at 202.2°F at a rate of 157,895 lbm/hleaves it at 282.4°F and 225.8 psia as saturated vapor. Thissaturated vapor expands in the turbine to 95.8°F and 19.0psia and produces 1271 kW of electric power. About 200 kWof this power is used by the pumps, the auxiliaries, and thesix fans of the condenser. Subsequently, the organic workingfluid is condensed in an air-cooled condenser by air thatenters the condenser at 55°F at a rate of 4,195,100 lbm/h andleaves at 84.5°F. The working fluid is pumped and then pre-heated in a preheater to 202.2°F by absorbing 11.14 MBtu/hof heat from the geothermal water (coming from the vapor-izer of Level II) entering the preheater at 211.8°F and leavingat 154.0°F.

Taking the average specific heat of the geothermal water tobe 1.03 Btu/lbm · °F, determine (a) the exit temperature ofthe geothermal water from the vaporizer, (b) the rate of heat

hcc � hg � hs � hghs

hcc � hg � hs � hghs

rejection from the working fluid to the air in the condenser,(c) the mass flow rate of the geothermal water at the pre-heater, and (d ) the thermal efficiency of the Level I cycle ofthis geothermal power plant. Answers: (a) 267.4°F, (b) 29.7MBtu/h, (c) 187,120 lbm/h, (d) 10.8 percent

Air

Condenser

Vapor

Generator Turbine

Electricity

Vaporizer

Vapor

Working fluid

Fluid pump

Land surface

Production pump

Hot geothermal brine

Cooler geothermal brine

Injectionpump

Preheater6

5

3

4

12

89

7

mgeo

FIGURE P10–90ESchematic of a binary geothermal power plant.

Courtesy of ORMAT Energy Systems, Inc.

10–91 Steam enters the turbine of a steam power plant thatoperates on a simple ideal Rankine cycle at a pressure of 6MPa, and it leaves as a saturated vapor at 7.5 kPa. Heat istransferred to the steam in the boiler at a rate of 40,000 kJ/s.Steam is cooled in the condenser by the cooling water from anearby river, which enters the condenser at 15°C. Show thecycle on a T-s diagram with respect to saturation lines, anddetermine (a) the turbine inlet temperature, (b) the net poweroutput and thermal efficiency, and (c) the minimum massflow rate of the cooling water required.

10–92 A steam power plant operates on an ideal Rankinecycle with two stages of reheat and has a net power output of

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120 MW. Steam enters all three stages of the turbine at500°C. The maximum pressure in the cycle is 15 MPa, andthe minimum pressure is 5 kPa. Steam is reheated at 5 MPathe first time and at 1 MPa the second time. Show thecycle on a T-s diagram with respect to saturation lines, anddetermine (a) the thermal efficiency of the cycle and (b) themass flow rate of the steam. Answers: (a) 45.5 percent,(b) 64.4 kg/s

10–93 Consider a steam power plant that operates on aregenerative Rankine cycle and has a net power output of 150MW. Steam enters the turbine at 10 MPa and 500°C and thecondenser at 10 kPa. The isentropic efficiency of the turbineis 80 percent, and that of the pumps is 95 percent. Steam isextracted from the turbine at 0.5 MPa to heat the feedwater inan open feedwater heater. Water leaves the feedwater heateras a saturated liquid. Show the cycle on a T-s diagram, anddetermine (a) the mass flow rate of steam through the boilerand (b) the thermal efficiency of the cycle. Also, determinethe exergy destruction associated with the regenerationprocess. Assume a source temperature of 1300 K and a sinktemperature of 303 K.

600 | Thermodynamics

10–96 Repeat Prob. 10–95 assuming an isentropic effi-ciency of 84 percent for the turbines and 100 percent for thepumps.

10–97 A steam power plant operates on an ideal reheat–regenerative Rankine cycle with one reheater and two feed-water heaters, one open and one closed. Steam enters thehigh-pressure turbine at 15 MPa and 600°C and the low-pressure turbine at 1 MPa and 500°C. The condenser pressureis 5 kPa. Steam is extracted from the turbine at 0.6 MPa forthe closed feedwater heater and at 0.2 MPa for the open feed-water heater. In the closed feedwater heater, the feedwater isheated to the condensation temperature of the extractedsteam. The extracted steam leaves the closed feedwater heateras a saturated liquid, which is subsequently throttled to theopen feedwater heater. Show the cycle on a T-s diagram withrespect to saturation lines. Determine (a) the fraction ofsteam extracted from the turbine for the open feedwaterheater, (b) the thermal efficiency of the cycle, and (c) the netpower output for a mass flow rate of 42 kg/s through theboiler.

P I

Turbine Boiler

CondenserP II

5

2

76

OpenFWH

4

1

3

y 1 – y

FIGURE P10–93

11

12

Pump I

High-PTurbine Boiler

Condenser

Pump II

5OpenFWH

y

1 – y – z

Low-PTurbine

ClosedFWH

6

z

43

7

21

13

10

9

8

Reheater

FIGURE P10–9710–94 Repeat Prob. 10–93 assuming both the pump and theturbine are isentropic.

10–95 Consider an ideal reheat–regenerative Rankine cyclewith one open feedwater heater. The boiler pressure is 10MPa, the condenser pressure is 15 kPa, the reheater pressureis 1 MPa, and the feedwater pressure is 0.6 MPa. Steamenters both the high- and low-pressure turbines at 500°C.Show the cycle on a T-s diagram with respect to saturationlines, and determine (a) the fraction of steam extracted forregeneration and (b) the thermal efficiency of the cycle.Answers: (a) 0.144, (b) 42.1 percent

10–98 Consider a cogeneration power plant that is modifiedwith reheat and that produces 3 MW of power and supplies 7MW of process heat. Steam enters the high-pressure turbineat 8 MPa and 500°C and expands to a pressure of 1 MPa. Atthis pressure, part of the steam is extracted from the turbineand routed to the process heater, while the remainder isreheated to 500°C and expanded in the low-pressure turbineto the condenser pressure of 15 kPa. The condensate from thecondenser is pumped to 1 MPa and is mixed with theextracted steam, which leaves the process heater as a com-

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pressed liquid at 120°C. The mixture is then pumped to theboiler pressure. Assuming the turbine to be isentropic, showthe cycle on a T-s diagram with respect to saturation lines,and disregarding pump work, determine (a) the rate of heatinput in the boiler and (b) the fraction of steam extracted forprocess heating.

10–102 Steam is to be supplied from a boiler to a high-pressure turbine whose isentropic efficiency is 75 percentat conditions to be determined. The steam is to leave thehigh-pressure turbine as a saturated vapor at 1.4 MPa, and theturbine is to produce 1 MW of power. Steam at the turbineexit is extracted at a rate of 1000 kg/min and routed to aprocess heater while the rest of the steam is supplied to alow-pressure turbine whose isentropic efficiency is 60 per-cent. The low-pressure turbine allows the steam to expand to10 kPa pressure and produces 0.8 MW of power. Determinethe temperature, pressure, and the flow rate of steam at theinlet of the high-pressure turbine.

10–103 A textile plant requires 4 kg/s of saturated steam at2 MPa, which is extracted from the turbine of a cogenerationplant. Steam enters the turbine at 8 MPa and 500°C at a rateof 11 kg/s and leaves at 20 kPa. The extracted steam leavesthe process heater as a saturated liquid and mixes with thefeedwater at constant pressure. The mixture is pumped to theboiler pressure. Assuming an isentropic efficiency of 88 per-cent for both the turbine and the pumps, determine (a) therate of process heat supply, (b) the net power output, and(c) the utilization factor of the plant. Answers: (a) 8.56 MW,(b) 8.60 MW, (c) 53.8 percent

High-PTurbine Boiler

Condenser

P I

P II

5

Low-PTurbine

3

1

8

7

6

9Processheater

3 MW

24

7 MW

FIGURE P10–98

10–99 The gas-turbine cycle of a combined gas–steampower plant has a pressure ratio of 8. Air enters the compres-sor at 290 K and the turbine at 1400 K. The combustiongases leaving the gas turbine are used to heat the steam at 15MPa to 450°C in a heat exchanger. The combustion gasesleave the heat exchanger at 247°C. Steam expands in a high-pressure turbine to a pressure of 3 MPa and is reheated in thecombustion chamber to 500°C before it expands in a low-pressure turbine to 10 kPa. The mass flow rate of steam is 30kg/s. Assuming all the compression and expansion processesto be isentropic, determine (a) the mass flow rate of air in thegas-turbine cycle, (b) the rate of total heat input, and (c) thethermal efficiency of the combined cycle.Answers: (a) 263 kg/s, (b) 2.80 � 105 kJ/s, (c) 55.6 percent

10–100 Repeat Prob. 10–99 assuming isentropic efficien-cies of 100 percent for the pump, 80 percent for the compres-sor, and 85 percent for the gas and steam turbines.

10–101 Starting with Eq. 10–20, show that the exergydestruction associated with a simple ideal Rankine cycle canbe expressed as i � qin(hth,Carnot � hth), where hth is effi-ciency of the Rankine cycle and hth,Carnot is the efficiency ofthe Carnot cycle operating between the same temperaturelimits.

Boiler

Turbine

6

7

4 21

8

Condenser

P IP II

5

3

Processheater

FIGURE P10–103

10–104 Using EES (or other) software, investigate theeffect of the condenser pressure on the perfor-

mance of a simple ideal Rankine cycle. Turbine inlet condi-tions of steam are maintained constant at 5 MPa and 500°Cwhile the condenser pressure is varied from 5 to 100 kPa.Determine the thermal efficiency of the cycle and plot itagainst the condenser pressure, and discuss the results.

10–105 Using EES (or other) software, investigate theeffect of the boiler pressure on the perfor-

mance of a simple ideal Rankine cycle. Steam enters the tur-bine at 500°C and exits at 10 kPa. The boiler pressure is

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602 | Thermodynamics

10–112 A simple ideal Rankine cycle operates between thepressure limits of 10 kPa and 3 MPa, with a turbine inlettemperature of 600°C. Disregarding the pump work, the cycleefficiency is

(a) 24 percent (b) 37 percent (c) 52 percent(d ) 63 percent (e) 71 percent

10–113 A simple ideal Rankine cycle operates between thepressure limits of 10 kPa and 5 MPa, with a turbine inlettemperature of 600°C. The mass fraction of steam that con-denses at the turbine exit is

(a) 6 percent (b) 9 percent (c) 12 percent(d ) 15 percent (e) 18 percent

10–114 A steam power plant operates on the simple idealRankine cycle between the pressure limits of 10 kPa and 10MPa, with a turbine inlet temperature of 600°C. The rate ofheat transfer in the boiler is 800 kJ/s. Disregarding the pumpwork, the power output of this plant is

(a) 243 kW (b) 284 kW (c) 508 kW(d ) 335 kW (e) 800 kW

10–115 Consider a combined gas-steam power plant. Waterfor the steam cycle is heated in a well-insulated heatexchanger by the exhaust gases that enter at 800 K at a rateof 60 kg/s and leave at 400 K. Water enters the heatexchanger at 200°C and 8 MPa and leaves at 350°C and 8MPa. If the exhaust gases are treated as air with constant spe-cific heats at room temperature, the mass flow rate of waterthrough the heat exchanger becomes

(a) 11 kg/s (b) 24 kg/s (c) 46 kg/s(d ) 53 kg/s (e) 60 kg/s

10–116 An ideal reheat Rankine cycle operates between thepressure limits of 10 kPa and 8 MPa, with reheat occurring at4 MPa. The temperature of steam at the inlets of both tur-bines is 500°C, and the enthalpy of steam is 3185 kJ/kg at theexit of the high-pressure turbine, and 2247 kJ/kg at the exitof the low-pressure turbine. Disregarding the pump work, thecycle efficiency is

(a) 29 percent (b) 32 percent (c) 36 percent(d ) 41 percent (e) 49 percent

10–117 Pressurized feedwater in a steam power plant is tobe heated in an ideal open feedwater heater that operates at apressure of 0.5 MPa with steam extracted from the turbine. Ifthe enthalpy of feedwater is 252 kJ/kg and the enthalpy ofextracted steam is 2665 kJ/kg, the mass fraction of steamextracted from the turbine is

(a) 4 percent (b) 10 percent (c) 16 percent(d ) 27 percent (e) 12 percent

10–118 Consider a steam power plant that operates on theregenerative Rankine cycle with one open feedwater heater.The enthalpy of the steam is 3374 kJ/kg at the turbine inlet,2797 kJ/kg at the location of bleeding, and 2346 kJ/kg at the

varied from 0.5 to 20 MPa. Determine the thermal efficiencyof the cycle and plot it against the boiler pressure, and discussthe results.

10–106 Using EES (or other) software, investigate theeffect of superheating the steam on the perfor-

mance of a simple ideal Rankine cycle. Steam enters the turbineat 3 MPa and exits at 10 kPa. The turbine inlet temperature isvaried from 250 to 1100°C. Determine the thermal efficiencyof the cycle and plot it against the turbine inlet temperature,and discuss the results.

10–107 Using EES (or other) software, investigate theeffect of reheat pressure on the performance

of an ideal Rankine cycle. The maximum and minimum pres-sures in the cycle are 15 MPa and 10 kPa, respectively, andsteam enters both stages of the turbine at 500°C. The reheatpressure is varied from 12.5 to 0.5 MPa. Determine the ther-mal efficiency of the cycle and plot it against the reheat pres-sure, and discuss the results.

10–108 Using EES (or other) software, investigate theeffect of number of reheat stages on the per-

formance of an ideal Rankine cycle. The maximum and mini-mum pressures in the cycle are 15 MPa and 10 kPa,respectively, and steam enters all stages of the turbine at500°C. For each case, maintain roughly the same pressureratio across each turbine stage. Determine the thermal effi-ciency of the cycle and plot it against the number of reheatstages 1, 2, 4, and 8, and discuss the results.

10–109 Using EES (or other) software, investigate theeffect of extraction pressure on the perfor-

mance of an ideal regenerative Rankine cycle with one openfeedwater heater. Steam enters the turbine at 15 MPa and600°C and the condenser at 10 kPa. Determine the thermalefficiency of the cycle, and plot it against extraction pressuresof 12.5, 10, 7, 5, 2, 1, 0.5, 0.1, and 0.05 MPa, and discuss theresults.

10–110 Using EES (or other) software, investigate theeffect of the number of regeneration stages on

the performance of an ideal regenerative Rankine cycle.Steam enters the turbine at 15 MPa and 600°C and the con-denser at 5 kPa. For each case, maintain about the same tem-perature difference between any two regeneration stages.Determine the thermal efficiency of the cycle, and plot itagainst the number of regeneration stages for 1, 2, 3, 4, 5, 6,8, and 10 regeneration stages.

Fundamentals of Engineering (FE) Exam Problems

10–111 Consider a steady-flow Carnot cycle with water asthe working fluid executed under the saturation domebetween the pressure limits of 8 MPa and 20 kPa. Waterchanges from saturated liquid to saturated vapor during theheat addition process. The net work output of this cycle is

(a) 494 kJ/kg (b) 975 kJ/kg (c) 596 kJ/kg(d ) 845 kJ/kg (e) 1148 kJ/kg

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Chapter 10 | 603

is cooled and condensed by the cooling water from anearby river, which enters the adiabatic condenser at a rate of463 kg/s.

1. The total power output of the turbine is

(a) 17.0 MW (b) 8.4 MW (c) 12.2 MW(d ) 20.0 MW (e) 3.4 MW

2. The temperature rise of the cooling water from the river inthe condenser is

(a) 8.0°C (b) 5.2°C (c) 9.6°C(d ) 12.9°C (e) 16.2°C

3. The mass flow rate of steam through the process heater is

(a) 1.6 kg/s (b) 3.8 kg/s (c) 5.2 kg/s(d ) 7.6 kg/s (e) 10.4 kg/s

4. The rate of heat supply from the process heater per unitmass of steam passing through it is

(a) 246 kJ/kg (b) 893 kJ/kg (c) 1344 kJ/kg(d ) 1891 kJ/kg (e) 2060 kJ/kg

5. The rate of heat transfer to the steam in the boiler is

(a) 26.0 MJ/s (b) 53.8 MJ/s (c) 39.5 MJ/s(d ) 62.8 MJ/s (e) 125.4 MJ/s

turbine exit. The net power output of the plant is 120 MW,and the fraction of steam bled off the turbine for regenerationis 0.172. If the pump work is negligible, the mass flow rate ofsteam at the turbine inlet is

(a) 117 kg/s (b) 126 kg/s (c) 219 kg/s(d ) 268 kg/s (e) 679 kg/s

10–119 Consider a simple ideal Rankine cycle. If the con-denser pressure is lowered while keeping turbine inlet statethe same,(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease.(c) the cycle efficiency will decrease.(d ) the moisture content at turbine exit will decrease.(e) the pump work input will decrease.

10–120 Consider a simple ideal Rankine cycle with fixedboiler and condenser pressures. If the steam is superheated toa higher temperature,(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease.(c) the cycle efficiency will decrease.(d ) the moisture content at turbine exit will decrease.(e) the amount of heat input will decrease.

10–121 Consider a simple ideal Rankine cycle with fixedboiler and condenser pressures. If the cycle is modified withreheating,(a) the turbine work output will decrease.(b) the amount of heat rejected will decrease.(c) the pump work input will decrease.(d ) the moisture content at turbine exit will decrease.(e) the amount of heat input will decrease.

10–122 Consider a simple ideal Rankine cycle with fixedboiler and condenser pressures. If the cycle is modified withregeneration that involves one open feedwater heater (selectthe correct statement per unit mass of steam flowing throughthe boiler),(a) the turbine work output will decrease.(b) the amount of heat rejected will increase.(c) the cycle thermal efficiency will decrease.(d) the quality of steam at turbine exit will decrease.(e) the amount of heat input will increase.

10–123 Consider a cogeneration power plant modified withregeneration. Steam enters the turbine at 6 MPa and 450°C ata rate of 20 kg/s and expands to a pressure of 0.4 MPa. Atthis pressure, 60 percent of the steam is extracted from theturbine, and the remainder expands to a pressure of 10 kPa.Part of the extracted steam is used to heat feedwater inan open feedwater heater. The rest of the extracted steam isused for process heating and leaves the process heater asa saturated liquid at 0.4 MPa. It is subsequently mixed withthe feedwater leaving the feedwater heater, and the mixtureis pumped to the boiler pressure. The steam in the condenser

P I

Turbine Boiler

CondenserCondenserCondenser

P II

9

6

4 3

8

h7 � h8 � h10 � 2665.6 kJ/kg

h3 � h4 � h9 � 604.66

h2 � 192.20

h11 � 2128.8

∆Th5 � 610.73

h1 � 191.81

h6 � 3302.9 kJ/kg

20 kg/s

463 kg/s1

2FWH

11

7

10Processheater

5

FIGURE P10–123

Design and Essay Problems

10–124 Design a steam power cycle that can achieve a cyclethermal efficiency of at least 40 percent under the conditionsthat all turbines have isentropic efficiencies of 85 percent andall pumps have isentropic efficiencies of 60 percent. Prepare

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an engineering report describing your design. Your designreport must include, but is not limited to, the following:(a) Discussion of various cycles attempted to meet the goal as

well as the positive and negative aspects of your design.(b) System figures and T-s diagrams with labeled states and

temperature, pressure, enthalpy, and entropy informationfor your design.

(c) Sample calculations.

10–125 Contact your power company and obtain informa-tion on the thermodynamic aspects of their most recently builtpower plant. If it is a conventional power plant, find out whyit is preferred over a highly efficient combined power plant.

10–126 Several geothermal power plants are in operation inthe United States and more are being built since the heatsource of a geothermal plant is hot geothermal water, whichis “free energy.” An 8-MW geothermal power plant is beingconsidered at a location where geothermal water at 160°C isavailable. Geothermal water is to serve as the heat source fora closed Rankine power cycle with refrigerant-134a as theworking fluid. Specify suitable temperatures and pressuresfor the cycle, and determine the thermal efficiency of thecycle. Justify your selections.

10–127 A 10-MW geothermal power plant is being consid-ered at a site where geothermal water at 230°C is available.Geothermal water is to be flashed into a chamber to a lowerpressure where part of the water evaporates. The liquid isreturned to the ground while the vapor is used to drive thesteam turbine. The pressures at the turbine inlet and the tur-bine exit are to remain above 200 kPa and 8 kPa, respec-tively. High-pressure flash chambers yield a small amount ofsteam with high exergy whereas lower-pressure flash cham-bers yield considerably more steam but at a lower exergy. Bytrying several pressures, determine the optimum pressure of

604 | Thermodynamics

the flash chamber to maximize the power production per unitmass of geothermal water withdrawn. Also, determine thethermal efficiency for each case assuming 10 percent of thepower produced is used to drive the pumps and other auxil-iary equipment.

10–128 A natural gas–fired furnace in a textile plant is usedto provide steam at 130°C. At times of high demand, the fur-nace supplies heat to the steam at a rate of 30 MJ/s. The plantalso uses up to 6 MW of electrical power purchased from thelocal power company. The plant management is consideringconverting the existing process plant into a cogenerationplant to meet both their process-heat and power requirements.Your job is to come up with some designs. Designs based ona gas turbine or a steam turbine are to be considered. Firstdecide whether a system based on a gas turbine or a steamturbine will best serve the purpose, considering the cost andthe complexity. Then propose your design for the cogenera-tion plant complete with pressures and temperatures and themass flow rates. Show that the proposed design meets thepower and process-heat requirements of the plant.

10–129E A photographic equipment manufacturer uses aflow of 64,500 lbm/h of steam in its manufacturing process.Presently the spent steam at 3.8 psig and 224°F is exhaustedto the atmosphere. Do the preliminary design of a system touse the energy in the waste steam economically. If electricityis produced, it can be generated about 8000 h/yr and its valueis $0.05/kWh. If the energy is used for space heating, thevalue is also $0.05/kWh, but it can only be used about 3000h/yr (only during the “heating season”). If the steam is con-densed and the liquid H2O is recycled through the process, itsvalue is $0.50/100 gal. Make all assumptions as realistic aspossible. Sketch the system you propose. Make a separate listof required components and their specifications (capacity,efficiency, etc.). The final result will be the calculated annualdollar value of the energy use plan (actually a saving becauseit will replace electricity or heat and/or water that would oth-erwise have to be purchased).

10–130 Design the condenser of a steam power plant thathas a thermal efficiency of 40 percent and generates 10 MWof net electric power. Steam enters the condenser as saturatedvapor at 10 kPa, and it is to be condensed outside horizontaltubes through which cooling water from a nearby river flows.The temperature rise of the cooling water is limited to 8°C,and the velocity of the cooling water in the pipes is limited to6 m/s to keep the pressure drop at an acceptable level. Fromprior experience, the average heat flux based on the outer sur-face of the tubes can be taken to be 12,000 W/m2. Specify thepipe diameter, total pipe length, and the arrangement of thepipes to minimize the condenser volume.

10–131 Water-cooled steam condensers are commonly usedin steam power plants. Obtain information about water-cooledsteam condensers by doing a literature search on the topic and

Turbine

Flashchamber

230°CGeothermal

water

FIGURE P10–127

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Chapter 10 | 605

also by contacting some condenser manufacturers. In a report,describe the various types, the way they are designed, the lim-itation on each type, and the selection criteria.

10–132 Steam boilers have long been used to provideprocess heat as well as to generate power. Write anessay on the history of steam boilers and the evolution ofmodern supercritical steam power plants. What was the roleof the American Society of Mechanical Engineers in thisdevelopment?

10–133 The technology for power generation using geother-mal energy is well established, and numerous geothermalpower plants throughout the world are currently generatingelectricity economically. Binary geothermal plants utilize avolatile secondary fluid such as isobutane, n-pentane, and R-114 in a closed loop. Consider a binary geothermal plant

with R-114 as the working fluid that is flowing at a rate of 600kg/s. The R-114 is vaporized in a boiler at 115°C by the geo-thermal fluid that enters at 165°C, and is condensed at 30°Coutside the tubes by cooling water that enters the tubes at18°C. Based on prior experience, the average heat flux basedon the outer surface of the tubes can be taken to be 4600W/m2. The enthalpy of vaporization of R-114 at 30°C is hfg �121.5 kJ/kg.

Specify (a) the length, diameter, and number of tubes andtheir arrangement in the condenser to minimize overall vol-ume of the condenser; (b) the mass flow rate of coolingwater; and (c) the flow rate of make-up water needed if acooling tower is used to reject the waste heat from the cool-ing water. The liquid velocity is to remain under 6 m/s andthe length of the tubes is limited to 8 m.

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Chapter 11REFRIGERATION CYCLES

| 607

Amajor application area of thermodynamics is refrigera-tion, which is the transfer of heat from a lower temper-ature region to a higher temperature one. Devices that

produce refrigeration are called refrigerators, and the cycles onwhich they operate are called refrigeration cycles. The mostfrequently used refrigeration cycle is the vapor-compressionrefrigeration cycle in which the refrigerant is vaporized andcondensed alternately and is compressed in the vapor phase.Another well-known refrigeration cycle is the gas refrigerationcycle in which the refrigerant remains in the gaseous phasethroughout. Other refrigeration cycles discussed in this chapterare cascade refrigeration, where more than one refrigerationcycle is used; absorption refrigeration, where the refrigerant isdissolved in a liquid before it is compressed; and, as a Topic ofSpecial Interest, thermoelectric refrigeration, where refrigera-tion is produced by the passage of electric current through twodissimilar materials.

ObjectivesThe objectives of Chapter 11 are to:

• Introduce the concepts of refrigerators and heat pumps andthe measure of their performance.

• Analyze the ideal vapor-compression refrigeration cycle.

• Analyze the actual vapor-compression refrigeration cycle.

• Review the factors involved in selecting the right refrigerantfor an application.

• Discuss the operation of refrigeration and heat pumpsystems.

• Evaluate the performance of innovative vapor-compressionrefrigeration systems.

• Analyze gas refrigeration systems.

• Introduce the concepts of absorption-refrigeration systems.

• Review the concepts of thermoelectric power generationand refrigeration.

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11–1 ■ REFRIGERATORS AND HEAT PUMPSWe all know from experience that heat flows in the direction of decreasingtemperature, that is, from high-temperature regions to low-temperature ones.This heat-transfer process occurs in nature without requiring any devices.The reverse process, however, cannot occur by itself. The transfer of heatfrom a low-temperature region to a high-temperature one requires specialdevices called refrigerators.

Refrigerators are cyclic devices, and the working fluids used in the refrig-eration cycles are called refrigerants. A refrigerator is shown schematicallyin Fig. 11–1a. Here QL is the magnitude of the heat removed from the refrig-erated space at temperature TL ,QH is the magnitude of the heat rejected tothe warm space at temperature TH , and Wnet,in is the net work input to therefrigerator. As discussed in Chap. 6, QL and QH represent magnitudes andthus are positive quantities.

Another device that transfers heat from a low-temperature medium to ahigh-temperature one is the heat pump. Refrigerators and heat pumps areessentially the same devices; they differ in their objectives only. The objec-tive of a refrigerator is to maintain the refrigerated space at a low tempera-ture by removing heat from it. Discharging this heat to a higher-temperaturemedium is merely a necessary part of the operation, not the purpose. Theobjective of a heat pump, however, is to maintain a heated space at a hightemperature. This is accomplished by absorbing heat from a low-temperaturesource, such as well water or cold outside air in winter, and supplying thisheat to a warmer medium such as a house (Fig. 11–1b).

The performance of refrigerators and heat pumps is expressed in terms ofthe coefficient of performance (COP), defined as

(11–1)

(11–2)

These relations can also be expressed in the rate form by replacing thequantities QL, QH, and Wnet,in by Q

.L, Q

.H, and W

.net,in, respectively. Notice that

both COPR and COPHP can be greater than 1. A comparison of Eqs. 11–1and 11–2 reveals that

(11–3)

for fixed values of QL and QH. This relation implies that COPHP � 1 sinceCOPR is a positive quantity. That is, a heat pump functions, at worst, as aresistance heater, supplying as much energy to the house as it consumes. Inreality, however, part of QH is lost to the outside air through piping andother devices, and COPHP may drop below unity when the outside air tem-perature is too low. When this happens, the system normally switches to thefuel (natural gas, propane, oil, etc.) or resistance-heating mode.

The cooling capacity of a refrigeration system—that is, the rate of heatremoval from the refrigerated space—is often expressed in terms of tons ofrefrigeration. The capacity of a refrigeration system that can freeze 1 ton(2000 lbm) of liquid water at 0°C (32°F) into ice at 0°C in 24 h is said to be

COPHP � COPR � 1

COPHP �Desired output

Required input�

Heating effect

Work input�

QH

Wnet,in

COPR �Desired output

Required input�

Cooling effect

Work input�

QL

Wnet,in

608 | Thermodynamics

WARMhouse

WARMenvironment

COLDrefrigerated

spaceCOLD

environment

(a) Refrigerator (b) Heat pump

QH(desiredoutput)

HPR

QH

QL(desiredoutput)

QL

Wnet,in(requiredinput)

Wnet,in(required input)

FIGURE 11–1The objective of a refrigerator is toremove heat (QL) from the coldmedium; the objective of a heat pumpis to supply heat (QH) to a warmmedium.

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1 ton. One ton of refrigeration is equivalent to 211 kJ/min or 200 Btu/min.The cooling load of a typical 200-m2 residence is in the 3-ton (10-kW)range.

11–2 ■ THE REVERSED CARNOT CYCLERecall from Chap. 6 that the Carnot cycle is a totally reversible cycle thatconsists of two reversible isothermal and two isentropic processes. It has themaximum thermal efficiency for given temperature limits, and it serves as astandard against which actual power cycles can be compared.

Since it is a reversible cycle, all four processes that comprise the Carnotcycle can be reversed. Reversing the cycle does also reverse the directionsof any heat and work interactions. The result is a cycle that operates in thecounterclockwise direction on a T-s diagram, which is called the reversedCarnot cycle. A refrigerator or heat pump that operates on the reversedCarnot cycle is called a Carnot refrigerator or a Carnot heat pump.

Consider a reversed Carnot cycle executed within the saturation dome of arefrigerant, as shown in Fig. 11–2. The refrigerant absorbs heat isothermallyfrom a low-temperature source at TL in the amount of QL (process 1-2), iscompressed isentropically to state 3 (temperature rises to TH), rejects heatisothermally to a high-temperature sink at TH in the amount of QH (process3-4), and expands isentropically to state 1 (temperature drops to TL). Therefrigerant changes from a saturated vapor state to a saturated liquid state inthe condenser during process 3-4.

Chapter 11 | 609

QH

2

THCondenser

WARM mediumat TH

COLD mediumat TL

QL

QH

QL

4 3

21

T

s

EvaporatorTL

Turbine Compressor

1

34

FIGURE 11–2Schematic of a Carnot refrigerator and T-s diagram of the reversed Carnot cycle.

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The coefficients of performance of Carnot refrigerators and heat pumpsare expressed in terms of temperatures as

(11–4)

and

(11–5)

Notice that both COPs increase as the difference between the two tempera-tures decreases, that is, as TL rises or TH falls.

The reversed Carnot cycle is the most efficient refrigeration cycle operatingbetween two specified temperature levels. Therefore, it is natural to look at itfirst as a prospective ideal cycle for refrigerators and heat pumps. If we could,we certainly would adapt it as the ideal cycle. As explained below, however,the reversed Carnot cycle is not a suitable model for refrigeration cycles.

The two isothermal heat transfer processes are not difficult to achieve inpractice since maintaining a constant pressure automatically fixes the tem-perature of a two-phase mixture at the saturation value. Therefore, processes1-2 and 3-4 can be approached closely in actual evaporators and condensers.However, processes 2-3 and 4-1 cannot be approximated closely in practice.This is because process 2-3 involves the compression of a liquid–vapor mix-ture, which requires a compressor that will handle two phases, and process4-1 involves the expansion of high-moisture-content refrigerant in a turbine.

It seems as if these problems could be eliminated by executing thereversed Carnot cycle outside the saturation region. But in this case we havedifficulty in maintaining isothermal conditions during the heat-absorptionand heat-rejection processes. Therefore, we conclude that the reversed Car-not cycle cannot be approximated in actual devices and is not a realisticmodel for refrigeration cycles. However, the reversed Carnot cycle can serveas a standard against which actual refrigeration cycles are compared.

11–3 ■ THE IDEAL VAPOR-COMPRESSIONREFRIGERATION CYCLE

Many of the impracticalities associated with the reversed Carnot cycle canbe eliminated by vaporizing the refrigerant completely before it is com-pressed and by replacing the turbine with a throttling device, such as anexpansion valve or capillary tube. The cycle that results is called the idealvapor-compression refrigeration cycle, and it is shown schematically andon a T-s diagram in Fig. 11–3. The vapor-compression refrigeration cycle isthe most widely used cycle for refrigerators, air-conditioning systems, andheat pumps. It consists of four processes:

1-2 Isentropic compression in a compressor2-3 Constant-pressure heat rejection in a condenser3-4 Throttling in an expansion device4-1 Constant-pressure heat absorption in an evaporator

In an ideal vapor-compression refrigeration cycle, the refrigerant enters thecompressor at state 1 as saturated vapor and is compressed isentropically tothe condenser pressure. The temperature of the refrigerant increases during

COPHP,Carnot �1

1 � TL >TH

COPR,Carnot �1

TH>TL � 1

610 | Thermodynamics

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this isentropic compression process to well above the temperature of the sur-rounding medium. The refrigerant then enters the condenser as superheatedvapor at state 2 and leaves as saturated liquid at state 3 as a result of heatrejection to the surroundings. The temperature of the refrigerant at this stateis still above the temperature of the surroundings.

The saturated liquid refrigerant at state 3 is throttled to the evaporatorpressure by passing it through an expansion valve or capillary tube. Thetemperature of the refrigerant drops below the temperature of the refriger-ated space during this process. The refrigerant enters the evaporator at state4 as a low-quality saturated mixture, and it completely evaporates byabsorbing heat from the refrigerated space. The refrigerant leaves the evapo-rator as saturated vapor and reenters the compressor, completing the cycle.

In a household refrigerator, the tubes in the freezer compartment whereheat is absorbed by the refrigerant serves as the evaporator. The coils behindthe refrigerator, where heat is dissipated to the kitchen air, serve as the con-denser (Fig. 11–4).

Remember that the area under the process curve on a T-s diagram repre-sents the heat transfer for internally reversible processes. The area under theprocess curve 4-1 represents the heat absorbed by the refrigerant in the evapo-rator, and the area under the process curve 2-3 represents the heat rejected inthe condenser. A rule of thumb is that the COP improves by 2 to 4 percent foreach °C the evaporating temperature is raised or the condensing temperatureis lowered.

Chapter 11 | 611

QH

QL4

3

2

1

T

s

4'

Saturated vapor

Saturatedliquid

Compressor

QH

2

Condenser

WARMenvironment

QL

Evaporator

1

COLD refrigeratedspace

WinExpansionvalve

4

3

Win

FIGURE 11–3Schematic and T-s diagram for the ideal vapor-compression refrigeration cycle.

Compressor

Condensercoils

Kitchen air25°C

Capillarytube

Evaporatorcoils

Freezercompartment

–18°C

3°C

QH

QL

FIGURE 11–4An ordinary household refrigerator.

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Another diagram frequently used in the analysis of vapor-compressionrefrigeration cycles is the P-h diagram, as shown in Fig. 11–5. On this dia-gram, three of the four processes appear as straight lines, and the heat trans-fer in the condenser and the evaporator is proportional to the lengths of thecorresponding process curves.

Notice that unlike the ideal cycles discussed before, the ideal vapor-compression refrigeration cycle is not an internally reversible cycle since itinvolves an irreversible (throttling) process. This process is maintained inthe cycle to make it a more realistic model for the actual vapor-compressionrefrigeration cycle. If the throttling device were replaced by an isentropicturbine, the refrigerant would enter the evaporator at state 4� instead of state4. As a result, the refrigeration capacity would increase (by the area underprocess curve 4�-4 in Fig. 11–3) and the net work input would decrease (bythe amount of work output of the turbine). Replacing the expansion valveby a turbine is not practical, however, since the added benefits cannot justifythe added cost and complexity.

All four components associated with the vapor-compression refrigerationcycle are steady-flow devices, and thus all four processes that make up thecycle can be analyzed as steady-flow processes. The kinetic and potentialenergy changes of the refrigerant are usually small relative to the work andheat transfer terms, and therefore they can be neglected. Then the steady-flow energy equation on a unit–mass basis reduces to

(11–6)

The condenser and the evaporator do not involve any work, and the com-pressor can be approximated as adiabatic. Then the COPs of refrigeratorsand heat pumps operating on the vapor-compression refrigeration cycle canbe expressed as

(11–7)

and

(11–8)

where and for the ideal case.Vapor-compression refrigeration dates back to 1834 when the Englishman

Jacob Perkins received a patent for a closed-cycle ice machine using etheror other volatile fluids as refrigerants. A working model of this machine wasbuilt, but it was never produced commercially. In 1850, Alexander Twiningbegan to design and build vapor-compression ice machines using ethylether, which is a commercially used refrigerant in vapor-compression sys-tems. Initially, vapor-compression refrigeration systems were large and weremainly used for ice making, brewing, and cold storage. They lacked auto-matic controls and were steam-engine driven. In the 1890s, electric motor-driven smaller machines equipped with automatic controls started to replacethe older units, and refrigeration systems began to appear in butcher shopsand households. By 1930, the continued improvements made it possible tohave vapor-compression refrigeration systems that were relatively efficient,reliable, small, and inexpensive.

h3 � hf @ P3h1 � hg @ P1

COPHP �qH

wnet,in�

h2 � h3

h2 � h1

COPR �qL

wnet,in�

h1 � h4

h2 � h1

1qin � qout 2 � 1win � wout 2 � he � hi

612 | Thermodynamics

1

h

23

4

P

QH

QL

Win

FIGURE 11–5The P-h diagram of an ideal vapor-compression refrigeration cycle.

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Chapter 11 | 613

EXAMPLE 11–1 The Ideal Vapor-Compression RefrigerationCycle

A refrigerator uses refrigerant-134a as the working fluid and operates on anideal vapor-compression refrigeration cycle between 0.14 and 0.8 MPa. If themass flow rate of the refrigerant is 0.05 kg/s, determine (a) the rate of heatremoval from the refrigerated space and the power input to the compressor,(b) the rate of heat rejection to the environment, and (c) the COP of therefrigerator.

Solution A refrigerator operates on an ideal vapor-compression refrigerationcycle between two specified pressure limits. The rate of refrigeration, thepower input, the rate of heat rejection, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–6.We note that this is an ideal vapor-compression refrigeration cycle, and thusthe compressor is isentropic and the refrigerant leaves the condenser as asaturated liquid and enters the compressor as saturated vapor. From therefrigerant-134a tables, the enthalpies of the refrigerant at all four states aredetermined as follows:

(a) The rate of heat removal from the refrigerated space and the power inputto the compressor are determined from their definitions:

and

(b) The rate of heat rejection from the refrigerant to the environment is

It could also be determined from

(c) The coefficient of performance of the refrigerator is

That is, this refrigerator removes about 4 units of thermal energy from therefrigerated space for each unit of electric energy it consumes.Discussion It would be interesting to see what happens if the throttling valvewere replaced by an isentropic turbine. The enthalpy at state 4s (the turbineexit with P4s � 0.14 MPa, and s4s � s3 � 0.35404 kJ/kg · K) is 88.94 kJ/kg,

COPR �Q#

L

W#

in

�7.18 kW

1.81 kW� 3.97

Q#

H � Q#

L � W#

in � 7.18 � 1.81 � 8.99 kW

Q#

H � m# 1h2 � h3 2 � 10.05 kg>s 2 3 1275.39 � 95.47 2 kJ>kg 4 � 9.0 kW

W#

in � m# 1h2 � h1 2 � 10.05 kg>s 2 3 1275.39 � 239.16 2 kJ>kg 4 � 1.81 kW

Q#

L � m# 1h1 � h4 2 � 10.05 kg>s 2 3 1239.16 � 95.47 2 kJ>kg 4 � 7.18 kW

h4 � h3 1throttling 2 ¡ h4 � 95.47 kJ>kg

P3 � 0.8 MPa ¡ h3 � hf @ 0.8 MPa � 95.47 kJ>kg

P2 � 0.8 MPa

s2 � s1f h2 � 275.39 kJ>kg

s1 � sg @ 0.14 MPa � 0.94456 kJ>kg # K

P1 � 0.14 MPa ¡ h1 � hg @ 0.14 MPa � 239.16 kJ>kg

T

s

QH

41

4s

3

2

0.14 MPa

0.8 MPaWin

QL

FIGURE 11–6T-s diagram of the ideal vapor-compression refrigeration cycledescribed in Example 11–1.

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11–4 ■ ACTUAL VAPOR-COMPRESSIONREFRIGERATION CYCLE

An actual vapor-compression refrigeration cycle differs from the ideal onein several ways, owing mostly to the irreversibilities that occur in variouscomponents. Two common sources of irreversibilities are fluid friction(causes pressure drops) and heat transfer to or from the surroundings. TheT-s diagram of an actual vapor-compression refrigeration cycle is shown inFig. 11–7.

In the ideal cycle, the refrigerant leaves the evaporator and enters thecompressor as saturated vapor. In practice, however, it may not be possibleto control the state of the refrigerant so precisely. Instead, it is easier todesign the system so that the refrigerant is slightly superheated at the com-pressor inlet. This slight overdesign ensures that the refrigerant is com-pletely vaporized when it enters the compressor. Also, the line connecting

614 | Thermodynamics

and the turbine would produce 0.33 kW of power. This would decrease thepower input to the refrigerator from 1.81 to 1.48 kW and increase the rate ofheat removal from the refrigerated space from 7.18 to 7.51 kW. As a result,the COP of the refrigerator would increase from 3.97 to 5.07, an increase of28 percent.

45

2

1

T

s

6 7 8

3

2'

4 3

7 8

Compressor

QH

2

Condenser

WARMenvironment

QL

Evaporator

1

COLD refrigeratedspace

WinExpansionvalve

6

5

FIGURE 11–7Schematic and T-s diagram for the actual vapor-compression refrigeration cycle.

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the evaporator to the compressor is usually very long; thus the pressure dropcaused by fluid friction and heat transfer from the surroundings to therefrigerant can be very significant. The result of superheating, heat gain inthe connecting line, and pressure drops in the evaporator and the connectingline is an increase in the specific volume, thus an increase in the powerinput requirements to the compressor since steady-flow work is proportionalto the specific volume.

The compression process in the ideal cycle is internally reversible andadiabatic, and thus isentropic. The actual compression process, however,involves frictional effects, which increase the entropy, and heat transfer,which may increase or decrease the entropy, depending on the direction.Therefore, the entropy of the refrigerant may increase (process 1-2) ordecrease (process 1-2�) during an actual compression process, depending onwhich effects dominate. The compression process 1-2� may be even moredesirable than the isentropic compression process since the specific volumeof the refrigerant and thus the work input requirement are smaller in thiscase. Therefore, the refrigerant should be cooled during the compressionprocess whenever it is practical and economical to do so.

In the ideal case, the refrigerant is assumed to leave the condenser as sat-urated liquid at the compressor exit pressure. In reality, however, it isunavoidable to have some pressure drop in the condenser as well as in thelines connecting the condenser to the compressor and to the throttling valve.Also, it is not easy to execute the condensation process with such precisionthat the refrigerant is a saturated liquid at the end, and it is undesirable toroute the refrigerant to the throttling valve before the refrigerant is com-pletely condensed. Therefore, the refrigerant is subcooled somewhat beforeit enters the throttling valve. We do not mind this at all, however, since therefrigerant in this case enters the evaporator with a lower enthalpy and thuscan absorb more heat from the refrigerated space. The throttling valve andthe evaporator are usually located very close to each other, so the pressuredrop in the connecting line is small.

Chapter 11 | 615

EXAMPLE 11–2 The Actual Vapor-Compression Refrigeration Cycle

Refrigerant-134a enters the compressor of a refrigerator as superheated vaporat 0.14 MPa and �10°C at a rate of 0.05 kg/s and leaves at 0.8 MPa and50°C. The refrigerant is cooled in the condenser to 26°C and 0.72 MPa andis throttled to 0.15 MPa. Disregarding any heat transfer and pressure dropsin the connecting lines between the components, determine (a) the rate ofheat removal from the refrigerated space and the power input to the com-pressor, (b) the isentropic efficiency of the compressor, and (c) the coeffi-cient of performance of the refrigerator.

Solution A refrigerator operating on a vapor-compression cycle is consid-ered. The rate of refrigeration, the power input, the compressor efficiency,and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible.

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11–5 ■ SELECTING THE RIGHT REFRIGERANTWhen designing a refrigeration system, there are several refrigerants fromwhich to choose, such as chlorofluorocarbons (CFCs), ammonia, hydrocarbons(propane, ethane, ethylene, etc.), carbon dioxide, air (in the air-conditioning ofaircraft), and even water (in applications above the freezing point). The right

616 | Thermodynamics

Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–8.We note that the refrigerant leaves the condenser as a compressed liquidand enters the compressor as superheated vapor. The enthalpies of therefrigerant at various states are determined from the refrigerant tables to be

h1 � 246.36 kJ/kg

h2 � 286.69 kJ/kg

h3 � h f @ 26°C � 87.83 kJ/kg

h4 � h3 (throttling) ⎯→ h4 � 87.83 kJ/kg

(a) The rate of heat removal from the refrigerated space and the power inputto the compressor are determined from their definitions:

and

(b) The isentropic efficiency of the compressor is determined from

where the enthalpy at state 2s (P2s � 0.8 MPa and s2s � s1 � 0.9724kJ/kg · K) is 284.21 kJ/kg. Thus,

(c) The coefficient of performance of the refrigerator is

Discussion This problem is identical to the one worked out in Example11–1, except that the refrigerant is slightly superheated at the compressorinlet and subcooled at the condenser exit. Also, the compressor is not isen-tropic. As a result, the heat removal rate from the refrigerated spaceincreases (by 10.4 percent), but the power input to the compressor increaseseven more (by 11.6 percent). Consequently, the COP of the refrigeratordecreases from 3.97 to 3.93.

COPR �Q#

L

W#

in

�7.93 kW

2.02 kW� 3.93

hC �284.21 � 246.36

286.69 � 246.36� 0.939 or 93.9%

hC �h2s � h1

h2 � h1

W#

in � m# 1h2 � h1 2 � 10.05 kg>s 2 3 1286.69 � 246.36 2 kJ>kg 4 � 2.02 kW

Q#

L � m# 1h1 � h4 2 � 10.05 kg>s 2 3 1246.36 � 87.83 2 kJ>kg 4 � 7.93 kW

P3 �T3 �

0.72 MPa26°C f

P2 �T2 �

0.8 MPa50°C f

P1 �T1 �

0.14 MPa�10°C f

T

s

3

0.72 MPa26°C

4

QH

Win

QL

0.15 MPa

20.8 MPa50°C2s

10.14 MPa–10°C

FIGURE 11–8T-s diagram for Example 11–2.

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choice of refrigerant depends on the situation at hand. Of these, refrigerantssuch as R-11, R-12, R-22, R-134a, and R-502 account for over 90 percent ofthe market in the United States.

Ethyl ether was the first commercially used refrigerant in vapor-compressionsystems in 1850, followed by ammonia, carbon dioxide, methyl chloride,sulphur dioxide, butane, ethane, propane, isobutane, gasoline, and chlorofluo-rocarbons, among others.

The industrial and heavy-commercial sectors were very satisfied withammonia, and still are, although ammonia is toxic. The advantages ofammonia over other refrigerants are its low cost, higher COPs (and thuslower energy cost), more favorable thermodynamic and transport propertiesand thus higher heat transfer coefficients (requires smaller and lower-costheat exchangers), greater detectability in the event of a leak, and no effecton the ozone layer. The major drawback of ammonia is its toxicity, whichmakes it unsuitable for domestic use. Ammonia is predominantly used infood refrigeration facilities such as the cooling of fresh fruits, vegetables,meat, and fish; refrigeration of beverages and dairy products such as beer,wine, milk, and cheese; freezing of ice cream and other foods; ice produc-tion; and low-temperature refrigeration in the pharmaceutical and otherprocess industries.

It is remarkable that the early refrigerants used in the light-commercial andhousehold sectors such as sulfur dioxide, ethyl chloride, and methyl chloridewere highly toxic. The widespread publicity of a few instances of leaks thatresulted in serious illnesses and death in the 1920s caused a public cry to banor limit the use of these refrigerants, creating a need for the development of asafe refrigerant for household use. At the request of Frigidaire Corporation,General Motors’ research laboratory developed R-21, the first member of theCFC family of refrigerants, within three days in 1928. Of several CFCs devel-oped, the research team settled on R-12 as the refrigerant most suitable forcommercial use and gave the CFC family the trade name “Freon.” Commercialproduction of R-11 and R-12 was started in 1931 by a company jointly formedby General Motors and E. I. du Pont de Nemours and Co., Inc. The versatilityand low cost of CFCs made them the refrigerants of choice. CFCs werealso widely used in aerosols, foam insulations, and the electronic industry assolvents to clean computer chips.

R-11 is used primarily in large-capacity water chillers serving air-conditioning systems in buildings. R-12 is used in domestic refrigeratorsand freezers, as well as automotive air conditioners. R-22 is used in windowair conditioners, heat pumps, air conditioners of commercial buildings, andlarge industrial refrigeration systems, and offers strong competition toammonia. R-502 (a blend of R-115 and R-22) is the dominant refrigerantused in commercial refrigeration systems such as those in supermarketsbecause it allows low temperatures at evaporators while operating at single-stage compression.

The ozone crisis has caused a major stir in the refrigeration and air-conditioning industry and has triggered a critical look at the refrigerants inuse. It was realized in the mid-1970s that CFCs allow more ultraviolet radi-ation into the earth’s atmosphere by destroying the protective ozone layerand thus contributing to the greenhouse effect that causes global warming.As a result, the use of some CFCs is banned by international treaties. Fully

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halogenated CFCs (such as R-11, R-12, and R-115) do the most damage tothe ozone layer. The nonfully halogenated refrigerants such as R-22 haveabout 5 percent of the ozone-depleting capability of R-12. Refrigerants thatare friendly to the ozone layer that protects the earth from harmful ultravioletrays have been developed. The once popular refrigerant R-12 has largelybeen replaced by the recently developed chlorine-free R-134a.

Two important parameters that need to be considered in the selection of arefrigerant are the temperatures of the two media (the refrigerated space andthe environment) with which the refrigerant exchanges heat.

To have heat transfer at a reasonable rate, a temperature difference of 5 to10°C should be maintained between the refrigerant and the medium withwhich it is exchanging heat. If a refrigerated space is to be maintained at�10°C, for example, the temperature of the refrigerant should remain atabout �20°C while it absorbs heat in the evaporator. The lowest pressure in arefrigeration cycle occurs in the evaporator, and this pressure should be aboveatmospheric pressure to prevent any air leakage into the refrigeration system.Therefore, a refrigerant should have a saturation pressure of 1 atm or higher at�20°C in this particular case. Ammonia and R-134a are two such substances.

The temperature (and thus the pressure) of the refrigerant on the con-denser side depends on the medium to which heat is rejected. Lower tem-peratures in the condenser (thus higher COPs) can be maintained if therefrigerant is cooled by liquid water instead of air. The use of water coolingcannot be justified economically, however, except in large industrial refrig-eration systems. The temperature of the refrigerant in the condenser cannotfall below the temperature of the cooling medium (about 20°C for a house-hold refrigerator), and the saturation pressure of the refrigerant at this tem-perature should be well below its critical pressure if the heat rejectionprocess is to be approximately isothermal. If no single refrigerant can meetthe temperature requirements, then two or more refrigeration cycles withdifferent refrigerants can be used in series. Such a refrigeration system iscalled a cascade system and is discussed later in this chapter.

Other desirable characteristics of a refrigerant include being nontoxic,noncorrosive, nonflammable, and chemically stable; having a high enthalpyof vaporization (minimizes the mass flow rate); and, of course, being avail-able at low cost.

In the case of heat pumps, the minimum temperature (and pressure) forthe refrigerant may be considerably higher since heat is usually extractedfrom media that are well above the temperatures encountered in refrigera-tion systems.

11–6 ■ HEAT PUMP SYSTEMSHeat pumps are generally more expensive to purchase and install than otherheating systems, but they save money in the long run in some areas becausethey lower the heating bills. Despite their relatively higher initial costs, thepopularity of heat pumps is increasing. About one-third of all single-familyhomes built in the United States in the last decade are heated by heat pumps.

The most common energy source for heat pumps is atmospheric air (air-to-air systems), although water and soil are also used. The major problemwith air-source systems is frosting, which occurs in humid climates whenthe temperature falls below 2 to 5°C. The frost accumulation on the evapo-

618 | Thermodynamics

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rator coils is highly undesirable since it seriously disrupts heat transfer. Thecoils can be defrosted, however, by reversing the heat pump cycle (runningit as an air conditioner). This results in a reduction in the efficiency of thesystem. Water-source systems usually use well water from depths of up to80 m in the temperature range of 5 to 18°C, and they do not have a frostingproblem. They typically have higher COPs but are more complex andrequire easy access to a large body of water such as underground water.Ground-source systems are also rather involved since they require long tub-ing placed deep in the ground where the soil temperature is relatively con-stant. The COP of heat pumps usually ranges between 1.5 and 4, dependingon the particular system used and the temperature of the source. A new classof recently developed heat pumps that use variable-speed electric motordrives are at least twice as energy efficient as their predecessors.

Both the capacity and the efficiency of a heat pump fall significantly atlow temperatures. Therefore, most air-source heat pumps require a supple-mentary heating system such as electric resistance heaters or an oil or gasfurnace. Since water and soil temperatures do not fluctuate much, supple-mentary heating may not be required for water-source or ground-source sys-tems. However, the heat pump system must be large enough to meet themaximum heating load.

Heat pumps and air conditioners have the same mechanical components.Therefore, it is not economical to have two separate systems to meet theheating and cooling requirements of a building. One system can be used asa heat pump in winter and an air conditioner in summer. This is accom-plished by adding a reversing valve to the cycle, as shown in Fig. 11–9. As

Chapter 11 | 619

HEAT PUMP OPERATION—COOLING MODE

Outdoor coil Reversing valve

Indoor coil

Fan

Fan

CompressorExpansionvalve

HEAT PUMP OPERATION—HEATING MODE

Outdoor coil Reversing valve

Indoor coil

Fan

Fan

CompressorExpansionvalve

High-pressure liquidLow-pressure liquid–vaporLow-pressure vaporHigh-pressure vapor

FIGURE 11–9A heat pump can be used to heat ahouse in winter and to cool it insummer.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 619

a result of this modification, the condenser of the heat pump (locatedindoors) functions as the evaporator of the air conditioner in summer. Also,the evaporator of the heat pump (located outdoors) serves as the condenserof the air conditioner. This feature increases the competitiveness of the heatpump. Such dual-purpose units are commonly used in motels.

Heat pumps are most competitive in areas that have a large cooling loadduring the cooling season and a relatively small heating load during theheating season, such as in the southern parts of the United States. In theseareas, the heat pump can meet the entire cooling and heating needs of resi-dential or commercial buildings. The heat pump is least competitive in areaswhere the heating load is very large and the cooling load is small, such as inthe northern parts of the United States.

11–7 ■ INNOVATIVE VAPOR-COMPRESSIONREFRIGERATION SYSTEMS

The simple vapor-compression refrigeration cycle discussed above is themost widely used refrigeration cycle, and it is adequate for most refrigera-tion applications. The ordinary vapor-compression refrigeration systems aresimple, inexpensive, reliable, and practically maintenance-free (when wasthe last time you serviced your household refrigerator?). However, for largeindustrial applications efficiency, not simplicity, is the major concern. Also,for some applications the simple vapor-compression refrigeration cycle isinadequate and needs to be modified. We now discuss a few such modifica-tions and refinements.

Cascade Refrigeration SystemsSome industrial applications require moderately low temperatures, and thetemperature range they involve may be too large for a single vapor-compression refrigeration cycle to be practical. A large temperature rangealso means a large pressure range in the cycle and a poor performance for areciprocating compressor. One way of dealing with such situations is to per-form the refrigeration process in stages, that is, to have two or more refrig-eration cycles that operate in series. Such refrigeration cycles are calledcascade refrigeration cycles.

A two-stage cascade refrigeration cycle is shown in Fig. 11–10. The twocycles are connected through the heat exchanger in the middle, which servesas the evaporator for the topping cycle (cycle A) and the condenser for thebottoming cycle (cycle B). Assuming the heat exchanger is well insulatedand the kinetic and potential energies are negligible, the heat transfer fromthe fluid in the bottoming cycle should be equal to the heat transfer to thefluid in the topping cycle. Thus, the ratio of mass flow rates through eachcycle should be

(11–9)

Also,

(11–10)COPR,cascade �Q#

L

W#

net,in

�m#

B 1h1 � h4 2m#

A 1h6 � h5 2 � m#

B 1h2 � h1 2

m#

A 1h5 � h8 2 � m#

B 1h2 � h3 2 ¡ m#

A

m#

B

�h2 � h3

h5 � h8

620 | Thermodynamics

cen84959_ch11.qxd 4/4/05 4:48 PM Page 620

In the cascade system shown in the figure, the refrigerants in both cyclesare assumed to be the same. This is not necessary, however, since there is nomixing taking place in the heat exchanger. Therefore, refrigerants with moredesirable characteristics can be used in each cycle. In this case, there wouldbe a separate saturation dome for each fluid, and the T-s diagram for one ofthe cycles would be different. Also, in actual cascade refrigeration systems,the two cycles would overlap somewhat since a temperature differencebetween the two fluids is needed for any heat transfer to take place.

It is evident from the T-s diagram in Fig. 11–10 that the compressor workdecreases and the amount of heat absorbed from the refrigerated spaceincreases as a result of cascading. Therefore, cascading improves the COPof a refrigeration system. Some refrigeration systems use three or fourstages of cascading.

Chapter 11 | 621

4

5

2

1

T

s

6

7

83

8 5

QH

Condenser

WARMenvironment

QL

Evaporator

Decrease incompressorwork

QH

QLIncrease inrefrigerationcapacity

Compressor

COLD refrigeratedspace

Expansionvalve

7 6

Compressor

Expansionvalve

3 2Condenser

Evaporator

A

B

QL

4

Heat exchanger

A

B

Heat

1

FIGURE 11–10A two-stage cascade refrigeration system with the same refrigerant in both stages.

EXAMPLE 11–3 A Two-Stage Cascade Refrigeration Cycle

Consider a two-stage cascade refrigeration system operating between the pres-sure limits of 0.8 and 0.14 MPa. Each stage operates on an ideal vapor-compression refrigeration cycle with refrigerant-134a as the working fluid. Heatrejection from the lower cycle to the upper cycle takes place in an adiabaticcounterflow heat exchanger where both streams enter at about 0.32 MPa.

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622 | Thermodynamics

(In practice, the working fluid of the lower cycle is at a higher pressure andtemperature in the heat exchanger for effective heat transfer.) If the mass flowrate of the refrigerant through the upper cycle is 0.05 kg/s, determine (a) themass flow rate of the refrigerant through the lower cycle, (b) the rate of heatremoval from the refrigerated space and the power input to the compressor,and (c) the coefficient of performance of this cascade refrigerator.

Solution A cascade refrigeration system operating between the specifiedpressure limits is considered. The mass flow rate of the refrigerant throughthe lower cycle, the rate of refrigeration, the power input, and the COP are tobe determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible. 3 The heat exchanger is adiabatic.Properties The enthalpies of the refrigerant at all eight states are deter-mined from the refrigerant tables and are indicated on the T-s diagram.Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–11.The topping cycle is labeled cycle A and the bottoming one, cycle B. Forboth cycles, the refrigerant leaves the condenser as a saturated liquid andenters the compressor as saturated vapor.

(a) The mass flow rate of the refrigerant through the lower cycle is deter-mined from the steady-flow energy balance on the adiabatic heat exchanger,

(b) The rate of heat removal by a cascade cycle is the rate of heat absorptionin the evaporator of the lowest stage. The power input to a cascade cycle isthe sum of the power inputs to all of the compressors:

W#

in � W#

comp I,in � W#

comp II,in � m#

A 1h6 � h5 2 � m#

B 1h2 � h1 2 Q#

L � m#

B 1h1 � h4 2 � 10.0390 kg>s 2 3 1239.16 � 55.16 2 kJ>kg 4 � 7.18 kW

m#

B � 0.0390 kg/s 10.05 kg>s 2 3 1251.88 � 95.47 2 kJ>kg 4 � m

#B 3 1255.93 � 55.16 2 kJ>kg 4

m#

A 1h5 � h8 2 � m#

B 1h2 � h3 2 E#

out � E#

in ¡ m#

Ah5 � m#

Bh3 � m#

Ah8 � m#

Bh2

4

3

2

1

T

s

6

7

85

h3 = 55.16

h7 = 95.47

h6 = 270.92 kJ/kg

h2 = 255.93

h5 = 251.88

h1 = 239.16

h4 = 55.16

h8 = 95.47

0.8 MPa

0.32 MPa

0.14 MPa

A

B

FIGURE 11–11T-s diagram of the cascaderefrigeration cycle described inExample 11–3.

cen84959_ch11.qxd 4/5/05 12:42 PM Page 622

Multistage Compression Refrigeration SystemsWhen the fluid used throughout the cascade refrigeration system is the same,the heat exchanger between the stages can be replaced by a mixing chamber(called a flash chamber) since it has better heat transfer characteristics. Suchsystems are called multistage compression refrigeration systems. A two-stage compression refrigeration system is shown in Fig. 11–12.

Chapter 11 | 623

8

5 2

1

T

s

76 3

9

QH

Condenser

WARMenvironment

High-pressurecompressor

COLD

refrigerated space

Expansionvalve

5 4

Expansionvalve

3

Evaporator

QL

4

9

2

1

6

7

8

Flashchamber

Low-pressurecompressor

FIGURE 11–12A two-stage compression refrigeration system with a flash chamber.

(c) The COP of a refrigeration system is the ratio of the refrigeration rate tothe net power input:

Discussion This problem was worked out in Example 11–1 for a single-stagerefrigeration system. Notice that the COP of the refrigeration systemincreases from 3.97 to 4.46 as a result of cascading. The COP of the systemcan be increased even more by increasing the number of cascade stages.

COPR �Q#

L

W#

net,in

�7.18 kW

1.61 kW� 4.46

� 1.61 kW

� 10.039 kg>s 2 3 1255.93 � 239.16 2 kJ>kg 4 � 10.05 kg>s 2 3 1270.92 � 251.88 2 kJ>kg 4

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624 | Thermodynamics

EXAMPLE 11–4 A Two-Stage Refrigeration Cycle with a Flash Chamber

Consider a two-stage compression refrigeration system operating between thepressure limits of 0.8 and 0.14 MPa. The working fluid is refrigerant-134a.The refrigerant leaves the condenser as a saturated liquid and is throttled toa flash chamber operating at 0.32 MPa. Part of the refrigerant evaporatesduring this flashing process, and this vapor is mixed with the refrigerantleaving the low-pressure compressor. The mixture is then compressed to thecondenser pressure by the high-pressure compressor. The liquid in the flashchamber is throttled to the evaporator pressure and cools the refrigeratedspace as it vaporizes in the evaporator. Assuming the refrigerant leaves theevaporator as a saturated vapor and both compressors are isentropic, deter-mine (a) the fraction of the refrigerant that evaporates as it is throttled tothe flash chamber, (b) the amount of heat removed from the refrigeratedspace and the compressor work per unit mass of refrigerant flowing throughthe condenser, and (c) the coefficient of performance.

Solution A two-stage compression refrigeration system operating betweenspecified pressure limits is considered. The fraction of the refrigerant thatevaporates in the flash chamber, the refrigeration and work input per unitmass, and the COP are to be determined.Assumptions 1 Steady operating conditions exist. 2 Kinetic and potentialenergy changes are negligible. 3 The flash chamber is adiabatic.Properties The enthalpies of the refrigerant at various states are determinedfrom the refrigerant tables and are indicated on the T-s diagram.Analysis The T-s diagram of the refrigeration cycle is shown in Fig. 11–13.We note that the refrigerant leaves the condenser as saturated liquid andenters the low-pressure compressor as saturated vapor.

(a) The fraction of the refrigerant that evaporates as it is throttled to theflash chamber is simply the quality at state 6, which is

(b) The amount of heat removed from the refrigerated space and the compres-sor work input per unit mass of refrigerant flowing through the condenser are

� 11 � 0.2049 2 3 1239.16 � 55.16 2 kJ>kg 4 � 146.3 kJ/kg

qL � 11 � x6 2 1h1 � h8 2

x6 �h6 � hf

hfg

�95.47 � 55.16

196.71� 0.2049

In this system, the liquid refrigerant expands in the first expansion valveto the flash chamber pressure, which is the same as the compressor inter-stage pressure. Part of the liquid vaporizes during this process. This satu-rated vapor (state 3) is mixed with the superheated vapor from thelow-pressure compressor (state 2), and the mixture enters the high-pressurecompressor at state 9. This is, in essence, a regeneration process. The satu-rated liquid (state 7) expands through the second expansion valve into theevaporator, where it picks up heat from the refrigerated space.

The compression process in this system resembles a two-stage compres-sion with intercooling, and the compressor work decreases. Care should beexercised in the interpretations of the areas on the T-s diagram in this casesince the mass flow rates are different in different parts of the cycle.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 624

Chapter 11 | 625

and

The enthalpy at state 9 is determined from an energy balance on the mixingchamber,

Also, s9 � 0.9416 kJ/kg · K. Thus the enthalpy at state 4 (0.8 MPa, s4 �s9) is h4 � 274.48 kJ/kg. Substituting,

(c) The coefficient of performance is

Discussion This problem was worked out in Example 11–1 for a single-stagerefrigeration system (COP � 3.97) and in Example 11–3 for a two-stage cas-cade refrigeration system (COP � 4.46). Notice that the COP of the refriger-ation system increased considerably relative to the single-stage compressionbut did not change much relative to the two-stage cascade compression.

COPR �qL

win�

146.3 kJ>kg

32.71 kJ>kg� 4.47

� 32.71 kJ/kg

win � 11 � 0.2049 2 3 1255.93 � 239.16 2 kJ>kg 4 � 1274.48 � 255.10 2 kJ>kg

h9 � 10.2049 2 1251.88 2 � 11 � 0.2049 2 1255.93 2 � 255.10 kJ>kg

11 2h9 � x6h3 � 11 � x6 2h2

E#out � E

#in

win � wcomp I,in � wcomp II,in � 11 � x6 2 1h2 � h1 2 � 11 2 1h4 � h9 2

8

7

2

1

T

s

4

5

6

9h7 = 55.16 h6 = 95.47

h4 = 274.48 kJ/kg

h2 = 255.93

h9 = 255.10

h1 = 239.16

h8 = 55.16

h3 = 251.883

h5 = 95.47

FIGURE 11–13T-s diagram of the two-stagecompression refrigeration cycledescribed in Example 11–4.

Multipurpose Refrigeration Systems with a Single CompressorSome applications require refrigeration at more than one temperature. Thiscould be accomplished by using a separate throttling valve and a separatecompressor for each evaporator operating at different temperatures. However,such a system is bulky and probably uneconomical. A more practical and

cen84959_ch11.qxd 4/4/05 4:48 PM Page 625

economical approach would be to route all the exit streams from the evapora-tors to a single compressor and let it handle the compression process for theentire system.

Consider, for example, an ordinary refrigerator–freezer unit. A simplifiedschematic of the unit and the T-s diagram of the cycle are shown inFig. 11–14. Most refrigerated goods have a high water content, and therefrigerated space must be maintained above the ice point to prevent freez-ing. The freezer compartment, however, is maintained at about �18°C.Therefore, the refrigerant should enter the freezer at about �25°C to haveheat transfer at a reasonable rate in the freezer. If a single expansion valveand evaporator were used, the refrigerant would have to circulate in bothcompartments at about �25°C, which would cause ice formation in theneighborhood of the evaporator coils and dehydration of the produce. Thisproblem can be eliminated by throttling the refrigerant to a higher pressure(hence temperature) for use in the refrigerated space and then throttling it tothe minimum pressure for use in the freezer. The entire refrigerant leavingthe freezer compartment is subsequently compressed by a single compressorto the condenser pressure.

Liquefaction of GasesThe liquefaction of gases has always been an important area of refrigerationsince many important scientific and engineering processes at cryogenic tem-peratures (temperatures below about �100°C) depend on liquefied gases.Some examples of such processes are the separation of oxygen and nitrogenfrom air, preparation of liquid propellants for rockets, the study of materialproperties at low temperatures, and the study of some exciting phenomenasuch as superconductivity.

626 | Thermodynamics

QH

QL,F

4

3

2

1

T

s

Compressor

QH

2

Kitchen air

Condenser

QL,F

Freezer

Expansionvalve

4

A 6

QL,R

5

Expansionvalve

QL,R1

3

6

Refrigerator

(Alternative path)A

5

FIGURE 11–14Schematic and T-s diagram for a refrigerator–freezer unit with one compressor.

cen84959_ch11.qxd 4/5/05 12:42 PM Page 626

At temperatures above the critical-point value, a substance exists in thegas phase only. The critical temperatures of helium, hydrogen, and nitrogen(three commonly used liquefied gases) are �268, �240, and �147°C,respectively. Therefore, none of these substances exist in liquid form atatmospheric conditions. Furthermore, low temperatures of this magnitudecannot be obtained by ordinary refrigeration techniques. Then the questionthat needs to be answered in the liquefaction of gases is this: How can welower the temperature of a gas below its critical-point value?

Several cycles, some complex and others simple, are used successfully forthe liquefaction of gases. Below we discuss the Linde-Hampson cycle,which is shown schematically and on a T-s diagram in Fig. 11–15.

Makeup gas is mixed with the uncondensed portion of the gas from theprevious cycle, and the mixture at state 2 is compressed by a multistagecompressor to state 3. The compression process approaches an isothermalprocess due to intercooling. The high-pressure gas is cooled in an after-cooler by a cooling medium or by a separate external refrigeration system tostate 4. The gas is further cooled in a regenerative counter-flow heatexchanger by the uncondensed portion of gas from the previous cycle tostate 5, and it is throttled to state 6, which is a saturated liquid–vapor mix-ture state. The liquid (state 7) is collected as the desired product, and thevapor (state 8) is routed through the regenerator to cool the high-pressuregas approaching the throttling valve. Finally, the gas is mixed with freshmakeup gas, and the cycle is repeated.

Chapter 11 | 627

4

52

1

T

s

78

3

Multistagecompressor

Q

6

9

4

6

5

8

7

Heatexchanger 3

2

Liquid removed

Vaporrecirculated

Makeup gas

Regenerator

1

9

FIGURE 11–15Linde-Hampson system for liquefying gases.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 627

This and other refrigeration cycles used for the liquefaction of gases canalso be used for the solidification of gases.

11–8 ■ GAS REFRIGERATION CYCLESAs explained in Sec. 11–2, the Carnot cycle (the standard of comparison forpower cycles) and the reversed Carnot cycle (the standard of comparisonfor refrigeration cycles) are identical, except that the reversed Carnot cycleoperates in the reverse direction. This suggests that the power cycles dis-cussed in earlier chapters can be used as refrigeration cycles by simplyreversing them. In fact, the vapor-compression refrigeration cycle is essen-tially a modified Rankine cycle operating in reverse. Another example is thereversed Stirling cycle, which is the cycle on which Stirling refrigeratorsoperate. In this section, we discuss the reversed Brayton cycle, better knownas the gas refrigeration cycle.

Consider the gas refrigeration cycle shown in Fig. 11–16. The surround-ings are at T0, and the refrigerated space is to be maintained at TL. The gasis compressed during process 1-2. The high-pressure, high-temperature gasat state 2 is then cooled at constant pressure to T0 by rejecting heat to thesurroundings. This is followed by an expansion process in a turbine, duringwhich the gas temperature drops to T4. (Can we achieve the cooling effectby using a throttling valve instead of a turbine?) Finally, the cool gasabsorbs heat from the refrigerated space until its temperature rises to T1.

628 | Thermodynamics

4

WARMenvironment

COLDrefrigerated space

QH

Heatexchanger

QL

3

2

1

T

s

4

QH

QL

3 2

Compressor

Wnet,in

Heatexchanger

Turbine

1

FIGURE 11–16Simple gas refrigeration cycle.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 628

All the processes described are internally reversible, and the cycle exe-cuted is the ideal gas refrigeration cycle. In actual gas refrigeration cycles,the compression and expansion processes deviate from the isentropic ones,and T3 is higher than T0 unless the heat exchanger is infinitely large.

On a T-s diagram, the area under process curve 4-1 represents the heatremoved from the refrigerated space, and the enclosed area 1-2-3-4-1 repre-sents the net work input. The ratio of these areas is the COP for the cycle,which may be expressed as

(11–11)

where

The gas refrigeration cycle deviates from the reversed Carnot cyclebecause the heat transfer processes are not isothermal. In fact, the gas tem-perature varies considerably during heat transfer processes. Consequently, thegas refrigeration cycles have lower COPs relative to the vapor-compressionrefrigeration cycles or the reversed Carnot cycle. This is also evident fromthe T-s diagram in Fig. 11–17. The reversed Carnot cycle consumes a frac-tion of the net work (rectangular area 1A3B) but produces a greater amountof refrigeration (triangular area under B1).

Despite their relatively low COPs, the gas refrigeration cycles have twodesirable characteristics: They involve simple, lighter components, whichmake them suitable for aircraft cooling, and they can incorporate regenera-tion, which makes them suitable for liquefaction of gases and cryogenicapplications. An open-cycle aircraft cooling system is shown in Fig. 11–18.Atmospheric air is compressed by a compressor, cooled by the surroundingair, and expanded in a turbine. The cool air leaving the turbine is thendirectly routed to the cabin.

wcomp,in � h2 � h1

wturb,out � h3 � h4

qL � h1 � h4

COPR �qL

wnet,in�

qL

wcomp,in � wturb,out

Chapter 11 | 629

3

2

1

T

s

4

Gasrefrigerationcycle

A

B

ReversedCarnotcycle

FIGURE 11–17A reserved Carnot cycle producesmore refrigeration (area under B1)with less work input (area 1A3B).

3

Compressor

Wnet,in

Heatexchanger

Cool air out

Warm airin

4

2

1

Turbine

Q

FIGURE 11–18An open-cycle aircraft cooling system.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 629

The regenerative gas cycle is shown in Fig. 11–19. Regenerative coolingis achieved by inserting a counter-flow heat exchanger into the cycle. With-out regeneration, the lowest turbine inlet temperature is T0, the temperatureof the surroundings or any other cooling medium. With regeneration, thehigh-pressure gas is further cooled to T4 before expanding in the turbine.Lowering the turbine inlet temperature automatically lowers the turbine exittemperature, which is the minimum temperature in the cycle. Extremely lowtemperatures can be achieved by repeating this process.

630 | Thermodynamics

1

45 WARMenvironment

Compressor

Wnet,in

Turbine

QHeat

exchanger

Heatexchanger

2

3

6

Regenerator

4

2

1

T

s

5

QH

QL

3

6

COLDrefrigerated space

FIGURE 11–19Gas refrigeration cycle with regeneration.

EXAMPLE 11–5 The Simple Ideal Gas Refrigeration Cycle

An ideal gas refrigeration cycle using air as the working medium is to maintaina refrigerated space at 0°F while rejecting heat to the surrounding medium at80°F. The pressure ratio of the compressor is 4. Determine (a) the maximumand minimum temperatures in the cycle, (b) the coefficient of performance,and (c) the rate of refrigeration for a mass flow rate of 0.1 lbm/s.

Solution An ideal gas refrigeration cycle using air as the working fluid isconsidered. The maximum and minimum temperatures, the COP, and therate of refrigeration are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air is an ideal gas withvariable specific heats. 3 Kinetic and potential energy changes are negligible.Analysis The T-s diagram of the gas refrigeration cycle is shown inFig. 11–20. We note that this is an ideal gas-compression refrigerationcycle, and thus, both the compressor and the turbine are isentropic, and theair is cooled to the environment temperature before it enters the turbine.

3

2

1

T, °F

s

4

QH

QL

Tmax

Tmin

800

·

·

FIGURE 11–20T-s diagram of the ideal-gasrefrigeration cycle described inExample 11–5.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 630

11–9 ■ ABSORPTION REFRIGERATION SYSTEMSAnother form of refrigeration that becomes economically attractive whenthere is a source of inexpensive thermal energy at a temperature of 100 to200°C is absorption refrigeration. Some examples of inexpensive thermalenergy sources include geothermal energy, solar energy, and waste heatfrom cogeneration or process steam plants, and even natural gas when it isavailable at a relatively low price.

As the name implies, absorption refrigeration systems involve the absorp-tion of a refrigerant by a transport medium. The most widely used absorp-tion refrigeration system is the ammonia–water system, where ammonia(NH3) serves as the refrigerant and water (H2O) as the transport medium.Other absorption refrigeration systems include water–lithium bromideand water–lithium chloride systems, where water serves as the refrigerant.The latter two systems are limited to applications such as air-conditioningwhere the minimum temperature is above the freezing point of water.

Chapter 11 | 631

(a) The maximum and minimum temperatures in the cycle are determinedfrom the isentropic relations of ideal gases for the compression and expansionprocesses. From Table A–17E,

T1 � 460 R ⎯→ h1 � 109.90 Btu/lbm and Pr1 � 0.7913

Pr2 � Pr1 � (4)(0.7913) � 3.165 ⎯→

T3 � 540 R ⎯→ h3 � 129.06 Btu/lbm and Pr3 � 1.3860

Pr4 � Pr3 � (0.25)(1.386) � 0.3465 ⎯→

Therefore, the highest and the lowest temperatures in the cycle are 223 and�97°F, respectively.

(b) The COP of this ideal gas refrigeration cycle is

where

Thus,

(c) The rate of refrigeration is

Discussion It is worth noting that an ideal vapor-compression cycle workingunder similar conditions would have a COP greater than 3.

Q#

refrig � m# 1qL 2 � 10.1 lbm>s 2 123.2 Btu>lbm 2 � 2.32 Btu/s

COPR �23.2

53.6 � 42.36� 2.06

Wcomp,in � h2 � h1 � 163.5 � 109.9 � 53.6 Btu>lbm

Wturb,out � h3 � h4 � 129.06 � 86.7 � 42.36 Btu>lbm

qL � h1 � h4 � 109.9 � 86.7 � 23.2 Btu>lbm

COPR �qL

wnet,in�

qL

wcomp,in � Wturb,out

eh4 �T4 �

86.7 Btu/lbm363 R (or �97°F)

P4

P3

eh2 �T2 �

163.5 Btu/lbm683 R (or 223°F)

P2

P1

cen84959_ch11.qxd 4/20/05 1:04 PM Page 631

To understand the basic principles involved in absorption refrigeration, weexamine the NH3–H2O system shown in Fig. 11–21. The ammonia–waterrefrigeration machine was patented by the Frenchman Ferdinand Carre in1859. Within a few years, the machines based on this principle were beingbuilt in the United States primarily to make ice and store food. You willimmediately notice from the figure that this system looks very much likethe vapor-compression system, except that the compressor has beenreplaced by a complex absorption mechanism consisting of an absorber, apump, a generator, a regenerator, a valve, and a rectifier. Once the pressureof NH3 is raised by the components in the box (this is the only thing theyare set up to do), it is cooled and condensed in the condenser by rejectingheat to the surroundings, is throttled to the evaporator pressure, and absorbsheat from the refrigerated space as it flows through the evaporator. So, thereis nothing new there. Here is what happens in the box:

Ammonia vapor leaves the evaporator and enters the absorber, where itdissolves and reacts with water to form NH3 · H2O. This is an exothermicreaction; thus heat is released during this process. The amount of NH3 thatcan be dissolved in H2O is inversely proportional to the temperature. There-fore, it is necessary to cool the absorber to maintain its temperature as lowas possible, hence to maximize the amount of NH3 dissolved in water. Theliquid NH3 � H2O solution, which is rich in NH3, is then pumped to thegenerator. Heat is transferred to the solution from a source to vaporize someof the solution. The vapor, which is rich in NH3, passes through a rectifier,which separates the water and returns it to the generator. The high-pressurepure NH3 vapor then continues its journey through the rest of the cycle. The

632 | Thermodynamics

QH

QL

Wpump

Qgen

Qcool

NH3 + H2O

WARMenvironment

Expansionvalve

Expansionvalve

Pump

Cooling water

Condenser

Evaporator

COLDrefrigerated space

Pure NH3

Pure NH3

Rectifier Generator

H2O

Solarenergy

NH3 + H2O

Absorber

RegeneratorQ

FIGURE 11–21Ammonia absorption refrigerationcycle.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 632

hot NH3 � H2O solution, which is weak in NH3, then passes through aregenerator, where it transfers some heat to the rich solution leaving thepump, and is throttled to the absorber pressure.

Compared with vapor-compression systems, absorption refrigeration sys-tems have one major advantage: A liquid is compressed instead of a vapor.The steady-flow work is proportional to the specific volume, and thus thework input for absorption refrigeration systems is very small (on the orderof one percent of the heat supplied to the generator) and often neglected inthe cycle analysis. The operation of these systems is based on heat transferfrom an external source. Therefore, absorption refrigeration systems areoften classified as heat-driven systems.

The absorption refrigeration systems are much more expensive than thevapor-compression refrigeration systems. They are more complex andoccupy more space, they are much less efficient thus requiring much largercooling towers to reject the waste heat, and they are more difficult to servicesince they are less common. Therefore, absorption refrigeration systemsshould be considered only when the unit cost of thermal energy is low andis projected to remain low relative to electricity. Absorption refrigerationsystems are primarily used in large commercial and industrial installations.

The COP of absorption refrigeration systems is defined as

(11–12)

The maximum COP of an absorption refrigeration system is determined byassuming that the entire cycle is totally reversible (i.e., the cycle involves noirreversibilities and any heat transfer is through a differential temperature dif-ference). The refrigeration system would be reversible if the heat from thesource (Qgen) were transferred to a Carnot heat engine, and the work outputof this heat engine (W � hth,rev Qgen) is supplied to a Carnot refrigerator toremove heat from the refrigerated space. Note that QL � W � COPR,rev �hth,revQgenCOPR,rev. Then the overall COP of an absorption refrigeration sys-tem under reversible conditions becomes (Fig. 11–22)

(11–13)

where TL, T0, and Ts are the thermodynamic temperatures of the refrigeratedspace, the environment, and the heat source, respectively. Any absorptionrefrigeration system that receives heat from a source at Ts and removes heatfrom the refrigerated space at TL while operating in an environment at T0has a lower COP than the one determined from Eq. 11–13. For example,when the source is at 120°C, the refrigerated space is at �10°C, and theenvironment is at 25°C, the maximum COP that an absorption refrigerationsystem can have is 1.8. The COP of actual absorption refrigeration systemsis usually less than 1.

Air-conditioning systems based on absorption refrigeration, calledabsorption chillers, perform best when the heat source can supply heat at ahigh temperature with little temperature drop. The absorption chillers aretypically rated at an input temperature of 116°C (240°F). The chillers per-form at lower temperatures, but their cooling capacity decreases sharply with

COPrev,absorption �QL

Q gen� hth,revCOPR,rev � a1 �

T0

Ts

b a TL

T0 � TL

b

COPabsorption �Desired output

Required input�

QL

Q gen � Wpump,in�

QL

Q gen

Chapter 11 | 633

SourceTs

T0environment

Reversibleheat

engine

Qgen

W = hrev Qgen

QL = COPR,rev × W

EnvironmentT0

TLRefrigerated

space

Reversiblerefrigerator

W = hrev Qgen = (1 – )Qgen

QL = COPR,revW = ( )W

COPrev,absorption = = (1 – )( )

T0Ts

TLT0 – TL

Qgen

QL T0Ts

TLT0 – TL

FIGURE 11–22Determining the maximum COP of anabsorption refrigeration system.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 633

decreasing source temperature, about 12.5 percent for each 6°C (10°F) dropin the source temperature. For example, the capacity goes down to 50 per-cent when the supply water temperature drops to 93°C (200°F). In that case,one needs to double the size (and thus the cost) of the chiller to achieve thesame cooling. The COP of the chiller is affected less by the decline of thesource temperature. The COP drops by 2.5 percent for each 6°C (10°F)drop in the source temperature. The nominal COP of single-stage absorptionchillers at 116°C (240°F) is 0.65 to 0.70. Therefore, for each ton of refriger-ation, a heat input of (12,000 Btu/h)/0.65 � 18,460 Btu/h is required. At 88°C (190°F), the COP drops by 12.5 percent and thus the heat inputincreases by 12.5 percent for the same cooling effect. Therefore, the eco-nomic aspects must be evaluated carefully before any absorption refrigera-tion system is considered, especially when the source temperature is below93°C (200°F).

Another absorption refrigeration system that is quite popular withcampers is a propane-fired system invented by two Swedish undergraduatestudents. In this system, the pump is replaced by a third fluid (hydrogen),which makes it a truly portable unit.

634 | Thermodynamics

All the refrigeration systems discussed above involve many moving parts andbulky, complex components. Then this question comes to mind: Is it reallynecessary for a refrigeration system to be so complex? Can we not achievethe same effect in a more direct way? The answer to this question is yes. It ispossible to use electric energy more directly to produce cooling withoutinvolving any refrigerants and moving parts. Below we discuss one such sys-tem, called thermoelectric refrigerator.

Consider two wires made from different metals joined at both ends (junc-tions), forming a closed circuit. Ordinarily, nothing will happen. However,when one of the ends is heated, something interesting happens: A currentflows continuously in the circuit, as shown in Fig. 11–23. This is called theSeebeck effect, in honor of Thomas Seebeck, who made this discovery in1821. The circuit that incorporates both thermal and electrical effects is calleda thermoelectric circuit, and a device that operates on this circuit is called athermoelectric device.

The Seebeck effect has two major applications: temperature measurementand power generation. When the thermoelectric circuit is broken, as shown inFig. 11–24, the current ceases to flow, and we can measure the driving force(the electromotive force) or the voltage generated in the circuit by a voltmeter.The voltage generated is a function of the temperature difference and thematerials of the two wires used. Therefore, temperature can be measured bysimply measuring voltages. The two wires used to measure the temperature in

TOPIC OF SPECIAL INTEREST* Thermoelectric Power Generation and Refrigeration Systems

*This section can be skipped without a loss in continuity.

Metal A

Metal B

II

FIGURE 11–23When one of the junctions of twodissimilar metals is heated, a current Iflows through the closed circuit.

+ –

Metal A

Metal B

I = 0

V

FIGURE 11–24When a thermoelectric circuit isbroken, a potential difference isgenerated.

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Chapter 11 | 635

this manner form a thermocouple, which is the most versatile and mostwidely used temperature measurement device. A common T-type thermocou-ple, for example, consists of copper and constantan wires, and it producesabout 40 mV per °C difference.

The Seebeck effect also forms the basis for thermoelectric power genera-tion. The schematic diagram of a thermoelectric generator is shown in Fig.11–25. Heat is transferred from a high-temperature source to the hot junctionin the amount of QH, and it is rejected to a low-temperature sink from thecold junction in the amount of QL. The difference between these two quanti-ties is the net electrical work produced, that is, We � QH � QL. It is evidentfrom Fig. 11–25 that the thermoelectric power cycle closely resembles anordinary heat engine cycle, with electrons serving as the working fluid.Therefore, the thermal efficiency of a thermoelectric generator operatingbetween the temperature limits of TH and TL is limited by the efficiency of aCarnot cycle operating between the same temperature limits. Thus, in theabsence of any irreversibilities (such as I2R heating, where R is the totalelectrical resistance of the wires), the thermoelectric generator will have theCarnot efficiency.

The major drawback of thermoelectric generators is their low efficiency.The future success of these devices depends on finding materials with moredesirable characteristics. For example, the voltage output of thermoelectricdevices has been increased several times by switching from metal pairs tosemiconductors. A practical thermoelectric generator using n-type (heavilydoped to create excess electrons) and p-type (heavily doped to create a defi-ciency of electrons) materials connected in series is shown in Fig. 11–26.Despite their low efficiencies, thermoelectric generators have definite weightand reliability advantages and are presently used in rural areas and in spaceapplications. For example, silicon–germanium-based thermoelectric genera-tors have been powering Voyager spacecraft since 1980 and are expected tocontinue generating power for many more years.

If Seebeck had been fluent in thermodynamics, he would probably havetried reversing the direction of flow of electrons in the thermoelectric circuit(by externally applying a potential difference in the reverse direction) to cre-ate a refrigeration effect. But this honor belongs to Jean Charles AthanasePeltier, who discovered this phenomenon in 1834. He noticed during hisexperiments that when a small current was passed through the junction of twodissimilar wires, the junction was cooled, as shown in Fig. 11–27. This iscalled the Peltier effect, and it forms the basis for thermoelectric refrigera-tion. A practical thermoelectric refrigeration circuit using semiconductormaterials is shown in Fig. 11–28. Heat is absorbed from the refrigerated spacein the amount of QL and rejected to the warmer environment in the amount ofQH. The difference between these two quantities is the net electrical work thatneeds to be supplied; that is, We � QH � QL. Thermoelectric refrigeratorspresently cannot compete with vapor-compression refrigeration systemsbecause of their low coefficient of performance. They are available in themarket, however, and are preferred in some applications because of theirsmall size, simplicity, quietness, and reliability.

Wnet

High-temperature sourceTH

Low-temperature sinkTL

QH

QL

I

I

Hot junction

Cold junction

FIGURE 11–25Schematic of a simple thermoelectricpower generator.

SOURCE

SINK

Hot plate

Cold plate

QH

QL

+ –

I

p n p n p n

Wnet

FIGURE 11–26A thermoelectric power generator.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 635

636 | Thermodynamics

EXAMPLE 11–6 Cooling of a Canned Drink by a Thermoelectric Refrigerator

A thermoelectric refrigerator that resembles a small ice chest is powered bya car battery and has a COP of 0.1. If the refrigerator cools a 0.350-Lcanned drink from 20 to 4°C in 30 min, determine the average electricpower consumed by the thermoelectric refrigerator.

Solution A thermoelectric refrigerator with a specified COP is used to coolcanned drinks. The power consumption of the refrigerator is to be determined.Assumptions Heat transfer through the walls of the refrigerator is negligibleduring operation.Properties The properties of canned drinks are the same as those of waterat room temperature, r � 1 kg/L and c � 4.18 kJ/kg · °C (Table A–3).Analysis The cooling rate of the refrigerator is simply the rate of decrease ofthe energy of the canned drinks,

Then the average power consumed by the refrigerator becomes

Discussion In reality, the power consumption will be larger because of theheat gain through the walls of the refrigerator.

W#

in �Q#

cooling

COPR�

13 W

0.10� 130 W

Q#cooling �

Qcooling

¢t�

23.4 kJ

30 � 60 s� 0.0130 kW � 13 W

Q cooling � mc ¢T � 10.350 kg 2 14.18 kJ>kg # °C 2 120 � 4 2°C � 23.4 kJ

m � rV � 11 kg>L 2 10.350 L 2 � 0.350 kg

The transfer of heat from lower temperature regions to highertemperature ones is called refrigeration. Devices that producerefrigeration are called refrigerators, and the cycles on whichthey operate are called refrigeration cycles. The working flu-ids used in refrigerators are called refrigerants. Refrigeratorsused for the purpose of heating a space by transferring heatfrom a cooler medium are called heat pumps.

The performance of refrigerators and heat pumps isexpressed in terms of coefficient of performance (COP),defined as

COPHP �Desired output

Required input�

Heating effect

Work input�

QH

Wnet,in

COPR �Desired output

Required output�

Cooling effect

Work input�

QL

Wnet,in

SUMMARY

The standard of comparison for refrigeration cycles is thereversed Carnot cycle. A refrigerator or heat pump that oper-ates on the reversed Carnot cycle is called a Carnot refrigera-tor or a Carnot heat pump, and their COPs are

The most widely used refrigeration cycle is the vapor-compression refrigeration cycle. In an ideal vapor-compressionrefrigeration cycle, the refrigerant enters the compressor as asaturated vapor and is cooled to the saturated liquid state inthe condenser. It is then throttled to the evaporator pressureand vaporizes as it absorbs heat from the refrigerated space.

COPHP,Carnot �1

1 � TL >TH

COPR,Carnot �1

TH>TL � 1

+–

Heat rejected

Heat absorbed

FIGURE 11–27When a current is passed through thejunction of two dissimilar materials,the junction is cooled.

WARMenvironment

Refrigeratedspace

Hot plate

Cold plate

QH

QL

+–I

p n p n p n

FIGURE 11–28A thermoelectric refrigerator.

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Chapter 11 | 637

Very low temperatures can be achieved by operating two ormore vapor-compression systems in series, called cascading.The COP of a refrigeration system also increases as a resultof cascading. Another way of improving the performance of avapor-compression refrigeration system is by using multi-stage compression with regenerative cooling. A refrigeratorwith a single compressor can provide refrigeration at severaltemperatures by throttling the refrigerant in stages. Thevapor-compression refrigeration cycle can also be used to liq-uefy gases after some modifications.

The power cycles can be used as refrigeration cycles bysimply reversing them. Of these, the reversed Brayton cycle,which is also known as the gas refrigeration cycle, is usedto cool aircraft and to obtain very low (cryogenic) tempera-tures after it is modified with regeneration. The work outputof the turbine can be used to reduce the work input require-ments to the compressor. Thus the COP of a gas refrigerationcycle is

COPabsorption �qL

wnet,in�

qL

wcomp,in � wturb,out

Another form of refrigeration that becomes economicallyattractive when there is a source of inexpensive thermal energyat a temperature of 100 to 200°C is absorption refrigeration,where the refrigerant is absorbed by a transport medium andcompressed in liquid form. The most widely used absorptionrefrigeration system is the ammonia–water system, whereammonia serves as the refrigerant and water as the transportmedium. The work input to the pump is usually very small,and the COP of absorption refrigeration systems is defined as

The maximum COP an absorption refrigeration system canhave is determined by assuming totally reversible conditions,which yields

where T0, TL, and Ts are the thermodynamic temperatures ofthe environment, the refrigerated space, and the heat source,respectively.

COPrev,absorption � hth,rev COPR,rev � a1 �T0

Ts

b a TL

T0 � TL

b

COPabsorption �Desired output

Required input�

QL

Q gen � Wpump,in�

QL

Q gen

1. ASHRAE, Handbook of Fundamentals. Atlanta: AmericanSociety of Heating, Refrigerating, and Air-ConditioningEngineers, 1985.

2. Heat Pump Systems—A Technology Review. OECDReport, Paris, 1982.

3. B. Nagengast. “A Historical Look at CFC Refrigerants.”ASHRAE Journal 30, no. 11 (November 1988), pp. 37–39.

4. W. F. Stoecker. “Growing Opportunities for AmmoniaRefrigeration.” Proceedings of the Meeting of the

REFERENCES AND SUGGESTED READINGS

International Institute of Ammonia Refrigeration, Austin,Texas, 1989.

5. W. F. Stoecker and J. W. Jones. Refrigeration and AirConditioning. 2nd ed. New York: McGraw-Hill, 1982.

6. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

The Reversed Carnot Cycle

11–1C Why is the reversed Carnot cycle executed within thesaturation dome not a realistic model for refrigeration cycles?

PROBLEMS*

11–2 A steady-flow Carnot refrigeration cycle uses refriger-ant-134a as the working fluid. The refrigerant changes fromsaturated vapor to saturated liquid at 30°C in the condenseras it rejects heat. The evaporator pressure is 160 kPa. Showthe cycle on a T-s diagram relative to saturation lines, anddetermine (a) the coefficient of performance, (b) the amountof heat absorbed from the refrigerated space, and (c) the network input. Answers: (a) 5.64, (b) 147 kJ/kg, (c) 26.1 kJ/kg

11–3E Refrigerant-134a enters the condenser of a steady-flow Carnot refrigerator as a saturated vapor at 90 psia, and itleaves with a quality of 0.05. The heat absorption from therefrigerated space takes place at a pressure of 30 psia. Show

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

cen84959_ch11.qxd 4/20/05 1:04 PM Page 637

the cycle on a T-s diagram relative to saturation lines, anddetermine (a) the coefficient of performance, (b) the qualityat the beginning of the heat-absorption process, and (c) thenet work input.

Ideal and Actual Vapor-Compression Refrigeration Cycles

11–4C Does the ideal vapor-compression refrigerationcycle involve any internal irreversibilities?

11–5C Why is the throttling valve not replaced by an isen-tropic turbine in the ideal vapor-compression refrigerationcycle?

11–6C It is proposed to use water instead of refrigerant-134a as the working fluid in air-conditioning applicationswhere the minimum temperature never falls below the freez-ing point. Would you support this proposal? Explain.

11–7C In a refrigeration system, would you recommend con-densing the refrigerant-134a at a pressure of 0.7 or 1.0 MPa ifheat is to be rejected to a cooling medium at 15°C? Why?

11–8C Does the area enclosed by the cycle on a T-s diagramrepresent the net work input for the reversed Carnot cycle?How about for the ideal vapor-compression refrigeration cycle?

11–9C Consider two vapor-compression refrigeration cycles.The refrigerant enters the throttling valve as a saturated liquidat 30°C in one cycle and as subcooled liquid at 30°C in theother one. The evaporator pressure for both cycles is the same.Which cycle do you think will have a higher COP?

11–10C The COP of vapor-compression refrigeration cyclesimproves when the refrigerant is subcooled before it enters thethrottling valve. Can the refrigerant be subcooled indefinitelyto maximize this effect, or is there a lower limit? Explain.

11–11 A commercial refrigerator with refrigerant-134a asthe working fluid is used to keep the refrigerated space at�30°C by rejecting its waste heat to cooling water that enters

638 | Thermodynamics

the condenser at 18°C at a rate of 0.25 kg/s and leaves at26°C. The refrigerant enters the condenser at 1.2 MPa and65°C and leaves at 42°C. The inlet state of the compressor is60 kPa and �34°C and the compressor is estimated to gain anet heat of 450 W from the surroundings. Determine (a) thequality of the refrigerant at the evaporator inlet, (b) the refrig-eration load, (c) the COP of the refrigerator, and (d) the theo-retical maximum refrigeration load for the same power inputto the compressor.

11–12 A refrigerator uses refrigerant-134a as the workingfluid and operates on an ideal vapor-compression refrigerationcycle between 0.12 and 0.7 MPa. The mass flow rate of therefrigerant is 0.05 kg/s. Show the cycle on a T-s diagram withrespect to saturation lines. Determine (a) the rate of heatremoval from the refrigerated space and the power input to thecompressor, (b) the rate of heat rejection to the environment,and (c) the coefficient of performance. Answers: (a) 7.41 kW,1.83 kW, (b) 9.23 kW, (c) 4.06

11–13 Repeat Prob. 11–12 for a condenser pressure of0.9 MPa.

11–14 If the throttling valve in Prob. 11–12 is replaced byan isentropic turbine, determine the percentage increase inthe COP and in the rate of heat removal from the refrigeratedspace. Answers: 4.2 percent, 4.2 percent

11–15 Consider a 300 kJ/min refrigeration systemthat operates on an ideal vapor-compression

refrigeration cycle with refrigerant-134a as the working fluid.The refrigerant enters the compressor as saturated vapor at140 kPa and is compressed to 800 kPa. Show the cycle on aT-s diagram with respect to saturation lines, and determine(a) the quality of the refrigerant at the end of the throttlingprocess, (b) the coefficient of performance, and (c) the powerinput to the compressor.

11–16 Reconsider Prob. 11–15. Using EES (or other)software, investigate the effect of evaporator

pressure on the COP and the power input. Let the evaporatorpressure vary from 100 to 400 kPa. Plot the COP and thepower input as functions of evaporator pressure, and discussthe results.

11–17 Repeat Prob. 11–15 assuming an isentropic effi-ciency of 85 percent for the compressor. Also, determine therate of exergy destruction associated with the compressionprocess in this case. Take T0 � 298 K.

11–18 Refrigerant-134a enters the compressor of a refrigera-tor as superheated vapor at 0.14 MPa and �10°C at a rate of0.12 kg/s, and it leaves at 0.7 MPa and 50°C. The refrigerant iscooled in the condenser to 24°C and 0.65 MPa, and it is throt-tled to 0.15 MPa. Disregarding any heat transfer and pressuredrops in the connecting lines between the components, showthe cycle on a T-s diagram with respect to saturation lines, anddetermine (a) the rate of heat removal from the refrigerated

Compressor

QH

Condenser

1.2 MPa65°C

60 kPa–34°C

Water18°C

26°C

42°C

QL

Evaporator

Win

Qin

Expansionvalve

3

4 1

2

·

·

·

·

FIGURE P11–11

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Chapter 11 | 639

space and the power input to the compressor, (b) the isentropicefficiency of the compressor, and (c) the COP of the refrigera-tor. Answers: (a) 19.4 kW, 5.06 kW, (b) 82.5 percent, (c) 3.83

11–19E An ice-making machine operates on the idealvapor-compression cycle, using refrigerant-134a. The refrig-erant enters the compressor as saturated vapor at 20 psia andleaves the condenser as saturated liquid at 80 psia. Waterenters the ice machine at 55°F and leaves as ice at 25°F. Foran ice production rate of 15 lbm/h, determine the power inputto the ice machine (169 Btu of heat needs to be removedfrom each lbm of water at 55°F to turn it into ice at 25°F).

11–20 Refrigerant-134a enters the compressor of a refrigera-tor at 140 kPa and �10°C at a rate of 0.3 m3/min and leavesat 1 MPa. The isentropic efficiency of the compressor is 78percent. The refrigerant enters the throttling valve at 0.95 MPaand 30°C and leaves the evaporator as saturated vapor at�18.5°C. Show the cycle on a T-s diagram with respect tosaturation lines, and determine (a) the power input to the com-pressor, (b) the rate of heat removal from the refrigeratedspace, and (c) the pressure drop and rate of heat gain in theline between the evaporator and the compressor. Answers:(a) 1.88 kW, (b) 4.99 kW, (c) 1.65 kPa, 0.241 kW

11–21 Reconsider Prob. 11–20. Using EES (or other)software, investigate the effects of varying the

compressor isentropic efficiency over the range 60 to 100percent and the compressor inlet volume flow rate from 0.1to 1.0 m3/min on the power input and the rate of refrigera-tion. Plot the rate of refrigeration and the power input to thecompressor as functions of compressor efficiency for com-pressor inlet volume flow rates of 0.1, 0.5, and 1.0 m3/min,and discuss the results.

11–22 A refrigerator uses refrigerant-134a as the workingfluid and operates on the ideal vapor-compression refrigera-tion cycle. The refrigerant enters the evaporator at 120 kPawith a quality of 30 percent and leaves the compressor at

60°C. If the compressor consumes 450 W of power, deter-mine (a) the mass flow rate of the refrigerant, (b) the con-denser pressure, and (c) the COP of the refrigerator.Answers: (a) 0.00727 kg/s, (b) 672 kPa, (c) 2.43

Selecting the Right Refrigerant

11–23C When selecting a refrigerant for a certain applica-tion, what qualities would you look for in the refrigerant?

11–24C Consider a refrigeration system using refrigerant-134a as the working fluid. If this refrigerator is to operate inan environment at 30°C, what is the minimum pressure towhich the refrigerant should be compressed? Why?

11–25C A refrigerant-134a refrigerator is to maintain therefrigerated space at �10°C. Would you recommend an evap-orator pressure of 0.12 or 0.14 MPa for this system? Why?

11–26 A refrigerator that operates on the ideal vapor-compression cycle with refrigerant-134a is to maintain therefrigerated space at �10°C while rejecting heat to the envi-ronment at 25°C. Select reasonable pressures for the evapora-tor and the condenser, and explain why you chose thosevalues.

11–27 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is used to heat ahouse and maintain it at 22°C by using underground water at10°C as the heat source. Select reasonable pressures for theevaporator and the condenser, and explain why you chosethose values.

Heat Pump Systems

11–28C Do you think a heat pump system will be morecost-effective in New York or in Miami? Why?

11–29C What is a water-source heat pump? How does theCOP of a water-source heat pump system compare to that ofan air-source system?

11–30E A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is used to heat ahouse and maintain it at 75°F by using underground water at50°F as the heat source. The house is losing heat at a rate of60,000 Btu/h. The evaporator and condenser pressures are 50and 120 psia, respectively. Determine the power input to theheat pump and the electric power saved by using a heat pumpinstead of a resistance heater. Answers: 2.46 hp, 21.1 hp

11–31 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is used to heat waterfrom 15 to 45°C at a rate of 0.12 kg/s. The condenser andevaporator pressures are 1.4 and 0.32 MPa, respectively.Determine the power input to the heat pump.

11–32 A heat pump using refrigerant-134a heats a house byusing underground water at 8°C as the heat source. The houseis losing heat at a rate of 60,000 kJ/h. The refrigerant entersthe compressor at 280 kPa and 0°C, and it leaves at 1 MPa

Compressor

QH

Condenser60°C

120 kPax = 0.3

QL

Evaporator

WinExpansionvalve

3

4 1

2

·

·

·

FIGURE P11–22

cen84959_ch11.qxd 4/4/05 4:48 PM Page 639

and 60°C. The refrigerant exits the condenser at 30°C. Deter-mine (a) the power input to the heat pump, (b) the rate of heatabsorption from the water, and (c) the increase in electricpower input if an electric resistance heater is used instead of aheat pump. Answers: (a) 3.55 kW, (b) 13.12 kW, (c) 13.12 kW

11–33 Reconsider Prob. 11–32. Using EES (or other)software, investigate the effect of varying the

compressor isentropic efficiency over the range 60 to 100percent. Plot the power input to the compressor and the elec-tric power saved by using a heat pump rather than electricresistance heating as functions of compressor efficiency, anddiscuss the results.

11–34 Refrigerant-134a enters the condenser of a residen-tial heat pump at 800 kPa and 55°C at a rate of 0.018 kg/sand leaves at 750 kPa subcooled by 3°C. The refrigerantenters the compressor at 200 kPa superheated by 4°C. Deter-mine (a) the isentropic efficiency of the compressor, (b) therate of heat supplied to the heated room, and (c) the COP ofthe heat pump. Also, determine (d ) the COP and the rate ofheat supplied to the heated room if this heat pump operatedon the ideal vapor-compression cycle between the pressurelimits of 200 and 800 kPa.

640 | Thermodynamics

subcooling of the refrigerant in the condenser, (b) the massflow rate of the refrigerant, (c) the heating load and the COPof the heat pump, and (d) the theoretical minimum powerinput to the compressor for the same heating load. Answers:(a) 3.8°C, (b) 0.0194 kg/s, (c) 3.07 kW, 4.68, (d) 0.238 kW

Innovative Refrigeration Systems

11–36C What is cascade refrigeration? What are the advan-tages and disadvantages of cascade refrigeration?

11–37C How does the COP of a cascade refrigeration sys-tem compare to the COP of a simple vapor-compressioncycle operating between the same pressure limits?

11–38C A certain application requires maintaining therefrigerated space at �32°C. Would you recommend a simplerefrigeration cycle with refrigerant-134a or a two-stage cas-cade refrigeration cycle with a different refrigerant at the bot-toming cycle? Why?

11–39C Consider a two-stage cascade refrigeration cycle anda two-stage compression refrigeration cycle with a flash cham-ber. Both cycles operate between the same pressure limits anduse the same refrigerant. Which system would you favor?Why?

11–40C Can a vapor-compression refrigeration system witha single compressor handle several evaporators operating atdifferent pressures? How?

11–41C In the liquefaction process, why are gases com-pressed to very high pressures?

11–42 Consider a two-stage cascade refrigeration systemoperating between the pressure limits of 0.8 and 0.14 MPa.

·

·

·

Compressor

QH

Condenser

750 kPa 800 kPa55°C

QL

Evaporator

WinExpansionvalve

3

4 1

2

FIGURE P11–34

11–35 A heat pump with refrigerant-134a as the workingfluid is used to keep a space at 25°C by absorbing heat fromgeothermal water that enters the evaporator at 50°C at a rateof 0.065 kg/s and leaves at 40°C. The refrigerant enters theevaporator at 20°C with a quality of 23 percent and leaves atthe inlet pressure as saturated vapor. The refrigerant loses 300W of heat to the surroundings as it flows through the com-pressor and the refrigerant leaves the compressor at 1.4 MPaat the same entropy as the inlet. Determine (a) the degrees of

·

·

·

Compressor

QH

Condenser1.4 MPas2 = s1

20°Cx = 0.23

Sat.QL

Evaporator

WinExpansionvalve

3

4 1

2

Water50°C

40°C

FIGURE P11–35

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Chapter 11 | 641

Each stage operates on the ideal vapor-compression refrigera-tion cycle with refrigerant-134a as the working fluid. Heatrejection from the lower cycle to the upper cycle takes placein an adiabatic counterflow heat exchanger where bothstreams enter at about 0.4 MPa. If the mass flow rate of therefrigerant through the upper cycle is 0.24 kg/s, determine (a)the mass flow rate of the refrigerant through the lower cycle,(b) the rate of heat removal from the refrigerated space andthe power input to the compressor, and (c) the coefficient ofperformance of this cascade refrigerator.Answers: (a) 0.195 kg/s, (b) 34.2 kW, 7.63 kW, (c) 4.49

11–43 Repeat Prob. 11–42 for a heat exchanger pressure of0.55 MPa.

11–44 A two-stage compression refrigeration systemoperates with refrigerant-134a between the

pressure limits of 1 and 0.14 MPa. The refrigerant leaves thecondenser as a saturated liquid and is throttled to a flashchamber operating at 0.5 MPa. The refrigerant leaving thelow-pressure compressor at 0.5 MPa is also routed to theflash chamber. The vapor in the flash chamber is then com-pressed to the condenser pressure by the high-pressure com-pressor, and the liquid is throttled to the evaporator pressure.Assuming the refrigerant leaves the evaporator as saturatedvapor and both compressors are isentropic, determine (a) thefraction of the refrigerant that evaporates as it is throttled tothe flash chamber, (b) the rate of heat removed from therefrigerated space for a mass flow rate of 0.25 kg/s throughthe condenser, and (c) the coefficient of performance.

11–45 Reconsider Prob. 11–44. Using EES (or other)software, investigate the effect of the various

refrigerants for compressor efficiencies of 80, 90, and 100percent. Compare the performance of the refrigeration systemwith different refrigerants.

11–46 Repeat Prob. 11–44 for a flash chamber pres-sure of 0.32 MPa.

11–47 Consider a two-stage cascade refrigeration systemoperating between the pressure limits of 1.2 MPa and 200kPa with refrigerant-134a as the working fluid. Heat rejectionfrom the lower cycle to the upper cycle takes place in an adi-abatic counterflow heat exchanger where the pressure in theupper and lower cycles are 0.4 and 0.5 MPa, respectively. Inboth cycles, the refrigerant is a saturated liquid at the con-denser exit and a saturated vapor at the compressor inlet, andthe isentropic efficiency of the compressor is 80 percent. Ifthe mass flow rate of the refrigerant through the lower cycleis 0.15 kg/s, determine (a) the mass flow rate of the refriger-ant through the upper cycle, (b) the rate of heat removal fromthe refrigerated space, and (c) the COP of this refrigerator.Answers: (a) 0.212 kg/s, (b) 25.7 kW, (c) 2.68

11–48 Consider a two-stage cascade refrigeration systemoperating between the pressure limits of 1.2 MPa and 200 kPawith refrigerant-134a as the working fluid. The refrigerantleaves the condenser as a saturated liquid and is throttled to aflash chamber operating at 0.45 MPa. Part of the refrigerantevaporates during this flashing process, and this vapor is mixedwith the refrigerant leaving the low-pressure compressor. Themixture is then compressed to the condenser pressure by thehigh-pressure compressor. The liquid in the flash chamber isthrottled to the evaporator pressure and cools the refrigeratedspace as it vaporizes in the evaporator. The mass flow rate ofthe refrigerant through the low-pressure compressor is 0.15kg/s. Assuming the refrigerant leaves the evaporator as a satu-rated vapor and the isentropic efficiency is 80 percent for bothcompressors, determine (a) the mass flow rate of the refriger-ant through the high-pressure compressor, (b) the rate of heatremoval from the refrigerated space, and (c) the COP of thisrefrigerator. Also, determine (d) the rate of heat removal andthe COP if this refrigerator operated on a single-stage cyclebetween the same pressure limits with the same compressorefficiency and the same flow rate as in part (a).

·

·

·

·

Compressor

QH

Condenser

WinExpansionvalve

7

8 5

6

Compressor

Evaporator

WinExpansionvalve

3

4 1

2

QL

Evaporator

Condenser

Heat

FIGURE P11–47

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Gas Refrigeration Cycle

11–49C How does the ideal-gas refrigeration cycle differfrom the Brayton cycle?

11–50C Devise a refrigeration cycle that works on thereversed Stirling cycle. Also, determine the COP for thiscycle.

11–51C How does the ideal-gas refrigeration cycle differfrom the Carnot refrigeration cycle?

11–52C How is the ideal-gas refrigeration cycle modifiedfor aircraft cooling?

11–53C In gas refrigeration cycles, can we replace the tur-bine by an expansion valve as we did in vapor-compressionrefrigeration cycles? Why?

11–54C How do we achieve very low temperatures withgas refrigeration cycles?

11–55 An ideal gas refrigeration cycle using air as theworking fluid is to maintain a refrigerated space at �23°Cwhile rejecting heat to the surrounding medium at 27°C. Ifthe pressure ratio of the compressor is 3, determine (a) themaximum and minimum temperatures in the cycle, (b) thecoefficient of performance, and (c) the rate of refrigerationfor a mass flow rate of 0.08 kg/s.

11–56 Air enters the compressor of an ideal gasrefrigeration cycle at 12°C and 50 kPa and the

turbine at 47°C and 250 kPa. The mass flow rate of airthrough the cycle is 0.08 kg/s. Assuming variable specific

642 | Thermodynamics

heats for air, determine (a) the rate of refrigeration, (b) thenet power input, and (c) the coefficient of performance.Answers: (a) 6.67 kW, (b) 3.88 kW, (c) 1.72

11–57 Reconsider Prob. 11–56. Using EES (or other)software, study the effects of compressor and

turbine isentropic efficiencies as they are varied from 70 to100 percent on the rate of refrigeration, the net power input,and the COP. Plot the T-s diagram of the cycle for the isen-tropic case.

11–58E Air enters the compressor of an ideal gas refrigera-tion cycle at 40°F and 10 psia and the turbine at 120°F and30 psia. The mass flow rate of air through the cycle is 0.5lbm/s. Determine (a) the rate of refrigeration, (b) the netpower input, and (c) the coefficient of performance.

11–59 Repeat Prob. 11–56 for a compressor isen-tropic efficiency of 80 percent and a turbine

isentropic efficiency of 85 percent.

11–60 A gas refrigeration cycle with a pressure ratio of 3uses helium as the working fluid. The temperature of thehelium is �10°C at the compressor inlet and 50°C at the tur-bine inlet. Assuming adiabatic efficiencies of 80 percent forboth the turbine and the compressor, determine (a) the mini-mum temperature in the cycle, (b) the coefficient of perfor-mance, and (c) the mass flow rate of the helium for arefrigeration rate of 18 kW.

11–61 A gas refrigeration system using air as the workingfluid has a pressure ratio of 4. Air enters the compressor at�7°C. The high-pressure air is cooled to 27°C by rejectingheat to the surroundings. It is further cooled to �15°C byregenerative cooling before it enters the turbine. Assumingboth the turbine and the compressor to be isentropic andusing constant specific heats at room temperature, determine(a) the lowest temperature that can be obtained by this cycle,(b) the coefficient of performance of the cycle, and (c) themass flow rate of air for a refrigeration rate of 12 kW.Answers: (a) �99.4°C, (b) 1.12, (c) 0.237 kg/s

11–62 Repeat Prob. 11–61 assuming isentropic efficien-cies of 75 percent for the compressor and 80 percent for theturbine.

11–63 A gas refrigeration system using air as the workingfluid has a pressure ratio of 5. Air enters the compressor at0°C. The high-pressure air is cooled to 35°C by rejecting heatto the surroundings. The refrigerant leaves the turbine at�80°C and then it absorbs heat from the refrigerated spacebefore entering the regenerator. The mass flow rate of air is0.4 kg/s. Assuming isentropic efficiencies of 80 percent forthe compressor and 85 percent for the turbine and using con-stant specific heats at room temperature, determine (a) theeffectiveness of the regenerator, (b) the rate of heat removalfrom the refrigerated space, and (c) the COP of the cycle.Also, determine (d) the refrigeration load and the COP if thissystem operated on the simple gas refrigeration cycle. Use the

7

High-pressurecompressor

Low-pressurecompressor

QH

Condenser

Flashchamber

Expansionvalve

Expansionvalve

5

6

9

23

8 1

4

Evaporator

QL

·

·

·

FIGURE P11–48

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Chapter 11 | 643

same compressor inlet temperature as given, the same turbineinlet temperature as calculated, and the same compressor andturbine efficiencies. Answers: (a) 0.434, (b) 21.4 kW, (c) 0.478,(d) 24.7 kW, 0.599

Absorption Refrigeration Systems

11–64C What is absorption refrigeration? How does anabsorption refrigeration system differ from a vapor-compressionrefrigeration system?

11–65C What are the advantages and disadvantages ofabsorption refrigeration?

11–66C Can water be used as a refrigerant in air-conditioningapplications? Explain.

11–67C In absorption refrigeration cycles, why is the fluidin the absorber cooled and the fluid in the generator heated?

11–68C How is the coefficient of performance of anabsorption refrigeration system defined?

11–69C What are the functions of the rectifier and theregenerator in an absorption refrigeration system?

11–70 An absorption refrigeration system that receives heatfrom a source at 130°C and maintains the refrigerated spaceat �5°C is claimed to have a COP of 2. If the environmenttemperature is 27°C, can this claim be valid? Justify youranswer.

11–71 An absorption refrigeration system receives heat froma source at 120°C and maintains the refrigerated space at 0°C.If the temperature of the environment is 25°C, what is themaximum COP this absorption refrigeration system can have?

11–72 Heat is supplied to an absorption refrigeration systemfrom a geothermal well at 130°C at a rate of 5 � 105 kJ/h.The environment is at 25°C, and the refrigerated space ismaintained at �30°C. Determine the maximum rate at whichthis system can remove heat from the refrigerated space.Answer: 5.75 � 105 kJ/h

11–73E Heat is supplied to an absorption refrigeration sys-tem from a geothermal well at 250°F at a rate of 105 Btu/h.The environment is at 80°F, and the refrigerated space ismaintained at 0°F. If the COP of the system is 0.55, deter-mine the rate at which this system can remove heat from therefrigerated space.

11–74 A reversible absorption refrigerator consists of areversible heat engine and a reversible refrigerator. The sys-tem removes heat from a cooled space at �10°C at a rate of22 kW. The refrigerator operates in an environment at 25°C.If the heat is supplied to the cycle by condensing saturatedsteam at 200°C, determine (a) the rate at which the steamcondenses and (b) the power input to the reversible refrigera-tor. (c) If the COP of an actual absorption chiller at the sametemperature limits has a COP of 0.7, determine the secondlaw efficiency of this chiller. Answers: (a) 0.00408 kg/s,(b) 2.93 kW, (c) 0.252

Ts

Rev.HE

T0

T0

Rev.Ref.

TL

FIGURE P11–74

Heatexch.

Heatexch.

Regenerator

Turbine Compressor

1

6

3

5

4

QL

QH 2

·

·

FIGURE P11–63

Special Topic: Thermoelectric Power Generation andRefrigeration Systems

11–75C What is a thermoelectric circuit?

11–76C Describe the Seebeck and the Peltier effects.

11–77C Consider a circular copper wire formed by con-necting the two ends of a copper wire. The connection pointis now heated by a burning candle. Do you expect any cur-rent to flow through the wire?

11–78C An iron and a constantan wire are formed into aclosed circuit by connecting the ends. Now both junctions areheated and are maintained at the same temperature. Do youexpect any electric current to flow through this circuit?

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11–79C A copper and a constantan wire are formed into aclosed circuit by connecting the ends. Now one junction isheated by a burning candle while the other is maintained atroom temperature. Do you expect any electric current to flowthrough this circuit?

11–80C How does a thermocouple work as a temperaturemeasurement device?

11–81C Why are semiconductor materials preferable tometals in thermoelectric refrigerators?

11–82C Is the efficiency of a thermoelectric generator lim-ited by the Carnot efficiency? Why?

11–83E A thermoelectric generator receives heat from asource at 340°F and rejects the waste heat to the environmentat 90°F. What is the maximum thermal efficiency this ther-moelectric generator can have? Answer: 31.3 percent

11–84 A thermoelectric refrigerator removes heat from arefrigerated space at �5°C at a rate of 130 W and rejects it toan environment at 20°C. Determine the maximum coefficientof performance this thermoelectric refrigerator can have andthe minimum required power input. Answers: 10.72, 12.1 W

11–85 A thermoelectric cooler has a COP of 0.15 andremoves heat from a refrigerated space at a rate of 180 W.Determine the required power input to the thermoelectriccooler, in W.

11–86E A thermoelectric cooler has a COP of 0.15 andremoves heat from a refrigerated space at a rate of 20Btu/min. Determine the required power input to the thermo-electric cooler, in hp.

11–87 A thermoelectric refrigerator is powered by a 12-Vcar battery that draws 3 A of current when running. Therefrigerator resembles a small ice chest and is claimed to coolnine canned drinks, 0.350-L each, from 25 to 3°C in 12 h.Determine the average COP of this refrigerator.

644 | Thermodynamics

in the cooling mode, determine (a) the average rate of heatremoval from the drink, (b) the average rate of heat supply tothe coffee, and (c) the electric power drawn from the batteryof the car, all in W.

11–89 It is proposed to run a thermoelectric generator inconjunction with a solar pond that can supply heat at a rate of106 kJ/h at 80°C. The waste heat is to be rejected to the envi-ronment at 30°C. What is the maximum power this thermo-electric generator can produce?

Review Problems

11–90 Consider a steady-flow Carnot refrigeration cycle thatuses refrigerant-134a as the working fluid. The maximum andminimum temperatures in the cycle are 30 and �20°C,respectively. The quality of the refrigerant is 0.15 at the begin-ning of the heat absorption process and 0.80 at the end. Showthe cycle on a T-s diagram relative to saturation lines, anddetermine (a) the coefficient of performance, (b) the con-denser and evaporator pressures, and (c) the net work input.

11–91 A large refrigeration plant is to be maintained at�15°C, and it requires refrigeration at a rate of 100 kW. Thecondenser of the plant is to be cooled by liquid water, whichexperiences a temperature rise of 8°C as it flows over the coilsof the condenser. Assuming the plant operates on the idealvapor-compression cycle using refrigerant-134a between thepressure limits of 120 and 700 kPa, determine (a) the massflow rate of the refrigerant, (b) the power input to the com-pressor, and (c) the mass flow rate of the cooling water.

11–92 Reconsider Prob. 11–91. Using EES (or other)software, investigate the effect of evaporator

pressure on the COP and the power input. Let the evaporatorpressure vary from 120 to 380 kPa. Plot the COP and thepower input as functions of evaporator pressure, and discussthe results.

11–93 Repeat Prob. 11–91 assuming the compressor has anisentropic efficiency of 75 percent. Also, determine the rateof exergy destruction associated with the compressionprocess in this case. Take T0 � 25°C.

11–94 A heat pump that operates on the ideal vapor-compression cycle with refrigerant-134a is used to heat ahouse. The mass flow rate of the refrigerant is 0.32 kg/s. Thecondenser and evaporator pressures are 900 and 200 kPa,respectively. Show the cycle on a T-s diagram with respect tosaturation lines, and determine (a) the rate of heat supply tothe house, (b) the volume flow rate of the refrigerant at thecompressor inlet, and (c) the COP of this heat pump.

11–95 Derive a relation for the COP of the two-stagerefrigeration system with a flash chamber as shown in Fig.11–12 in terms of the enthalpies and the quality at state 6.Consider a unit mass in the condenser.

11–96 Consider a two-stage compression refrigeration system operating between the pressure limits of 0.8 and

FIGURE P11–87

11–88E Thermoelectric coolers that plug into the cigarettelighter of a car are commonly available. One such cooler isclaimed to cool a 12-oz (0.771-lbm) drink from 78 to 38°F orto heat a cup of coffee from 75 to 130°F in about 15 min in awell-insulated cup holder. Assuming an average COP of 0.2

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Chapter 11 | 645

0.14 MPa. The working fluid is refrigerant-134a. The refrig-erant leaves the condenser as a saturated liquid and is throt-tled to a flash chamber operating at 0.4 MPa. Part of therefrigerant evaporates during this flashing process, and thisvapor is mixed with the refrigerant leaving the low-pressurecompressor. The mixture is then compressed to the condenserpressure by the high-pressure compressor. The liquid in theflash chamber is throttled to the evaporator pressure, and itcools the refrigerated space as it vaporizes in the evaporator.Assuming the refrigerant leaves the evaporator as saturatedvapor and both compressors are isentropic, determine (a) thefraction of the refrigerant that evaporates as it is throttled tothe flash chamber, (b) the amount of heat removed from therefrigerated space and the compressor work per unit mass ofrefrigerant flowing through the condenser, and (c) the coeffi-cient of performance. Answers: (a) 0.165, (b) 146.4 kJ/kg,32.6 kJ/kg, (c) 4.49

11–97 An aircraft on the ground is to be cooled by a gasrefrigeration cycle operating with air on an open cycle. Airenters the compressor at 30°C and 100 kPa and is com-pressed to 250 kPa. Air is cooled to 70°C before it enters theturbine. Assuming both the turbine and the compressor to beisentropic, determine the temperature of the air leaving theturbine and entering the cabin. Answer: �9°C

11–98 Consider a regenerative gas refrigeration cycle usinghelium as the working fluid. Helium enters the compressor at100 kPa and �10°C and is compressed to 300 kPa. Helium isthen cooled to 20°C by water. It then enters the regeneratorwhere it is cooled further before it enters the turbine. Heliumleaves the refrigerated space at �25°C and enters the regen-erator. Assuming both the turbine and the compressor to beisentropic, determine (a) the temperature of the helium at theturbine inlet, (b) the coefficient of performance of the cycle,and (c) the net power input required for a mass flow rate of0.45 kg/s.

11–99 An absorption refrigeration system is to remove heatfrom the refrigerated space at �10°C at a rate of 12 kWwhile operating in an environment at 25°C. Heat is to be sup-plied from a solar pond at 85°C. What is the minimum rate ofheat supply required? Answer: 9.53 kW

11–100 Reconsider Prob. 11–99. Using EES (or other)software, investigate the effect of the source

temperature on the minimum rate of heat supply. Let thesource temperature vary from 50 to 250°C. Plot the minimumrate of heat supply as a function of source temperature, anddiscuss the results.

11–101 A typical 200-m2 house can be cooled adequately bya 3.5-ton air conditioner whose COP is 4.0. Determine the rateof heat gain of the house when the air conditioner is runningcontinuously to maintain a constant temperature in the house.

11–102 Rooms with floor areas of up to 15-m2 are cooledadequately by window air conditioners whose cooling capacity

is 5000 Btu/h. Assuming the COP of the air conditioner to be3.5, determine the rate of heat gain of the room, in Btu/h, whenthe air conditioner is running continuously to maintain a con-stant room temperature.

11–103 A gas refrigeration system using air as the workingfluid has a pressure ratio of 5. Air enters the compressor at0°C. The high-pressure air is cooled to 35°C by rejecting heatto the surroundings. The refrigerant leaves the turbine at�80°C and enters the refrigerated space where it absorbs heatbefore entering the regenerator. The mass flow rate of air is0.4 kg/s. Assuming isentropic efficiencies of 80 percent forthe compressor and 85 percent for the turbine and using vari-able specific heats, determine (a) the effectiveness of theregenerator, (b) the rate of heat removal from the refrigeratedspace, and (c) the COP of the cycle. Also, determine (d) therefrigeration load and the COP if this system operated on thesimple gas refrigeration cycle. Use the same compressor inlettemperature as given, the same turbine inlet temperature ascalculated, and the same compressor and turbine efficiencies.

Heatexch.

Heatexch.

Regenerator

Turbine Compressor

1

6

3

5

4

QL

QH 2

·

·

FIGURE P11–103

11–104 An air conditioner with refrigerant-134a as theworking fluid is used to keep a room at 26°C by rejecting thewaste heat to the outside air at 34°C. The room is gaining heatthrough the walls and the windows at a rate of 250 kJ/minwhile the heat generated by the computer, TV, and lightsamounts to 900 W. An unknown amount of heat is also gener-ated by the people in the room. The condenser and evaporatorpressures are 1200 and 500 kPa, respectively. The refrigerantis saturated liquid at the condenser exit and saturated vapor atthe compressor inlet. If the refrigerant enters the compressorat a rate of 100 L/min and the isentropic efficiency of thecompressor is 75 percent, determine (a) the temperature ofthe refrigerant at the compressor exit, (b) the rate of heatgeneration by the people in the room, (c) the COP of the air

cen84959_ch11.qxd 4/4/05 4:48 PM Page 645

conditioner, and (d ) the minimum volume flow rate of therefrigerant at the compressor inlet for the same compressorinlet and exit conditions.Answers: (a) 54.5°C, (b) 670 W, (c) 5.87, (d) 15.7 L/min

11–105 A heat pump water heater (HPWH) heats water byabsorbing heat from the ambient air and transferring it towater. The heat pump has a COP of 2.2 and consumes 2 kWof electricity when running. Determine if this heat pump canbe used to meet the cooling needs of a room most of the timefor “free” by absorbing heat from the air in the room. Therate of heat gain of a room is usually less than 5000 kJ/h.

646 | Thermodynamics

11–106 The vortex tube (also known as a Ranque or Hirschtube) is a device that produces a refrigeration effect byexpanding pressurized gas such as air in a tube (instead of aturbine as in the reversed Brayton cycle). It was invented andpatented by Ranque in 1931 and improved by Hirsch in 1945,and is commercially available in various sizes.

The vortex tube is simply a straight circular tube equippedwith a nozzle, as shown in the figure. The compressed gas attemperature T1 and pressure P1 is accelerated in the nozzle byexpanding it to nearly atmospheric pressure and is introducedinto the tube tangentially at a very high (typically supersonic)velocity to produce a swirling motion (vortex) within thetube. The rotating gas is allowed to exit through the full-sizetube that extends to the right, and the mass flow rate is con-trolled by a valve located about 30 diameters downstream.A smaller amount of air at the core region is allowed toescape to the left through a small aperture at the center. It isobserved that the gas that is in the core region and escapesthrough the central aperture is cold while the gas that is inthe peripheral region and escapes through the full-size tube ishot. If the temperature and the mass flow rate of the coldstream are Tc and m.c, respectively, the rate of refrigeration inthe vortex tube can be expressed as

where cp is the specific heat of the gas and T1 � Tc is thetemperature drop of the gas in the vortex tube (the coolingeffect). Temperature drops as high as 60°C (or 108°F)are obtained at high pressure ratios of about 10. The coeffi-cient of performance of a vortex tube can be defined as theratio of the refrigeration rate as given above to the powerused to compress the gas. It ranges from about 0.1 to 0.15,which is well below the COPs of ordinary vapor compressionrefrigerators.

This interesting phenomenon can be explained as follows:the centrifugal force creates a radial pressure gradient in thevortex, and thus the gas at the periphery is pressurized andheated by the gas at the core region, which is cooled as aresult. Also, energy is transferred from the inner layerstoward the outer layers as the outer layers slow down theinner layers because of fluid viscosity that tends to produce asolid vortex. Both of these effects cause the energy and thusthe temperature of the gas in the core region to decline. Theconservation of energy requires the energy of the fluid at theouter layers to increase by an equivalent amount.

The vortex tube has no moving parts, and thus it is inher-ently reliable and durable. The ready availability of the com-pressed air at pressures up to 10 atm in most industrialfacilities makes the vortex tube particularly attractive in suchsettings. Despite its low efficiency, the vortex tube has foundapplication in small-scale industrial spot-cooling operationssuch as cooling of soldered parts or critical electronic compo-nents, cooling drinking water, and cooling the suits of work-ers in hot environments.

Q#

refrig,vortex tube � m#

c 1h1 � hc 2 � m#

ccp 1T1 � Tc 2

Compressor

QH

Condenser1200 kPa

500 kPa

26°C

34°C

QL

Evaporator

WinExpansionvalve

3

4 1

2

·

·

·

FIGURE P11–104

Coldwater

in

Waterheater

Cool airto the room

Warm airfrom the room

Hotwaterout

FIGURE P11–105

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Chapter 11 | 647

Consider a vortex tube that receives compressed air at 500kPa and 300 K and supplies 25 percent of it as cold air at 100kPa and 278 K. The ambient air is at 300 K and 100 kPa, andthe compressor has an isentropic efficiency of 80 percent.The air suffers a pressure drop of 35 kPa in the aftercoolerand the compressed air lines between the compressor and thevortex tube.

(a) Without performing any calculations, explain how theCOP of the vortex tube would compare to the COP of anactual air refrigeration system based on the reversed Braytoncycle for the same pressure ratio. Also, compare the mini-mum temperatures that can be obtained by the two systemsfor the same inlet temperature and pressure.

(b) Assuming the vortex tube to be adiabatic and usingspecific heats at room temperature, determine the exit tem-perature of the hot fluid stream.

(c) Show, with calculations, that this process does not vio-late the second law of thermodynamics.

(d ) Determine the coefficient of performance of thisrefrigeration system, and compare it to the COP of a Carnotrefrigerator.

Fundamentals of Engineering (FE) Exam Problems

11–110 Consider a heat pump that operates on the reversedCarnot cycle with R-134a as the working fluid executedunder the saturation dome between the pressure limits of 140and 800 kPa. R-134a changes from saturated vapor to satu-rated liquid during the heat rejection process. The net workinput for this cycle is

(a) 28 kJ/kg (b) 34 kJ/kg (c) 49 kJ/kg(d ) 144 kJ/kg (e) 275 kJ/kg

11–111 A refrigerator removes heat from a refrigeratedspace at �5°C at a rate of 0.35 kJ/s and rejects it to an envi-ronment at 20°C. The minimum required power input is

(a) 30 W (b) 33 W (c) 56 W(d ) 124 W (e) 350 W

11–112 A refrigerator operates on the ideal vapor compres-sion refrigeration cycle with R-134a as the working fluidbetween the pressure limits of 120 and 800 kPa. If the rate ofheat removal from the refrigerated space is 32 kJ/s, the massflow rate of the refrigerant is

(a) 0.19 kg/s (b) 0.15 kg/s (c) 0.23 kg/s(d ) 0.28 kg/s (e) 0.81 kg/s

11–113 A heat pump operates on the ideal vapor compres-sion refrigeration cycle with R-134a as the working fluidbetween the pressure limits of 0.32 and 1.2 MPa. If the massflow rate of the refrigerant is 0.193 kg/s, the rate of heat sup-ply by the heat pump to the heated space is

(a) 3.3 kW (b) 23 kW (c) 26 kW(d ) 31 kW (e) 45 kW

11–114 An ideal vapor compression refrigeration cycle withR-134a as the working fluid operates between the pressure lim-its of 120 kPa and 1000 kPa. The mass fraction of the refriger-ant that is in the liquid phase at the inlet of the evaporator is

(a) 0.65 (b) 0.60 (c) 0.40(d ) 0.55 (e) 0.35

11–115 Consider a heat pump that operates on the idealvapor compression refrigeration cycle with R-134a as theworking fluid between the pressure limits of 0.32 and1.2 MPa. The coefficient of performance of this heat pump is

(a) 0.17 (b) 1.2 (c) 3.1(d ) 4.9 (e) 5.9

11–116 An ideal gas refrigeration cycle using air as theworking fluid operates between the pressure limits of 80 and280 kPa. Air is cooled to 35°C before entering the turbine.The lowest temperature of this cycle is

(a) �58°C (b) �26°C (c) 5°C(d ) 11°C (e) 24°C

11–117 Consider an ideal gas refrigeration cycle usinghelium as the working fluid. Helium enters the compressor at100 kPa and �10°C and compressed to 250 kPa. Helium is

Coldair

Warmair

Compressedair

FIGURE P11–106

11–107 Repeat Prob. 11–106 for a pressure of 600 kPa atthe vortex tube intake.

11–108 Using EES (or other) software, investigate theeffect of the evaporator pressure on the COP

of an ideal vapor-compression refrigeration cycle with R-134aas the working fluid. Assume the condenser pressure is keptconstant at 1 MPa while the evaporator pressure is variedfrom 100 kPa to 500 kPa. Plot the COP of the refrigerationcycle against the evaporator pressure, and discuss the results.

11–109 Using EES (or other) software, investigate theeffect of the condenser pressure on the COP of

an ideal vapor-compression refrigeration cycle with R-134a asthe working fluid. Assume the evaporator pressure is keptconstant at 120 kPa while the condenser pressure is variedfrom 400 to 1400 kPa. Plot the COP of the refrigeration cycleagainst the condenser pressure, and discuss the results.

cen84959_ch11.qxd 4/4/05 4:48 PM Page 647

then cooled to 20°C before it enters the turbine. For a massflow rate of 0.2 kg/s, the net power input required is

(a) 9.3 kW (b) 27.6 kW (c) 48.8 kW(d ) 93.5 kW (e) 119 kW

11–118 An absorption air-conditioning system is to removeheat from the conditioned space at 20°C at a rate of 150 kJ/swhile operating in an environment at 35°C. Heat is to be sup-plied from a geothermal source at 140°C. The minimum rateof heat supply is

(a) 86 kJ/s (b) 21 kJ/s (c) 30 kJ/s(d ) 61 kJ/s (e) 150 kJ/s

11–119 Consider a refrigerator that operates on the vaporcompression refrigeration cycle with R-134a as the workingfluid. The refrigerant enters the compressor as saturatedvapor at 160 kPa, and exits at 800 kPa and 50°C, and leavesthe condenser as saturated liquid at 800 kPa. The coefficientof performance of this refrigerator is

(a) 2.6 (b) 1.0 (c) 4.2(d ) 3.2 (e) 4.4

Design and Essay Problems

11–120 Design a vapor-compression refrigeration systemthat will maintain the refrigerated space at �15°C whileoperating in an environment at 20°C using refrigerant-134aas the working fluid.

11–121 Write an essay on air-, water-, and soil-based heatpumps. Discuss the advantages and the disadvantages of eachsystem. For each system identify the conditions under whichthat system is preferable over the other two. In what situationswould you not recommend a heat pump heating system?

11–122 Consider a solar pond power plant operating on aclosed Rankine cycle. Using refrigerant-134a as the workingfluid, specify the operating temperatures and pressures in thecycle, and estimate the required mass flow rate of refrigerant-134a for a net power output of 50 kW. Also, estimate the sur-face area of the pond for this level of continuous powerproduction. Assume that the solar energy is incident on thepond at a rate of 500 W per m2 of pond area at noontime, andthat the pond is capable of storing 15 percent of the incidentsolar energy in the storage zone.

11–123 Design a thermoelectric refrigerator that is capableof cooling a canned drink in a car. The refrigerator is to bepowered by the cigarette lighter of the car. Draw a sketch ofyour design. Semiconductor components for building thermo-electric power generators or refrigerators are available fromseveral manufacturers. Using data from one of these manu-facturers, determine how many of these components you needin your design, and estimate the coefficient of performance ofyour system. A critical problem in the design of thermoelec-tric refrigerators is the effective rejection of waste heat. Dis-cuss how you can enhance the rate of heat rejection withoutusing any devices with moving parts such as a fan.

648 | Thermodynamics

11–124 It is proposed to use a solar-powered thermoelectricsystem installed on the roof to cool residential buildings. Thesystem consists of a thermoelectric refrigerator that is pow-ered by a thermoelectric power generator whose top surfaceis a solar collector. Discuss the feasibility and the cost ofsuch a system, and determine if the proposed system installedon one side of the roof can meet a significant portion of thecooling requirements of a typical house in your area.

Thermoelectricrefrigerator

Thermoelectricgenerator

Electriccurrent

Solarenergy

Wasteheat

SUN

FIGURE P11–124

11–125 A refrigerator using R-12 as the working fluidkeeps the refrigerated space at �15°C in an environment at30°C. You are asked to redesign this refrigerator by replacingR-12 with the ozone-friendly R-134a. What changes in thepressure levels would you suggest in the new system? Howdo you think the COP of the new system will compare to theCOP of the old system?

11–126 In the 1800s, before the development of modernair-conditioning, it was proposed to cool air for buildingswith the following procedure using a large piston–cylinderdevice [“John Gorrie: Pioneer of Cooling and Ice Making,”ASHRAE Journal 33, no. 1 (Jan. 1991)]:

1. Pull in a charge of outdoor air.2. Compress it to a high pressure.3. Cool the charge of air using outdoor air.4. Expand it back to atmospheric pressure.5. Discharge the charge of air into the space to be

cooled.

Suppose the goal is to cool a room 6 m � 10 m � 2.5 m.Outdoor air is at 30°C, and it has been determined that 10 airchanges per hour supplied to the room at 10°C could provideadequate cooling. Do a preliminary design of the system and

cen84959_ch11.qxd 4/4/05 4:48 PM Page 648

Chapter 11 | 649

do calculations to see if it would be feasible. (You may makeoptimistic assumptions for the analysis.)

(a) Sketch the system showing how you will drive it andhow step 3 will be accomplished.

(b) Determine what pressure will be required (step 2).(c) Estimate (guess) how long step 3 will take and what

size will be needed for the piston–cylinder to provide therequired air changes and temperature.

(d ) Determine the work required in step 2 for one cycleand per hour.

(e) Discuss any problems you see with the concept of yourdesign. (Include discussion of any changes that may berequired to offset optimistic assumptions.)

11–127 Solar or photovoltaic (PV) cells convert sunlight toelectricity and are commonly used to power calculators, satel-lites, remote communication systems, and even pumps. Theconversion of light to electricity is called the photoelectriceffect. It was first discovered in 1839 by Frenchman EdmondBecquerel, and the first PV module, which consisted of sev-eral cells connected to each other, was built in 1954 by BellLaboratories. The PV modules today have conversion efficien-cies of about 12 to 15 percent. Noting that the solar energyincident on a normal surface on earth at noontime is about1000 W/m2 during a clear day, PV modules on a 1-m2 surfacecan provide as much as 150 W of electricity. The annual aver-age daily solar energy incident on a horizontal surface in theUnited States ranges from about 2 to 6 kWh/m2.

A PV-powered pump is to be used in Arizona to pump waterfor wildlife from a depth of 180 m at an average rate of 400L/day. Assuming a reasonable efficiency for the pumping sys-

tem, which can be defined as the ratio of the increase in thepotential energy of the water to the electrical energy consumedby the pump, and taking the conversion efficiency of the PVcells to be 0.13 to be on the conservative side, determine thesize of the PV module that needs to be installed, in m2.

11–128 The temperature in a car parked in the sun canapproach 100°C when the outside air temperature is just 25°C,and it is desirable to ventilate the parked car to avoid such hightemperatures. However, the ventilating fans may run down thebattery if they are powered by it. To avoid that happening, it isproposed to use the PV cells discussed in the preceding prob-lem to power the fans. It is determined that the air in the carshould be replaced once every minute to avoid excessive rise inthe interior temperature. Determine if this can be accomplishedby installing PV cells on part of the roof of the car. Also, findout if any car is currently ventilated this way.

11–129 A company owns a refrigeration system whoserefrigeration capacity is 200 tons (1 ton of refrigeration � 211kJ/min), and you are to design a forced-air cooling system forfruits whose diameters do not exceed 7 cm under the follow-ing conditions: The fruits are to be cooled from 28°C to anaverage temperature of 8°C. The air temperature is to remainabove �2°C and below 10°C at all times, and the velocity ofair approaching the fruits must remain under 2 m/s. The cool-ing section can be as wide as 3.5 m and as high as 2 m.

Assuming reasonable values for the average fruit density,specific heat, and porosity (the fraction of air volume in a box),recommend reasonable values for (a) the air velocity approach-ing the cooling section, (b) the product-cooling capacity of thesystem, in kg · fruit/h, and (c) the volume flow rate of air.

Sun

Water

PV-poweredpump

PV panel

FIGURE P11–127

Solar energy

Solar panels

Solar-poweredexhaust fan

FIGURE P11–128

cen84959_ch11.qxd 4/4/05 4:48 PM Page 649

cen84959_ch11.qxd 4/4/05 4:48 PM Page 650

Chapter 12THERMODYNAMIC PROPERTY RELATIONS

| 651

In the preceding chapters we made extensive use of theproperty tables. We tend to take the property tables forgranted, but thermodynamic laws and principles are of little

use to engineers without them. In this chapter, we focus ourattention on how the property tables are prepared and howsome unknown properties can be determined from limitedavailable data.

It will come as no surprise that some properties such astemperature, pressure, volume, and mass can be measureddirectly. Other properties such as density and specific volumecan be determined from these using some simple relations.However, properties such as internal energy, enthalpy, andentropy are not so easy to determine because they cannot bemeasured directly or related to easily measurable propertiesthrough some simple relations. Therefore, it is essential thatwe develop some fundamental relations between commonlyencountered thermodynamic properties and express theproperties that cannot be measured directly in terms of easilymeasurable properties.

By the nature of the material, this chapter makes extensiveuse of partial derivatives. Therefore, we start by reviewingthem. Then we develop the Maxwell relations, which formthe basis for many thermodynamic relations. Next we discussthe Clapeyron equation, which enables us to determine theenthalpy of vaporization from P, v, and T measurementsalone, and we develop general relations for cv, cp, du, dh,and ds that are valid for all pure substances under all condi-tions. Then we discuss the Joule-Thomson coefficient, whichis a measure of the temperature change with pressure duringa throttling process. Finally, we develop a method of evaluat-ing the �h, �u, and �s of real gases through the use of gen-eralized enthalpy and entropy departure charts.

ObjectivesThe objectives of Chapter 12 are to:

• Develop fundamental relations between commonlyencountered thermodynamic properties and express theproperties that cannot be measured directly in terms ofeasily measurable properties.

• Develop the Maxwell relations, which form the basis formany thermodynamic relations.

• Develop the Clapeyron equation and determine theenthalpy of vaporization from P, v, and T measurementsalone.

• Develop general relations for cv, cp, du, dh, and ds that arevalid for all pure substances.

• Discuss the Joule-Thomson coefficient.

• Develop a method of evaluating the �h, �u, and �s of realgases through the use of generalized enthalpy and entropydeparture charts.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 651

12–1 ■ A LITTLE MATH—PARTIAL DERIVATIVESAND ASSOCIATED RELATIONS

Many of the expressions developed in this chapter are based on the state pos-tulate, which expresses that the state of a simple, compressible substance iscompletely specified by any two independent, intensive properties. All otherproperties at that state can be expressed in terms of those two properties.Mathematically speaking,

where x and y are the two independent properties that fix the state and z rep-resents any other property. Most basic thermodynamic relations involve dif-ferentials. Therefore, we start by reviewing the derivatives and variousrelations among derivatives to the extent necessary in this chapter.

Consider a function f that depends on a single variable x, that is, f � f (x).Figure 12–1 shows such a function that starts out flat but gets rather steep as xincreases. The steepness of the curve is a measure of the degree of depen-dence of f on x. In our case, the function f depends on x more strongly atlarger x values. The steepness of a curve at a point is measured by the slope ofa line tangent to the curve at that point, and it is equivalent to the derivativeof the function at that point defined as

(12–1)

Therefore, the derivative of a function f(x) with respect to x represents therate of change of f with x.

df

dx� lim

¢xS0 ¢f

¢x� lim

¢xS0 f 1x � ¢x 2 � f 1x 2

¢x

z � z 1x, y 2

652 | Thermodynamics

f ( x )

f ( x )

( x + ∆ x )

x + ∆ x

∆ x

∆ f

x

Slope

x

FIGURE 12–1The derivative of a function at aspecified point represents the slope ofthe function at that point.

h(T ), kJ/kg

T, K

305.22

295.17

295 300 305

Slope = cp(T )

FIGURE 12–2Schematic for Example 12–1.

EXAMPLE 12–1 Approximating Differential Quantities by Differences

The cp of ideal gases depends on temperature only, and it is expressed ascp(T ) � dh(T )/dT. Determine the cp of air at 300 K, using the enthalpy datafrom Table A–17, and compare it to the value listed in Table A–2b.

Solution The cp value of air at a specified temperature is to be determinedusing enthalpy data.Analysis The cp value of air at 300 K is listed in Table A–2b to be 1.005kJ/kg · K. This value could also be determined by differentiating the functionh(T ) with respect to T and evaluating the result at T � 300 K. However, thefunction h(T ) is not available. But, we can still determine the cp value approx-imately by replacing the differentials in the cp(T ) relation by differences inthe neighborhood of the specified point (Fig. 12–2):

Discussion Note that the calculated cp value is identical to the listed value.Therefore, differential quantities can be viewed as differences. They can

�1305.22 � 295.17 2 kJ>kg

1305 � 295 2 K � 1.005 kJ/kg # K

cp 1300 K 2 � c dh 1T 2dTd

T � 300 K � c ¢h 1T 2

¢Td

T � 300 K�

h 1305 K 2 � h 1295 K 21305 � 295 2 K

cen84959_ch12.qxd 4/20/05 2:07 PM Page 652

even be replaced by differences, whenever necessary, to obtain approximateresults. The widely used finite difference numerical method is based on thissimple principle.

Partial DifferentialsNow consider a function that depends on two (or more) variables, such asz � z(x, y). This time the value of z depends on both x and y. It is sometimesdesirable to examine the dependence of z on only one of the variables. Thisis done by allowing one variable to change while holding the others constantand observing the change in the function. The variation of z(x, y) with xwhen y is held constant is called the partial derivative of z with respect tox, and it is expressed as

(12–2)

This is illustrated in Fig. 12–3. The symbol � represents differentialchanges, just like the symbol d. They differ in that the symbol d representsthe total differential change of a function and reflects the influence of allvariables, whereas � represents the partial differential change due to thevariation of a single variable.

Note that the changes indicated by d and � are identical for independentvariables, but not for dependent variables. For example, (�x)y � dx but (�z)y� dz. [In our case, dz � (�z)x � (�z)y.] Also note that the value of the par-tial derivative (�z /�x)y, in general, is different at different y values.

To obtain a relation for the total differential change in z(x, y) for simulta-neous changes in x and y, consider a small portion of the surface z(x, y)shown in Fig. 12–4. When the independent variables x and y change by �xand �y, respectively, the dependent variable z changes by �z, which can beexpressed as

Adding and subtracting z(x, y � �y), we get

or

Taking the limits as �x → 0 and �y → 0 and using the definitions of partialderivatives, we obtain

(12–3)

Equation 12–3 is the fundamental relation for the total differential of adependent variable in terms of its partial derivatives with respect to theindependent variables. This relation can easily be extended to include moreindependent variables.

dz � a 0z

0xb

y

dx � a 0z

0yb

x

dy

¢z �z 1x � ¢x, y � ¢y 2 � z 1x, y � ¢y 2

¢x ¢x �

z 1x, y � ¢y 2 � z 1x, y 2¢y

¢y

¢z � z 1x � ¢x, y � ¢y 2 � z 1x, y � ¢y 2 � z 1x, y � ¢y 2 � z 1x, y 2

¢z � z 1x � ¢x, y � ¢y 2 � z 1x, y 2

a 0z

0xb

y

� lim¢xS0

a ¢z

¢xb

y

� lim¢xS0

z 1x � ¢x, y 2 � z 1x, y 2¢x

Chapter 12 | 653

x y

z

z(x + ∆x, y + ∆y)

x + ∆x, y + ∆y

x, y + ∆yx + ∆x, y

z(x, y)

FIGURE 12–4Geometric representation of totalderivative dz for a function z(x, y).

z

x

y

∂z––∂x( ) y

FIGURE 12–3Geometric representation of partialderivative (�z/�x)y.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 653

654 | Thermodynamics

EXAMPLE 12–2 Total Differential versus Partial Differential

Consider air at 300 K and 0.86 m3/kg. The state of air changes to 302 Kand 0.87 m3/kg as a result of some disturbance. Using Eq. 12–3, estimatethe change in the pressure of air.

Solution The temperature and specific volume of air changes slightly dur-ing a process. The resulting change in pressure is to be determined.Assumptions Air is an ideal gas.Analysis Strictly speaking, Eq. 12–3 is valid for differential changes in vari-ables. However, it can also be used with reasonable accuracy if these changesare small. The changes in T and v, respectively, can be expressed as

and

An ideal gas obeys the relation Pv � RT. Solving for P yields

Note that R is a constant and P � P(T, v). Applying Eq. 12–3 and usingaverage values for T and v,

Therefore, the pressure will decrease by 0.491 kPa as a result of this distur-bance. Notice that if the temperature had remained constant (dT � 0), thepressure would decrease by 1.155 kPa as a result of the 0.01 m3/kgincrease in specific volume. However, if the specific volume had remainedconstant (dv � 0), the pressure would increase by 0.664 kPa as a result ofthe 2-K rise in temperature (Fig. 12–5). That is,

and

Discussion Of course, we could have solved this problem easily (and exactly)by evaluating the pressure from the ideal-gas relation P � RT/v at the finalstate (302 K and 0.87 m3/kg) and the initial state (300 K and 0.86 m3/kg)and taking their difference. This yields �0.491 kPa, which is exactly thevalue obtained above. Thus the small finite quantities (2 K, 0.01 m3/kg) canbe approximated as differential quantities with reasonable accuracy.

dP � 10P 2 v � 10P 2T � 0.664 � 1.155 � �0.491 kPa

a 0P

0vb

T

dv � 10P 2T � �1.155 kPa

a 0P

0Tb

v dT � 10P 2 v � 0.664 kPa

� �0.491 kPa

� 0.664 kPa � 1.155 kPa

� 10.287 kPa # m3>kg # K 2 c 2 K

0.865 m3>kg�1301 K 2 10.01 m3>kg 210.865 m3>kg 2 2 d

dP � a 0P

0Tb

v dT � a 0P

0vb

T

dv �R dT

v�

RT dvv 2

P �RT

v

dv � ¢v � 10.87 � 0.86 2 m3>kg � 0.01 m3>kg

dT � ¢T � 1302 � 300 2 K � 2 K

P, kPa(∂P)v = 0.664

(∂P)T = –1.155

dP = –0.491

T, K

302

300

0.86 0.87

v, m3/kg

FIGURE 12–5Geometric representation of thedisturbance discussed in Example 12–2.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 654

Partial Differential RelationsNow let us rewrite Eq. 12–3 as

(12–4)

where

Taking the partial derivative of M with respect to y and of N with respect tox yields

The order of differentiation is immaterial for properties since they are con-tinuous point functions and have exact differentials. Therefore, the two rela-tions above are identical:

(12–5)

This is an important relation for partial derivatives, and it is used in calculusto test whether a differential dz is exact or inexact. In thermodynamics, thisrelation forms the basis for the development of the Maxwell relations dis-cussed in the next section.

Finally, we develop two important relations for partial derivatives—thereciprocity and the cyclic relations. The function z � z(x, y) can also beexpressed as x � x(y, z) if y and z are taken to be the independent variables.Then the total differential of x becomes, from Eq. 12–3,

(12–6)

Eliminating dx by combining Eqs. 12–3 and 12–6, we have

Rearranging,

(12–7)

The variables y and z are independent of each other and thus can be variedindependently. For example, y can be held constant (dy � 0), and z can bevaried over a range of values (dz � 0). Therefore, for this equation to bevalid at all times, the terms in the brackets must equal zero, regardless ofthe values of y and z. Setting the terms in each bracket equal to zero gives

(12–8)

(12–9) a 0z

0xb

y

a 0x

0yb

z

� � a 0x

0yb

x

S a 0x

0yb

z

a 0y

0zb

x

a 0z

0xb

y

� �1

a 0x

0zb

y

a 0z

0xb

y

� 1 S a 0x

0zb

y

�1

10z>0x 2 y

c a 0z

0xb

y

a 0x

0yb

z

� a 0z

0yb

x

d dy � c1 � a 0x

0zb

y

a 0z

0xb

y

d dz

dz � c a 0z

0xb

y

a 0x

0yb

z

� a 0z

0yb

x

d dy � a 0x

0zb

y

a 0z

0xb

y

dz

dx � a 0x

0yb

z

dy � a 0x

0zb

y

dz

a 0M

0yb

x

� a 0N

0xb

y

a 0M

0yb

x

�02z

0x 0yand a 0N

0xb

y

�02z

0y 0x

M � a 0z

0xb

y

and N � a 0z

0yb

x

dz � M dx � N dy

Chapter 12 | 655

cen84959_ch12.qxd 4/5/05 3:58 PM Page 655

The first relation is called the reciprocity relation, and it shows that theinverse of a partial derivative is equal to its reciprocal (Fig. 12–6). The sec-ond relation is called the cyclic relation, and it is frequently used in ther-modynamics (Fig. 12–7).

656 | Thermodynamics

Function: z + 2xy – 3y2z = 0

1) z = ==

= 1––––––Thus,

2xy—––––3y2 – 1

2y—––––3y2 – 1

�z––�x( ) y

2) x = ==3y2z – z—––––

2y

3y2 – 1—––––

2y

�x––�z( ) y

�x––�z( ) y

�z––�x( ) y

FIGURE 12–6Demonstration of the reciprocityrelation for the function z � 2xy � 3y2z � 0.

FIGURE 12–7Partial differentials are powerful toolsthat are supposed to make life easier,not harder.

© Reprinted with special permission of KingFeatures Syndicate.

EXAMPLE 12–3 Verification of Cyclic and Reciprocity Relations

Using the ideal-gas equation of state, verify (a) the cyclic relation and (b)the reciprocity relation at constant P.

Solution The cyclic and reciprocity relations are to be verified for an ideal gas.Analysis The ideal-gas equation of state Pv � RT involves the three vari-ables P, v, and T. Any two of these can be taken as the independent vari-ables, with the remaining one being the dependent variable.

(a) Replacing x, y, and z in Eq. 12–9 by P, v, and T, respectively, we canexpress the cyclic relation for an ideal gas as

where

Substituting yields

which is the desired result.

(b) The reciprocity rule for an ideal gas at P � constant can be expressed as

Performing the differentiations and substituting, we have

Thus the proof is complete.

R

P�

1

P>R SR

P�

R

P

a 0v0Tb

P

�1

10T>0v 2P

a�RT

v 2 b a R

Pb a v

Rb � �

RT

Pv� �1

T � T 1P, v 2 �PvR

S a 0T

0Pb

v�

vR

v � v 1P, T 2 �RT

PS a 0v

0Tb

P

�R

P

P � P 1v, T 2 �RT

vS a 0P

0vb

T

� �RT

v 2

a 0P

0vb

T

a 0v0Tb

P

a 0T

0Pb

v� �1

12–2 ■ THE MAXWELL RELATIONSThe equations that relate the partial derivatives of properties P, v, T, and sof a simple compressible system to each other are called the Maxwell rela-tions. They are obtained from the four Gibbs equations by exploiting theexactness of the differentials of thermodynamic properties.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 656

Two of the Gibbs relations were derived in Chap. 7 and expressed as

(12–10)

(12–11)

The other two Gibbs relations are based on two new combination proper-ties—the Helmholtz function a and the Gibbs function g, defined as

(12–12)

(12–13)

Differentiating, we get

Simplifying the above relations by using Eqs. 12–10 and 12–11, we obtainthe other two Gibbs relations for simple compressible systems:

(12–14)

(12–15)

A careful examination of the four Gibbs relations reveals that they are of theform

(12–4)

with

(12–5)

since u, h, a, and g are properties and thus have exact differentials. Apply-ing Eq. 12–5 to each of them, we obtain

(12–16)

(12–17)

(12–18)

(12–19)

These are called the Maxwell relations (Fig. 12–8). They are extremelyvaluable in thermodynamics because they provide a means of determiningthe change in entropy, which cannot be measured directly, by simply mea-suring the changes in properties P, v, and T. Note that the Maxwell relationsgiven above are limited to simple compressible systems. However, othersimilar relations can be written just as easily for nonsimple systems such asthose involving electrical, magnetic, and other effects.

a 0s

0Pb

T

� � a 0v0Tb

P

a 0s

0vb

T

� a 0P

0Tb

v

a 0T

0Pb

s

� a 0v0sb

P

a 0T

0vb

s

� � a 0P

0sb

v

a 0M

0yb

x

� a 0N

0xb

y

dz � M dx � N dy

dg � �s dT � v dP

da � �s dT � P dv

dg � dh � T ds � s dT

da � du � T ds � s dT

g � h � Ts

a � u � Ts

dh � T ds � v dP

du � T ds � P dv

Chapter 12 | 657

= –∂P–– ( )∂s

∂T–– ( ) s∂v

= ∂T–– ( ) s∂P

∂s–– ( ) T∂P

= ∂P–– ( )∂T

∂s–– ( ) T∂v

= –

∂v–– ( )∂s P

∂v–– ( )∂T P

v

v

FIGURE 12–8Maxwell relations are extremelyvaluable in thermodynamic analysis.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 657

EXAMPLE 12–4 Verification of the Maxwell Relations

Verify the validity of the last Maxwell relation (Eq. 12–19) for steam at250°C and 300 kPa.

Solution The validity of the last Maxwell relation is to be verified for steamat a specified state.Analysis The last Maxwell relation states that for a simple compressiblesubstance, the change in entropy with pressure at constant temperature isequal to the negative of the change in specific volume with temperature atconstant pressure.

If we had explicit analytical relations for the entropy and specific volumeof steam in terms of other properties, we could easily verify this by perform-ing the indicated derivations. However, all we have for steam are tables ofproperties listed at certain intervals. Therefore, the only course we can taketo solve this problem is to replace the differential quantities in Eq. 12–19with corresponding finite quantities, using property values from the tables(Table A–6 in this case) at or about the specified state.

T

?� �

T � 250°C

?� �P � 300 kPa

T � 250°C

?� �P � 300 kPa

?� �

�0.00165 m3/kg � K � �0.00159 m3/kg � K

since kJ � kPa · m3 and K � °C for temperature differences. The two valuesare within 4 percent of each other. This difference is due to replacing thedifferential quantities by relatively large finite quantities. Based on the closeagreement between the two values, the steam seems to satisfy Eq. 12–19 atthe specified state.Discussion This example shows that the entropy change of a simple com-pressible system during an isothermal process can be determined from aknowledge of the easily measurable properties P, v, and T alone.

(0.87535 � 0.71643) m3�kg

(300 � 200)°C(7.3804 � 7.7100) kJ�kg # K

(400 � 200) kPa

c v300°C � v200°C

(300 � 200)°Cd� s400 kPa � s200 kPa

(400 � 200) kPa�

a 0v0Tb��s

�P�

a 0v0Tb

P��s�P�

12–3 ■ THE CLAPEYRON EQUATIONThe Maxwell relations have far-reaching implications in thermodynamicsand are frequently used to derive useful thermodynamic relations. TheClapeyron equation is one such relation, and it enables us to determine theenthalpy change associated with a phase change (such as the enthalpy ofvaporization hfg) from a knowledge of P, v, and T data alone.

Consider the third Maxwell relation, Eq. 12–18:

During a phase-change process, the pressure is the saturation pressure,which depends on the temperature only and is independent of the specific

a 0P

0Tb

v� a 0s

0vb

T

658 | Thermodynamics

cen84959_ch12.qxd 4/5/05 3:58 PM Page 658

volume. That is, Psat � f (Tsat). Therefore, the partial derivative (�P/�T )v canbe expressed as a total derivative (dP/dT )sat, which is the slope of the satu-ration curve on a P-T diagram at a specified saturation state (Fig. 12–9).This slope is independent of the specific volume, and thus it can be treatedas a constant during the integration of Eq. 12–18 between two saturationstates at the same temperature. For an isothermal liquid–vapor phase-changeprocess, for example, the integration yields

(12–20)

or

(12–21)

During this process the pressure also remains constant. Therefore, fromEq. 12–11,

Substituting this result into Eq. 12–21, we obtain

(12–22)

which is called the Clapeyron equation after the French engineer andphysicist E. Clapeyron (1799–1864). This is an important thermodynamicrelation since it enables us to determine the enthalpy of vaporization hfg at agiven temperature by simply measuring the slope of the saturation curve ona P-T diagram and the specific volume of saturated liquid and saturatedvapor at the given temperature.

The Clapeyron equation is applicable to any phase-change process thatoccurs at constant temperature and pressure. It can be expressed in a generalform as

(12–23)

where the subscripts 1 and 2 indicate the two phases.

a dP

dTb

sat�

h12

Tv12

a dP

dTb

sat�

hfg

Tvfg

dh � T ds � v dP S �g

f

dh � �g

f

T ds S hfg � Tsfg

a dP

dTb

sat�

sfg

vfg

sg � sf � a dP

dTb

sat

1vg � vf 2

Chapter 12 | 659

P

TT

= const.

LIQUID

SOLID

VAPOR

∂P––( ) sat∂T

FIGURE 12–9The slope of the saturation curve on aP-T diagram is constant at a constant T or P.

EXAMPLE 12–5 Evaluating the hfg of a Substance from the P-v-T Data

Using the Clapeyron equation, estimate the value of the enthalpy of vaporiza-tion of refrigerant-134a at 20°C, and compare it with the tabulated value.

Solution The hfg of refrigerant-134a is to be determined using the Clapeyronequation.Analysis From Eq. 12–22,

hfg � Tvfg a dP

dTb

sat

→0

cen84959_ch12.qxd 4/5/05 3:58 PM Page 659

The Clapeyron equation can be simplified for liquid–vapor and solid–vaporphase changes by utilizing some approximations. At low pressures vg vf ,and thus vfg � vg. By treating the vapor as an ideal gas, we have vg � RT/P.Substituting these approximations into Eq. 12–22, we find

or

For small temperature intervals hfg can be treated as a constant at some aver-age value. Then integrating this equation between two saturation states yields

(12–24)

This equation is called the Clapeyron–Clausius equation, and it can beused to determine the variation of saturation pressure with temperature. Itcan also be used in the solid–vapor region by replacing hfg by hig (theenthalpy of sublimation) of the substance.

ln a P2

P1b

sat�

hfg

Ra 1

T1�

1

T2b

sat

a dP

Pb

sat�

hfg

Ra dT

T 2 bsat

a dP

dTb

sat�

Phfg

RT 2

660 | Thermodynamics

where, from Table A–11,

since �T(°C) � �T(K). Substituting, we get

The tabulated value of hfg at 20°C is 182.27 kJ/kg. The small differencebetween the two values is due to the approximation used in determining theslope of the saturation curve at 20°C.

� 182.40 kJ/kg

hfg � 1293.15 K 2 10.035153 m3>kg 2 117.70 kPa>K 2 a 1 kJ

1 kPa # m3 b

�646.18 � 504.58 kPa

8°C� 17.70 kPa>K

a dP

d Tb

sat,20°C� a ¢P

¢Tb

sat,20°C�

Psat @ 24°C � Psat @ 16°C

24°C � 16°C

vfg � 1vg � vf 2@ 20°C � 0.035969 � 0.0008161 � 0.035153 m3>kg

EXAMPLE 12–6 Extrapolating Tabular Data with the Clapeyron Equation

Estimate the saturation pressure of refrigerant-134a at �50°F, using thedata available in the refrigerant tables.

Solution The saturation pressure of refrigerant-134a is to be determinedusing other tabulated data.Analysis Table A–11E lists saturation data at temperatures �40°F andabove. Therefore, we should either resort to other sources or use extrapolation

cen84959_ch12.qxd 4/5/05 3:58 PM Page 660

12–4 ■ GENERAL RELATIONS FOR du, dh, ds, cv, AND cp

The state postulate established that the state of a simple compressible systemis completely specified by two independent, intensive properties. Therefore,at least theoretically, we should be able to calculate all the properties of asystem at any state once two independent, intensive properties are available.This is certainly good news for properties that cannot be measured directlysuch as internal energy, enthalpy, and entropy. However, the calculation ofthese properties from measurable ones depends on the availability of simpleand accurate relations between the two groups.

In this section we develop general relations for changes in internal energy,enthalpy, and entropy in terms of pressure, specific volume, temperature, andspecific heats alone. We also develop some general relations involving specificheats. The relations developed will enable us to determine the changes in theseproperties. The property values at specified states can be determined only afterthe selection of a reference state, the choice of which is quite arbitrary.

Internal Energy ChangesWe choose the internal energy to be a function of T and v; that is, u �u(T, v) and take its total differential (Eq. 12–3):

Using the definition of cv, we have

(12–25)du � cv dT � a 0u

0vb

T

dv

du � a 0u

0Tb

vdT � a 0u

0vb

T

dv

Chapter 12 | 661

to obtain saturation data at lower temperatures. Equation 12–24 provides anintelligent way to extrapolate:

In our case T1 � �40°F and T2 � �50°F. For refrigerant-134a, R � 0.01946Btu/lbm · R. Also from Table A–11E at �40°F, we read hfg � 97.100 Btu/lbmand P1 � Psat @ �40°F � 7.432 psia. Substituting these values into Eq. 12–24gives

Therefore, according to Eq. 12–24, the saturation pressure of refrigerant-134aat �50°F is 5.56 psia. The actual value, obtained from another source,is 5.506 psia. Thus the value predicted by Eq. 12–24 is in error by about1 percent, which is quite acceptable for most purposes. (If we had used linearextrapolation instead, we would have obtained 5.134 psia, which is in error by7 percent.)

P2 � 5.56 psia

ln a P2

7.432 psiab �

97.100 Btu>lbm

0.01946 Btu>lbm # Ra 1

420 R�

1

410 Rb

ln a P2

P1b

sat�

hfg

Ra 1

T1�

1

T2b

sat

cen84959_ch12.qxd 4/5/05 3:58 PM Page 661

Now we choose the entropy to be a function of T and v; that is, s � s(T, v)and take its total differential,

(12–26)

Substituting this into the T ds relation du � T ds � P dv yields

(12–27)

Equating the coefficients of dT and dv in Eqs. 12–25 and 12–27 gives

(12–28)

Using the third Maxwell relation (Eq. 12–18), we get

Substituting this into Eq. 12–25, we obtain the desired relation for du:

(12–29)

The change in internal energy of a simple compressible system associatedwith a change of state from (T1, v1) to (T2, v2) is determined by integration:

(12–30)

Enthalpy ChangesThe general relation for dh is determined in exactly the same manner. Thistime we choose the enthalpy to be a function of T and P, that is, h � h(T, P),and take its total differential,

Using the definition of cp, we have

(12–31)

Now we choose the entropy to be a function of T and P; that is, we takes � s(T, P) and take its total differential,

(12–32)

Substituting this into the T ds relation dh � T ds � v dP gives

(12–33)dh � T a 0s

0Tb

P

dT � cv � T a 0s

0Pb

T

d dP

ds � a 0s

0Tb

P

dT � a 0s

0Pb

T

dP

dh � cp dT � a 0h

0Pb

T

dP

dh � a 0h

0Tb

P

dT � a 0h

0Pb

T

dP

u2 � u1 � �T2

T1

cv dT � �v2

v1

cT a 0P

0Tb

v� P d dv

du � cv dT � cT a 0P

0Tb

v� P d dv

a 0u

0vb

T

� T a 0P

0Tb

v� P

a 0u

0vb

T

� T a 0s

0vb

T

� P

a 0s

0Tb

v�

cv

T

du � T a 0s

0Tb

v dT � cT a 0s

0vb

T

� P d dv

ds � a 0s

0Tb

v dT � a 0s

0vb

T

dv

662 | Thermodynamics

cen84959_ch12.qxd 4/5/05 3:58 PM Page 662

Equating the coefficients of dT and dP in Eqs. 12–31 and 12–33, we obtain

(12–34)

Using the fourth Maxwell relation (Eq. 12–19), we have

Substituting this into Eq. 12–31, we obtain the desired relation for dh:

(12–35)

The change in enthalpy of a simple compressible system associated with achange of state from (T1, P1) to (T2, P2) is determined by integration:

(12–36)

In reality, one needs only to determine either u2 � u1 from Eq. 12–30 orh2 � h1 from Eq. 12–36, depending on which is more suitable to the data athand. The other can easily be determined by using the definition of enthalpyh � u � Pv:

(12–37)

Entropy ChangesBelow we develop two general relations for the entropy change of a simplecompressible system.

The first relation is obtained by replacing the first partial derivative in thetotal differential ds (Eq. 12–26) by Eq. 12–28 and the second partial deriva-tive by the third Maxwell relation (Eq. 12–18), yielding

(12–38)

and

(12–39)

The second relation is obtained by replacing the first partial derivative in thetotal differential of ds (Eq. 12–32) by Eq. 12–34, and the second partialderivative by the fourth Maxwell relation (Eq. 12–19), yielding

(12–40)

and

(12–41)

Either relation can be used to determine the entropy change. The properchoice depends on the available data.

s2 � s1 � �T2

T1

cp

T dT � �

P2

P1

a 0v0Tb

P

dP

ds �cP

T dT � a 0v

0Tb

P

dP

s2 � s1 � �T2

T1

cv

T dT � �

v2

v1

a 0P

0Tb

v dv

ds �cv

T dT � a 0P

0Tb

vdv

h2 � h1 � u2 � u1 � 1P2v2 � P1v1 2

h2 � h1 � �T2

T1

cp dT � �P2

P1

cv � T a 0v0Tb

P

d dP

dh � cp d T � cv � T a 0v0Tb

P

d dP

a 0h

0Pb

T

� v � T a 0v0Tb

P

a 0h

0Pb

T

� v � T a 0s

0Pb

T

a 0s

0Tb

P

�cp

T

Chapter 12 | 663

cen84959_ch12.qxd 4/5/05 3:58 PM Page 663

Specific Heats cv and cpRecall that the specific heats of an ideal gas depend on temperature only.For a general pure substance, however, the specific heats depend on specificvolume or pressure as well as the temperature. Below we develop some gen-eral relations to relate the specific heats of a substance to pressure, specificvolume, and temperature.

At low pressures gases behave as ideal gases, and their specific heatsessentially depend on temperature only. These specific heats are called zeropressure, or ideal-gas, specific heats (denoted cv 0 and cp0), and they are rel-atively easier to determine. Thus it is desirable to have some general rela-tions that enable us to calculate the specific heats at higher pressures (orlower specific volumes) from a knowledge of cv 0 or cp 0 and the P-v-Tbehavior of the substance. Such relations are obtained by applying the testof exactness (Eq. 12–5) on Eqs. 12–38 and 12–40, which yields

(12–42)

and

(12–43)

The deviation of cp from cp 0 with increasing pressure, for example, is deter-mined by integrating Eq. 12–43 from zero pressure to any pressure P alongan isothermal path:

(12–44)

The integration on the right-hand side requires a knowledge of the P-v-Tbehavior of the substance alone. The notation indicates that v should be dif-ferentiated twice with respect to T while P is held constant. The resultingexpression should be integrated with respect to P while T is held constant.

Another desirable general relation involving specific heats is one that relatesthe two specific heats cp and cv. The advantage of such a relation is obvious:We will need to determine only one specific heat (usually cp) and calculatethe other one using that relation and the P-v-T data of the substance. Westart the development of such a relation by equating the two ds relations(Eqs. 12–38 and 12–40) and solving for dT:

Choosing T � T(v, P) and differentiating, we get

Equating the coefficient of either dv or dP of the above two equations givesthe desired result:

(12–45)cp � cv � T a 0v0Tb

P

a 0P

0Tb

v

dT � a 0T

0vb

P

dv � a 0T

0Pb

v dP

dT �T 10P>0T 2 v

cp � cv dv �

T 10v>0T 2Pcp � cv

dP

1cp � cp0 2T � �T �P

0

a 02v0T 2b

P

dP

a 0cp

0Pb

T

� �T a 02v0T 2 b

P

a 0cv

0vb

T

� T a 02P

0T 2 bv

664 | Thermodynamics

cen84959_ch12.qxd 4/5/05 3:58 PM Page 664

An alternative form of this relation is obtained by using the cyclic relation:

Substituting the result into Eq. 12–45 gives

(12–46)

This relation can be expressed in terms of two other thermodynamic proper-ties called the volume expansivity b and the isothermal compressibility a,which are defined as (Fig. 12–10)

(12–47)

and

(12–48)

Substituting these two relations into Eq. 12–46, we obtain a third generalrelation for cp � cv:

(12–49)

It is called the Mayer relation in honor of the German physician and physicistJ. R. Mayer (1814–1878). We can draw several conclusions from this equation:

1. The isothermal compressibility a is a positive quantity for all sub-stances in all phases. The volume expansivity could be negative for somesubstances (such as liquid water below 4°C), but its square is always positiveor zero. The temperature T in this relation is thermodynamic temperature,which is also positive. Therefore we conclude that the constant-pressure spe-cific heat is always greater than or equal to the constant-volume specific heat:

(12–50)

2. The difference between cp and cv approaches zero as the absolutetemperature approaches zero.

3. The two specific heats are identical for truly incompressible sub-stances since v � constant. The difference between the two specific heats isvery small and is usually disregarded for substances that are nearly incom-pressible, such as liquids and solids.

cp cv

cp � cv �vTb2

a

a � �1

va 0v

0Pb

T

b �1

va 0v

0Tb

P

cp � cv � �T a 0v0Tb 2

P a 0P

0vb

T

a 0P

0Tb

va 0T

0vb

P

a 0v0Pb

T

� �1 S a 0P

0Tb

v� � a 0v

0Tb

P

a 0P

0vb

T

Chapter 12 | 665

20°C100 kPa

1 kg

21°C100 kPa

1 kg

20°C100 kPa

1 kg

21°C100 kPa

1 kg

(a) A substance with a large β

(b) A substance with a small β

∂v–– ( )∂T P

∂–– ( )∂T P

v

FIGURE 12–10The volume expansivity (also calledthe coefficient of volumetricexpansion) is a measure of the changein volume with temperature atconstant pressure.

EXAMPLE 12–7 Internal Energy Change of a van der Waals Gas

Derive a relation for the internal energy change as a gas that obeys the vander Waals equation of state. Assume that in the range of interest cv variesaccording to the relation cv � c1 � c2T, where c1 and c2 are constants.

Solution A relation is to be obtained for the internal energy change of avan der Waals gas.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 665

666 | Thermodynamics

Analysis The change in internal energy of any simple compressible systemin any phase during any process can be determined from Eq. 12–30:

The van der Waals equation of state is

Then

Thus,

Substituting gives

Integrating yields

which is the desired relation.

u2 � u1 � c1 1T2 � T1 2 �c2

21T 2

2 � T 21 2 � a a 1

v1�

1

v2b

u2 � u1 � �T2

T1

1c1 � c2T 2 dT � �v2

v1

a

v 2 dv

T a 0P

0Tb

v� P �

RT

v � b�

RT

v � b�

a

v 2 �a

v 2

a 0P

0Tb

v�

R

v � b

P �RT

v � b�

a

v 2

u2 � u1 � �T2

T1

cv dT � �v2

v1

cT a 0P

0Tb

v� P d dv

EXAMPLE 12–8 Internal Energy as a Function of Temperature Alone

Show that the internal energy of (a) an ideal gas and (b) an incompressiblesubstance is a function of temperature only, u � u(T).

Solution It is to be shown that u � u(T) for ideal gases and incompressiblesubstances.Analysis The differential change in the internal energy of a general simplecompressible system is given by Eq. 12–29 as

(a) For an ideal gas Pv � RT. Then

Thus,

To complete the proof, we need to show that cv is not a function of v either.This is done with the help of Eq. 12–42:

a 0cv

0vb

T

� T a 02P

0T 2bv

du � cv dT

T a 0P

0Tb

v� P � T a R

vb � P � P � P � 0

du � cv dT� cT a 0P

0Tb

v� P d dv

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Chapter 12 | 667

For an ideal gas P � RT/v. Then

Thus,

which states that cv does not change with specific volume. That is, cv is nota function of specific volume either. Therefore we conclude that the internalenergy of an ideal gas is a function of temperature only (Fig. 12–11).

(b) For an incompressible substance, v � constant and thus dv � 0. Alsofrom Eq. 12–49, cp � cv � c since a � b � 0 for incompressible substances.Then Eq. 12–29 reduces to

Again we need to show that the specific heat c depends on temperature onlyand not on pressure or specific volume. This is done with the help ofEq. 12–43:

since v � constant. Therefore, we conclude that the internal energy of atruly incompressible substance depends on temperature only.

a 0cp

0Pb

T

� �T a 02v0T 2 b

P

� 0

du � c dT

a 0cv

0vb

T

� 0

a 0P

0Tb

v�

R

vand a 02P

0T 2bv

� c 0 1R>v 20T

dv

� 0

EXAMPLE 12–9 The Specific Heat Difference of an Ideal Gas

Show that cp � cv � R for an ideal gas.

Solution It is to be shown that the specific heat difference for an ideal gasis equal to its gas constant.Analysis This relation is easily proved by showing that the right-hand sideof Eq. 12–46 is equivalent to the gas constant R of the ideal gas:

Substituting,

Therefore,cp � cv � R

�T a 0v0Tb 2

P

a 0P

0vb

T

� �T a R

Pb 2 a�

P

vb � R

v �RT

PS a 0v

0Tb 2

P

� a R

Pb 2

P �RT

vS a 0P

0vb

T

� �RT

v 2 �P

v

cp � cv � �T a 0v0Tb 2

P

a 0P

0vb

T

AIR

LAKE

u = u(T )u = u(T )cv = cv (T )cp = cp(T )

u = u(T )c = c(T )

FIGURE 12–11The internal energies and specificheats of ideal gases andincompressible substances depend ontemperature only.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 667

12–5 ■ THE JOULE-THOMSON COEFFICIENTWhen a fluid passes through a restriction such as a porous plug, a capillarytube, or an ordinary valve, its pressure decreases. As we have shown inChap. 5, the enthalpy of the fluid remains approximately constant duringsuch a throttling process. You will remember that a fluid may experience alarge drop in its temperature as a result of throttling, which forms the basisof operation for refrigerators and air conditioners. This is not always thecase, however. The temperature of the fluid may remain unchanged, or itmay even increase during a throttling process (Fig. 12–12).

The temperature behavior of a fluid during a throttling (h � constant)process is described by the Joule-Thomson coefficient, defined as

(12–51)

Thus the Joule-Thomson coefficient is a measure of the change in tempera-ture with pressure during a constant-enthalpy process. Notice that if

during a throttling process.A careful look at its defining equation reveals that the Joule-Thomson

coefficient represents the slope of h � constant lines on a T-P diagram.Such diagrams can be easily constructed from temperature and pressuremeasurements alone during throttling processes. A fluid at a fixed tempera-ture and pressure T1 and P1 (thus fixed enthalpy) is forced to flow through aporous plug, and its temperature and pressure downstream (T2 and P2) aremeasured. The experiment is repeated for different sizes of porous plugs,each giving a different set of T2 and P2. Plotting the temperatures againstthe pressures gives us an h � constant line on a T-P diagram, as shown inFig. 12–13. Repeating the experiment for different sets of inlet pressure andtemperature and plotting the results, we can construct a T-P diagram for asubstance with several h � constant lines, as shown in Fig. 12–14.

Some constant-enthalpy lines on the T-P diagram pass through a point ofzero slope or zero Joule-Thomson coefficient. The line that passes throughthese points is called the inversion line, and the temperature at a pointwhere a constant-enthalpy line intersects the inversion line is called theinversion temperature. The temperature at the intersection of the P � 0line (ordinate) and the upper part of the inversion line is called the maxi-mum inversion temperature. Notice that the slopes of the h � constantlines are negative (mJT � 0) at states to the right of the inversion line andpositive (mJT 0) to the left of the inversion line.

A throttling process proceeds along a constant-enthalpy line in the direc-tion of decreasing pressure, that is, from right to left. Therefore, the tempera-ture of a fluid increases during a throttling process that takes place on theright-hand side of the inversion line. However, the fluid temperaturedecreases during a throttling process that takes place on the left-hand side ofthe inversion line. It is clear from this diagram that a cooling effect cannotbe achieved by throttling unless the fluid is below its maximum inversion

m JT •6 0 temperature increases

� 0 temperature remains constant

7 0 temperature decreases

m � a 0T

0Pb

h

668 | Thermodynamics

T1 = 20°C T2 = 20°C><

P1 = 800 kPa P2 = 200 kPa

FIGURE 12–12The temperature of a fluid mayincrease, decrease, or remain constantduring a throttling process.

T

P

Exit states

Inletstate

h = constant line

22 2

2

2

1

P1

P2, T2(varied)

P1, T1(fixed)

FIGURE 12–13The development of an h � constantline on a P-T diagram.

Maximum inversiontemperature

Inversion line

h = const.

µJT > 0 µJT < 0

T

P

FIGURE 12–14Constant-enthalpy lines of a substanceon a T-P diagram.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 668

temperature. This presents a problem for substances whose maximum inver-sion temperature is well below room temperature. For hydrogen, for example,the maximum inversion temperature is �68°C. Thus hydrogen must becooled below this temperature if any further cooling is to be achieved bythrottling.

Next we would like to develop a general relation for the Joule-Thomsoncoefficient in terms of the specific heats, pressure, specific volume, andtemperature. This is easily accomplished by modifying the generalized rela-tion for enthalpy change (Eq. 12–35)

For an h � constant process we have dh � 0. Then this equation can berearranged to give

(12–52)

which is the desired relation. Thus, the Joule-Thomson coefficient can bedetermined from a knowledge of the constant-pressure specific heat and theP-v-T behavior of the substance. Of course, it is also possible to predict theconstant-pressure specific heat of a substance by using the Joule-Thomsoncoefficient, which is relatively easy to determine, together with the P-v-Tdata for the substance.

�1cp

cv � T a 0v0Tb

P

d � a 0T

0Pb

h

� m JT

dh � cp dT � cv � T a 0v0Tb

P

d dP

Chapter 12 | 669

EXAMPLE 12–10 Joule-Thomson Coefficient of an Ideal Gas

Show that the Joule-Thomson coefficient of an ideal gas is zero.

Solution It is to be shown that mJT � 0 for an ideal gas.Analysis For an ideal gas v � RT/P, and thus

Substituting this into Eq. 12–52 yields

Discussion This result is not surprising since the enthalpy of an ideal gas is afunction of temperature only, h � h(T), which requires that the temperatureremain constant when the enthalpy remains constant. Therefore, a throttlingprocess cannot be used to lower the temperature of an ideal gas (Fig. 12–15).

mJT ��1cpcv � T a 0v

0Tb

P

d ��1cpcv � T

R

Pd � �

1cp1v � v 2 � 0

a 0v0Tb

P

�R

P

12–6 ■ THE �h, �u, AND �s OF REAL GASESWe have mentioned many times that gases at low pressures behave as idealgases and obey the relation Pv � RT. The properties of ideal gases are rela-tively easy to evaluate since the properties u, h, cv, and cp depend on tem-perature only. At high pressures, however, gases deviate considerably fromideal-gas behavior, and it becomes necessary to account for this deviation.

T

P1 P2 P

h = constant line

FIGURE 12–15The temperature of an ideal gasremains constant during a throttlingprocess since h � constant and T �constant lines on a T-P diagramcoincide.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 669

In Chap. 3 we accounted for the deviation in properties P, v, and T by eitherusing more complex equations of state or evaluating the compressibility fac-tor Z from the compressibility charts. Now we extend the analysis to evalu-ate the changes in the enthalpy, internal energy, and entropy of nonideal(real) gases, using the general relations for du, dh, and ds developed earlier.

Enthalpy Changes of Real GasesThe enthalpy of a real gas, in general, depends on the pressure as well as onthe temperature. Thus the enthalpy change of a real gas during a process canbe evaluated from the general relation for dh (Eq. 12–36)

where P1, T1 and P2, T2 are the pressures and temperatures of the gas at theinitial and the final states, respectively. For an isothermal process dT � 0,and the first term vanishes. For a constant-pressure process, dP � 0, and thesecond term vanishes.

Properties are point functions, and thus the change in a property betweentwo specified states is the same no matter which process path is followed.This fact can be exploited to greatly simplify the integration of Eq. 12–36.Consider, for example, the process shown on a T-s diagram in Fig. 12–16.The enthalpy change during this process h2 � h1 can be determined by per-forming the integrations in Eq. 12–36 along a path that consists oftwo isothermal (T1 � constant and T2 � constant) lines and one isobaric(P0 � constant) line instead of the actual process path, as shown inFig. 12–16.

Although this approach increases the number of integrations, it also sim-plifies them since one property remains constant now during each part ofthe process. The pressure P0 can be chosen to be very low or zero, so thatthe gas can be treated as an ideal gas during the P0 � constant process.Using a superscript asterisk (*) to denote an ideal-gas state, we can expressthe enthalpy change of a real gas during process 1-2 as

(12–53)

where, from Eq. 12–36,

(12–54)

(12–55)

(12–56)

The difference between h and h* is called the enthalpy departure, and itrepresents the variation of the enthalpy of a gas with pressure at a fixedtemperature. The calculation of enthalpy departure requires a knowledge ofthe P-v-T behavior of the gas. In the absence of such data, we can use therelation Pv � ZRT, where Z is the compressibility factor. Substituting

h*1 � h1 � 0 � �P1*

P1

cv � T a 0v0Tb

P

dT�T1

dP � � �P1

P0

cv � T a 0v0Tb

P

dT�T1

dP

h*2 � h*1 � �T2

T1

cp dT � 0 � �T2

T1

cp0 1T 2 dT

h2 � h*2 � 0 � �P2

P*2

cv � T a 0v0Tb

P

dT�T2

dP � �P2

P0

cv � T a 0v0Tb

P

dT�T2

dP

h2 � h1 � 1h2 � h*2 2 � 1h*2 � h*1 2 � 1h*1 � h1 2

h2 � h1 � �T2

T1

cp dT � �P2

P1

cv � T a 0v0Tb

P

d dP

670 | Thermodynamics

T

s

Actualprocesspath

Alternativeprocesspath

T2

T11

1*

2*2

P 0 =

0P 2P 1

FIGURE 12–16An alternative process path to evaluatethe enthalpy changes of real gases.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 670

v � ZRT/P and simplifying Eq. 12–56, we can write the enthalpy departureat any temperature T and pressure P as

The above equation can be generalized by expressing it in terms of the reducedcoordinates, using T � TcrTR and P � PcrPR. After some manipulations, theenthalpy departure can be expressed in a nondimensionalized form as

(12–57)

where Zh is called the enthalpy departure factor. The integral in the aboveequation can be performed graphically or numerically by employing data fromthe compressibility charts for various values of PR and TR. The values of Zh arepresented in graphical form as a function of PR and TR in Fig. A–29. Thisgraph is called the generalized enthalpy departure chart, and it is used todetermine the deviation of the enthalpy of a gas at a given P and T from theenthalpy of an ideal gas at the same T. By replacing h* by hideal for clarity, Eq.12–53 for the enthalpy change of a gas during a process 1-2 can be rewritten as

(12–58)

or

(12–59)

where the values of Zh are determined from the generalized enthalpy depar-ture chart and (h

–2 � h

–1)ideal is determined from the ideal-gas tables. Notice

that the last terms on the right-hand side are zero for an ideal gas.

Internal Energy Changes of Real GasesThe internal energy change of a real gas is determined by relating it to theenthalpy change through the definition h

–� u– � Pv– � u– � ZRuT:

(12–60)

Entropy Changes of Real GasesThe entropy change of a real gas is determined by following an approachsimilar to that used above for the enthalpy change. There is some differencein derivation, however, owing to the dependence of the ideal-gas entropy onpressure as well as the temperature.

The general relation for ds was expressed as (Eq. 12–41)

where P1, T1 and P2, T2 are the pressures and temperatures of the gas at theinitial and the final states, respectively. The thought that comes to mind atthis point is to perform the integrations in the previous equation first along aT1 � constant line to zero pressure, then along the P � 0 line to T2, and

s2 � s1 � �T2

T1

cp

T dT � �

P2

P1

a 0v0Tb

P

dP

u�2 � u�1 � 1h2 � h1 2 � Ru 1Z 2 T2 � Z 1T1 2

h2 � h1 � 1h2 � h1 2 ideal � RTcr 1Zh2� Zh1

2

h2 � h1 � 1h2 � h1 2 ideal � RuTcr 1Zh2� Zh1

2

Zh �1h* � h 2T

RuTcr� T 2

R �PR

0 a 0Z

0TR

bPR

d 1ln PR 2

1h* � h 2T � �RT 2 �P

0

a 0Z

0Tb

P

dP

P

Chapter 12 | 671

cen84959_ch12.qxd 4/5/05 3:58 PM Page 671

finally along the T2 � constant line to P2, as we did for the enthalpy. Thisapproach is not suitable for entropy-change calculations, however, since itinvolves the value of entropy at zero pressure, which is infinity. We canavoid this difficulty by choosing a different (but more complex) pathbetween the two states, as shown in Fig. 12–17. Then the entropy changecan be expressed as

(12–61)

States 1 and 1* are identical (T1 � T1* and P1 � P1

*) and so are states 2and 2*. The gas is assumed to behave as an ideal gas at the imaginary states1* and 2* as well as at the states between the two. Therefore, the entropychange during process 1*-2* can be determined from the entropy-changerelations for ideal gases. The calculation of entropy change between anactual state and the corresponding imaginary ideal-gas state is moreinvolved, however, and requires the use of generalized entropy departurecharts, as explained below.

Consider a gas at a pressure P and temperature T. To determine how muchdifferent the entropy of this gas would be if it were an ideal gas at the sametemperature and pressure, we consider an isothermal process from the actualstate P, T to zero (or close to zero) pressure and back to the imaginary ideal-gas state P*, T* (denoted by superscript *), as shown in Fig. 12–17. Theentropy change during this isothermal process can be expressed as

where v � ZRT/P and v* � videal � RT/P. Performing the differentiationsand rearranging, we obtain

By substituting T � TcrTR and P � PcrPR and rearranging, the entropydeparture can be expressed in a nondimensionalized form as

(12–62)

The difference (s–* � s–)T,P is called the entropy departure and Zs is calledthe entropy departure factor. The integral in the above equation can beperformed by using data from the compressibility charts. The values of Zsare presented in graphical form as a function of PR and TR in Fig. A–30.This graph is called the generalized entropy departure chart, and it isused to determine the deviation of the entropy of a gas at a given P and Tfrom the entropy of an ideal gas at the same P and T. Replacing s* by sidealfor clarity, we can rewrite Eq. 12–61 for the entropy change of a gas duringa process 1-2 as

(12–63)s�2 � s�1 � 1 s�2 � s�1 2 ideal � Ru 1Zs2� Zs1

2

Zs �1 s�* � s� 2T,P

Ru

� �PR

0

cZ � 1 � TR a 0Z

0TR

bPR

d d 1ln PR 2

1sP � sP* 2T � �

P

0

c 11 � Z 2RP

�RT

Pa 0Zr

0Tb

P

d dP

� � �P

0

a 0v0Tb

P

dP � �0

P

a 0v*

0Tb

P

dP

1sP � sP* 2T � 1sP � s0* 2T � 1s 0* � sP

* 2T

s2 � s1 � 1s2 � sb*2 � 1sb

* � s2* 2 � 1s2* � s1* 2 � 1s1* � sa* 2 � 1sa

* � s1 2

672 | Thermodynamics

s

T

2*

b*

a*

1*

1

2

Alternativeprocess path

Actualprocess path

P2

P1

P 0

T2

T1

FIGURE 12–17An alternative process path to evaluatethe entropy changes of real gasesduring process 1-2.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 672

or

(12–64)

where the values of Zs are determined from the generalized entropy depar-ture chart and the entropy change (s2 � s1)ideal is determined from the ideal-gas relations for entropy change. Notice that the last terms on the right-handside are zero for an ideal gas.

EXAMPLE 12–11 The �h and �s of Oxygen at High Pressures

Determine the enthalpy change and the entropy change of oxygen per unitmole as it undergoes a change of state from 220 K and 5 MPa to 300 K and10 MPa (a) by assuming ideal-gas behavior and (b) by accounting for thedeviation from ideal-gas behavior.

Solution Oxygen undergoes a process between two specified states. Theenthalpy and entropy changes are to be determined by assuming ideal-gasbehavior and by accounting for the deviation from ideal-gas behavior.Analysis The critical temperature and pressure of oxygen are Tcr � 154.8 Kand Pcr � 5.08 MPa (Table A–1), respectively. The oxygen remains above itscritical temperature; therefore, it is in the gas phase, but its pressure isquite high. Therefore, the oxygen will deviate from ideal-gas behavior andshould be treated as a real gas.(a) If the O2 is assumed to behave as an ideal gas, its enthalpy will dependon temperature only, and the enthalpy values at the initial and the final tem-peratures can be determined from the ideal-gas table of O2 (Table A–19) atthe specified temperatures:

The entropy depends on both temperature and pressure even for ideal gases.Under the ideal-gas assumption, the entropy change of oxygen is determinedfrom

(b) The deviation from the ideal-gas behavior can be accounted for by deter-mining the enthalpy and entropy departures from the generalized charts ateach state:

TR1�

T1

Tcr�

220 K

154.8 K� 1.42

PR1�

P1

Pcr�

5 MPa

5.08 MPa� 0.98

∂ Zh1� 0.53, Zs1

� 0.25

� 3.28 kJ/kmol # K

� 1205.213 � 196.171 2 kJ>kmol # K � 18.314 kJ>kmol # K 2 ln 10 MPa

5 MPa

1 s�2 � s�1 2 ideal � s�°2 � s�°1 � Ru ln P2

P1

� 2332 kJ/kmol

� 18736 � 6404 2 kJ>kmol

1h2 � h1 2 ideal � h2,ideal � h1,ideal

s2 � s1 � 1s2 � s1 2 ideal � R 1Zs2� Zs1

2

Chapter 12 | 673

cen84959_ch12.qxd 4/5/05 3:58 PM Page 673

674 | Thermodynamics

and

Then the enthalpy and entropy changes of oxygen during this process aredetermined by substituting the values above into Eqs. 12–58 and 12–63,

and

Discussion Note that the ideal-gas assumption would underestimate theenthalpy change of the oxygen by 2.7 percent and the entropy change by11.4 percent.

� 3.70 kJ/kmol # K

� 3.28 kJ>kmol # K � 18.314 kJ>kmol # K 2 10.20 � 0.25 2 s�2 � s�1 � 1 s�2 � s�1 2 ideal � Ru 1Zs2

� Zs12

� 2396 kJ/kmol

� 2332 kJ>kmol � 18.314 kJ>kmol # K 2 3154.8 K 10.48 � 0.53 2 4 h2 � h1 � 1h2 � h1 2 ideal � RuTcr 1Zh2

� Zh12

TR2�

T2

Tcr�

300 K

154.8 K� 1.94

PR2�

P2

Pcr�

10 MPa

5.08 MPa� 1.97

∂ Zh2� 0.48, Zs2

� 0.20

Some thermodynamic properties can be measured directly, butmany others cannot. Therefore, it is necessary to developsome relations between these two groups so that the propertiesthat cannot be measured directly can be evaluated. The deriva-tions are based on the fact that properties are point functions,and the state of a simple, compressible system is completelyspecified by any two independent, intensive properties.

The equations that relate the partial derivatives of proper-ties P, v, T, and s of a simple compressible substance to eachother are called the Maxwell relations. They are obtainedfrom the four Gibbs equations, expressed as

dg � �s dT � v dP

da � �s dT � P dv

dh � T ds � v dP

du � T ds � P dv

SUMMARY

The Maxwell relations are

The Clapeyron equation enables us to determine the enthalpychange associated with a phase change from a knowledge ofP, v, and T data alone. It is expressed as

a dP

dTb

sat�

hfg

T vfg

a 0s

0Pb

T

� � a 0v0Tb

P

a 0s

0vb

T

� a 0P

0Tb

v

a 0T

0Pb

s

� a 0v0sb

P

a 0T

0vb

s

� � a 0P

0sb

v

cen84959_ch12.qxd 4/5/05 3:58 PM Page 674

Chapter 12 | 675

For liquid–vapor and solid–vapor phase-change processes atlow pressures, it can be approximated as

The changes in internal energy, enthalpy, and entropy of asimple compressible substance can be expressed in terms ofpressure, specific volume, temperature, and specific heatsalone as

or

For specific heats, we have the following general relations:

cp � cv � �T a 0v0Tb 2

P

a 0P

0vb

T

cp,T � cp0,T � �T �P

0

a 02v0T 2 b

P

dP

a 0cp

0Pb

T

� �T a 02v0T 2 b

P

a 0cv

0vb

T

� T a 02P

0T 2 bv

ds �cp

T dT � a 0v

0Tb

P

dP

ds �cv

T dT � a 0P

0Tb

v dv

dh � cp dT � cv � T a 0v0Tb

P

d dP

du � cv dT � cT a 0P

0Tb

v� P d dv

ln a P2

P1b

sat�

hfg

Ra T2 � T1

T1T2b

sat

where b is the volume expansivity and a is the isothermalcompressibility, defined as

The difference cp � cv is equal to R for ideal gases and tozero for incompressible substances.

The temperature behavior of a fluid during a throttling (h �constant) process is described by the Joule-Thomson coefficient,defined as

The Joule-Thomson coefficient is a measure of the change intemperature of a substance with pressure during a constant-enthalpy process, and it can also be expressed as

The enthalpy, internal energy, and entropy changes of real gasescan be determined accurately by utilizing generalized enthalpyor entropy departure charts to account for the deviation fromthe ideal-gas behavior by using the following relations:

where the values of Zh and Zs are determined from the gener-alized charts.

s�2 � s�1 � 1 s�2 � s�1 2 ideal � Ru 1Zs2� Zs1

2 u�2 � u�1 � 1h2 � h1 2 � Ru 1Z2T2 � Z1T1 2 h2 � h1 � 1h2 � h1 2 ideal � RuTcr 1Zh2

� Zh12

mJT � �1cp

cv � T a 0v0Tb

P

d

mJT � a 0T

0Pb

h

b �1

v a 0v

0Tb

P

and a � �1

v a 0v

0Pb

T

cp � cv �vTb2

a

1. G. J. Van Wylen and R. E. Sonntag. Fundamentals ofClassical Thermodynamics. 3rd ed. New York: John Wiley & Sons, 1985.

REFERENCES AND SUGGESTED READINGS

2. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

Partial Derivatives and Associated Relations

12–1C Consider the function z(x, y). Plot a differential sur-face on x-y-z coordinates and indicate �x, dx, �y, dy, (�z)x,(�z)y, and dz.

12–2C What is the difference between partial differentialsand ordinary differentials?

PROBLEMS*

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 675

12–3C Consider the function z(x, y), its partial derivatives(�z/�x)y and (�z/�y)x, and the total derivative dz/dx.(a) How do the magnitudes (�x)y and dx compare?(b) How do the magnitudes (�z)y and dz compare?(c) Is there any relation among dz, (�z)x, and (�z)y?

12–4C Consider a function z(x, y) and its partial derivative(�z/�y)x. Under what conditions is this partial derivative equalto the total derivative dz/dy?

12–5C Consider a function z(x, y) and its partial derivative(�z/�y)x. If this partial derivative is equal to zero for all valuesof x, what does it indicate?

12–6C Consider a function z(x, y) and its partial derivative(�z/�y)x. Can this partial derivative still be a function of x?

12–7C Consider a function f(x) and its derivative df/dx.Can this derivative be determined by evaluating dx/df andtaking its inverse?

12–8 Consider air at 400 K and 0.90 m3/kg. Using Eq. 12–3,determine the change in pressure corresponding to an increaseof (a) 1 percent in temperature at constant specific volume,(b) 1 percent in specific volume at constant temperature, and(c) 1 percent in both the temperature and specific volume.

12–9 Repeat Problem 12–8 for helium.

12–10 Prove for an ideal gas that (a) the P � constant lineson a T-v diagram are straight lines and (b) the high-pressurelines are steeper than the low-pressure lines.

12–11 Derive a relation for the slope of the v � constantlines on a T-P diagram for a gas that obeys the van der Waalsequation of state. Answer: (v � b)/R

12–12 Nitrogen gas at 400 K and 300 kPa behaves as anideal gas. Estimate the cp and cv of the nitrogen at this state,using enthalpy and internal energy data from Table A–18, andcompare them to the values listed in Table A–2b.

12–13E Nitrogen gas at 600 R and 30 psia behaves as anideal gas. Estimate the cp and cv of the nitrogen at this state,using enthalpy and internal energy data from Table A–18E,and compare them to the values listed in Table A–2Eb.Answers: 0.249 Btu/lbm · R, 0.178 Btu/lbm · R

12–14 Consider an ideal gas at 400 K and 100 kPa. As aresult of some disturbance, the conditions of the gas changeto 404 K and 96 kPa. Estimate the change in the specific vol-ume of the gas using (a) Eq. 12–3 and (b) the ideal-gas rela-tion at each state.

12–15 Using the equation of state P(v � a) � RT, verify(a) the cyclic relation and (b) the reciprocity relation at constant v.

The Maxwell Relations

12–16 Verify the validity of the last Maxwell relation(Eq. 12–19) for refrigerant-134a at 80°C and 1.2 MPa.

676 | Thermodynamics

12–17 Reconsider Prob. 12–16. Using EES (or other)software, verify the validity of the last Maxwell

relation for refrigerant-134a at the specified state.

12–18E Verify the validity of the last Maxwell relation(Eq. 12–19) for steam at 800°F and 400 psia.

12–19 Using the Maxwell relations, determine a relation for(�s/�P)T for a gas whose equation of state is P(v � b) � RT.Answer: �R/P

12–20 Using the Maxwell relations, determine a relationfor (�s/�v)T for a gas whose equation of state is (P � a/v2)(v � b) � RT.

12–21 Using the Maxwell relations and the ideal-gas equa-tion of state, determine a relation for (�s/�v)T for an idealgas. Answer: R/v

The Clapeyron Equation

12–22C What is the value of the Clapeyron equation inthermodynamics?

12–23C Does the Clapeyron equation involve any approxi-mations, or is it exact?

12–24C What approximations are involved in the Clapeyron-Clausius equation?

12–25 Using the Clapeyron equation, estimate the enthalpyof vaporization of refrigerant-134a at 40°C, and compare it tothe tabulated value.

12–26 Reconsider Prob. 12–25. Using EES (or other)software, plot the enthalpy of vaporization of

refrigerant-134a as a function of temperature over the tempera-ture range �20 to 80°C by using the Clapeyron equation andthe refrigerant-134a data in EES. Discuss your results.

12–27 Using the Clapeyron equation, estimate the enthalpyof vaporization of steam at 300 kPa, and compare it to thetabulated value.

12–28 Calculate the hfg and sfg of steam at 120°C fromthe Clapeyron equation, and compare them to the tabulatedvalues.

12–29E Determine the hfg of refrigerant-134a at 50°Fon the basis of (a) the Clapeyron equation

and (b) the Clapeyron-Clausius equation. Compare yourresults to the tabulated hfg value.

12–30 Plot the enthalpy of vaporization of steam as afunction of temperature over the temperature

range 10 to 200°C by using the Clapeyron equation andsteam data in EES.

12–31 Using the Clapeyron-Clausius equation and the triple-point data of water, estimate the sublimation pressure ofwater at �30°C and compare to the value in Table A–8.

cen84959_ch12.qxd 4/5/05 3:58 PM Page 676

Chapter 12 | 677

General Relations for du, dh, ds, cv, and cp

12–32C Can the variation of specific heat cp with pressureat a given temperature be determined from a knowledge of P-v-T data alone?

12–33 Show that the enthalpy of an ideal gas is a functionof temperature only and that for an incompressible substanceit also depends on pressure.

12–34 Derive expressions for (a) �u, (b) �h, and (c) �s fora gas that obeys the van der Waals equation of state for anisothermal process.

12–35 Derive expressions for (a) �u, (b) �h, and (c) �s fora gas whose equation of state is P(v � a) � RT for an isother-mal process. Answers: (a) 0, (b) a(P2 � P1), (c) �R ln (P2/P1)

12–36 Derive expressions for (�u/�P)T and (�h/�v)T interms of P, v, and T only.

12–37 Derive an expression for the specific-heat differencecp � cv for (a) an ideal gas, (b) a van der Waals gas, and(c) an incompressible substance.

12–38 Estimate the specific-heat difference cp � cv for liquidwater at 15 MPa and 80°C. Answer: 0.32 kJ/kg · K

12–39E Estimate the specific-heat difference cp � cv for liq-uid water at 1000 psia and 150°F. Answer: 0.057 Btu/lbm · R

12–40 Derive a relation for the volume expansivity b andthe isothermal compressibility a (a) for an ideal gas and(b) for a gas whose equation of state is P(v � a) � RT.

12–41 Estimate the volume expansivity b and the isothermalcompressibility a of refrigerant-134a at 200 kPa and 30°C.

The Joule-Thomson Coefficient

12–42C What does the Joule-Thomson coefficient represent?

12–43C Describe the inversion line and the maximuminversion temperature.

12–44C The pressure of a fluid always decreases duringan adiabatic throttling process. Is this also the case for thetemperature?

12–45C Does the Joule-Thomson coefficient of a substancechange with temperature at a fixed pressure?

12–46C Will the temperature of helium change if it is throt-tled adiabatically from 300 K and 600 kPa to 150 kPa?

12–47 Consider a gas whose equation of state is P(v � a) �RT, where a is a positive constant. Is it possible to cool thisgas by throttling?

12–48 Derive a relation for the Joule-Thomson coefficientand the inversion temperature for a gas whose equation ofstate is (P � a/v2)v � RT.

12–49 Estimate the Joule-Thomson coefficient of steam at(a) 3 MPa and 300°C and (b) 6 MPa and 500°C.

12–50E Estimate the Joule-Thomson coefficient ofnitrogen at (a) 200 psia and 500 R and

(b) 2000 psia and 400 R. Use nitrogen properties from EESor other source.

12–51E Reconsider Prob. 12–50E. Using EES (orother) software, plot the Joule-Thomson coef-

ficient for nitrogen over the pressure range 100 to 1500 psiaat the enthalpy values 100, 175, and 225 Btu/lbm. Discussthe results.

12–52 Estimate the Joule-Thomson coefficient of refriger-ant-134a at 0.7 MPa and 50°C.

12–53 Steam is throttled slightly from 1 MPa and 300°C.Will the temperature of the steam increase, decrease, orremain the same during this process?

The dh, du, and ds of Real Gases

12–54C What is the enthalpy departure?

12–55C On the generalized enthalpy departure chart, thenormalized enthalpy departure values seem to approach zeroas the reduced pressure PR approaches zero. How do youexplain this behavior?

12–56C Why is the generalized enthalpy departure chart pre-pared by using PR and TR as the parameters instead of P and T ?

12–57 Determine the enthalpy of nitrogen, in kJ/kg, at175 K and 8 MPa using (a) data from the ideal-gas nitrogentable and (b) the generalized enthalpy departure chart. Com-pare your results to the actual value of 125.5 kJ/kg. Answers:(a) 181.5 kJ/kg, (b) 121.6 kJ/kg

12–58E Determine the enthalpy of nitrogen, in Btu/lbm, at400 R and 2000 psia using (a) data from the ideal-gas nitro-gen table and (b) the generalized enthalpy chart. Compareyour results to the actual value of 177.8 Btu/lbm.

12–59 What is the error involved in the (a) enthalpy and(b) internal energy of CO2 at 350 K and 10 MPa if it isassumed to be an ideal gas? Answers: (a) 50%, (b) 49%

12–60 Determine the enthalpy change and the entropychange of nitrogen per unit mole as it undergoes a change ofstate from 225 K and 6 MPa to 320 K and 12 MPa, (a) byassuming ideal-gas behavior and (b) by accounting for thedeviation from ideal-gas behavior through the use of general-ized charts.

12–61 Determine the enthalpy change and the entropychange of CO2 per unit mass as it undergoes a change of statefrom 250 K and 7 MPa to 280 K and 12 MPa, (a) by assum-ing ideal-gas behavior and (b) by accounting for the deviationfrom ideal-gas behavior.

12–62 Methane is compressed adiabatically by a steady-flowcompressor from 2 MPa and �10°C to 10 MPa and 110°C at arate of 0.55 kg/s. Using the generalized charts, determine therequired power input to the compressor. Answer: 133 kW

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12–63 Propane is compressed isothermally by a piston–cylinder device from 100°C and 1 MPa to 4

MPa. Using the generalized charts, determine the work doneand the heat transfer per unit mass of propane.

12–64 Reconsider Prob. 12–63. Using EES (or other)software, extend the problem to compare the

solutions based on the ideal-gas assumption, generalized chartdata, and real fluid data. Also extend the solution to methane.

12–65E Propane is compressed isothermally by a piston–cylinder device from 200°F and 200 psia to 800 psia. Usingthe generalized charts, determine the work done and the heattransfer per unit mass of the propane.Answers: 45.3 Btu/lbm, 141 Btu/lbm

12–66 Determine the exergy destruction associated with theprocess described in Prob. 12–63. Assume T0 � 30°C.

12–67 Carbon dioxide enters an adiabatic nozzle at 8 MPaand 450 K with a low velocity and leaves at 2 MPa and 350 K.Using the generalized enthalpy departure chart, determine theexit velocity of the carbon dioxide. Answer: 384 m/s

12–68 Reconsider Prob. 12–67. Using EES (or other)software, compare the exit velocity to the noz-

zle assuming ideal-gas behavior, the generalized chart data,and EES data for carbon dioxide.

12–69 A 0.08-m3 well-insulated rigid tank contains oxygenat 220 K and 10 MPa. A paddle wheel placed in the tank isturned on, and the temperature of the oxygen rises to 250 K.Using the generalized charts, determine (a) the final pressurein the tank and (b) the paddle-wheel work done during thisprocess. Answers: (a) 12,190 kPa, (b) 393 kJ

12–70 Carbon dioxide is contained in a constant-volume tankand is heated from 100°C and 1 MPa to 8 MPa. Determine theheat transfer and entropy change per unit mass of the carbondioxide using (a) the ideal-gas assumption, (b) the generalizedcharts, and (c) real fluid data from EES or other sources.

Review Problems

12–71 For b 0, prove that at every point of a single-phase region of an h-s diagram, the slope of a constant-pressure (P � constant) line is greater than the slope of a

678 | Thermodynamics

constant-temperature (T � constant) line, but less than theslope of a constant-volume (v � constant) line.

12–72 Using the cyclic relation and the first Maxwell rela-tion, derive the other three Maxwell relations.

12–73 Starting with the relation dh � T ds + v dP, showthat the slope of a constant-pressure line on an h-s diagram(a) is constant in the saturation region and (b) increases withtemperature in the superheated region.

12–74 Derive relations for (a) �u, (b) �h, and (c) �s of agas that obeys the equation of state (P + a/v 2)v � RT for anisothermal process.

12–75 Show that

12–76 Estimate the cp of nitrogen at 300 kPa and 400 K,using (a) the relation in the above problem and (b) its defini-tion. Compare your results to the value listed in Table A–2b.

12–77 Steam is throttled from 4.5 MPa and 300°C to 2.5MPa. Estimate the temperature change of the steam duringthis process and the average Joule-Thomson coefficient.Answers: �26.3°C, 13.1°C/MPa

12–78 A rigid tank contains 1.2 m3 of argon at �100°C and1 MPa. Heat is now transferred to argon until the temperaturein the tank rises to 0°C. Using the generalized charts, deter-mine (a) the mass of the argon in the tank, (b) the final pres-sure, and (c) the heat transfer.Answers: (a) 35.1 kg, (b) 1531 kPa, (c) 1251 kJ

12–79 Argon gas enters a turbine at 7 MPa and 600 K witha velocity of 100 m/s and leaves at 1 MPa and 280 K with avelocity of 150 m/s at a rate of 5 kg/s. Heat is being lost tothe surroundings at 25°C at a rate of 60 kW. Using the gener-alized charts, determine (a) the power output of the turbineand (b) the exergy destruction associated with the process.

cv � �T a 0v0Tb

s

a 0P

0Tb

vand cp � T a 0P

0Tb

s

a 0v0Tb

P

10 MPa110°C

2 MPa–10°C

CH4

m· = 0.55 kg/s

FIGURE P12–62

60 kW

W·

7 MPa600 K

100 m/s

1 MPa280 K

150 m/s

T0 = 25°C

Arm· = 5 kg/s

FIGURE P12–79

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Chapter 12 | 679

12–80 Reconsider Prob. 12–79. Using EES (or other)software, solve the problem assuming steam is

the working fluid by using the generalized chart method andEES data for steam. Plot the power output and the exergydestruction rate for these two calculation methods against theturbine exit pressure as it varies over the range 0.1 to 1 MPawhen the turbine exit temperature is 455 K.

12–81E Argon gas enters a turbine at 1000 psia and 1000 Rwith a velocity of 300 ft/s and leaves at 150 psia and 500 Rwith a velocity of 450 ft/s at a rate of 12 lbm/s. Heat is beinglost to the surroundings at 75°F at a rate of 80 Btu/s. Usingthe generalized charts, determine (a) the power output of theturbine and (b) the exergy destruction associated with theprocess. Answers: (a) 922 hp, (b) 121.5 Btu/s

12–82 An adiabatic 0.2-m3 storage tank that is initiallyevacuated is connected to a supply line that carries nitrogenat 225 K and 10 MPa. A valve is opened, and nitrogen flowsinto the tank from the supply line. The valve is closed whenthe pressure in the tank reaches 10 MPa. Determine the finaltemperature in the tank (a) treating nitrogen as an ideal gasand (b) using generalized charts. Compare your results to theactual value of 293 K.

12–85 The volume expansivity of water at 20°C is b �0.207 � 10�6 K�1. Treating this value as a constant, deter-mine the change in volume of 1 m3 of water as it is heatedfrom 10°C to 30°C at constant pressure.

12–86 The volume expansivity b values of copper at 300 Kand 500 K are 49.2 � 10�6 K�1 and 54.2 � 10�6 K�1,respectively, and b varies almost linearly in this temperaturerange. Determine the percent change in the volume of a cop-per block as it is heated from 300 K to 500 K at atmosphericpressure.

12–87 Starting with mJT � (1/cp) [T(�v/�T )p � v] and not-ing that Pv � ZRT, where Z � Z(P, T ) is the compressibilityfactor, show that the position of the Joule-Thomson coeffi-cient inversion curve on the T-P plane is given by the equa-tion (�Z/�T)P � 0.

12–88 Consider an infinitesimal reversible adiabatic com-pression or expansion process. By taking s � s(P, v) andusing the Maxwell relations, show that for this process Pv k �constant, where k is the isentropic expansion exponentdefined as

Also, show that the isentropic expansion exponent k reducesto the specific heat ratio cp /cv for an ideal gas.

12–89 Refrigerant-134a undergoes an isothermal pro-cess at 60°C from 3 to 0.1 MPa in a closed sys-

tem. Determine the work done by the refrigerant-134a byusing the tabular (EES) data and the generalized charts, inkJ/kg.

12–90 Methane is contained in a piston–cylinder device andis heated at constant pressure of 4 MPa from 100 to 350°C.Determine the heat transfer, work and entropy change perunit mass of the methane using (a) the ideal-gas assumption,(b) the generalized charts, and (c) real fluid data from EES orother sources.

Fundamentals of Engineering (FE) Exam Problems

12–91 A substance whose Joule-Thomson coefficient isnegative is throttled to a lower pressure. During this process,(select the correct statement)(a) the temperature of the substance will increase.(b) the temperature of the substance will decrease.(c) the entropy of the substance will remain constant.(d) the entropy of the substance will decrease.(e) the enthalpy of the substance will decrease.

12–92 Consider the liquid–vapor saturation curve of a puresubstance on the P-T diagram. The magnitude of the slope ofthe tangent line to this curve at a temperature T (in Kelvin) is

k �vPa 0P

0vb

s

0.2 m3

Initiallyevacuated

N2

225 K10 MPa

FIGURE P12–82

12–83 For a homogeneous (single-phase) simple pure sub-stance, the pressure and temperature are independent proper-ties, and any property can be expressed as a function of thesetwo properties. Taking v � v(P, T ), show that the change inspecific volume can be expressed in terms of the volumeexpansivity b and isothermal compressibility a as

Also, assuming constant average values for b and a, obtain arelation for the ratio of the specific volumes v2/v1 as a homo-geneous system undergoes a process from state 1 to state 2.

12–84 Repeat Prob. 12–83 for an isobaric process.

dvv

� b dT � a dP

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(a) proportional to the enthalpy of vaporization hfg at thattemperature.

(b) proportional to the temperature T.(c) proportional to the square of the temperature T.(d) proportional to the volume change vfg at that temperature.(e) inversely proportional to the entropy change sfg at that

temperature.

12–93 Based on the generalized charts, the error involvedin the enthalpy of CO2 at 350 K and 8 MPa if it is assumed tobe an ideal gas is

(a) 0 (b) 20% (c) 35% (d ) 26% (e) 65%

12–94 Based on data from the refrigerant-134a tables, theJoule-Thompson coefficient of refrigerant-134a at 0.8 MPaand 100°C is approximately

(a) 0 (b) �5°C/MPa (c) 11°C/MPa(d ) 8°C/MPa (e) 26°C/MPa

12–95 For a gas whose equation of state is P(v � b) � RT,the specified heat difference cp � cv is equal to

(a) R (b) R � b (c) R � b (d ) 0 (e) R(1 + v/b)

Design and Essay Problems

12–96 Consider the function z � z(x, y). Write an essay onthe physical interpretation of the ordinary derivative dz /dxand the partial derivative (�z /�x)y. Explain how these twoderivatives are related to each other and when they becomeequivalent.

12–97 There have been several attempts to represent the ther-modynamic relations geometrically, the best known of these

680 | Thermodynamics

being Koenig’s thermodynamic square shown in the figure.There is a systematic way of obtaining the four Maxwell rela-tions as well as the four relations for du, dh, dg, and da fromthis figure. By comparing these relations to Koenig’s diagram,come up with the rules to obtain these eight thermodynamicrelations from this diagram.

12–98 Several attempts have been made to express the par-tial derivatives of the most common thermodynamic prop-erties in a compact and systematic manner in terms ofmeasurable properties. The work of P. W. Bridgman is per-haps the most fruitful of all, and it resulted in the well-knownBridgman’s table. The 28 entries in that table are sufficient toexpress the partial derivatives of the eight common propertiesP, T, v, s, u, h, f, and g in terms of the six properties P, v, T,cp, b, and a, which can be measured directly or indirectlywith relative ease. Obtain a copy of Bridgman’s table andexplain, with examples, how it is used.

a T

g

Phs

u

v

FIGURE P12–97

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Chapter 13GAS MIXTURES

| 681

Up to this point, we have limited our consideration tothermodynamic systems that involve a single puresubstance such as water. Many important thermody-

namic applications, however, involve mixtures of several puresubstances rather than a single pure substance. Therefore, itis important to develop an understanding of mixtures andlearn how to handle them.

In this chapter, we deal with nonreacting gas mixtures.A nonreacting gas mixture can be treated as a pure sub-stance since it is usually a homogeneous mixture of differentgases. The properties of a gas mixture obviously depend onthe properties of the individual gases (called components orconstituents) as well as on the amount of each gas in themixture. Therefore, it is possible to prepare tables of proper-ties for mixtures. This has been done for common mixturessuch as air. It is not practical to prepare property tables forevery conceivable mixture composition, however, since thenumber of possible compositions is endless. Therefore, weneed to develop rules for determining mixture properties froma knowledge of mixture composition and the properties of theindividual components. We do this first for ideal-gas mixturesand then for real-gas mixtures. The basic principles involvedare also applicable to liquid or solid mixtures, called solutions.

ObjectivesThe objectives of Chapter 13 are to:

• Develop rules for determining nonreacting gas mixtureproperties from knowledge of mixture composition and theproperties of the individual components.

• Define the quantities used to describe the composition of amixture, such as mass fraction, mole fraction, and volumefraction.

• Apply the rules for determining mixture properties to ideal-gas mixtures and real-gas mixtures.

• Predict the P-v-T behavior of gas mixtures based onDalton’s law of additive pressures and Amagat’s law ofadditive volumes.

• Perform energy and exergy analysis of mixing processes.

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13–1 � COMPOSITION OF A GAS MIXTURE: MASS AND MOLE FRACTIONS

To determine the properties of a mixture, we need to know the compositionof the mixture as well as the properties of the individual components. Thereare two ways to describe the composition of a mixture: either by specifyingthe number of moles of each component, called molar analysis, or by spec-ifying the mass of each component, called gravimetric analysis.

Consider a gas mixture composed of k components. The mass of the mix-ture mm is the sum of the masses of the individual components, and themole number of the mixture Nm is the sum of the mole numbers of the indi-vidual components* (Figs. 13–1 and 13–2). That is,

(13–1a, b)

The ratio of the mass of a component to the mass of the mixture is calledthe mass fraction mf, and the ratio of the mole number of a component tothe mole number of the mixture is called the mole fraction y:

(13–2a, b)

Dividing Eq. 13–1a by mm or Eq. 13–1b by Nm, we can easily show thatthe sum of the mass fractions or mole fractions for a mixture is equal to 1(Fig. 13–3):

The mass of a substance can be expressed in terms of the mole number Nand molar mass M of the substance as m � NM. Then the apparent (oraverage) molar mass and the gas constant of a mixture can be expressed as

(13–3a, b)

The molar mass of a mixture can also be expressed as

(13–4)

Mass and mole fractions of a mixture are related by

(13–5)mfi �mi

mm�

Ni Mi

Nm Mm

� yi Mi

Mm

Mm �mm

Nm

�mm

a mi >Mi

�1

a mi > 1mm Mi 2 �1

ak

i�1

mfi

Mi

Mm �mm

Nm

�a mi

Nm

�a Ni Mi

Nm

�ak

i�1yi Mi and Rm �

Ru

Mm

ak

i�1mfi � 1 and a

k

i�1yi � 1

mfi �mi

mmand yi �

Ni

Nm

mm �ak

i�1mi and Nm �a

k

i�1Ni

682 | Thermodynamics

H2

6 kg

O2

32 kg

H2 + O2

38 kg+

FIGURE 13–1The mass of a mixture is equal to thesum of the masses of its components.

H2

3 kmol

O2

1 kmol

H2 + O2

4 kmol+

FIGURE 13–2The number of moles of a nonreactingmixture is equal to the sum of thenumber of moles of its components.

H2 + O2

yH2 =

yO2 =

0.750.251.00

FIGURE 13–3The sum of the mole fractions of amixture is equal to 1.

*Throughout this chapter, the subscript m denotes the gas mixture and the subscript idenotes any single component of the mixture.

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Chapter 13 | 683

EXAMPLE 13–1 Mass and Mole Fractions of a Gas Mixture

Consider a gas mixture that consists of 3 kg of O2, 5 kg of N2, and 12 kg ofCH4, as shown in Fig. 13–4. Determine (a) the mass fraction of each compo-nent, (b) the mole fraction of each component, and (c) the average molarmass and gas constant of the mixture.

Solution The masses of components of a gas mixture are given. The massfractions, the mole fractions, the molar mass, and the gas constant of themixture are to be determined.Analysis (a) The total mass of the mixture is

Then the mass fraction of each component becomes

(b) To find the mole fractions, we need to determine the mole numbers ofeach component first:

Thus,

and

(c) The average molar mass and gas constant of the mixture are determinedfrom their definitions,

Mm �mm

Nm

�20 kg

1.023 kmol� 19.6 kg/kmol

yCH4�

NCH4

Nm

�0.750 kmol

1.023 kmol� 0.733

yN2�

NN2

Nm

�0.179 kmol

1.023 kmol� 0.175

yO2�

NO2

Nm

�0.094 kmol

1.023 kmol� 0.092

Nm � NO2� NN2

� NCH4� 0.094 � 0.179 � 0.750 � 1.023 kmol

NCH4�

mCH4

MCH4

�12 kg

16 kg>kmol� 0.750 kmol

NN2�

mN2

MN2

�5 kg

28 kg>kmol� 0.179 kmol

NO2�

mO2

MO2

�3 kg

32 kg>kmol� 0.094 kmol

mfCH4�

mCH4

mm�

12 kg

20 kg� 0.60

mfN2�

mN2

mm�

5 kg

20 kg� 0.25

mfO2�

mO2

mm�

3 kg

20 kg� 0.15

mm � mO2� mN2

� mCH4� 3 � 5 � 12 � 20 kg

3 kg O2

5 kg N2

12 kg CH4

FIGURE 13–4Schematic for Example 13–1.

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13–2 � P-v-T BEHAVIOR OF GAS MIXTURES: IDEAL AND REAL GASES

An ideal gas is defined as a gas whose molecules are spaced far apart sothat the behavior of a molecule is not influenced by the presence of othermolecules—a situation encountered at low densities. We also mentionedthat real gases approximate this behavior closely when they are at a lowpressure or high temperature relative to their critical-point values. The P-v-Tbehavior of an ideal gas is expressed by the simple relation Pv � RT, whichis called the ideal-gas equation of state. The P-v-T behavior of real gases isexpressed by more complex equations of state or by Pv � ZRT, where Z isthe compressibility factor.

When two or more ideal gases are mixed, the behavior of a molecule nor-mally is not influenced by the presence of other similar or dissimilar mole-cules, and therefore a nonreacting mixture of ideal gases also behaves as anideal gas. Air, for example, is conveniently treated as an ideal gas in therange where nitrogen and oxygen behave as ideal gases. When a gas mix-ture consists of real (nonideal) gases, however, the prediction of the P-v-Tbehavior of the mixture becomes rather involved.

The prediction of the P-v-T behavior of gas mixtures is usually based ontwo models: Dalton’s law of additive pressures and Amagat’s law of addi-tive volumes. Both models are described and discussed below.

Dalton’s law of additive pressures: The pressure of a gas mixture is equalto the sum of the pressures each gas would exert if it existed alone at themixture temperature and volume (Fig. 13–5).

Amagat’s law of additive volumes: The volume of a gas mixture is equalto the sum of the volumes each gas would occupy if it existed alone at themixture temperature and pressure (Fig. 13–6).

Dalton’s and Amagat’s laws hold exactly for ideal-gas mixtures, but onlyapproximately for real-gas mixtures. This is due to intermolecular forcesthat may be significant for real gases at high densities. For ideal gases, thesetwo laws are identical and give identical results.

684 | Thermodynamics

or

Also,

Discussion When mass fractions are available, the molar mass and molefractions could also be determined directly from Eqs. 13–4 and 13–5.

Rm �Ru

Mm

�8.314 kJ> 1kmol # K 2

19.6 kg>kmol� 0.424 kJ/kg # K

� 19.6 kg>kmol

� 10.092 2 132 2 � 10.175 2 128 2 � 10.733 2 116 2 Mm �a yi Mi � yO2

MO2� yN2

MN2� yCH4

MCH4

+

GasmixtureA + BV, T

PA + PB

Gas BV, T

PB

Gas AV, T

PA

FIGURE 13–5Dalton’s law of additive pressures fora mixture of two ideal gases.

+

Gas mixtureA + BP , T

VA + VB

Gas A P, T

VA

Gas B P, T

VB

FIGURE 13–6Amagat’s law of additive volumes fora mixture of two ideal gases.

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Dalton’s and Amagat’s laws can be expressed as follows:

Dalton’s law: (13–6)

Amagat’s law: (13–7)

In these relations, Pi is called the component pressure and Vi is called thecomponent volume (Fig. 13–7). Note that Vi is the volume a componentwould occupy if it existed alone at Tm and Pm, not the actual volume occu-pied by the component in the mixture. (In a vessel that holds a gas mix-ture, each component fills the entire volume of the vessel. Therefore, thevolume of each component is equal to the volume of the vessel.) Also, theratio Pi /Pm is called the pressure fraction and the ratio Vi /Vm is calledthe volume fraction of component i.

Ideal-Gas MixturesFor ideal gases, Pi and Vi can be related to yi by using the ideal-gas relationfor both the components and the gas mixture:

Therefore,

(13–8)

Equation 13–8 is strictly valid for ideal-gas mixtures since it is derived byassuming ideal-gas behavior for the gas mixture and each of its components.The quantity yi Pm is called the partial pressure (identical to the componentpressure for ideal gases), and the quantity yiVm is called the partial volume(identical to the component volume for ideal gases). Note that for an ideal-gasmixture, the mole fraction, the pressure fraction, and the volume fraction of acomponent are identical.

The composition of an ideal-gas mixture (such as the exhaust gases leav-ing a combustion chamber) is frequently determined by a volumetric analy-sis (called the Orsat Analysis) and Eq. 13–8. A sample gas at a knownvolume, pressure, and temperature is passed into a vessel containingreagents that absorb one of the gases. The volume of the remaining gas isthen measured at the original pressure and temperature. The ratio of thereduction in volume to the original volume (volume fraction) represents themole fraction of that particular gas.

Real-Gas MixturesDalton’s law of additive pressures and Amagat’s law of additive volumescan also be used for real gases, often with reasonable accuracy. This time,however, the component pressures or component volumes should be evalu-ated from relations that take into account the deviation of each component

Pi

Pm

�Vi

Vm

�Ni

Nm

� yi

Vi 1Tm, Pm 2

Vm

�NiRuTm>Pm

NmRuTm>Pm

�Ni

Nm

� yi

Pi 1Tm, Vm 2

Pm

�NiRuTm>Vm

NmRuTm>Vm

�Ni

Nm

� yi

Vm �ak

i�1Vi 1Tm, Pm 2

Pm �ak

i�1Pi 1Tm, Vm 2

Chapter 13 | 685

O2 + N2

100 kPa400 K1 m3

O2

100 kPa400 K0.3 m3

N2

100 kPa400 K0.7 m3

FIGURE 13–7The volume a component wouldoccupy if it existed alone at themixture T and P is called thecomponent volume (for ideal gases, itis equal to the partial volume yiVm).

exact for ideal gases,approximatefor real gases

∂

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from ideal-gas behavior. One way of doing that is to use more exact equa-tions of state (van der Waals, Beattie–Bridgeman, Benedict–Webb–Rubin,etc.) instead of the ideal-gas equation of state. Another way is to use thecompressibility factor (Fig. 13–8) as

(13–9)

The compressibility factor of the mixture Zm can be expressed in terms of thecompressibility factors of the individual gases Zi by applying Eq. 13–9 to bothsides of Dalton’s law or Amagat’s law expression and simplifying. We obtain

(13–10)

where Zi is determined either at Tm and Vm (Dalton’s law) or at Tm and Pm(Amagat’s law) for each individual gas. It may seem that using either lawgives the same result, but it does not.

The compressibility-factor approach, in general, gives more accurate resultswhen the Zi’s in Eq. 13–10 are evaluated by using Amagat’s law instead ofDalton’s law. This is because Amagat’s law involves the use of mixturepressure Pm, which accounts for the influence of intermolecular forcesbetween the molecules of different gases. Dalton’s law disregards the influ-ence of dissimilar molecules in a mixture on each other. As a result, it tendsto underpredict the pressure of a gas mixture for a given Vm and Tm. There-fore, Dalton’s law is more appropriate for gas mixtures at low pressures.Amagat’s law is more appropriate at high pressures.

Note that there is a significant difference between using the compressibil-ity factor for a single gas and for a mixture of gases. The compressibility fac-tor predicts the P-v-T behavior of single gases rather accurately, as discussedin Chapter 3, but not for mixtures of gases. When we use compressibilityfactors for the components of a gas mixture, we account for the influence oflike molecules on each other; the influence of dissimilar molecules remainslargely unaccounted for. Consequently, a property value predicted by thisapproach may be considerably different from the experimentally determinedvalue.

Another approach for predicting the P-v-T behavior of a gas mixture isto treat the gas mixture as a pseudopure substance (Fig. 13–9). One suchmethod, proposed by W. B. Kay in 1936 and called Kay’s rule, involves theuse of a pseudocritical pressure P�cr,m and pseudocritical temperature T�cr,mfor the mixture, defined in terms of the critical pressures and temperaturesof the mixture components as

(13–11a, b)

The compressibility factor of the mixture Zm is then easily determined byusing these pseudocritical properties. The result obtained by using Kay’srule is accurate to within about 10 percent over a wide range of tempera-tures and pressures, which is acceptable for most engineering purposes.

Another way of treating a gas mixture as a pseudopure substance is touse a more accurate equation of state such as the van der Waals, Beattie–Bridgeman, or Benedict–Webb–Rubin equation for the mixture, and to deter-mine the constant coefficients in terms of the coefficients of the components.

P ¿cr,m �ak

i�1yi Pcr,i and T ¿cr,m �a

k

i�1yi Tcr,i

Zm �ak

i�1yiZi

PV � ZNRuT

686 | Thermodynamics

Pm Vm = Z = Zm Nm Ru Tm

Zm m = = ∑ y yi Z Zi

k

i = i = 1

FIGURE 13–8One way of predicting the P-v-Tbehavior of a real-gas mixture is to use compressibility factor.

Pseudopure substance

P'cr,m = ∑ yi Pcr,i

k

i = 1

T'cr,m = ∑ yi Tcr,i

k

i = 1

FIGURE 13–9Another way of predicting the P-v-Tbehavior of a real-gas mixture is totreat it as a pseudopure substance withcritical properties P�cr and T�cr.

cen84959_ch13.qxd 4/6/05 9:35 AM Page 686

In the van der Waals equation, for example, the two constants for the mixtureare determined from

(13–12a, b)

where expressions for ai and bi are given in Chapter 3.

am � aak

i�1yi a

1>2i b

2

and bm �ak

i�1yi bi

Chapter 13 | 687

EXAMPLE 13–2 P-v-T Behavior of Nonideal Gas Mixtures

A rigid tank contains 2 kmol of N2 and 6 kmol of CO2 gases at 300 K and15 MPa (Fig. 13–10). Estimate the volume of the tank on the basis of(a) the ideal-gas equation of state, (b) Kay’s rule, (c) compressibility factorsand Amagat’s law, and (d ) compressibility factors and Dalton’s law.

Solution The composition of a mixture in a rigid tank is given. The volumeof the tank is to be determined using four different approaches.Assumptions Stated in each section.Analysis (a) When the mixture is assumed to behave as an ideal gas, thevolume of the mixture is easily determined from the ideal-gas relation for themixture:

since

(b) To use Kay’s rule, we need to determine the pseudocritical temperatureand pseudocritical pressure of the mixture by using the critical-point proper-ties of N2 and CO2 from Table A–1. However, first we need to determine themole fraction of each component:

Then,

Thus,

Vm �Zm Nm Ru Tm

Pm

� ZmVideal � 10.49 2 11.330 m3 2 � 0.652 m3

TR �Tm

T ¿cr,m�

300 K

259.7 K� 1.16

PR �Pm

P ¿cr,m�

15 MPa

6.39 MPa� 2.35

∂ Zm � 0.49 1Fig. A–15b 2

� 10.25 2 13.39 MPa 2 � 10.75 2 17.39 MPa 2 � 6.39 MPa

P ¿cr,m �a yi Pcr,i � yN2Pcr,N2

� yCO2Pcr,CO2

� 10.25 2 1126.2 K 2 � 10.75 2 1304.2 K 2 � 259.7 K

T ¿cr,m �a yi Tcr,i � yN2Tcr,N2

� yCO2Tcr,CO2

yN2�

NN2

Nm

�2 kmol

8 kmol� 0.25 and yCO2

�NCO2

Nm

�6 kmol

8 kmol� 0.75

Nm � NN2� NCO2

� 2 � 6 � 8 kmol

Vm �Nm Ru Tm

Pm

�18 kmol 2 18.314 kPa # m3>kmol # K 2 1300 K 2

15,000 kPa� 1.330 m3

2 kmol N26 kmol CO2

300 K15 MPaVm = ?

FIGURE 13–10Schematic for Example 13–2.

cen84959_ch13.qxd 4/6/05 9:35 AM Page 687

688 | Thermodynamics

(c) When Amagat’s law is used in conjunction with compressibility factors, Zmis determined from Eq. 13–10. But first we need to determine the Z of eachcomponent on the basis of Amagat’s law:

N2:

CO2:

Mixture:

Thus,

The compressibility factor in this case turned out to be almost the same asthe one determined by using Kay’s rule.

(d) When Dalton’s law is used in conjunction with compressibility factors, Zmis again determined from Eq. 13–10. However, this time the Z of each com-ponent is to be determined at the mixture temperature and volume, which isnot known. Therefore, an iterative solution is required. We start the calcula-tions by assuming that the volume of the gas mixture is 1.330 m3, the valuedetermined by assuming ideal-gas behavior.

The TR values in this case are identical to those obtained in part (c) andremain constant. The pseudoreduced volume is determined from its defini-tion in Chap. 3:

Similarly,

From Fig. A–15, we read and . Thus,

and

Vm �Zm Nm RTm

Pm

� ZmVideal � 10.67 2 11.330 m3 2 � 0.891 m3

Zm � yN2ZN2

� yCO2ZCO2

� 10.25 2 10.99 2 � 10.75 2 10.56 2 � 0.67

Z CO2� 0.56Z N2

� 0.99

vR,CO2�

11.33 m3 2 > 16 kmol 218.314 kPa # m3>kmol # K 2 1304.2 K 2 > 17390 kPa 2 � 0.648

�11.33 m3 2 > 12 kmol 2

18.314 kPa # m3>kmol # K 2 1126.2 K 2 > 13390 kPa 2 � 2.15

vR,N2�

v�N2

RuTcr,N2 >Pcr,N2

�Vm >NN2

RuTcr,N2 >Pcr,N2

Vm �Zm Nm Ru Tm

Pm

� ZmVideal � 10.48 2 11.330 m3 2 � 0.638 m3

� 10.25 2 11.02 2 � 10.75 2 10.30 2 � 0.48

Zm �a yi Zi � yN2 ZN2

� yCO2 ZCO2

TR,CO2 �Tm

Tcr,CO2

�300 K

304.2 K� 0.99

PR,CO2�

Pm

Pcr,CO2

�15 MPa

7.39 MPa� 2.03

∂ ZCO2� 0.30 1Fig. A–15b 2

TR,N2 �Tm

Tcr,N2

�300 K

126.2 K� 2.38

PR,N2�

Pm

Pcr,N2

�15 MPa

3.39 MPa� 4.42

∂ ZN2� 1.02 1Fig. A–15b 2

cen84959_ch13.qxd 4/6/05 9:35 AM Page 688

13–3 � PROPERTIES OF GAS MIXTURES: IDEAL AND REAL GASES

Consider a gas mixture that consists of 2 kg of N2 and 3 kg of CO2. Thetotal mass (an extensive property) of this mixture is 5 kg. How did we do it?Well, we simply added the mass of each component. This example suggestsa simple way of evaluating the extensive properties of a nonreacting ideal-or real-gas mixture: Just add the contributions of each component of themixture (Fig. 13–11). Then the total internal energy, enthalpy, and entropyof a gas mixture can be expressed, respectively, as

(13–13)

(13–14)

(13–15)

By following a similar logic, the changes in internal energy, enthalpy, andentropy of a gas mixture during a process can be expressed, respectively, as

(13–16)

(13–17)

(13–18)

Now reconsider the same mixture, and assume that both N2 and CO2 are at25°C. The temperature (an intensive property) of the mixture is, as you wouldexpect, also 25°C. Notice that we did not add the component temperatures todetermine the mixture temperature. Instead, we used some kind of averagingscheme, a characteristic approach for determining the intensive properties ofa mixture. The internal energy, enthalpy, and entropy of a mixture per unitmass or per unit mole of the mixture can be determined by dividing the equa-tions above by the mass or the mole number of the mixture (mm or Nm). Weobtain (Fig. 13–12)

¢Sm �ak

i�1¢Si �a

k

i�1mi ¢si �a

k

i�1Ni ¢ s�i 1kJ>K 2

¢Hm �ak

i�1¢Hi �a

k

i�1mi ¢hi �a

k

i�1Ni ¢h�i 1kJ 2

¢Um �ak

i�1¢Ui �a

k

i�1mi ¢ui �a

k

i�1Ni ¢u�i 1kJ 2

Sm �ak

i�1Si �a

k

i�1misi �a

k

i�1Ni s�i 1kJ>K 2

Hm �ak

i�1Hi �a

k

i�1mihi �a

k

i�1Nihi 1kJ 2

Um �ak

i�1Ui �a

k

i�1miui �a

k

i�1Niu�i 1kJ 2

Chapter 13 | 689

This is 33 percent lower than the assumed value. Therefore, we shouldrepeat the calculations, using the new value of Vm. When the calculationsare repeated we obtain 0.738 m3 after the second iteration, 0.678 m3 afterthe third iteration, and 0.648 m3 after the fourth iteration. This value doesnot change with more iterations. Therefore,

Discussion Notice that the results obtained in parts (b), (c), and (d ) arevery close. But they are very different from the ideal-gas values. Therefore,treating a mixture of gases as an ideal gas may yield unacceptable errors athigh pressures.

Vm � 0.648 m3

2 kmol A6 kmol B

UA=1000 kJ

UB=1800 kJ

Um=2800 kJ

FIGURE 13–11The extensive properties of a mixtureare determined by simply adding theproperties of the components.

2 kmol A3 kmol B

u-A=500 kJ/kmol

u-B=600 kJ/kmol

u-m=560 kJ/kmol

FIGURE 13–12The intensive properties of a mixtureare determined by weighted averaging.

cen84959_ch13.qxd 4/20/05 2:22 PM Page 689

(13–19)

(13–20)

(13–21)

Similarly, the specific heats of a gas mixture can be expressed as

(13–22)

(13–23)

Notice that properties per unit mass involve mass fractions (mfi) and prop-erties per unit mole involve mole fractions (yi).

The relations given above are exact for ideal-gas mixtures, and approxi-mate for real-gas mixtures. (In fact, they are also applicable to nonreactingliquid and solid solutions especially when they form an “ideal solution.”)The only major difficulty associated with these relations is the determina-tion of properties for each individual gas in the mixture. The analysis can besimplified greatly, however, by treating the individual gases as ideal gases,if doing so does not introduce a significant error.

Ideal-Gas MixturesThe gases that comprise a mixture are often at a high temperature and lowpressure relative to the critical-point values of individual gases. In such cases,the gas mixture and its components can be treated as ideal gases with negligi-ble error. Under the ideal-gas approximation, the properties of a gas are notinfluenced by the presence of other gases, and each gas component in themixture behaves as if it exists alone at the mixture temperature Tm and mix-ture volume Vm. This principle is known as the Gibbs–Dalton law, which isan extension of Dalton’s law of additive pressures. Also, the h, u, cv, and cp ofan ideal gas depend on temperature only and are independent of the pressureor the volume of the ideal-gas mixture. The partial pressure of a component inan ideal-gas mixture is simply Pi � yi Pm, where Pm is the mixture pressure.

Evaluation of �u or �h of the components of an ideal-gas mixture duringa process is relatively easy since it requires only a knowledge of the initialand final temperatures. Care should be exercised, however, in evaluating the�s of the components since the entropy of an ideal gas depends on the pres-sure or volume of the component as well as on its temperature. The entropychange of individual gases in an ideal-gas mixture during a process can bedetermined from

(13–24)

or

(13–25)¢ s�i � s�°i,2 � s�°i,1 � Ru ln Pi,2

Pi,1� c�p,i ln

Ti,2

Ti,1� Ru ln

Pi,2

Pi,1

¢si � s°i,2 � s°i,1 � Ri ln Pi,2

Pi,1� cp,i ln

Ti,2

Ti,1� Ri ln

Pi,2

Pi,1

cp,m �ak

i�1mfi cp,i 1kJ>kg # K 2 and c�p,m �a

k

i�1yi c�p,i 1kJ>kmol # K 2

cv,m �ak

i�1mfi cv,i 1kJ>kg # K 2 and c�v,m �a

k

i�1yi c

�v,i 1kJ>kmol # K 2

sm �ak

i�1mfi si 1kJ>kg # K 2 and s�m �a

k

i�1yi s�i 1kJ>kmol # K 2

hm �ak

i�1mfi hi 1kJ>kg 2 and h�m �a

k

i�1yi h

�i 1kJ>kmol 2

um �ak

i�1mfi ui 1kJ>kg 2 and u�m �a

k

i�1yiu

�i 1kJ>kmol 2

690 | Thermodynamics

cen84959_ch13.qxd 4/6/05 9:35 AM Page 690

where Pi,2 � yi,2Pm,2 and Pi,1 � yi,1Pm,1. Notice that the partial pressure Pi ofeach component is used in the evaluation of the entropy change, not themixture pressure Pm (Fig. 13–13).

Chapter 13 | 691

Partial pressurePartial pressureof component of component i

at state 2 at state 2

Partial pressurePartial pressureof component of component i

at state 1 at state 1

∆si° = = si°,2 – si°,1 – Ri lnlnPi,i,2—––—––Pi,i,1

FIGURE 13–13Partial pressures (not the mixturepressure) are used in the evaluation ofentropy changes of ideal-gas mixtures.

EXAMPLE 13–3 Mixing Two Ideal Gases in a Tank

An insulated rigid tank is divided into two compartments by a partition, asshown in Fig. 13–14. One compartment contains 7 kg of oxygen gas at 40°Cand 100 kPa, and the other compartment contains 4 kg of nitrogen gas at20°C and 150 kPa. Now the partition is removed, and the two gases areallowed to mix. Determine (a) the mixture temperature and (b) the mixturepressure after equilibrium has been established.

Solution A rigid tank contains two gases separated by a partition. The pres-sure and temperature of the mixture are to be determined after the partitionis removed.Assumptions 1 We assume both gases to be ideal gases, and their mixtureto be an ideal-gas mixture. This assumption is reasonable since both the oxygen and nitrogen are well above their critical temperatures and well belowtheir critical pressures. 2 The tank is insulated and thus there is no heattransfer. 3 There are no other forms of work involved.Properties The constant-volume specific heats of N2 and O2 at room temper-ature are cv,N2

� 0.743 kJ/kg · K and cv,O2� 0.658 kJ/kg · K (Table A–2a).

Analysis We take the entire contents of the tank (both compartments) asthe system. This is a closed system since no mass crosses the boundary dur-ing the process. We note that the volume of a rigid tank is constant and thusthere is no boundary work done.

(a) Noting that there is no energy transfer to or from the tank, the energybalance for the system can be expressed as

By using cv values at room temperature, the final temperature of the mixtureis determined to be

(b) The final pressure of the mixture is determined from the ideal-gas relation

where

Nm � NO2� NN2

� 0.219 � 0.143 � 0.362 kmol

NN2�

mN2

MN2

�4 kg

28 kg>kmol� 0.143 kmol

NO2�

mO2

MO2

�7 kg

32 kg>kmol� 0.219 kmol

PmVm � Nm Ru Tm

Tm � 32.2°C

14 kg 2 10.743 kJ>kg # K 2 1Tm � 20°C 2 � 17 kg 2 10.658 kJ>kg # K 2 1Tm � 40°C 2 � 0

3mcv 1Tm � T1 2 4N2� 3mcv 1Tm � T1 2 4O2

� 0

0 � ¢U � ¢UN2� ¢UO2

Ein � Eout � ¢Esystem

O2

7 kg40°C

100 kPa

N2

4 kg20°C

150 kPa

Partition

FIGURE 13–14Schematic for Example 13–3.

cen84959_ch13.qxd 4/6/05 9:35 AM Page 691

692 | Thermodynamics

and

Thus,

Discussion We could also determine the mixture pressure by using PmVm �mmRmTm, where Rm is the apparent gas constant of the mixture. Thiswould require a knowledge of mixture composition in terms of mass or molefractions.

Pm �Nm Ru Tm

Vm

�10.362 kmol 2 18.314 kPa # m3>kmol # K 2 1305.2 K 2

8.02 m3 � 114.5 kPa

Vm � VO2� VN2

� 5.70 � 2.32 � 8.02 m3

VN2� aNRuT1

P1b

N2

�10.143 kmol 2 18.314 kPa # m3>kmol # K 2 1293 K 2

150 kPa� 2.32 m3

VO2� aNRuT1

P1b

O2

�10.219 kmol 2 18.314 kPa # m3>kmol # K 2 1313 K 2

100 kPa� 5.70 m3

EXAMPLE 13–4 Exergy Destruction during Mixing of Ideal Gases

An insulated rigid tank is divided into two compartments by a partition, asshown in Fig. 13–15. One compartment contains 3 kmol of O2, and theother compartment contains 5 kmol of CO2. Both gases are initially at 25°Cand 200 kPa. Now the partition is removed, and the two gases are allowedto mix. Assuming the surroundings are at 25°C and both gases behave asideal gases, determine the entropy change and exergy destruction associatedwith this process.

Solution A rigid tank contains two gases separated by a partition. Theentropy change and exergy destroyed after the partition is removed are to bedetermined.Assumptions Both gases and their mixture are ideal gases.Analysis We take the entire contents of the tank (both compartments) asthe system. This is a closed system since no mass crosses the boundary dur-ing the process. We note that the volume of a rigid tank is constant, andthere is no energy transfer as heat or work. Also, both gases are initially atthe same temperature and pressure.

When two ideal gases initially at the same temperature and pressure aremixed by removing a partition between them, the mixture will also be at thesame temperature and pressure. (Can you prove it? Will this be true for non-ideal gases?) Therefore, the temperature and pressure in the tank will still be25°C and 200 kPa, respectively, after the mixing. The entropy change ofeach component gas can be determined from Eqs. 13–18 and 13–25:

since Pm,2 � Pi,1 � 200 kPa. It is obvious that the entropy change is inde-pendent of the composition of the mixture in this case and depends on only

� �Ru a Ni ln yi,2Pm,2

Pi,1� �Ru a Ni ln yi,2

¢Sm �a ¢Si �a Ni ¢ s�i �a Ni a c�p,i ln Ti,2

Ti,1� Ru ln

Pi,2

Pi,1b

O2

25°C200 kPa

CO2

25°C200 kPa

FIGURE 13–15Schematic for Example 13–4.

→0

cen84959_ch13.qxd 4/6/05 9:35 AM Page 692

Real-Gas MixturesWhen the components of a gas mixture do not behave as ideal gases, theanalysis becomes more complex because the properties of real (nonideal)gases such as u, h, cv , and cp depend on the pressure (or specific volume) aswell as on the temperature. In such cases, the effects of deviation fromideal-gas behavior on the mixture properties should be accounted for.

Consider two nonideal gases contained in two separate compartments ofan adiabatic rigid tank at 100 kPa and 25°C. The partition separating thetwo gases is removed, and the two gases are allowed to mix. What do youthink the final pressure in the tank will be? You are probably tempted to say100 kPa, which would be true for ideal gases. However, this is not true fornonideal gases because of the influence of the molecules of different gaseson each other (deviation from Dalton’s law, Fig. 13–16).

When real-gas mixtures are involved, it may be necessary to account for theeffect of nonideal behavior on the mixture properties such as enthalpy andentropy. One way of doing that is to use compressibility factors in conjunctionwith generalized equations and charts developed in Chapter 12 for real gases.

Consider the following T ds relation for a gas mixture:

It can also be expressed as

d aa mfihi b � Tm d aa mfi si b � aa mfi vi b dPm

dhm � Tm dsm � vm dPm

Chapter 13 | 693

the mole fraction of the gases in the mixture. What is not so obvious is thatif the same gas in two different chambers is mixed at constant temperatureand pressure, the entropy change is zero.

Substituting the known values, the entropy change becomes

The exergy destruction associated with this mixing process is determinedfrom

Discussion This large value of exergy destruction shows that mixing pro-cesses are highly irreversible.

� 13.1 MJ

� 1298 K 2 144.0 kJ>K 2 X destroyed � T0Sgen � T0 ¢Ssys

� 44.0 kJ/K � � 18.314 kJ>kmol # K 2 3 13 kmol 2 1ln 0.375 2 � 15 kmol 2 1ln 0.625 2 4

¢Sm � �Ru 1NO2 ln yO2

� NCO2 ln yCO2

2 yCO2

�NCO2

Nm

�5 kmol

8 kmol� 0.625

yO2�

NO2

Nm

�3 kmol

8 kmol� 0.375

Nm � NO2� NCO2

� 13 � 5 2 kmol � 8 kmol

Real gasA

25°C0.4 m3

100 kPa

Real gasmixtureA + B25°C1 m3

102 kPa ?

Real gasB

25°C0.6 m3

100 kPa

FIGURE 13–16It is difficult to predict the behavior ofnonideal-gas mixtures because of theinfluence of dissimilar molecules oneach other.

cen84959_ch13.qxd 4/20/05 2:22 PM Page 693

or

which yields

(13–26)

This is an important result because Eq. 13–26 is the starting equation in thedevelopment of the generalized relations and charts for enthalpy and entropy.It suggests that the generalized property relations and charts for real gasesdeveloped in Chapter 12 can also be used for the components of real-gasmixtures. But the reduced temperature TR and reduced pressure PR for eachcomponent should be evaluated by using the mixture temperature Tm andmixture pressure Pm. This is because Eq. 13–26 involves the mixture pres-sure Pm, not the component pressure Pi.

The approach described above is somewhat analogous to Amagat’s law ofadditive volumes (evaluating mixture properties at the mixture pressure andtemperature), which holds exactly for ideal-gas mixtures and approximatelyfor real-gas mixtures. Therefore, the mixture properties determined with thisapproach are not exact, but they are sufficiently accurate.

What if the mixture volume and temperature are specified instead of themixture pressure and temperature? Well, there is no need to panic. Just eval-uate the mixture pressure, using Dalton’s law of additive pressures, and thenuse this value (which is only approximate) as the mixture pressure.

Another way of evaluating the properties of a real-gas mixture is to treatthe mixture as a pseudopure substance having pseudocritical properties,determined in terms of the critical properties of the component gases byusing Kay’s rule. The approach is quite simple, and the accuracy is usuallyacceptable.

dhi � Tm dsi � vi dPm

a mfi 1dhi � Tm dsi � vi dPm 2 � 0

694 | Thermodynamics

AIR79% N221% O2

T1 = 220 K T2 = 160 K

P1 = 10 MPa P2 = 10 MPa

Heat

FIGURE 13–17Schematic for Example 13–5.

EXAMPLE 13–5 Cooling of a Nonideal Gas Mixture

Air is a mixture of N2, O2, and small amounts of other gases, and it can beapproximated as 79 percent N2 and 21 percent O2 on mole basis. During asteady-flow process, air is cooled from 220 to 160 K at a constant pressureof 10 MPa (Fig. 13–17). Determine the heat transfer during this processper kmol of air, using (a) the ideal-gas approximation, (b) Kay’s rule, and(c) Amagat’s law.

Solution Air at a low temperature and high pressure is cooled at constantpressure. The heat transfer is to be determined using three differentapproaches.Assumptions 1 This is a steady-flow process since there is no change withtime at any point and thus �mCV � 0 and �ECV � 0. 2 The kinetic andpotential energy changes are negligible.Analysis We take the cooling section as the system. This is a control volumesince mass crosses the system boundary during the process. We note thatheat is transferred out of the system.

The critical properties are Tcr � 126.2 K and Pcr � 3.39 MPa for N2 andTcr � 154.8 K and Pcr � 5.08 MPa for O2. Both gases remain above their

cen84959_ch13.qxd 4/6/05 9:35 AM Page 694

critical temperatures, but they are also above their critical pressures. There-fore, air will probably deviate from ideal-gas behavior, and thus it should betreated as a real-gas mixture.

The energy balance for this steady-flow system can be expressed on a unitmole basis as

where the enthalpy change for either component can be determined from thegeneralized enthalpy departure chart (Fig. A–29) and Eq. 12–58:

The first two terms on the right-hand side of this equation represent theideal-gas enthalpy change of the component. The terms in parentheses repre-sent the deviation from the ideal-gas behavior, and their evaluation requires aknowledge of reduced pressure PR and reduced temperature TR, which arecalculated at the mixture temperature Tm and mixture pressure Pm.

(a) If the N2 and O2 mixture is assumed to behave as an ideal gas, theenthalpy of the mixture will depend on temperature only, and the enthalpyvalues at the initial and the final temperatures can be determined from theideal-gas tables of N2 and O2 (Tables A–18 and A–19):

(b) Kay’s rule is based on treating a gas mixture as a pseudopure substancewhose critical temperature and pressure are

and

Then,

TR,2 �Tm,2

Tcr,m�

160 K

132.2 K� 1.21

PR �Pm

Pcr,m�

10 MPa

3.74 MPa� 2.67

TR,1 �Tm,1

Tcr,m�

220 K

132.2 K� 1.66

� 10.79 2 13.39 MPa 2 � 10.21 2 15.08 MPa 2 � 3.74 MPa

P ¿cr,m �a yi Pcr,i � yN2 Pcr,N2

� yO2 Pcr,O2

� 10.79 2 1126.2 K 2 � 10.21 2 1154.8 K 2 � 132.2 K

T ¿cr,m �a yiTcr,i � yN2Tcr,N2

� yO2Tcr,O2

� 1744 kJ/kmol

� 10.79 2 16391 � 4648 2 kJ>kmol � 10.21 2 16404 � 4657 2 kJ>kmol

q�out � yN21h1 � h2 2N2

� yO21h1 � h2 2O2

h2,ideal,O2� 4657 kJ>kmol

T2 � 160 K S h2,ideal,N2� 4648 kJ>kmol

h1,ideal,O2� 6404 kJ>kmol

T1 � 220 K S h1,ideal,N2� 6391 kJ>kmol

h1 � h2 � h1,ideal � h2,ideal � RuTcr 1Zh1 � Zh2 2

q�out � h1 � h2 � yN21h1 � h2 2N2

� yO21h1 � h2 2O2

ein � eout � ¢esystem � 0 S ein � eout S h1 � h2 � q�out

Chapter 13 | 695

f Zh1,m � 1.0

f Zh2,m � 2.6

→0

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696 | Thermodynamics

Also,

Therefore,

(c) The reduced temperatures and pressures for both N2 and O2 at the initialand final states and the corresponding enthalpy departure factors are, fromFig. A–29,

N2:

O2:

From Eq. 12–58,

TR1,O2�

Tm,2

Tcr,O2

�160 K

154.8 K� 1.03

PR,O2�

Pm

Pcr,O2

�10 MPa

5.08 MPa� 1.97

TR1,O2�

Tm,1

Tcr,O2

�220 K

154.8 K� 1.42

TR2,N2�

Tm,2

Tcr,N2

�160 K

126.2 K� 1.27

PR,N2�

Pm

Pcr,N2

�10 MPa

3.39 MPa� 2.95

TR1,N2�

Tm,1

Tcr,N2

�220 K

126.2 K� 1.74

� 3503 kJ/kmol

� 3 16394 � 4650 2 kJ>kmol 4 � 18.314 kJ>kmol # K 2 1132.2 K 2 11.0 � 2.6 2 q�out � 1hm1,ideal � hm2,ideal 2 � RuTcr 1Zh1

� Zh22m

� 4650 kJ>kmol

� 10.79 2 14648 kJ>kmol 2 � 10.21 2 14657 kJ>kmol 2 hm2,ideal � yN2

h2,ideal,N2� yO2

h2,ideal,O2

� 6394 kJ>kmol

� 10.79 2 16391 kJ>kmol 2 � 10.21 2 16404 kJ>kmol 2 hm1,ideal � yN2

h1,ideal,N2� yO2

h1,ideal,O2

f Zh1,N2� 0.9

f Zh2,N2� 2.4

f Zh1,O2� 1.3

f Zh2,O2� 4.0

� 5222 kJ>kmol

� 3 16404 � 4657 2 kJ>kmol 4 � 18.314 kJ>kmol # K 2 1154.8 K 2 11.3 � 4.0 2 1h1 � h2 2O2

� 1h1,ideal � h2,ideal 2O2� RuTcr 1Zh1

� Zh22O2

� 3317 kJ>kmol

� 3 16391 � 4648 2 kJ>kmol 4 � 18.314 kJ>kmol # K 2 1126.2 K 2 10.9 � 2.4 2 1h1 � h2 2N2

� 1h1,ideal � h2,ideal 2N2� RuTcr 1Zh1

� Zh22N2

Therefore,

� 3717 kJ/kmol

� 10.79 2 13317 kJ>kmol 2 � 10.21 2 15222 kJ>kmol 2 q�out � yN2

1h1 � h2 2N2� yO2

1h1 � h2 2O2

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Chapter 13 | 697

Discussion This result is about 6 percent greater than the result obtained inpart (b) by using Kay’s rule. But it is more than twice the result obtained byassuming the mixture to be an ideal gas.

When two gases or two miscible liquids are brought into contact, they mixand form a homogeneous mixture or solution without requiring any workinput. That is, the natural tendency of miscible substances brought into con-tact is to mix with each other. As such, these are irreversible processes, andthus it is impossible for the reverse process of separation to occur sponta-neously. For example, pure nitrogen and oxygen gases readily mix whenbrought into contact, but a mixture of nitrogen and oxygen (such as air)never separates into pure nitrogen and oxygen when left unattended.

Mixing and separation processes are commonly used in practice. Separa-tion processes require a work (or, more generally, exergy) input, and mini-mizing this required work input is an important part of the design process ofseparation plants. The presence of dissimilar molecules in a mixture affecteach other, and therefore the influence of composition on the properties mustbe taken into consideration in any thermodynamic analysis. In this sectionwe analyze the general mixing processes, with particular emphasis on idealsolutions, and determine the entropy generation and exergy destruction. Wethen consider the reverse process of separation, and determine the minimum(or reversible) work input needed for separation.

The specific Gibbs function (or Gibbs free energy) g is defined as thecombination property g � h � Ts. Using the relation dh � v dP � T ds, thedifferential change of the Gibbs function of a pure substance is obtained bydifferentiation to be

(13–27)

For a mixture, the total Gibbs function is a function of two independentintensive properties as well as the composition, and thus it can be expressedas G � G(P, T, N1, N2, . . . , Ni). Its differential is

(13–28)

where the subscript Nj indicates that the mole numbers of all components inthe mixture other than component i are to be held constant during differentiation. For a pure substance, the last term drops out since the com-position is fixed, and the equation above must reduce to the one for a puresubstance. Comparing Eqs. 13–27 and 13–28 gives

(13–29)dG � V dP � S dT �ai

m i dNi or dg� � v� dP � s� dT �ai

m i dyi

dG � a 0G

0Pb

T,N dP � a 0G

0Tb

P,N dT �a

i

a 0G

0Ni

bP,T,Nj

dNi 1mixture 2

dg � v dP � s dT or dG � V dP � S dT 1pure substance 2

TOPIC OF SPECIAL INTEREST* Chemical Potential and the Separation Work of Mixtures

*This section can be skipped without a loss in continuity.

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698 | Thermodynamics

i

�

� �

� h � Ts � g

� gi,mixture

� hi,mixture �Tsi,mixture

Mixture:

Pure substance:

FIGURE 13–18For a pure substance, the chemicalpotential is equivalent to the Gibbsfunction.

where yi � Ni /Nm is the mole fraction of component i (Nm is the total num-ber of moles of the mixture) and

(13–30)

is the chemical potential, which is the change in the Gibbs function of themixture in a specified phase when a unit amount of component i in the samephase is added as pressure, temperature, and the amounts of all other com-ponents are held constant. The symbol tilde (as in v�, h

�, and s�) is used to

denote the partial molar properties of the components. Note that the sum-mation term in Eq. 13–29 is zero for a single component system and thus thechemical potential of a pure system in a given phase is equivalent to themolar Gibbs function (Fig. 13–18) since G � Ng � Nm, where

(13–31)

Therefore, the difference between the chemical potential and the Gibbsfunction is due to the effect of dissimilar molecules in a mixture on eachother. It is because of this molecular effect that the volume of the mixture oftwo miscible liquids may be more or less than the sum of the initial volumesof the individual liquids. Likewise, the total enthalpy of the mixture of twocomponents at the same pressure and temperature, in general, is not equal tothe sum of the total enthalpies of the individual components before mixing,the difference being the enthalpy (or heat) of mixing, which is the heatreleased or absorbed as two or more components are mixed isothermally. Forexample, the volume of an ethyl alcohol–water mixture is a few percent lessthan the sum of the volumes of the individual liquids before mixing. Also,when water and flour are mixed to make dough, the temperature of the doughrises noticeably due to the enthalpy of mixing released.

For reasons explained above, the partial molar properties of the compo-nents (denoted by an tilde) should be used in the evaluation of the extensiveproperties of a mixture instead of the specific properties of the pure compo-nents. For example, the total volume, enthalpy, and entropy of a mixtureshould be determined from, respectively,

(13–32)

instead of

(13–33)

Then the changes in these extensive properties during mixing become

(13–34)

where �Hmixing is the enthalpy of mixing and �Smixing is the entropy ofmixing (Fig. 13–19). The enthalpy of mixing is positive for exothermic mix-

¢Vmixing �ai

Ni 1v�i � v�i 2 , ¢Hmixing �ai

Ni 1h�i � hi 2 , ¢Smixing �ai

Ni 1 s�i � s�i 2

V * �ai

Niv�i H* �ai

Ni hi and S* �ai

Ni s�i

V �ai

Niv�i H �ai

Ni h�

i and S �ai

Nis�

i 1mixture 2

m � a 0G

0Nb

P,T� g� � h � Ts� 1pure substance 2

mi � a 0G

0Ni

bP,T,Nj

� g�i � h�i � T s�i 1for component i of a mixture 2

Mixingchamber

A

B

yB

yA A + B

mixture

FIGURE 13–19The amount of heat released orabsorbed during a mixing process iscalled the enthalpy (or heat) ofmixing, which is zero for idealsolutions.

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Chapter 13 | 699

ing processes, negative for endothermic mixing processes, and zero forisothermal mixing processes during which no heat is absorbed or released.Note that mixing is an irreversible process, and thus the entropy of mixingmust be a positive quantity during an adiabatic process. The specific volume,enthalpy, and entropy of a mixture are determined from

(13–35)

where yi is the mole fraction of component i in the mixture.Reconsider Eq. 13–29 for dG. Recall that properties are point functions, and

they have exact differentials. Therefore, the test of exactness can be applied tothe right-hand side of Eq. 13–29 to obtain some important relations. For the dif-ferential dz � M dx � N dy of a function z(x, y), the test of exactness isexpressed as (�M/�y)x � (�N/�x)y. When the amount of component i in a mix-ture is varied at constant pressure or temperature while other components (indi-cated by j ) are held constant, Eq. 13–29 simplifies to

(13–36)

(13–37)

Applying the test of exactness to both of these relations gives

(13–38)

where the subscript N indicates that the mole numbers of all components(and thus the composition of the mixture) is to remain constant. Taking thechemical potential of a component to be a function of temperature, pressure,and composition and thus mi � mi (P, T, y1, y2, . . . , yj . . .), its total differ-ential can be expressed as

(13–39)

where the subscript y indicates that the mole fractions of all components(and thus the composition of the mixture) is to remain constant. SubstitutingEqs. 13–38 into the above relation gives

(13–40)

For a mixture of fixed composition undergoing an isothermal process, it sim-plifies to

(13–41)

Ideal-Gas Mixtures and Ideal SolutionsWhen the effect of dissimilar molecules in a mixture on each other is negli-gible, the mixture is said to be an ideal mixture or ideal solution and thechemical potential of a component in such a mixture equals the Gibbs function

dm i � v�i dP 1T � constant, yi � constant 2

dmi � v�i dP � s�i dT �ai

a 0mi

0yi

bP,T,yj

dyi

dmi � dg�i � a 0mi

0Pb

T,y dP � a 0mi

0Tb

P,y dT �a

i

a 0mi

0yi

bP,T,yj

dyi

a 0mi

0Tb

P,N� � a 0S

0Ni

bT,P,Nj

� � s�i and a 0mi

0Pb

T,N� a 0V

0Ni

bT,P,Nj

� v�i

dG � V dP � m i dNi 1for T � constant and Nj � constant 2dG � �S dT � m i dNi 1for P � constant and Nj � constant 2

v� �ai

yiv�i h �ai

yih�

i and s� �ai

yis�

i

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700 | Thermodynamics

of the pure component. Many liquid solutions encountered in practice, espe-cially dilute ones, satisfy this condition very closely and can be considered tobe ideal solutions with negligible error. As expected, the ideal solutionapproximation greatly simplifies the thermodynamic analysis of mixtures. Inan ideal solution, a molecule treats the molecules of all components in themixture the same way—no extra attraction or repulsion for the molecules ofother components. This is usually the case for mixtures of similar substancessuch as those of petroleum products. Very dissimilar substances such aswater and oil won’t even mix at all to form a solution.

For an ideal-gas mixture at temperature T and total pressure P, the partialmolar volume of a component i is v�i � vi � RuT/P. Substituting this relationinto Eq. 13–41 gives

(13–42)

since, from Dalton’s law of additive pressures, Pi � yi P for an ideal gasmixture and

(13–43)

for constant yi. Integrating Eq. 13–42 at constant temperature from the totalmixture pressure P to the component pressure Pi of component i gives

(13–44)

For yi � 1 (i.e., a pure substance of component i alone), the last term in theabove equation drops out and we end up with mi(T, Pi) � mi(T, P), which isthe value for the pure substance i. Therefore, the term mi(T, P) is simply thechemical potential of the pure substance i when it exists alone at total mix-ture pressure and temperature, which is equivalent to the Gibbs functionsince the chemical potential and the Gibbs function are identical for puresubstances. The term mi(T, P) is independent of mixture composition andmole fractions, and its value can be determined from the property tables ofpure substances. Then Eq. 13–44 can be rewritten more explicitly as

(13–45)

Note that the chemical potential of a component of an ideal gas mixturedepends on the mole fraction of the components as well as the mixture tem-perature and pressure, and is independent of the identity of the other con-stituent gases. This is not surprising since the molecules of an ideal gasbehave like they exist alone and are not influenced by the presence of othermolecules.

Eq. 13–45 is developed for an ideal-gas mixture, but it is also applicable tomixtures or solutions that behave the same way—that is, mixtures or solutionsin which the effects of molecules of different components on each other arenegligible. The class of such mixtures is called ideal solutions (or ideal mix-tures), as discussed before. The ideal-gas mixture described is just one cate-

mi,mixture,ideal 1T, Pi 2 � mi,pure 1T, P 2 � RuT ln yi

m i 1T, Pi 2 � m i 1T, P 2 � RuT ln Pi

P� m i 1T, P 2 � RuT ln yi 1ideal gas 2

d ln Pi � d ln 1yiP 2 � d 1ln yi � ln P 2 � d ln P 1yi � constant 2

dm i �RuT

P dP � RuTd ln P � RuTd ln Pi 1T � constant, yi � constant, ideal gas 2

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Chapter 13 | 701

gory of ideal solutions. Another major category of ideal solutions is the diluteliquid solutions, such as the saline water. It can be shown that the enthalpy ofmixing and the volume change due to mixing are zero for ideal solutions (seeWark, 1995). That is,

(13–46)

Then it follows that v�i � v–i and h�

i � h–

i. That is, the partial molar volumeand the partial molar enthalpy of a component in a solution equal the specificvolume and enthalpy of that component when it existed alone as a pure sub-stance at the mixture temperature and pressure. Therefore, the specific volumeand enthalpy of individual components do not change during mixing if theyform an ideal solution. Then the specific volume and enthalpy of an idealsolution can be expressed as (Fig. 13–20)

(13–47)

Note that this is not the case for entropy and the properties that involveentropy such as the Gibbs function, even for ideal solutions. To obtain a rela-tion for the entropy of a mixture, we differentiate Eq. 13–45 with respect totemperature at constant pressure and mole fraction,

(13–48)

We note from Eq. 13–38 that the two partial derivatives above are simply thenegative of the partial molar entropies. Substituting,

(13–49)

Note that ln yi is a negative quantity since yi � 1, and thus �Ru ln yi is alwayspositive. Therefore, the entropy of a component in a mixture is always greaterthan the entropy of that component when it exists alone at the mixture temper-ature and pressure. Then the entropy of mixing of an ideal solution is deter-mined by substituting Eq. 13–49 into Eq. 13–34 to be

(13–50a)

or, dividing by the total number of moles of the mixture Nm,

(13–50b)

Minimum Work of Separation of MixturesThe entropy balance for a steady-flow system simplifies to Sin � Sout � Sgen

� 0. Noting that entropy can be transferred by heat and mass only, the

¢ s�mixing,ideal �ai

yi 1 s�i � s�i 2 � �Ru ai

yi ln yi 1per unit mole of mixture 2

¢Smixing,ideal �ai

Ni 1 s�i � s�i 2 � �Ru ai

Ni ln yi 1ideal solution 2

s�i,mixture,ideal 1T, Pi 2 � s�i,pure 1T, P 2 � Ru ln y1 1ideal solution 2

a 0m i,mixing 1T, Pi 20T

bP,y

� a 0m i,pure 1T, P 20T

bP,y

� Ru ln yi

v�mixing,ideal �ai

yiv�

i �ai

yiv�i,pure and hmixture,ideal �ai

yih�

i �ai

yihi,pure

¢Vmixing,ideal �ai

Ni 1v�i � v�i 2 � 0 and ¢Hmixing,ideal �ai

Ni 1h�i � hi 2 � 0

� �

�V mixing,ideal � 0

�H mixing,ideal � 0

v mixture � �yivi,pure

� �vi,mixture � vi,pure

� �h mixture � �yihi,pure

� �hi,mixing � hi,pure

i

i

FIGURE 13–20The specific volume and enthalpy ofindividual components do not changeduring mixing if they form an idealsolution (this is not the case forentropy).

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702 | Thermodynamics

entropy generation during an adiabatic mixing process that forms an idealsolution becomes

(13–51a)

or

(13–51b)

Also noting that Xdestroyed � T0 Sgen, the exergy destroyed during this (and anyother) process is obtained by multiplying the entropy generation by the tem-perature of the environment T0. It gives

(13–52a)

or

(13–52b)

Exergy destroyed represents the wasted work potential—the work that wouldbe produced if the mixing process occurred reversibly. For a reversible or“thermodynamically perfect” process, the entropy generation and thus theexergy destroyed is zero. Also, for reversible processes, the work output is amaximum (or, the work input is a minimum if the process does not occurnaturally and requires input). The difference between the reversible work andthe actual useful work is due to irreversibilities and is equal to the exergydestruction. Therefore, Xdestroyed � Wrev � Wactual. Then it follows that for anaturally occurring process during which no work is produced, the reversiblework is equal to the exergy destruction (Fig. 13–21). Therefore, for the adia-batic mixing process that forms an ideal solution, the reversible work (totaland per unit mole of mixture) is, from Eq. 13–52,

(13–53)

A reversible process, by definition, is a process that can be reversed withoutleaving a net effect on the surroundings. This requires that the direction of allinteractions be reversed while their magnitudes remain the same when theprocess is reversed. Therefore, the work input during a reversible separationprocess must be equal to the work output during the reverse process of mix-ing. A violation of this requirement will be a violation of the second law ofthermodynamics. The required work input for a reversible separation processis the minimum work input required to accomplish that separation since thework input for reversible processes is always less than the work input of cor-responding irreversible processes. Then the minimum work input required forthe separation process can be expressed as

(13–54)Wmin,in � �RuT0 ai

Ni ln yi and w�min,in � �RuT0 ai

yi ln yi

Wrev � �RuT0 ai

Ni ln yi and w�rev � �RuT0 ai

yi ln yi

x�destroyed � T0 s�gen � �RuT0 ai

yi ln yi 1per unit mole of mixture 2

X destroyed � T0 Sgen � �RuT0 ai

Ni ln yi 1ideal soluton 2

s�gen � s�out � s�in � ¢smixing � �Ru ai

yi ln yi 1per unit mole of mixture 2

Sgen � Sout � Sin � ¢Smixing � �Ru ai

Ni ln yi 1ideal solution 2

Mixingchamber T0

Wrev � Xdestruction � T0Sgen

A

B

A + B

mixture

FIGURE 13–21For a naturally occurring processduring which no work is produced orconsumed, the reversible work is equalto the exergy destruction.

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Chapter 13 | 703

It can also be expressed in the rate form as

(13–55)

where W.

min,in is the minimum power input required to separate a solution thatapproaches at a rate of (or ) into its compo-nents. The work of separation per unit mass of mixture can be determinedfrom , where Mm is the apparent molar mass of themixture.

The minimum work relations above are for complete separation of thecomponents in the mixture. The required work input will be less if the exit-ing streams are not pure. The reversible work for incomplete separation canbe determined by calculating the minimum separation work for the incomingmixture and the minimum separation works for the outgoing mixtures, andthen taking their difference.

Reversible Mixing ProcessesThe mixing processes that occur naturally are irreversible, and all the workpotential is wasted during such processes. For example, when the fresh waterfrom a river mixes with the saline water in an ocean, an opportunity to pro-duce work is lost. If this mixing is done reversibly (through the use of semi-permeable membranes, for example) some work can be produced. Themaximum amount of work that can be produced during a mixing process isequal to the minimum amount of work input needed for the correspondingseparation process (Fig. 13–22). That is,

(13–56)

Therefore, the minimum work input relations given above for separation canalso be used to determine the maximum work output for mixing.

The minimum work input relations are independent of any hardware orprocess. Therefore, the relations developed above are applicable to any sepa-ration process regardless of actual hardware, system, or process, and can beused for a wide range of separation processes including the desalination ofsea or brackish water.

Second-Law EfficiencyThe second-law efficiency is a measure of how closely a process approxi-mates a corresponding reversible process, and it indicates the range availablefor potential improvements. Noting that the second-law efficiency rangesfrom 0 for a totally irreversible process to 100 percent for a totally reversibleprocess, the second-law efficiency for separation and mixing processes canbe defined as

(13–57)

where W.act,in is the actual power input (or exergy consumption) of the separa-

tion plant and W.act,out is the actual power produced during mixing. Note that

hII,separation �W#

min,in

W#

act,in

�wmin,in

wact,inand hII,mixing �

W#act,out

W#

max,out

�wact,out

wmax,out

Wmax,out,mixing � Wmin,in,separation

wmin,in � w�min,in>Mm

m#

m � N#mMm kg>sN

#m kmol>s

W#min,in � �RuT0 a

i

N#

i ln yi � �N#mRuT0 a

i

yi ln yi 1kW 2

Mixingchamber

A

B

(a) Mixing

(b) Separation

A + B

mixture

A + B

mixture

yA

yB

A

B

yA

yB

Separationunit

Wmax,out = 5 kJ/kg mixture

Wmax,in = 5 kJ/kg mixture

FIGURE 13–22Under reversible conditions, the workconsumed during separation is equalto the work produced during thereverse process of mixing.

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704 | Thermodynamics

the second-law efficiency is always less than 1 since the actual separationprocess requires a greater amount of work input because of irreversibilities.Therefore, the minimum work input and the second-law efficiency provide abasis for comparison of actual separation processes to the “idealized” onesand for assessing the thermodynamic performance of separation plants.

A second-law efficiency for mixing processes can also be defined as theactual work produced during mixing divided by the maximum work potentialavailable. This definition does not have much practical value, however, sinceno effort is done to produce work during most mixing processes and thus thesecond-law efficiency is zero.

Special Case: Separation of a Two Component MixtureConsider a mixture of two components A and B whose mole fractions are yA

and yB. Noting that yB � 1 � yA, the minimum work input required to sepa-rate 1 kmol of this mixture at temperature T0 completely into pure A andpure B is, from Eq. 13–54,

(13–58a)

or

(13–58b)

or, from Eq. 13–55,

(13–58c)

Some separation processes involve the extraction of just one of the compo-nents from a large amount of mixture so that the composition of the remainingmixture remains practically the same. Consider a mixture of two componentsA and B whose mole fractions are yA and yB, respectively. The minimum workrequired to separate 1 kmol of pure component A from the mixture of Nm � NA

� NB kmol (with NA �� 1) is determined by subtracting the minimum workrequired to separate the remaining mixture �RuT0[(NA � 1)ln yA � NB ln yB]from the minimum work required to separate the initial mixture Wmin,in ��RuT0(NA ln yA � NB ln yB). It gives (Fig. 13–23)

(13–59)

The minimum work needed to separate a unit mass (1 kg) of component A isdetermined from the above relation by replacing Ru by RA (or by dividing therelation above by the molar mass of component A) since RA � Ru/MA. Eq.13–59 also gives the maximum amount of work that can be done as one unitof pure component A mixes with a large amount of A � B mixture.

An Application: Desalination ProcessesThe potable water needs of the world is increasing steadily due to populationgrowth, rising living standards, industrialization, and irrigation in agriculture.There are over 10,000 desalination plants in the world, with a total desalted

w�min,in � �RuT0 ln yA � RuT0 ln 11>yA 2 1kJ>kmol A 2

� �m#

mRmT0 1yA ln yA � yB ln yB 2 1kW 2W#

min,in � �N#mRuT0 1yA ln yA � yB ln yB 2

Wmin,in � �RuT0 1NA ln yA � NB ln yB 2 1kJ 2

w�min,in � �RuT0 1yA ln yA � yB ln yB 2 1kJ>kmol mixture 2�

Separationunit

A + B

Separationunit

A + B

yA, yB

yA, yB

1 kmol

pure A

(1 kmol)

A + B

pure A

pure B

(a) Separating 1 kmol of A from a large body of mixture

(b) Complete separation of 1 kmol mixture into its components A and B

wmin,in = –RuT0 ln yA (kJ/kmol A)

�wmin,in = –RuT0 ( yA ln yA + yB ln yB) (kJ/kmol mixture)

FIGURE 13–23The minimum work required toseparate a two-component mixture forthe two limiting cases.

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Chapter 13 | 705

water capacity of over 5 billion gallons a day. Saudi Arabia is the largest userof desalination with about 25 percent of the world capacity, and the UnitedStates is the second largest user with 10 percent. The major desalinationmethods are distillation and reverse osmosis. The relations can be useddirectly for desalination processes, by taking the water (the solvent) to becomponent A and the dissolved salts (the solute) to be component B. Thenthe minimum work needed to produce 1 kg of pure water from a large reser-voir of brackish or seawater at temperature T0 in an environment at T0 is,from Eq. 13–59,

Desalination: (13–60)

where Rw � 0.4615 kJ/kg � K is the gas constant of water and yw is the molefraction of water in brackish or seawater. The relation above also gives themaximum amount of work that can be produced as 1 kg of fresh water (froma river, for example) mixes with seawater whose water mole fraction is yw.

The reversible work associated with liquid flow can also be expressed interms of pressure difference �P and elevation difference �z (potentialenergy) as wmin,in � �P/r � g �z where r is the density of the liquid. Com-bining these relations with Eq. 13–60 gives

(13–61)

and

(13–62)

where �Pmin is the osmotic pressure, which represents the pressure differ-ence across a semipermeable membrane that separates fresh water from thesaline water under equilibrium conditions, r is the density of saline water,and �zmin is the osmotic rise, which represents the vertical distance thesaline water would rise when separated from the fresh water by a membranethat is permeable to water molecules alone (again at equilibrium). For desali-nation processes, �Pmin represents the minimum pressure that the salinewater must be compressed in order to force the water molecules in salinewater through the membrane to the fresh water side during a reverse osmosisdesalination process. Alternately, �zmin represents the minimum height abovethe fresh water level that the saline water must be raised to produce therequired osmotic pressure difference across the membrane to produce freshwater. The �zmin also represents the height that the water with dissolvedorganic matter inside the roots will rise through a tree when the roots aresurrounded by fresh water with the roots acting as semipermeable mem-branes. The reverse osmosis process with semipermeable membranes is alsoused in dialysis machines to purify the blood of patients with failed kidneys.

¢zmin � wmin,in>g � RwT0 ln 11>yw 2 >g 1m 2

¢Pmin � rwmin,in � rRwT0 ln 11>yw 2 1kPa 2

wmin,in � �RwT0 ln 11>yw 2 1kJ>kg pure water 2

EXAMPLE 13–6 Obtaining Fresh Water from Seawater

Fresh water is to be obtained from seawater at 15°C with a salinity of 3.48percent on mass basis (or TDS � 34,800 ppm). Determine (a) the molefractions of the water and the salts in the seawater, (b) the minimum workinput required to separate 1 kg of seawater completely into pure water andpure salts, (c) the minimum work input required to obtain 1 kg of fresh

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706 | Thermodynamics

water from the sea, and (d ) the minimum gauge pressure that the seawatermust be raised if fresh water is to be obtained by reverse osmosis usingsemipermeable membranes.

Solution Fresh water is to be obtained from seawater. The mole fractions ofseawater, the minimum works of separation needed for two limiting cases,and the required pressurization of seawater for reverse osmosis are to bedetermined.Assumptions 1 The seawater is an ideal solution since it is dilute. 2 Thetotal dissolved solids in water can be treated as table salt (NaCl). 3 The envi-ronment temperature is also 15°C.Properties The molar masses of water and salt are Mw � 18.0 kg/kmoland Ms � 58.44 kg/kmol. The gas constant of pure water is Rw � 0.4615kJ/kg · K (Table A–1). The density of seawater is 1028 kg/m3.Analysis (a) Noting that the mass fractions of salts and water in seawaterare mfs � 0.0348 and mfw � 1 � mfs � 0.9652, the mole fractions aredetermined from Eqs. 13–4 and 13–5 to be

(b) The minimum work input required to separate 1 kg of seawater com-pletely into pure water and pure salts is

Therefore, it takes a minimum of 7.98 kJ of work input to separate 1 kg ofseawater into 0.0348 kg of salt and 0.9652 kg (nearly 1 kg) of fresh water.

(c) The minimum work input required to produce 1 kg of fresh water fromseawater is

Note that it takes about 5 times more work to separate 1 kg of seawatercompletely into fresh water and salt than it does to produce 1 kg of freshwater from a large amount of seawater.

� 1.50 kJ/kg fresh water

� 10.4615 kJ>kg # K 2 1288.15 K 2 ln 11>0.9888 2 wmin,in � RwT0 ln 11>yw 2

wmin,in �w�min,in

Mm

�147.2 kJ>kmol

18.44 kg>kmol� 7.98 kJ/kg seawater

� 147.2 kJ>kmol

� � 18.314 kJ>kmol # K2 1288.15 K2 10.9888 ln 0.9888 � 0.0112 ln 0.01122 w�min,in � �RuT0 1yA ln yA � yB ln yB 2 � �RuT0 1yw ln yw � ys ln ys 2

ys � 1 � yw � 1 � 0.9888 � 0.0112 � 1.12%

yw � mfw

Mm

Mw

� 0.9652 18.44 kg>kmol

18.0 kg>kmol� 0.9888

Mm �1

amfi

Mi

�1

mfs

Ms

�mfw

Mw

�1

0.0348

58.44�

0.9652

18.0

� 18.44 kg>kmol

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Chapter 13 | 707

(d ) The osmotic pressure in this case is

which is equal to the minimum gauge pressure to which seawater must becompressed if the fresh water is to be discharged at the local atmosphericpressure. As an alternative to pressurizing, the minimum height above thefresh water level that the seawater must be raised to produce fresh water is(Fig. 13–24)

Discussion The minimum separation works determined above also representthe maximum works that can be produced during the reverse process of mix-ing. Therefore, 7.98 kJ of work can be produced when 0.0348 kg of salt ismixed with 0.9652 kg of water reversibly to produce 1 kg of saline water,and 1.50 kJ of work can be produced as 1 kg of fresh water is mixed withseawater reversibly. Therefore, the power that can be generated as a riverwith a flow rate of 106 m3/s mixes reversibly with seawater through semiper-meable membranes is (Fig. 13–25)

which shows the tremendous amount of power potential wasted as the riversdischarge into the seas.

� 1.5 106 MW

W#

max,out � rV#wmax,out � 11000 kg>m3 2 1106 m3>s 2 11.50 kJ>kg 2 a 1 MW

103 kJ>s b

¢zmin �wmin,in

g�

1.50 kJ>kg

9.81 m>s2 a 1 kg # m>s2

1 Nb a 1000 N # m

1 kJb � 153 m

� 1540 kPa

� 11028 kg>m3 2 10.4615 kPa # m3>kg # K 2 1288.15 K 2 ln 11>0.9888 2 ¢Pmin � rm Rw T0 ln 11>yw 2

Salinewater

∆z

Membrane

P2 P1

Purewater

∆P � P2 � P1

FIGURE 13–24The osmotic pressure and the osmoticrise of saline water.

Fresh and saline water mixing irreversibly

Fresh river water

Sea water salinity � 3.48%

�z � 153 m

Fresh and saline water mixing reversiblythrough semi-permeable membranes, and producing power

FIGURE 13–25Power can be produced by mixing solutions of different concentrations reversibly.

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708 | Thermodynamics

A mixture of two or more gases of fixed chemical composi-tion is called a nonreacting gas mixture The composition of agas mixture is described by specifying either the mole frac-tion or the mass fraction of each component, defined as

where

The apparent (or average) molar mass and gas constant of amixture are expressed as

Also,

Dalton’s law of additive pressures states that the pressureof a gas mixture is equal to the sum of the pressures each gaswould exert if it existed alone at the mixture temperature andvolume. Amagat’s law of additive volumes states that the vol-ume of a gas mixture is equal to the sum of the volumes eachgas would occupy if it existed alone at the mixture tempera-ture and pressure. Dalton’s and Amagat’s laws hold exactlyfor ideal-gas mixtures, but only approximately for real-gasmixtures. They can be expressed as

Dalton’s law:

Amagat’s law:

Here Pi is called the component pressure and Vi is calledthe component volume. Also, the ratio Pi /Pm is called thepressure fraction and the ratio Vi /Vm is called the volumefraction of component i. For ideal gases, Pi and Vi can berelated to yi by

The quantity yiPm is called the partial pressure and the quan-tity yiVm is called the partial volume. The P-v-T behaviorof real-gas mixtures can be predicted by using generalized

Pi

Pm

�Vi

Vm

�Ni

Nm

� yi

Vm �ak

i�1Vi 1Tm, Pm 2

Pm �ak

i�1Pi 1Tm, Vm 2

mfi � yi Mi

Mm

and Mm �1

ak

i�1

mfi

Mi

Mm �mm

Nm

�ak

i�1yiMi and Rm �

Ru

Mm

mm �ak

i�1mi and Nm �a

k

i�1Ni

mfi �mi

mmand yi �

Ni

Nm

SUMMARY

compressibility charts. The compressibility factor of the mix-ture can be expressed in terms of the compressibility factorsof the individual gases as

where Zi is determined either at Tm and Vm (Dalton’s law) or atTm and Pm (Amagat’s law) for each individual gas. The P-v-Tbehavior of a gas mixture can also be predicted approximatelyby Kay’s rule, which involves treating a gas mixture as a puresubstance with pseudocritical properties determined from

The extensive properties of a gas mixture, in general, canbe determined by summing the contributions of each compo-nent of the mixture. The evaluation of intensive properties ofa gas mixture, however, involves averaging in terms of massor mole fractions:

and

These relations are exact for ideal-gas mixtures and approxi-mate for real-gas mixtures. The properties or property changesof individual components can be determined by using ideal-gas or real-gas relations developed in earlier chapters.

cp,m �ak

i�1mficp,i and c�p,m �a

k

i�1yic�p,i

cv,m �ak

i�1mficv,i and c�v,m �a

k

i�1yic�v,i

sm �ak

i�1mfisi and s�m �a

k

i�1yi s�i

hm �ak

i�1mfihi and hm �a

k

i�1yihi

um �ak

i�1mfiui and u�m �a

k

i�1yiu�i

Sm �ak

i�1Si �a

k

i�1mi si �a

k

i�1Ni s�i

Hm �ak

i�1Hi �a

k

i�1mi hi �a

k

i�1Ni hi

Um �ak

i�1Ui �a

k

i�1miui �a

k

i�1Niu�i

P ¿cr,m �ak

i�1yi Pcr,i and T ¿cr,m � a

k

i�1yi Tcr,i

Zm �ak

i�1yi Zi

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Chapter 13 | 709

1. A. Bejan. Advanced Engineering Thermodynamics. 2nded. New York: Wiley Interscience, 1997.

2. Y. A. Çengel, Y. Cerci, and B. Wood, “Second LawAnalysis of Separation Processes of Mixtures,” ASMEInternational Mechanical Engineering Congress andExposition, Nashville, Tennessee, 1999.

3. Y. Cerci, Y. A. Çengel, and B. Wood, “The MinimumSeparation Work for Desalination Processes,” ASME

REFERENCES AND SUGGESTED READING

International Mechanical Engineering Congress andExposition, Nashville, Tennessee, 1999.

4. J. P. Holman. Thermodynamics. 3rd ed. New York:McGraw-Hill, 1980.

5. K. Wark, Jr. Advanced Thermodynamics for Engineers.New York: McGraw-Hill, 1995.

Composition of Gas Mixtures

13–1C What is the apparent gas constant for a gas mixture?Can it be larger than the largest gas constant in the mixture?

13–2C Consider a mixture of two gases. Can the apparentmolar mass of this mixture be determined by simply takingthe arithmetic average of the molar masses of the individualgases? When will this be the case?

13–3C What is the apparent molar mass for a gas mixture?Does the mass of every molecule in the mixture equal theapparent molar mass?

13–4C Consider a mixture of several gases of identicalmasses. Will all the mass fractions be identical? How aboutthe mole fractions?

13–5C The sum of the mole fractions for an ideal-gas mix-ture is equal to 1. Is this also true for a real-gas mixture?

13–6C What are mass and mole fractions?

13–7C Using the definitions of mass and mole fractions,derive a relation between them.

13–8C Somebody claims that the mass and mole fractionsfor a mixture of CO2 and N2O gases are identical. Is thistrue? Why?

13–9C Consider a mixture of two gases A and B. Show thatwhen the mass fractions mfA and mfB are known, the molefractions can be determined from

where MA and MB are the molar masses of A and B.

yA �MB

MA 11>mfA � 1 2 � MB

and yB � 1 � yA

PROBLEMS*

13–10 The composition of moist air is given on a molarbasis to be 78 percent N2, 20 percent O2, and 2 percent watervapor. Determine the mass fractions of the constituents of air.

13–11 A gas mixture has the following composition on amole basis: 60 percent N2 and 40 percent CO2. Determine thegravimetric analysis of the mixture, its molar mass, and gasconstant.

13–12 Repeat Prob. 13–11 by replacing N2 by O2.

13–13 A gas mixture consists of 5 kg of O2, 8 kg of N2,and 10 kg of CO2. Determine (a) the mass fraction of eachcomponent, (b) the mole fraction of each component, and(c) the average molar mass and gas constant of the mixture.

13–14 Determine the mole fractions of a gas mixture thatconsists of 75 percent CH4 and 25 percent CO2 by mass.Also, determine the gas constant of the mixture.

13–15 A gas mixture consists of 8 kmol of H2 and 2 kmol ofN2. Determine the mass of each gas and the apparent gas con-stant of the mixture. Answers: 16 kg, 56 kg, 1.155 kJ/kg · K

13–16E A gas mixture consists of 5 lbmol of H2 and4 lbmol of N2. Determine the mass of each gas and the appar-ent gas constant of the mixture.

13–17 A gas mixture consists of 20 percent O2, 30 percentN2, and 50 percent CO2 on mass basis. Determine the volu-metric analysis of the mixture and the apparent gas constant.

P-v-T Behavior of Gas Mixtures

13–18C Is a mixture of ideal gases also an ideal gas? Givean example.

13–19C Express Dalton’s law of additive pressures. Does thislaw hold exactly for ideal-gas mixtures? How about nonideal-gas mixtures?

13–20C Express Amagat’s law of additive volumes. Does thislaw hold exactly for ideal-gas mixtures? How about nonideal-gas mixtures?

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

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13–21C How is the P-v-T behavior of a component in anideal-gas mixture expressed? How is the P-v-T behavior of acomponent in a real-gas mixture expressed?

13–22C What is the difference between the component pres-sure and the partial pressure? When are these two equivalent?

13–23C What is the difference between the component vol-ume and the partial volume? When are these two equivalent?

13–24C In a gas mixture, which component will have thehigher partial pressure—the one with the higher mole numberor the one with the larger molar mass?

13–25C Consider a rigid tank that contains a mixture oftwo ideal gases. A valve is opened and some gas escapes. Asa result, the pressure in the tank drops. Will the partial pres-sure of each component change? How about the pressurefraction of each component?

13–26C Consider a rigid tank that contains a mixture oftwo ideal gases. The gas mixture is heated, and the pressureand temperature in the tank rise. Will the partial pressure ofeach component change? How about the pressure fraction ofeach component?

13–27C Is this statement correct? The volume of an ideal-gas mixture is equal to the sum of the volumes of each indi-vidual gas in the mixture. If not, how would you correct it?

13–28C Is this statement correct? The temperature of anideal-gas mixture is equal to the sum of the temperatures ofeach individual gas in the mixture. If not, how would youcorrect it?

13–29C Is this statement correct? The pressure of an ideal-gas mixture is equal to the sum of the partial pressures ofeach individual gas in the mixture. If not, how would youcorrect it?

13–30C Explain how a real-gas mixture can be treated as apseudopure substance using Kay’s rule.

13–31 A rigid tank contains 8 kmol of O2 and 10 kmol ofCO2 gases at 290 K and 150 kPa. Estimate the volume of thetank. Answer: 289 m3

13–32 Repeat Prob. 13–31 for a temperature of 400 K.

13–33 A rigid tank contains 0.5 kmol of Ar and 2 kmol of N2at 250 kPa and 280 K. The mixture is now heated to 400 K.Determine the volume of the tank and the final pressure of themixture.

13–34 A gas mixture at 300 K and 200 kPa consists of 1 kgof CO2 and 3 kg of CH4. Determine the partial pressure ofeach gas and the apparent molar mass of the gas mixture.

13–35E A gas mixture at 600 R and 20 psia consists of1 lbm of CO2 and 3 lbm of CH4. Determine the partial pressureof each gas and the apparent molar mass of the gas mixture.

13–36 A 0.3-m3 rigid tank contains 0.6 kg of N2 and 0.4 kgof O2 at 300 K. Determine the partial pressure of each gas

710 | Thermodynamics

and the total pressure of the mixture. Answers: 178.1 kPa,103.9 kPa, 282.0 kPa

13–37 A gas mixture at 350 K and 300 kPa has the follow-ing volumetric analysis: 65 percent N2, 20 percent O2, and15 percent CO2. Determine the mass fraction and partial pres-sure of each gas.

13–38 A rigid tank that contains 1 kg of N2 at 25°C and300 kPa is connected to another rigid tank that contains 3 kgof O2 at 25°C and 500 kPa. The valve connecting the twotanks is opened, and the two gases are allowed to mix. If thefinal mixture temperature is 25°C, determine the volume ofeach tank and the final mixture pressure. Answers: 0.295 m3,0.465 m3, 422 kPa

O2

3 kg25°C

500 kPa

N2

1 kg25°C

300 kPa

FIGURE P13–38

Ar1 kmol220 K5 MPa

N2

3 kmol190 K8 MPa

FIGURE P13–40

13–39 A volume of 0.3 m3 of O2 at 200 K and 8 MPa ismixed with 0.5 m3 of N2 at the same temperature and pres-sure, forming a mixture at 200 K and 8 MPa. Determine thevolume of the mixture, using (a) the ideal-gas equation ofstate, (b) Kay’s rule, and (c) the compressibility chart andAmagat’s law. Answers: (a) 0.8 m3, (b) 0.79 m3, (c) 0.80 m3

13–40 A rigid tank contains 1 kmol of Ar gas at 220 Kand 5 MPa. A valve is now opened, and 3 kmol

of N2 gas is allowed to enter the tank at 190 K and 8 MPa.The final mixture temperature is 200 K. Determine the pres-sure of the mixture, using (a) the ideal-gas equation of stateand (b) the compressibility chart and Dalton’s law.

13–41 Reconsider Prob. 13–40. Using EES (or other)software, study the effect of varying the moles

of nitrogen supplied to the tank over the range of 1 to 10kmol of N2. Plot the final pressure of the mixture as a func-tion of the amount of nitrogen supplied using the ideal-gasequation of state and the compressibility chart with Dalton’slaw.

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Chapter 13 | 711

13–42E A rigid tank contains 1 lbmol of argon gas at 400 Rand 750 psia. A valve is now opened, and 3 lbmol of N2 gasis allowed to enter the tank at 340 R and 1200 psia. The finalmixture temperature is 360 R. Determine the pressure of themixture, using (a) the ideal-gas equation of state and (b) thecompressibility chart and Dalton’s law. Answers: (a) 2700psia, (b) 2507 psia

Properties of Gas Mixtures

13–43C Is the total internal energy of an ideal-gas mixtureequal to the sum of the internal energies of each individual gasin the mixture? Answer the same question for a real-gas mixture.

13–44C Is the specific internal energy of a gas mixtureequal to the sum of the specific internal energies of each indi-vidual gas in the mixture?

13–45C Answer Prob. 13–43C and 13–44C for entropy.

13–46C Is the total internal energy change of an ideal-gasmixture equal to the sum of the internal energy changes ofeach individual gas in the mixture? Answer the same questionfor a real-gas mixture.

13–47C When evaluating the entropy change of the compo-nents of an ideal-gas mixture, do we have to use the partialpressure of each component or the total pressure of the mixture?

13–48C Suppose we want to determine the enthalpy changeof a real-gas mixture undergoing a process. The enthalpychange of each individual gas is determined by using the gen-eralized enthalpy chart, and the enthalpy change of the mixtureis determined by summing them. Is this an exact approach?Explain.

13–49 A process requires a mixture that is 21 percent oxygen, 78 percent nitrogen, and 1 percent argon by volume.All three gases are supplied from separate tanks to an adia-batic, constant-pressure mixing chamber at 200 kPa but at dif-ferent temperatures. The oxygen enters at 10°C, the nitrogen at60°C, and the argon at 200°C. Determine the total entropychange for the mixing process per unit mass of mixture.

13–50 A mixture that is 15 percent carbon dioxide, 5 percentcarbon monoxide, 10 percent oxygen, and 70 percent nitrogenby volume undergoes an adiabatic compression process havinga compression ratio of 8:1. If the initial state of the mixture is300 K and 100 kPa, determine the makeup of the mixture on amass basis and the internal energy change per unit mass ofmixture.

13–51 Propane and air are supplied to an internal combus-tion engine such that the air-fuel ratio is 16:1 when the pres-sure is 95 kPa and the temperature is 30°C. The compressionratio of the engine is 9.5:1. If the compression process isisentropic, determine the required work input for this com-pression process, in kJ/kg of mixture.

13–52 An insulated rigid tank is divided into two compart-ments by a partition. One compartment contains 2.5 kmol of

CO2 at 27°C and 200 kPa, and the other compartment con-tains 7.5 kmol of H2 gas at 40°C and 400 kPa. Now the parti-tion is removed, and the two gases are allowed to mix.Determine (a) the mixture temperature and (b) the mixturepressure after equilibrium has been established. Assume con-stant specific heats at room temperature for both gases.

13–53 A 0.9-m3 rigid tank is divided into two equal com-partments by a partition. One compartment contains Ne at20°C and 100 kPa, and the other compartment contains Ar at50°C and 200 kPa. Now the partition is removed, and the twogases are allowed to mix. Heat is lost to the surrounding airduring this process in the amount of 15 kJ. Determine (a) thefinal mixture temperature and (b) the final mixture pressure.Answers: (a) 16.2°C, (b) 138.9 kPa

13–54 Repeat Prob. 13–53 for a heat loss of 8 kJ.

13–55 Ethane (C2H6) at 20°C and 200 kPa and meth-ane (CH4) at 45°C and 200 kPa enter an adia-

batic mixing chamber. The mass flow rate of ethane is 9 kg/s,which is twice the mass flow rate of methane. Determine(a) the mixture temperature and (b) the rate of entropy gener-ation during this process, in kW/K. Take T0 � 25°C.

13–56 Reconsider Prob. 13–55. Using EES (or other)software, determine the effect of the mass frac-

tion of methane in the mixture on the mixture temperatureand the rate of exergy destruction. The total mass flow rate ismaintained constant at 13.5 kg/s, and the mass fraction ofmethane is varied from 0 to 1. Plot the mixture temperatureand the rate of exergy destruction against the mass fraction,and discuss the results.

13–57 An equimolar mixture of helium and argon gases is tobe used as the working fluid in a closed-loop gas-turbine cycle.

H2

7.5 kmol40°C

400 kPa

CO2

2.5 kmol27°C

200 kPa

FIGURE P13–52

2.5 MPa1300 K

200 kPa

He - Arturbine

FIGURE P13–57

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The mixture enters the turbine at 2.5 MPa and 1300 K andexpands isentropically to a pressure of 200 kPa. Determine thework output of the turbine per unit mass of the mixture.

13–58E A mixture of 80 percent N2 and 20 percentCO2 gases (on a mass basis) enters the nozzle

of a turbojet engine at 90 psia and 1800 R with a low veloc-ity, and it expands to a pressure of 12 psia. If the isentropicefficiency of the nozzle is 92 percent, determine (a) the exittemperature and (b) the exit velocity of the mixture. Assumeconstant specific heats at room temperature.

13–59E Reconsider Prob. 13–58E. Using EES (orother) software, first solve the stated problem

and then, for all other conditions being the same, resolve theproblem to determine the composition of the nitrogen andcarbon dioxide that is required to have an exit velocity of2600 ft /s at the nozzle exit.

13–60 A piston–cylinder device contains a mixture of0.5 kg of H2 and 1.6 kg of N2 at 100 kPa and 300 K. Heat isnow transferred to the mixture at constant pressure until thevolume is doubled. Assuming constant specific heats at theaverage temperature, determine (a) the heat transfer and(b) the entropy change of the mixture.

13–61 An insulated tank that contains 1 kg of O2 at 15°Cand 300 kPa is connected to a 2-m3 uninsulated tank thatcontains N2 at 50°C and 500 kPa. The valve connecting thetwo tanks is opened, and the two gases form a homogeneousmixture at 25°C. Determine (a) the final pressure in the tank,(b) the heat transfer, and (c) the entropy generated during thisprocess. Assume T0 � 25°C.Answers: (a) 444.6 kPa, (b) 187.2 kJ, (c) 0.962 kJ/K

712 | Thermodynamics

during this process by treating the mixture (a) as an ideal gasand (b) as a nonideal gas and using Amagat’s law.Answers: (a) 4273 kJ, (b) 4745 kJ

13–64 Determine the total entropy change and exergydestruction associated with the process described in Prob.13–63 by treating the mixture (a) as an ideal gas and (b) as anonideal gas and using Amagat’s law. Assume constant spe-cific heats at room temperature and take T0 � 30°C.

13–65 Air, which may be considered as a mixture of 79 per-cent N2 and 21 percent O2 by mole numbers, is compressedisothermally at 200 K from 4 to 8 MPa in a steady-flowdevice. The compression process is internally reversible, andthe mass flow rate of air is 2.9 kg/s. Determine the powerinput to the compressor and the rate of heat rejection by treat-ing the mixture (a) as an ideal gas and (b) as a nonideal gasand using Amagat’s law. Answers: (a) 115.3 kW, 115.3 kW,(b) 143.6 kW, 94.2 kW

O2

1 kg15°C

300 kPa

N2

2 m3

50°C500 kPa

FIGURE P13–61

Heat

6 kg H2

21 kg N2

160 K5 MPa

FIGURE P13–63

200 K8 MPa

200 K4 MPa

79% N221% O2

W·

FIGURE P13–6513–62 Reconsider Prob. 13–61. Using EES (or other)

software, compare the results obtained assum-ing ideal-gas behavior with constant specific heats at theaverage temperature, and using real-gas data obtained fromEES by assuming variable specific heats over the temperaturerange.

13–63 A piston–cylinder device contains 6 kg of H2 and21 kg of N2 at 160 K and 5 MPa. Heat is now transferred tothe device, and the mixture expands at constant pressure untilthe temperature rises to 200 K. Determine the heat transfer

13–66 Reconsider Prob. 13–65. Using EES (or other)software, compare the results obtained by

assuming ideal behavior, real gas behavior with Amagat’slaw, and real gas behavior with EES data.

13–67 The combustion of a hydrocarbon fuel with air resultsin a mixture of products of combustion having the compositionon a volume basis as follows: 4.89 percent carbon dioxide,

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Chapter 13 | 713

6.50 percent water vapor, 12.20 percent oxygen, and 76.41 per-cent nitrogen. Determine the average molar mass of the mix-ture, the average specific heat at constant pressure of themixture at 600 K, in kJ/kmol � K, and the partial pressure ofthe water vapor in the mixture for a mixture pressure of200 kPa.

13–68 A mixture that is 20 percent carbon dioxide, 10 per-cent oxygen, and 70 percent nitrogen by volume undergoes aprocess from 300 K and 100 kPa to 500 K and 400 kPa.Determine the makeup of the mixture on a mass basis and theenthalpy change per unit mass of mixture.

Special Topic: Chemical Potential and the Separation Work of Mixtures

13–69C It is common experience that two gases brought intocontact mix by themselves. In the future, could it be possibleto invent a process that will enable a mixture to separate intoits components by itself without any work (or exergy) input?

13–70C A 2-L liquid is mixed with 3 L of another liquid,forming a homogeneous liquid solution at the same tempera-ture and pressure. Can the volume of the solution be more orless than the 5 L? Explain.

13–71C A 2-L liquid at 20°C is mixed with 3 L of anotherliquid at the same temperature and pressure in an adiabaticcontainer, forming a homogeneous liquid solution. Someoneclaims that the temperature of the mixture rose to 22°C aftermixing. Another person refutes the claim, saying that thiswould be a violation of the first law of thermodynamics. Whodo you think is right?

13–72C What is an ideal solution? Comment on the vol-ume change, enthalpy change, entropy change, and chemicalpotential change during the formation of ideal and nonidealsolutions.

13–73 Brackish water at 12°C with total dissolved solid con-tent of TDS � 780 ppm (a salinity of 0.078 percent on massbasis) is to be used to produce fresh water with negligible saltcontent at a rate of 280 L/s. Determine the minimum powerinput required. Also, determine the minimum height to whichthe brackish water must be pumped if fresh water is to beobtained by reverse osmosis using semipermeable membranes.

13–74 A river is discharging into the ocean at a rate of400,000 m3/s. Determine the amount of power that can be gen-erated if the river water mixes with the ocean water reversibly.Take the salinity of the ocean to be 3.5 percent on mass basis,and assume both the river and the ocean are at 15°C.

13–75 Reconsider Prob. 13–74. Using EES (or other)software, investigate the effect of the salinity of

the ocean on the maximum power generated. Let the salinityvary from 0 to 5 percent. Plot the power produced versus thesalinity of the ocean, and discuss the results.

13–76E Fresh water is to be obtained from brackish waterat 65°F with a salinity of 0.12 percent on mass basis (orTDS � 1200 ppm). Determine (a) the mole fractions of thewater and the salts in the brackish water, (b) the minimumwork input required to separate 1 lbm of brackish water com-pletely into pure water and pure salts, and (c) the minimumwork input required to obtain 1 lbm of fresh water.

13–77 A desalination plant produces fresh water from sea-water at 10°C with a salinity of 3.2 percent on mass basis at arate of 1.4 m3/s while consuming 8.5 MW of power. The saltcontent of the fresh water is negligible, and the amount offresh water produced is a small fraction of the seawater used.Determine the second-law efficiency of this plant.

13–78 Fresh water is obtained from seawater at a rate of0.5 m3/s by a desalination plant that consumes 3.3 MW ofpower and has a second-law efficiency of 18 percent. Deter-mine the power that can be produced if the fresh water pro-duced is mixed with the seawater reversibly.

Review Problems

13–79 Air has the following composition on a mole basis:21 percent O2, 78 percent N2, and 1 percent Ar. Determinethe gravimetric analysis of air and its molar mass. Answers:23.2 percent O2, 75.4 percent N2, 1.4 percent Ar, 28.96 kg/kmol

13–80 Using Amagat’s law, show that

for a real-gas mixture of k gases, where Z is the compressibil-ity factor.

13–81 Using Dalton’s law, show that

for a real-gas mixture of k gases, where Z is the compressibil-ity factor.

13–82 A mixture of carbon dioxide and nitrogen flowsthrough a converging nozzle. The mixture leaves the nozzleat a temperature of 500 K with a velocity of 360 m/s. If thevelocity is equal to the speed of sound at the exit tempera-ture, determine the required makeup of the mixture on a massbasis.

13–83 A piston–cylinder device contains products of com-bustion from the combustion of a hydrocarbon fuel with air.The combustion process results in a mixture that has the com-position on a volume basis as follows: 4.89 percent carbondioxide, 6.50 percent water vapor, 12.20 percent oxygen, and76.41 percent nitrogen. This mixture is initially at 1800 K and1 MPa and expands in an adiabatic, reversible process to200 kPa. Determine the work done on the piston by the gas,in kJ/kg of mixture. Treat the water vapor as an ideal gas.

Zm �ak

i�1yi Zi

Zm �ak

i�1yi Zi

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13–84 A rigid tank contains 2 kmol of N2 and 6 kmol ofCH4 gases at 200 K and 12 MPa. Estimate the volume of thetank, using (a) the ideal-gas equation of state, (b) Kay’s rule,and (c) the compressibility chart and Amagat’s law.

13–85 A steady stream of equimolar N2 and CO2 mixture at100 kPa and 18°C is to be separated into N2 and CO2 gases at100 kPa and 18°C. Determine the minimum work requiredper unit mass of mixture to accomplish this separationprocess. Assume T0 � 18°C.

13–86 A gas mixture consists of O2 and N2. The ratio of themole numbers of N2 to O2 is 3:1. This mixture is heated dur-ing a steady-flow process from 180 to 210 K at a constantpressure of 8 MPa. Determine the heat transfer during thisprocess per mole of the mixture, using (a) the ideal-gasapproximation and (b) Kay’s rule.

13–87 Reconsider Prob. 13–86. Using EES (or other)software, investigate the effect of the mole

fraction of oxygen in the mixture on heat transfer using real-gas behavior with EES data. Let the mole fraction of oxygenvary from 0 to 1. Plot the heat transfer against the mole frac-tion, and discuss the results.

13–88 Determine the total entropy change and exergydestruction associated with the process described in Prob.13–86, using (a) the ideal-gas approximation and (b) Kay’srule. Assume constant specific heats and T0 � 30°C.

13–89 A rigid tank contains a mixture of 4 kg of He and8 kg of O2 at 170 K and 7 MPa. Heat is now transferred to thetank, and the mixture temperature rises to 220 K. Treating theHe as an ideal gas and the O2 as a nonideal gas, determine(a) the final pressure of the mixture and (b) the heat transfer.

13–90 A mixture of 60 percent carbon dioxide and 40 per-cent methane on a mole basis expands through a turbine from1600 K and 800 kPa to 100 kPa. The volume flow rate at theturbine entrance is 10 L/s. Determine the rate of work done bythe mixture using (a) ideal-gas approximation and (b) Kay’srule.

13–91 A pipe fitted with a closed valve connects two tanks.One tank contains a 5-kg mixture of 62.5 percent CO2 and37.5 percent O2 on a mole basis at 30°C and 125 kPa. Thesecond tank contains 10 kg of N2 at 15°C and 200 kPa. Thevalve in the pipe is opened and the gases are allowed to mix.During the mixing process 100 kJ of heat energy is suppliedto the combined tanks. Determine the final pressure and tem-perature of the mixture and the total volume of the mixture.

13–92 Using EES (or other) software, write a programto determine the mole fractions of the compo-

nents of a mixture of three gases with known molar masseswhen the mass fractions are given, and to determine the massfractions of the components when the mole fractions are given.Run the program for a sample case, and give the results.

714 | Thermodynamics

13–93 Using EES (or other) software, write a programto determine the apparent gas constant, con-

stant volume specific heat, and internal energy of a mixtureof three ideal gases when the mass fractions and other prop-erties of the constituent gases are given. Run the program fora sample case, and give the results.

13–94 Using EES (or other) software, write a programto determine the entropy change of a mixture

of three ideal gases when the mass fractions and other prop-erties of the constituent gases are given. Run the program fora sample case, and give the results.

Fundamentals of Engineering (FE) Exam Problems

13–95 An ideal-gas mixture whose apparent molar mass is36 kg/kmol consists of N2 and three other gases. If the molefraction of nitrogen is 0.30, its mass fraction is

(a) 0.15 (b) 0.23 (c) 0.30 (d ) 0.39 (e) 0.70

13–96 An ideal-gas mixture consists of 2 kmol of N2 and6 kmol of CO2. The mass fraction of CO2 in the mixture is

(a) 0.175 (b) 0.250 (c) 0.500 (d ) 0.750 (e) 0.875

13–97 An ideal-gas mixture consists of 2 kmol of N2 and4 kmol of CO2. The apparent gas constant of the mixture is

(a) 0.215 kJ/kg � K (b) 0.225 kJ/kg � K (c) 0.243 kJ/kg � K(d) 0.875 kJ/kg � K (e) 1.24 kJ/kg � K

13–98 A rigid tank is divided into two compartments by apartition. One compartment contains 3 kmol of N2 at 600 kPaand the other compartment contains 7 kmol of CO2 at 200kPa. Now the partition is removed, and the two gases form ahomogeneous mixture at 300 kPa. The partial pressure of N2in the mixture is

(a) 75 kPa (b) 90 kPa (c) 150 kPa (d ) 175 kPa (e) 225 kPa

13–99 An 80-L rigid tank contains an ideal-gas mixture of5 g of N2 and 5 g of CO2 at a specified pressure and temper-ature. If N2 were separated from the mixture and stored atmixture temperature and pressure, its volume would be

(a) 32 L (b) 36 L (c) 40 L (d ) 49 L (e) 80 L

13–100 An ideal-gas mixture consists of 3 kg of Ar and6 kg of CO2 gases. The mixture is now heated at constantvolume from 250 K to 350 K. The amount of heat transfer is

(a) 374 kJ (b) 436 kJ (c) 488 kJ(d ) 525 kJ (e) 664 kJ

13–101 An ideal-gas mixture consists of 30 percent heliumand 70 percent argon gases by mass. The mixture is nowexpanded isentropically in a turbine from 400°C and 1.2 MPato a pressure of 200 kPa. The mixture temperature at turbineexit is

(a) 195°C (b) 56°C (c) 112°C(d ) 130°C (e) 400°C

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Chapter 13 | 715

13–102 One compartment of an insulated rigid tank con-tains 2 kmol of CO2 at 20°C and 150 kPa while the othercompartment contains 5 kmol of H2 gas at 35°C and 300 kPa.Now the partition between the two gases is removed, and thetwo gases form a homogeneous ideal-gas mixture. The tem-perature of the mixture is

(a) 25°C (b) 29°C (c) 22°C (d ) 32°C (e) 34°C

13–103 A piston–cylinder device contains an ideal-gas mix-ture of 3 kmol of He gas and 7 kmol of Ar gas at 50°C and400 kPa. Now the gas expands at constant pressure until itsvolume doubles. The amount of heat transfer to the gas mix-ture is

(a) 6.2 MJ (b) 4.2 MJ (c) 27 MJ(d ) 10 MJ (e) 67 MJ

13–104 An ideal-gas mixture of helium and argon gaseswith identical mass fractions enters a turbine at 1200 K and1 MPa at a rate of 0.3 kg/s, and expands isentropically to100 kPa. The power output of the turbine is

(a) 478 kW (b) 619 kW (c) 926 KW(d ) 729 kW (e) 564 kW

Design and Essay Problem

13–105 Prolonged exposure to mercury even at relativelylow but toxic concentrations in the air is known to cause per-manent mental disorders, insomnia, and pain and numbnessin the hands and the feet, among other things. Therefore, themaximum allowable concentration of mercury vapor in theair at work places is regulated by federal agencies. These reg-ulations require that the average level of mercury concentra-tion in the air does not exceed 0.1 mg/m3.

Consider a mercury spill that occurs in an airtight storageroom at 20°C in San Francisco during an earthquake. Calcu-late the highest level of mercury concentration in the air thatcan occur in the storage room, in mg/m3, and determine if itis within the safe level. The vapor pressure of mercury at20°C is 0.173 Pa. Propose some guidelines to safeguardagainst the formation of toxic concentrations of mercuryvapor in air in storage rooms and laboratories.

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Chapter 14GAS–VAPOR MIXTURES AND AIR-CONDITIONING

| 717

At temperatures below the critical temperature, the gasphase of a substance is frequently referred to as avapor. The term vapor implies a gaseous state that is

close to the saturation region of the substance, raising thepossibility of condensation during a process.

In Chap. 13, we discussed mixtures of gases that are usu-ally above their critical temperatures. Therefore, we were notconcerned about any of the gases condensing during aprocess. Not having to deal with two phases greatly simplifiedthe analysis. When we are dealing with a gas–vapor mixture,however, the vapor may condense out of the mixture during aprocess, forming a two-phase mixture. This may complicatethe analysis considerably. Therefore, a gas–vapor mixtureneeds to be treated differently from an ordinary gas mixture.

Several gas–vapor mixtures are encountered in engineer-ing. In this chapter, we consider the air–water-vapor mixture,which is the most commonly encountered gas–vapor mixturein practice. We also discuss air-conditioning, which is the pri-mary application area of air–water-vapor mixtures.

ObjectivesThe objectives of Chapter 14 are to:

• Differentiate between dry air and atmospheric air.

• Define and calculate the specific and relative humidity ofatmospheric air.

• Calculate the dew-point temperature of atmospheric air.

• Relate the adiabatic saturation temperature and wet-bulbtemperatures of atmospheric air.

• Use the psychrometric chart as a tool to determine theproperties of atmospheric air.

• Apply the principles of the conservation of mass and energyto various air-conditioning processes.

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14–1 � DRY AND ATMOSPHERIC AIRAir is a mixture of nitrogen, oxygen, and small amounts of some othergases. Air in the atmosphere normally contains some water vapor (or mois-ture) and is referred to as atmospheric air. By contrast, air that contains nowater vapor is called dry air. It is often convenient to treat air as a mixtureof water vapor and dry air since the composition of dry air remains rela-tively constant, but the amount of water vapor changes as a result of con-densation and evaporation from oceans, lakes, rivers, showers, and even thehuman body. Although the amount of water vapor in the air is small, it playsa major role in human comfort. Therefore, it is an important considerationin air-conditioning applications.

The temperature of air in air-conditioning applications ranges from about�10 to about 50°C. In this range, dry air can be treated as an ideal gas witha constant cp value of 1.005 kJ/kg · K [0.240 Btu/lbm · R] with negligibleerror (under 0.2 percent), as illustrated in Fig. 14–1. Taking 0°C as the ref-erence temperature, the enthalpy and enthalpy change of dry air can bedetermined from

(14–1a)

and

(14–1b)

where T is the air temperature in °C and �T is the change in temperature. Inair-conditioning processes we are concerned with the changes in enthalpy�h, which is independent of the reference point selected.

It certainly would be very convenient to also treat the water vapor in theair as an ideal gas and you would probably be willing to sacrifice someaccuracy for such convenience. Well, it turns out that we can have the con-venience without much sacrifice. At 50°C, the saturation pressure of wateris 12.3 kPa. At pressures below this value, water vapor can be treated as anideal gas with negligible error (under 0.2 percent), even when it is a satu-rated vapor. Therefore, water vapor in air behaves as if it existed alone andobeys the ideal-gas relation Pv � RT. Then the atmospheric air can betreated as an ideal-gas mixture whose pressure is the sum of the partial pres-sure of dry air* Pa and that of water vapor Pv:

(14–2)

The partial pressure of water vapor is usually referred to as the vapor pres-sure. It is the pressure water vapor would exert if it existed alone at thetemperature and volume of atmospheric air.

Since water vapor is an ideal gas, the enthalpy of water vapor is a functionof temperature only, that is, h � h(T ). This can also be observed from the T-s diagram of water given in Fig. A–9 and Fig. 14–2 where the constant-enthalpy lines coincide with constant-temperature lines at temperatures

P � Pa � Pv 1kPa 2

¢hdry air � cp¢T � 11.005 kJ>kg # °C 2 ¢T 1kJ>kg 2

hdry air � cpT � 11.005 kJ>kg # °C 2T 1kJ>kg 2

718 | Thermodynamics

T, °C

s

h = const.

50

FIGURE 14–2At temperatures below 50°C, the h � constant lines coincide with theT � constant lines in the superheatedvapor region of water.

DRY AIR

T,°C cp,kJ/kg ·°C

–100

1020304050

1.00381.00411.00451.00491.00541.00591.0065

FIGURE 14–1The cp of air can be assumed to beconstant at 1.005 kJ/kg · °C in thetemperature range �10 to 50°C withan error under 0.2 percent.

*Throughout this chapter, the subscript a denotes dry air and the subscript v denoteswater vapor.

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below 50°C. Therefore, the enthalpy of water vapor in air can be taken to beequal to the enthalpy of saturated vapor at the same temperature. That is,

(14–3)

The enthalpy of water vapor at 0°C is 2500.9 kJ/kg. The average cp value ofwater vapor in the temperature range �10 to 50°C can be taken to be 1.82kJ/kg · °C. Then the enthalpy of water vapor can be determined approxi-mately from

(14–4)

or

(14–5)

in the temperature range �10 to 50°C (or 15 to 120°F), with negligibleerror, as shown in Fig. 14–3.

14–2 � SPECIFIC AND RELATIVE HUMIDITY OF AIRThe amount of water vapor in the air can be specified in various ways.Probably the most logical way is to specify directly the mass of water vaporpresent in a unit mass of dry air. This is called absolute or specific humid-ity (also called humidity ratio) and is denoted by v:

(14–6)

The specific humidity can also be expressed as

(14–7)

or

(14–8)

where P is the total pressure.Consider 1 kg of dry air. By definition, dry air contains no water vapor,

and thus its specific humidity is zero. Now let us add some water vapor tothis dry air. The specific humidity will increase. As more vapor or moistureis added, the specific humidity will keep increasing until the air can hold nomore moisture. At this point, the air is said to be saturated with moisture,and it is called saturated air. Any moisture introduced into saturated airwill condense. The amount of water vapor in saturated air at a specifiedtemperature and pressure can be determined from Eq. 14–8 by replacing Pvby Pg, the saturation pressure of water at that temperature (Fig. 14–4).

The amount of moisture in the air has a definite effect on how comfort-able we feel in an environment. However, the comfort level depends moreon the amount of moisture the air holds (mv) relative to the maximumamount of moisture the air can hold at the same temperature (mg). The ratioof these two quantities is called the relative humidity f (Fig. 14–5)

(14–9)f �mv

mg�

PvV>RvT

PgV>RvT�

Pv

Pg

v �0.622Pv

P � Pv

1kg water vapor>kg dry air 2

v �mv

ma�

PvV>RvT

PaV>RaT�

Pv >Rv

Pa >Ra

� 0.622 Pv

Pa

v �mv

ma1kg water vapor>kg dry air 2

hg 1T 2 � 1060.9 � 0.435T 1Btu>lbm 2 T in °F

hg 1T 2 � 2500.9 � 1.82T 1kJ>kg 2 T in °C

hv 1T, low P 2 � hg 1T 2

Chapter 14 | 719

WATER VAPOR

–100

1020304050

2482.12500.92519.22537.42555.62573.52591.3

2482.72500.92519.12537.32555.52573.72591.9

–0.60.00.10.10.1

–0.2–0.6

hg,kJ/kg

T,°C Table A-4 Eq. 14-4 Difference,

kJ/kg

FIGURE 14–3In the temperature range �10 to 50°C,the hg of water can be determinedfrom Eq. 14–4 with negligible error.

AIRAIR 2525°C,100 kC,100 kPaPa

(Psat,Hsat,H 2O @ 25O @ 25°C = 3.1698 k = 3.1698 kPa)Pa)

Pv = = 0Pv < 3.1698 k < 3.1698 kPaPaPv = 3.1698 k = 3.1698 kPaPa

dry airdry airunsaturated airunsaturated airsaturated airsaturated air

FIGURE 14–4For saturated air, the vapor pressure isequal to the saturation pressure ofwater.

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where

(14–10)

Combining Eqs. 14–8 and 14–9, we can also express the relative humidity as

(14–11a, b)

The relative humidity ranges from 0 for dry air to 1 for saturated air. Notethat the amount of moisture air can hold depends on its temperature. There-fore, the relative humidity of air changes with temperature even when itsspecific humidity remains constant.

Atmospheric air is a mixture of dry air and water vapor, and thusthe enthalpy of air is expressed in terms of the enthalpies of the dry air andthe water vapor. In most practical applications, the amount of dry air in theair–water-vapor mixture remains constant, but the amount of water vaporchanges. Therefore, the enthalpy of atmospheric air is expressed per unitmass of dry air instead of per unit mass of the air–water vapor mixture.

The total enthalpy (an extensive property) of atmospheric air is the sum ofthe enthalpies of dry air and the water vapor:

Dividing by ma gives

or

(14–12)

since hv � hg (Fig. 14–6).Also note that the ordinary temperature of atmospheric air is frequently

referred to as the dry-bulb temperature to differentiate it from other formsof temperatures that shall be discussed.

h � ha � vhg 1kJ>kg dry air 2

h �Hma

� ha �mv

ma hv � ha � vhv

H � Ha � Hv � maha � mv hv

f �vP

10.622 � v 2Pg

and v �0.622fPg

P � fPg

Pg � Psat @ T

720 | Thermodynamics

AIRAIR2525°C,1 atmC,1 atm

ma = = 1 kg1 kgmv = =

mv, , maxmax = = 0.01 kg0.01 kg0.02 kg0.02 kg

Specific humidity: Specific humidity: ω = 0.01 = 0.01

Relative humidity: Relative humidity: φ = 50% = 50%

kg Hkg H2O

kg dry air kg dry air

FIGURE 14–5Specific humidity is the actual amountof water vapor in 1 kg of dry air,whereas relative humidity is the ratioof the actual amount of moisture in the air at a given temperature to themaximum amount of moisture air canhold at the same temperature.

moistureω kg

hg

Dry air1 kgha

(1 + ω) kg ofmoist air

h = ha + ωhg,kJ/kg dry air

FIGURE 14–6The enthalpy of moist (atmospheric)air is expressed per unit mass of dryair, not per unit mass of moist air.

T = 25°CP = 100 kPaφ = 75%

ROOM

5 m × 5 m × 3 m

FIGURE 14–7Schematic for Example 14–1.

EXAMPLE 14–1 The Amount of Water Vapor in Room Air

A 5-m � 5-m � 3-m room shown in Fig. 14–7 contains air at 25°C and 100kPa at a relative humidity of 75 percent. Determine (a) the partial pressureof dry air, (b) the specific humidity, (c) the enthalpy per unit mass of the dryair, and (d ) the masses of the dry air and water vapor in the room.

Solution The relative humidity of air in a room is given. The dry air pres-sure, specific humidity, enthalpy, and the masses of dry air and water vaporin the room are to be determined.Assumptions The dry air and the water vapor in the room are ideal gases.Properties The constant-pressure specific heat of air at room temperature iscp�1.005 kJ/kg · K (Table A–2a). For water at 25°C, we have Tsat � 3.1698kPa and hg � 2546.5 kJ/kg (Table A–4).

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14–3 � DEW-POINT TEMPERATUREIf you live in a humid area, you are probably used to waking up most summermornings and finding the grass wet. You know it did not rain the night before.So what happened? Well, the excess moisture in the air simply condensed onthe cool surfaces, forming what we call dew. In summer, a considerableamount of water vaporizes during the day. As the temperature falls during the

Chapter 14 | 721

Analysis (a) The partial pressure of dry air can be determined from Eq. 14–2:

where

Thus,

(b) The specific humidity of air is determined from Eq. 14–8:

(c) The enthalpy of air per unit mass of dry air is determined fromEq. 14–12:

The enthalpy of water vapor (2546.5 kJ/kg) could also be determined fromthe approximation given by Eq. 14–4:

which is almost identical to the value obtained from Table A–4.

(d ) Both the dry air and the water vapor fill the entire room completely.Therefore, the volume of each gas is equal to the volume of the room:

The masses of the dry air and the water vapor are determined from the ideal-gas relation applied to each gas separately:

The mass of the water vapor in the air could also be determined fromEq. 14–6:

mv � vma � 10.0152 2 185.61 kg 2 � 1.30 kg

mv �PvVv

RvT�

12.38 kPa 2 175 m3 210.4615 kPa # m3>kg # K 2 1298 K 2 � 1.30 kg

ma �PaVa

RaT�

197.62 kPa 2 175 m3 210.287 kPa # m3>kg # K 2 1298 K 2 � 85.61 kg

Va � Vv � Vroom � 15 m 2 15 m 2 13 m 2 � 75 m3

hg @ 25°C � 2500.9 � 1.82 125 2 � 2546.4 kJ>kg

� 63.8 kJ/kg dry air

� 11.005 kJ>kg # °C 2 125°C 2 � 10.0152 2 12546.5 kJ>kg 2 h � ha � vhv � cpT � vhg

v �0.622Pv

P � Pv

�10.622 2 12.38 kPa 21100 � 2.38 2 kPa

� 0.0152 kg H2O/kg dry air

Pa � 1100 � 2.38 2 kPa � 97.62 kPa

Pv � fPg � fPsat @ 25°C � 10.75 2 13.1698 kPa 2 � 2.38 kPa

Pa � P � Pv

cen84959_ch14.qxd 4/26/05 4:00 PM Page 721

night, so does the “moisture capacity” of air, which is the maximum amountof moisture air can hold. (What happens to the relative humidity during thisprocess?) After a while, the moisture capacity of air equals its moisture con-tent. At this point, air is saturated, and its relative humidity is 100 percent.Any further drop in temperature results in the condensation of some of themoisture, and this is the beginning of dew formation.

The dew-point temperature Tdp is defined as the temperature at whichcondensation begins when the air is cooled at constant pressure. In otherwords, Tdp is the saturation temperature of water corresponding to the vaporpressure:

(14–13)

This is also illustrated in Fig. 14–8. As the air cools at constant pressure, thevapor pressure Pv remains constant. Therefore, the vapor in the air (state 1)undergoes a constant-pressure cooling process until it strikes the saturatedvapor line (state 2). The temperature at this point is Tdp, and if the tempera-ture drops any further, some vapor condenses out. As a result, the amount ofvapor in the air decreases, which results in a decrease in Pv. The air remainssaturated during the condensation process and thus follows a path of100 percent relative humidity (the saturated vapor line). The ordinary temperature and the dew-point temperature of saturated air are identical.

You have probably noticed that when you buy a cold canned drink from avending machine on a hot and humid day, dew forms on the can. The for-mation of dew on the can indicates that the temperature of the drink isbelow the dew-point temperature of the surrounding air (Fig. 14–9).

The dew-point temperature of room air can be determined easily by cool-ing some water in a metal cup by adding small amounts of ice and stirring.The temperature of the outer surface of the cup when dew starts to form onthe surface is the dew-point temperature of the air.

Tdp � Tsat @ Pv

722 | Thermodynamics

T

s

T1

Tdp2

1

P v =

con

st.

FIGURE 14–8Constant-presssure cooling of moistair and the dew-point temperature onthe T-s diagram of water.

MOISTAIR

Liquid waterdroplets(dew)

T < Tdp

FIGURE 14–9When the temperature of a cold drinkis below the dew-point temperature ofthe surrounding air, it “sweats.”

COLDOUTDOORS10°C

AIR20°C, 75%

Typical temperaturedistribution

16°C18°C

20°C 20°C 20°C18°C

16°C

FIGURE 14–10Schematic for Example 14–2.

EXAMPLE 14–2 Fogging of the Windows in a House

In cold weather, condensation frequently occurs on the inner surfaces of thewindows due to the lower air temperatures near the window surface. Considera house, shown in Fig. 14–10, that contains air at 20°C and 75 percent rela-tive humidity. At what window temperature will the moisture in the air startcondensing on the inner surfaces of the windows?

Solution The interior of a house is maintained at a specified temperatureand humidity. The window temperature at which fogging starts is to bedetermined.Properties The saturation pressure of water at 20°C is Psat � 2.3392 kPa(Table A–4).Analysis The temperature distribution in a house, in general, is not uniform.When the outdoor temperature drops in winter, so does the indoor tempera-ture near the walls and the windows. Therefore, the air near the walls andthe windows remains at a lower temperature than at the inner parts of ahouse even though the total pressure and the vapor pressure remain constant throughout the house. As a result, the air near the walls and the windowsundergoes a Pv � constant cooling process until the moisture in the air

cen84959_ch14.qxd 4/26/05 4:00 PM Page 722

14–4 � ADIABATIC SATURATION AND WET-BULB TEMPERATURES

Relative humidity and specific humidity are frequently used in engineeringand atmospheric sciences, and it is desirable to relate them to easily measur-able quantities such as temperature and pressure. One way of determiningthe relative humidity is to determine the dew-point temperature of air,as discussed in the last section. Knowing the dew-point temperature, wecan determine the vapor pressure Pv and thus the relative humidity. Thisapproach is simple, but not quite practical.

Another way of determining the absolute or relative humidity is related toan adiabatic saturation process, shown schematically and on a T-s diagramin Fig. 14–11. The system consists of a long insulated channel that containsa pool of water. A steady stream of unsaturated air that has a specifichumidity of v1 (unknown) and a temperature of T1 is passed through thischannel. As the air flows over the water, some water evaporates and mixeswith the airstream. The moisture content of air increases during this process,and its temperature decreases, since part of the latent heat of vaporization ofthe water that evaporates comes from the air. If the channel is long enough,the airstream exits as saturated air (f � 100 percent) at temperature T2,which is called the adiabatic saturation temperature.

If makeup water is supplied to the channel at the rate of evaporation attemperature T2, the adiabatic saturation process described above can be ana-lyzed as a steady-flow process. The process involves no heat or work inter-actions, and the kinetic and potential energy changes can be neglected. Thenthe conservation of mass and conservation of energy relations for this two-inlet, one-exit steady-flow system reduces to the following:

Mass balance:

or

m#

av1 � m#

f � m#

av2

m#

w1� m

#f � m

#w2

m#

a1� m

#a2

� m#

a

Chapter 14 | 723

starts condensing. This happens when the air reaches its dew-point tempera-ture Tdp, which is determined from Eq. 14–13 to be

where

Thus,

Discussion Note that the inner surface of the window should be maintainedabove 15.4°C if condensation on the window surfaces is to be avoided.

Tdp � Tsat @ 1.754 kPa � 15.4 °C

Pv � fPg @ 20°C � 10.75 2 12.3392 kPa 2 � 1.754 kPa

Tdp � Tsat @ Pv

T

s

2

1

Adiabaticsaturationtemperature

Dew-pointtemperature

Unsaturated air T1, ω1f1

Saturated air T2, ω2

f2 � 100%

1 2

Liquid waterat T2

Liquid water

P v1

FIGURE 14–11The adiabatic saturation process andits representation on a T-s diagram ofwater.

(The mass flow rate of dry airremains constant)

(The mass flow rate of vapor in theair increases by an amount equalto the rate of evaporation m. f )

cen84959_ch14.qxd 4/26/05 4:00 PM Page 723

Thus,

Energy balance:

or

Dividing by m.

a gives

or

which yields

(14–14)

where, from Eq. 14–11b,

(14–15)

since f2 � 100 percent. Thus we conclude that the specific humidity (andrelative humidity) of air can be determined from Eqs. 14–14 and 14–15 bymeasuring the pressure and temperature of air at the inlet and the exit of anadiabatic saturator.

If the air entering the channel is already saturated, then the adiabatic satu-ration temperature T2 will be identical to the inlet temperature T1, in whichcase Eq. 14–14 yields v1 � v2. In general, the adiabatic saturation tempera-ture is between the inlet and dew-point temperatures.

The adiabatic saturation process discussed above provides a means ofdetermining the absolute or relative humidity of air, but it requires a longchannel or a spray mechanism to achieve saturation conditions at the exit. Amore practical approach is to use a thermometer whose bulb is covered witha cotton wick saturated with water and to blow air over the wick, as shown inFig. 14–12. The temperature measured in this manner is called the wet-bulbtemperature Twb, and it is commonly used in air-conditioning applications.

The basic principle involved is similar to that in adiabatic saturation.When unsaturated air passes over the wet wick, some of the water in thewick evaporates. As a result, the temperature of the water drops, creating atemperature difference (which is the driving force for heat transfer) betweenthe air and the water. After a while, the heat loss from the water by evapora-tion equals the heat gain from the air, and the water temperature stabilizes.The thermometer reading at this point is the wet-bulb temperature. The wet-bulb temperature can also be measured by placing the wet-wicked ther-mometer in a holder attached to a handle and rotating the holder rapidly,that is, by moving the thermometer instead of the air. A device that works

v2 �0.622Pg2

P2 � Pg2

v1 �cp 1T2 � T1 2 � v2hfg2

hg1� hf2

1cpT1 � v1hg12 � 1v2 � v1 2hf2

� 1cpT2 � v2hg22

h1 � 1v2 � v1 2hf2� h2

m#

a h1 � m#

a 1v2 � v1 2hf2� m

#a h2

m#

a h1 � m#

f hf2� m

#a h2

E#in � E

#out 1since Q

#� 0 and W

#� 0 2

m#

f � m#

a 1v2 � v1 2

724 | Thermodynamics

Ordinarythermometer

Wet-bulbthermometer

Airflow

Liquidwater

Wick

FIGURE 14–12A simple arrangement to measure thewet-bulb temperature.

cen84959_ch14.qxd 4/26/05 4:00 PM Page 724

on this principle is called a sling psychrometer and is shown in Fig. 14–13.Usually a dry-bulb thermometer is also mounted on the frame of this deviceso that both the wet- and dry-bulb temperatures can be read simultaneously.

Advances in electronics made it possible to measure humidity directly in afast and reliable way. It appears that sling psychrometers and wet-wicked ther-mometers are about to become things of the past. Today, hand-held electronichumidity measurement devices based on the capacitance change in a thin poly-mer film as it absorbs water vapor are capable of sensing and digitally display-ing the relative humidity within 1 percent accuracy in a matter of seconds.

In general, the adiabatic saturation temperature and the wet-bulb tempera-ture are not the same. However, for air–water vapor mixtures at atmosphericpressure, the wet-bulb temperature happens to be approximately equal to theadiabatic saturation temperature. Therefore, the wet-bulb temperature Twbcan be used in Eq. 14–14 in place of T2 to determine the specific humidityof air.

Chapter 14 | 725

Dry-bulbthermometer

Wet-bulbthermometerwick

Wet-bulbthermometer

FIGURE 14–13Sling psychrometer.

EXAMPLE 14–3 The Specific and Relative Humidity of Air

The dry- and the wet-bulb temperatures of atmospheric air at 1 atm (101.325kPa) pressure are measured with a sling psychrometer and determined to be25 and 15°C, respectively. Determine (a) the specific humidity, (b) the rela-tive humidity, and (c) the enthalpy of the air.

Solution Dry- and wet-bulb temperatures are given. The specific humidity,relative humidity, and enthalpy are to be determined.Properties The saturation pressure of water is 1.7057 kPa at 15°C, and3.1698 kPa at 25°C (Table A–4). The constant-pressure specific heat of airat room temperature is cp � 1.005 kJ/kg · K (Table A–2a).Analysis (a) The specific humidity v1 is determined from Eq. 14–14,

where T2 is the wet-bulb temperature and v2 is

Thus,

(b) The relative humidity f1 is determined from Eq. 14–11a to be

f1 �v1P2

10.622 � v1 2Pg1

�10.00653 2 1101.325 kPa 2

10.622 � 0.00653 2 13.1698 kPa 2 � 0.332 or 33.2%

� 0.00653 kg H2O/kg dry air

v1 �11.005 kJ>kg # °C 2 3 115 � 25 2°C 4 � 10.01065 2 12465.4 kJ>kg 2

12546.5 � 62.982 2 kJ>kg

� 0.01065 kg H2O>kg dry air

v2 �0.622Pg2

P2 � Pg2

�10.622 2 11.7057 kPa 21101.325 � 1.7057 2 kPa

v1 �cp 1T2 � T1 2 � v2hfg2

hg1� hf2

cen84959_ch14.qxd 4/26/05 4:00 PM Page 725

14–5 � THE PSYCHROMETRIC CHARTThe state of the atmospheric air at a specified pressure is completely speci-fied by two independent intensive properties. The rest of the properties canbe calculated easily from the previous relations. The sizing of a typical air-conditioning system involves numerous such calculations, which may even-tually get on the nerves of even the most patient engineers. Therefore, thereis clear motivation to computerize calculations or to do these calculationsonce and to present the data in the form of easily readable charts. Suchcharts are called psychrometric charts, and they are used extensively inair-conditioning applications. A psychrometric chart for a pressure of 1 atm(101.325 kPa or 14.696 psia) is given in Fig. A–31 in SI units and in Fig.A–31E in English units. Psychrometric charts at other pressures (for use atconsiderably higher elevations than sea level) are also available.

The basic features of the psychrometric chart are illustrated in Fig. 14–14.The dry-bulb temperatures are shown on the horizontal axis, and the spe-cific humidity is shown on the vertical axis. (Some charts also show thevapor pressure on the vertical axis since at a fixed total pressure P there is aone-to-one correspondence between the specific humidity v and the vaporpressure Pv, as can be seen from Eq. 14–8.) On the left end of the chart,there is a curve (called the saturation line) instead of a straight line. All thesaturated air states are located on this curve. Therefore, it is also the curveof 100 percent relative humidity. Other constant relative-humidity curveshave the same general shape.

Lines of constant wet-bulb temperature have a downhill appearance to theright. Lines of constant specific volume (in m3/kg dry air) look similar, exceptthey are steeper. Lines of constant enthalpy (in kJ/kg dry air) lie very nearlyparallel to the lines of constant wet-bulb temperature. Therefore, the constant-wet-bulb-temperature lines are used as constant-enthalpy lines in some charts.

For saturated air, the dry-bulb, wet-bulb, and dew-point temperatures areidentical (Fig. 14–15). Therefore, the dew-point temperature of atmosphericair at any point on the chart can be determined by drawing a horizontal line (aline of v � constant or Pv � constant) from the point to the saturated curve.The temperature value at the intersection point is the dew-point temperature.

The psychrometric chart also serves as a valuable aid in visualizing the air-conditioning processes. An ordinary heating or cooling process, for example,appears as a horizontal line on this chart if no humidification or dehumidifica-tion is involved (that is, v � constant). Any deviation from a horizontal lineindicates that moisture is added or removed from the air during the process.

726 | Thermodynamics

(c) The enthalpy of air per unit mass of dry air is determined from Eq. 14–12:

Discussion The previous property calculations can be performed easily usingEES or other programs with built-in psychrometric functions.

� 41.8 kJ/kg dry air

� 11.005 kJ>kg # °C 2 125°C 2 � 10.00653 2 12546.5 kJ>kg 2 h1 � ha1

� v1hv1� cpT1 � v1hg1

Dry-bulb temperature

Spec

ific

hum

idity

,ω

Satu

ratio

n lin

e,

φ

= 10

0%φ

= co

nst.

Twb = const.

h = const.

v = const.

FIGURE 14–14Schematic for a psychrometric chart.

Saturation line

Tdp = 15°C

T db

= 1

5°C

Tw

b = 15°C

15°C

15°C

FIGURE 14–15For saturated air, the dry-bulb,wet-bulb, and dew-point temperatures are identical.

cen84959_ch14.qxd 4/26/05 4:00 PM Page 726

14–6 � HUMAN COMFORT AND AIR-CONDITIONINGHuman beings have an inherent weakness—they want to feel comfortable.They want to live in an environment that is neither hot nor cold, neitherhumid nor dry. However, comfort does not come easily since the desires ofthe human body and the weather usually are not quite compatible. Achievingcomfort requires a constant struggle against the factors that cause discomfort,such as high or low temperatures and high or low humidity. As engineers, itis our duty to help people feel comfortable. (Besides, it keeps us employed.)

Chapter 14 | 727

EXAMPLE 14–4 The Use of the Psychrometric Chart

Consider a room that contains air at 1 atm, 35°C, and 40 percent relativehumidity. Using the psychrometric chart, determine (a) the specific humidity,(b) the enthalpy, (c) the wet-bulb temperature, (d ) the dew-point tempera-ture, and (e) the specific volume of the air.

Solution The relative humidity of air in a room is given. The specific humid-ity, enthalpy, wet-bulb temperature, dew-point temperature, and specific vol-ume of the air are to be determined using the psychrometric chart.Analysis At a given total pressure, the state of atmospheric air is completelyspecified by two independent properties such as the dry-bulb temperatureand the relative humidity. Other properties are determined by directly read-ing their values at the specified state.

(a) The specific humidity is determined by drawing a horizontal line from thespecified state to the right until it intersects with the v axis, as shown inFig. 14–16. At the intersection point we read

(b) The enthalpy of air per unit mass of dry air is determined by drawing aline parallel to the h � constant lines from the specific state until it inter-sects the enthalpy scale, giving

(c) The wet-bulb temperature is determined by drawing a line parallel to theTwb � constant lines from the specified state until it intersects the satura-tion line, giving

(d ) The dew-point temperature is determined by drawing a horizontal line fromthe specified state to the left until it intersects the saturation line, giving

(e) The specific volume per unit mass of dry air is determined by noting thedistances between the specified state and the v � constant lines on both sidesof the point. The specific volume is determined by visual interpolation to be

Discussion Values read from the psychrometric chart inevitably involve read-ing errors, and thus are of limited accuracy.

v � 0.893 m3/kg dry air

Tdp � 19.4°C

Twb � 24°C

h � 71.5 kJ/kg dry air

v � 0.0142 kg H2O/kg dry air

T = 35°C

Tdp

Twb

h

φ

= 40

%

ωv

FIGURE 14–16Schematic for Example 14–4.

cen84959_ch14.qxd 4/26/05 4:00 PM Page 727

It did not take long for people to realize that they could not change theweather in an area. All they can do is change it in a confined space such as ahouse or a workplace (Fig. 14–17). In the past, this was partially accomplishedby fire and simple indoor heating systems. Today, modern air-conditioningsystems can heat, cool, humidify, dehumidify, clean, and even deodorize theair–in other words, condition the air to peoples’ desires. Air-conditioning sys-tems are designed to satisfy the needs of the human body; therefore, it isessential that we understand the thermodynamic aspects of the body.

The human body can be viewed as a heat engine whose energy input isfood. As with any other heat engine, the human body generates waste heatthat must be rejected to the environment if the body is to continue operat-ing. The rate of heat generation depends on the level of the activity. For anaverage adult male, it is about 87 W when sleeping, 115 W when resting ordoing office work, 230 W when bowling, and 440 W when doing heavyphysical work. The corresponding numbers for an adult female are about15 percent less. (This difference is due to the body size, not the body temperature. The deep-body temperature of a healthy person is maintainedconstant at about 37°C.) A body will feel comfortable in environments inwhich it can dissipate this waste heat comfortably (Fig. 14–18).

Heat transfer is proportional to the temperature difference. Therefore incold environments, a body loses more heat than it normally generates,which results in a feeling of discomfort. The body tries to minimize theenergy deficit by cutting down the blood circulation near the skin (causing apale look). This lowers the skin temperature, which is about 34°C for anaverage person, and thus the heat transfer rate. A low skin temperaturecauses discomfort. The hands, for example, feel painfully cold when theskin temperature reaches 10°C (50°F). We can also reduce the heat lossfrom the body either by putting barriers (additional clothes, blankets, etc.)in the path of heat or by increasing the rate of heat generation within thebody by exercising. For example, the comfort level of a resting persondressed in warm winter clothing in a room at 10°C (50°F) is roughly equalto the comfort level of an identical person doing moderate work in a roomat about �23°C (�10°F). Or we can just cuddle up and put our handsbetween our legs to reduce the surface area through which heat flows.

In hot environments, we have the opposite problem—we do not seem tobe dissipating enough heat from our bodies, and we feel as if we are goingto burst. We dress lightly to make it easier for heat to get away from ourbodies, and we reduce the level of activity to minimize the rate of wasteheat generation in the body. We also turn on the fan to continuously replacethe warmer air layer that forms around our bodies as a result of body heatby the cooler air in other parts of the room. When doing light work or walk-ing slowly, about half of the rejected body heat is dissipated through perspi-ration as latent heat while the other half is dissipated through convection andradiation as sensible heat. When resting or doing office work, most of theheat (about 70 percent) is dissipated in the form of sensible heat whereaswhen doing heavy physical work, most of the heat (about 60 percent) is dis-sipated in the form of latent heat. The body helps out by perspiring or sweat-ing more. As this sweat evaporates, it absorbs latent heat from the body andcools it. Perspiration is not much help, however, if the relative humidity of

728 | Thermodynamics

FIGURE 14–17We cannot change the weather, but wecan change the climate in a confinedspace by air-conditioning.

© Vol. 77/PhotoDisc

23°C

Waste heat

37°C

FIGURE 14–18A body feels comfortable when it canfreely dissipate its waste heat, and nomore.

cen84959_ch14.qxd 4/27/05 10:46 AM Page 728

the environment is close to 100 percent. Prolonged sweating without anyfluid intake causes dehydration and reduced sweating, which may lead to arise in body temperature and a heat stroke.

Another important factor that affects human comfort is heat transfer byradiation between the body and the surrounding surfaces such as walls andwindows. The sun’s rays travel through space by radiation. You warm up infront of a fire even if the air between you and the fire is quite cold. Likewise,in a warm room you feel chilly if the ceiling or the wall surfaces are at aconsiderably lower temperature. This is due to direct heat transfer betweenyour body and the surrounding surfaces by radiation. Radiant heaters arecommonly used for heating hard-to-heat places such as car repair shops.

The comfort of the human body depends primarily on three factors: the(dry-bulb) temperature, relative humidity, and air motion (Fig. 14–19). Thetemperature of the environment is the single most important index of com-fort. Most people feel comfortable when the environment temperature isbetween 22 and 27°C (72 and 80°F). The relative humidity also has a con-siderable effect on comfort since it affects the amount of heat a body candissipate through evaporation. Relative humidity is a measure of air’s abilityto absorb more moisture. High relative humidity slows down heat rejectionby evaporation, and low relative humidity speeds it up. Most people prefer arelative humidity of 40 to 60 percent.

Air motion also plays an important role in human comfort. It removes thewarm, moist air that builds up around the body and replaces it with freshair. Therefore, air motion improves heat rejection by both convection andevaporation. Air motion should be strong enough to remove heat and mois-ture from the vicinity of the body, but gentle enough to be unnoticed. Mostpeople feel comfortable at an airspeed of about 15 m/min. Very-high-speedair motion causes discomfort instead of comfort. For example, an environ-ment at 10°C (50°F) with 48 km/h winds feels as cold as an environment at�7°C (20°F) with 3 km/h winds as a result of the body-chilling effect of theair motion (the wind-chill factor). Other factors that affect comfort are aircleanliness, odor, noise, and radiation effect.

14–7 � AIR-CONDITIONING PROCESSESMaintaining a living space or an industrial facility at the desired temperatureand humidity requires some processes called air-conditioning processes.These processes include simple heating (raising the temperature), simple cool-ing (lowering the temperature), humidifying (adding moisture), and dehumidi-fying (removing moisture). Sometimes two or more of these processes areneeded to bring the air to a desired temperature and humidity level.

Various air-conditioning processes are illustrated on the psychrometricchart in Fig. 14–20. Notice that simple heating and cooling processes appearas horizontal lines on this chart since the moisture content of the air remainsconstant (v � constant) during these processes. Air is commonly heatedand humidified in winter and cooled and dehumidified in summer. Noticehow these processes appear on the psychrometric chart.

Chapter 14 | 729

23°Cf = 50%

Air motion15 m/min

FIGURE 14–19A comfortable environment.

© Reprinted with special permission of KingFeatures Syndicate.

Cooling

Heating

Hum

idif

ying

Deh

umid

ifyi

ng

Coolin

g an

d

dehu

mid

ifyin

g

Heatin

g an

d

hum

idify

ing

FIGURE 14–20Various air-conditioning processes.

cen84959_ch14.qxd 4/26/05 4:00 PM Page 729

Most air-conditioning processes can be modeled as steady-flow processes,and thus the mass balance relation m

.in � m

.out can be expressed for dry air

and water as

Mass balance for dry air: (14–16)

Mass balance for water: (14–17)

Disregarding the kinetic and potential energy changes, the steady-flowenergy balance relation E

.in � E

.out can be expressed in this case as

(14–18)

The work term usually consists of the fan work input, which is small rela-tive to the other terms in the energy balance relation. Next we examinesome commonly encountered processes in air-conditioning.

Simple Heating and Cooling (V � constant)Many residential heating systems consist of a stove, a heat pump, or an elec-tric resistance heater. The air in these systems is heated by circulating itthrough a duct that contains the tubing for the hot gases or the electric resis-tance wires, as shown in Fig. 14–21. The amount of moisture in the airremains constant during this process since no moisture is added to orremoved from the air. That is, the specific humidity of the air remains con-stant (v � constant) during a heating (or cooling) process with no humidifi-cation or dehumidification. Such a heating process proceeds in the directionof increasing dry-bulb temperature following a line of constant specifichumidity on the psychrometric chart, which appears as a horizontal line.

Notice that the relative humidity of air decreases during a heating processeven if the specific humidity v remains constant. This is because the relativehumidity is the ratio of the moisture content to the moisture capacity of airat the same temperature, and moisture capacity increases with temperature.Therefore, the relative humidity of heated air may be well below comfort-able levels, causing dry skin, respiratory difficulties, and an increase in static electricity.

A cooling process at constant specific humidity is similar to the heatingprocess discussed above, except the dry-bulb temperature decreases and therelative humidity increases during such a process, as shown in Fig. 14–22.Cooling can be accomplished by passing the air over some coils throughwhich a refrigerant or chilled water flows.

The conservation of mass equations for a heating or cooling process thatinvolves no humidification or dehumidification reduce to m

.a1

� m.

a2� m

.a for

dry air and v1 � v2 for water. Neglecting any fan work that may be present,the conservation of energy equation in this case reduces to

where h1 and h2 are enthalpies per unit mass of dry air at the inlet and theexit of the heating or cooling section, respectively.

Q#

� m#

a 1h2 � h1 2 or q � h2 � h1

Q#

in � W#in �a

inm#h � Q

#out � W

#out �a

outm#h

ain

m#

w �aout

m#

w or ain

m#

a v �aout

m#

a v

ain

m#

a �aout

m#

a 1kg>s 2

730 | Thermodynamics

ω2 = ω1

Heating coils

Heat

Air T2

T1, ω1, f1f2 < f1

FIGURE 14–21During simple heating, specifichumidity remains constant, but relativehumidity decreases.

12

12°C 30°C

v = constantCooling

f2 = 80% f1 = 30%

FIGURE 14–22During simple cooling, specifichumidity remains constant, but relativehumidity increases.

cen84959_ch14.qxd 4/27/05 10:46 AM Page 730

Heating with HumidificationProblems associated with the low relative humidity resulting from simpleheating can be eliminated by humidifying the heated air. This is accom-plished by passing the air first through a heating section (process 1-2) andthen through a humidifying section (process 2-3), as shown in Fig. 14–23.

The location of state 3 depends on how the humidification is accom-plished. If steam is introduced in the humidification section, this will resultin humidification with additional heating (T3 � T2). If humidification isaccomplished by spraying water into the airstream instead, part of the latentheat of vaporization comes from the air, which results in the cooling of theheated airstream (T3 � T2). Air should be heated to a higher temperature inthe heating section in this case to make up for the cooling effect during thehumidification process.

EXAMPLE 14–5 Heating and Humidification of Air

An air-conditioning system is to take in outdoor air at 10°C and 30 percentrelative humidity at a steady rate of 45 m3/min and to condition it to 25°Cand 60 percent relative humidity. The outdoor air is first heated to 22°C inthe heating section and then humidified by the injection of hot steam in thehumidifying section. Assuming the entire process takes place at a pressureof 100 kPa, determine (a) the rate of heat supply in the heating section and(b) the mass flow rate of the steam required in the humidifying section.

Solution Outdoor air is first heated and then humidified by steam injec-tion. The rate of heat transfer and the mass flow rate of steam are to bedetermined.Assumptions 1 This is a steady-flow process and thus the mass flow rate ofdry air remains constant during the entire process. 2 Dry air and water vaporare ideal gases. 3 The kinetic and potential energy changes are negligible.Properties The constant-pressure specific heat of air at room temperature iscp � 1.005 kJ/kg · K, and its gas constant is Ra � 0.287 kJ/kg · K (TableA–2a). The saturation pressure of water is 1.2281 kPa at 10°C, and 3.1698kPa at 25°C. The enthalpy of saturated water vapor is 2519.2 kJ/kg at 10°C,and 2541.0 kJ/kg at 22°C (Table A–4).Analysis We take the system to be the heating or the humidifying section,as appropriate. The schematic of the system and the psychrometric chart ofthe process are shown in Fig. 14–24. We note that the amount of watervapor in the air remains constant in the heating section (v1 � v2) butincreases in the humidifying section (v3 � v2).

(a) Applying the mass and energy balances on the heating section gives

Dry air mass balance:

Water mass balance:

Energy balance:

The psychrometric chart offers great convenience in determining the propertiesof moist air. However, its use is limited to a specified pressure only, which is 1atm (101.325 kPa) for the one given in the appendix. At pressures other than

Q#

in � m#

a h1 � m#

a h2 S Q#

in � m#

a 1h2 � h1 2 m#

a1v1 � m

#a2v2 S v1 � v2

m#

a1� m

#a2

� m#

a

Chapter 14 | 731

Air

Heating coils

ω2 = ω1

1 2 3

Heatingsection

Humidifyingsection

ω3 > ω2

Humidifier

FIGURE 14–23Heating with humidification.

·Air

1 2 3

V1 = 45 m3/min

10°C 22°C 25°C

1 2

3

1 = 30

%

3= 60

%

T1 = 10°C

T2 = 22°C

T3 = 25°C

Humidifier

f1 = 30% f3 = 60%

Heating coils

f

f

FIGURE 14–24Schematic and psychrometric chart forExample 14–5.

cen84959_ch14.qxd 4/26/05 4:01 PM Page 731

1 atm, either other charts for that pressure or the relations developed earliershould be used. In our case, the choice is clear:

since v2 � v1. Then the rate of heat transfer to air in the heating sectionbecomes

(b) The mass balance for water in the humidifying section can be expressed as

or

where

Thus,

Discussion The result 0.539 kg/min corresponds to a water requirement ofclose to one ton a day, which is significant.

Cooling with DehumidificationThe specific humidity of air remains constant during a simple coolingprocess, but its relative humidity increases. If the relative humidity reachesundesirably high levels, it may be necessary to remove some moisture fromthe air, that is, to dehumidify it. This requires cooling the air below its dew-point temperature.

� 0.539 kg/min

m#

w � 155.2 kg>min 2 10.01206 � 0.0023 2

� 0.01206 kg H2O>kg dry air

v3 �0.622f3Pg3

P3 � f3Pg3

�0.622 10.60 2 13.1698 kPa 23100 � 10.60 2 13.1698 2 4 kPa

m#

w � m#

a 1v3 � v2 2

m#

a2v2 � m

#w � m

#a3v3

� 673 kJ/min

Q#

in � m#

a 1h2 � h1 2 � 155.2 kg>min 2 3 128.0 � 15.8 2 kJ>kg 4

� 28.0 kJ>kg dry air

h2 � cpT2 � v2hg2� 11.005 kJ>kg # °C 2 122°C 2 � 10.0023 2 12541.0 kJ>kg 2

� 15.8 kJ>kg dry air

h1 � cpT1 � v1hg1� 11.005 kJ>kg # °C 2 110°C 2 � 10.0023 2 12519.2 kJ>kg 2

v1 �0.622Pv1

P1 � Pv1

�0.622 10.368 kPa 21100 � 0.368 2 kPa

� 0.0023 kg H2O>kg dry air

m#

a �V#1

v1�

45 m3>min

0.815 m3>kg� 55.2 kg>min

v1 �RaT1

Pa

�10.287 kPa # m3>kg # K 2 1283 K 2

99.632 kPa� 0.815 m3>kg dry air

Pa1� P1 � Pv1

� 1100 � 0.368 2 kPa � 99.632 kPa

Pv1� f1Pg1

� fPsat @ 10°C � 10.3 2 11.2281 kPa 2 � 0.368 kPa

732 | Thermodynamics

cen84959_ch14.qxd 4/26/05 4:01 PM Page 732

The cooling process with dehumidifying is illustrated schematically andon the psychrometric chart in Fig. 14–25 in conjunction with Example14–6. Hot, moist air enters the cooling section at state 1. As it passesthrough the cooling coils, its temperature decreases and its relative humidityincreases at constant specific humidity. If the cooling section is sufficientlylong, air reaches its dew point (state x, saturated air). Further cooling of airresults in the condensation of part of the moisture in the air. Air remains sat-urated during the entire condensation process, which follows a line of 100percent relative humidity until the final state (state 2) is reached. The watervapor that condenses out of the air during this process is removed from thecooling section through a separate channel. The condensate is usuallyassumed to leave the cooling section at T2.

The cool, saturated air at state 2 is usually routed directly to the room,where it mixes with the room air. In some cases, however, the air at state 2may be at the right specific humidity but at a very low temperature. In suchcases, air is passed through a heating section where its temperature is raisedto a more comfortable level before it is routed to the room.

EXAMPLE 14–6 Cooling and Dehumidification of Air

Air enters a window air conditioner at 1 atm, 30°C, and 80 percent relativehumidity at a rate of 10 m3/min, and it leaves as saturated air at 14°C. Partof the moisture in the air that condenses during the process is also removedat 14°C. Determine the rates of heat and moisture removal from the air.

Solution Air is cooled and dehumidified by a window air conditioner. Therates of heat and moisture removal are to be determined.Assumptions 1 This is a steady-flow process and thus the mass flow rate of dryair remains constant during the entire process. 2 Dry air and the water vaporare ideal gases. 3 The kinetic and potential energy changes are negligible.Properties The enthalpy of saturated liquid water at 14°C is 58.8 kJ/kg(Table A–4). Also, the inlet and the exit states of the air are completely spec-ified, and the total pressure is 1 atm. Therefore, we can determine the prop-erties of the air at both states from the psychrometric chart to be

h1 � 85.4 kJ/kg dry air h2 � 39.3 kJ/kg dry air

v1 � 0.0216 kg H2O/kg dry air and v2 � 0.0100 kg H2O/kg dry air

v1 � 0.889 m3/kg dry air

Analysis We take the cooling section to be the system. The schematic ofthe system and the psychrometric chart of the process are shown in Fig.14–25. We note that the amount of water vapor in the air decreases duringthe process (v2 � v1) due to dehumidification. Applying the mass andenergy balances on the cooling and dehumidification section gives

Dry air mass balance:

Water mass balance:

Energy balance: ain

m#h � Q

#out �a

outm#h S Q

#out � m

# 1h1 � h2 2 � m#

w hw

m#

a1v1 � m

#a2v2 � m

#w S m

#w � m

#a 1v1 � v2 2

m#

a1� m

#a2

� m#

a

Chapter 14 | 733

·

Air

Cooling coils

2 1

14°C 30°C

1

2

f1 = 80%

f2 = 100%

T2 = 14°Cf2 = 100%

T1 = 30°Cf1 = 80%V1 = 10 m3/minCondensate

removal

14°C

Condensate

x

FIGURE 14–25Schematic and psychrometric chart forExample 14–6.

cen84959_ch14.qxd 4/26/05 4:01 PM Page 733

Then,

Therefore, this air-conditioning unit removes moisture and heat from the airat rates of 0.131 kg/min and 511 kJ/min, respectively.

Evaporative CoolingConventional cooling systems operate on a refrigeration cycle, and they canbe used in any part of the world. But they have a high initial and operatingcost. In desert (hot and dry) climates, we can avoid the high cost of coolingby using evaporative coolers, also known as swamp coolers.

Evaporative cooling is based on a simple principle: As water evaporates,the latent heat of vaporization is absorbed from the water body and the sur-rounding air. As a result, both the water and the air are cooled during theprocess. This approach has been used for thousands of years to cool water.A porous jug or pitcher filled with water is left in an open, shaded area.A small amount of water leaks out through the porous holes, and the pitcher“sweats.” In a dry environment, this water evaporates and cools the remain-ing water in the pitcher (Fig. 14–26).

You have probably noticed that on a hot, dry day the air feels a lot coolerwhen the yard is watered. This is because water absorbs heat from the air asit evaporates. An evaporative cooler works on the same principle. The evap-orative cooling process is shown schematically and on a psychrometric chartin Fig. 14–27. Hot, dry air at state 1 enters the evaporative cooler, where itis sprayed with liquid water. Part of the water evaporates during this processby absorbing heat from the airstream. As a result, the temperature of theairstream decreases and its humidity increases (state 2). In the limiting case,the air leaves the evaporative cooler saturated at state 2�. This is the lowesttemperature that can be achieved by this process.

The evaporative cooling process is essentially identical to the adiabatic satu-ration process since the heat transfer between the airstream and the surround-ings is usually negligible. Therefore, the evaporative cooling process follows aline of constant wet-bulb temperature on the psychrometric chart. (Note thatthis will not exactly be the case if the liquid water is supplied at a temperaturedifferent from the exit temperature of the airstream.) Since the constant-wet-bulb-temperature lines almost coincide with the constant-enthalpy lines, theenthalpy of the airstream can also be assumed to remain constant. That is,

(14–19)

and(14–20)

during an evaporative cooling process. This is a reasonably accurate approx-imation, and it is commonly used in air-conditioning calculations.

h � constant

Twb � constant

� 511 kJ/min

Q#

out � 111.25 kg>min 2 3 185.4 � 39.3 2 kJ>kg 4 � 10.131 kg>min 2 158.8 kJ>kg 2 m#

w � 111.25 kg>min 2 10.0216 � 0.0100 2 � 0.131 kg/min

m#

a �V#1

v1�

10 m3>min

0.889 m3>kg dry air� 11.25 kg>min

734 | Thermodynamics

Water thatleaks out

Hot, dryair

FIGURE 14–26Water in a porous jug left in an open,breezy area cools as a result ofevaporative cooling.

HOT, DRYAIR

2 1

1

2

2'

COOL, MOISTAIR

Liquidwater

Twb = const. h = const.

~~

FIGURE 14–27Evaporative cooling.

cen84959_ch14.qxd 4/26/05 4:01 PM Page 734

EXAMPLE 14–7 Evaporative Cooling of Air by a Swamp Cooler

Air enters an evaporative (or swamp) cooler at 14.7 psi, 95°F, and 20 percentrelative humidity, and it exits at 80 percent relative humidity. Determine(a) the exit temperature of the air and (b) the lowest temperature to whichthe air can be cooled by this evaporative cooler.

Solution Air is cooled steadily by an evaporative cooler. The temperatureof discharged air and the lowest temperature to which the air can be cooledare to be determined.Analysis The schematic of the evaporative cooler and the psychrometricchart of the process are shown in Fig. 14–28.

(a) If we assume the liquid water is supplied at a temperature not much dif-ferent from the exit temperature of the airstream, the evaporative coolingprocess follows a line of constant wet-bulb temperature on the psychrometricchart. That is,

The wet-bulb temperature at 95°F and 20 percent relative humidity is deter-mined from the psychrometric chart to be 66.0°F. The intersection point ofthe Twb � 66.0°F and the f � 80 percent lines is the exit state of the air.The temperature at this point is the exit temperature of the air, and it isdetermined from the psychrometric chart to be

(b) In the limiting case, air leaves the evaporative cooler saturated (f � 100percent), and the exit state of the air in this case is the state where the Twb� 66.0°F line intersects the saturation line. For saturated air, the dry- andthe wet-bulb temperatures are identical. Therefore, the lowest temperature towhich air can be cooled is the wet-bulb temperature, which is

Discussion Note that the temperature of air drops by as much as 30°F inthis case by evaporative cooling.

Adiabatic Mixing of AirstreamsMany air-conditioning applications require the mixing of two airstreams.This is particularly true for large buildings, most production and processplants, and hospitals, which require that the conditioned air be mixed with acertain fraction of fresh outside air before it is routed into the living space.The mixing is accomplished by simply merging the two airstreams, asshown in Fig. 14–29.

The heat transfer with the surroundings is usually small, and thus the mix-ing processes can be assumed to be adiabatic. Mixing processes normallyinvolve no work interactions, and the changes in kinetic and potential ener-gies, if any, are negligible. Then the mass and energy balances for the adia-batic mixing of two airstreams reduce to

Mass of dry air: (14–21)

Mass of water vapor: (14–22)

Energy: (14–23) m#

a1h1 � m

#a2

h2 � m#

a3h3

v1m#

a1� v2m

#a2

� v3m#

a3

m#

a1� m

#a2

� m#

a3

Tmin � T2¿ � 66.0°F

T2 � 70.4°F

Twb � constant

Chapter 14 | 735

2' 1

1

22'

AIR

Tmin T2 95°F

2

T1 = 95°F

f = 20%

P = 14.7 psia

f 2 =

80%

f 1 =

20%

FIGURE 14–28Schematic and psychrometric chart forExample 14–7.

cen84959_ch14.qxd 4/26/05 4:01 PM Page 735

Eliminating m.

a3from the relations above, we obtain

(14–24)

This equation has an instructive geometric interpretation on the psychro-metric chart. It shows that the ratio of v2 � v3 to v3 � v1 is equal to theratio of m

.a1

to m.

a2. The states that satisfy this condition are indicated by the

dashed line AB. The ratio of h2 � h3 to h3 � h1 is also equal to the ratio ofm.

a1to m

.a2

, and the states that satisfy this condition are indicated by the dashedline CD. The only state that satisfies both conditions is the intersection pointof these two dashed lines, which is located on the straight line connectingstates 1 and 2. Thus we conclude that when two airstreams at two differentstates (states 1 and 2) are mixed adiabatically, the state of the mixture (state 3)lies on the straight line connecting states 1 and 2 on the psychrometric chart,and the ratio of the distances 2-3 and 3-1 is equal to the ratio of mass flowrates m

.a1

and m.

a2.

The concave nature of the saturation curve and the conclusion above leadto an interesting possibility. When states 1 and 2 are located close to the sat-uration curve, the straight line connecting the two states will cross the satu-ration curve, and state 3 may lie to the left of the saturation curve. In thiscase, some water will inevitably condense during the mixing process.

EXAMPLE 14–8 Mixing of Conditioned Air with Outdoor Air

Saturated air leaving the cooling section of an air-conditioning system at14°C at a rate of 50 m3/min is mixed adiabatically with the outside air at32°C and 60 percent relative humidity at a rate of 20 m3/min. Assuming thatthe mixing process occurs at a pressure of 1 atm, determine the specifichumidity, the relative humidity, the dry-bulb temperature, and the volumeflow rate of the mixture.

Solution Conditioned air is mixed with outside air at specified rates. Thespecific and relative humidities, dry-bulb temperature, and the flow rate ofthe mixture are to be determined.Assumptions 1 Steady operating conditions exist. 2 Dry air and water vaporare ideal gases. 3 The kinetic and potential energy changes are negligible.4 The mixing section is adiabatic.Properties The properties of each inlet stream are determined from the psy-chrometric chart to be

and

v2 � 0.889 m3>kg dry air

v2 � 0.0182 kg H2O>kg dry air

h2 � 79.0 kJ>kg dry air

v1 � 0.826 m3>kg dry air

v1 � 0.010 kg H2O>kg dry air

h1 � 39.4 kJ>kg dry air

m#

a1

m#

a2

�v2 � v3

v3 � v1�

h2 � h3

h3 � h1

736 | Thermodynamics

1

2

3A

C

h2

h3

h1

D

ω2 – ω3

ω3 – ω1

ω2

ω3

ω1

B

2

1

3Mixingsection

ω1 h1

ω2 h2

ω 3 h3

h2 – h3

h3 – h1

FIGURE 14–29When two airstreams at states 1 and 2are mixed adiabatically, the state ofthe mixture lies on the straight lineconnecting the two states.

cen84959_ch14.qxd 4/26/05 4:01 PM Page 736

Chapter 14 | 737

Analysis We take the mixing section of the streams as the system. Theschematic of the system and the psychrometric chart of the process areshown in Fig. 14–30. We note that this is a steady-flow mixing process.

The mass flow rates of dry air in each stream are

From the mass balance of dry air,

The specific humidity and the enthalpy of the mixture can be determinedfrom Eq. 14–24,

which yield

These two properties fix the state of the mixture. Other properties of the mix-ture are determined from the psychrometric chart:

Finally, the volume flow rate of the mixture is determined from

Discussion Notice that the volume flow rate of the mixture is approximatelyequal to the sum of the volume flow rates of the two incoming streams. Thisis typical in air-conditioning applications.

Wet Cooling TowersPower plants, large air-conditioning systems, and some industries generatelarge quantities of waste heat that is often rejected to cooling water fromnearby lakes or rivers. In some cases, however, the cooling water supply islimited or thermal pollution is a serious concern. In such cases, the wasteheat must be rejected to the atmosphere, with cooling water recirculatingand serving as a transport medium for heat transfer between the source andthe sink (the atmosphere). One way of achieving this is through the use ofwet cooling towers.

A wet cooling tower is essentially a semienclosed evaporative cooler. Aninduced-draft counterflow wet cooling tower is shown schematically in

V#3 � m

#a3v3 � 183 kg>min 2 10.844 m3>kg 2 � 70.1 m3/min

v3 � 0.844 m3>kg dry air

f3 � 89%

T3 � 19.0°C

h3 � 50.1 kJ>kg dry air

v3 � 0.0122 kg H2O/kg dry air

60.5

22.5�

0.0182 � v3

v3 � 0.010�

79.0 � h3

h3 � 39.4

m#

a1

m#

a2

�v2 � v3

v3 � v1�

h2 � h3

h3 � h1

m#

a3� m

#a1

� m#

a2� 160.5 � 22.5 2 kg>min � 83 kg>min

m#

a2�

V#2

v2�

20 m3>min

0.889 m3>kg dry air� 22.5 kg>min

m#

a1�

V#1

v1�

50 m3>min

0.826 m3>kg dry air� 60.5 kg>min

·

·

T2 = 32°Cf2 = 60%

V2 = 20 m3/min

Saturated airT1 = 14°CV1 = 50 m3/min

2

1

3MixingsectionP = 1 atm

V3v3f3T3

231

14°C 32°C

f 1 =

100%

f2

= 60

%

·

FIGURE 14–30Schematic and psychrometric chart forExample 14–8.

cen84959_ch14.qxd 4/26/05 4:01 PM Page 737

Fig. 14–31. Air is drawn into the tower from the bottom and leaves throughthe top. Warm water from the condenser is pumped to the top of the towerand is sprayed into this airstream. The purpose of spraying is to expose alarge surface area of water to the air. As the water droplets fall under theinfluence of gravity, a small fraction of water (usually a few percent) evapo-rates and cools the remaining water. The temperature and the moisture contentof the air increase during this process. The cooled water collects at the bottomof the tower and is pumped back to the condenser to absorb additional wasteheat. Makeup water must be added to the cycle to replace the water lost byevaporation and air draft. To minimize water carried away by the air, drifteliminators are installed in the wet cooling towers above the spray section.

The air circulation in the cooling tower described is provided by a fan,and therefore it is classified as a forced-draft cooling tower. Another populartype of cooling tower is the natural-draft cooling tower, which looks like alarge chimney and works like an ordinary chimney. The air in the tower hasa high water-vapor content, and thus it is lighter than the outside air. Conse-quently, the light air in the tower rises, and the heavier outside air fills thevacant space, creating an airflow from the bottom of the tower to the top.The flow rate of air is controlled by the conditions of the atmospheric air.Natural-draft cooling towers do not require any external power to induce theair, but they cost a lot more to build than forced-draft cooling towers. Thenatural-draft cooling towers are hyperbolic in profile, as shown in Fig.14–32, and some are over 100 m high. The hyperbolic profile is for greaterstructural strength, not for any thermodynamic reason.

The idea of a cooling tower started with the spray pond, where the warmwater is sprayed into the air and is cooled by the air as it falls into the pond,as shown in Fig. 14–33. Some spray ponds are still in use today. However,they require 25 to 50 times the area of a cooling tower, water loss due to airdrift is high, and they are unprotected against dust and dirt.

We could also dump the waste heat into a still cooling pond, which isbasically a large artificial lake open to the atmosphere. Heat transfer fromthe pond surface to the atmosphere is very slow, however, and we wouldneed about 20 times the area of a spray pond in this case to achieve thesame cooling.

EXAMPLE 14–9 Cooling of a Power Plant by a Cooling Tower

Cooling water leaves the condenser of a power plant and enters a wet coolingtower at 35°C at a rate of 100 kg/s. Water is cooled to 22°C in the coolingtower by air that enters the tower at 1 atm, 20°C, and 60 percent relativehumidity and leaves saturated at 30°C. Neglecting the power input to thefan, determine (a) the volume flow rate of air into the cooling tower and (b)the mass flow rate of the required makeup water.

Solution Warm cooling water from a power plant is cooled in a wet coolingtower. The flow rates of makeup water and air are to be determined.Assumptions 1 Steady operating conditions exist and thus the mass flowrate of dry air remains constant during the entire process. 2 Dry air and thewater vapor are ideal gases. 3 The kinetic and potential energy changes arenegligible. 4 The cooling tower is adiabatic.

738 | Thermodynamics

WARM

WATER

COOL

WATER

AIRINLET

FIGURE 14–32A natural-draft cooling tower.

FIGURE 14–33A spray pond.

Photo by Yunus Çengel.

COOL

WATER

AIR EXIT

WARM

WATER

AIRINLET

FAN

FIGURE 14–31An induced-draft counterflow coolingtower.

cen84959_ch14.qxd 4/28/05 2:58 PM Page 738

Chapter 14 | 739

COOLWATER

WARMWATER

AIR

35°C100 kg/s

100 kg/s

Systemboundary

4

1

3

2

1 atm20°Cf1 = 60%

30°Cf2 = 100%

V1

Makeup water

22°C

·

FIGURE 14–34Schematic for Example 14–9.

Properties The enthalpy of saturated liquid water is 92.28 kJ/kg at 22°Cand 146.64 kJ/kg at 35°C (Table A–4). From the psychrometric chart,

h1 � 42.2 kJ/kg dry air h2 � 100.0 kJ/kg dry air

v1 � 0.0087 kg H2O/kg dry air v2 � 0.0273 kg H2O/kg dry air

v1 � 0.842 m3/kg dry air

Analysis We take the entire cooling tower to be the system, which is shownschematically in Fig. 14–34. We note that the mass flow rate of liquid waterdecreases by an amount equal to the amount of water that vaporizes in thetower during the cooling process. The water lost through evaporation must bemade up later in the cycle to maintain steady operation.(a) Applying the mass and energy balances on the cooling tower gives

Dry air mass balance:

Water mass balance:

or

Energy balance:

or

Solving for m·a gives

Substituting,

Then the volume flow rate of air into the cooling tower becomes

(b) The mass flow rate of the required makeup water is determined from

Discussion Note that over 98 percent of the cooling water is saved andrecirculated in this case.

m#

makeup � m#

a 1v2 � v1 2 � 196.9 kg>s 2 10.0273 � 0.0087 2 � 1.80 kg/s

V#1 � m

#av1 � 196.9 kg>s 2 10.842 m3>kg 2 � 81.6 m3/s

m#

a �1100 kg>s 2 3 1146.64 � 92.28 2 kJ>kg 4

3 1100.0 � 42.2 2 kJ>kg 4 � 3 10.0273 � 0.0087 2 192.28 2 kJ>kg 4 � 96.9 kg>s

m#

a �m#

3 1h3 � h4 21h2 � h1 2 � 1v2 � v1 2h4

m#

3 h3 � m#

a 1h2 � h1 2 � 1m# 3 � m# makeup 2h4

ain

m#h �a

outm#h S m

#a1

h1 � m#

3 h3 � m#

a2h2 � m

#4 h4

m#

3 � m#

4 � m#

a 1v 2 � v1 2 � m#

makeup

m#

3 � m#

a1v1 � m

#4 � m

#a2v2

m#

a1� m

#a2

� m#

a

SUMMARY

In this chapter we discussed the air–water-vapor mixture,which is the most commonly encountered gas–vapor mixturein practice. The air in the atmosphere normally contains somewater vapor, and it is referred to as atmospheric air. By con-trast, air that contains no water vapor is called dry air. In thetemperature range encountered in air-conditioning applica-tions, both the dry air and the water vapor can be treated asideal gases. The enthalpy change of dry air during a processcan be determined from

The atmospheric air can be treated as an ideal-gas mixturewhose pressure is the sum of the partial pressure of dry air Paand that of the water vapor Pv,

P � Pa � Pv

¢hdry air � cp ¢T � 11.005 kJ>kg # °C 2 ¢T

cen84959_ch14.qxd 4/26/05 4:01 PM Page 739

740 | Thermodynamics

The enthalpy of water vapor in the air can be taken to be equalto the enthalpy of the saturated vapor at the same temperature:

hv(T, low P) � hg(T) � 2500.9 � 1.82T (kJ/kg) T in °C� 1060.9 � 0.435T (Btu/lbm) T in °F

in the temperature range �10 to 50°C (15 to 120°F).The mass of water vapor present per unit mass of dry air is

called the specific or absolute humidity v,

where P is the total pressure of air and Pv is the vapor pres-sure. There is a limit on the amount of vapor the air can holdat a given temperature. Air that is holding as much moistureas it can at a given temperature is called saturated air. Theratio of the amount of moisture air holds (mv) to the maxi-mum amount of moisture air can hold at the same tempera-ture (mg) is called the relative humidity f,

where Pg � Psat @ T. The relative and specific humidities canalso be expressed as

Relative humidity ranges from 0 for dry air to 1 for saturatedair.

The enthalpy of atmospheric air is expressed per unit massof dry air, instead of per unit mass of the air–water-vapormixture, as

The ordinary temperature of atmospheric air is referred toas the dry-bulb temperature to differentiate it from other formsof temperatures. The temperature at which condensation beginsif the air is cooled at constant pressure is called the dew-pointtemperature Tdp:

Relative humidity and specific humidity of air can be deter-mined by measuring the adiabatic saturation temperature ofair, which is the temperature air attains after flowing overwater in a long adiabatic channel until it is saturated,

v1 �cp 1T2 � T1 2 � v2hfg2

hg1� hf2

Tdp � Tsat @ Pv

h � ha � vhg 1kJ>kg dry air 2

f �vP

10.622 � v 2Pg

and v �0.622fPg

P � fPg

f �mv

mg�

Pv

Pg

v �mv

ma�

0.622Pv

P � Pv

1kg H2O>kg dry air 2

where

and T2 is the adiabatic saturation temperature. A more practi-cal approach in air-conditioning applications is to use a ther-mometer whose bulb is covered with a cotton wick saturatedwith water and to blow air over the wick. The temperaturemeasured in this manner is called the wet-bulb temperatureTwb, and it is used in place of the adiabatic saturation temper-ature. The properties of atmospheric air at a specified totalpressure are presented in the form of easily readable charts,called psychrometric charts. The lines of constant enthalpyand the lines of constant wet-bulb temperature are verynearly parallel on these charts.

The needs of the human body and the conditions of theenvironment are not quite compatible. Therefore, it oftenbecomes necessary to change the conditions of a living spaceto make it more comfortable. Maintaining a living space oran industrial facility at the desired temperature and humiditymay require simple heating (raising the temperature), simplecooling (lowering the temperature), humidifying (addingmoisture), or dehumidifying (removing moisture). Sometimestwo or more of these processes are needed to bring the air tothe desired temperature and humidity level.

Most air-conditioning processes can be modeled as steady-flow processes, and therefore they can be analyzed by apply-ing the steady-flow mass (for both dry air and water) andenergy balances,

Dry air mass:

Water mass:

Energy:

The changes in kinetic and potential energies are assumed tobe negligible.

During a simple heating or cooling process, the specifichumidity remains constant, but the temperature and the rela-tive humidity change. Sometimes air is humidified after it isheated, and some cooling processes include dehumidification.In dry climates, air can be cooled via evaporative cooling bypassing it through a section where it is sprayed with water. Inlocations with limited cooling water supply, large amounts ofwaste heat can be rejected to the atmosphere with minimumwater loss through the use of cooling towers.

Q#

in � W#

in �ain

m#

h � Q#

out � W#

out �aout

m#h

ain

m#

w �aout

m#

w or ain

m#

av �aout

m#

av

ain

m#

a �aout

m#

a

v2 �0.622Pg2

P2 � Pg2

cen84959_ch14.qxd 4/26/05 4:01 PM Page 740

Chapter 14 | 741

1. ASHRAE. 1981 Handbook of Fundamentals. Atlanta,GA: American Society of Heating, Refrigerating, and Air-Conditioning Engineers, 1981.

2. S. M. Elonka. “Cooling Towers.” Power, March 1963.

3. W. F. Stoecker and J. W. Jones. Refrigeration and AirConditioning. 2nd ed. New York: McGraw-Hill, 1982.

REFERENCES AND SUGGESTED READINGS

4. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

5. L. D. Winiarski and B. A. Tichenor. “Model of NaturalDraft Cooling Tower Performance.” Journal of theSanitary Engineering Division, Proceedings of theAmerican Society of Civil Engineers, August 1970.

Dry and Atmospheric Air: Specific and Relative Humidity

14–1C Is it possible to obtain saturated air from unsatu-rated air without adding any moisture? Explain.

14–2C Is the relative humidity of saturated air necessarily100 percent?

14–3C Moist air is passed through a cooling section whereit is cooled and dehumidified. How do (a) the specific humid-ity and (b) the relative humidity of air change during thisprocess?

14–4C What is the difference between dry air and atmo-spheric air?

14–5C Can the water vapor in air be treated as an idealgas? Explain.

14–6C What is vapor pressure?

14–7C How would you compare the enthalpy of watervapor at 20°C and 2 kPa with the enthalpy of water vapor at20°C and 0.5 kPa?

14–8C What is the difference between the specific humid-ity and the relative humidity?

14–9C How will (a) the specific humidity and (b) the rela-tive humidity of the air contained in a well-sealed roomchange as it is heated?

PROBLEMS*

14–10C How will (a) the specific humidity and (b) the rela-tive humidity of the air contained in a well-sealed roomchange as it is cooled?

14–11C Consider a tank that contains moist air at 3 atmand whose walls are permeable to water vapor. The surround-ing air at 1 atm pressure also contains some moisture. Is itpossible for the water vapor to flow into the tank from sur-roundings? Explain.

14–12C Why are the chilled water lines always wrappedwith vapor barrier jackets?

14–13C Explain how vapor pressure of the ambient air isdetermined when the temperature, total pressure, and the rel-ative humidity of air are given.

14–14 An 8 m3-tank contains saturated air at 30°C, 105 kPa.Determine (a) the mass of dry air, (b) the specific humidity,and (c) the enthalpy of the air per unit mass of the dry air.

14–15 A tank contains 21 kg of dry air and 0.3 kg of watervapor at 30°C and 100 kPa total pressure. Determine (a) thespecific humidity, (b) the relative humidity, and (c) the vol-ume of the tank.

14–16 Repeat Prob. 14–15 for a temperature of 24°C.

14–17 A room contains air at 20°C and 98 kPa at a relativehumidity of 85 percent. Determine (a) the partial pressure ofdry air, (b) the specific humidity of the air, and (c) theenthalpy per unit mass of dry air.

14–18 Repeat Prob. 14–17 for a pressure of 85 kPa.

14–19E A room contains air at 70°F and 14.6 psia at a relative humidity of 85 percent. Determine (a) the partialpressure of dry air, (b) the specific humidity, and (c) theenthalpy per unit mass of dry air. Answers: (a) 14.291 psia,(b) 0.0134 lbm H2O/lbm dry air, (c) 31.43 Btu/lbm dry air

14–20 Determine the masses of dry air and the water vaporcontained in a 240-m3 room at 98 kPa, 23°C, and 50 percentrelative humidity. Answers: 273 kg, 2.5 kg

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

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Dew-Point, Adiabatic Saturation, and Wet-BulbTemperatures

14–21C What is the dew-point temperature?

14–22C Andy and Wendy both wear glasses. On a coldwinter day, Andy comes from the cold outside and enters thewarm house while Wendy leaves the house and goes outside.Whose glasses are more likely to be fogged? Explain.

14–23C In summer, the outer surface of a glass filled withiced water frequently “sweats.” How can you explain thissweating?

14–24C In some climates, cleaning the ice off the wind-shield of a car is a common chore on winter mornings.Explain how ice forms on the windshield during some nightseven when there is no rain or snow.

14–25C When are the dry-bulb and dew-point temperaturesidentical?

14–26C When are the adiabatic saturation and wet-bulbtemperatures equivalent for atmospheric air?

14–27 A house contains air at 25°C and 65 percent relativehumidity. Will any moisture condense on the inner surfacesof the windows when the temperature of the window drops to10°C?

14–28 After a long walk in the 8°C outdoors, a personwearing glasses enters a room at 25°C and 40 percent relativehumidity. Determine whether the glasses will become fogged.

14–29 Repeat Prob. 14–28 for a relative humidity of 30percent.

14–30E A thirsty woman opens the refrigerator and picksup a cool canned drink at 40°F. Do you think the can will“sweat” as she enjoys the drink in a room at 80°F and 50 per-cent relative humidity?

14–31 The dry- and wet-bulb temperatures of atmosphericair at 95 kPa are 25 and 17°C, respectively. Determine (a) thespecific humidity, (b) the relative humidity, and (c) theenthalpy of the air, in kJ/kg dry air.

14–32 The air in a room has a dry-bulb temperature of 22°Cand a wet-bulb temperature of 16°C. Assuming a pressure of100 kPa, determine (a) the specific humidity, (b) the relativehumidity, and (c) the dew-point temperature. Answers:(a) 0.0090 kg H2O/kg dry air, (b) 54.1 percent, (c) 12.3°C

14–33 Reconsider Prob. 14–32. Determine the requiredproperties using EES (or other) software. What

would the property values be at a pressure of 300 kPa?

14–34E The air in a room has a dry-bulb temperature of80°F and a wet-bulb temperature of 65°F. Assuming a pres-sure of 14.7 psia, determine (a) the specific humidity, (b) therelative humidity, and (c) the dew-point temperature.Answers: (a) 0.0097 lbm H2O/lbm dry air, (b) 44.7 percent,(c) 56.6°F

742 | Thermodynamics

14–35 Atmospheric air at 35°C flows steadily into an adia-batic saturation device and leaves as a saturated mixture at25°C. Makeup water is supplied to the device at 25°C. Atmo-spheric pressure is 98 kPa. Determine the relative humidityand specific humidity of the air.

Psychrometric Chart

14–36C How do constant-enthalpy and constant-wet-bulb-temperature lines compare on the psychrometric chart?

14–37C At what states on the psychrometric chart are thedry-bulb, wet-bulb, and dew-point temperatures identical?

14–38C How is the dew-point temperature at a specifiedstate determined on the psychrometric chart?

14–39C Can the enthalpy values determined from a psy-chrometric chart at sea level be used at higher elevations?

14–40 The air in a room is at 1 atm, 32°C, and60 percent relative humidity. Determine (a) the

specific humidity, (b) the enthalpy (in kJ/kg dry air), (c) thewet-bulb temperature, (d ) the dew-point temperature, and(e) the specific volume of the air (in m3/kg dry air). Use thepsychrometric chart or available software.

14–41 Reconsider Prob. 14–40. Determine the requiredproperties using EES (or other) software instead

of the psychrometric chart. What would the property valuesbe at a location at 1500 m altitude?

14–42 A room contains air at 1 atm, 26°C, and 70 percentrelative humidity. Using the psychrometric chart, determine(a) the specific humidity, (b) the enthalpy (in kJ/kg dry air),(c) the wet-bulb temperature, (d ) the dew-point temperature,and (e) the specific volume of the air (in m3/kg dry air).

14–43 Reconsider Prob. 14–42. Determine the requiredproperties using EES (or other) software instead

of the psychrometric chart. What would the property valuesbe at a location at 2000 m altitude?

14–44E A room contains air at 1 atm, 82°F, and 70 percentrelative humidity. Using the psychrometric chart, determine(a) the specific humidity, (b) the enthalpy (in Btu/lbmdry air), (c) the wet-bulb temperature, (d ) the dew-point tem-perature, and (e) the specific volume of the air (in ft3/lbmdry air).

14–45E Reconsider Prob. 14–44E. Determine therequired properties using EES (or other) soft-

ware instead of the psychrometric chart. What would theproperty values be at a location at 5000 ft altitude?

14–46 The air in a room has a pressure of 1 atm, a dry-bulbtemperature of 24°C, and a wet-bulb temperature of 17°C.Using the psychrometric chart, determine (a) the specifichumidity, (b) the enthalpy (in kJ/kg dry air), (c) the relativehumidity, (d ) the dew-point temperature, and (e) the specificvolume of the air (in m3/kg dry air).

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Chapter 14 | 743

14–47 Reconsider Prob. 14–46. Determine the requiredproperties using EES (or other) software instead

of the psychrometric chart. What would the property valuesbe at a location at 3000 m altitude?

Human Comfort and Air-Conditioning

14–48C What does a modern air-conditioning system dobesides heating or cooling the air?

14–49C How does the human body respond to (a) hotweather, (b) cold weather, and (c) hot and humid weather?

14–50C What is the radiation effect? How does it affecthuman comfort?

14–51C How does the air motion in the vicinity of thehuman body affect human comfort?

14–52C Consider a tennis match in cold weather whereboth players and spectators wear the same clothes. Whichgroup of people will feel colder? Why?

14–53C Why do you think little babies are more suscepti-ble to cold?

14–54C How does humidity affect human comfort?

14–55C What are humidification and dehumidification?

14–56C What is metabolism? What is the range of meta-bolic rate for an average man? Why are we interested in themetabolic rate of the occupants of a building when we dealwith heating and air-conditioning?

14–57C Why is the metabolic rate of women, in general,lower than that of men? What is the effect of clothing on theenvironmental temperature that feels comfortable?

14–58C What is sensible heat? How is the sensible heatloss from a human body affected by the (a) skin temperature,(b) environment temperature, and (c) air motion?

14–59C What is latent heat? How is the latent heat loss fromthe human body affected by the (a) skin wettedness and(b) relative humidity of the environment? How is the rate ofevaporation from the body related to the rate of latent heatloss?

14–60 An average person produces 0.25 kg of moisturewhile taking a shower and 0.05 kg while bathing in a tub.Consider a family of four who each shower once a day in abathroom that is not ventilated. Taking the heat of vaporiza-tion of water to be 2450 kJ/kg, determine the contribution ofshowers to the latent heat load of the air conditioner per dayin summer.

14–61 An average (1.82 kg or 4.0 lbm) chicken has a basalmetabolic rate of 5.47 W and an average metabolic rate of10.2 W (3.78 W sensible and 6.42 W latent) during normalactivity. If there are 100 chickens in a breeding room, deter-mine the rate of total heat generation and the rate of moistureproduction in the room. Take the heat of vaporization ofwater to be 2430 kJ/kg.

14–62 A department store expects to have 120 customersand 15 employees at peak times in summer. Determine thecontribution of people to the total cooling load of the store.

14–63E In a movie theater in winter, 500 people, each gen-erating sensible heat at a rate of 70 W, are watching a movie.The heat losses through the walls, windows, and the roof areestimated to be 130,000 Btu/h. Determine if the theater needsto be heated or cooled.

14–64 For an infiltration rate of 1.2 air changes per hour(ACH), determine sensible, latent, and total infiltration heatload of a building at sea level, in kW, that is 20 m long, 13 mwide, and 3 m high when the outdoor air is at 32°C and 50percent relative humidity. The building is maintained at 24°Cand 50 percent relative humidity at all times.

14–65 Repeat Prob. 14–64 for an infiltration rate of1.8 ACH.

Simple Heating and Cooling

14–66C How do relative and specific humidities changeduring a simple heating process? Answer the same questionfor a simple cooling process.

14–67C Why does a simple heating or cooling processappear as a horizontal line on the psychrometric chart?

14–68 Air enters a heating section at 95 kPa, 12°C, and 30percent relative humidity at a rate of 6 m3/min, and it leavesat 25°C. Determine (a) the rate of heat transfer in the heatingsection and (b) the relative humidity of the air at the exit.Answers: (a) 91.1 kJ/min, (b) 13.3 percent

14–69E A heating section consists of a 15-in.-diameter ductthat houses a 4-kW electric resistance heater. Air enters theheating section at 14.7 psia, 50°F, and 40 percent relativehumidity at a velocity of 25 ft/s. Determine (a) the exit tem-perature, (b) the exit relative humidity of the air, and (c) theexit velocity. Answers: (a) 56.6°F, (b) 31.4 percent, (c) 25.4 ft/s

14–70 Air enters a 40-cm-diameter cooling section at1 atm, 32°C, and 30 percent relative humidity at 18 m/s. Heatis removed from the air at a rate of 1200 kJ/min. Determine(a) the exit temperature, (b) the exit relative humidity of theair, and (c) the exit velocity. Answers: (a) 24.4°C, (b) 46.6percent, (c) 17.6 m/s

1200 kJ/min

1 atm

AIR32°C, 30%

18 m/s

FIGURE P14–70

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14–71 Repeat Prob. 14–70 for a heat removal rate of 800kJ/min.

Heating with Humidification

14–72C Why is heated air sometimes humidified?

14–73 Air at 1 atm, 15°C, and 60 percent relative humidityis first heated to 20°C in a heating section and then humidi-fied by introducing water vapor. The air leaves the humidify-ing section at 25°C and 65 percent relative humidity.Determine (a) the amount of steam added to the air, and(b) the amount of heat transfer to the air in the heating sec-tion. Answers: (a) 0.0065 kg H2O/kg dry air, (b) 5.1 kJ/kg dry air

14–74E Air at 14.7 psia, 50°F, and 60 percent relativehumidity is first heated to 72°F in a heating section and thenhumidified by introducing water vapor. The air leaves thehumidifying section at 75°F and 55 percent relative humidity.Determine (a) the amount of steam added to the air, in lbmH2O/lbm dry air, and (b) the amount of heat transfer to the airin the heating section, in Btu/lbm dry air.

14–75 An air-conditioning system operates at a total pres-sure of 1 atm and consists of a heating section and a humidi-fier that supplies wet steam (saturated water vapor) at 100°C.Air enters the heating section at 10°C and 70 percent relativehumidity at a rate of 35 m3/min, and it leaves the humidify-ing section at 20°C and 60 percent relative humidity. Deter-mine (a) the temperature and relative humidity of air when itleaves the heating section, (b) the rate of heat transfer in theheating section, and (c) the rate at which water is added tothe air in the humidifying section.

744 | Thermodynamics

that condenses during the process is also removed at 15°C.Determine the rates of heat and moisture removal from theair. Answers: 97.7 kJ/min, 0.023 kg/min

14–79 An air-conditioning system is to take in air at 1 atm,34°C, and 70 percent relative humidity and deliver it at 22°Cand 50 percent relative humidity. The air flows first over thecooling coils, where it is cooled and dehumidified, and thenover the resistance heating wires, where it is heated to thedesired temperature. Assuming that the condensate is removedfrom the cooling section at 10°C, determine (a) the tempera-ture of air before it enters the heating section, (b) the amountof heat removed in the cooling section, and (c) the amount ofheat transferred in the heating section, both in kJ/kg dry air.

14–80 Air enters a 30-cm-diameter cooling section at1 atm, 35°C, and 60 percent relative humidity

at 120 m/min. The air is cooled by passing it over a coolingcoil through which cold water flows. The water experiences atemperature rise of 8°C. The air leaves the cooling sectionsaturated at 20°C. Determine (a) the rate of heat transfer,(b) the mass flow rate of the water, and (c) the exit velocityof the airstream.

AIR

Heating coils Humidifier

10°C70%35 m3/min P = 1 atm

20°C60%

Sat. vapor100°C

FIGURE P14–75

14–76 Repeat Prob. 14–75 for a total pressure of 95 kPa forthe airstream. Answers: (a) 19.5°C, 37.7 percent, (b) 391 kJ/min,(c) 0.147 kg/min

Cooling with Dehumidification

14–77C Why is cooled air sometimes reheated in summerbefore it is discharged to a room?

14–78 Air enters a window air conditioner at 1 atm, 32°C,and 70 percent relative humidity at a rate of 2 m3/min, and itleaves as saturated air at 15°C. Part of the moisture in the air

Cooling coils

AIR

T + 8°CT

20°CSaturated

35°C60%120 m/min

Water

FIGURE P14–80

14–81 Reconsider Prob. 14–80. Using EES (or other)software, develop a general solution of the

problem in which the input variables may be supplied andparametric studies performed. For each set of input variablesfor which the pressure is atmospheric, show the process onthe psychrometric chart.

14–82 Repeat Prob. 14–80 for a total pressure of 95 kPa forair. Answers: (a) 293.2 kJ/min, (b) 8.77 kg/min, (c) 113 m/min

14–83E Air enters a 1-ft-diameter cooling section at 14.7 psia,90°F, and 60 percent relative humidity at 600 ft/min. The air iscooled by passing it over a cooling coil through which coldwater flows. The water experiences a temperature rise of 14°F.The air leaves the cooling section saturated at 70°F. Determine(a) the rate of heat transfer, (b) the mass flow rate of the water,and (c) the exit velocity of the airstream.

14–84E Reconsider Prob. 14–83E. Using EES (orother) software, study the effect of the total

pressure of the air over the range 14.3 to 15.2 psia on the

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Chapter 14 | 745

required results. Plot the required results as functions of airtotal pressure.

14–85E Repeat Prob. 14–83E for a total pressure of 14.4 psiafor air.

14–86 Atmospheric air from the inside of an automobileenters the evaporator section of the air conditioner at 1 atm,27°C and 50 percent relative humidity. The air returns to theautomobile at 10°C and 90 percent relative humidity. The pas-senger compartment has a volume of 2 m3 and 5 air changesper minute are required to maintain the inside of the automo-bile at the desired comfort level. Sketch the psychrometricdiagram for the atmospheric air flowing through the air condi-tioning process. Determine the dew point and wet bulb tem-peratures at the inlet to the evaporator section, in °C.Determine the required heat transfer rate from the atmosphericair to the evaporator fluid, in kW. Determine the rate of con-densation of water vapor in the evaporator section, in kg/min.

14–89 Air from a workspace enters an air conditioner unitat 30°C dry bulb and 25°C wet bulb. The air leaves the airconditioner and returns to the space at 25°C dry-bulb and6.5°C dew-point temperature. If there is any, the condensateleaves the air conditioner at the temperature of the air leavingthe cooling coils. The volume flow rate of the air returned tothe workspace is 1000 m3/min. Atmospheric pressure is98 kPa. Determine the heat transfer rate from the air, in kW,and the mass flow rate of condensate water, if any, in kg/h.

Evaporative Cooling

14–90C Does an evaporation process have to involve heattransfer? Describe a process that involves both heat and masstransfer.

14–91C During evaporation from a water body to air, underwhat conditions will the latent heat of vaporization be equalto the heat transfer from the air?

14–92C What is evaporative cooling? Will it work in humidclimates?

14–93 Air enters an evaporative cooler at 1 atm, 36°C, and20 percent relative humidity at a rate of 4 m3/min, and itleaves with a relative humidity of 90 percent. Determine(a) the exit temperature of the air and (b) the required rate ofwater supply to the evaporative cooler.

Cooling coils

Condensate

AIR

FIGURE P14–86

14–87 Two thousand cubic meters per hour of atmosphericair at 28°C with a dew point temperature of 25°C flows intoan air conditioner that uses chilled water as the cooling fluid.The atmospheric air is to be cooled to 18°C. Sketch the sys-tem hardware and the psychrometric diagram for the process.Determine the mass flow rate of the condensate water, if any,leaving the air conditioner, in kg/h. If the cooling water has a10°C temperature rise while flowing through the air condi-tioner, determine the volume flow rate of chilled water sup-plied to the air conditioner heat exchanger, in m3/min. The airconditioning process takes place at 100 kPa.

14–88 An automobile air conditioner uses refrigerant-134aas the cooling fluid. The evaporator operates at 275 kPagauge and the condenser operates at 1.7 MPa gage. The com-pressor requires a power input of 6 kW and has an isentropicefficiency of 85 percent. Atmospheric air at 22°C and 50 per-cent relative humidity enters the evaporator and leaves at 8°Cand 90 percent relative humidity. Determine the volume flowrate of the atmospheric air entering the evaporator of the airconditioner, in m3/min.

AIR

Humidifier

1 atm36°Cf1 = 20%

Watermω

f2 = 90%

FIGURE P14–93

14–94E Air enters an evaporative cooler at 14.7 psia, 90°F,and 20 percent relative humidity at a rate of 150 ft3/min, andit leaves with a relative humidity of 90 percent. Determine(a) the exit temperature of air and (b) the required rate ofwater supply to the evaporative cooler. Answers: (a) 65°F,(b) 0.06 lbm/min

14–95 Air enters an evaporative cooler at 95 kPa, 40°C, and25 percent relative humidity and exits saturated. Determinethe exit temperature of air. Answer: 23.1°C

14–96E Air enters an evaporative cooler at 14.5 psia, 93°F,and 30 percent relative humidity and exits saturated. Deter-mine the exit temperature of air.

14–97 Air enters an evaporative cooler at 1 atm, 32°C, and30 percent relative humidity at a rate of 5 m3/min and leaves

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at 22°C. Determine (a) the final relative humidity and (b) theamount of water added to air.

14–98 What is the lowest temperature that air can attainin an evaporative cooler if it enters at 1 atm, 29°C, and 40percent relative humidity? Answer: 19.3°C

14–99 Air at 1 atm, 15°C, and 60 percent relative humidityis first heated to 30°C in a heating section and then passedthrough an evaporative cooler where its temperature drops to25°C. Determine (a) the exit relative humidity and (b) theamount of water added to air, in kg H2O/kg dry air.

Adiabatic Mixing of Airstreams

14–100C Two unsaturated airstreams are mixed adiabati-cally. It is observed that some moisture condenses during themixing process. Under what conditions will this be the case?

14–101C Consider the adiabatic mixing of two airstreams.Does the state of the mixture on the psychrometric chart haveto be on the straight line connecting the two states?

14–102 Two airstreams are mixed steadily and adiabatically.The first stream enters at 32°C and 40 percent relative humid-ity at a rate of 20 m3/min, while the second stream enters at12°C and 90 percent relative humidity at a rate of 25 m3/min.Assuming that the mixing process occurs at a pressure of1 atm, determine the specific humidity, the relative humidity,the dry-bulb temperature, and the volume flow rate of the mix-ture. Answers: 0.0096 kg H2O/kg dry air, 63.4 percent, 20.6°C,45.0 m3/min

746 | Thermodynamics

14–105E Reconsider Prob. 14–104E. Using EES (orother) software, develop a general solution

of the problem in which the input variables may be suppliedand parametric studies performed. For each set of input vari-ables for which the pressure is atmospheric, show the processon the psychrometric chart.

14–106 A stream of warm air with a dry-bulb temperatureof 40°C and a wet-bulb temperature of 32°C is mixed adia-batically with a stream of saturated cool air at 18°C. The dryair mass flow rates of the warm and cool airstreams are 8 and6 kg/s, respectively. Assuming a total pressure of 1 atm,determine (a) the temperature, (b) the specific humidity, and(c) the relative humidity of the mixture.

14–107 Reconsider Prob. 14–106. Using EES (orother) software, determine the effect of the

mass flow rate of saturated cool air stream on the mixturetemperature, specific humidity, and relative humidity. Varythe mass flow rate of saturated cool air from 0 to 16 kg/swhile maintaining the mass flow rate of warm air constant at8 kg/s. Plot the mixture temperature, specific humidity, andrelative humidity as functions of the mass flow rate of coolair, and discuss the results.

Wet Cooling Towers

14–108C How does a natural-draft wet cooling towerwork?

14–109C What is a spray pond? How does its performancecompare to the performance of a wet cooling tower?

14–110 The cooling water from the condenser of a powerplant enters a wet cooling tower at 40°C at a rate of 90 kg/s.The water is cooled to 25°C in the cooling tower by air thatenters the tower at 1 atm, 23°C, and 60 percent relativehumidity and leaves saturated at 32°C. Neglecting the powerinput to the fan, determine (a) the volume flow rate of airinto the cooling tower and (b) the mass flow rate of therequired makeup water.

14–111E The cooling water from the condenser of a powerplant enters a wet cooling tower at 110°F at a rate of 100lbm/s. Water is cooled to 80°F in the cooling tower by air thatenters the tower at 1 atm, 76°F, and 60 percent relative humid-ity and leaves saturated at 95°F. Neglecting the power input tothe fan, determine (a) the volume flow rate of air into thecooling tower and (b) the mass flow rate of the requiredmakeup water. Answers: (a) 1325 ft3/s, (b) 2.42 lbm/s

14–112 A wet cooling tower is to cool 60 kg/s of waterfrom 40 to 26°C. Atmospheric air enters the tower at 1 atmwith dry- and wet-bulb temperatures of 22 and 16°C, respec-tively, and leaves at 34°C with a relative humidity of 90 per-cent. Using the psychrometric chart, determine (a) thevolume flow rate of air into the cooling tower and (b) themass flow rate of the required makeup water. Answers:(a) 44.9 m3/s, (b) 1.16 kg/s

2

3P = 1 atm

AIRT3

132°C40%

12°C90%

f3

v3

FIGURE P14–102

14–103 Repeat Prob. 14–102 for a total mixing-chamberpressure of 90 kPa.

14–104E During an air-conditioning process, 900 ft3/min ofconditioned air at 65°F and 30 percent relative humidity ismixed adiabatically with 300 ft3/min of outside air at 80°Fand 90 percent relative humidity at a pressure of 1 atm.Determine (a) the temperature, (b) the specific humidity, and(c) the relative humidity of the mixture. Answers: (a) 68.7°F,(b) 0.0078 lbm H2O/lbm dry air, (c) 52.1 percent

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14–113 A wet cooling tower is to cool 25 kg/s of coolingwater from 40 to 30°C at a location where the atmosphericpressure is 96 kPa. Atmospheric air enters the tower at 20°Cand 70 percent relative humidity and leaves saturated at35°C. Neglecting the power input to the fan, determine(a) the volume flow rate of air into the cooling tower and(b) the mass flow rate of the required makeup water.Answers: (a) 11.2 m3/s, (b) 0.35 kg/s

14–114 A natural-draft cooling tower is to remove waste heatfrom the cooling water flowing through the condenser of asteam power plant. The turbine in the steam power plantreceives 42 kg/s of steam from the steam generator. Eighteenpercent of the steam entering the turbine is extracted for vari-ous feedwater heaters. The condensate of the higher pressurefeedwater heaters is trapped to the next lowest pressure feed-water heater. The last feedwater heater operates at 0.2 MPa andall of the steam extracted for the feedwater heaters is throttledfrom the last feedwater heater exit to the condenser operatingat a pressure of 10 kPa. The remainder of the steam produceswork in the turbine and leaves the lowest pressure stage of theturbine at 10 kPa with an entropy of 7.962 kJ/kg � K. The cool-ing tower supplies the cooling water at 26°C to the condenser,and cooling water returns from the condenser to the coolingtower at 40°C. Atmospheric air enters the tower at 1 atm withdry- and wet-bulb temperatures of 23 and 18°C, respectively,and leaves saturated at 37°C. Determine (a) the mass flow rateof the cooling water, (b) the volume flow rate of air into thecooling tower, and (c) the mass flow rate of the requiredmakeup water.

Review Problems

14–115 The condensation of the water vapor in compressed-air lines is a major concern in industrial facilities, and thecompressed air is often dehumidified to avoid the problemsassociated with condensation. Consider a compressor thatcompresses ambient air from the local atmospheric pressureof 92 kPa to a pressure of 800 kPa (absolute). The com-pressed air is then cooled to the ambient temperature as itflows through the compressed-air lines. Disregarding anypressure losses, determine if there will be any condensationin the compressed-air lines on a day when the ambient air isat 20°C and 50 percent relative humidity.

14–116E The relative humidity of air at 80°F and 14.7 psiais increased from 25 to 75 percent during a humidificationprocess at constant temperature and pressure. Determine thepercent error involved in assuming the density of air to haveremained constant.

14–117 Dry air whose molar analysis is 78.1 percent N2,20.9 percent O2, and 1 percent Ar flows over a water bodyuntil it is saturated. If the pressure and temperature of airremain constant at 1 atm and 25°C during the process, deter-mine (a) the molar analysis of the saturated air and (b) thedensity of air before and after the process. What do you con-clude from your results?

14–118E Determine the mole fraction of the water vapor atthe surface of a lake whose surface temperature is 60°F, andcompare it to the mole fraction of water in the lake, which isvery nearly 1.0. The air at the lake surface is saturated, and theatmospheric pressure at lake level can be taken to be 13.8 psia.

14–119 Determine the mole fraction of dry air at the sur-face of a lake whose temperature is 18°C. The air at the lakesurface is saturated, and the atmospheric pressure at lakelevel can be taken to be 100 kPa.

14–120E Consider a room that is cooled adequately by an airconditioner whose cooling capacity is 7500 Btu/h. If the roomis to be cooled by an evaporative cooler that removes heat at thesame rate by evaporation, determine how much water needs tobe supplied to the cooler per hour at design conditions.

14–121E The capacity of evaporative coolers is usuallyexpressed in terms of the flow rate of air in ft3/min (or cfm),and a practical way of determining the required size of anevaporative cooler for an 8-ft-high house is to multiply thefloor area of the house by 4 (by 3 in dry climates and by 5 inhumid climates). For example, the capacity of an evaporativecooler for a 30-ft-long, 40-ft-wide house is 1200 � 4 � 4800cfm. Develop an equivalent rule of thumb for the selection ofan evaporative cooler in SI units for 2.4-m-high houseswhose floor areas are given in m2.

14–122 A cooling tower with a cooling capacity of 100 tons(440 kW) is claimed to evaporate 15,800 kg of water per day.Is this a reasonable claim?

COOLWATER

34°C90%

WARMWATER

AIRINLET

60 kg/s40°C

1 atmTdb = 22°CTwb = 16°C

Makeup water

26°C

AIREXIT

FIGURE P14–112

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14–123E The U.S. Department of Energy estimates that190,000 barrels of oil would be saved per day if every house-hold in the United States raised the thermostat setting in sum-mer by 6°F (3.3°C). Assuming the average cooling season tobe 120 days and the cost of oil to be $20/barrel, determinehow much money would be saved per year.

14–124E The thermostat setting of a house can be loweredby 2°F by wearing a light long-sleeved sweater, or by 4°F bywearing a heavy long-sleeved sweater for the same level ofcomfort. If each °F reduction in thermostat setting reducesthe heating cost of a house by 4 percent at a particular loca-tion, determine how much the heating costs of a house can bereduced by wearing heavy sweaters if the annual heating costof the house is $600.

14–125 The air-conditioning costs of a house can bereduced by up to 10 percent by installing the outdoor unit(the condenser) of the air conditioner at a location shaded bytrees and shrubs. If the air-conditioning costs of a house are$500 a year, determine how much the trees will save thehome owner in the 20-year life of the system.

14–126 A 3-m3 tank contains saturated air at 25°C and97 kPa. Determine (a) the mass of the dry air, (b) the specifichumidity, and (c) the enthalpy of the air per unit mass of thedry air. Answers: (a) 3.29 kg, (b) 0.0210 kg H2O/kg dry air,(c) 78.6 kJ/kg dry air

14–127 Reconsider Prob. 14–126. Using EES (orother) software, determine the properties of

the air at the initial state. Study the effect of heating the air atconstant volume until the pressure is 110 kPa. Plot therequired heat transfer, in kJ, as a function of pressure.

14–128E Air at 15 psia, 60°F, and 50 percent relative humid-ity flows in an 8-in.-diameter duct at a velocity of 50 ft/s.Determine (a) the dew-point temperature, (b) the volume flowrate of air, and (c) the mass flow rate of dry air.

14–129 Air enters a cooling section at 97 kPa, 35°C, and30 percent relative humidity at a rate of 6 m3/min, where it iscooled until the moisture in the air starts condensing. Deter-mine (a) the temperature of the air at the exit and (b) the rateof heat transfer in the cooling section.

14–130 Outdoor air enters an air-conditioning system at10°C and 40 percent relative humidity at a steady rate of22 m3/min, and it leaves at 25°C and 55 percent relativehumidity. The outdoor air is first heated to 22°C in the heat-ing section and then humidified by the injection of hot steamin the humidifying section. Assuming the entire process takesplace at a pressure of 1 atm, determine (a) the rate of heatsupply in the heating section and (b) the mass flow rate ofsteam required in the humidifying section.

14–131 Air enters an air-conditioning system that usesrefrigerant-134a at 30°C and 70 percent relative humidity at arate of 4 m3/min. The refrigerant enters the cooling section at

748 | Thermodynamics

700 kPa with a quality of 20 percent and leaves as saturatedvapor. The air is cooled to 20°C at a pressure of 1 atm. Deter-mine (a) the rate of dehumidification, (b) the rate of heattransfer, and (c) the mass flow rate of the refrigerant.

14–132 Repeat Prob. 14–131 for a total pressure of 95 kPafor air.

14–133 An air-conditioning system operates at a total pres-sure of 1 atm and consists of a heating section and an evapo-rative cooler. Air enters the heating section at 10°C and70 percent relative humidity at a rate of 30 m3/min, and itleaves the evaporative cooler at 20°C and 60 percent rela-tively humidity. Determine (a) the temperature and relativehumidity of the air when it leaves the heating section, (b) therate of heat transfer in the heating section, and (c) the rateof water added to air in the evaporative cooler. Answers:(a) 28.3°C, 22.3 percent, (b) 696 kJ/min, (c) 0.13 kg/min

14–134 Reconsider Prob. 14–133. Using EES (orother) software, study the effect of total pres-

sure in the range 94 to 100 kPa on the results required in theproblem. Plot the results as functions of total pressure.

14–135 Repeat Prob. 14–133 for a total pressure of 96 kPa.

14–136 Conditioned air at 13°C and 90 percent relativehumidity is to be mixed with outside air at 34°C and 40 percentrelative humidity at 1 atm. If it is desired that the mixture havea relative humidity of 60 percent, determine (a) the ratio of thedry air mass flow rates of the conditioned air to the outside airand (b) the temperature of the mixture.

14–137 Reconsider Prob. 14–136. Determine thedesired quantities using EES (or other) soft-

ware instead of the psychrometric chart. What would theanswers be at a location at an atmospheric pressure of 80 kPa?

14–138 A natural-draft cooling tower is to remove50 MW of waste heat from the cooling water

that enters the tower at 42°C and leaves at 27°C. Atmo-spheric air enters the tower at 1 atm with dry- and wet-bulbtemperatures of 23 and 18°C, respectively, and leaves satu-rated at 37°C. Determine (a) the mass flow rate of the cool-ing water, (b) the volume flow rate of air into the coolingtower, and (c) the mass flow rate of the required makeupwater.

14–139 Reconsider Prob. 14–138. Using EES (orother) software, investigate the effect of air

inlet wet-bulb temperature on the required air volume flowrate and the makeup water flow rate when the other inputdata are the stated values. Plot the results as functions of wet-bulb temperature.

14–140 Atmospheric air enters an air-conditioning system at30°C and 70 percent relative humidity with a volume flow rateof 4 m3/min and is cooled to 20°C and 20 percent relativehumidity at a pressure of 1 atm. The system uses refrigerant-134a as the cooling fluid that enters the cooling section at

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Chapter 14 | 749

350 kPa with a quality of 20 percent and leaves as a saturatedvapor. Draw a schematic and show the process on the psychro-metric chart. What is the heat transfer from the air to the cool-ing coils, in kW? If any water is condensed from the air, howmuch water will be condensed from the atmospheric air permin? Determine the mass flow rate of the refrigerant, inkg/min.

14–141 An uninsulated tank having a volume of 0.5 m3

contains air at 35°C, 130 kPa, and 20 percent relative humid-ity. The tank is connected to a water supply line in whichwater flows at 50°C. Water is sprayed into the tank until therelative humidity of the air–vapor mixture is 90 percent.Determine the amount of water supplied to the tank, in kg,the final pressure of the air–vapor mixture in the tank, in kPa,and the heat transfer required during the process to maintainthe air– vapor mixture in the tank at 35°C.

14–142 Air flows steadily through an isentropic nozzle.The air enters the nozzle at 35°C, 200 kPa and 50 percent rel-ative humidity. If no condensation is to occur during theexpansion process, determine the pressure, temperature, andvelocity of the air at the nozzle exit.

Fundamentals of Engineering (FE) Exam Problems

14–143 A room is filled with saturated moist air at 25°Cand a total pressure of 100 kPa. If the mass of dry air in theroom is 100 kg, the mass of water vapor is

(a) 0.52 kg (b) 1.97 kg (c) 2.96 kg(d ) 2.04 kg (e) 3.17 kg

14–144 A room contains 50 kg of dry air and 0.6 kg ofwater vapor at 25°C and 95 kPa total pressure. The relativehumidity of air in the room is

(a) 1.2% (b) 18.4% (c) 56.7%(d ) 65.2% (e) 78.0%

14–145 A 40-m3 room contains air at 30°C and a total pres-sure of 90 kPa with a relative humidity of 75 percent. Themass of dry air in the room is

(a) 24.7 kg (b) 29.9 kg (c) 39.9 kg(d ) 41.4 kg (e) 52.3 kg

14–146 A room contains air at 30°C and a total pressure of96.0 kPa with a relative humidity of 75 percent. The partialpressure of dry air is

(a) 82.0 kPa (b) 85.8 kPa (c) 92.8 kPa(d ) 90.6 kPa (e) 72.0 kPa

14–147 The air in a house is at 20°C and 50 percent rela-tive humidity. Now the air is cooled at constant pressure.The temperature at which the moisture in the air will startcondensing is

(a) 8.7°C (b) 11.3°C (c) 13.8°C(d ) 9.3°C (e) 10.0°C

14–148 On the psychrometric chart, a cooling and dehu-midification process appears as a line that is(a) horizontal to the left(b) vertical downward(c) diagonal upwards to the right (NE direction)(d ) diagonal upwards to the left (NW direction)(e) diagonal downwards to the left (SW direction)

14–149 On the psychrometric chart, a heating and humidifi-cation process appears as a line that is(a) horizontal to the right(b) vertical upward(c) diagonal upwards to the right (NE direction)(d ) diagonal upwards to the left (NW direction)(e) diagonal downwards to the right (SE direction)

14–150 An air stream at a specified temperature and rela-tive humidity undergoes evaporative cooling by sprayingwater into it at about the same temperature. The lowest tem-perature the air stream can be cooled to is(a) the dry bulb temperature at the given state(b) the wet bulb temperature at the given state(c) the dew point temperature at the given state(d ) the saturation temperature corresponding to the

humidity ratio at the given state(e) the triple point temperature of water

14–151 Air is cooled and dehumidified as it flows over thecoils of a refrigeration system at 85 kPa from 30°C and ahumidity ratio of 0.023 kg/kg dry air to 15°C and a humidityratio of 0.015 kg/kg dry air. If the mass flow rate of dry air is0.7 kg/s, the rate of heat removal from the air is

(a) 5 kJ/s (b) 10 kJ/s (c) 15 kJ/s(d ) 20 kJ/s (e) 25 kJ/s

14–152 Air at a total pressure of 90 kPa, 15°C, and 75 per-cent relative humidity is heated and humidified to 25°C and75 percent relative humidity by introducing water vapor. Ifthe mass flow rate of dry air is 4 kg/s, the rate at which steamis added to the air is

(a) 0.032 kg/s (b) 0.013 kg/s (c) 0.019 kg/s(d ) 0.0079 kg/s (e) 0 kg/s

Cooling coils

Condensate

AIR

FIGURE P14–140

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Design and Essay Problems

14–153 The condensation and even freezing of moisture inbuilding walls without effective vapor retarders are of realconcern in cold climates as they undermine the effectivenessof the insulation. Investigate how the builders in your area arecoping with this problem, whether they are using vaporretarders or vapor barriers in the walls, and where they arelocated in the walls. Prepare a report on your findings, andexplain the reasoning for the current practice.

14–154 The air-conditioning needs of a large building canbe met by a single central system or by several individualwindow units. Considering that both approaches are com-monly used in practice, the right choice depends on the situa-tion on hand. Identify the important factors that need to beconsidered in decision making, and discuss the conditions

750 | Thermodynamics

under which an air-conditioning system that consists of sev-eral window units is preferable over a large single centralsystem, and vice versa.

14–155 Identify the major sources of heat gain in yourhouse in summer, and propose ways of minimizing them andthus reducing the cooling load.

14–156 Write an essay on different humidity measurementdevices, including electronic ones, and discuss the advantagesand disadvantages of each device.

14–157 Design an inexpensive evaporative cooling systemsuitable for use in your house. Show how you would obtain awater spray, how you would provide airflow, and how youwould prevent water droplets from drifting into the livingspace.

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Chapter 15CHEMICAL REACTIONS

| 751

In the preceding chapters we limited our consideration tononreacting systems—systems whose chemical composi-tion remains unchanged during a process. This was the

case even with mixing processes during which a homoge-neous mixture is formed from two or more fluids without theoccurrence of any chemical reactions. In this chapter, wespecifically deal with systems whose chemical compositionchanges during a process, that is, systems that involve chem-ical reactions.

When dealing with nonreacting systems, we need to con-sider only the sensible internal energy (associated with tem-perature and pressure changes) and the latent internalenergy (associated with phase changes). When dealing withreacting systems, however, we also need to consider thechemical internal energy, which is the energy associated withthe destruction and formation of chemical bonds between theatoms. The energy balance relations developed for nonreact-ing systems are equally applicable to reacting systems, butthe energy terms in the latter case should include the chemi-cal energy of the system.

In this chapter we focus on a particular type of chemicalreaction, known as combustion, because of its importance inengineering. But the reader should keep in mind, however,that the principles developed are equally applicable to otherchemical reactions.

We start this chapter with a general discussion of fuels andcombustion. Then we apply the mass and energy balances toreacting systems. In this regard we discuss the adiabaticflame temperature, which is the highest temperature a react-ing mixture can attain. Finally, we examine the second-lawaspects of chemical reactions.

ObjectivesThe objectives of Chapter 15 are to:

• Give an overview of fuels and combustion.

• Apply the conservation of mass to reacting systems todetermine balanced reaction equations.

• Define the parameters used in combustion analysis, suchas air–fuel ratio, percent theoretical air, and dew-pointtemperature.

• Apply energy balances to reacting systems for both steady-flow control volumes and fixed mass systems.

• Calculate the enthalpy of reaction, enthalpy of combustion,and the heating values of fuels.

• Determine the adiabatic flame temperature for reactingmixtures.

• Evaluate the entropy change of reacting systems.

• Analyze reacting systems from the second-law perspective.

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15–1 � FUELS AND COMBUSTIONAny material that can be burned to release thermal energy is called a fuel.Most familiar fuels consist primarily of hydrogen and carbon. They arecalled hydrocarbon fuels and are denoted by the general formula CnHm.Hydrocarbon fuels exist in all phases, some examples being coal, gasoline,and natural gas.

The main constituent of coal is carbon. Coal also contains varyingamounts of oxygen, hydrogen, nitrogen, sulfur, moisture, and ash. It is diffi-cult to give an exact mass analysis for coal since its composition variesconsiderably from one geographical area to the next and even within thesame geographical location. Most liquid hydrocarbon fuels are a mixtureof numerous hydrocarbons and are obtained from crude oil by distillation(Fig. 15–1). The most volatile hydrocarbons vaporize first, forming what weknow as gasoline. The less volatile fuels obtained during distillation arekerosene, diesel fuel, and fuel oil. The composition of a particular fueldepends on the source of the crude oil as well as on the refinery.

Although liquid hydrocarbon fuels are mixtures of many different hydro-carbons, they are usually considered to be a single hydrocarbon for conve-nience. For example, gasoline is treated as octane, C8H18, and the dieselfuel as dodecane, C12H26. Another common liquid hydrocarbon fuel ismethyl alcohol, CH3OH, which is also called methanol and is used in somegasoline blends. The gaseous hydrocarbon fuel natural gas, which is a mix-ture of methane and smaller amounts of other gases, is often treated asmethane, CH4, for simplicity.

Natural gas is produced from gas wells or oil wells rich in natural gas. Itis composed mainly of methane, but it also contains small amounts ofethane, propane, hydrogen, helium, carbon dioxide, nitrogen, hydrogen sul-fate, and water vapor. On vehicles, it is stored either in the gas phase atpressures of 150 to 250 atm as CNG (compressed natural gas), or in the liq-uid phase at �162°C as LNG (liquefied natural gas). Over a million vehi-cles in the world, mostly buses, run on natural gas. Liquefied petroleum gas(LPG) is a byproduct of natural gas processing or the crude oil refining. Itconsists mainly of propane and thus LPG is usually referred to as propane.However, it also contains varying amounts of butane, propylene, andbutylenes. Propane is commonly used in fleet vehicles, taxis, school buses,and private cars. Ethanol is obtained from corn, grains, and organic waste.Methonal is produced mostly from natural gas, but it can also be obtainedfrom coal and biomass. Both alcohols are commonly used as additives inoxygenated gasoline and reformulated fuels to reduce air pollution.

Vehicles are a major source of air pollutants such as nitric oxides, carbonmonoxide, and hydrocarbons, as well as the greenhouse gas carbon dioxide,and thus there is a growing shift in the transportation industry from the tra-ditional petroleum-based fuels such as gaoline and diesel fuel to the cleanerburning alternative fuels friendlier to the environment such as natural gas,alcohols (ethanol and methanol), liquefied petroleum gas (LPG), andhydrogen. The use of electric and hybrid cars is also on the rise. A compari-son of some alternative fuels for transportation to gasoline is given in Table15–1. Note that the energy contents of alternative fuels per unit volume arelower than that of gasoline or diesel fuel, and thus the driving range of a

752 | Thermodynamics

Gasoline

Kerosene

Diesel fuel

Fuel oil

CRUDEOIL

FIGURE 15–1Most liquid hydrocarbon fuels areobtained from crude oil by distillation.

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vehicle on a full tank is lower when running on an alternative fuel. Also,when comparing cost, a realistic measure is the cost per unit energy ratherthan cost per unit volume. For example, methanol at a unit cost of $1.20/Lmay appear cheaper than gasoline at $1.80/L, but this is not the case sincethe cost of 10,000 kJ of energy is $0.57 for gasoline and $0.66 formethanol.

A chemical reaction during which a fuel is oxidized and a large quantityof energy is released is called combustion (Fig. 15–2). The oxidizer mostoften used in combustion processes is air, for obvious reasons—it is freeand readily available. Pure oxygen O2 is used as an oxidizer only in somespecialized applications, such as cutting and welding, where air cannot beused. Therefore, a few words about the composition of air are in order.

On a mole or a volume basis, dry air is composed of 20.9 percent oxygen,78.1 percent nitrogen, 0.9 percent argon, and small amounts of carbon diox-ide, helium, neon, and hydrogen. In the analysis of combustion processes,the argon in the air is treated as nitrogen, and the gases that exist in traceamounts are disregarded. Then dry air can be approximated as 21 percentoxygen and 79 percent nitrogen by mole numbers. Therefore, each mole ofoxygen entering a combustion chamber is accompanied by 0.79/0.21 � 3.76mol of nitrogen (Fig. 15–3). That is,

(15–1)

During combustion, nitrogen behaves as an inert gas and does not react withother elements, other than forming a very small amount of nitric oxides.However, even then the presence of nitrogen greatly affects the outcome ofa combustion process since nitrogen usually enters a combustion chamber inlarge quantities at low temperatures and exits at considerably higher tempera-tures, absorbing a large proportion of the chemical energy released duringcombustion. Throughout this chapter, nitrogen is assumed to remain perfectly

1 kmol O2 � 3.76 kmol N2 � 4.76 kmol air

Chapter 15 | 753

FIGURE 15–2Combustion is a chemical reactionduring which a fuel is oxidized and alarge quantity of energy is released.

© Reprinted with special permission of KingFeatures Syndicate.

AIRAIR

( )( )21% O21% O279% N79% N2

1 kmol O1 kmol O23.76 kmol N3.76 kmol N2

FIGURE 15–3Each kmol of O2 in air is accompaniedby 3.76 kmol of N2.

TABLE 15–1A comparison of some alternative fuels to the traditional petroleum-based fuelsused in transportation

Energy content Gasoline equivalence,*Fuel kJ/L L/L-gasoline

Gasoline 31,850 1Light diesel 33,170 0.96Heavy diesel 35,800 0.89LPG (Liquefied petroleum gas,

primarily propane) 23,410 1.36Ethanol (or ethyl alcohol) 29,420 1.08Methanol (or methyl alcohol) 18,210 1.75CNG (Compressed natural gas,

primarily methane, at 200 atm) 8,080 3.94LNG (Liquefied natural gas,

primarily methane) 20,490 1.55

*Amount of fuel whose energy content is equal to the energy content of 1-L gasoline.

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inert. Keep in mind, however, that at very high temperatures, such as thoseencountered in internal combustion engines, a small fraction of nitrogenreacts with oxygen, forming hazardous gases such as nitric oxide.

Air that enters a combustion chamber normally contains some watervapor (or moisture), which also deserves consideration. For most combus-tion processes, the moisture in the air and the H2O that forms during com-bustion can also be treated as an inert gas, like nitrogen. At very hightemperatures, however, some water vapor dissociates into H2 and O2 as wellas into H, O, and OH. When the combustion gases are cooled below thedew-point temperature of the water vapor, some moisture condenses. It isimportant to be able to predict the dew-point temperature since the waterdroplets often combine with the sulfur dioxide that may be present in thecombustion gases, forming sulfuric acid, which is highly corrosive.

During a combustion process, the components that exist before the reac-tion are called reactants and the components that exist after the reaction arecalled products (Fig. 15–4). Consider, for example, the combustion of1 kmol of carbon with 1 kmol of pure oxygen, forming carbon dioxide,

(15–2)

Here C and O2 are the reactants since they exist before combustion, andCO2 is the product since it exists after combustion. Note that a reactant doesnot have to react chemically in the combustion chamber. For example, ifcarbon is burned with air instead of pure oxygen, both sides of the combus-tion equation will include N2. That is, the N2 will appear both as a reactantand as a product.

We should also mention that bringing a fuel into intimate contact withoxygen is not sufficient to start a combustion process. (Thank goodness it isnot. Otherwise, the whole world would be on fire now.) The fuel must bebrought above its ignition temperature to start the combustion. The mini-mum ignition temperatures of various substances in atmospheric air areapproximately 260°C for gasoline, 400°C for carbon, 580°C for hydrogen,610°C for carbon monoxide, and 630°C for methane. Moreover, the propor-tions of the fuel and air must be in the proper range for combustion tobegin. For example, natural gas does not burn in air in concentrations lessthan 5 percent or greater than about 15 percent.

As you may recall from your chemistry courses, chemical equations arebalanced on the basis of the conservation of mass principle (or the massbalance), which can be stated as follows: The total mass of each element isconserved during a chemical reaction (Fig. 15–5). That is, the total mass ofeach element on the right-hand side of the reaction equation (the products)must be equal to the total mass of that element on the left-hand side (thereactants) even though the elements exist in different chemical compoundsin the reactants and products. Also, the total number of atoms of each ele-ment is conserved during a chemical reaction since the total number ofatoms is equal to the total mass of the element divided by its atomic mass.

For example, both sides of Eq. 15–2 contain 12 kg of carbon and 32 kg ofoxygen, even though the carbon and the oxygen exist as elements in thereactants and as a compound in the product. Also, the total mass of reactantsis equal to the total mass of products, each being 44 kg. (It is commonpractice to round the molar masses to the nearest integer if great accuracy is

C � O2 S CO2

754 | Thermodynamics

Reactionchamber

ReactantsProducts

FIGURE 15–4In a steady-flow combustion process,the components that enter the reactionchamber are called reactants and thecomponents that exit are calledproducts.

H2 � 2

2 kg hydrogen2 kg hydrogen

16 kg oxygen16 kg oxygen

2 kg hydrogen2 kg hydrogen16 kg oxygen16 kg oxygen

1 O2 → H2O

FIGURE 15–5The mass (and number of atoms) ofeach element is conserved during achemical reaction.

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EXAMPLE 15–1 Balancing the Combustion Equation

One kmol of octane (C8H18) is burned with air that contains 20 kmol of O2,as shown in Fig. 15–7. Assuming the products contain only CO2, H2O, O2,and N2, determine the mole number of each gas in the products and theair–fuel ratio for this combustion process.

Solution The amount of fuel and the amount of oxygen in the air are given.The amount of the products and the AF are to be determined.Assumptions The combustion products contain CO2, H2O, O2, and N2 only.Properties The molar mass of air is Mair � 28.97 kg/kmol � 29.0 kg/kmol(Table A–1).Analysis The chemical equation for this combustion process can be written as

where the terms in the parentheses represent the composition of dry air thatcontains 1 kmol of O2 and x, y, z, and w represent the unknown mole num-bers of the gases in the products. These unknowns are determined by apply-ing the mass balance to each of the elements—that is, by requiring that thetotal mass or mole number of each element in the reactants be equal to thatin the products:

C:

H:

O:

N2:

Substituting yields

Note that the coefficient 20 in the balanced equation above represents thenumber of moles of oxygen, not the number of moles of air. The latter isobtained by adding 20 � 3.76 � 75.2 moles of nitrogen to the 20 moles of

C8H18 � 20 1O2 � 3.76N2 2 S 8CO2 � 9H2O � 7.5O2 � 75.2N2

120 2 13.76 2 � w S w � 75.2

20 � 2 � 2x � y � 2z S z � 7.5

18 � 2y S y � 9

8 � x S x � 8

C8H18 � 20 1O2 � 3.76N2 2 S xCO2 � yH2O � zO2 � wN2

not required.) However, notice that the total mole number of the reactants(2 kmol) is not equal to the total mole number of the products (1 kmol). Thatis, the total number of moles is not conserved during a chemical reaction.

A frequently used quantity in the analysis of combustion processes toquantify the amounts of fuel and air is the air–fuel ratio AF. It is usuallyexpressed on a mass basis and is defined as the ratio of the mass of air tothe mass of fuel for a combustion process (Fig. 15–6). That is,

(15–3)

The mass m of a substance is related to the number of moles N through therelation m � NM, where M is the molar mass.

The air–fuel ratio can also be expressed on a mole basis as the ratio of themole numbers of air to the mole numbers of fuel. But we will use the for-mer definition. The reciprocal of air–fuel ratio is called the fuel–air ratio.

AF �m air

m fuel

Chapter 15 | 755

Combustionchamber

Air

Products

AF = 1717 kg

Fuel

1 kg

18 kg

FIGURE 15–6The air–fuel ratio (AF) represents theamount of air used per unit mass offuel during a combustion process.

Combustion

chamberAIR

C8H18

1 kmolx CO2

y H2Oz O2

w N2

FIGURE 15–7Schematic for Example 15–1.

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15–2 � THEORETICAL AND ACTUALCOMBUSTION PROCESSES

It is often instructive to study the combustion of a fuel by assuming that thecombustion is complete. A combustion process is complete if all the carbonin the fuel burns to CO2, all the hydrogen burns to H2O, and all the sulfur (ifany) burns to SO2. That is, all the combustible components of a fuel areburned to completion during a complete combustion process (Fig. 15–8).Conversely, the combustion process is incomplete if the combustion prod-ucts contain any unburned fuel or components such as C, H2, CO, or OH.

Insufficient oxygen is an obvious reason for incomplete combustion, but itis not the only one. Incomplete combustion occurs even when more oxygenis present in the combustion chamber than is needed for complete combus-tion. This may be attributed to insufficient mixing in the combustion cham-ber during the limited time that the fuel and the oxygen are in contact.Another cause of incomplete combustion is dissociation, which becomesimportant at high temperatures.

Oxygen has a much greater tendency to combine with hydrogen than itdoes with carbon. Therefore, the hydrogen in the fuel normally burns tocompletion, forming H2O, even when there is less oxygen than needed forcomplete combustion. Some of the carbon, however, ends up as CO or justas plain C particles (soot) in the products.

The minimum amount of air needed for the complete combustion of a fuelis called the stoichiometric or theoretical air. Thus, when a fuel is com-pletely burned with theoretical air, no uncombined oxygen is present in theproduct gases. The theoretical air is also referred to as the chemically cor-rect amount of air, or 100 percent theoretical air. A combustion processwith less than the theoretical air is bound to be incomplete. The ideal com-bustion process during which a fuel is burned completely with theoreticalair is called the stoichiometric or theoretical combustion of that fuel (Fig.15–9). For example, the theoretical combustion of methane is

Notice that the products of the theoretical combustion contain no unburnedmethane and no C, H2, CO, OH, or free O2.

CH4 � 2 1O2 � 3.76N2 2 S CO2 � 2H2O � 7.52N2

756 | Thermodynamics

oxygen, giving a total of 95.2 moles of air. The air–fuel ratio (AF) is deter-mined from Eq. 15–3 by taking the ratio of the mass of the air and the massof the fuel,

That is, 24.2 kg of air is used to burn each kilogram of fuel during thiscombustion process.

� 24.2 kg air/kg fuel

�120 � 4.76 kmol 2 129 kg>kmol 2

18 kmol 2 112 kg>kmol 2 � 19 kmol 2 12 kg>kmol 2

AF �m air

m fuel�

1NM 2 air

1NM 2C � 1NM 2H2

CH4 + 2(O2 + 3.76N2) →

• no unburned fuel• no free oxygen in products

CO2 + 2H2O + 7.52N2

FIGURE 15–9The complete combustion processwith no free oxygen in the products iscalled theoretical combustion.

CombustionchamberAIR

CnHm

Fuel n CO2m H2O

Excess O2N2

2

FIGURE 15–8A combustion process is complete ifall the combustible components of thefuel are burned to completion.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 756

In actual combustion processes, it is common practice to use more airthan the stoichiometric amount to increase the chances of complete combus-tion or to control the temperature of the combustion chamber. The amountof air in excess of the stoichiometric amount is called excess air. Theamount of excess air is usually expressed in terms of the stoichiometric airas percent excess air or percent theoretical air. For example, 50 percentexcess air is equivalent to 150 percent theoretical air, and 200 percentexcess air is equivalent to 300 percent theoretical air. Of course, the stoi-chiometric air can be expressed as 0 percent excess air or 100 percent theo-retical air. Amounts of air less than the stoichiometric amount are calleddeficiency of air and are often expressed as percent deficiency of air. Forexample, 90 percent theoretical air is equivalent to 10 percent deficiency ofair. The amount of air used in combustion processes is also expressed interms of the equivalence ratio, which is the ratio of the actual fuel–air ratioto the stoichiometric fuel–air ratio.

Predicting the composition of the products is relatively easy when thecombustion process is assumed to be complete and the exact amounts of thefuel and air used are known. All one needs to do in this case is simply applythe mass balance to each element that appears in the combustion equation,without needing to take any measurements. Things are not so simple, how-ever, when one is dealing with actual combustion processes. For one thing,actual combustion processes are hardly ever complete, even in the presenceof excess air. Therefore, it is impossible to predict the composition of theproducts on the basis of the mass balance alone. Then the only alternative wehave is to measure the amount of each component in the products directly.

A commonly used device to analyze the composition of combustion gasesis the Orsat gas analyzer. In this device, a sample of the combustion gasesis collected and cooled to room temperature and pressure, at which point itsvolume is measured. The sample is then brought into contact with a chemi-cal that absorbs the CO2. The remaining gases are returned to the room tem-perature and pressure, and the new volume they occupy is measured. Theratio of the reduction in volume to the original volume is the volume frac-tion of the CO2, which is equivalent to the mole fraction if ideal-gas behav-ior is assumed (Fig. 15–10). The volume fractions of the other gases aredetermined by repeating this procedure. In Orsat analysis the gas sample iscollected over water and is maintained saturated at all times. Therefore, thevapor pressure of water remains constant during the entire test. For this rea-son the presence of water vapor in the test chamber is ignored and data arereported on a dry basis. However, the amount of H2O formed during com-bustion is easily determined by balancing the combustion equation.

Chapter 15 | 757

EXAMPLE 15–2 Dew-Point Temperature of Combustion Products

Ethane (C2H6) is burned with 20 percent excess air during a combustionprocess, as shown in Fig. 15–11. Assuming complete combustion and a totalpressure of 100 kPa, determine (a) the air–fuel ratio and (b) the dew-pointtemperature of the products.

Solution The fuel is burned completely with excess air. The AF and thedew point of the products are to be determined.

BEFOREAFTER

100 kPa25°C

Gas sampleincluding CO2

1 liter

100 kPa25°C

Gas samplewithout CO2

0.9 liter

yCO2

=V

CO2

V0.11

= = 0.1

FIGURE 15–10Determining the mole fraction of theCO2 in combustion gases by using theOrsat gas analyzer.

Combustionchamber

AIR

C2H6CO2

H2OO2

N2(20% excess)100 kPa

FIGURE 15–11Schematic for Example 15–2.

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758 | Thermodynamics

Assumptions 1 Combustion is complete. 2 Combustion gases are ideal gases.Analysis The combustion products contain CO2, H2O, N2, and some excessO2 only. Then the combustion equation can be written as

where ath is the stoichiometric coefficient for air. We have automaticallyaccounted for the 20 percent excess air by using the factor 1.2ath instead of athfor air. The stoichiometric amount of oxygen (athO2) is used to oxidize the fuel,and the remaining excess amount (0.2athO2) appears in the products as unusedoxygen. Notice that the coefficient of N2 is the same on both sides of the equa-tion, and that we wrote the C and H balances directly since they are so obvi-ous. The coefficient ath is determined from the O2 balance to be

Substituting,

(a) The air–fuel ratio is determined from Eq. 15–3 by taking the ratio of themass of the air to the mass of the fuel,

That is, 19.3 kg of air is supplied for each kilogram of fuel during this com-bustion process.

(b) The dew-point temperature of the products is the temperature at whichthe water vapor in the products starts to condense as the products arecooled at constant pressure. Recall from Chap. 14 that the dew-point tem-perature of a gas–vapor mixture is the saturation temperature of the watervapor corresponding to its partial pressure. Therefore, we need to determinethe partial pressure of the water vapor Pv in the products first. Assumingideal-gas behavior for the combustion gases, we have

Thus,

(Table A–5)Tdp � Tsat @ 13.96 kPa � 52.3°C

Pv � a Nv

Nprodb 1Pprod 2 � a 3 kmol

21.49 kmolb 1100 kPa 2 � 13.96 kPa

� 19.3 kg air/kg fuel

AF �mair

m fuel�

14.2 � 4.76 kmol 2 129 kg>kmol 212 kmol 2 112 kg>kmol 2 � 13 kmol 2 12 kg>kmol 2

C2H6 � 4.2 1O2 � 3.76N2 2 S 2CO2 � 3H2O � 0.7O2 � 15.79N2

O2: 1.2ath � 2 � 1.5 � 0.2ath S ath � 3.5

C2H6 � 1.2ath 1O2 � 3.76N2 2 S 2CO2 � 3H2O � 0.2athO2 � 11.2 � 3.76 2athN2

EXAMPLE 15–3 Combustion of a Gaseous Fuel with Moist Air

A certain natural gas has the following volumetric analysis: 72 percent CH4,9 percent H2, 14 percent N2, 2 percent O2, and 3 percent CO2. This gas isnow burned with the stoichiometric amount of air that enters the combustionchamber at 20°C, 1 atm, and 80 percent relative humidity, as shown inFig. 15–12. Assuming complete combustion and a total pressure of 1 atm,determine the dew-point temperature of the products.

Solution A gaseous fuel is burned with the stoichiometric amount of moistair. The dew point temperature of the products is to be determined.

Combustionchamber

AIR

FUEL

CO2

H2ON2

20°C, = 80%φ

1 atm

CH4, O2, H2,

N2, CO2

FIGURE 15–12Schematic for Example 15–3.

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Chapter 15 | 759

Assumptions 1 The fuel is burned completely and thus all the carbon in thefuel burns to CO2 and all the hydrogen to H2O. 2 The fuel is burned with thestoichiometric amount of air and thus there is no free O2 in the productgases. 3 Combustion gases are ideal gases.Properties The saturation pressure of water at 20°C is 2.3392 kPa (Table A–4).Analysis We note that the moisture in the air does not react with anything;it simply shows up as additional H2O in the products. Therefore, for simplic-ity, we balance the combustion equation by using dry air and then add themoisture later to both sides of the equation.

Considering 1 kmol of fuel,

fuel dry air

The unknown coefficients in the above equation are determined from massbalances on various elements,

C:

H:

O2:

N2:

Next we determine the amount of moisture that accompanies 4.76ath �(4.76)(1.465) � 6.97 kmol of dry air. The partial pressure of the moisturein the air is

Assuming ideal-gas behavior, the number of moles of the moisture in theair is

which yields

The balanced combustion equation is obtained by substituting the coeffi-cients determined earlier and adding 0.131 kmol of H2O to both sides of theequation:

fuel dry air

moisture includes moisture

The dew-point temperature of the products is the temperature at whichthe water vapor in the products starts to condense as the products are cooled.Again, assuming ideal-gas behavior, the partial pressure of the water vapor inthe combustion gases is

Pv,prod � aNv,prod

NprodbPprod � a 1.661 kmol

8.059 kmolb 1101.325 kPa 2 � 20.88 kPa

� 0.131H2O S 0.75CO2 � 1.661H2O � 5.648N2

10.72CH4 � 0.09H2 � 0.14N2 � 0.02O2 � 0.03CO2 2 � 1.465 1O2 � 3.76N2 2

Nv,air � 0.131 kmol

Nv,air � a Pv,air

Ptotalb Ntotal � a 1.871 kPa

101.325 kPab 16.97 � Nv,air 2

Pv,air � fair Psat @ 20°C � 10.80 2 12.3392 kPa 2 � 1.871 kPa

0.14 � 3.76a th � z S z � 5.648

0.02 � 0.03 � a th � x �y

2S a th � 1.465

0.72 � 4 � 0.09 � 2 � 2y S y � 1.53

0.72 � 0.03 � x S x � 0.75

xCO2 � yH2O � zN2

10.72CH4 � 0.09H2 � 0.14N2 � 0.02O2 � 0.03CO2 2 � ath 1O2 � 3.76N2 2 S

1555555555555552555555555555553 155525553

155555555555552555555555555553 1555255553

14243 14243

cen84959_ch15.qxd 4/20/05 3:23 PM Page 759

760 | Thermodynamics

Thus,

Discussion If the combustion process were achieved with dry air instead ofmoist air, the products would contain less moisture, and the dew-point tem-perature in this case would be 59.5°C.

Tdp � Tsat @ 20.88 kPa � 60.9°C

EXAMPLE 15–4 Reverse Combustion Analysis

Octane (C8H18) is burned with dry air. The volumetric analysis of the prod-ucts on a dry basis is (Fig. 15–13)

CO2: 10.02 percentO2: 5.62 percentCO: 0.88 percentN2: 83.48 percent

Determine (a) the air–fuel ratio, (b) the percentage of theoretical air used,and (c) the amount of H2O that condenses as the products are cooled to25°C at 100 kPa.

Solution Combustion products whose composition is given are cooled to25°C. The AF, the percent theoretical air used, and the fraction of watervapor that condenses are to be determined.Assumptions Combustion gases are ideal gases.Properties The saturation pressure of water at 25°C is 3.1698 kPa (Table A–4).Analysis Note that we know the relative composition of the products, butwe do not know how much fuel or air is used during the combustion process.However, they can be determined from mass balances. The H2O in the com-bustion gases will start condensing when the temperature drops to the dew-point temperature.

For ideal gases, the volume fractions are equivalent to the mole fractions.Considering 100 kmol of dry products for convenience, the combustionequation can be written as

The unknown coefficients x, a, and b are determined from mass balances,

N2:

C:

H:

O2:

The O2 balance is not necessary, but it can be used to check the valuesobtained from the other mass balances, as we did previously. Substituting,we get

10.02CO2 � 0.88CO � 5.62O2 � 83.48N2 � 12.24H2O1.36C8H18 � 22.2 1O2 � 3.76N2 2 S

a � 10.02 � 0.44 � 5.62 �b

2S 22.20 � 22.20

18x � 2b S b � 12.24

8x � 10.02 � 0.88 S x � 1.36

3.76a � 83.48 S a � 22.20

xC8H18 � a 1O2 � 3.76N2 2 S 10.02CO2 � 0.88CO � 5.62O2 � 83.48N2 � bH2O

Combustionchamber

AIR

C8H1810.02% CO2

5.62% O2

0.88% CO 83.48% N2

FIGURE 15–13Schematic for Example 15–4.

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Chapter 15 | 761

The combustion equation for 1 kmol of fuel is obtained by dividing theabove equation by 1.36,

(a) The air–fuel ratio is determined by taking the ratio of the mass of the airto the mass of the fuel (Eq. 15–3),

(b) To find the percentage of theoretical air used, we need to know the theo-retical amount of air, which is determined from the theoretical combustionequation of the fuel,

O2:

Then,

That is, 31 percent excess air was used during this combustion process.Notice that some carbon formed carbon monoxide even though there wasconsiderably more oxygen than needed for complete combustion.

(c) For each kmol of fuel burned, 7.37 � 0.65 � 4.13 � 61.38 � 9 �82.53 kmol of products are formed, including 9 kmol of H2O. Assuming thatthe dew-point temperature of the products is above 25°C, some of the watervapor will condense as the products are cooled to 25°C. If Nw kmol of H2Ocondenses, there will be (9 � Nw) kmol of water vapor left in the products.The mole number of the products in the gas phase will also decrease to82.53 � Nw as a result. By treating the product gases (including the remain-ing water vapor) as ideal gases, Nw is determined by equating the mole frac-tion of the water vapor to its pressure fraction,

Therefore, the majority of the water vapor in the products (73 percent of it)condenses as the product gases are cooled to 25°C.

Nw � 6.59 kmol

9 � Nw

82.53 � Nw

�3.1698 kPa

100 kPa

Nv

Nprod,gas�

Pv

Pprod

� 131%

�116.32 2 14.76 2 kmol

112.50 2 14.76 2 kmol

Percentage of theoretical air �mair,act

mair,th�

Nair,act

Nair,th

ath � 8 � 4.5 S ath � 12.5

C8H18 � ath 1O2 � 3.76N2 2 S 8CO2 � 9H2O � 3.76athN2

� 19.76 kg air/kg fuel

AF �m air

m fuel�

116.32 � 4.76 kmol 2 129 kg>kmol 218 kmol 2 112 kg>kmol 2 � 19 kmol 2 12 kg>kmol 2

7.37CO2 � 0.65CO � 4.13O2 � 61.38N2 � 9H2O

C8H18 � 16.32 1O2 � 3.76N2 2 S

cen84959_ch15.qxd 4/20/05 3:23 PM Page 761

15–3 � ENTHALPY OF FORMATION AND ENTHALPY OF COMBUSTION

We mentioned in Chap. 2 that the molecules of a system possess energy invarious forms such as sensible and latent energy (associated with a changeof state), chemical energy (associated with the molecular structure),and nuclear energy (associated with the atomic structure), as illustrated inFig. 15–14. In this text we do not intend to deal with nuclear energy. Wealso ignored chemical energy until now since the systems considered in pre-vious chapters involved no changes in their chemical structure, and thus nochanges in chemical energy. Consequently, all we needed to deal with werethe sensible and latent energies.

During a chemical reaction, some chemical bonds that bind the atoms intomolecules are broken, and new ones are formed. The chemical energy asso-ciated with these bonds, in general, is different for the reactants and theproducts. Therefore, a process that involves chemical reactions involveschanges in chemical energies, which must be accounted for in an energybalance (Fig. 15–15). Assuming the atoms of each reactant remain intact (nonuclear reactions) and disregarding any changes in kinetic and potentialenergies, the energy change of a system during a chemical reaction is due toa change in state and a change in chemical composition. That is,

(15–4)

Therefore, when the products formed during a chemical reaction exit thereaction chamber at the inlet state of the reactants, we have �Estate � 0 andthe energy change of the system in this case is due to the changes in itschemical composition only.

In thermodynamics we are concerned with the changes in the energy of asystem during a process, and not the energy values at the particular states.Therefore, we can choose any state as the reference state and assign a valueof zero to the internal energy or enthalpy of a substance at that state. Whena process involves no changes in chemical composition, the reference statechosen has no effect on the results. When the process involves chemicalreactions, however, the composition of the system at the end of a process isno longer the same as that at the beginning of the process. In this case itbecomes necessary to have a common reference state for all substances. Thechosen reference state is 25°C (77°F) and 1 atm, which is known as thestandard reference state. Property values at the standard reference stateare indicated by a superscript (°) (such as h° and u°).

When analyzing reacting systems, we must use property values relative to thestandard reference state. However, it is not necessary to prepare a new set ofproperty tables for this purpose. We can use the existing tables by subtractingthe property values at the standard reference state from the values at the speci-fied state. The ideal-gas enthalpy of N2 at 500 K relative to the standard refer-ence state, for example, is h

–500 K � h

–° � 14,581 � 8669 � 5912 kJ/kmol.

Consider the formation of CO2 from its elements, carbon and oxygen,during a steady-flow combustion process (Fig. 15–16). Both the carbon andthe oxygen enter the combustion chamber at 25°C and 1 atm. The CO2formed during this process also leaves the combustion chamber at 25°C and1 atm. The combustion of carbon is an exothermic reaction (a reaction dur-

¢Esys � ¢Estate � ¢Echem

762 | Thermodynamics

Nuclear energy

Chemical energy

Latent energy

Sensibleenergy

MOLECULE

MOLECULE

ATOM

ATOM

FIGURE 15–14The microscopic form of energy of asubstance consists of sensible, latent,chemical, and nuclear energies.

Combustionchamber

CO2

393,520 kJ

25°C, 1 atm1 kmol O2

25°C, 1 atm

1 kmol C

25°C, 1 atm

FIGURE 15–16The formation of CO2 during a steady-flow combustion process at 25�C and1 atm.

Brokenchemical bond

Sensibleenergy

ATOM ATOM

ATOM

FIGURE 15–15When the existing chemical bonds aredestroyed and new ones are formedduring a combustion process, usually alarge amount of sensible energy isabsorbed or released.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 762

ing which chemical energy is released in the form of heat). Therefore, someheat is transferred from the combustion chamber to the surroundings duringthis process, which is 393,520 kJ/kmol CO2 formed. (When one is dealingwith chemical reactions, it is more convenient to work with quantities perunit mole than per unit time, even for steady-flow processes.)

The process described above involves no work interactions. Therefore,from the steady-flow energy balance relation, the heat transfer during thisprocess must be equal to the difference between the enthalpy of the productsand the enthalpy of the reactants. That is,

(15–5)

Since both the reactants and the products are at the same state, the enthalpychange during this process is solely due to the changes in the chemical com-position of the system. This enthalpy change is different for different reac-tions, and it is very desirable to have a property to represent the changes inchemical energy during a reaction. This property is the enthalpy of reac-tion hR, which is defined as the difference between the enthalpy of the prod-ucts at a specified state and the enthalpy of the reactants at the same statefor a complete reaction.

For combustion processes, the enthalpy of reaction is usually referred toas the enthalpy of combustion hC, which represents the amount of heatreleased during a steady-flow combustion process when 1 kmol (or 1 kg)of fuel is burned completely at a specified temperature and pressure(Fig. 15–17). It is expressed as

(15–6)

which is �393,520 kJ/kmol for carbon at the standard reference state. Theenthalpy of combustion of a particular fuel is different at different tempera-tures and pressures.

The enthalpy of combustion is obviously a very useful property for ana-lyzing the combustion processes of fuels. However, there are so many dif-ferent fuels and fuel mixtures that it is not practical to list hC values for allpossible cases. Besides, the enthalpy of combustion is not of much usewhen the combustion is incomplete. Therefore a more practical approachwould be to have a more fundamental property to represent the chemicalenergy of an element or a compound at some reference state. This propertyis the enthalpy of formation h

–f , which can be viewed as the enthalpy of a

substance at a specified state due to its chemical composition.To establish a starting point, we assign the enthalpy of formation of all

stable elements (such as O2, N2, H2, and C) a value of zero at the standardreference state of 25°C and 1 atm. That is, h

–f � 0 for all stable elements.

(This is no different from assigning the internal energy of saturated liquidwater a value of zero at 0.01°C.) Perhaps we should clarify what we meanby stable. The stable form of an element is simply the chemically stableform of that element at 25°C and 1 atm. Nitrogen, for example, exists indiatomic form (N2) at 25°C and 1 atm. Therefore, the stable form of nitro-gen at the standard reference state is diatomic nitrogen N2, not monatomicnitrogen N. If an element exists in more than one stable form at 25°C and1 atm, one of the forms should be specified as the stable form. For carbon,for example, the stable form is assumed to be graphite, not diamond.

hR � hC � Hprod � Hreact

Q � Hprod � Hreact � �393,520 kJ>kmol

Chapter 15 | 763

Combustionprocess

1 kmol CO2

hC = Q = –393,520 kJ/kmol C

25°C, 1 atm1 kmol O2

25°C, 1 atm

1 kmol C

25°C, 1 atm

FIGURE 15–17The enthalpy of combustion representsthe amount of energy released as afuel is burned during a steady-flowprocess at a specified state.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 763

Now reconsider the formation of CO2 (a compound) from its elements Cand O2 at 25°C and 1 atm during a steady-flow process. The enthalpychange during this process was determined to be �393,520 kJ/kmol. How-ever, Hreact � 0 since both reactants are elements at the standard referencestate, and the products consist of 1 kmol of CO2 at the same state. There-fore, the enthalpy of formation of CO2 at the standard reference state is�393,520 kJ/kmol (Fig. 15–18). That is,

The negative sign is due to the fact that the enthalpy of 1 kmol of CO2 at25°C and 1 atm is 393,520 kJ less than the enthalpy of 1 kmol of C and1 kmol of O2 at the same state. In other words, 393,520 kJ of chemicalenergy is released (leaving the system as heat) when C and O2 combine toform 1 kmol of CO2. Therefore, a negative enthalpy of formation for a com-pound indicates that heat is released during the formation of that compoundfrom its stable elements. A positive value indicates heat is absorbed.

You will notice that two h–°f values are given for H2O in Table A–26, one

for liquid water and the other for water vapor. This is because both phasesof H2O are encountered at 25°C, and the effect of pressure on the enthalpyof formation is small. (Note that under equilibrium conditions, water existsonly as a liquid at 25°C and 1 atm.) The difference between the twoenthalpies of formation is equal to the hfg of water at 25°C, which is 2441.7kJ/kg or 44,000 kJ/kmol.

Another term commonly used in conjunction with the combustion of fuelsis the heating value of the fuel, which is defined as the amount of heatreleased when a fuel is burned completely in a steady-flow process and theproducts are returned to the state of the reactants. In other words, the heat-ing value of a fuel is equal to the absolute value of the enthalpy of combus-tion of the fuel. That is,

The heating value depends on the phase of the H2O in the products. Theheating value is called the higher heating value (HHV) when the H2O inthe products is in the liquid form, and it is called the lower heating value(LHV) when the H2O in the products is in the vapor form (Fig. 15–19). Thetwo heating values are related by

(15–7)

where m is the mass of H2O in the products per unit mass of fuel and hfg isthe enthalpy of vaporization of water at the specified temperature. Higherand lower heating values of common fuels are given in Table A–27.

The heating value or enthalpy of combustion of a fuel can be determinedfrom a knowledge of the enthalpy of formation for the compounds involved.This is illustrated with the following example.

EXAMPLE 15–5 Evaluation of the Enthalpy of Combustion

Determine the enthalpy of combustion of liquid octane (C8H18) at 25°C and1 atm, using enthalpy-of-formation data from Table A–26. Assume the waterin the products is in the liquid form.

HHV � LHV � 1mhfg 2H2O 1kJ>kg fuel 2

Heating value � 0hC 0 1kJ>kg fuel 2

h°f,CO2� �393,520 kJ>kmol

764 | Thermodynamics

Combustionchamber

1 kmol CO2

h �f = Q = –393,520 kJ/kmol CO2

25°C, 1 atm1 kmol O2

25°C, 1 atm

1 kmol C

25°C, 1 atm

FIGURE 15–18The enthalpy of formation of acompound represents the amount ofenergy absorbed or released as thecomponent is formed from its stableelements during a steady-flow processat a specified state.

Combustion Products(vapor H2O)

Products(liquid H2O)

chamber

(mhfg)H2O

Air

Fuel

H2O

LHV = Qout

HHV = LHV + (mhfg)

1 kg

FIGURE 15–19The higher heating value of a fuel isequal to the sum of the lower heatingvalue of the fuel and the latent heat ofvaporization of the H2O in theproducts.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 764

Solution The enthalpy of combustion of a fuel is to be determined usingenthalpy of formation data.Properties The enthalpy of formation at 25°C and 1 atm is �393,520kJ/kmol for CO2, �285,830 kJ/kmol for H2O(�), and �249,950 kJ/kmol forC8H18(�) (Table A–26).Analysis The combustion of C8H18 is illustrated in Fig. 15–20. The stoi-chiometric equation for this reaction is

Both the reactants and the products are at the standard reference state of25°C and 1 atm. Also, N2 and O2 are stable elements, and thus theirenthalpy of formation is zero. Then the enthalpy of combustion of C8H18becomes (Eq. 15–6)

Substituting,

which is practially identical to the listed value of 47,890 kJ/kg in TableA–27. Since the water in the products is assumed to be in the liquid phase,this hC value corresponds to the HHV of liquid C8H18.Discussion It can be shown that the result for gaseous octane is�5,512,200 kJ/kmol or �48,255 kJ/kg.

When the exact composition of the fuel is known, the enthalpy of combus-tion of that fuel can be determined using enthalpy of formation data asshown above. However, for fuels that exhibit considerable variations incomposition depending on the source, such as coal, natural gas, and fuel oil,it is more practical to determine their enthalpy of combustion experimen-tally by burning them directly in a bomb calorimeter at constant volume orin a steady-flow device.

15–4 � FIRST-LAW ANALYSIS OF REACTING SYSTEMS

The energy balance (or the first-law) relations developed in Chaps. 4 and 5are applicable to both reacting and nonreacting systems. However, chemi-cally reacting systems involve changes in their chemical energy, and thus itis more convenient to rewrite the energy balance relations so that thechanges in chemical energies are explicitly expressed. We do this first forsteady-flow systems and then for closed systems.

Steady-Flow SystemsBefore writing the energy balance relation, we need to express the enthalpyof a component in a form suitable for use for reacting systems. That is, weneed to express the enthalpy such that it is relative to the standard reference

� �5,471,000 kJ>kmol C8H18 � �47,891 kJ>kg C8H18

� 11 kmol 2 1�249,950 kJ>kmol 2 hC � 18 kmol 2 1�393,520 kJ>kmol 2 � 19 kmol 2 1�285,830 kJ>kmol 2

�a Nph°f,p �a Nr h°f,r � 1Nh°f 2CO2� 1Nh°f 2H2O � 1Nh°f 2C8H18

hC � Hprod � Hreact

C8H18 � ath 1O2 � 3.76N2 2 S 8CO2 � 9H2O 1� 2 � 3.76athN2

Chapter 15 | 765

CO2

hC = Hprod – Hreact

25°C

AIR

25°C, 1 atm

C8H18

25°C, 1 atm

1 atmH2O(�)

N2

(�)

FIGURE 15–20Schematic for Example 15–5.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 765

state and the chemical energy term appears explicitly. When expressed prop-erly, the enthalpy term should reduce to the enthalpy of formation h

–°f at the

standard reference state. With this in mind, we express the enthalpy of acomponent on a unit mole basis as (Fig. 15–21)

where the term in the parentheses represents the sensible enthalpy relative tothe standard reference state, which is the difference between h

–the sensible

enthalpy at the specified state) and h–° (the sensible enthalpy at the standard

reference state of 25°C and 1 atm). This definition enables us to use enthalpyvalues from tables regardless of the reference state used in their construction.

When the changes in kinetic and potential energies are negligible, thesteady-flow energy balance relation E

.in � E

.out can be expressed for a chem-

ically reacting steady-flow system more explicitly as

(15–8)

Rate of net energy transfer in Rate of net energy transfer outby heat, work, and mass by heat, work, and mass

where n.

p and n.

r represent the molal flow rates of the product p and the reac-tant r, respectively.

In combustion analysis, it is more convenient to work with quantitiesexpressed per mole of fuel. Such a relation is obtained by dividing eachterm of the equation above by the molal flow rate of the fuel, yielding

(15–9)

Energy transfer in per mole of fuel Energy transfer out per mole of fuelby heat, work, and mass by heat, work, and mass

where Nr and Np represent the number of moles of the reactant r and theproduct p, respectively, per mole of fuel. Note that Nr � 1 for the fuel, andthe other Nr and Np values can be picked directly from the balancedcombustion equation. Taking heat transfer to the system and work done bythe system to be positive quantities, the energy balance relation just dis-cussed can be expressed more compactly as

(15–10)

or as

(15–11)

where

If the enthalpy of combustion h–°C for a particular reaction is available, the

steady-flow energy equation per mole of fuel can be expressed as

(15–12)

The energy balance relations above are sometimes written without the workterm since most steady-flow combustion processes do not involve any workinteractions.

Q � W � h°C �a Np 1h � h° 2 p�a Nr 1h � h° 2 r 1kJ>kmol 2

Hreact �a Nr 1h°f � h � h° 2 r 1kJ>kmol fuel 2 Hprod �a Np 1h°f � h � h° 2 p 1kJ>kmol fuel 2

Q � W � Hprod � Hreact 1kJ>kmol fuel 2

Q � W �a Np 1h°f � h � h° 2 p �a Nr 1h°f � h � h° 2 r

Q in � Win �a Nr 1h°f � h � h° 2 r � Q out � Wout �a Np 1h°f � h � h° 2 p

Q#in � W

#in �a n

#r 1h°f � h � h° 2 r � Q

#out � W

#out �a n

#p 1h°f � h � h° 2 p

Enthalpy � h°f � 1h � h° 2 1kJ>kmol 2

766 | Thermodynamics

Enthalpy atEnthalpy at2525°C, 1 atmC, 1 atm

SensibleSensibleenthalpy relative enthalpy relative

to 25to 25°C, 1 atmC, 1 atm

H H = = N (hf + h + h – h h )° °

FIGURE 15–21The enthalpy of a chemical componentat a specified state is the sum of theenthalpy of the component at 25�C, 1atm (hf°), and the sensible enthalpy ofthe component relative to 25�C, 1 atm.

1555555552555555553 15555555525555555553

1555555552555555553 15555555525555555553

cen84959_ch15.qxd 4/20/05 3:23 PM Page 766

A combustion chamber normally involves heat output but no heat input.Then the energy balance for a typical steady-flow combustion processbecomes

(15–13)

Energy in by mass Energy out by massper mole of fuel per mole of fuel

It expresses that the heat output during a combustion process is simply thedifference between the energy of the reactants entering and the energy ofthe products leaving the combustion chamber.

Closed SystemsThe general closed-system energy balance relation Ein � Eout � �Esystem canbe expressed for a stationary chemically reacting closed system as

(15–14)

where Uprod represents the internal energy of the products and Ureact repre-sents the internal energy of the reactants. To avoid using another property—the internal energy of formation u–f°—we utilize the definition of enthalpy (u– � h

–� Pv– or u–f° � u– � u–° � h

–°f � h

–� h

–° � Pv) and express the above

equation as (Fig. 15–22)

(15–15)

where we have taken heat transfer to the system and work done by the sys-tem to be positive quantities. The Pv– terms are negligible for solids and liq-uids, and can be replaced by RuT for gases that behave as an ideal gas. Also,if desired, the terms in Eq. 15–15 can be replaced by u–.

The work term in Eq. 15–15 represents all forms of work, including theboundary work. It was shown in Chap. 4 that �U � Wb � �H for nonreact-ing closed systems undergoing a quasi-equilibrium P � constant expansion orcompression process. This is also the case for chemically reacting systems.

There are several important considerations in the analysis of reacting sys-tems. For example, we need to know whether the fuel is a solid, a liquid, ora gas since the enthalpy of formation hf° of a fuel depends on the phase ofthe fuel. We also need to know the state of the fuel when it enters the com-bustion chamber in order to determine its enthalpy. For entropy calculationsit is especially important to know if the fuel and air enter the combustionchamber premixed or separately. When the combustion products are cooledto low temperatures, we need to consider the possibility of condensation ofsome of the water vapor in the product gases.

EXAMPLE 15–6 First-Law Analysis of Steady-Flow Combustion

Liquid propane (C3H8) enters a combustion chamber at 25°C at a rate of0.05 kg/min where it is mixed and burned with 50 percent excess air thatenters the combustion chamber at 7°C, as shown in Fig. 15–23. An analysisof the combustion gases reveals that all the hydrogen in the fuel burnsto H2O but only 90 percent of the carbon burns to CO2, with the remaining10 percent forming CO. If the exit temperature of the combustion gases is

h � Pv�

Q � W �a Np 1h°f � h � h° � Pv� 2 p �a Nr 1h°f � h � h° � Pv� 2 r

1Qin � Qout 2 � 1Win � Wout 2 � Uprod � Ureact 1kJ>kmol fuel 2

Qout �a Nr 1h°f � h � h° 2 r �a Np 1h°f � h � h° 2 p

Chapter 15 | 767

15555255553 15555255553

U = H – PV= N(hf + h – h°) – PV°

°= N(hf + h – h° – Pv )

– – –

– – – –

FIGURE 15–22An expression for the internal energyof a chemical component in terms ofthe enthalpy.

C3H8 (�)

CO2

Q = ?

1500 K

AIR

7°C

25°C, 0.05 kg/minCO

N2

Combustionchamber

H2O

O2

·

FIGURE 15–23Schematic for Example 15–6.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 767

1500 K, determine (a) the mass flow rate of air and (b) the rate of heattransfer from the combustion chamber.

Solution Liquid propane is burned steadily with excess air. The mass flowrate of air and the rate of heat transfer are to be determined.Assumptions 1 Steady operating conditions exist. 2 Air and the combustiongases are ideal gases. 3 Kinetic and potential energies are negligible.Analysis We note that all the hydrogen in the fuel burns to H2O but 10percent of the carbon burns incompletely and forms CO. Also, the fuel isburned with excess air and thus there is some free O2 in the product gases.

The theoretical amount of air is determined from the stoichiometric reac-tion to be

O2 balance:

Then the balanced equation for the actual combustion process with50 percent excess air and some CO in the products becomes

(a) The air–fuel ratio for this combustion process is

Thus,

(b) The heat transfer for this steady-flow combustion process is determinedfrom the steady-flow energy balance Eout � Ein applied on the combustionchamber per unit mole of the fuel,

or

Assuming the air and the combustion products to be ideal gases, we haveh � h(T ), and we form the following minitable using data from the propertytables:

h–

f° h–

280 K h–

298 K h–

1500 K

Substance kJ/kmol kJ/kmol kJ/kmol kJ/kmol

C3H8(�) �118,910 — — —O2 0 8150 8682 49,292N2 0 8141 8669 47,073H2O(g) �241,820 — 9904 57,999CO2 �393,520 — 9364 71,078CO �110,530 — 8669 47,517

Qout �a Nr 1h°f � h � h° 2 r �a Np 1h°f � h � h° 2 p

Qout �a Np 1h°f � h � h° 2 p �a Nr 1h°f � h � h° 2 r

� 1.18 kg air/min

� 123.53 kg air>kg fuel 2 10.05 kg fuel>min 2m#

air � 1AF 2 1m# fuel 2

� 25.53 kg air>kg fuel

AF �m air

m fuel�

17.5 � 4.76 kmol 2 129 kg>kmol 213 kmol 2 112 kg>kmol 2 � 14 kmol 2 12 kg>kmol 2

C3H8 1� 2 � 7.5 1O2 � 3.76N2 2 S 2.7CO2 � 0.3CO � 4H2O � 2.65O2 � 28.2N2

ath � 3 � 2 � 5

C3H8 1� 2 � ath 1O2 � 3.76N2 2 S 3CO2 � 4H2O � 3.76athN2

768 | Thermodynamics

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Chapter 15 | 769

1 lbmol CH4

CO2

1800 R1 atm77°F

P2

BEFOREREACTION

H2O

O23 lbmol O2

AFTERREACTION

FIGURE 15–24Schematic for Example 15–7.

The h–

f° of liquid propane is obtained by subtracting the h–

fg of propane at25°C from the h

–f° of gas propane. Substituting gives

Thus 363,880 kJ of heat is transferred from the combustion chamber foreach kmol (44 kg) of propane. This corresponds to 363,880/44 � 8270 kJof heat loss per kilogram of propane. Then the rate of heat transfer for amass flow rate of 0.05 kg/min for the propane becomes

EXAMPLE 15–7 First-Law Analysis of Combustion in a Bomb

The constant-volume tank shown in Fig. 15–24 contains 1 lbmol of methane(CH4) gas and 3 lbmol of O2 at 77°F and 1 atm. The contents of the tankare ignited, and the methane gas burns completely. If the final temperatureis 1800 R, determine (a) the final pressure in the tank and (b) the heattransfer during this process.

Solution Methane is burned in a rigid tank. The final pressure in the tankand the heat transfer are to be determined.Assumptions 1 The fuel is burned completely and thus all the carbon in thefuel burns to CO2 and all the hydrogen to H2O. 2 The fuel, the air, and thecombustion gases are ideal gases. 3 Kinetic and potential energies are negli-gible. 4 There are no work interactions involved.Analysis The balanced combustion equation is

(a) At 1800 R, water exists in the gas phase. Using the ideal-gas relation forboth the reactants and the products, the final pressure in the tank is deter-mined to be

Substituting, we get

Pprod � 11 atm 2 a 4 lbmol

4 lbmolb a 1800 R

537 Rb � 3.35 atm

PreactV � NreactRuTreact

PprodV � NprodRuTprodf Pprod � Preact a Nprod

Nreactb a Tprod

Treactb

CH4 1g 2 � 3O2 S CO2 � 2H2O � O2

Q#

out � m#qout � 10.05 kg>min 2 18270 kJ>kg 2 � 413.5 kJ>min � 6.89 kW

� 363,880 kJ>kmol of C3H8

� 128.2 kmol N2 2 3 10 � 47,073 � 8669 2 kJ>kmol N2 4 � 12.65 kmol O2 2 3 10 � 49,292 � 8682 2 kJ>kmol O2 4 � 14 kmol H2O 2 3 1�241,820 � 57,999 � 9904 2 kJ>kmol H2O 4 � 10.3 kmol CO 2 3 1�110,530 � 47,517 � 8669 2 kJ>kmol CO 4 � 12.7 kmol CO2 2 3 1�393,520 � 71,078 � 9364 2 kJ>kmol CO2 4 � 128.2 kmol N2 2 3 10 � 8141 � 8669 2 kJ>kmol N2 4 � 17.5 kmol O2 2 3 10 � 8150 � 8682 2 kJ>kmol O2 4

Q out � 11 kmol C3H8 2 3 1�118,910 � h 298 � h 298 2 kJ>kmol C3H8 4

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770 | Thermodynamics

Insulation

Tmax

Combustionchamber

Air

Products

Fuel

FIGURE 15–25The temperature of a combustionchamber becomes maximum whencombustion is complete and no heat is lost to the surroundings (Q � 0).

(b) Noting that the process involves no work interactions, the heat transferduring this constant-volume combustion process can be determined from theenergy balance Ein � Eout � �Esystem applied to the tank,

Since both the reactants and the products are assumed to be ideal gases, allthe internal energy and enthalpies depend on temperature only, and the Pv–

terms in this equation can be replaced by RuT. It yields

since the reactants are at the standard reference temperature of 537 R.From h

–f° and ideal-gas tables in the Appendix,

h–

f° h–

537 R h–

1800 R

Substance Btu/lbmol Btu/lbmol Btu/lbmol

CH4 �32,210 — —O2 0 3725.1 13,485.8CO2 �169,300 4027.5 18,391.5H2O(g) �104,040 4258.0 15,433.0

Substituting, we have

Discussion On a mass basis, the heat transfer from the tank would be308,730/16 � 19,300 Btu/lbm of methane.

15–5 � ADIABATIC FLAME TEMPERATUREIn the absence of any work interactions and any changes in kinetic or poten-tial energies, the chemical energy released during a combustion processeither is lost as heat to the surroundings or is used internally to raise thetemperature of the combustion products. The smaller the heat loss, thelarger the temperature rise. In the limiting case of no heat loss to the sur-roundings (Q � 0), the temperature of the products reaches a maximum,which is called the adiabatic flame or adiabatic combustion temperatureof the reaction (Fig. 15–25).

� 308,730 Btu/lbmol CH4

� 11 lbmol O2 2 3 10 � 13,485.8 � 3725.1 � 1.986 � 1800 2 Btu>lbmol O2 4 Btu>lbmol H2O 4 � 12 lbmol H2O 2 3 1�104,040 � 15,433.0 � 4258.0 � 1.986 � 1800 2 Btu>lbmol CO2 4 � 11 lbmol CO2 2 3 1�169,300 � 18,391.5 � 4027.5 � 1.986 � 1800 2 � 13 lbmol O2 2 3 10 � 1.986 � 537 2 Btu>lbmol O2 4

Q out � 11 lbmol CH4 2 3 1�32,210 � 1.986 � 537 2 Btu>lbmol CH4 4

Qout �a Nr 1h°f � Ru T 2 r �a Np 1h°f � h 1800 R � h 537 R � RuT 2 p

�Qout �a Np 1h°f � h � h° � Pv� 2 p �a Nr 1h°f � h � h° � Pv� 2 r

cen84959_ch15.qxd 4/27/05 10:55 AM Page 770

The adiabatic flame temperature of a steady-flow combustion process isdetermined from Eq. 15–11 by setting Q � 0 and W � 0. It yields

(15–16)

or

(15–17)

Once the reactants and their states are specified, the enthalpy of the reactantsHreact can be easily determined. The calculation of the enthalpy of the productsHprod is not so straightforward, however, because the temperature of the prod-ucts is not known prior to the calculations. Therefore, the determination of theadiabatic flame temperature requires the use of an iterative technique unlessequations for the sensible enthalpy changes of the combustion products areavailable. A temperature is assumed for the product gases, and the Hprod isdetermined for this temperature. If it is not equal to Hreact, calculations arerepeated with another temperature. The adiabatic flame temperature is thendetermined from these two results by interpolation. When the oxidant is air,the product gases mostly consist of N2, and a good first guess for the adiabaticflame temperature is obtained by treating the entire product gases as N2.

In combustion chambers, the highest temperature to which a materialcan be exposed is limited by metallurgical considerations. Therefore, the adi-abatic flame temperature is an important consideration in the design of com-bustion chambers, gas turbines, and nozzles. The maximum temperaturesthat occur in these devices are considerably lower than the adiabatic flametemperature, however, since the combustion is usually incomplete, some heatloss takes place, and some combustion gases dissociate at high temperatures(Fig. 15–26). The maximum temperature in a combustion chamber can becontrolled by adjusting the amount of excess air, which serves as a coolant.

Note that the adiabatic flame temperature of a fuel is not unique. Its valuedepends on (1) the state of the reactants, (2) the degree of completion of thereaction, and (3) the amount of air used. For a specified fuel at a specifiedstate burned with air at a specified state, the adiabatic flame temperatureattains its maximum value when complete combustion occurs with the theo-retical amount of air.

EXAMPLE 15–8 Adiabatic Flame Temperature in Steady Combustion

Liquid octane (C8H18) enters the combustion chamber of a gas turbinesteadily at 1 atm and 25°C, and it is burned with air that enters the com-bustion chamber at the same state, as shown in Fig. 15–27. Determine theadiabatic flame temperature for (a) complete combustion with 100 percenttheoretical air, (b) complete combustion with 400 percent theoretical air,and (c) incomplete combustion (some CO in the products) with 90 percenttheoretical air.

Solution Liquid octane is burned steadily. The adiabatic flame temperatureis to be determined for different cases.

a Np 1h°f � h � h° 2 p �a Nr 1h°f � h � h° 2 r

Hprod � Hreact

Chapter 15 | 771

Heat loss

• Incomplete combustion

Air

Products

Fuel

• DissociationTprod < Tmax

FIGURE 15–26The maximum temperatureencountered in a combustion chamberis lower than the theoretical adiabaticflame temperature.

Combustionchamber

Air

C8H18

25°C, 1 atm

25°C, 1 atmCO2

N2

H2O

O2

TP

1 atm

FIGURE 15–27Schematic for Example 15–8.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 771

Assumptions 1 This is a steady-flow combustion process. 2 The combustionchamber is adiabatic. 3 There are no work interactions. 4 Air and the com-bustion gases are ideal gases. 5 Changes in kinetic and potential energiesare negligible.Analysis (a) The balanced equation for the combustion process with thetheoretical amount of air is

The adiabatic flame temperature relation Hprod � Hreact in this case reduces to

since all the reactants are at the standard reference state and h–

f° � 0 for O2and N2. The h

–f° and h values of various components at 298 K are

h–

f° h–

298 K

Substance kJ/kmol kJ/kmol

C8H18(�) �249,950 —O2 0 8682N2 0 8669H2O(g) �241,820 9904CO2 �393,520 9364

Substituting, we have

which yields

It appears that we have one equation with three unknowns. Actually we haveonly one unknown—the temperature of the products Tprod—since h � h(T )for ideal gases. Therefore, we have to use an equation solver such as EES ora trial-and-error approach to determine the temperature of the products.

A first guess is obtained by dividing the right-hand side of the equation bythe total number of moles, which yields 5,646,081/(8 + 9 + 47) � 88,220kJ/kmol. This enthalpy value corresponds to about 2650 K for N2, 2100 K forH2O, and 1800 K for CO2. Noting that the majority of the moles are N2, wesee that Tprod should be close to 2650 K, but somewhat under it. Therefore,a good first guess is 2400 K. At this temperature,

This value is higher than 5,646,081 kJ. Therefore, the actual temperature isslightly under 2400 K. Next we choose 2350 K. It yields

8 � 122,091 � 9 � 100,846 � 47 � 77,496 � 5,526,654

� 5,660,828 kJ

8hCO2� 9hH2O � 47hN2

� 8 � 125,152 � 9 � 103,508 � 47 � 79,320

8hCO2� 9hH2O � 47hN2

� 5,646,081 kJ

� 11 kmol C8H18 2 1�249,950 kJ>kmol C8H18 2 � 147 kmol N2 2 3 10 � h N2

� 8669 2 kJ>kmol N2 4 � 19 kmol H2O 2 3 1�241,820 � h H2O � 9904 2 kJ>kmol H2O 4 18 kmol CO2 2 3 1�393,520 � h CO2

� 9364 2 kJ>kmol CO2 4

a Np 1h°f � h � h° 2 p �a Nr h°f,r � 1Nh°f 2C8H18

C8H18 1� 2 � 12.5 1O2 � 3.76N2 2 S 8CO2 � 9H2O � 47N2

772 | Thermodynamics

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Chapter 15 | 773

which is lower than 5,646,081 kJ. Therefore, the actual temperature of theproducts is between 2350 and 2400 K. By interpolation, it is found to beTprod � 2395 K.

(b) The balanced equation for the complete combustion process with 400percent theoretical air is

By following the procedure used in (a), the adiabatic flame temperature inthis case is determined to be Tprod � 962 K.

Notice that the temperature of the products decreases significantly as aresult of using excess air.

(c) The balanced equation for the incomplete combustion process with90 percent theoretical air is

Following the procedure used in (a), we find the adiabatic flame temperaturein this case to be Tprod � 2236 K.Discussion Notice that the adiabatic flame temperature decreases as aresult of incomplete combustion or using excess air. Also, the maximum adi-abatic flame temperature is achieved when complete combustion occurs withthe theoretical amount of air.

15–6 � ENTROPY CHANGE OF REACTING SYSTEMSSo far we have analyzed combustion processes from the conservation ofmass and the conservation of energy points of view. The thermodynamicanalysis of a process is not complete, however, without the examination ofthe second-law aspects. Of particular interest are the exergy and exergydestruction, both of which are related to entropy.

The entropy balance relations developed in Chap. 7 are equally applicableto both reacting and nonreacting systems provided that the entropies of indi-vidual constituents are evaluated properly using a common basis. Theentropy balance for any system (including reacting systems) undergoingany process can be expressed as

(15–18)

Net entropy transfer Entropy Changeby heat and mass generation in entropy

Using quantities per unit mole of fuel and taking the positive direction ofheat transfer to be to the system, the entropy balance relation can beexpressed more explicitly for a closed or steady-flow reacting system as(Fig. 15–28)

(15–19)

where Tk is temperature at the boundary where Qk crosses it. For an adia-batic process (Q � 0), the entropy transfer term drops out and Eq. 15–19reduces to

(15–20)Sgen,adiabatic � Sprod � Sreact � 0

aQk

Tk

� Sgen � Sprod � Sreact 1kJ>K 2

Sin � Sout � Sgen � ¢Ssystem 1kJ>K 2

C8H18 1� 2 � 11.25 1O2 � 3.76N2 2 S 5.5CO2 � 2.5CO � 9H2O � 42.3N2

C8H18 1� 2 � 50 1O2 � 3.76N2 2 S 8CO2 � 9H2O � 37.5O2 � 188N2

15253 123 123

Reactionchamber

ProductsSprod

ReactantsSreact

Surroundings

∆Ssys

FIGURE 15–28The entropy change associated with achemical relation.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 773

The total entropy generated during a process can be determined by apply-ing the entropy balance to an extended system that includes the system itselfand its immediate surroundings where external irreversibilities might beoccurring. When evaluating the entropy transfer between an extended sys-tem and the surroundings, the boundary temperature of the extended systemis simply taken to be the environment temperature, as explained in Chap. 7.

The determination of the entropy change associated with a chemical reac-tion seems to be straightforward, except for one thing: The entropy relationsfor the reactants and the products involve the entropies of the components,not entropy changes, which was the case for nonreacting systems. Thus weare faced with the problem of finding a common base for the entropy of allsubstances, as we did with enthalpy. The search for such a common base ledto the establishment of the third law of thermodynamics in the early partof this century. The third law was expressed in Chap. 7 as follows: Theentropy of a pure crystalline substance at absolute zero temperature is zero.

Therefore, the third law of thermodynamics provides an absolute base forthe entropy values for all substances. Entropy values relative to this base arecalled the absolute entropy. The s–° values listed in Tables A–18 throughA–25 for various gases such as N2, O2, CO, CO2, H2, H2O, OH, and O arethe ideal-gas absolute entropy values at the specified temperature and at apressure of 1 atm. The absolute entropy values for various fuels are listed inTable A–26 together with the h

–°f values at the standard reference state of

25°C and 1 atm.Equation 15–20 is a general relation for the entropy change of a reacting

system. It requires the determination of the entropy of each individual com-ponent of the reactants and the products, which in general is not very easyto do. The entropy calculations can be simplified somewhat if the gaseouscomponents of the reactants and the products are approximated as idealgases. However, entropy calculations are never as easy as enthalpy or inter-nal energy calculations, since entropy is a function of both temperature andpressure even for ideal gases.

When evaluating the entropy of a component of an ideal-gas mixture, weshould use the temperature and the partial pressure of the component. Notethat the temperature of a component is the same as the temperature of themixture, and the partial pressure of a component is equal to the mixturepressure multiplied by the mole fraction of the component.

Absolute entropy values at pressures other than P0 � 1 atm for any tem-perature T can be obtained from the ideal-gas entropy change relation writ-ten for an imaginary isothermal process between states (T,P0) and (T,P), asillustrated in Fig. 15–29:

(15–21)

For the component i of an ideal-gas mixture, this relation can be written as

(15–22)

where P0 � 1 atm, Pi is the partial pressure, yi is the mole fraction of thecomponent, and Pm is the total pressure of the mixture.

s�i 1T,Pi 2 � s�°i 1T,P0 2 � Ru ln yi Pm

P01kJ>kmol # K 2

s� 1T,P 2 � s�° 1T,P0 2 � Ru ln P

P0

774 | Thermodynamics

∆s = – Ru lnP0

T

s

P

T

P 0 = 1 atm

P

s°(T,P0)

(Tabulated)

s(T,P)

FIGURE 15–29At a specified temperature, theabsolute entropy of an ideal gas atpressures other than P0 � 1 atm can be determined by subtracting Ru ln (P/P0) from the tabulated value at 1 atm.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 774

If a gas mixture is at a relatively high pressure or low temperature, thedeviation from the ideal-gas behavior should be accounted for by incorpo-rating more accurate equations of state or the generalized entropy charts.

15–7 � SECOND-LAW ANALYSIS OF REACTING SYSTEMS

Once the total entropy change or the entropy generation is evaluated, theexergy destroyed Xdestroyed associated with a chemical reaction can be deter-mined from

(15–23)

where T0 is the thermodynamic temperature of the surroundings.When analyzing reacting systems, we are more concerned with the

changes in the exergy of reacting systems than with the values of exergy atvarious states (Fig. 15–30). Recall from Chap. 8 that the reversible workWrev represents the maximum work that can be done during a process. In theabsence of any changes in kinetic and potential energies, the reversible workrelation for a steady-flow combustion process that involves heat transferwith only the surroundings at T0 can be obtained by replacing the enthalpyterms by h

–°f � h

–� h

–°, yielding

(15–24)

An interesting situation arises when both the reactants and the products areat the temperature of the surroundings T0. In that case, h

–� T0s– � (h

–� T0s–)T0

� g–0, which is, by definition, the Gibbs function of a unit mole of a sub-stance at temperature T0. The Wrev relation in this case can be written as

(15–25)

or

(15–26)

where g–°f is the Gibbs function of formation (g–°f � 0 for stable elements likeN2 and O2 at the standard reference state of 25°C and 1 atm, just like theenthalpy of formation) and g–T0

� g–° represents the value of the sensibleGibbs function of a substance at temperature T0 relative to the standardreference state.

For the very special case of Treact � Tprod � T0 � 25°C (i.e., the reactants,the products, and the surroundings are at 25°C) and the partial pressurePi � 1 atm for each component of the reactants and the products, Eq. 15–26reduces to

(15–27)

We can conclude from the above equation that the �g–°f value (the negativeof the Gibbs function of formation at 25°C and 1 atm) of a compoundrepresents the reversible work associated with the formation of that com-pound from its stable elements at 25°C and 1 atm in an environment at25°C and 1 atm (Fig. 15–31). The g–°f values of several substances are listedin Table A–26.

Wrev �a Nrg�°f,r �a npg�°f,p 1kJ 2

Wrev �a Nr 1g�°f � g�T0� g�° 2 r �a Np 1g�°f � g�T0

� g�° 2 p

Wrev �a Nrg�0,r �a Npg�0,p

Wrev �a Nr 1h°f � h � h° � T0 s� 2 r �a Np 1h°f � h � h° � T0 s� 2 p

Xdestroyed � T0 Sgen 1kJ 2

Chapter 15 | 775

T, P State

Reacta

nts

Reversiblework

Products

Exergy

FIGURE 15–30The difference between the exergy of the reactants and of the productsduring a chemical reaction is thereversible work associated with thatreaction.

2525°C,C,1 atm1 atm

C + OC + O2 → CO CO2

Wrevrev = = – gf, , COCO2 2 = 394,360 k= 394,360 kJ/J/kmolkmol–

StableStable

T0 = 25 = 25°C

elementselements CompoundCompound

2525°C,C,1 atm1 atm

2525°C,C,1 atm1 atm

°

FIGURE 15–31The negative of the Gibbs function offormation of a compound at 25�C, 1atm represents the reversible workassociated with the formation of thatcompound from its stable elements at25�C, 1 atm in an environment that isat 25�C, 1 atm.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 775

EXAMPLE 15–9 Reversible Work Associated with a Combustion Process

One lbmol of carbon at 77°F and 1 atm is burned steadily with 1 lbmol ofoxygen at the same state as shown in Fig. 15–32. The CO2 formed duringthe process is then brought to 77°F and 1 atm, the conditions of the sur-roundings. Assuming the combustion is complete, determine the reversiblework for this process.

Solution Carbon is burned steadily with pure oxygen. The reversible workassociated with this process is to be determined.Assumptions 1 Combustion is complete. 2 Steady-flow conditions exist duringcombustion. 3 Oxygen and the combustion gases are ideal gases. 4 Changesin kinetic and potential energies are negligible.Properties The Gibbs function of formation at 77°F and 1 atm is 0 for Cand O2, and �169,680 Btu/lbmol for CO2. The enthalpy of formation is 0 forC and O2, and �169,300 Btu/lbmol for CO2. The absolute entropy is 1.36Btu/lbmol · R for C, 49.00 Btu/lbmol · R for O2, and 51.07 Btu/lbmol · R forCO2 (Table A–26E).Analysis The combustion equation is

The C, O2, and CO2 are at 77°F and 1 atm, which is the standard referencestate and also the state of the surroundings. Therefore, the reversible work inthis case is simply the difference between the Gibbs function of formation ofthe reactants and that of the products (Eq. 15–27):

since the g–f° of stable elements at 77°F and 1 atm is zero. Therefore,169,680 Btu of work could be done as 1 lbmol of C is burned with 1 lbmolof O2 at 77°F and 1 atm in an environment at the same state. The reversiblework in this case represents the exergy of the reactants since the product(the CO2) is at the state of the surroundings.Discussion We could also determine the reversible work without involvingthe Gibbs function by using Eq. 15–24:

Substituting the enthalpy of formation and absolute entropy values, we obtain

which is identical to the result obtained before.

� 169,680 Btu

� 11 lbmol CO22 3�169,300 Btu>lbmol � 1537 R 2 151.07 Btu>lbmol # R 2 4 � 11 lbmol O2 2 30 � 1537 R 2 149.00 Btu>lbmol # R 2 4

Wrev � 11 lbmol C 2 30 � 1537 R 2 11.36 Btu>lbmol # R 2 4

� NC 1h°f � T0 s�° 2C � NO21h°f � T0 s�° 2O2

� NCO21h°f � T0 s�° 2CO2

�a Nr 1h°f � T0 s� 2 r �a Np 1h°f � T0 s� 2 p Wrev �a Nr 1h°f � h � h° � T0 s� 2 r �a Np 1h°f � h � h° � T0 s� 2 p

� 169,680 Btu

� 1�1 lbmol 2 1�169,680 Btu>lbmol 2 � NCg�°f,C � NO2

g�°f,O2 � NCO2

g�°f,CO2� �NCO2

g�°f,CO2

Wrev �a Nrg�°f,r �a Npg�°f,p

C � O2 S CO2

776 | Thermodynamics

Combustionchamber

C

77°F, 1 atm

CO2

T0 = 77°F

O2

77°F, 1 atm

77°F, 1 atm

P0 = 1 atm

FIGURE 15–32Schematic for Example 15–9.

0¡

0¡

cen84959_ch15.qxd 4/20/05 3:23 PM Page 776

Chapter 15 | 777

Adiabaticcombustion

chamber25°C, 1 atm

CO2

T0 = 25°C

AIR

CH4

25°C, 1 atmH2O

O2N2

FIGURE 15–33Schematic for Example 15–10.

EXAMPLE 15–10 Second-Law Analysis of Adiabatic Combustion

Methane (CH4) gas enters a steady-flow adiabatic combustion chamber at25°C and 1 atm. It is burned with 50 percent excess air that also enters at25°C and 1 atm, as shown in Fig. 15–33. Assuming complete combustion,determine (a) the temperature of the products, (b) the entropy generation,and (c) the reversible work and exergy destruction. Assume that T0 � 298 Kand the products leave the combustion chamber at 1 atm pressure.

Solution Methane is burned with excess air in a steady-flow combustionchamber. The product temperature, entropy generated, reversible work, andexergy destroyed are to be determined.Assumptions 1 Steady-flow conditions exist during combustion. 2 Air andthe combustion gases are ideal gases. 3 Changes in kinetic and potentialenergies are negligible. 4 The combustion chamber is adiabatic and thusthere is no heat transfer. 5 Combustion is complete.Analysis (a) The balanced equation for the complete combustion processwith 50 percent excess air is

Under steady-flow conditions, the adiabatic flame temperature is determinedfrom Hprod � Hreact, which reduces to

since all the reactants are at the standard reference state and h–

f° � O for O2

and N2. Assuming ideal-gas behavior for air and for the products, the h–

f° andh values of various components at 298 K can be listed as

h–

f° h–

298 K

Substance kJ/kmol kJ/kmol

CH4(g) �74,850 —O2 0 8682N2 0 8669H2O(g) �241,820 9904CO2 �393,520 9364

Substituting, we have

which yields

By trial and error, the temperature of the products is found to be

Tprod � 1789 K

h CO2� 2hH2O � h O2

� 11.28h N2� 937,950 kJ

� 11 kmol CH4 2 1�74,850 kJ>kmol CH4 2 � 11 kmol O2 2 3 10 � h O2

� 8682 2 kJ>kmol O2 4 � 111.28 kmol N2 2 3 10 � h N2

� 8669 2 kJ>kmol N2 4 � 12 kmol H2O 2 3 1�241,820 � h H2O � 9904 2 kJ>kmol H2O 4 11 kmol CO2 2 3 1�393,520 � h CO2

� 9364 2 kJ>kmol CO2 4

a Np 1h°f � h � h° 2 p �a Nrh°f,r � 1Nh°f 2CH4

CH4 1g 2 � 3 1O2 � 3.76N2 2 S CO2 � 2H2O � O2 � 11.28N2

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778 | Thermodynamics

Combustionchamber

25°C, 1 atm

CO2

T0 = 25°C

AIR

CH4

25°C, 1 atm H2O

O2

N2

25°C,

1 atm

FIGURE 15–34Schematic for Example 15–11.

(b) Noting that combustion is adiabatic, the entropy generation during thisprocess is determined from Eq. 15–20:

The CH4 is at 25°C and 1 atm, and thus its absolute entropy is s–CH4�

186.16 kJ/kmol � K (Table A–26). The entropy values listed in the ideal-gastables are for 1 atm pressure. Both the air and the product gases are at atotal pressure of 1 atm, but the entropies are to be calculated at the partialpressure of the components, which is equal to Pi � yiPtotal, where yi is themole fraction of component i. From Eq. 15–22:

The entropy calculations can be represented in tabular form as follows:

Ni yi s–°i (T, 1 atm) �Ru ln yiPm Nis–

i

CH4 1 1.00 186.16 — 186.16O2 3 0.21 205.04 12.98 654.06N2 11.28 0.79 191.61 1.96 2183.47

Sreact � 3023.69

CO2 1 0.0654 302.517 22.674 325.19H2O 2 0.1309 258.957 16.905 551.72O2 1 0.0654 264.471 22.674 287.15N2 11.28 0.7382 247.977 2.524 2825.65

Sprod � 3989.71

Thus,

(c) The exergy destruction or irreversibility associated with this process isdetermined from Eq. 15–23,

That is, 288 MJ of work potential is wasted during this combustion processfor each kmol of methane burned. This example shows that even completecombustion processes are highly irreversible.

This process involves no actual work. Therefore, the reversible work andexergy destroyed are identical:

That is, 288 MJ of work could be done during this process but is not.Instead, the entire work potential is wasted.

EXAMPLE 15–11 Second-Law Analysis of Isothermal Combustion

Methane (CH4) gas enters a steady-flow combustion chamber at 25°C and1 atm and is burned with 50 percent excess air, which also enters at 25°Cand 1 atm, as shown in Fig. 15–34. After combustion, the productsare allowed to cool to 25°C. Assuming complete combustion, determine

Wrev � 288 MJ/kmol CH4

� 288 MJ/kmol CH4

X destroyed � T0 Sgen � 1298 K 2 1966.0 kJ>kmol # K 2

� 966.0 kJ/kmol # K

Sgen � Sprod � Sreact � 13989.71 � 3023.69 2 kJ>kmol # K CH4

Si � Ni s�i 1T, Pi 2 � Ni 3 s�°i 1T, P0 2 � Ru ln yiPm 4

Sgen � Sprod � Sreact �a Np s�p �a Nr s�r

cen84959_ch15.qxd 4/20/05 3:23 PM Page 778

Chapter 15 | 779

(a) the heat transfer per kmol of CH4, (b) the entropy generation, and (c) thereversible work and exergy destruction. Assume that T0 � 298 K and theproducts leave the combustion chamber at 1 atm pressure.

Solution This is the same combustion process we discussed in Example15–10, except that the combustion products are brought to the state of thesurroundings by transferring heat from them. Thus the combustion equationremains the same:

At 25°C, part of the water will condense. The amount of water vapor thatremains in the products is determined from (see Example 15–3)

and

Therefore, 1.57 kmol of the H2O formed is in the liquid form, which isremoved at 25°C and 1 atm. When one is evaluating the partial pressures ofthe components in the product gases, the only water molecules that need tobe considered are those that are in the vapor phase. As before, all thegaseous reactants and products are treated as ideal gases.

(a) Heat transfer during this steady-flow combustion process is determinedfrom the steady-flow energy balance Eout � Ein on the combustion chamber,

since all the reactants and products are at the standard reference of 25°Cand the enthalpy of ideal gases depends on temperature only. Solving forQout and substituting the h

–f° values, we have

(b) The entropy of the reactants was evaluated in Example 15–10 and wasdetermined to be Sreact � 3023.69 kJ/kmol · K CH4. By following a similarapproach, the entropy of the products is determined to be

Ni yi s–°i (T, 1 atm) �Ru ln yiPm Nis–

i

H2O(�) 1.57 1.0000 69.92 — 109.77H2O 0.43 0.0314 188.83 28.77 93.57CO2 1 0.0729 213.80 21.77 235.57O2 1 0.0729 205.04 21.77 226.81N2 11.28 0.8228 191.61 1.62 2179.63

Sprod � 2845.35

� 871,400 kJ/kmol CH4

� 31.57 kmol H2O 1� 2 4 3�285.830 kJ>kmol H2O 1� 2 4 � 30.43 kmol H2O 1g 2 4 3�241,820 kJ>kmol H2O 1g 2 4 � 11 kmol CO2 2 1�393,520 kJ>kmol CO2 2

Q out � 11 kmol CH4 2 1�74,850 kJ>kmol CH4 2

Q out �a Nph�°f,p �a Nr h�°f,r

Nv � a Pv

Ptotal b Ngas � 10.03128 2 113.28 � Nv 2 S Nv � 0.43 kmol

Nv

Ngas�

Pv

Ptotal�

3.1698 kPa

101.325 kPa� 0.03128

CH4 1g 2 � 3 1O2 � 3.76N2 2 S CO2 � 2H2O � O2 � 11.28N2

cen84959_ch15.qxd 4/20/05 3:23 PM Page 779

780 | Thermodynamics

Then the total entropy generation during this process is determined from anentropy balance applied on an extended system that includes the immediatesurroundings of the combustion chamber

(c) The exergy destruction and reversible work associated with this processare determined from

and

since this process involves no actual work. Therefore, 818 MJ of work couldbe done during this process but is not. Instead, the entire work potential iswasted. The reversible work in this case represents the exergy of the reac-tants before the reaction starts since the products are in equilibrium withthe surroundings, that is, they are at the dead state.Discussion Note that, for simplicity, we calculated the entropy of the prod-uct gases before they actually entered the atmosphere and mixed with theatmospheric gases. A more complete analysis would consider the composi-tion of the atmosphere and the mixing of the product gases with the gases inthe atmosphere, forming a homogeneous mixture. There is additional entropygeneration during this mixing process, and thus additional wasted workpotential.

Wrev � X destroyed � 818 MJ/kmol CH4

� 818 MJ/kmol CH4

X destroyed � T0 Sgen � 1298 K 2 12746 kJ>kmol # K 2

� 2746 kJ/kmol # K CH4

� 12845.35 � 3023.69 2 kJ>kmol �871,400 kJ>kmol

298 K

Sgen � Sprod � Sreact �Q out

Tsurr

Fuels like methane are commonly burned to provide thermal energy at hightemperatures for use in heat engines. However, a comparison of thereversible works obtained in the last two examples reveals that the exergy ofthe reactants (818 MJ/kmol CH4) decreases by 288 MJ/kmol as a result ofthe irreversible adiabatic combustion process alone. That is, the exergy of thehot combustion gases at the end of the adiabatic combustion process is 818� 288 � 530 MJ/kmol CH4. In other words, the work potential of the hotcombustion gases is about 65 percent of the work potential of the reactants.It seems that when methane is burned, 35 percent of the work potential islost before we even start using the thermal energy (Fig. 15–35).

Thus, the second law of thermodynamics suggests that there should be abetter way of converting the chemical energy to work. The better way is, ofcourse, the less irreversible way, the best being the reversible case. In chemi-

TOPIC OF SPECIAL INTEREST* Fuel Cells

*This section can be skipped without a loss in continuity.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 780

Chapter 15 | 781

cal reactions, the irreversibility is due to uncontrolled electron exchangebetween the reacting components. The electron exchange can be controlledby replacing the combustion chamber by electrolytic cells, like car batteries.(This is analogous to replacing unrestrained expansion of a gas in mechani-cal systems by restrained expansion.) In the electrolytic cells, the electronsare exchanged through conductor wires connected to a load, and the chemi-cal energy is directly converted to electric energy. The energy conversiondevices that work on this principle are called fuel cells. Fuel cells are notheat engines, and thus their efficiencies are not limited by the Carnot effi-ciency. They convert chemical energy to electric energy essentially in anisothermal manner.

A fuel cell functions like a battery, except that it produces its own electric-ity by combining a fuel with oxygen in a cell electrochemically withoutcombustion, and discards the waste heat. A fuel cell consists of two elec-trodes separated by an electrolyte such as a solid oxide, phosphoric acid, ormolten carbonate. The electric power generated by a single fuel cell is usu-ally too small to be of any practical use. Therefore, fuel cells are usuallystacked in practical applications. This modularity gives the fuel cells consid-erable flexibility in applications: The same design can be used to generate asmall amount of power for a remote switching station or a large amount ofpower to supply electricity to an entire town. Therefore, fuel cells are termedthe “microchip of the energy industry.”

The operation of a hydrogen–oxygen fuel cell is illustrated in Fig. 15–36.Hydrogen is ionized at the surface of the anode, and hydrogen ions flowthrough the electrolyte to the cathode. There is a potential differencebetween the anode and the cathode, and free electrons flow from the anodeto the cathode through an external circuit (such as a motor or a generator).Hydrogen ions combine with oxygen and the free electrons at the surface ofthe cathode, forming water. Therefore, the fuel cell operates like an electrol-ysis system working in reverse. In steady operation, hydrogen and oxygencontinuously enter the fuel cell as reactants, and water leaves as the product.Therefore, the exhaust of the fuel cell is drinkable quality water.

The fuel cell was invented by William Groves in 1839, but it did notreceive serious attention until the 1960s, when they were used to produceelectricity and water for the Gemini and Apollo spacecraft during their mis-sions to the moon. Today they are used for the same purpose in the spaceshuttle missions. Despite the irreversible effects such as internal resistance toelectron flow, fuel cells have a great potential for much higher conversionefficiencies. Currently fuel cells are available commercially, but they arecompetitive only in some niche markets because of their higher cost. Fuelcells produce high-quality electric power efficiently and quietly while gener-ating low emissions using a variety of fuels such as hydrogen, natural gas,propane, and biogas. Recently many fuel cells have been installed to gener-ate electricity. For example, a remote police station in Central Park in NewYork City is powered by a 200-kW phosphoric acid fuel cell that has anefficiency of 40 percent with negligible emissions (it emits 1 ppm NOx and5 ppm CO).

Adiabatic

25°CREACTANTS

(CH4, air)Exergy = 818 MJ

(100%)

1789 KPRODUCTS

Exergy = 530 MJ(65%)

combustionchamber

FIGURE 15–35The availability of methane decreasesby 35 percent as a result of irreversiblecombustion process.

2e–

O2

Porousanode

Load

O2

H2

H22H+

Porouscathode

H2O

2e– 2e–

FIGURE 15–36The operation of a hydrogen–oxygenfuel cell.

cen84959_ch15.qxd 4/20/05 3:23 PM Page 781

Hybrid power systems (HPS) that combine high-temperature fuel cells andgas turbines have the potential for very high efficiency in converting naturalgas (or even coal) to electricity. Also, some car manufacturers are planningto introduce cars powered by fuel-cell engines, thus more than doubling theefficiency from less than 30 percent for the gasoline engines to up to 60 per-cent for fuel cells. In 1999, DaimlerChrysler unveiled its hydrogen fuel-cellpowered car called NECAR IV that has a refueling range of 280 miles andcan carry 4 passengers at 90 mph. Some research programs to develop suchhybrid systems with an efficiency of at least 70 percent by 2010 are underway.

782 | Thermodynamics

Any material that can be burned to release energy is called afuel, and a chemical reaction during which a fuel is oxidizedand a large quantity of energy is released is called combus-tion. The oxidizer most often used in combustion processes isair. The dry air can be approximated as 21 percent oxygenand 79 percent nitrogen by mole numbers. Therefore,

During a combustion process, the components that existbefore the reaction are called reactants and the componentsthat exist after the reaction are called products. Chemicalequations are balanced on the basis of the conservation ofmass principle, which states that the total mass of each ele-ment is conserved during a chemical reaction. The ratio ofthe mass of air to the mass of fuel during a combustionprocess is called the air–fuel ratio AF:

where mair � (NM)air and mfuel � �(Ni Mi)fuel.A combustion process is complete if all the carbon in the

fuel burns to CO2, all the hydrogen burns to H2O, and allthe sulfur (if any) burns to SO2. The minimum amount of airneeded for the complete combustion of a fuel is calledthe stoichiometric or theoretical air. The theoretical air isalso referred to as the chemically correct amount of air or100 percent theoretical air. The ideal combustion process dur-ing which a fuel is burned completely with theoretical air iscalled the stoichiometric or theoretical combustion of thatfuel. The air in excess of the stoichiometric amount is calledthe excess air. The amount of excess air is usually expressedin terms of the stoichiometric air as percent excess air or per-cent theoretical air.

During a chemical reaction, some chemical bonds are bro-ken and others are formed. Therefore, a process that involveschemical reactions involves changes in chemical energies.Because of the changed composition, it is necessary to have a

AF �m air

m fuel

1 kmol O2 � 3.76 kmol N2 � 4.76 kmol air

standard reference state for all substances, which is chosen tobe 25°C (77°F) and 1 atm.

The difference between the enthalpy of the products at aspecified state and the enthalpy of the reactants at thesame state for a complete reaction is called the enthalpy ofreaction hR. For combustion processes, the enthalpy of reac-tion is usually referred to as the enthalpy of combustion hC,which represents the amount of heat released during a steady-flow combustion process when 1 kmol (or 1 kg) of fuel isburned completely at a specified temperature and pressure.The enthalpy of a substance at a specified state due to itschemical composition is called the enthalpy of formation .The enthalpy of formation of all stable elements is assigned avalue of zero at the standard reference state of 25°C and 1atm. The heating value of a fuel is defined as the amount ofheat released when a fuel is burned completely in a steady-flow process and the products are returned to the state of thereactants. The heating value of a fuel is equal to the absolutevalue of the enthalpy of combustion of the fuel,

Taking heat transfer to the system and work done by thesystem to be positive quantities, the conservation of energyrelation for chemically reacting steady-flow systems can beexpressed per unit mole of fuel as

where the superscript ° represents properties at the standard ref-erence state of 25°C and 1 atm. For a closed system, it becomes

The terms are negligible for solids and liquids and can bereplaced by RuT for gases that behave as ideal gases.

Pv��a Nr 1h°f � h � h° � Pv�2 r

Q � W �a Np 1h°f � h � h° � Pv�2 p

Q � W �a Np 1h°f � h � h° 2 p �a Nr 1h°f � h � h° 2 r

Heating value � 0hC 0 1kJ>kg fuel 2

hf

SUMMARY

cen84959_ch15.qxd 4/20/05 3:23 PM Page 782

Chapter 15 | 783

In the absence of any heat loss to the surroundings (Q � 0),the temperature of the products will reach a maximum, whichis called the adiabatic flame temperature of the reaction.The adiabatic flame temperature of a steady-flow combus-tion process is determined from Hprod � Hreact or

Taking the positive direction of heat transfer to be to thesystem, the entropy balance relation can be expressed for aclosed system or steady-flow combustion chamber as

For an adiabatic process it reduces to

The third law of thermodynamics states that the entropy ofa pure crystalline substance at absolute zero temperature iszero. The third law provides a common base for the entropyof all substances, and the entropy values relative to this baseare called the absolute entropy. The ideal-gas tables list theabsolute entropy values over a wide range of temperatures butat a fixed pressure of P0 � 1 atm. Absolute entropy values atother pressures P for any temperature T are determined from

Sgen,adiabatic � Sprod � Sreact � 0

a Qk

Tk

� Sgen � Sprod � Sreact

a Np 1h°f � h � h° 2 p �a Nr 1h°f � h � h° 2 r

For component i of an ideal-gas mixture, this relation can bewritten as

where Pi is the partial pressure, yi is the mole fraction of thecomponent, and Pm is the total pressure of the mixture inatmospheres.

The exergy destruction and the reversible work associatedwith a chemical reaction are determined from

and

When both the reactants and the products are at the tempera-ture of the surroundings T0, the reversible work can beexpressed in terms of the Gibbs functions as

Wrev �a Nr 1g�°f � g�T0� g�° 2 r �a Np 1g�°f � g�T0

� g�° 2 p

Wrev �a Nr 1h°f � h � h° � T0 s� 2 r �a Np 1h°f � h � h° � T0 s� 2 p

X destroyed � Wrev � Wact � T0 Sgen

s�i 1T, Pi 2 � s�°i 1T, P0 2 � Ru ln yi Pm

P0

s� 1T, P 2 � s�° 1T, P0 2 � Ru ln P

P0

1. S. W. Angrist. Direct Energy Conversion. 4th ed. Boston:Allyn and Bacon, 1982.

2. W. Z. Black and J. G. Hartley. Thermodynamics. NewYork: Harper & Row, 1985.

3. I. Glassman. Combustion. New York: Academic Press,1977.

4. R. Strehlow. Fundamentals of Combustion. Scranton,PA: International Textbook Co., 1968.

5. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

REFERENCES AND SUGGESTED READINGS

PROBLEMS*

Fuels and Combustion

15–1C What are the approximate chemical compositions ofgasoline, diesel fuel, and natural gas?

15–2C How does the presence of N2 in air affect the out-come of a combustion process?

15–3C How does the presence of moisture in air affect theoutcome of a combustion process?

15–4C What does the dew-point temperature of the productgases represent? How is it determined?

15–5C Is the number of atoms of each element conservedduring a chemical reaction? How about the total number ofmoles?

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

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15–6C What is the air–fuel ratio? How is it related to thefuel–air ratio?

15–7C Is the air–fuel ratio expressed on a mole basis iden-tical to the air–fuel ratio expressed on a mass basis?

Theoretical and Actual Combustion Processes

15–8C What are the causes of incomplete combustion?

15–9C Which is more likely to be found in the products ofan incomplete combustion of a hydrocarbon fuel, CO or OH?Why?

15–10C What does 100 percent theoretical air represent?

15–11C Are complete combustion and theoretical combus-tion identical? If not, how do they differ?

15–12C Consider a fuel that is burned with (a) 130 percenttheoretical air and (b) 70 percent excess air. In which case isthe fuel burned with more air?

15–13 Methane (CH4) is burned with stoichiometricamount of air during a combustion process. Assuming com-plete combustion, determine the air–fuel and fuel–air ratios.

15–14 Propane (C3H8) is burned with 75 percent excess airduring a combustion process. Assuming complete combus-tion, determine the air–fuel ratio. Answer: 27.5 kg air/kg fuel

15–15 Acetylene (C2H2) is burned with stoichiometricamount of air during a combustion process. Assuming com-plete combustion, determine the air–fuel ratio on a mass andon a mole basis.

15–16 One kmol of ethane (C2H6) is burned with anunknown amount of air during a combustion process. Ananalysis of the combustion products reveals that the combus-tion is complete, and there are 3 kmol of free O2 in the prod-ucts. Determine (a) the air–fuel ratio and (b) the percentageof theoretical air used during this process.

15–17E Ethylene (C2H4) is burned with 200 percent theo-retical air during a combustion process. Assuming completecombustion and a total pressure of 14.5 psia, determine(a) the air–fuel ratio and (b) the dew-point temperature ofthe products. Answers: (a) 29.6 lbm air/lbm fuel, (b) 101°F

15–18 Propylene (C3H6) is burned with 50 percent excessair during a combustion process. Assuming complete com-bustion and a total pressure of 105 kPa, determine (a) theair–fuel ratio and (b) the temperature at which the watervapor in the products will start condensing.

15–19 Propal alcohol (C3H7OH) is burned with 50 percentexcess air. Write the balanced reaction equation for completecombustion and determine the air-to-fuel ratio.Answer: 15.5 kg air/kg fuel

15–20 Butane (C4H10) is burned in 200 percent theoreticalair. For complete combustion, how many kmol of water mustbe sprayed into the combustion chamber per kmol of fuel if

the products of combustion are to have a dew-point tempera-ture of 60°C when the product pressure is 100 kPa?

15–21 A fuel mixture of 20 percent by mass methane (CH4)and 80 percent by mass ethanol (C2H6O), is burned completelywith theoretical air. If the total flow rate of the fuel is 31 kg/s,determine the required flow rate of air. Answer: 330 kg/s

15–22 Octane (C8H18) is burned with 250 percent theoreti-cal air, which enters the combustion chamber at 25°C.Assuming complete combustion and a total pressure of 1 atm,determine (a) the air–fuel ratio and (b) the dew-point temper-ature of the products.

Combustionchamber

AIR

C8H18

25°CP = 1 atm

Products

FIGURE P15–22

15–23 Gasoline (assumed C8H18) is burned steadily with airin a jet engine. If the air–fuel ratio is 18 kg air/kg fuel, deter-mine the percentage of theoretical air used during thisprocess.

15–24 In a combustion chamber, ethane (C2H6) is burned ata rate of 8 kg/h with air that enters the combustion chamberat a rate of 176 kg/h. Determine the percentage of excess airused during this process. Answer: 37 percent

15–25 One kilogram of butane (C4H10) is burned with25 kg of air that is at 30°C and 90 kPa. Assuming that thecombustion is complete and the pressure of the products is90 kPa, determine (a) the percentage of theoretical air usedand (b) the dew-point temperature of the products.

15–26E One lbm of butane (C4H10) is burned with 25 lbmof air that is at 90°F and 14.7 psia. Assuming that thecombustion is complete and the pressure of the products is14.7 psia, determine (a) the percentage of theoretical airused and (b) the dew-point temperature of the products.Answers: (a) 161 percent, (b) 113°F

15–27 A certain natural gas has the following volumetricanalysis: 65 percent CH4, 8 percent H2, 18 percent N2, 3 per-cent O2, and 6 percent CO2. This gas is now burned com-pletely with the stoichiometric amount of dry air. What is theair–fuel ratio for this combustion process?

15–28 Repeat Prob. 15–27 by replacing the dry air by moistair that enters the combustion chamber at 25°C, 1 atm, and85 percent relative humidity.

15–29 A gaseous fuel with a volumetric analysis of 60 per-cent CH4, 30 percent H2, and 10 percent N2 is burned to com-pletion with 130 percent theoretical air. Determine (a) the

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air–fuel ratio and (b) the fraction of water vapor that wouldcondense if the product gases were cooled to 20°C at 1 atm.Answers: (a) 18.6 kg air/kg fuel, (b) 88 percent

15–30 Reconsider Prob. 15–29. Using EES (or other)software, study the effects of varying the per-

centages of CH4, H2, and N2 making up the fuel and theproduct gas temperature in the range 5 to 150°C.

15–31 A certain coal has the following analysis on a massbasis: 82 percent C, 5 percent H2O, 2 percent H2, 1 percentO2, and 10 percent ash. The coal is burned with 50 percentexcess air. Determine the air–fuel ratio. Answer: 15.1 kgair/kg coal

15–32 Octane (C8H18) is burned with dry air. The volumet-ric analysis of the products on a dry basis is 9.21 percentCO2, 0.61 percent CO, 7.06 percent O2, and 83.12 percentN2. Determine (a) the air–fuel ratio and (b) the percentage oftheoretical air used.

15–33 Carbon (C) is burned with dry air. The volumetricanalysis of the products is 10.06 percent CO2, 0.42 percentCO, 10.69 percent O2, and 78.83 percent N2. Determine (a)the air–fuel ratio and (b) the percentage of theoretical airused.

15–34 Methane (CH4) is burned with dry air. The volumetricanalysis of the products on a dry basis is 5.20 percent CO2,0.33 percent CO, 11.24 percent O2, and 83.23 percent N2.Determine (a) the air–fuel ratio and (b) the percentage of theo-retical air used. Answers: (a) 34.5 kg air/kg fuel, (b) 200 percent

Enthalpy of Formation and Enthalpy of Combustion

15–35C What is enthalpy of combustion? How does it dif-fer from the enthalpy of reaction?

15–36C What is enthalpy of formation? How does it differfrom the enthalpy of combustion?

15–37C What are the higher and the lower heating valuesof a fuel? How do they differ? How is the heating value of afuel related to the enthalpy of combustion of that fuel?

15–38C When are the enthalpy of formation and theenthalpy of combustion identical?

15–39C Does the enthalpy of formation of a substancechange with temperature?

15–40C The of N2 is listed as zero. Does this mean thatN2 contains no chemical energy at the standard referencestate?

15–41C Which contains more chemical energy, 1 kmol ofH2 or 1 kmol of H2O?

15–42 Determine the enthalpy of combustion of methane(CH4) at 25°C and 1 atm, using the enthalpy of formationdata from Table A–26. Assume that the water in the productsis in the liquid form. Compare your result to the value listedin Table A–27. Answer: –890,330 kJ/kmol

h°f

15–43 Reconsider Prob. 15–42. Using EES (or other)software, study the effect of temperature on the

enthalpy of combustion. Plot the enthalpy of combustion as afunction of temperature over the range 25 to 600°C.

15–44 Repeat Prob. 15–42 for gaseous ethane (C2H6).

15–45 Repeat Prob. 15–42 for liquid octane (C8H18).

First-Law Analysis of Reacting Systems

15–46C Derive an energy balance relation for a reactingclosed system undergoing a quasi-equilibrium constant pres-sure expansion or compression process.

15–47C Consider a complete combustion process duringwhich both the reactants and the products are maintained atthe same state. Combustion is achieved with (a) 100 percenttheoretical air, (b) 200 percent theoretical air, and (c) thechemically correct amount of pure oxygen. For which casewill the amount of heat transfer be the highest? Explain.

15–48C Consider a complete combustion process duringwhich the reactants enter the combustion chamber at 20°Cand the products leave at 700°C. Combustion is achievedwith (a) 100 percent theoretical air, (b) 200 percent theoreti-cal air, and (c) the chemically correct amount of pure oxygen.For which case will the amount of heat transfer be the low-est? Explain.

15–49 Methane (CH4) is burned completely with the stoi-chiometric amount of air during a steady-flow combustionprocess. If both the reactants and the products are maintainedat 25°C and 1 atm and the water in the products exists in theliquid form, determine the heat transfer from the combustionchamber during this process. What would your answer be ifcombustion were achieved with 100 percent excess air?Answer: 890,330 kJ/kmol

15–50 Hydrogen (H2) is burned completely with the stoi-chiometric amount of air during a steady-flow combustionprocess. If both the reactants and the products are maintainedat 25°C and 1 atm and the water in the products exists in theliquid form, determine the heat transfer from the combustionchamber during this process. What would your answer be ifcombustion were achieved with 50 percent excess air?

15–51 Liquid propane (C3H8) enters a combustion chamberat 25°C at a rate of 1.2 kg/min where it is mixed and burnedwith 150 percent excess air that enters the combustionchamber at 12°C. If the combustion is complete and the exit

Combustionchamber

AIR

C3H8

12°C

Products25°C

1200 K

Qout·

FIGURE P15–51

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temperature of the combustion gases is 1200 K, determine(a) the mass flow rate of air and (b) the rate of heat transferfrom the combustion chamber. Answers: (a) 47.1 kg/min,(b) 5194 kJ/min

15–52E Liquid propane (C3H8) enters a combustion cham-ber at 77°F at a rate of 0.75 lbm/min where it is mixed andburned with 150 percent excess air that enters the combustionchamber at 40°F. if the combustion is complete and the exittemperature of the combustion gases is 1800 R, determine(a) the mass flow rate of air and (b) the rate of heat transferfrom the combustion chamber. Answers: (a) 29.4 lbm/min,(b) 4479 Btu/min

15–53 Acetylene gas (C2H2) is burned completely with20 percent excess air during a steady-flow combustionprocess. The fuel and air enter the combustion chamber at25°C, and the products leave at 1500 K. Determine (a) theair–fuel ratio and (b) the heat transfer for this process.

15–54E Liquid octane (C8H18) at 77°F is burned com-pletely during a steady-flow combustion process with 180percent theoretical air that enters the combustion chamber at77°F. If the products leave at 2500 R, determine (a) theair–fuel ratio and (b) the heat transfer from the combustionchamber during this process.

15–55 Benzene gas (C6H6) at 25°C is burned during asteady-flow combustion process with 95 percent theoreticalair that enters the combustion chamber at 25°C. All thehydrogen in the fuel burns to H2O, but part of the carbonburns to CO. If the products leave at 1000 K, determine(a) the mole fraction of the CO in the products and (b) theheat transfer from the combustion chamber during thisprocess. Answers: (a) 2.1 percent, (b) 2,112,800 kJ/kmol C6H6

15–56 Diesel fuel (C12H26) at 25°C is burned in a steady-flow combustion chamber with 20 percent excess air that alsoenters at 25°C. The products leave the combustion chamberat 500 K. Assuming combustion is complete, determine therequired mass flow rate of the diesel fuel to supply heat at arate of 2000 kJ/s. Answer: 49.5 g/s

15–57E Diesel fuel (C12H26) at 77°F is burned in a steady-flow combustion chamber with 20 percent excess air that alsoenters at 77°F. The products leave the combustion chamber at800 R. Assuming combustion is complete, determine therequired mass flow rate of the diesel fuel to supply heat at arate of 1800 Btu/s. Answer: 0.1 lbm/s

15–58 Octane gas (C8H18) at 25°C is burned steadilywith 30 percent excess air at 25°C, 1 atm, and

60 percent relative humidity. Assuming combustion iscomplete and the products leave the combustion chamber at600 K, determine the heat transfer for this process per unitmass of octane.

15–59 Reconsider Prob. 15–58. Using EES (or other)software, investigate the effect of the amount

of excess air on the heat transfer for the combustion process.Let the excess air vary from 0 to 200 percent. Plot the heattransfer against excess air, and discuss the results.

15–60 Ethane gas (C2H6) at 25°C is burned in a steady-flowcombustion chamber at a rate of 5 kg/h with the stoichiomet-ric amount of air, which is preheated to 500 K before enter-ing the combustion chamber. An analysis of the combustiongases reveals that all the hydrogen in the fuel burns to H2Obut only 95 percent of the carbon burns to CO2, the remain-ing 5 percent forming CO. If the products leave the combus-tion chamber at 800 K, determine the rate of heat transferfrom the combustion chamber. Answer: 200,170 kJ/h

CombustionchamberAIR

C2H6

500 K

Qout

25°C

800 K

H2OCO2COO2

N2

·

FIGURE P15–60

15–61 A constant-volume tank contains a mixture of120 g of methane (CH4) gas and 600 g of O2 at

25°C and 200 kPa. The contents of the tank are now ignited,and the methane gas burns completely. If the final tempera-ture is 1200 K, determine (a) the final pressure in the tankand (b) the heat transfer during this process.

15–62 Reconsider Prob. 15–61. Using EES (or other)software, investigate the effect of the final tem-

perature on the final pressure and the heat transfer for thecombustion process. Let the final temperature vary from 500to 1500 K. Plot the final pressure and heat transfer againstthe final temperature, and discuss the results.

15–63 A closed combustion chamber is designed so that itmaintains a constant pressure of 300 kPa during a combus-tion process. The combustion chamber has an initial volumeof 0.5 m3 and contains a stoichiometric mixture of octane(C8H18) gas and air at 25°C. The mixture is now ignited, andthe product gases are observed to be at 1000 K at the end ofthe combustion process. Assuming complete combustion, andtreating both the reactants and the products as ideal gases,determine the heat transfer from the combustion chamberduring this process. Answer: 3610 kJ

15–64 A constant-volume tank contains a mixture of1 kmol of benzene (C6H6) gas and 30 percent excess air at25°C and 1 atm. The contents of the tank are now ignited,and all the hydrogen in the fuel burns to H2O but only 92percent of the carbon burns to CO2, the remaining 8 percentforming CO. If the final temperature in the tank is 1000 K,determine the heat transfer from the combustion chamberduring this process.

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15–65E A constant-volume tank contains a mixture of 1 lb-mol of benzene (C6H6) gas and 30 percent excess air at 77°Fand 1 atm. The contents of the tank are now ignited, and allthe hydrogen in the fuel burns to H2O but only 92 percent ofthe carbon burns to CO2, the remaining 8 percent formingCO. If the final temperature in the tank is 1800 R, determinethe heat transfer from the combustion chamber during thisprocess. Answer: 946,870 Btu

15–66 To supply heated air to a house, a high-efficiencygas furnace burns gaseous propane (C3H8) with a combustionefficiency of 96 percent. Both the fuel and 140 percent theo-retical air are supplied to the combustion chamber at 25°Cand 100 kPa, and the combustion is complete. Because this isa high-efficiency furnace, the product gases are cooled to25°C and 100 kPa before leaving the furnace. To maintainthe house at the desired temperature, a heat transfer rate of31,650 kJ/h is required from the furnace. Determine thevolume of water condensed from the product gases per day.Answer: 8.7 L/day

15–67 Liquid ethyl alcohol (C2H5OH(�)) at 25°C is burnedin a steady-flow combustion chamber with 40 percent excessair that also enters at 25°C. The products leave the combus-tion chamber at 600 K. Assuming combustion is complete,determine the required volume flow rate of the liquid ethylalcohol, to supply heat at a rate of 2000 kJ/s. At 25°C thedensity of liquid ethyl alcohol is 790 kg/m3, the specific heatat a constant pressure is 114.08 kJ/kmol � K, and the enthalpyof vaporization is 42,340 kJ/kmol. Answer: 6.81 L/min

Adiabatic Flame Temperature

15–68C A fuel is completely burned first with the stoichio-metric amount of air and then with the stoichiometric amountof pure oxygen. For which case will the adiabatic flame tem-perature be higher?

15–69C A fuel at 25°C is burned in a well-insulatedsteady-flow combustion chamber with air that is also at 25°C.Under what conditions will the adiabatic flame temperatureof the combustion process be a maximum?

15–70 Hydrogen (H2) at 7°C is burned with 20 per-cent excess air that is also at 7°C during an

adiabatic steady-flow combustion process. Assuming com-plete combustion, determine the exit temperature of the prod-uct gases. Answer: 2251.4 K

15–71 Reconsider Prob. 15–70. Using EES (or other)software, modify this problem to include the

fuels butane, ethane, methane, and propane as well as H2; toinclude the effects of inlet air and fuel temperatures; and thepercent theoretical air supplied. Select a range of input param-eters and discuss the results for your choices.

15–72E Hydrogen (H2) at 40°F is burned with 20 percentexcess air that is also at 40°F during an adiabatic steady-flowcombustion process. Assuming complete combustion, find theexit temperature of the product gases.

15–73 Acetylene gas (C2H2) at 25°C is burned during asteady-flow combustion process with 30 percent excess air at27°C. It is observed that 75,000 kJ of heat is being lost fromthe combustion chamber to the surroundings per kmol ofacetylene. Assuming combustion is complete, determine theexit temperature of the product gases. Answer: 2301 K

15–74 An adiabatic constant-volume tank contains a mix-ture of 1 kmol of hydrogen (H2) gas and the stoichiometricamount of air at 25°C and 1 atm. The contents of the tank arenow ignited. Assuming complete combustion, determine thefinal temperature in the tank.

15–75 Octane gas (C8H18) at 25°C is burned steadily with30 percent excess air at 25°C, 1 atm, and 60 percent relativehumidity. Assuming combustion is complete and adiabatic,calculate the exit temperature of the product gases.

15–76 Reconsider Prob. 15–75. Using EES (or other)software, investigate the effect of the relative

humidity on the exit temperature of the product gases. Plotthe exit temperature of the product gases as a function of rel-ative humidity for 0 � f � 100 percent.

Entropy Change and Second-Law Analysis of Reacting Systems

15–77C Express the increase of entropy principle for chem-ically reacting systems.

15–78C How are the absolute entropy values of ideal gasesat pressures different from 1 atm determined?

15–79C What does the Gibbs function of formation of acompound represent?

15–80 One kmol of H2 at 25°C and 1 atm is burned steadilywith 0.5 kmol of O2 at the same state. The H2O formed duringthe process is then brought to 25°C and 1 atm, the conditions

g°f

C6H630% excess air

25°C1 atm

Qout

FIGURE P15–64

CombustionchamberAIR

H2

7°C Products

Tprod

7°C

FIGURE P15–70

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of the surroundings. Assuming combustion is complete, deter-mine the reversible work and exergy destruction for thisprocess.

15–81 Ethylene (C2H4) gas enters an adiabatic combustionchamber at 25°C and 1 atm and is burned with 20 percentexcess air that enters at 25°C and 1 atm. The combustion iscomplete, and the products leave the combustion chamber at1 atm pressure. Assuming T0 � 25°C, determine (a) the tem-perature of the products, (b) the entropy generation, and(c) the exergy destruction. Answers: (a) 2269.6 K, (b) 1311.3kJ/kmol · K, (c) 390,760 kJ/kmol

15–82 Liquid octane (C8H18) enters a steady-flow combus-tion chamber at 25°C and 1 atm at a rate of 0.25 kg/min. It isburned with 50 percent excess air that also enters at 25°Cand 1 atm. After combustion, the products are allowed tocool to 25°C. Assuming complete combustion and that all theH2O in the products is in liquid form, determine (a) the heattransfer rate from the combustion chamber, (b) the entropygeneration rate, and (c) the exergy destruction rate. Assumethat T0 � 298 K and the products leave the combustionchamber at 1 atm pressure.

15–83 Acetylene gas (C2H2) is burned completely with20 percent excess air during a steady-flow combustionprocess. The fuel and the air enter the combustion chamberseparately at 25°C and 1 atm, and heat is being lost from thecombustion chamber to the surroundings at 25°C at a rate of300,000 kJ/kmol C2H2. The combustion products leave thecombustion chamber at 1 atm pressure. Determine (a) thetemperature of the products, (b) the total entropy change perkmol of C2H2, and (c) the exergy destruction during thisprocess.

15–84 A steady-flow combustion chamber is supplied withCO gas at 37°C and 110 kPa at a rate of 0.4 m3/min and airat 25°C and 110 kPa at a rate of 1.5 kg/min. Heat is trans-ferred to a medium at 800 K, and the combustion productsleave the combustion chamber at 900 K. Assuming the com-bustion is complete and T0 � 25°C, determine (a) the rate ofheat transfer from the combustion chamber and (b) the rate ofexergy destruction. Answers: (a) 3567 kJ/min, (b) 1610 kJ/min

15–85E Benzene gas (C6H6) at 1 atm and 77°F is burnedduring a steady-flow combustion process with 95 percent

theoretical air that enters the combustion chamber at 77°Fand 1 atm. All the hydrogen in the fuel burns to H2O, but partof the carbon burns to CO. Heat is lost to the surroundingsat 77°F, and the products leave the combustion chamber at1 atm and 1500 R. Determine (a) the heat transfer from thecombustion chamber and (b) the exergy destruction.

15–86 Liquid propane (C3H8) enters a steady-flowcombustion chamber at 25°C and 1 atm at a

rate of 0.4 kg/min where it is mixed and burned with 150percent excess air that enters the combustion chamber at12°C. If the combustion products leave at 1200 K and 1 atm,determine (a) the mass flow rate of air, (b) the rate of heattransfer from the combustion chamber, and (c) the rate ofentropy generation during this process. Assume T0 � 25°C.Answers: (a) 15.7 kg/min, (b) 1732 kJ/min, (c) 34.2 kJ/min · K

15–87 Reconsider Prob. 15–86. Using EES (or other)software, study the effect of varying the sur-

roundings temperature from 0 to 38°C on the rate of exergydestruction, and plot it as a function of surroundings tem-perature.

Review Problems

15–88 A 1-g sample of a certain fuel is burned in a bombcalorimeter that contains 2 kg of water in the presence of 100g of air in the reaction chamber. If the water temperaturerises by 2.5°C when equilibrium is established, determine theheating value of the fuel, in kJ/kg.

15–89E Hydrogen (H2) is burned with 100 percent excessair that enters the combustion chamber at 90°F, 14.5 psia, and60 percent relative humidity. Assuming complete combustion,determine (a) the air–fuel ratio and (b) the volume flow rateof air required to burn the hydrogen at a rate of 25 lbm/h.

15–90 A gaseous fuel with 80 percent CH4, 15 percent N2,and 5 percent O2 (on a mole basis) is burned to completionwith 120 percent theoretical air that enters the combustionchamber at 30°C, 100 kPa, and 60 percent relative humidity.Determine (a) the air–fuel ratio and (b) the volume flow rateof air required to burn fuel at a rate of 2 kg/min.

15–91 A gaseous fuel with 80 percent CH4, 15 percent N2,and 5 percent O2 (on a mole basis) is burned with dry air thatenters the combustion chamber at 25°C and 100 kPa. The vol-umetric analysis of the products on a dry basis is 3.36 percentCO2, 0.09 percent CO, 14.91 percent O2, and 81.64 percentN2. Determine (a) the air–fuel ratio, (b) the percent theoretical

Combustionchamber

AIR

C8H18(�)Qout

25°C Products

1 atm25°C

25°C

T0 = 298 K·

FIGURE P15–82

Combustionchamber

AIR

80% CH43.36% CO20.09% CO14.91% O2

81.64% N2

15% N2

5% O2

FIGURE P15–91

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air used, and (c) the volume flow rate of air used to burn fuelat a rate of 1.4 kg/min.

15–92 A steady-flow combustion chamber is supplied withCO gas at 37°C and 110 kPa at a rate of 0.4 m3/min and airat 25°C and 110 kPa at a rate of 1.5 kg/min. The combustionproducts leave the combustion chamber at 900 K. Assumingcombustion is complete, determine the rate of heat transferfrom the combustion chamber.

15–93 Methane gas (CH4) at 25°C is burned steadily withdry air that enters the combustion chamber at 17°C. The vol-umetric analysis of the products on a dry basis is 5.20 percentCO2, 0.33 percent CO, 11.24 percent O2, and 83.23 percentN2. Determine (a) the percentage of theoretical air used and(b) the heat transfer from the combustion chamber per kmolof CH4 if the combustion products leave at 700 K.

15–94 A 6-m3 rigid tank initially contains a mixture of1 kmol of hydrogen (H2) gas and the stoichiometric amountof air at 25°C. The contents of the tank are ignited, and allthe hydrogen in the fuel burns to H2O. If the combustionproducts are cooled to 25°C, determine (a) the fraction ofthe H2O that condenses and (b) the heat transfer from thecombustion chamber during this process.

15–95 Propane gas (C3H8) enters a steady-flow combustionchamber at 1 atm and 25°C and is burned with air that entersthe combustion chamber at the same state. Determine the adi-abatic flame temperature for (a) complete combustion with100 percent theoretical air, (b) complete combustion with 300percent theoretical air, and (c) incomplete combustion (someCO in the products) with 95 percent theoretical air.

15–96 Determine the highest possible temperature that canbe obtained when liquid gasoline (assumed C8H18) at 25°C isburned steadily with air at 25°C and 1 atm. What would youranswer be if pure oxygen at 25°C were used to burn the fuelinstead of air?

15–97E Determine the work potential of 1 lbmol of dieselfuel (C12H26) at 77°F and 1 atm in an environment at thesame state. Answer: 3,375,000 Btu

15–98 Liquid octane (C8H18) enters a steady-flow combus-tion chamber at 25°C and 8 atm at a rate of 0.8 kg/min. It isburned with 200 percent excess air that is compressed andpreheated to 500 K and 8 atm before entering the combustionchamber. After combustion, the products enter an adiabaticturbine at 1300 K and 8 atm and leave at 950 K and 2 atm.Assuming complete combustion and T0 � 25°C, determine(a) the heat transfer rate from the combustion chamber,(b) the power output of the turbine, and (c) the reversiblework and exergy destruction for the entire process. Answers:(a) 770 kJ/min, (b) 263 kW, (c) 514 kW, 251 kW

15–99 The combustion of a fuel usually results in anincrease in pressure when the volume is held constant, or anincrease in volume when the pressure is held constant,

because of the increase in the number of moles and the tem-perature. The increase in pressure or volume will be maxi-mum when the combustion is complete and when it occursadiabatically with the theoretical amount of air.

Consider the combustion of methyl alcohol vapor(CH3OH(g)) with the stoichiometric amount of air in an 0.8-Lcombustion chamber. Initially, the mixture is at 25°C and 98kPa. Determine (a) the maximum pressure that can occur inthe combustion chamber if the combustion takes place at con-stant volume and (b) the maximum volume of the combustionchamber if the combustion occurs at constant pressure.

15–100 Reconsider Prob. 15–99. Using EES (or other)software, investigate the effect of the initial

volume of the combustion chamber over the range 0.1 to 2.0liters on the results. Plot the maximum pressure of the cham-ber for constant volume combustion or the maximum volumeof the chamber for constant pressure combustion as functionsof the initial volume.

15–101 Repeat Prob. 15–99 using methane (CH4(g)) as thefuel instead of methyl alcohol.

15–102 A mixture of 40 percent by volume methane (CH4),and 60 percent by volume propane (C3H8), is burned com-pletely with theoretical air and leaves the combustion cham-ber at 100°C. The products have a pressure of 100 kPa andare cooled at constant pressure to 39°C. Sketch the T-s dia-gram for the water vapor that does not condense, if any. Howmuch of the water formed during the combustion process willbe condensed, in kmol H2O/kmol fuel? Answer: 1.96

15–103 Liquid propane (C3H8(�)) enters a combustion cham-ber at 25°C and 1 atm at a rate of 0.4 kg/min where it ismixed and burned with 150 percent excess air that enters thecombustion chamber at 25°C. The heat transfer from thecombustion process is 53 kW. Write the balanced combustionequation and determine (a) the mass flow rate of air; (b) theaverage molar mass (molecular weight) of the product gases;(c) the average specific heat at constant pressure of the prod-uct gases; and (d ) the temperature of the products of combus-tion. Answers: (a) 15.63 kg/min, (b) 28.63 kg/kmol, (c) 36.06kJ/kmol � K, (d ) 1282 K

15–104 A gaseous fuel mixture of 30 percent propane(C3H8), and 70 percent butane (C4H10), on a volume basis isburned in air such that the air–fuel ratio is 20 kg air/kg fuelwhen the combustion process is complete. Determine (a) themoles of nitrogen in the air supplied to the combustionprocess, in kmol/kmol fuel; (b) the moles of water formed inthe combustion process, in kmol/kmol fuel; and (c) the molesof oxygen in the product gases. Answers: (a) 29.41, (b) 4.7,(c) 1.77

15–105 A liquid–gas fuel mixture consists of 90 percentoctane (C8H18), and 10 percent alcohol (C2H5OH), by moles.This fuel is burned with 200 percent theoretical dry air. Writethe balanced reaction equation for complete combustion of

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790 | Thermodynamics

this fuel mixture. Determine (a) the theoretical air–fuel ratiofor this reaction; (b) the product–fuel ratio for this reaction;(c) the air-flow rate for a fuel mixture flow rate of 5 kg/s; and(d) the lower heating value of the fuel mixture with 200 per-cent theoretical air at 25°C. Answers: (a) 14.83 kg air/kg fuel,(b) 30.54 kg product/kg fuel, (c) 148.3 kg/s, (d) 43,672 kJ/kg fuel

15–106 The furnace of a particular power plant can be con-sidered to consist of two chambers: an adiabatic combustionchamber where the fuel is burned completely and adiabati-cally, and a heat exchanger where heat is transferred to aCarnot heat engine isothermally. The combustion gases in theheat exchanger are well-mixed so that the heat exchanger isat a uniform temperature at all times that is equal to the tem-perature of the exiting product gases, Tp. The work output ofthe Carnot heat engine can be expressed as

where Q is the magnitude of the heat transfer to the heatengine and T0 is the temperature of the environment. Thework output of the Carnot engine will be zero either whenTp � Taf (which means the product gases will enter and exitthe heat exchanger at the adiabatic flame temperature Taf, andthus Q � 0) or when Tp � T0 (which means the temperature

W � QhC � Q a1 �T0

Tp

b

of the product gases in the heat exchanger will be T0, andthus hC � 0), and will reach a maximum somewhere inbetween. Treating the combustion products as ideal gaseswith constant specific heats and assuming no change in theircomposition in the heat exchanger, show that the work outputof the Carnot heat engine will be maximum when

Also, show that the maximum work output of the Carnotengine in this case becomes

where C is a constant whose value depends on the composi-tion of the product gases and their specific heats.

15–107 The furnace of a particular power plant can be con-sidered to consist of two chambers: an adiabatic combustionchamber where the fuel is burned completely and adiabati-cally and a counterflow heat exchanger where heat is trans-ferred to a reversible heat engine. The mass flow rate of theworking fluid of the heat engine is such that the workingfluid is heated from T0 (the temperature of the environment)to Taf (the adiabatic flame temperature) while the combustionproducts are cooled from Taf to T0. Treating the combustionproducts as ideal gases with constant specific heats andassuming no change in their composition in the heatexchanger, show that the work output of this reversible heatengine is

W � CT0 a Taf

T0� 1 � ln

Taf

T0b

Wmax � CTaf a1 � BT0

Tafb 2

Tp � 2TafT0

FIGURE P15–106

HeatexchangerTp = const.

Q

Fuel

Air

Adiabaticcombustion

chamber

SurroundingsT0

W

T0

Tp

Adiabaticcombustion

chamber

W

T0

T0

Taf

T0Heat

exchanger

Surroundings

Fuel Air

T0

FIGURE P15–107

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Chapter 15 | 791

where C is a constant whose value depends on the composi-tion of the product gases and their specific heats.

Also, show that the effective flame temperature Te of thisfurnace is

That is, the work output of the reversible engine would be thesame if the furnace above is considered to be an isothermalfurnace at a constant temperature Te.

15–108 Using EES (or other) software, determine theeffect of the amount of air on the adiabatic

flame temperature of liquid octane (C8H18). Assume both theair and the octane are initially at 25°C. Determine the adia-batic flame temperature for 75, 90, 100, 120, 150, 200, 300,500, and 800 percent theoretical air. Assume the hydrogen inthe fuel always burns H2O and the carbon CO2, except whenthere is a deficiency of air. In the latter case, assume that partof the carbon forms CO. Plot the adiabatic flame temperatureagainst the percent theoretical air, and discuss the results.

15–109 Using EES (or other) software, write a gen-eral program to determine the heat transfer

during the complete combustion of a hydrocarbon fuel(CnHm) at 25°C in a steady-flow combustion chamber whenthe percent of excess air and the temperatures of air and theproducts are specified. As a sample case, determine the heattransfer per unit mass of fuel as liquid propane (C3H8) isburned steadily with 50 percent excess air at 25°C and the com-bustion products leave the combustion chamber at 1800 K.

15–110 Using EES (or other) software, write a gen-eral program to determine the adiabatic flame

temperature during the complete combustion of a hydrocar-bon fuel (CnHm) at 25°C in a steady-flow combustion cham-ber when the percent of excess air and its temperature arespecified. As a sample case, determine the adiabatic flametemperature of liquid propane (C3H8) as it is burned steadilywith 50 percent excess air at 25°C.

15–111 Using EES (or other) software, determine theadiabatic flame temperature of the fuels

CH4(g), C2H2(g), CH3OH(g), C3H8(g), C8H18(�). Assumeboth the fuel and the air enter the steady-flow combustionchamber at 25°C.

15–112 Using EES (or other) software, determine theminimum percent of excess air that needs to

be used for the fuels CH4(g), C2H2(g), CH3OH(g), C3H8(g),C8H18(�) if the adiabatic flame temperature is not to exceed1500 K. Assume both the fuel and the air enter the steady-flow combustion chamber at 25°C.

15–113 Using EES (or other) software, repeat Prob.15–112 for adiabatic flame temperatures of

(a) 1200 K, (b) 1750 K, and (c) 2000 K.

Te �Taf � T0

ln 1Taf>T0 2

15–114 Using EES (or other) software, determine theadiabatic flame temperature of CH4(g) when

both the fuel and the air enter the combustion chamber at25°C for the cases of 0, 20, 40, 60, 80, 100, 200, 500, and1000 percent excess air.

15–115 Using EES (or other) software, determine therate of heat transfer for the fuels CH4(g),

C2H2(g), CH3OH(g), C3H8(g), and C8H18(�) when they areburned completely in a steady-flow combustion chamberwith the theoretical amount of air. Assume the reactants enterthe combustion chamber at 298 K and the products leave at1200 K.

15–116 Using EES (or other) software, repeat Prob.15–115 for (a) 50, (b) 100, and (c) 200 per-

cent excess air.

15–117 Using EES (or other) software, determine thefuel among CH4(g), C2H2(g), C2H6(g),

C3H8(g), C8H18(�) that gives the highest temperature whenburned completely in an adiabatic constant-volume chamberwith the theoretical amount of air. Assume the reactants are atthe standard reference state.

Fundamentals of Engineering (FE) Exam Problems

15–118 A fuel is burned with 90 percent theoretical air.This is equivalent to

(a) 10% excess air (b) 90% excess air(c) 10% deficiency of air (d) 90% deficiency of air(e) stoichiometric amount of air

15–119 Propane (C3H8) is burned with 150 percent theoret-ical air. The air–fuel mass ratio for this combustion process is

(a) 5.3 (b) 10.5 (c) 15.7(d) 23.4 (e) 39.3

15–120 One kmol of methane (CH4) is burned with anunknown amount of air during a combustion process. If thecombustion is complete and there are 2 kmol of free O2 in theproducts, the air–fuel mass ratio is

(a) 34.3 (b) 17.2 (c) 19.0(d) 14.9 (e) 12.1

15–121 A fuel is burned steadily in a combustion chamber.The combustion temperature will be the highest except when(a) the fuel is preheated.(b) the fuel is burned with a deficiency of air.(c) the air is dry.(d) the combustion chamber is well insulated.(e) the combustion is complete.

15–122 An equimolar mixture of carbon dioxide and watervapor at 1 atm and 60°C enter a dehumidifying section wherethe entire water vapor is condensed and removed from themixture, and the carbon dioxide leaves at 1 atm and 60°C.

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792 | Thermodynamics

The entropy change of carbon dioxide in the dehumidifyingsection is

(a) �2.8 kJ/kg · K (b) �0.13 kJ/kg · K(c) 0 (d) 0.13 kJ/kg · K(e) 2.8 kJ/kg · K

15–123 Methane (CH4) is burned completely with 80 per-cent excess air during a steady-flow combustion process. Ifboth the reactants and the products are maintained at 25°Cand 1 atm and the water in the products exists in the liquidform, the heat transfer from the combustion chamber per unitmass of methane is

(a) 890 MJ/kg (b) 802 MJ/kg (c) 75 MJ/kg(d) 56 MJ/kg (e) 50 MJ/kg

15–124 The higher heating value of a hydrocarbon fuelCnHm with m � 8 is given to be 1560 MJ/kmol of fuel. Thenits lower heating value is

(a) 1384 MJ/kmol (b) 1208 MJ/kmol(c) 1402 MJ/kmol (d) 1514 MJ/kmol(e) 1551 MJ/kmol

15–125 Acetylene gas (C2H2) is burned completely duringa steady-flow combustion process. The fuel and the air enterthe combustion chamber at 25°C, and the products leave at1500 K. If the enthalpy of the products relative to the stan-dard reference state is �404 MJ/kmol of fuel, the heat trans-fer from the combustion chamber is

(a) 177 MJ/kmol (b) 227 MJ/kmol (c) 404 MJ/kmol(d) 631 MJ/kmol (e) 751 MJ/kmol

15–126 Benzene gas (C6H6) is burned with 90 percent theo-retical air during a steady-flow combustion process. The molefraction of the CO in the products is

(a) 1.6% (b) 4.4% (c) 2.5%(d) 10% (e) 16.7%

15–127 A fuel is burned during a steady-flow combustionprocess. Heat is lost to the surroundings at 300 K at a rate of1120 kW. The entropy of the reactants entering per unit timeis 17 kW/K and that of the products is 15 kW/K. The totalrate of exergy destruction during this combustion process is

(a) 520 kW (b) 600 kW (c) 1120 kW(d) 340 kW (e) 739 kW

Design and Essay Problems

15–128 Design a combustion process suitable for use in agas-turbine engine. Discuss possible fuel selections for theseveral applications of the engine.

15–129 Constant-volume vessels that contain flammablemixtures of hydrocarbon vapors and air at low pressures are

frequently used. Although the ignition of such mixtures isvery unlikely as there is no source of ignition in the tank, theSafety and Design Codes require that the tank withstand fourtimes the pressure that may occur should an explosion takeplace in the tank. For operating gauge pressures under 25kPa, determine the pressure for which these vessels must bedesigned in order to meet the requirements of the codes for(a) acetylene C2H2(g), (b) propane C3H8(g), and (c) n-octaneC8H18(g). Justify any assumptions that you make.

15–130 The safe disposal of hazardous waste material is amajor environmental concern for industrialized societies andcreates challenging problems for engineers. The disposalmethods commonly used include landfilling, burying in theground, recycling, and incineration or burning. Incineration isfrequently used as a practical means for the disposal of com-bustible waste such as organic materials. The EPA regulationsrequire that the waste material be burned almost completelyabove a specified temperature without polluting the environ-ment. Maintaining the temperature above a certain level, typ-ically about 1100°C, necessitates the use of a fuel when thecombustion of the waste material alone is not sufficient toobtain the minimum specified temperature.

A certain industrial process generates a liquid solution ofethanol and water as the waste product at a rate of 10 kg/s.The mass fraction of ethanol in the solution is 0.2. This solu-tion is to be burned using methane (CH4) in a steady-flowcombustion chamber. Propose a combustion process that willaccomplish this task with a minimal amount of methane.State your assumptions.

15–131 Obtain the following information about a powerplant that is closest to your town: the net power output; thetype and amount of fuel; the power consumed by the pumps,fans, and other auxiliary equipment; stack gas losses; and therate of heat rejection at the condenser. Using these data,determine the rate of heat loss from the pipes and other com-ponents, and calculate the thermal efficiency of the plant.

15–132 What is oxygenated fuel? How would the heatingvalue of oxygenated fuels compare to those of comparablehydrocarbon fuels on a unit-mass basis? Why is the use ofoxygenated fuels mandated in some major cities in wintermonths?

15–133 A promising method of power generation by directenergy conversion is through the use of magnetohydrody-namic (MHD) generators. Write an essay on the current sta-tus of MHD generators. Explain their operation principlesand how they differ from conventional power plants. Discussthe problems that need to be overcome before MHD genera-tors can become economical.

cen84959_ch15.qxd 4/20/05 3:24 PM Page 792

Chapter 16CHEMICAL AND PHASE EQUILIBRIUM

| 793

In Chapter 15 we analyzed combustion processes underthe assumption that combustion is complete when there issufficient time and oxygen. Often this is not the case,

however. A chemical reaction may reach a state of equilib-rium before reaching completion even when there is sufficienttime and oxygen.

A system is said to be in equilibrium if no changes occurwithin the system when it is isolated from its surroundings.An isolated system is in mechanical equilibrium if no changesoccur in pressure, in thermal equilibrium if no changes occurin temperature, in phase equilibrium if no transformationsoccur from one phase to another, and in chemical equilib-rium if no changes occur in the chemical composition of thesystem. The conditions of mechanical and thermal equilib-rium are straightforward, but the conditions of chemical andphase equilibrium can be rather involved.

The equilibrium criterion for reacting systems is based onthe second law of thermodynamics; more specifically, theincrease of entropy principle. For adiabatic systems, chemicalequilibrium is established when the entropy of the reactingsystem reaches a maximum. Most reacting systems encoun-tered in practice are not adiabatic, however. Therefore, weneed to develop an equilibrium criterion applicable to anyreacting system.

In this chapter, we develop a general criterion for chemicalequilibrium and apply it to reacting ideal-gas mixtures. Wethen extend the analysis to simultaneous reactions. Finally,we discuss phase equilibrium for nonreacting systems.

ObjectivesThe objectives of Chapter 16 are to:

• Develop the equilibrium criterion for reacting systems basedon the second law of thermodynamics.

• Develop a general criterion for chemical equilibriumapplicable to any reacting system based on minimizing theGibbs function for the system.

• Define and evaluate the chemical equilibrium constant.

• Apply the general criterion for chemical equilibrium analysisto reacting ideal-gas mixtures.

• Apply the general criterion for chemical equilibrium analysisto simultaneous reactions.

• Relate the chemical equilibrium constant to the enthalpy ofreaction.

• Establish the phase equilibrium for nonreacting systems interms of the specific Gibbs function of the phases of a puresubstance.

• Apply the Gibbs phase rule to determine the number ofindependent variables associated with a multicomponent,multiphase system.

• Apply Henry’s law and Raoult’s law for gases dissolved inliquids.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 793

16–1 ■ CRITERION FOR CHEMICAL EQUILIBRIUMConsider a reaction chamber that contains a mixture of CO, O2, and CO2 ata specified temperature and pressure. Let us try to predict what will happenin this chamber (Fig. 16–1). Probably the first thing that comes to mind is achemical reaction between CO and O2 to form more CO2:

This reaction is certainly a possibility, but it is not the only possibility. It isalso possible that some CO2 in the combustion chamber dissociated into COand O2. Yet a third possibility would be to have no reactions among thethree components at all, that is, for the system to be in chemical equi-librium. It appears that although we know the temperature, pressure, andcomposition (thus the state) of the system, we are unable to predict whetherthe system is in chemical equilibrium. In this chapter we develop the neces-sary tools to correct this.

Assume that the CO, O2, and CO2 mixture mentioned above is in chemicalequilibrium at the specified temperature and pressure. The chemical compo-sition of this mixture does not change unless the temperature or the pressureof the mixture is changed. That is, a reacting mixture, in general, has differ-ent equilibrium compositions at different pressures and temperatures. There-fore, when developing a general criterion for chemical equilibrium, weconsider a reacting system at a fixed temperature and pressure.

Taking the positive direction of heat transfer to be to the system, theincrease of entropy principle for a reacting or nonreacting system wasexpressed in Chapter 7 as

(16–1)

A system and its surroundings form an adiabatic system, and for such systemsEq. 16–1 reduces to dSsys � 0. That is, a chemical reaction in an adiabaticchamber proceeds in the direction of increasing entropy. When the entropyreaches a maximum, the reaction stops (Fig. 16–2). Therefore, entropy is avery useful property in the analysis of reacting adiabatic systems.

When a reacting system involves heat transfer, the increase of entropyprinciple relation (Eq. 16–1) becomes impractical to use, however, since itrequires a knowledge of heat transfer between the system and its surround-ings. A more practical approach would be to develop a relation for theequilibrium criterion in terms of the properties of the reacting system only.Such a relation is developed below.

Consider a reacting (or nonreacting) simple compressible system of fixedmass with only quasi-equilibrium work modes at a specified temperature Tand pressure P (Fig. 16–3). Combining the first- and the second-lawrelations for this system gives

(16–2)dQ � P dV � dU

dS �dQ

T

¶ dU � P dV � T ds � 0

dSsys �dQ

T

CO � 12 O2 S CO2

794 | Thermodynamics

CO2

CO

O2

O2

O2

CO2

CO2

CO

CO

FIGURE 16–1A reaction chamber that contains amixture of CO2, CO, and O2 at aspecified temperature and pressure.

100%products

Violation ofsecond law

S

Equilibriumcomposition

100%reactants

dS = 0

dS > 0

dS < 0

FIGURE 16–2Equilibrium criteria for a chemicalreaction that takes place adiabatically.

Wb

REACTIONCHAMBER

δ

δ

Controlmass

T, P

Q

FIGURE 16–3A control mass undergoing a chemicalreaction at a specified temperature andpressure.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 794

The differential of the Gibbs function (G � H � TS) at constant tempera-ture and pressure is

(16–3)

From Eqs. 16–2 and 16–3, we have (dG)T,P � 0. Therefore, a chemical reac-tion at a specified temperature and pressure proceeds in the direction of adecreasing Gibbs function. The reaction stops and chemical equilibrium isestablished when the Gibbs function attains a minimum value (Fig. 16–4).Therefore, the criterion for chemical equilibrium can be expressed as

(16–4)

A chemical reaction at a specified temperature and pressure cannot proceedin the direction of the increasing Gibbs function since this will be a viola-tion of the second law of thermodynamics. Notice that if the temperature orthe pressure is changed, the reacting system will assume a different equilib-rium state, which is the state of the minimum Gibbs function at the newtemperature or pressure.

To obtain a relation for chemical equilibrium in terms of the properties ofthe individual components, consider a mixture of four chemical componentsA, B, C, and D that exist in equilibrium at a specified temperature and pres-sure. Let the number of moles of the respective components be NA, NB, NC,and ND. Now consider a reaction that occurs to an infinitesimal extentduring which differential amounts of A and B (reactants) are converted to Cand D (products) while the temperature and the pressure remain constant(Fig. 16–5):

The equilibrium criterion (Eq. 16–4) requires that the change in the Gibbsfunction of the mixture during this process be equal to zero. That is,

(16–5)

or

(16–6)

where the g–’s are the molar Gibbs functions (also called the chemical poten-tials) at the specified temperature and pressure and the dN’s are the differen-tial changes in the number of moles of the components.

To find a relation between the dN’s, we write the corresponding stoichio-metric (theoretical) reaction

(16–7)

where the n’s are the stoichiometric coefficients, which are evaluated easilyonce the reaction is specified. The stoichiometric reaction plays an impor-tant role in the determination of the equilibrium composition of the reacting

nA A � nB B ∆ nC C � nD D

g�C dNC � g�D dND � g�A dNA � g�B dNB � 0

1dG 2T,P �a 1dGi 2T,P �a 1g�i dNi 2T,P � 0

dNA A � dNB B ¡ dNC C � dND D

1dG 2T,P � 0

� dU � P dV � T dS

� 1dU � P dV � V dP 2 � T dS � S dT

1dG 2T,P � dH � T dS � S dT

Chapter 16 | 795

100%products

Violation ofsecond law

G

Equilibriumcomposition

100%reactants

dG = 0

dG < 0 dG > 0

FIGURE 16–4Criteria for chemical equilibrium for afixed mass at a specified temperatureand pressure.

REACTIONCHAMBER

T, PNA moles of ANB moles of BNC moles of CND moles of D

dNAA + dNBB → dNCC + dNDD

FIGURE 16–5An infinitesimal reaction in a chamberat constant temperature and pressure.

→0

→0

cen84959_ch16.qxd 5/11/05 10:08 AM Page 795

mixtures because the changes in the number of moles of the components areproportional to the stoichiometric coefficients (Fig. 16–6). That is,

(16–8)

where e is the proportionality constant and represents the extent of a reac-tion. A minus sign is added to the first two terms because the number ofmoles of the reactants A and B decreases as the reaction progresses.

For example, if the reactants are C2H6 and O2 and the products are CO2and H2O, the reaction of 1 mmol (10�6 mol) of C2H6 results in a 2-mmolincrease in CO2, a 3-mmol increase in H2O, and a 3.5-mmol decrease in O2in accordance with the stoichiometric equation

That is, the change in the number of moles of a component is one-millionth(e � 10�6) of the stoichiometric coefficient of that component in this case.

Substituting the relations in Eq. 16–8 into Eq. 16–6 and canceling e, weobtain

(16–9)

This equation involves the stoichiometric coefficients and the molar Gibbsfunctions of the reactants and the products, and it is known as the criterionfor chemical equilibrium. It is valid for any chemical reaction regardlessof the phases involved.

Equation 16–9 is developed for a chemical reaction that involves tworeactants and two products for simplicity, but it can easily be modified tohandle chemical reactions with any number of reactants and products. Nextwe analyze the equilibrium criterion for ideal-gas mixtures.

16–2 ■ THE EQUILIBRIUM CONSTANT FOR IDEAL-GAS MIXTURES

Consider a mixture of ideal gases that exists in equilibrium at a specifiedtemperature and pressure. Like entropy, the Gibbs function of an ideal gasdepends on both the temperature and the pressure. The Gibbs function val-ues are usually listed versus temperature at a fixed reference pressure P0,which is taken to be 1 atm. The variation of the Gibbs function of an idealgas with pressure at a fixed temperature is determined by using the defini-tion of the Gibbs function and the entropy-change relationfor isothermal processes . It yields

Thus the Gibbs function of component i of an ideal-gas mixture at its partialpressure Pi and mixture temperature T can be expressed as

(16–10)g�i 1T, Pi 2 � g�*i 1T 2 � RuT ln Pi

1¢g� 2T � ¢h�

� T 1¢ s� 2T � �T 1¢ s� 2T � RuT ln P2

P1

3¢ s� � �Ru ln 1P2>P1 2 41g� � h� � T s� 2

nCg�C � nDg�D � nAg�A � nBg�B � 0

C2H6 � 3.5O2 S 2CO2 � 3H2O

dNA � �enA dNC � enCdNB � �enB dND � enD

796 | Thermodynamics

0.1H2 → 0.2H

H2 → 2H

H2 = 1

H = 2

0.01H2 → 0.02H

0.001H2 → 0.002H

n

n

FIGURE 16–6The changes in the number of molesof the components during a chemicalreaction are proportional to thestoichiometric coefficients regardlessof the extent of the reaction.

→0

cen84959_ch16.qxd 5/11/05 10:08 AM Page 796

where g–i* (T) represents the Gibbs function of component i at 1 atm pres-sure and temperature T, and Pi represents the partial pressure of componenti in atmospheres. Substituting the Gibbs function expression for each com-ponent into Eq. 16–9, we obtain

For convenience, we define the standard-state Gibbs function change as

(16–11)

Substituting, we get

(16–12)

Now we define the equilibrium constant KP for the chemical equilibriumof ideal-gas mixtures as

(16–13)

Substituting into Eq. 16–12 and rearranging, we obtain

(16–14)

Therefore, the equilibrium constant KP of an ideal-gas mixture at a specifiedtemperature can be determined from a knowledge of the standard-stateGibbs function change at the same temperature. The KP values for severalreactions are given in Table A–28.

Once the equilibrium constant is available, it can be used to determine theequilibrium composition of reacting ideal-gas mixtures. This is accom-plished by expressing the partial pressures of the components in terms oftheir mole fractions:

where P is the total pressure and Ntotal is the total number of moles presentin the reaction chamber, including any inert gases. Replacing the partialpressures in Eq. 16–13 by the above relation and rearranging, we obtain(Fig. 16–7)

(16–15)

where

Equation 16–15 is written for a reaction involving two reactants and twoproducts, but it can be extended to reactions involving any number of reac-tants and products.

¢n � nC � nD � nA � nB

KP �NCnCND

nD

NAnANB

nBa P

Ntotalb ¢n

Pi � yi P �Ni

Ntotal P

KP � e�¢G*1T2>RuT

KP �PCnCPD

nD

PAnAPBnB

¢G* 1T 2 � �RuT 1nC ln PC � nD ln PD � nA ln PA � nB ln PB 2 � �RuT ln P CnCP D

nD

P AnAP B

nB

¢G* 1T 2 � nCg�*C 1T 2 � nDg�*D 1T 2 � nAg�*A 1T 2 � nBg�*B 1T 2

�nA 3 g�*A 1T 2 � RuT ln PA 4 � nB 3 g�*B 1T 2 � RuT ln PB 4 � 0

nC 3 g�*C 1T 2 � RuT ln PC 4 � nD 3 g�D* 1T 2 � RuT ln PD 4

Chapter 16 | 797

(1) In terms of partial pressures

KP =PC PD

C D

PA PBA B

(3) In terms of the equilibrium composition

KP = NC ND (NA NB

A B (Ntotal

P ∆

(2) In terms of ∆G*(T )

KP= e–∆G*(T )/R T

C D

u

n n

n n

n n

n n

n

FIGURE 16–7Three equivalent KP relations forreacting ideal-gas mixtures.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 797

798 | Thermodynamics

EXAMPLE 16–1 Equilibrium Constant of a Dissociation Process

Using Eq. 16–14 and the Gibbs function data, determine the equilibriumconstant KP for the dissociation process N2 → 2N at 25°C. Compare yourresult to the KP value listed in Table A–28.

Solution The equilibrium constant of the reaction N2 → 2N is listed inTable A–28 at different temperatures. It is to be verified using Gibbs func-tion data.Assumptions 1 The constituents of the mixture are ideal gases. 2 The equi-librium mixture consists of N2 and N only.Properties The equilibrium constant of this reaction at 298 K is ln KP ��367.5 (Table A–28). The Gibbs function of formation at 25°C and 1 atm isO for N2 and 455,510 kJ/kmol for N (Table A–26).Analysis In the absence of KP tables, KP can be determined from the Gibbsfunction data and Eq. 16–14,

where, from Eq. 16–11,

Substituting, we find

or

The calculated KP value is in agreement with the value listed in Table A–28.The KP value for this reaction is practically zero, indicating that this reactionwill not occur at this temperature.Discussion Note that this reaction involves one product (N) and one reactant(N2), and the stoichiometric coefficients for this reaction are nN � 2 and nN2

� 1. Also note that the Gibbs function of all stable elements (such as N2)is assigned a value of zero at the standard reference state of 25°C and 1 atm.The Gibbs function values at other temperatures can be calculated from theenthalpy and absolute entropy data by using the definition of the Gibbs func-tion, , where .h� 1T 2 � h�*f � h�T � h�298 Kg�* 1T 2 � h� 1T 2 � T s�* 1T 2

KP � 2 � 10�160

� �367.5

ln KP � �911,020 kJ>kmol

18.314 kJ>kmol # K 2 1298.15 K 2

� 911,020 kJ>kmol

� 12 2 1455,510 kJ>kmol 2 � 0

¢G* 1T 2 � nNg�*N 1T 2 � nN2g�*

N21T 2

KP � e�¢G*1T2>RuT

EXAMPLE 16–2 Dissociation Temperature of Hydrogen

Determine the temperature at which 10 percent of diatomic hydrogen (H2)dissociates into monatomic hydrogen (H) at a pressure of 10 atm.

Solution The temperature at which 10 percent of H2 dissociates into 2H isto be determined.Assumptions 1 The constituents of the mixture are ideal gases. 2 The equi-librium mixture consists of H2 and H only.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 798

A double arrow is used in equilibrium equations as an indication that achemical reaction does not stop when chemical equilibrium is established;rather, it proceeds in both directions at the same rate. That is, at equilibrium,the reactants are depleted at exactly the same rate as they are replenishedfrom the products by the reverse reaction.

16–3 ■ SOME REMARKS ABOUT THE KPOF IDEAL-GAS MIXTURES

In the last section we developed three equivalent expressions for the equilib-rium constant KP of reacting ideal-gas mixtures: Eq. 16–13, which expressesKP in terms of partial pressures; Eq. 16–14, which expresses KP in terms ofthe standard-state Gibbs function change ∆G*(T); and Eq. 16–15, whichexpresses KP in terms of the number of moles of the components. All threerelations are equivalent, but sometimes one is more convenient to use thanthe others. For example, Eq. 16–15 is best suited for determining the equi-librium composition of a reacting ideal-gas mixture at a specified tem-perature and pressure. On the basis of these relations, we may draw thefollowing conclusions about the equilibrium constant KP of ideal-gasmixtures:

1. The KP of a reaction depends on temperature only. It is independent ofthe pressure of the equilibrium mixture and is not affected by the presenceof inert gases. This is because KP depends on ∆G*(T), which depends on

Chapter 16 | 799

Analysis This is a dissociation process that is significant at very high tem-peratures only. For simplicity we consider 1 kmol of H2, as shown in Fig.16–8. The stoichiometric and actual reactions in this case are as follows:

Stoichiometric:

Actual:

reactants products(leftover)

A double-headed arrow is used for the stoichiometric reaction to differentiateit from the actual reaction. This reaction involves one reactant (H2) and oneproduct (H). The equilibrium composition consists of 0.9 kmol of H2 (theleftover reactant) and 0.2 kmol of H (the newly formed product). Therefore,NH2 � 0.9 and NH � 0.2 and the equilibrium constant KP is determinedfrom Eq. 16–15 to be

From Table A–28, the temperature corresponding to this KP value is

Discussion We conclude that 10 percent of H2 dissociates into H when thetemperature is raised to 3535 K. If the temperature is increased further, thepercentage of H2 that dissociates into H will also increase.

T � 3535 K

KP �NHnH

NH2

nH2

a P

Ntotalb nH�nH2

�10.2 2 20.9a 10

0.9 � 0.2b2�1

� 0.404

H2 ¡ 0.9H2 � 0.2H

H2 ∆ 2H 1thus nH2� 1 and nH � 2 2

1 kmol H2

10 atm

Initialcomposition

0.9H2

0.2H

Equilibriumcomposition

FIGURE 16–8Schematic for Example 16–2.

123 123

cen84959_ch16.qxd 5/11/05 10:08 AM Page 799

temperature only, and the ∆G*(T) of inert gases is zero (see Eq. 16–14).Thus, at a specified temperature the following four reactions have the sameKP value:

at 1 atm

at 5 atm

at 3 atm

at 2 atm

2. The KP of the reverse reaction is 1/KP. This is easily seen from Eq.16–13. For reverse reactions, the products and reactants switch places, andthus the terms in the numerator move to the denominator and vice versa.Consequently, the equilibrium constant of the reverse reaction becomes1/KP. For example, from Table A–28,

3. The larger the KP, the more complete the reaction. This is also apparentfrom Fig. 16–9 and Eq. 16–13. If the equilibrium composition consistslargely of product gases, the partial pressures of the products (PC and PD)are considerably larger than the partial pressures of the reactants (PA andPB), which results in a large value of KP. In the limiting case of a completereaction (no leftover reactants in the equilibrium mixture), KP approachesinfinity. Conversely, very small values of KP indicate that a reaction doesnot proceed to any appreciable degree. Thus reactions with very small KPvalues at a specified temperature can be neglected.

A reaction with KP � 1000 (or ln KP � 7) is usually assumed to proceedto completion, and a reaction with KP � 0.001 (or ln KP � �7) is assumednot to occur at all. For example, ln KP � �6.8 for the reaction N2 ∆ 2Nat 5000 K. Therefore, the dissociation of N2 into monatomic nitrogen (N)can be disregarded at temperatures below 5000 K.4. The mixture pressure affects the equilibrium composition (although itdoes not affect the equilibrium constant KP). This can be seen fromEq. 16–15, which involves the term P∆n, where ∆n � � nP � � nR (the dif-ference between the number of moles of products and the number of molesof reactants in the stoichiometric reaction). At a specified temperature, theKP value of the reaction, and thus the right-hand side of Eq. 16–15, remainsconstant. Therefore, the mole numbers of the reactants and the productsmust change to counteract any changes in the pressure term. The directionof the change depends on the sign of ∆n. An increase in pressure at a speci-fied temperature increases the number of moles of the reactants anddecreases the number of moles of products if ∆n is positive, have the oppo-site effect if ∆n is negative, and have no effect if ∆n is zero.5. The presence of inert gases affects the equilibrium composition (althoughit does not affect the equilibrium constant KP). This can be seen from Eq.16–15, which involves the term (1/Ntotal)

∆n, where Ntotal is the total numberof moles of the ideal-gas mixture at equilibrium, including inert gases. Thesign of ∆n determines how the presence of inert gases influences the equi-librium composition (Fig. 16–10). An increase in the number of moles ofinert gases at a specified temperature and pressure decreases the number of

KP � 8.718 � 10�11 for H2O ∆ H2 � 12 O2 at 1000 K

KP � 0.1147 � 1011 for H2 � 12 O2 ∆ H2O at 1000 K

H2 � 2O2 � 5N2 ∆ H2O � 1.5O2 � 5N2

H2 � 12 O2 � 3N2 ∆ H2O � 3N2

H2 � 12 O2 ∆ H2O

H2 � 12 O2 ∆ H2O

800 | Thermodynamics

40005000

100020003000

T, KP = 1 atm

6000

5.17 � 10–18

2.65 � 10–6

2.54541.47

0.025

267.7

76.8097.70

0.000.16

14.63

99.63

KP % mol H

H2 → 2H

FIGURE 16–9The larger the KP, the more completethe reaction.

1 mol H2

Initialcomposition

Equilibriumcomposition at3000 K, 1 atm

(a)

(b)1 mol N2

1 mol H2

KP = 0.02510.158 mol H

0.921 mol H2

KP = 0.0251

1.240 mol H

0.380 mol H2

1 mol N2

FIGURE 16–10The presence of inert gases does notaffect the equilibrium constant, but itdoes affect the equilibriumcomposition.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 800

moles of the reactants and increases the number of moles of products if ∆nis positive, have the opposite effect if ∆n is negative, and have no effect if∆n is zero.6. When the stoichiometric coefficients are doubled, the value of KP issquared. Therefore, when one is using KP values from a table, the stoichio-metric coefficients (the n’s) used in a reaction must be exactly the sameones appearing in the table from which the KP values are selected. Multiply-ing all the coefficients of a stoichiometric equation does not affect the massbalance, but it does affect the equilibrium constant calculations since thestoichiometric coefficients appear as exponents of partial pressures inEq. 16–13. For example,

For

But for

7. Free electrons in the equilibrium composition can be treated as an idealgas. At high temperatures (usually above 2500 K), gas molecules start todissociate into unattached atoms (such as H2 ∆ 2H), and at even highertemperatures atoms start to lose electrons and ionize, for example,

(16–16)

The dissociation and ionization effects are more pronounced at low pres-sures. Ionization occurs to an appreciable extent only at very high tempera-tures, and the mixture of electrons, ions, and neutral atoms can be treated asan ideal gas. Therefore, the equilibrium composition of ionized gas mixturescan be determined from Eq. 16–15 (Fig. 16–11). This treatment may not beadequate in the presence of strong electric fields, however, since the elec-trons may be at a different temperature than the ions in this case.8. Equilibrium calculations provide information on the equilibrium compo-sition of a reaction, not on the reaction rate. Sometimes it may even takeyears to achieve the indicated equilibrium composition. For example, theequilibrium constant of the reaction at 298 K is about1040, which suggests that a stoichiometric mixture of H2 and O2 at roomtemperature should react to form H2O, and the reaction should go to com-pletion. However, the rate of this reaction is so slow that it practically doesnot occur. But when the right catalyst is used, the reaction goes to comple-tion rather quickly to the predicted value.

H2 � 12 O2 ∆ H2O

H ∆ H� � e�

2H2 � O2 ∆ 2H2O KP2�

P 2H2O

P 2H2

PO2

� 1KP12 2

H2 � 12 O2 ∆ H2O KP1

�PH2O

PH2P 1>2

O2

Chapter 16 | 801

EXAMPLE 16–3 Equilibrium Composition at a Specified Temperature

A mixture of 2 kmol of CO and 3 kmol of O2 is heated to 2600 K at a pres-sure of 304 kPa. Determine the equilibrium composition, assuming the mix-ture consists of CO2, CO, and O2 (Fig. 16–12).

Solution A reactive gas mixture is heated to a high temperature. The equi-librium composition at that temperature is to be determined.

2 kmol CO

Initialcomposition

x CO2

Equilibriumcomposition at

2600 K, 304 kPa

y COz O2

3 kmol O2

FIGURE 16–12Schematic for Example 16–3.

where

Ntotal = NH + NH+ + Ne–

KP =

H → H+ + e–

∆ = H+ + e– – H

= 1 + 1 – 1

= 1

NH+ Ne– ∆Ntotal

( (NH

H+ e –

PHn

nn

n

n n n n

FIGURE 16–11Equilibrium-constant relation for theionization reaction of hydrogen.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 801

802 | Thermodynamics

Assumptions 1 The equilibrium composition consists of CO2, CO, and O2.2 The constituents of the mixture are ideal gases.Analysis The stoichiometric and actual reactions in this case are as follows:

Stoichiometric:

Actual:

products reactants(leftover)

C balance:

O balance:

Total number of moles:

Pressure:

The closest reaction listed in Table A–28 is CO2 ∆ CO � 1–2O2, for whichln KP � �2.801 at 2600 K. The reaction we have is the inverse of this, andthus ln KP � �2.801, or KP � 16.461 in our case.

Assuming ideal-gas behavior for all components, the equilibrium constantrelation (Eq. 16–15) becomes

Substituting, we get

Solving for x yields

Then

Therefore, the equilibrium composition of the mixture at 2600 K and 304kPa is

Discussion In solving this problem, we disregarded the dissociation of O2into O according to the reaction O2 → 2O, which is a real possibility at hightemperatures. This is because ln KP � �7.521 at 2600 K for this reaction,which indicates that the amount of O2 that dissociates into O is negligible.(Besides, we have not learned how to deal with simultaneous reactions yet.We will do so in the next section.)

1.906CO2 � 0.094CO � 2.074O2

z � 3 �x

2� 2.047

y � 2 � x � 0.094

x � 1.906

16.461 �x

12 � x 2 13 � x>2 2 1>2 a 3

5 � x>2 b�1>2

KP �NCO2

nCO2

NCOnCONO2

nO2

a P

Ntotalb nCO2

�nCO�nO2

P � 304 kPa � 3.0 atm

Ntotal � x � y � z � 5 �x

2

8 � 2x � y � 2z or z � 3 �x

2

2 � x � y or y � 2 � x

2CO � 3O2 ¡ xCO2 � yCO � zO2

CO � 12 O2 ∆ CO2 1thus nCO2

� 1, nCO � 1, and nO2� 1

2 2123 1552553

cen84959_ch16.qxd 5/11/05 10:08 AM Page 802

Chapter 16 | 803

EXAMPLE 16–4 Effect of Inert Gases on EquilibriumComposition

A mixture of 3 kmol of CO, 2.5 kmol of O2, and 8 kmol of N2 is heated to2600 K at a pressure of 5 atm. Determine the equilibrium composition ofthe mixture (Fig. 16–13).

Solution A gas mixture is heated to a high temperature. The equilibriumcomposition at the specified temperature is to be determined.Assumptions 1 The equilibrium composition consists of CO2, CO, O2, andN2. 2 The constituents of the mixture are ideal gases.Analysis This problem is similar to Example 16–3, except that it involvesan inert gas N2. At 2600 K, some possible reactions are O2 ∆ 2O (ln KP� �7.521), N2 ∆ 2N (ln KP � �28.304), 1–2O2 � 1–2N2 ∆ NO (ln KP ��2.671), and CO � 1–2O2 ∆ CO2 (ln KP � 2.801 or KP � 16.461). Basedon these KP values, we conclude that the O2 and N2 will not dissociate toany appreciable degree, but a small amount will combine to form someoxides of nitrogen. (We disregard the oxides of nitrogen in this example, butthey should be considered in a more refined analysis.) We also conclude thatmost of the CO will combine with O2 to form CO2. Notice that despite thechanges in pressure, the number of moles of CO and O2 and the presence ofan inert gas, the KP value of the reaction is the same as that used in Exam-ple 16–3.

The stoichiometric and actual reactions in this case are

Stoichiometric:

Actual:

products reactants inert(leftover)

C balance:

O balance:

Total number of moles:

Assuming ideal-gas behavior for all components, the equilibrium constantrelation (Eq. 16–15) becomes

Substituting, we get

Solving for x yields

x � 2.754

16.461 �x

13 � x 2 12.5 � x>2 2 1>2 a5

13.5 � x>2 b�1>2

KP �NCO2

nCO2

NCOnCONO2

nO2

a P

Ntotalb nCO2

�nCO�nO2

Ntotal � x � y � z � 8 � 13.5 �x

2

8 � 2x � y � 2z or z � 2.5 �x

2

3 � x � y or y � 3 � x

3CO � 2.5O2 � 8N2 ¡ xCO2 � yCO � zO2 � 8N2

CO � 12 O2 ∆ CO2 1thus nCO2

� 1, nCO � 1, and nO2� 1

2 2

3 kmol CO

Initialcomposition

x CO2

Equilibriumcomposition at2600 K, 5 atm

y COz O28 kmol N2

2.5 kmol O2

8 N2

FIGURE 16–13Schematic for Example 16–4.

123 123152553

cen84959_ch16.qxd 5/11/05 10:08 AM Page 803

16–4 ■ CHEMICAL EQUILIBRIUM FOR SIMULTANEOUS REACTIONS

The reacting mixtures we have considered so far involved only one reaction,and writing a KP relation for that reaction was sufficient to determine theequilibrium composition of the mixture. However, most practical chemicalreactions involve two or more reactions that occur simultaneously, whichmakes them more difficult to deal with. In such cases, it becomes necessaryto apply the equilibrium criterion to all possible reactions that may occur inthe reaction chamber. When a chemical species appears in more than onereaction, the application of the equilibrium criterion, together with the massbalance for each chemical species, results in a system of simultaneous equa-tions from which the equilibrium composition can be determined.

We have shown earlier that a reacting system at a specified temperatureand pressure achieves chemical equilibrium when its Gibbs function reachesa minimum value, that is, (dG)T,P � 0. This is true regardless of the numberof reactions that may be occurring. When two or more reactions areinvolved, this condition is satisfied only when (dG)T,P � 0 for each reaction.Assuming ideal-gas behavior, the KP of each reaction can be determinedfrom Eq. 16–15, with Ntotal being the total number of moles present in theequilibrium mixture.

The determination of the equilibrium composition of a reacting mixturerequires that we have as many equations as unknowns, where the unknownsare the number of moles of each chemical species present in the equilibriummixture. The mass balance of each element involved provides one equation.The rest of the equations must come from the KP relations written for eachreaction. Thus we conclude that the number of KP relations needed to deter-mine the equilibrium composition of a reacting mixture is equal to thenumber of chemical species minus the number of elements present in equi-librium. For an equilibrium mixture that consists of CO2, CO, O2, and O,for example, two KP relations are needed to determine the equilibriumcomposition since it involves four chemical species and two elements(Fig. 16–14).

The determination of the equilibrium composition of a reacting mixture inthe presence of two simultaneous reactions is here with an example.

804 | Thermodynamics

Then

Therefore, the equilibrium composition of the mixture at 2600 K and 5 atm is

Discussion Note that the inert gases do not affect the KP value or the KPrelation for a reaction, but they do affect the equilibrium composition.

2.754CO2 � 0.246CO � 1.123O2 � 8N2

z � 2.5 �x

2� 1.123

y � 3 � x � 0.246

Composition: COComposition: CO2, CO, O, CO, O2, O, O

No. of components: 4No. of components: 4No. of elements: 2No. of elements: 2No. of No. of Kp relations needed: 4 relations needed: 4 – 2 = 2 2 = 2

FIGURE 16–14The number of KP relations needed todetermine the equilibrium compositionof a reacting mixture is the differencebetween the number of species and thenumber of elements.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 804

Chapter 16 | 805

EXAMPLE 16–5 Equilibrium Composition for Simultaneous Reactions

A mixture of 1 kmol of H2O and 2 kmol of O2 is heated to 4000 K at a pres-sure of 1 atm. Determine the equilibrium composition of this mixture,assuming that only H2O, OH, O2, and H2 are present (Fig. 16–15).

Solution A gas mixture is heated to a specified temperature at a specifiedpressure. The equilibrium composition is to be determined.Assumptions 1 The equilibrium composition consists of H2O, OH, O2, andH2. 2 The constituents of the mixture are ideal gases.Analysis The chemical reaction during this process can be expressed as

Mass balances for hydrogen and oxygen yield

H balance: (1)

O balance: (2)

The mass balances provide us with only two equations with four unknowns,and thus we need to have two more equations (to be obtained from the KPrelations) to determine the equilibrium composition of the mixture. Itappears that part of the H2O in the products is dissociated into H2 and OHduring this process, according to the stoichiometric reactions

The equilibrium constants for these two reactions at 4000 K are determinedfrom Table A–28 to be

The KP relations for these two simultaneous reactions are

where

Substituting yields

(3)

(4) 0.9570 �1w 2 1y 2 1>2

x a 1

x � y � z � wb 1>2

0.5816 �1y 2 1z 2 1>2

x a 1

x � y � z � wb 1>2

Ntotal � NH2O � NH2� NO2

� NOH � x � y � z � w

KP2�

NH2

nH2 NOHnOH

NH2OnH2O

a P

Ntotalb nH2

�nOH�nH2O

KP1�

NH2

nH2 NO2

nO2

NH2OnH2O

a P

Ntotalb nH2

�nO2�nH2O

ln KP2� �0.044 ¡ KP2

� 0.9570

ln KP1� �0.542 ¡ KP1

� 0.5816

H2O ∆ 12 H2 � OH 1reaction 2 2 H2O ∆ H2 � 1

2 O2 1reaction 1 2

5 � x � 2z � w

2 � 2x � 2y � w

H2O � 2O2 ¡ xH2O � yH2 � zO2 � wOH

1 kmol H2O

Initialcomposition

x H2O

Equilibriumcomposition at4000 K, 1 atm

y H2z O2

2 kmol O2

w OH

FIGURE 16–15Schematic for Example 16–5.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 805

Solving a system of simultaneous nonlinear equations is extremely tediousand time-consuming if it is done by hand. Thus it is often necessary to solvethese kinds of problems by using an equation solver such as EES.

16–5 ■ VARIATION OF KP WITH TEMPERATUREIt was shown in Section 16–2 that the equilibrium constant KP of an idealgas depends on temperature only, and it is related to the standard-stateGibbs function change ∆G*(T) through the relation (Eq. 16–14)

In this section we develop a relation for the variation of KP with temperaturein terms of other properties.

Substituting ∆G*(T) � ∆H*(T) � T ∆S*(T) into the above relation anddifferentiating with respect to temperature, we get

At constant pressure, the second T ds relation, T ds � dh � v dP, reduces toT ds � dh. Also, T d(∆S*) � d(∆H*) since ∆S* and ∆H* consist of entropyand enthalpy terms of the reactants and the products. Therefore, the last twoterms in the above relation cancel, and it reduces to

(16–17)

where is the enthalpy of reaction at temperature T. Notice that wedropped the superscript * (which indicates a constant pressure of 1 atm)from ∆H(T), since the enthalpy of an ideal gas depends on temperature onlyand is independent of pressure. Equation 16–17 is an expression of the vari-ation of KP with temperature in terms of , and it is known as the van’tHoff equation. To integrate it, we need to know how varies with T. Forsmall temperature intervals, can be treated as a constant and Eq. 16–17can be integrated to yield

(16–18)ln KP2

KP1

�h�R

Ru

a 1

T1�

1

T2b

h�R

hR

hR 1T 2

hR 1T 2

d 1ln Kp 2dT

�¢H* 1T 2

RuT2 �

h�

R 1T 2RuT

2

d 1ln Kp 2dT

�¢H* 1T 2

RuT2 �

d 3¢H* 1T 2 4RuT dT

�d 3¢S* 1T 2 4

Ru dT

ln KP � �¢G* 1T 2

RuT

806 | Thermodynamics

Solving Eqs. (1), (2), (3), and (4) simultaneously for the four unknowns x, y,z, and w yields

Therefore, the equilibrium composition of 1 kmol H2O and 2 kmol O2 at1 atm and 4000 K is

Discussion We could also solve this problem by using the KP relation for thestoichiometric reaction O2 ∆ 2O as one of the two equations.

0.271H2O � 0.213H2 � 1.849O2 � 1.032OH

z � 1.849 w � 1.032

x � 0.271 y � 0.213

cen84959_ch16.qxd 5/11/05 10:08 AM Page 806

This equation has two important implications. First, it provides a means ofcalculating the of a reaction from a knowledge of KP, which is easier todetermine. Second, it shows that exothermic reactions such ascombustion processes are less complete at higher temperatures since KPdecreases with temperature for such reactions (Fig. 16–16).

1hR 6 0 2hR

Chapter 16 | 807

EXAMPLE 16–6 The Enthalpy of Reaction of a Combustion Process

Estimate the enthalpy of reaction for the combustion process of hydrogenH2 + 0.5O2 → H2O at 2000 K, using (a) enthalpy data and (b) KP data.

Solution The at a specified temperature is to be determined using theenthalpy and Kp data.Assumptions Both the reactants and the products are ideal gases.Analysis (a) The of the combustion process of H2 at 2000 K is the amountof energy released as 1 kmol of H2 is burned in a steady-flow combustionchamber at a temperature of 2000 K. It can be determined from Eq. 15–6,

Substituting yields

(b) The value at 2000 K can be estimated by using KP values at 1800and 2200 K (the closest two temperatures to 2000 K for which KP data areavailable) from Table A–28. They are KP1

� 18,509 at T1 � 1800 K and KP2

� 869.6 at T2 � 2200 K. By substituting these values into Eq. 16–18,the value is determined to be

Discussion Despite the large temperature difference between T1 and T2(400 K), the two results are almost identical. The agreement between thetwo results would be even better if a smaller temperature interval were used.

hR � �251,698 kJ/kmol

ln 869.6

18,509�

hR

8.314 kJ>kmol # K a 1

1800 K�

1

2200 Kb

ln KP2

KP1

�hR

Ru

a 1

T1�

1

T2b

hR

hR

� �251,663 kJ/kmol

� 10.5 kmol O2 2 3 10 � 67,881 � 8682 2 kJ>kmol O2 4 � 11 kmol H2 2 3 10 � 61,400 � 8468 2 kJ>kmol H2 4

hR � 11 kmol H2O 2 3 1�241,820 � 82,593 � 9904 2 kJ>kmol H2O 4

�NO21hf° � h2000 K � h298 K 2O2

� NH2O 1hf° � h2000 K � h298 K 2H2O � NH21hf° � h2000 K � h298 K 2H2

hR �a Np 1hf° � h � h° 2 p �a Nr 1hf° � h � h° 2 r

hR

hR

hR

4000

1000

2000

3000

T, K KP

Reaction: C + O2 → CO2

4.78 � 1020

2.25 � 1010

7.80 � 106

1.41 � 105

FIGURE 16–16Exothermic reactions are lesscomplete at higher temperatures.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 807

16–6 ■ PHASE EQUILIBRIUMWe showed at the beginning of this chapter that the equilibrium state of asystem at a specified temperature and pressure is the state of the minimumGibbs function, and the equilibrium criterion for a reacting or nonreactingsystem was expressed as (Eq. 16–4)

In the preceding sections we applied the equilibrium criterion to reactingsystems. In this section, we apply it to nonreacting multiphase systems.

We know from experience that a wet T-shirt hanging in an open areaeventually dries, a small amount of water left in a glass evaporates, and theaftershave in an open bottle quickly disappears (Fig. 16–17). Theseexamples suggest that there is a driving force between the two phases of asubstance that forces the mass to transform from one phase to another. Themagnitude of this force depends, among other things, on the relativeconcentrations of the two phases. A wet T-shirt dries much quicker in dryair than it does in humid air. In fact, it does not dry at all if the relativehumidity of the environment is 100 percent. In this case, there is no trans-formation from the liquid phase to the vapor phase, and the two phases arein phase equilibrium. The conditions of phase equilibrium change, how-ever, if the temperature or the pressure is changed. Therefore, we examinephase equilibrium at a specified temperature and pressure.

Phase Equilibrium for a Single-Component SystemThe equilibrium criterion for two phases of a pure substance such as wateris easily developed by considering a mixture of saturated liquid and satu-rated vapor in equilibrium at a specified temperature and pressure, such asthat shown in Fig. 16–18. The total Gibbs function of this mixture is

where gf and gg are the Gibbs functions of the liquid and vapor phases perunit mass, respectively. Now imagine a disturbance during which a differen-tial amount of liquid dmf evaporates at constant temperature and pressure.The change in the total Gibbs function during this disturbance is

since gf and gg remain constant at constant temperature and pressure. Atequilibrium, (dG)T,P � 0. Also from the conservation of mass, dmg � �dmf.Substituting, we obtain

which must be equal to zero at equilibrium. It yields

(16–19)

Therefore, the two phases of a pure substance are in equilibrium when eachphase has the same value of specific Gibbs function. Also, at the triple point(the state at which all three phases coexist in equilibrium), the specificGibbs functions of all three phases are equal to each other.

gf � gg

1dG 2T,P � 1gf � gg 2 dmf

1dG 2T,P � gf dmf � gg dmg

G � mf gf � mg gg

1dG 2T,P � 0

808 | Thermodynamics

FIGURE 16–17Wet clothes hung in an open areaeventually dry as a result of masstransfer from the liquid phase to thevapor phase.

© Vol. OS36/PhotoDisc

T, PVAPOR

mg

mf

LIQUID

FIGURE 16–18A liquid–vapor mixture in equilibriumat a constant temperature and pressure.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 808

What happens if gf � gg? Obviously the two phases are not in equilibriumat that moment. The second law requires that (dG)T, P � (gf � gg) dmf � 0.Thus, dmf must be negative, which means that some liquid must vaporizeuntil gf � gg. Therefore, the Gibbs function difference is the driving forcefor phase change, just as the temperature difference is the driving force forheat transfer.

Chapter 16 | 809

EXAMPLE 16–7 Phase Equilibrium for a Saturated Mixture

Show that a mixture of saturated liquid water and saturated water vapor at120°C satisfies the criterion for phase equilibrium.

Solution It is to be shown that a saturated mixture satisfies the criterionfor phase equilibrium.Properties The properties of saturated water at 120°C are hf � 503.81 kJ/kg,sf � 1.5279 kJ/kg · K, hg � 2706.0 kJ/kg, and sg � 7.1292 kJ/kg · K (TableA–4).Analysis Using the definition of Gibbs function together with the enthalpyand entropy data, we have

and

Discussion The two results are in close agreement. They would matchexactly if more accurate property data were used. Therefore, the criterion forphase equilibrium is satisfied.

� �96.8 kJ>kg

gg � hg � Tsg � 2706.0 kJ>kg � 1393.15 K 2 17.1292 kJ>kg # K 2

� �96.9 kJ>kg

gf � hf � Tsf � 503.81 kJ>kg � 1393.15 K 2 11.5279 kJ>kg # K 2

The Phase RuleNotice that a single-component two-phase system may exist in equilibriumat different temperatures (or pressures). However, once the temperature isfixed, the system is locked into an equilibrium state and all intensive prop-erties of each phase (except their relative amounts) are fixed. Therefore, asingle-component two-phase system has one independent property, whichmay be taken to be the temperature or the pressure.

In general, the number of independent variables associated with a multicom-ponent, multiphase system is given by the Gibbs phase rule, expressed as

(16–20)

where IV � the number of independent variables, C � the number of com-ponents, and PH � the number of phases present in equilibrium. For thesingle-component (C � 1) two-phase (PH � 2) system discussed above, forexample, one independent intensive property needs to be specified (IV � 1,Fig. 16–19). At the triple point, however, PH � 3 and thus IV � 0. Thatis, none of the properties of a pure substance at the triple point can be var-ied. Also, based on this rule, a pure substance that exists in a single phase

IV � C � PH � 2

TWATER VAPOR

LIQUID WATER

100°C150°C200°C

...

FIGURE 16–19According to the Gibbs phase rule, asingle-component, two-phase systemcan have only one independentvariable.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 809

(PH � 1) has two independent variables. In other words, two independentintensive properties need to be specified to fix the equilibrium state of apure substance in a single phase.

Phase Equilibrium for a Multicomponent SystemMany multiphase systems encountered in practice involve two or more com-ponents. A multicomponent multiphase system at a specified temperatureand pressure is in phase equilibrium when there is no driving force betweenthe different phases of each component. Thus, for phase equilibrium, thespecific Gibbs function of each component must be the same in all phases(Fig. 16–20). That is,

We could also derive these relations by using mathematical vigor instead ofphysical arguments.

Some components may exist in more than one solid phase at the specifiedtemperature and pressure. In this case, the specific Gibbs function of eachsolid phase of a component must also be the same for phase equilibrium.

In this section we examine the phase equilibrium of two-component sys-tems that involve two phases (liquid and vapor) in equilibrium. For suchsystems, C � 2, PH � 2, and thus IV � 2. That is, a two-component, two-phase system has two independent variables, and such a system will not bein equilibrium unless two independent intensive properties are fixed.

In general, the two phases of a two-component system do not have thesame composition in each phase. That is, the mole fraction of a componentis different in different phases. This is illustrated in Fig. 16–21 for the two-phase mixture of oxygen and nitrogen at a pressure of 0.1 MPa. On this dia-gram, the vapor line represents the equilibrium composition of the vaporphase at various temperatures, and the liquid line does the same for the liq-uid phase. At 84 K, for example, the mole fractions are 30 percent nitrogenand 70 percent oxygen in the liquid phase and 66 percent nitrogen and34 percent oxygen in the vapor phase. Notice that

(16–21a)

(16–21b)

Therefore, once the temperature and pressure (two independent variables) ofa two-component, two-phase mixture are specified, the equilibrium compo-sition of each phase can be determined from the phase diagram, which isbased on experimental measurements.

It is interesting to note that temperature is a continuous function, but molefraction (which is a dimensionless concentration), in general, is not. Thewater and air temperatures at the free surface of a lake, for example, arealways the same. The mole fractions of air on the two sides of a water–airinterface, however, are obviously very different (in fact, the mole fraction ofair in water is close to zero). Likewise, the mole fractions of water on the

yg,N2� yg,O2

� 0.66 � 0.34 � 1

yf,N2� yf,O2

� 0.30 � 0.70 � 1

gf,N � gg,N � gs,N for component N

p p p p p p gf,2 � gg,2 � gs,2 for component 2

gf,1 � gg,1 � gs,1 for component 1

810 | Thermodynamics

T, P

NH3 + H2O VAPOR

gf,NH3 = gg,NH3

LIQUID NH3 + H2O

gf,H2O = gg,H2O

FIGURE 16–20A multicomponent multiphase systemis in phase equilibrium when thespecific Gibbs function of eachcomponent is the same in all phases.

100% O2

VAPOR

T, K

0

LIQUID + VAPOR

LIQUID

10 20 30 40 50 60 70 80 900% N2100 90 80 70 60 50 40 30 20 10

74

77.378

82

86

90

94

90.2

FIGURE 16–21Equilibrium diagram for the two-phasemixture of oxygen and nitrogen at0.1 MPa.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 810

two sides of a water–air interface are also different even when air is satu-rated (Fig. 16–22). Therefore, when specifying mole fractions in two-phasemixtures, we need to clearly specify the intended phase.

In most practical applications, the two phases of a mixture are not inphase equilibrium since the establishment of phase equilibrium requires thediffusion of species from higher concentration regions to lower concentra-tion regions, which may take a long time. However, phase equilibriumalways exists at the interface of two phases of a species. In the case ofair–water interface, the mole fraction of water vapor in the air is easilydetermined from saturation data, as shown in Example 16–8.

The situation is similar at solid–liquid interfaces. Again, at a given tem-perature, only a certain amount of solid can be dissolved in a liquid, and thesolubility of the solid in the liquid is determined from the requirement thatthermodynamic equilibrium exists between the solid and the solution at theinterface. The solubility represents the maximum amount of solid that canbe dissolved in a liquid at a specified temperature and is widely available inchemistry handbooks. In Table 16–1 we present sample solubility data forsodium chloride (NaCl) and calcium bicarbonate [Ca(HO3)2] at various tem-peratures. For example, the solubility of salt (NaCl) in water at 310 K is36.5 kg per 100 kg of water. Therefore, the mass fraction of salt in the satu-rated brine is simply

whereas the mass fraction of salt in the pure solid salt is mf � 1.0.Many processes involve the absorption of a gas into a liquid. Most gases are

weakly soluble in liquids (such as air in water), and for such dilute solutionsthe mole fractions of a species i in the gas and liquid phases at the interfaceare observed to be proportional to each other. That is, yi,gas side yi,liquid side orPi,gas side Pyi,liquid side since yi � Pi /P for ideal-gas mixtures. This is known asthe Henry’s law and is expressed as

(16–22)

where H is the Henry’s constant, which is the product of the total pressureof the gas mixture and the proportionality constant. For a given species, it isa function of temperature only and is practically independent of pressure forpressures under about 5 atm. Values of the Henry’s constant for a number ofaqueous solutions are given in Table 16–2 for various temperatures. Fromthis table and the equation above we make the following observations:

1. The concentration of a gas dissolved in a liquid is inversely proportionalto Henry’s constant. Therefore, the larger the Henry’s constant, thesmaller the concentration of dissolved gases in the liquid.

2. The Henry’s constant increases (and thus the fraction of a dissolved gas inthe liquid decreases) with increasing temperature. Therefore, the dissolvedgases in a liquid can be driven off by heating the liquid (Fig. 16–23).

3. The concentration of a gas dissolved in a liquid is proportional to thepartial pressure of the gas. Therefore, the amount of gas dissolved in aliquid can be increased by increasing the pressure of the gas. This can beused to advantage in the carbonation of soft drinks with CO2 gas.

yi,liquid side �Pi,gas side

H

mfsalt,liquid side �msalt

m�

36.5 kg

1100 � 36.5 2 kg� 0.267 1or 26.7 percent 2

Chapter 16 | 811

yH2O,liquid side � 1

Air

Water

yH2O,gas side

Jump inconcentration

Concentrationprofile

x

FIGURE 16–22Unlike temperature, the mole fractionof species on the two sides of aliquid–gas (or solid–gas orsolid–liquid) interface are usually notthe same.

TABLE 16–1Solubility of two inorganiccompounds in water at varioustemperatures, in kg (in 100 kg ofwater)(from Handbook of Chemistry, McGraw-Hill,1961)

Solute

CalciumTempera- Salt bicarbonateture, K NaCl Ca(HCO3)2

273.15 35.7 16.15280 35.8 16.30290 35.9 16.53300 36.2 16.75310 36.5 16.98320 36.9 17.20330 37.2 17.43340 37.6 17.65350 38.2 17.88360 38.8 18.10370 39.5 18.33373.15 39.8 18.40

cen84959_ch16.qxd 5/11/05 10:08 AM Page 811

Strictly speaking, the result obtained from Eq. 16–22 for the mole fractionof dissolved gas is valid for the liquid layer just beneath the interface,but not necessarily the entire liquid. The latter will be the case only whenthermodynamic phase equilibrium is established throughout the entire liq-uid body.

We mentioned earlier that the use of Henry’s law is limited to dilutegas–liquid solutions, that is, liquids with a small amount of gas dissolved inthem. Then the question that arises naturally is, what do we do when the gasis highly soluble in the liquid (or solid), such as ammonia in water? In thiscase, the linear relationship of Henry’s law does not apply, and the molefraction of a gas dissolved in the liquid (or solid) is usually expressed as afunction of the partial pressure of the gas in the gas phase and the tempera-ture. An approximate relation in this case for the mole fractions of a specieson the liquid and gas sides of the interface is given by Raoult’s law as

(16–23)

where Pi,sat(T) is the saturation pressure of the species i at the interface tem-perature and Ptotal is the total pressure on the gas phase side. Tabular dataare available in chemical handbooks for common solutions such asthe ammonia–water solution that is widely used in absorption-refrigerationsystems.

Gases may also dissolve in solids, but the diffusion process in this casecan be very complicated. The dissolution of a gas may be independent ofthe structure of the solid, or it may depend strongly on its porosity. Somedissolution processes (such as the dissolution of hydrogen in titanium, simi-lar to the dissolution of CO2 in water) are reversible, and thus maintainingthe gas content in the solid requires constant contact of the solid with areservoir of that gas. Some other dissolution processes are irreversible. Forexample, oxygen gas dissolving in titanium forms TiO2 on the surface, andthe process does not reverse itself.

The molar density of the gas species i in the solid at the interfaceis proportional to the partial pressure of the species i in the gas

Pi,gas side on the gas side of the interface and is expressed as

(16–24)r�i,solid side � � � Pi,gas side 1kmol>m3 2

r�i,solid side

Pi,gas side � yi,gas side Ptotal � yi,liquid side Pi,sat 1T 2

812 | Thermodynamics

TABLE 16–2

Henry’s constant H (in bars) for selected gases in water at low to moderatepressures (for gas i, H � Pi,gas side /yi,water side) (from Mills, Table A.21, p. 874)

Solute 290 K 300 K 310 K 320 K 330 K 340 K

H2S 440 560 700 830 980 1140CO2 1,280 1,710 2,170 2,720 3,220 —O2 38,000 45,000 52,000 57,000 61,000 65,000H2 67,000 72,000 75,000 76,000 77,000 76,000CO 51,000 60,000 67,000 74,000 80,000 84,000Air 62,000 74,000 84,000 92,000 99,000 104,000N2 76,000 89,000 101,000 110,000 118,000 124,000

yA,gas side yA,liquid side

yA,gas side

yA,liquid side

or

yA,liquid side

or

PA,gas side = HyA,liquid side

Gas: ALiquid: B

Gas A

PA,gas side————

P

FIGURE 16–23Dissolved gases in a liquid can bedriven off by heating the liquid.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 812

where � is the solubility. Expressing the pressure in bars and noting that theunit of molar concentration is kmol of species i per m3, the unit of solubilityis kmol/m3 · bar. Solubility data for selected gas–solid combinations aregiven in Table 16–3. The product of solubility of a gas and the diffusioncoefficient of the gas in a solid is referred to as the permeability, which is ameasure of the ability of the gas to penetrate a solid. Permeability isinversely proportional to thickness and has the unit kmol/s · m · bar.

Finally, if a process involves the sublimation of a pure solid such as ice orthe evaporation of a pure liquid such as water in a different medium such asair, the mole (or mass) fraction of the substance in the liquid or solid phaseis simply taken to be 1.0, and the partial pressure and thus the mole fractionof the substance in the gas phase can readily be determined from the satura-tion data of the substance at the specified temperature. Also, the assumptionof thermodynamic equilibrium at the interface is very reasonable for puresolids, pure liquids, and solutions except when chemical reactions areoccurring at the interface.

Chapter 16 | 813

TABLE 16–3Solubility of selected gases andsolids (from Barrer)

(for gas i, � � r–i,solid side/Pi,gas side)

�

Gas Solid T K kmol/m3 · bar

O2 Rubber 298 0.00312N2 Rubber 298 0.00156CO2 Rubber 298 0.04015He SiO2 298 0.00045H2 Ni 358 0.00901

EXAMPLE 16–8 Mole Fraction of Water Vapor Just over a Lake

Determine the mole fraction of the water vapor at the surface of a lake whosetemperature is 15°C, and compare it to the mole fraction of water in the lake(Fig. 16–24). Take the atmospheric pressure at lake level to be 92 kPa.

Solution The mole fraction of water vapor at the surface of a lake is to bedetermined and to be compared to the mole fraction of water in the lake.Assumptions 1 Both the air and water vapor are ideal gases. 2 The amountof air dissolved in water is negligible.Properties The saturation pressure of water at 15°C is 1.7057 kPa (Table A–4).Analysis There exists phase equilibrium at the free surface of the lake,and thus the air at the lake surface is always saturated at the interfacetemperature.

The air at the water surface is saturated. Therefore, the partial pressure ofwater vapor in the air at the lake surface will simply be the saturation pres-sure of water at 15°C,

The mole fraction of water vapor in the air at the surface of the lake is deter-mined from Eq. 16–22 to be

Water contains some dissolved air, but the amount is negligible. Therefore,we can assume the entire lake to be liquid water. Then its mole fractionbecomes

Discussion Note that the concentration of water on a molar basis is100 percent just beneath the air–water interface and less than 2 percentjust above it even though the air is assumed to be saturated (so this is thehighest value at 15°C). Therefore, large discontinuities can occur in the con-centrations of a species across phase boundaries.

ywater,liquid side � 1.0 or 100 percent

yv �Pv

P�

1.7057 kPa

92 kPa� 0.0185 or 1.85 percent

Pv � Psat @ 15°C � 1.7057 kPa

yH2O,liquid side ≅ 1.0Lake15°C

Air92 kPa

yH2O,air side = 0.0185

Saturated air

FIGURE 16–24Schematic for Example 16–8.

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814 | Thermodynamics

EXAMPLE 16–9 The Amount of Dissolved Air in Water

Determine the mole fraction of air at the surface of a lake whose temperature is17°C (Fig. 16–25). Take the atmospheric pressure at lake level to be 92 kPa.

Solution The mole fraction of air in lake water is to be determined.Assumptions Both the air and vapor are ideal gases.Properties The saturation pressure of water at 17°C is 1.96 kPa (TableA–4). The Henry’s constant for air dissolved in water at 290 K is H �62,000 bar (Table 16–2).Analysis This example is similar to the previous example. Again the air at thewater surface is saturated, and thus the partial pressure of water vapor in theair at the lake surface is the saturation pressure of water at 17°C,

The partial pressure of dry air is

Note that we could have ignored the vapor pressure since the amount ofvapor in air is so small with little loss in accuracy (an error of about 2 per-cent). The mole fraction of air in the water is, from Henry’s law,

Discussion This value is very small, as expected. Therefore, the concentrationof air in water just below the air–water interface is 1.45 moles per 100,000moles. But obviously this is enough oxygen for fish and other creatures in thelake. Note that the amount of air dissolved in water will decrease with increas-ing depth unless phase equilibrium exists throughout the entire lake.

ydry air,liquid side �Pdry air,gas side

H�

0.9004 bar

62,000 bar� 1.45 � 10�5

Pdry air � P � Pv � 92 � 1.96 � 90.04 kPa � 0.9004 bar

Pv � Psat @ 17°C � 1.96 kPa

EXAMPLE 16–10 Diffusion of Hydrogen Gas into a Nickel Plate

Consider a nickel plate that is placed into a tank filled with hydrogen gas at358 K and 300 kPa. Determine the molar and mass density of hydrogen inthe nickel plate when phase equilibrium is established (Fig. 16–26).

Solution A nickel plate is exposed to hydrogen gas. The density of hydro-gen in the plate is to be determined.Properties The molar mass of hydrogen H2 is M � 2 kg/kmol, and the solu-bility of hydrogen in nickel at the specified temperature is given in Table16–3 to be 0.00901 kmol/m3 · bar.Analysis Noting that 300 kPa � 3 bar, the molar density of hydrogen in thenickel plate is determined from Eq. 16–24 to be

It corresponds to a mass density of

That is, there will be 0.027 kmol (or 0.054 kg) of H2 gas in each m3 volumeof nickel plate when phase equilibrium is established.

� 10.027 kmol>m3 2 12 kg>kmol 2 � 0.054 kg/m3

rH2,solid side � r�H2,solid side MH2

� 10.00901 kmol>m3 # bar 2 13 bar 2 � 0.027 kmol/m3

r�H2,solid side � � � PH2,gas side

ydry air,liquid sideLake17°C

Air

Pdry air,gas sideSaturated air

FIGURE 16–25Schematic for Example 16–9.

Nickel plate

Hydrogen gas358 K, 300 kPa

H2

H2

FIGURE 16–26Schematic for Example 16–10.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 814

Chapter 16 | 815

EXAMPLE 16–11 Composition of Different Phases of a Mixture

In absorption refrigeration systems, a two-phase equilibrium mixture of liquidammonia (NH3) and water (H2O) is frequently used. Consider one such mix-ture at 40°C, shown in Fig. 16–27. If the composition of the liquid phase is70 percent NH3 and 30 percent H2O by mole numbers, determine the com-position of the vapor phase of this mixture.

Solution A two-phase mixture of ammonia and water at a specified temper-ature is considered. The composition of the liquid phase is given, and thecomposition of the vapor phase is to be determined.Assumptions The mixture is ideal and thus Raoult’s law is applicable.Properties The saturation pressures of H2O and NH3 at 40°C are PH2O,sat �7.3851 kPa and PNH3,sat � 1554.33 kPa.Analysis The vapor pressures are determined from

The total pressure of the mixture is

Then the mole fractions in the gas phase are

Discussion Note that the gas phase consists almost entirely of ammonia,making this mixture very suitable for absorption refrigeration.

yNH3,gas side �PNH3,gas side

Ptotal�

1088.03 kPa

1090.25 kPa� 0.9980

yH2O,gas side �PH2O,gas side

Ptotal�

2.22 kPa

1090.25 kPa� 0.0020

Ptotal � PH2O � PNH3� 2.22 � 1088.03 � 1090.25 kPa

PNH3,gas side � yNH3,liquid side PNH3,sat 1T 2 � 0.70 11554.33 kPa 2 � 1088.03 kPa

PH2O,gas side � yH2O,liquid side PH2O,sat 1T 2 � 0.30 17.3851 kPa 2 � 2.22 kPa

40°Cyg,H2O = ?

LIQUID

VAPORH2O + NH3

yg,NH3 = ?

yf,H2O = 0.30

yf,NH3 = 0.70

FIGURE 16–27Schematic for Example 16–11.

SUMMARY

An isolated system is said to be in chemical equilibrium if nochanges occur in the chemical composition of the system.The criterion for chemical equilibrium is based on the secondlaw of thermodynamics, and for a system at a specified tem-perature and pressure it can be expressed as

For the reaction

where the n’s are the stoichiometric coefficients, the equilib-rium criterion can be expressed in terms of the Gibbs func-tions as

which is valid for any chemical reaction regardless of thephases involved.

nCg�C � nDg�D � nAg�A � nBg�B � 0

nAA � nBB ∆ nCC � nDD

1dG 2T, P � 0

For reacting systems that consist of ideal gases only, theequilibrium constant KP can be expressed as

where the standard-state Gibbs function change ∆G*(T ) andthe equilibrium constant KP are defined as

and

Here, Pi’s are the partial pressures of the components in atm.The KP of ideal-gas mixtures can also be expressed in termsof the mole numbers of the components as

KP �N nC

C NnDD

N nAA N

nBB

a P

Ntotalb ¢n

KP �P nC

C PnDD

P nAA P

nBB

¢G* 1T 2 � nCg�*C1T 2 � nDg�*

D1T 2 � nAg�*

A1T 2 � nBg�*

B1T 2

KP � e�¢G*1T 2>RuT

cen84959_ch16.qxd 5/11/05 10:08 AM Page 815

where ∆n � nC � nD � nA � nB, P is the total pressure inatm, and Ntotal is the total number of moles present in thereaction chamber, including any inert gases. The equationabove is written for a reaction involving two reactants andtwo products, but it can be extended to reactions involvingany number of reactants and products.

The equilibrium constant KP of ideal-gas mixtures dependson temperature only. It is independent of the pressure of theequilibrium mixture, and it is not affected by the presence ofinert gases. The larger the KP, the more complete the reac-tion. Very small values of KP indicate that a reaction does notproceed to any appreciable degree. A reaction with KP �1000 is usually assumed to proceed to completion, and areaction with KP � 0.001 is assumed not to occur at all. Themixture pressure affects the equilibrium composition, althoughit does not affect the equilibrium constant KP.

The variation of KP with temperature is expressed in termsof other thermochemical properties through the van’t Hoffequation

where is the enthalpy of reaction at temperature T. Forsmall temperature intervals, it can be integrated to yield

This equation shows that combustion processes are less com-plete at higher temperatures since KP decreases with tempera-ture for exothermic reactions.

Two phases are said to be in phase equilibrium when thereis no transformation from one phase to the other. Two phases

ln KP2

KP1

�hR

Ru

a 1

T1�

1

T2b

hR 1T 2

d 1ln KP 2dT

�hR 1T 2RuT

2

816 | Thermodynamics

of a pure substance are in equilibrium when each phase hasthe same value of specific Gibbs function. That is,

In general, the number of independent variables associatedwith a multicomponent, multiphase system is given by theGibbs phase rule, expressed as

where IV � the number of independent variables, C � thenumber of components, and PH � the number of phases pres-ent in equilibrium.

A multicomponent, multiphase system at a specified tem-perature and pressure is in phase equilibrium when the spe-cific Gibbs function of each component is the same in allphases.

For a gas i that is weakly soluble in a liquid (such as airin water), the mole fraction of the gas in the liquid yi,liquid sideis related to the partial pressure of the gas Pi,gas side byHenry’s law

where H is Henry’s constant. When a gas is highly soluble ina liquid (such as ammonia in water), the mole fractions of thespecies of a two-phase mixture in the liquid and gas phasesare given approximately by Raoult’s law, expressed as

where Ptotal is the total pressure of the mixture, Pi,sat(T ) is thesaturation pressure of species i at the mixture temperature,and yi,liquid side and yi,gas side are the mole fractions of species iin the liquid and vapor phases, respectively.

Pi,gas side � yi,gas side Ptotal � yi,liquid side Pi,sat 1T 2

yi,liquid side �Pi,gas side

H

IV � C � PH � 2

gf � gg

REFERENCES AND SUGGESTED READINGS

1. R. M. Barrer. Diffusion in and through Solids. New York:Macmillan, 1941.

2. I. Glassman. Combustion. New York: Academic Press,1977.

3. A. M. Kanury. Introduction to Combustion Phenomena.New York: Gordon and Breach, 1975.

4. A. F. Mills. Basic Heat and Mass Transfer. Burr Ridge,IL: Richard D. Irwin, 1995.

5. J. M. Smith and H. C. Van Ness. Introduction to ChemicalEngineering Thermodynamics. 3rd ed. New York: JohnWiley & Sons, 1986.

6. K. Wark and D. E. Richards. Thermodynamics. 6th ed.New York: McGraw-Hill, 1999.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 816

Chapter 16 | 817

KP and the Equilibrium Composition of Ideal Gases

16–1C Why is the criterion for chemical equilibriumexpressed in terms of the Gibbs function instead of entropy?

16–2C Is a wooden table in chemical equilibrium with the air?

16–3C Write three different KP relations for reacting ideal-gas mixtures, and state when each relation should be used.

16–4C The equilibrium constant of the reaction CO �→ CO2 at 1000 K and 1 atm is . Express the equilibriumconstant of the following reactions at 1000 K in terms of :

(a) at 3 atm

(b) at 1 atm

(c) at 1 atm(d) CO � 2O2 � 5N2 CO2 � 1.5O2 � 5N2 at 4 atm(e) 2CO � O2 2CO2 at 1 atm

16–5C The equilibrium constant of the dissociation reac-tion H2 → 2H at 3000 K and 1 atm is . Express the equi-librium constants of the following reactions at 3000 K interms of :

(a) H2 2H at 2 atm(b) 2H H2 at 1 atm(c) 2H2 4H at 1 atm(d) H2 � 2N2 2H � 2N2 at 2 atm(e) 6H 3H2 at 4 atm

16–6C Consider a mixture of CO2, CO, and O2 in equilib-rium at a specified temperature and pressure. Now the pres-sure is doubled.

(a) Will the equilibrium constant KP change?(b) Will the number of moles of CO2, CO, and O2 change?

How?

16–7C Consider a mixture of NO, O2, and N2 in equilib-rium at a specified temperature and pressure. Now the pres-sure is tripled.

(a) Will the equilibrium constant KP change?(b) Will the number of moles of NO, O2, and N2 change?

How?

16–8C A reaction chamber contains a mixture of CO2, CO,and O2 in equilibrium at a specified temperature and pres-

∆∆∆∆∆

KP1

KP1

∆∆

CO � O2 ∆ CO2 � 12 O2

CO2 ∆ CO � 12 O2

CO � 12 O2 ∆ CO2

KP1

KP1

12O2

sure. How will (a) increasing the temperature at constantpressure and (b) increasing the pressure at constant tempera-ture affect the number of moles of CO2?

16–9C A reaction chamber contains a mixture of N2 and Nin equilibrium at a specified temperature and pressure. Howwill (a) increasing the temperature at constant pressure and(b) increasing the pressure at constant temperature affect thenumber of moles of N2?

16–10C A reaction chamber contains a mixture of CO2,CO, and O2 in equilibrium at a specified temperature andpressure. Now some N2 is added to the mixture while themixture temperature and pressure are kept constant. Will thisaffect the number of moles of O2? How?

16–11C Which element is more likely to dissociate into itsmonatomic form at 3000 K, H2 or N2? Why?

16–12 Using the Gibbs function data, determine the equi-librium constant KP for the reaction at(a) 298 K and (b) 2000 K. Compare your results with the KPvalues listed in Table A–28.

16–13E Using Gibbs function data, determine the equilib-rium constant KP for the reaction at(a) 537 R and (b) 3240 R. Compare your results with theKP values listed in Table A–28. Answers: (a) 1.12 � 1040,(b) 1.90 � 104

16–14 Determine the equilibrium constant KP for the pro-cess CO � O2 � CO2 at (a) 298 K and (b) 2000 K. Com-pare your results with the values for KP listed in Table A–28.

16–15 Study the effect of varying the percent excessair during the steady-flow combustion of

hydrogen at a pressure of 1 atm. At what temperature will 97percent of H2 burn into H2O? Assume the equilibrium mix-ture consists of H2O, H2, O2, and N2.

16–16 Determine the equilibrium constant KP for thereaction CH4 � 2O2 CO2 � 2H2O at 25°C.Answer: 1.96 � 10140

16–17 Using the Gibbs function data, determine the equi-librium constant KP for the dissociation process

at (a) 298 K and (b) 1800 K. Compare yourresults with the KP values listed in Table A–28.

16–18 Using the Gibbs function data, determine the equi-librium constant KP for the reaction at 25°C. Compare your result with the KP value listed inTable A–28.

16–19 Determine the temperature at which 5 percent ofdiatomic oxygen (O2) dissociates into monatomic oxygen (O)at a pressure of 3 atm. Answer: 3133 K

16–20 Repeat Prob. 16–19 for a pressure of 6 atm.

H2O ∆ 12 H2 � OH

CO � 12 O2

CO2 ∆

∆

12

H2 � 12 O2 ∆ H2O

H2 � 12 O2 ∆ H2O

PROBLEMS*

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

cen84959_ch16.qxd 5/11/05 10:08 AM Page 817

16–21 Carbon monoxide is burned with 100 percentexcess air during a steady-flow process at a

pressure of 1 atm. At what temperature will 97 percent of COburn to CO2? Assume the equilibrium mixture consists ofCO2, CO, O2, and N2. Answer: 2276 K

16–22 Reconsider Prob. 16–21. Using EES (or other)software, study the effect of varying the per-

cent excess air during the steady-flow process from 0 to 200percent on the temperature at which 97 percent of CO burnsinto CO2. Plot the temperature against the percent excess air,and discuss the results.

16–23E Repeat Prob. 16–21 using data in English units.

16–24 Hydrogen is burned with 150 percent theoretical airduring a steady-flow process at a pressure of 1 atm. At whattemperature will 98 percent of H2 burn to H2O? Assume theequilibrium mixture consists of H2O, H2, O2, and N2.

16–25 Air (79 percent N2 and 21 percent O2) is heated to2000 K at a constant pressure of 2 atm. Assuming the equi-librium mixture consists of N2, O2, and NO, determine theequilibrium composition at this state. Is it realistic to assumethat no monatomic oxygen or nitrogen will be present in theequilibrium mixture? Will the equilibrium compositionchange if the pressure is doubled at constant temperature?

16–26 Hydrogen (H2) is heated to 3200 K at a constantpressure of 8 atm. Determine the percentage of H2 that willdissociate into H during this process. Answer: 5.0 percent

16–27 Carbon dioxide (CO2) is heated to 2400 K at a con-stant pressure of 3 atm. Determine the percentage of CO2 thatwill dissociate into CO and O2 during this process.

16–28 A mixture of 1 mol of CO and 3 mol of O2 is heatedto 2200 K at a pressure of 2 atm. Determine the equilibriumcomposition, assuming the mixture consists of CO2, CO, andO2. Answers: 0.995CO2, 0.005CO, 2.5025O2

16–29E A mixture of 2 mol of CO, 2 mol of O2, and 6 molof N2 is heated to 4320 R at a pressure of 3 atm. Determinethe equilibrium composition of the mixture.Answers: 1.93CO2, 0.07CO, 1.035O2, 6N2

16–30 A mixture of 3 mol of N2, 1 mol of O2, and 0.1 molof Ar is heated to 2400 K at a constant pressure of 10 atm.Assuming the equilibrium mixture consists of N2, O2, Ar, andNO, determine the equilibrium composition.Answers: 0.0823NO, 2.9589N2, 0.9589O2, 0.1Ar

16–31 Determine the mole fraction of sodium that ionizesaccording to the reaction Na Na� � e� at 2000 K and0.8 atm (KP � 0.668 for this reaction). Answer: 67.5 percent

16–32 Liquid propane (C3H8) enters a combustion chamberat 25°C at a rate of 1.2 kg/min where it is mixed and burnedwith 150 percent excess air that enters the combustion cham-ber at 12°C. If the combustion gases consist of CO2, H2O,CO, O2, and N2 that exit at 1200 K and 2 atm, determine

∆

818 | Thermodynamics

(a) the equilibrium composition of the product gases and (b) therate of heat transfer from the combustion chamber. Is it real-istic to disregard the presence of NO in the product gases?Answers: (a) 3CO2, 7.5O2, 4H2O, 47N2, (b) 5066 kJ/min

Combustionchamber

2 atmAIR

C3H8

12°C

25°C 1200 KH2OCO2

CO

O2

N2

FIGURE P16–32

16–33 Reconsider Prob. 16–32. Using EES (or other)software, investigate if it is realistic to disre-

gard the presence of NO in the product gases?

16–34E A steady-flow combustion chamber is suppliedwith CO gas at 560 R and 16 psia at a rate of 12.5 ft3/minand with oxygen (O2) at 537 R and 16 psia at a rate of0.7 lbm/min. The combustion products leave the combustionchamber at 3600 R and 16 psia. If the combustion gases con-sist of CO2, CO, and O2, determine (a) the equilibrium com-position of the product gases and (b) the rate of heat transferfrom the combustion chamber.

16–35 Oxygen (O2) is heated during a steady-flow processat 1 atm from 298 to 3000 K at a rate of 0.5 kg/min. Deter-mine the rate of heat supply needed during this process,assuming (a) some O2 dissociates into O and (b) no dissocia-tion takes place.

·

3000 K

O2, O

Qin

298 K

O2

FIGURE P16–35

16–36 Estimate KP for the following equilibrium reaction at2500 K:

At 2000 K it is known that the enthalpy of reaction is�26176 kJ/kmol and KP is 0.2209. Compare your result withthe value obtained from the definition of the equilibriumconstant.

16–37 A constant-volume tank contains a mixture of 1kmol H2 and 1 kmol O2 at 25°C and 1 atm. The contents areignited. Determine the final temperature and pressure in thetank when the combustion gases are H2O, H2, and O2.

CO � H2O � CO2 � H2

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Chapter 16 | 819

Simultaneous Reactions

16–38C What is the equilibrium criterion for systems thatinvolve two or more simultaneous chemical reactions?

16–39C When determining the equilibrium composition ofa mixture involving simultaneous reactions, how would youdetermine the number of KP relations needed?

16–40 One mole of H2O is heated to 3400 K at a pressure of1 atm. Determine the equilibrium composition, assuming thatonly H2O, OH, O2, and H2 are present. Answers: 0.574H2O,0.308H2, 0.095O2, 0.236OH

16–41 A mixture of 2 mol of CO2 and 1 mol of O2 isheated to 3200 K at a pressure of 2 atm. Determine the equi-librium composition of the mixture, assuming that only CO2,CO, O2, and O are present.

16–42 Air (21 percent O2, 79 percent N2) is heated to 3000 Kat a pressure of 2 atm. Determine the equilibrium composi-tion, assuming that only O2, N2, O, and NO are present. Is itrealistic to assume that no N will be present in the final equi-librium mixture?

16–47 Ethyl alcohol (C2H5OH(g)) at 25°C is burnedin a steady-flow adiabatic combustion chamber

with 40 percent excess air that also enters at 25°C. Determinethe adiabatic flame temperature of the products at 1 atmassuming the significant equilibrium reactions are CO2 � CO� O2 and N2 � O2 � NO. Plot the adiabatic flame tem-perature and kmoles of CO2, CO, and NO at equilibrium forvalues of percent excess air between 10 and 100 percent.

Variations of KP with Temperature

16–48C What is the importance of the van’t Hoff equation?

16–49C Will a fuel burn more completely at 2000 or 2500 K?

16–50 Estimate the enthalpy of reaction h–

R for the combus-tion process of carbon monoxide at 2200 K, using (a) enthalpydata and (b) KP data.

16–51E Estimate the enthalpy of reaction h–

R for the com-bustion process of carbon monoxide at 3960 R, using(a) enthalpy data and (b) KP data. Answers: (a) �119,030Btu/lbmol, (b) �119,041 Btu/lbmol

16–52 Using the enthalpy of reaction h–

R data and the KPvalue at 2400 K, estimate the KP value of the combustionprocess at 2600 K. Answer: 104.1

16–53 Estimate the enthalpy of reaction h–

R for the dissocia-tion process at 2200 K, using(a) enthalpy data and (b) KP data.

16–54 Estimate the enthalpy of reaction h–

R for the dissociationprocess O2 2O at 3100 K, using (a) enthalpy data and(b) KP data. Answers: (a) 513,614 kJ/kmol, (b) 512,808 kJ/kmol

16–55 Estimate the enthalpy of reaction for the equilibriumreaction CH4 � 2O2 � CO2 � 2H2O at 2500 K, using(a) enthalpy data and (b) KP data.

Phase Equilibrium

16–56C Consider a tank that contains a saturated liquid–vapor mixture of water in equilibrium. Some vapor is nowallowed to escape the tank at constant temperature and pres-sure. Will this disturb the phase equilibrium and cause someof the liquid to evaporate?

16–57C Consider a two-phase mixture of ammonia andwater in equilibrium. Can this mixture exist in two phases atthe same temperature but at a different pressure?

16–58C Using the solubility data of a solid in a specified liq-uid, explain how you would determine the mole fraction of thesolid in the liquid at the interface at a specified temperature.

16–59C Using solubility data of a gas in a solid, explainhow you would determine the molar concentration of thegas in the solid at the solid–gas interface at a specifiedtemperature.

∆

CO2 ∆ CO � 12 O2

H2 � 12O2 ∆ H2O

12

12

12

Reactionchamber

AIR

3000 K

O2, N2, O, NO

Qin

FIGURE P16–42

16–43E Air (21 percent O2, 79 percent N2) is heatedto 5400 R at a pressure of 1 atm. Determine

the equilibrium composition, assuming that only O2, N2, O,and NO are present. Is it realistic to assume that no N will bepresent in the final equilibrium mixture?

16–44E Reconsider Prob. 16–43E. Use EES (or other)software to obtain the equilibrium solution.

Compare your solution technique with that used in Prob.16–43E.

16–45 Water vapor (H2O) is heated during a steady-flowprocess at 1 atm from 298 to 3000 K at a rate of 0.2 kg/min.Determine the rate of heat supply needed during this process,assuming (a) some H2O dissociates into H2, O2, and OH and(b) no dissociation takes place. Answers: (a) 2055 kJ/min,(b) 1404 kJ/min

16–46 Reconsider Prob. 16–45. Using EES (or other)software, study the effect of the final tempera-

ture on the rate of heat supplied for the two cases. Let thefinal temperature vary from 2500 to 3500 K. For each of thetwo cases, plot the rate of heat supplied as a function finaltemperature.

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16–60C Using the Henry’s constant data for a gas dissolvedin a liquid, explain how you would determine the mole frac-tion of the gas dissolved in the liquid at the interface at aspecified temperature.

16–61 Show that a mixture of saturated liquid water andsaturated water vapor at 100°C satisfies the criterion forphase equilibrium.

16–62 Show that a mixture of saturated liquid water andsaturated water vapor at 300 kPa satisfies the criterion forphase equilibrium.

16–63 Show that a saturated liquid–vapor mixture of refriger-ant-134a at �10°C satisfies the criterion for phase equilibrium.

16–64 Consider a mixture of oxygen and nitrogen in thegas phase. How many independent properties are needed tofix the state of the system? Answer: 3

16–65 In absorption refrigeration systems, a two-phaseequilibrium mixture of liquid ammonia (NH3) and water(H2O) is frequently used. Consider a liquid–vapor mixture ofammonia and water in equilibrium at 30°C. If the composi-tion of the liquid phase is 60 percent NH3 and 40 percentH2O by mole numbers, determine the composition of thevapor phase of this mixture. Saturation pressure of NH3 at30°C is 1167.4 kPa.

16–66 Consider a liquid–vapor mixture of ammonia andwater in equilibrium at 25°C. If the composition of the liquidphase is 50 percent NH3 and 50 percent H2O by mole num-bers, determine the composition of the vapor phase of thismixture. Saturation pressure of NH3 at 25°C is 1003.5 kPa.Answers: 0.31 percent, 99.69 percent

16–67 A two-phase mixture of ammonia and water is inequilibrium at 50°C. If the composition of the vapor phase is99 percent NH3 and 1 percent H2O by mole numbers, deter-mine the composition of the liquid phase of this mixture. Sat-uration pressure of NH3 at 50°C is 2033.5 kPa.

16–68 Using the liquid–vapor equilibrium diagram of anoxygen–nitrogen mixture, determine the composition of eachphase at 80 K and 100 kPa.

16–69 Using the liquid–vapor equilibrium diagram of anoxygen–nitrogen mixture, determine the composition of eachphase at 84 K and 100 kPa.

16–70 Using the liquid–vapor equilibrium diagram of anoxygen–nitrogen mixture at 100 kPa, determine the tempera-ture at which the composition of the vapor phase is 79 per-cent N2 and 21 percent O2. Answer: 82 K

16–71 Using the liquid–vapor equilibrium diagram of anoxygen–nitrogen mixture at 100 kPa, determine the tempera-ture at which the composition of the liquid phase is 30 per-cent N2 and 70 percent O2.

820 | Thermodynamics

16–72 Consider a rubber plate that is in contact with nitro-gen gas at 298 K and 250 kPa. Determine the molar and massdensity of nitrogen in the rubber at the interface.

16–73 A wall made of natural rubber separates O2 and N2gases at 25°C and 500 kPa. Determine the molar concentra-tions of O2 and N2 in the wall.

16–74 Consider a glass of water in a room at 27°C and 97kPa. If the relative humidity in the room is 100 percent andthe water and the air are in thermal and phase equilibrium,determine (a) the mole fraction of the water vapor in the airand (b) the mole fraction of air in the water.

16–75E Water is sprayed into air at 80°F and 14.3 psia, andthe falling water droplets are collected in a container on thefloor. Determine the mass and mole fractions of air dissolvedin the water.

16–76 Consider a carbonated drink in a bottle at 27°C and130 kPa. Assuming the gas space above the liquid consists ofa saturated mixture of CO2 and water vapor and treating thedrink as water, determine (a) the mole fraction of the watervapor in the CO2 gas and (b) the mass of dissolved CO2 in a300-ml drink.

Review Problems

16–77 Using the Gibbs function data, determine the equi-librium constant KP for the dissociation process O22O at 2000 K. Compare your result with the KP value listedin Table A–28. Answer: 4.4 � 10�7

16–78 A mixture of 1 mol of H2 and 1 mol of Ar is heatedat a constant pressure of 1 atm until 15 percent of H2 dissoci-ates into monatomic hydrogen (H). Determine the final tem-perature of the mixture.

16–79 A mixture of 1 mol of H2O, 2 mol of O2, and 5 mol ofN2 is heated to 2200 K at a pressure of 5 atm. Assuming theequilibrium mixture consists of H2O, O2, N2, and H2, deter-mine the equilibrium composition at this state. Is it realistic toassume that no OH will be present in the equilibrium mixture?

16–80 Determine the mole fraction of argon that ionizesaccording to the reaction Ar Ar� � e� at 10,000 Kand 0.35 atm (KP � 0.00042 for this reaction).

16–81 Methane gas (CH4) at 25°C is burned with thestoichiometric amount of air at 25°C during an

adiabatic steady-flow combustion process at 1 atm. Assumingthe product gases consist of CO2, H2O, CO, N2, and O2,determine (a) the equilibrium composition of the productgases and (b) the exit temperature.

16–82 Reconsider Prob. 16–81. Using EES (or other)software, study the effect of excess air on the

equilibrium composition and the exit temperature by varyingthe percent excess air from 0 to 200 percent. Plot the exittemperature against the percent excess air, and discuss theresults.

∆

∆

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Chapter 16 | 821

16–83 A constant-volume tank contains a mixture of 1 molof H2 and 0.5 mol of O2 at 25°C and 1 atm. The contents ofthe tank are ignited, and the final temperature and pressure inthe tank are 2800 K and 5 atm, respectively. If the combustiongases consist of H2O, H2, and O2, determine (a) the equilib-rium composition of the product gases and (b) the amount ofheat transfer from the combustion chamber. Is it realistic toassume that no OH will be present in the equilibrium mixture?Answers: (a) 0.944H2O, 0.056H2, 0.028O2, (b) 132,574 J/mol H2

16–84 A mixture of 2 mol of H2O and 3 mol of O2 isheated to 3600 K at a pressure of 8 atm. Determine the equi-librium composition of the mixture, assuming that only H2O,OH, O2, and H2 are present.

16–85 A mixture of 3 mol of CO2 and 3 mol of O2 isheated to 3400 K at a pressure of 2 atm. Determine the equi-librium composition of the mixture, assuming that only CO2,CO, O2, and O are present. Answers: 1.313CO2, 1.687CO,3.187O2, 1.314O

16–86 Reconsider Prob. 16–85. Using EES (or other)software, study the effect of pressure on the

equilibrium composition by varying pressure from 1 atm to10 atm. Plot the amount of CO present at equilibrium as afunction of pressure.

16–87 Estimate the enthalpy of reaction for the com-bustion process of hydrogen at 2400 K, using (a) enthalpydata and (b) KP data.Answers: (a) �252,377 kJ/kmol, (b) �252,047 kJ/kmol

16–88 Reconsider Prob. 16–87. Using EES (or other)software, investigate the effect of temperature

on the enthalpy of reaction using both methods by varyingthe temperature from 2000 to 3000 K.

16–89 Using the enthalpy of reaction data and the KPvalue at 2800 K, estimate the KP value of the dissociationprocess O2 2O at 3000 K.

16–90 Show that when the three phases of a pure substanceare in equilibrium, the specific Gibbs function of each phaseis the same.

16–91 Show that when the two phases of a two-componentsystem are in equilibrium, the specific Gibbs function of eachphase of each component is the same.

16–92 A constant-volume tank initially contains 1 kmol ofcarbon monoxide CO and 3 kmol of oxygen O2 (no nitrogen)at 25°C and 2 atm. Now the mixture is ignited and the COburns completely to carbon dioxide CO2. If the final tempera-ture in the tank is 500 K, determine the final pressure in thetank and the amount of heat transfer. Is it realistic to assumethat there will be no CO in the tank when chemical equilib-rium is reached?

16–93 Using Henry’s law, show that the dissolved gases ina liquid can be driven off by heating the liquid.

∆

hR

hR

16–94 Consider a glass of water in a room at 25°C and 100kPa. If the relative humidity in the room is 70 percent and thewater and the air are in thermal equilibrium, determine(a) the mole fraction of the water vapor in the room air,(b) the mole fraction of the water vapor in the air adjacent tothe water surface, and (c) the mole fraction of air in the waternear the surface.

16–95 Repeat Prob. 16–94 for a relative humidity of25 percent.

16–96 A carbonated drink is fully charged with CO2 gas at17°C and 600 kPa such that the entire bulk of the drink is inthermodynamic equilibrium with the CO2–water vapor mix-ture. Now consider a 2-L soda bottle. If the CO2 gas in thatbottle were to be released and stored in a container at 20°Cand 100 kPa, determine the volume of the container.

16–97 Ethyl alcohol (C2H5OH(g)) at 25°C is burnedin a steady-flow adiabatic combustion chamber

with 40 percent excess air that also enters at 25°C. Determinethe adiabatic flame temperature of the products at 1 atmassuming the only significant equilibrium reaction is CO2 �CO � O2. Plot the adiabatic flame temperature as the per-cent excess air varies from 10 to 100 percent.

16–98 Tabulate the natural log of the equilibrium con-stant as a function of temperature between 298

to 3000 K for the equilibrium reaction CO � H2O � CO2 �H2. Compare your results to those obtained by combining theln KP values for the two equilibrium reactions CO2 � CO �O2 and H2O � H2 � O2 given in Table A–28.

16–99 It is desired to control the amount of CO in theproducts of combustion of octane C8H18 so that

the volume fraction of CO in the products is less than 0.1percent. Determine the percent theoretical air required for thecombustion of octane at 5 atm such that the reactant andproduct temperatures are 298 K and 2000 K, respectively.Determine the heat transfer per kmol of octane for thisprocess if the combustion occurs in a steady-flow combustionchamber. Plot the percent theoretical air required for 0.1 per-cent CO in the products as a function of product pressuresbetween 100 and 2300 kPa.

Fundamentals of Engineering (FE) Exam Problems

16–100 If the equilibrium constant for the reaction H2 �

is K, the equilibrium constant for the reaction2H2O → 2H2 � O2 at the same temperature is

(a) 1/K (b) 1/(2K) (c) 2K (d ) K2 (e) 1/K2

16–101 If the equilibrium constant for the reaction CO �is K, the equilibrium constant for the reaction

at the same temperature is

(a) 1/K (b) 1/(K � 3) (c) 4K (d ) K (e) 1/K2

CO2 � 3N2 S CO � 12 O2 � 3N2

12 O2 S CO2

12 O2 S H2O

12

12

12

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16–102 The equilibrium constant for the reaction H2 �at 1 atm and 1500°C is given to be K. Of the

reactions given below, all at 1500°C, the reaction that has adifferent equilibrium constant is

(a) at 5 atm

(b) 2H2 � O2 → 2H2O at 1 atm

(c) at 2 atm

(d ) at 5 atm

(e) at 1 atm

16–103 Of the reactions given below, the reaction whoseequilibrium composition at a specified temperature is notaffected by pressure is

(a)

(b)

(c) N2 � O2 → 2NO

(d ) N2 → 2N

(e) all of the above.

16–104 Of the reactions given below, the reaction whosenumber of moles of products increases by the addition ofinert gases into the reaction chamber at constant pressure andtemperature is

(a)

(b)

(c) N2 � O2 → 2NO

(d ) N2 → 2N

(e) all of the above.

16–105 Moist air is heated to a very high temperature. Ifthe equilibrium composition consists of H2O, O2, N2, OH,H2, and NO, the number of equilibrium constant relationsneeded to determine the equilibrium composition of themixture is

(a) 1 (b) 2 (c) 3 (d ) 4 (e) 5

16–106 Propane C3H8 is burned with air, and the combus-tion products consist of CO2, CO, H2O, O2, N2, OH, H2, andNO. The number of equilibrium constant relations needed todetermine the equilibrium composition of the mixture is

(a) 1 (b) 2 (c) 3 (d ) 4 (e) 5

16–107 Consider a gas mixture that consists of three com-ponents. The number of independent variables that need to bespecified to fix the state of the mixture is

(a) 1 (b) 2 (c) 3 (d ) 4 (e) 5

CO � 12 O2 S CO2

H2 � 12 O2 S H2O

CO � 12 O2 S CO2

H2 � 12 O2 S H2O

H2 � 12 O2 � 3N2 S H2O � 3N2

H2 � 12 O2 � 3N2 S H2O � 3N2

H2 � O2 S H2O � 12 O2

H2 � 12 O2 S H2O

12 O2 S H2O

822 | Thermodynamics

16–108 The value of Henry’s constant for CO2 gas dis-solved in water at 290 K is 12.8 MPa. Consider waterexposed to atmospheric air at 100 kPa that contains 3 percentCO2 by volume. Under phase equilibrium conditions, themole fraction of CO2 gas dissolved in water at 290 K is

(a) 2.3 � 10�4 (b) 3.0 � 10�4 (c) 0.80 � 10�4

(d ) 2.2 � 10�4 (e) 5.6 � 10�4

16–109 The solubility of nitrogen gas in rubber at 25°C is0.00156 kmol/m3 · bar. When phase equilibrium is established,the density of nitrogen in a rubber piece placed in a nitrogengas chamber at 800 kPa is

(a) 0.012 kg/m3 (b) 0.35 kg/m3 (c) 0.42 kg/m3

(d ) 0.56 kg/m3 (e) 0.078 kg/m3

Design and Essay Problems

16–110 An article that appeared in the Reno Gazette-Journal on May 18, 1992, quotes an inventor as saying thathe has turned water into motor vehicle fuel in a breakthroughthat would increase engine efficiency, save gasoline, andreduce smog. There is also a picture of a car that the inventorhas modified to run on half water and half gasoline. Theinventor claims that sparks from catalytic poles in the con-verted engine break down the water into oxygen and hydro-gen, which is burned with the gasoline. He adds thathydrogen has a higher energy density than carbon and thehigh-energy density enables one to get more power. Theinventor states that the fuel efficiency of his car increasedfrom 20 mpg (miles per gallon) to more than 50 mpg ofgasoline as a result of conversion and notes that the conver-sion has sharply reduced emissions of hydrocarbons, carbonmonoxide, and other exhaust pollutants.

Evaluate the claims made by the inventor, and write areport that is to be submitted to a group of investors who areconsidering financing this invention.

16–111 Automobiles are major emitters of air pollutantssuch as NOx, CO, and hydrocarbons HC. Find out the legallimits of these pollutants in your area, and estimate the totalamount of each pollutant, in kg, that would be produced inyour town if all the cars were emitting pollutants at the legallimit. State your assumptions.

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Chapter 17COMPRESSIBLE FLOW

| 823

For the most part, we have limited our consideration sofar to flows for which density variations and thus com-pressibility effects are negligible. In this chapter we lift

this limitation and consider flows that involve significantchanges in density. Such flows are called compressible flows,and they are frequently encountered in devices that involvethe flow of gases at very high velocities. Compressible flowcombines fluid dynamics and thermodynamics in that bothare necessary to the development of the required theoreticalbackground. In this chapter, we develop the general relationsassociated with one-dimensional compressible flows for anideal gas with constant specific heats.

We start this chapter by introducing the concepts of stag-nation state, speed of sound, and Mach number for com-pressible flows. The relationships between the static andstagnation fluid properties are developed for isentropic flows ofideal gases, and they are expressed as functions of specific-heat ratios and the Mach number. The effects of areachanges for one-dimensional isentropic subsonic and super-sonic flows are discussed. These effects are illustrated byconsidering the isentropic flow through converging andconverging–diverging nozzles. The concept of shock wavesand the variation of flow properties across normal and obliqueshocks are discussed. Finally, we consider the effects of heattransfer on compressible flows and examine steam nozzles.

ObjectivesThe objectives of Chapter 17 are to:

• Develop the general relations for compressible flowsencountered when gases flow at high speeds.

• Introduce the concepts of stagnation state, speed of sound,and Mach number for a compressible fluid.

• Develop the relationships between the static and stagnationfluid properties for isentropic flows of ideal gases.

• Derive the relationships between the static and stagnationfluid properties as functions of specific-heat ratios andMach number.

• Derive the effects of area changes for one-dimensionalisentropic subsonic and supersonic flows.

• Solve problems of isentropic flow through converging andconverging–diverging nozzles.

• Discuss the shock wave and the variation of flow propertiesacross the shock wave.

• Develop the concept of duct flow with heat transfer andnegligible friction known as Rayleigh flow.

• Examine the operation of steam nozzles commonly used insteam turbines.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 823

17–1 ■ STAGNATION PROPERTIESWhen analyzing control volumes, we find it very convenient to combine theinternal energy and the flow energy of a fluid into a single term, enthalpy,defined per unit mass as h � u � Pv. Whenever the kinetic and potentialenergies of the fluid are negligible, as is often the case, the enthalpy repre-sents the total energy of a fluid. For high-speed flows, such as thoseencountered in jet engines (Fig. 17–1), the potential energy of the fluid isstill negligible, but the kinetic energy is not. In such cases, it is convenientto combine the enthalpy and the kinetic energy of the fluid into a singleterm called stagnation (or total) enthalpy h0, defined per unit mass as

(17–1)

When the potential energy of the fluid is negligible, the stagnation enthalpyrepresents the total energy of a flowing fluid stream per unit mass. Thus itsimplifies the thermodynamic analysis of high-speed flows.

Throughout this chapter the ordinary enthalpy h is referred to as the staticenthalpy, whenever necessary, to distinguish it from the stagnationenthalpy. Notice that the stagnation enthalpy is a combination property of afluid, just like the static enthalpy, and these two enthalpies become identicalwhen the kinetic energy of the fluid is negligible.

Consider the steady flow of a fluid through a duct such as a nozzle, dif-fuser, or some other flow passage where the flow takes place adiabaticallyand with no shaft or electrical work, as shown in Fig. 17–2. Assuming thefluid experiences little or no change in its elevation and its potential energy,the energy balance relation (E

.in � E

.out) for this single-stream steady-flow

system reduces to

(17–2)

or

(17–3)

That is, in the absence of any heat and work interactions and any changes inpotential energy, the stagnation enthalpy of a fluid remains constant duringa steady-flow process. Flows through nozzles and diffusers usually satisfythese conditions, and any increase in fluid velocity in these devices createsan equivalent decrease in the static enthalpy of the fluid.

If the fluid were brought to a complete stop, then the velocity at state 2would be zero and Eq. 17–2 would become

Thus the stagnation enthalpy represents the enthalpy of a fluid when it isbrought to rest adiabatically.

During a stagnation process, the kinetic energy of a fluid is converted toenthalpy (internal energy � flow energy), which results in an increase in thefluid temperature and pressure (Fig. 17–3). The properties of a fluid at thestagnation state are called stagnation properties (stagnation temperature,

h1 �V 2

1

2� h2 � h02

h01 � h02

h1 �V 2

1

2� h2 �

V 22

2

h0 � h �V 2

21kJ>kg 2

824 | Thermodynamics

FIGURE 17–1Aircraft and jet engines involve highspeeds, and thus the kinetic energyterm should always be consideredwhen analyzing them.

(a) Photo courtesy of NASA,http://lisar.larc.nasa.gov/IMAGES/SMALL/EL-1999-00108.jpeg, and (b) Figure courtesy of Prattand Whitney. Used by permission.

Controlvolume

h02 =

h1

V1 V2

h01 h01

h2

FIGURE 17–2Steady flow of a fluid through anadiabatic duct.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 824

stagnation pressure, stagnation density, etc.). The stagnation state and thestagnation properties are indicated by the subscript 0.

The stagnation state is called the isentropic stagnation state when thestagnation process is reversible as well as adiabatic (i.e., isentropic). Theentropy of a fluid remains constant during an isentropic stagnation process.The actual (irreversible) and isentropic stagnation processes are shown onthe h-s diagram in Fig. 17–4. Notice that the stagnation enthalpy of the fluid(and the stagnation temperature if the fluid is an ideal gas) is the same forboth cases. However, the actual stagnation pressure is lower than the isen-tropic stagnation pressure since entropy increases during the actual stagna-tion process as a result of fluid friction. The stagnation processes are oftenapproximated to be isentropic, and the isentropic stagnation properties aresimply referred to as stagnation properties.

When the fluid is approximated as an ideal gas with constant specificheats, its enthalpy can be replaced by cpT and Eq. 17–1 can be expressed as

or

(17–4)

Here T0 is called the stagnation (or total) temperature, and it representsthe temperature an ideal gas attains when it is brought to rest adiabatically.The term V 2/2cp corresponds to the temperature rise during such a processand is called the dynamic temperature. For example, the dynamic temper-ature of air flowing at 100 m/s is (100 m/s)2/(2 � 1.005 kJ/kg · K) � 5.0 K.Therefore, when air at 300 K and 100 m/s is brought to rest adiabatically (atthe tip of a temperature probe, for example), its temperature rises to thestagnation value of 305 K (Fig. 17–5). Note that for low-speed flows, thestagnation and static (or ordinary) temperatures are practically the same.But for high-speed flows, the temperature measured by a stationary probeplaced in the fluid (the stagnation temperature) may be significantly higherthan the static temperature of the fluid.

The pressure a fluid attains when brought to rest isentropically is calledthe stagnation pressure P0. For ideal gases with constant specific heats, P0is related to the static pressure of the fluid by

(17–5)

By noting that r � 1/v and using the isentropic relation , theratio of the stagnation density to static density can be expressed as

(17–6)

When stagnation enthalpies are used, there is no need to refer explicitly tokinetic energy. Then the energy balance for a single-stream,steady-flow device can be expressed as

(17–7)qin � win � 1h01 � gz1 2 � qout � wout � 1h02 � gz2 2E#in � E

#out

r0

r� a T0

Tb 1>1k�12

Pv k � P0vk0

P0

P� a T0

Tb k>1k�12

T0 � T �V 2

2cp

cpT0 � cpT �V 2

2

Chapter 17 | 825

FIGURE 17–3Kinetic energy is converted toenthalpy during a stagnation process.

© Reprinted with special permission of KingFeatures Syndicate.

s

Actual state

h

Isentropicstagnationstate

P 0P 0,

act

Actual stagnationstate

h

V 2

0

h

P

2

FIGURE 17–4The actual state, actual stagnationstate, and isentropic stagnation state of a fluid on an h-s diagram.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 825

where h01 and h02 are the stagnation enthalpies at states 1 and 2, respectively.When the fluid is an ideal gas with constant specific heats, Eq. 17–7 becomes

(17–8)

where T01 and T02 are the stagnation temperatures.Notice that kinetic energy terms do not explicitly appear in Eqs. 17–7 and

17–8, but the stagnation enthalpy terms account for their contribution.

1qin � qout 2 � 1win � wout 2 � cp 1T02 � T01 2 � g 1z2 � z1 2

826 | Thermodynamics

Temperaturerise duringstagnation

AIR100 m/s

305 K

300 K

FIGURE 17–5The temperature of an ideal gasflowing at a velocity V rises by V2/2cpwhen it is brought to a complete stop.

Compressor

T1 = 255.7 K

V1 = 250 m/s

P1 = 54.05 kPa

Diffuser

1 01 02

Aircraftengine

FIGURE 17–6Schematic for Example 17–1.

EXAMPLE 17–1 Compression of High-Speed Air in an Aircraft

An aircraft is flying at a cruising speed of 250 m/s at an altitude of 5000 mwhere the atmospheric pressure is 54.05 kPa and the ambient air tempera-ture is 255.7 K. The ambient air is first decelerated in a diffuser before itenters the compressor (Fig. 17–6). Assuming both the diffuser and the com-pressor to be isentropic, determine (a) the stagnation pressure at the com-pressor inlet and (b) the required compressor work per unit mass if thestagnation pressure ratio of the compressor is 8.

Solution High-speed air enters the diffuser and the compressor of an air-craft. The stagnation pressure of air and the compressor work input are to bedetermined.Assumptions 1 Both the diffuser and the compressor are isentropic. 2 Air isan ideal gas with constant specific heats at room temperature.Properties The constant-pressure specific heat cp and the specific heat ratiok of air at room temperature are (Table A–2a)

Analysis (a) Under isentropic conditions, the stagnation pressure at thecompressor inlet (diffuser exit) can be determined from Eq. 17–5. However,first we need to find the stagnation temperature T01 at the compressor inlet.Under the stated assumptions, T01 can be determined from Eq. 17–4 to be

Then from Eq. 17–5,

That is, the temperature of air would increase by 31.1°C and the pressure by26.72 kPa as air is decelerated from 250 m/s to zero velocity. Theseincreases in the temperature and pressure of air are due to the conversion ofthe kinetic energy into enthalpy.

(b) To determine the compressor work, we need to know the stagnation tem-perature of air at the compressor exit T02. The stagnation pressure ratioacross the compressor P02/P01 is specified to be 8. Since the compressionprocess is assumed to be isentropic, T02 can be determined from the ideal-gas isentropic relation (Eq. 17–5):

T02 � T01 a P02

P01b 1k�12>k

� 1286.8 K 2 18 2 11.4�12>1.4 � 519.5 K

� 80.77 kPa

P01 � P1 a T01

T1b k>1k�12

� 154.05 kPa 2 a 286.8 K

255.7 Kb 1.4>11.4�12

� 286.8 K

T01 � T1 �V 2

1

2cp

� 255.7 K �1250 m>s 2 2

12 2 11.005 kJ>kg # K 2 a1 kJ>kg

1000 m2>s2 b

cp � 1.005 kJ>kg # K and k � 1.4

cen84959_ch17.qxd 4/21/05 11:08 AM Page 826

17–2 ■ SPEED OF SOUND AND MACH NUMBERAn important parameter in the study of compressible flow is the speed ofsound (or the sonic speed), which is the speed at which an infinitesimallysmall pressure wave travels through a medium. The pressure wave may becaused by a small disturbance, which creates a slight rise in local pressure.

To obtain a relation for the speed of sound in a medium, consider a pipethat is filled with a fluid at rest, as shown in Fig. 17–7. A piston fitted in thepipe is now moved to the right with a constant incremental velocity dV, cre-ating a sonic wave. The wave front moves to the right through the fluid atthe speed of sound c and separates the moving fluid adjacent to the pistonfrom the fluid still at rest. The fluid to the left of the wave front experiencesan incremental change in its thermodynamic properties, while the fluid onthe right of the wave front maintains its original thermodynamic properties,as shown in Fig. 17–7.

To simplify the analysis, consider a control volume that encloses the wavefront and moves with it, as shown in Fig. 17–8. To an observer travelingwith the wave front, the fluid to the right will appear to be moving towardthe wave front with a speed of c and the fluid to the left to be moving awayfrom the wave front with a speed of c � dV. Of course, the observer willthink the control volume that encloses the wave front (and herself or him-self) is stationary, and the observer will be witnessing a steady-flow process.The mass balance for this single-stream, steady-flow process can beexpressed as

or

By canceling the cross-sectional (or flow) area A and neglecting the higher-order terms, this equation reduces to

(a)

No heat or work crosses the boundaries of the control volume during thissteady-flow process, and the potential energy change, if any, can beneglected. Then the steady-flow energy balance ein � eout becomes

h �c2

2� h � dh �

1c � dV 2 22

c dr � r dV � 0

rAc � 1r � dr 2A 1c � dV 2

m#

right � m#

left

Chapter 17 | 827

Disregarding potential energy changes and heat transfer, the compressorwork per unit mass of air is determined from Eq. 17–8:

Thus the work supplied to the compressor is 233.9 kJ/kg.Discussion Notice that using stagnation properties automatically accountsfor any changes in the kinetic energy of a fluid stream.

� 233.9 kJ/kg

� 11.005 kJ>kg # K 2 1519.5 K � 286.8 K 2 win � cp 1T02 � T01 2

x

dV

+ dr r r

Movingwave frontPiston

Stationaryfluid

P + dPh + dh

Ph

dV

V

x0

P + dP

P

P

c

FIGURE 17–7Propagation of a small pressure wavealong a duct.

dV

+ r r rd

Control volumetraveling withthe wave front

P + dPh + dh

Phc – c

FIGURE 17–8Control volume moving with the smallpressure wave along a duct.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 827

which yields

(b)

where we have neglected the second-order term dV 2. The amplitude of theordinary sonic wave is very small and does not cause any appreciablechange in the pressure and temperature of the fluid. Therefore, the propaga-tion of a sonic wave is not only adiabatic but also very nearly isentropic.Then the second T ds relation developed in Chapter 7 reduces to

or

(c)

Combining Eqs. a, b, and c yields the desired expression for the speed ofsound as

or

(17–9)

It is left as an exercise for the reader to show, by using thermodynamicproperty relations (see Chap. 12) that Eq. 17–9 can also be written as

(17–10)

where k is the specific heat ratio of the fluid. Note that the speed of soundin a fluid is a function of the thermodynamic properties of that fluid.

When the fluid is an ideal gas (P � rRT), the differentiation in Eq. 17–10can easily be performed to yield

or

(17–11)

Noting that the gas constant R has a fixed value for a specified ideal gas andthe specific heat ratio k of an ideal gas is, at most, a function of tempera-ture, we see that the speed of sound in a specified ideal gas is a function oftemperature alone (Fig. 17–9).

A second important parameter in the analysis of compressible fluid flowis the Mach number Ma, named after the Austrian physicist Ernst Mach(1838–1916). It is the ratio of the actual velocity of the fluid (or an object instill air) to the speed of sound in the same fluid at the same state:

(17–12)

Note that the Mach number depends on the speed of sound, which dependson the state of the fluid. Therefore, the Mach number of an aircraft cruising

Ma �Vc

c � 2kRT

c2 � k a 0P

0rb

T

� k c 0 1rRT 20r

dT

� kRT

c2 � k a 0P

0rb

T

c2 � a 0P

0rb

s

c2 �dP

drat s � constant

dh �dPr

T ds � dh �dPr

dh � c dV � 0

828 | Thermodynamics

AIR HELIUM

347 m/s

634 m/s

200 K

300 K

1000 K

284 m/s

1861 m/s

1019 m/s

832 m/s

FIGURE 17–9The speed of sound changes withtemperature and varies with the fluid.

0¡

cen84959_ch17.qxd 4/21/05 11:08 AM Page 828

at constant velocity in still air may be different at different locations(Fig. 17–10).

Fluid flow regimes are often described in terms of the flow Mach number.The flow is called sonic when Ma � 1, subsonic when Ma � 1, supersonicwhen Ma � 1, hypersonic when Ma �� 1, and transonic when Ma � 1.

Chapter 17 | 829

V = 320 m/sAIR200 K

300 KAIR V = 320 m/s

Ma = 0.92

Ma = 1.13

FIGURE 17–10The Mach number can be different atdifferent temperatures even if thevelocity is the same.

DiffuserV = 200 m/sT = 30°C

AIR

FIGURE 17–11Schematic for Example 17–2.

1400

Stagnation region:1400 kPa200°CCO2

1000

�3 kg/s

767 200P, kPa

m⋅

FIGURE 17–12Schematic for Example 17–3.

EXAMPLE 17–2 Mach Number of Air Entering a Diffuser

Air enters a diffuser shown in Fig. 17–11 with a velocity of 200 m/s. Deter-mine (a) the speed of sound and (b) the Mach number at the diffuser inletwhen the air temperature is 30°C.

Solution Air enters a diffuser with a high velocity. The speed of sound andthe Mach number are to be determined at the diffuser inlet.Assumptions Air at specified conditions behaves as an ideal gas.Properties The gas constant of air is R � 0.287 kJ/kg · K, and its specificheat ratio at 30°C is 1.4 (Table A–2a).Analysis We note that the speed of sound in a gas varies with temperature,which is given to be 30°C.

(a) The speed of sound in air at 30°C is determined from Eq. 17–11 to be

(b) Then the Mach number becomes

Discussion The flow at the diffuser inlet is subsonic since Ma � 1.

Ma �Vc

�200 m>s349 m>s � 0.573

c � 2kRT � B 11.4 2 10.287 kJ>kg # K 2 1303 K 2 a 1000 m2>s2

1 kJ>kgb � 349 m/s

17–3 ■ ONE-DIMENSIONAL ISENTROPIC FLOWDuring fluid flow through many devices such as nozzles, diffusers, and tur-bine blade passages, flow quantities vary primarily in the flow directiononly, and the flow can be approximated as one-dimensional isentropic flowwith good accuracy. Therefore, it merits special consideration. Before pre-senting a formal discussion of one-dimensional isentropic flow, we illustratesome important aspects of it with an example.

EXAMPLE 17–3 Gas Flow through a Converging–Diverging Duct

Carbon dioxide flows steadily through a varying cross-sectional-area ductsuch as a nozzle shown in Fig. 17–12 at a mass flow rate of 3 kg/s. The car-bon dioxide enters the duct at a pressure of 1400 kPa and 200°C with a lowvelocity, and it expands in the nozzle to a pressure of 200 kPa. The duct isdesigned so that the flow can be approximated as isentropic. Determine thedensity, velocity, flow area, and Mach number at each location along theduct that corresponds to a pressure drop of 200 kPa.

Solution Carbon dioxide enters a varying cross-sectional-area duct at speci-fied conditions. The flow properties are to be determined along the duct.

cen84959_ch17.qxd 4/21/05 12:19 PM Page 829

830 | Thermodynamics

Assumptions 1 Carbon dioxide is an ideal gas with constant specific heatsat room temperature. 2 Flow through the duct is steady, one-dimensional,and isentropic.Properties For simplicity we use cp � 0.846 kJ/kg · K and k � 1.289throughout the calculations, which are the constant-pressure specific heatand specific heat ratio values of carbon dioxide at room temperature. Thegas constant of carbon dioxide is R � 0.1889 kJ/kg � K (Table A–2a).Analysis We note that the inlet temperature is nearly equal to the stagna-tion temperature since the inlet velocity is small. The flow is isentropic, andthus the stagnation temperature and pressure throughout the duct remainconstant. Therefore,

and

To illustrate the solution procedure, we calculate the desired properties atthe location where the pressure is 1200 kPa, the first location that corre-sponds to a pressure drop of 200 kPa.

From Eq. 17–5,

From Eq. 17–4,

From the ideal-gas relation,

From the mass flow rate relation,

From Eqs. 17–11 and 17–12,

The results for the other pressure steps are summarized in Table 17–1 andare plotted in Fig. 17–13.Discussion Note that as the pressure decreases, the temperature and speedof sound decrease while the fluid velocity and Mach number increase in theflow direction. The density decreases slowly at first and rapidly later as thefluid velocity increases.

Ma �Vc

�164.5 m>s333.6 m>s � 0.493

c � 2kRT � B 11.289 2 10.1889 kJ>kg # K 2 1457 K 2 a1000 m2>s2

1 kJ>kgb � 333.6 m>s

A �m#

rV�

3 kg>s113.9 kg>m3 2 1164.5 m>s 2 � 13.1 � 10�4 m2 � 13.1 cm2

r �P

RT�

1200 kPa

10.1889 kPa # m3>kg # K 2 1457 K 2 � 13.9 kg/m3

� 164.5 m/s

� B2 10.846 kJ>kg # K 2 1473 K � 457 K 2 a 1000 m2>s3

1 kJ>kgb

V � 22cp 1T0 � T 2

T � T0 a P

P0b 1k�12>k

� 1473 K 2 a 1200 kPa

1400 kPab 11.289�12>1.289

� 457 K

P0 � P1 � 1400 kPa

T0 � T1 � 200°C � 473 K

cen84959_ch17.qxd 4/21/05 11:08 AM Page 830

We note from Example 17–3 that the flow area decreases with decreasingpressure up to a critical-pressure value where the Mach number is unity, andthen it begins to increase with further reductions in pressure. The Machnumber is unity at the location of smallest flow area, called the throat (Fig.17–14). Note that the velocity of the fluid keeps increasing after passing thethroat although the flow area increases rapidly in that region. This increasein velocity past the throat is due to the rapid decrease in the fluid density.The flow area of the duct considered in this example first decreases andthen increases. Such ducts are called converging–diverging nozzles. Thesenozzles are used to accelerate gases to supersonic speeds and should not beconfused with Venturi nozzles, which are used strictly for incompressibleflow. The first use of such a nozzle occurred in 1893 in a steam turbine

Chapter 17 | 831

TABLE 17–1Variation of fluid properties in flow direction in duct described inExample 17–3 for m

.� 3 kg/s � constant

P, kPa T, K V, m/s r, kg/m3 c, m/s A, cm2 Ma

1400 473 0 15.7 339.4 ∞ 01200 457 164.5 13.9 333.6 13.1 0.4931000 439 240.7 12.1 326.9 10.3 0.736

800 417 306.6 10.1 318.8 9.64 0.962767* 413 317.2 9.82 317.2 9.63 1.000600 391 371.4 8.12 308.7 10.0 1.203400 357 441.9 5.93 295.0 11.5 1.498200 306 530.9 3.46 272.9 16.3 1.946

* 767 kPa is the critical pressure where the local Mach number is unity.

Flow direction

2004006008001000

Ma

r

A, M

a, r

, T, V

12001400P, kPa

T

A

V

FIGURE 17–13Variation of normalized fluidproperties and cross-sectional areaalong a duct as the pressure dropsfrom 1400 to 200 kPa.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 831

designed by a Swedish engineer, Carl G. B. de Laval (1845–1913), andtherefore converging–diverging nozzles are often called Laval nozzles.

Variation of Fluid Velocity with Flow AreaIt is clear from Example 17–3 that the couplings among the velocity, den-sity, and flow areas for isentropic duct flow are rather complex. In theremainder of this section we investigate these couplings more thoroughly,and we develop relations for the variation of static-to-stagnation propertyratios with the Mach number for pressure, temperature, and density.

We begin our investigation by seeking relationships among the pressure,temperature, density, velocity, flow area, and Mach number for one-dimensional isentropic flow. Consider the mass balance for a steady-flowprocess:

Differentiating and dividing the resultant equation by the mass flow rate, weobtain

(17–13)

Neglecting the potential energy, the energy balance for an isentropic flow withno work interactions can be expressed in the differential form as (Fig. 17–15)

(17–14)

This relation is also the differential form of Bernoulli’s equation whenchanges in potential energy are negligible, which is a form of the conserva-tion of momentum principle for steady-flow control volumes. CombiningEqs. 17–13 and 17–14 gives

(17–15)

Rearranging Eq. 17–9 as (∂r/∂P)s � 1/c2 and substituting into Eq. 17–15 yield

(17–16)

This is an important relation for isentropic flow in ducts since it describesthe variation of pressure with flow area. We note that A, r, and V are positivequantities. For subsonic flow (Ma � 1), the term 1 � Ma2 is positive; andthus dA and dP must have the same sign. That is, the pressure of the fluidmust increase as the flow area of the duct increases and must decrease as theflow area of the duct decreases. Thus, at subsonic velocities, the pressuredecreases in converging ducts (subsonic nozzles) and increases in divergingducts (subsonic diffusers).

In supersonic flow (Ma � 1), the term 1 � Ma2 is negative, and thus dAand dP must have opposite signs. That is, the pressure of the fluid must

dA

A�

dP

rV 2 11 � Ma2 2

dA

A�

dPra 1

V 2 �dr

dPb

dPr

� V dV � 0

dr

r�

dA

A�

dV

V� 0

m#

� rAV � constant

832 | Thermodynamics

Converging nozzle

Converging–diverging nozzle

Throat

Throat

Fluid

Fluid

FIGURE 17–14The cross section of a nozzle at thesmallest flow area is called the throat.

0 (isentropic)

dP

CONSERVATION OF ENERGY(steady flow, w = 0, q = 0, ∆pe = 0)

h1 +V 2

21 = h2 +

V 2

22

or

h +V 2

2= constant

Differentiate,

dh + V dV = 0Also,

= dh – dP

dh = dP v r

r

v

= 1

Substitute,dP + V dV = 0

T ds

FIGURE 17–15Derivation of the differential form ofthe energy equation for steadyisentropic flow.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 832

increase as the flow area of the duct decreases and must decrease as theflow area of the duct increases. Thus, at supersonic velocities, the pressuredecreases in diverging ducts (supersonic nozzles) and increases in converg-ing ducts (supersonic diffusers).

Another important relation for the isentropic flow of a fluid is obtained bysubstituting rV � �dP/dV from Eq. 17–14 into Eq. 17–16:

(17–17)

This equation governs the shape of a nozzle or a diffuser in subsonic orsupersonic isentropic flow. Noting that A and V are positive quantities, weconclude the following:

Thus the proper shape of a nozzle depends on the highest velocity desiredrelative to the sonic velocity. To accelerate a fluid, we must use a converg-ing nozzle at subsonic velocities and a diverging nozzle at supersonic veloc-ities. The velocities encountered in most familiar applications are wellbelow the sonic velocity, and thus it is natural that we visualize a nozzle asa converging duct. However, the highest velocity we can achieve by a con-verging nozzle is the sonic velocity, which occurs at the exit of the nozzle.If we extend the converging nozzle by further decreasing the flow area, inhopes of accelerating the fluid to supersonic velocities, as shown inFig. 17–16, we are up for disappointment. Now the sonic velocity will occurat the exit of the converging extension, instead of the exit of the originalnozzle, and the mass flow rate through the nozzle will decrease because ofthe reduced exit area.

Based on Eq. 17–16, which is an expression of the conservation of massand energy principles, we must add a diverging section to a converging noz-zle to accelerate a fluid to supersonic velocities. The result is a converging–diverging nozzle. The fluid first passes through a subsonic (converging) sec-tion, where the Mach number increases as the flow area of the nozzledecreases, and then reaches the value of unity at the nozzle throat. The fluidcontinues to accelerate as it passes through a supersonic (diverging) section.Noting that m

.� rAV for steady flow, we see that the large decrease in den-

sity makes acceleration in the diverging section possible. An example of thistype of flow is the flow of hot combustion gases through a nozzle in a gasturbine.

The opposite process occurs in the engine inlet of a supersonic aircraft.The fluid is decelerated by passing it first through a supersonic diffuser,which has a flow area that decreases in the flow direction. Ideally, the flowreaches a Mach number of unity at the diffuser throat. The fluid is further

For sonic flow 1Ma � 1 2 , dA

dV� 0

For supersonic flow 1Ma 7 1 2 , dA

dV7 0

For subsonic flow 1Ma 6 1 2 , dA

dV6 0

dA

A� �

dV

V11 � Ma2 2

Chapter 17 | 833

P0, T0

Ma

MaA < 1

Ma

B = 1

(sonic)

Attachment

BA

P0, T0 A = 1

(sonic)

A

Convergingnozzle

Convergingnozzle

FIGURE 17–16We cannot obtain supersonic velocitiesby attaching a converging section to aconverging nozzle. Doing so will onlymove the sonic cross section fartherdownstream and decrease the massflow rate.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 833

decelerated in a subsonic diffuser, which has a flow area that increases inthe flow direction, as shown in Fig. 17–17.

Property Relations for Isentropic Flow of Ideal GasesNext we develop relations between the static properties and stagnation proper-ties of an ideal gas in terms of the specific heat ratio k and the Mach numberMa. We assume the flow is isentropic and the gas has constant specific heats.

The temperature T of an ideal gas anywhere in the flow is related to thestagnation temperature T0 through Eq. 17–4:

or

Noting that cp � kR/(k � 1), c2 � kRT, and Ma � V/c, we see that

Substituting yields

(17–18)

which is the desired relation between T0 and T.

T0

T� 1 � a k � 1

2bMa2

V2

2cpT�

V2

2 3kR> 1k � 1 2 4T � a k � 1

2b V 2

c2 � a k � 1

2bMa2

T0

T� 1 �

V 2

2cpT

T0 � T �V 2

2cp

834 | Thermodynamics

Subsonic nozzle

(a) Subsonic flow

Ma < 1

Supersonic diffuser

Ma > 1

Supersonic nozzle

Ma > 1

Subsonic diffuser

Ma < 1

(b) Supersonic flow

P decreasesV increases

Ma increasesT decreasesr decreases

P decreasesV increases

Ma increasesT decreasesr decreases

P increasesV decreases

Ma decreasesT increasesr increases

P increasesV decreases

Ma decreasesT increasesr increasesFIGURE 17–17

Variation of flow properties insubsonic and supersonic nozzles anddiffusers.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 834

The ratio of the stagnation to static pressure is obtained by substitutingEq. 17–18 into Eq. 17–5:

(17–19)

The ratio of the stagnation to static density is obtained by substitutingEq. 17–18 into Eq. 17–6:

(17–20)

Numerical values of T/T0, P/P0, and r/r0 are listed versus the Mach numberin Table A–32 for k � 1.4, which are very useful for practical compressibleflow calculations involving air.

The properties of a fluid at a location where the Mach number is unity (thethroat) are called critical properties, and the ratios in Eqs. (17–18) through(17–20) are called critical ratios (Fig. 17–18). It is common practice in theanalysis of compressible flow to let the superscript asterisk (*) represent thecritical values. Setting Ma � 1 in Eqs. 17–18 through 17–20 yields

(17–21)

(17–22)

(17–23)

These ratios are evaluated for various values of k and are listed in Table17–2. The critical properties of compressible flow should not be confusedwith the properties of substances at the critical point (such as the criticaltemperature Tc and critical pressure Pc).

r*

r0� a 2

k � 1b 1>1k�12

P*

P0� a 2

k � 1b k>1k�12

T*

T0�

2

k � 1

r0

r� c1 � a k � 1

2bMa2 d 1>1k�12

P0

P� c1 � a k � 1

2bMa2 d k>1k�12

Chapter 17 | 835

Subsonicnozzle

Supersonicnozzle

T*, P*, r*

TtP

r r

t

t

ToP

r

o

o

T *P *

*(if Mat = 1)

(Mat = 1)

Throat

Throat

FIGURE 17–18When Mat � 1, the properties at thenozzle throat become the criticalproperties.

TABLE 17–2The critical-pressure, critical-temperature, and critical-density ratios forisentropic flow of some ideal gases

Superheated Hot products Monatomicsteam, of combustion, Air, gases,k � 1.3 k � 1.33 k � 1.4 k � 1.667

0.5457 0.5404 0.5283 0.4871

0.8696 0.8584 0.8333 0.7499

0.6276 0.6295 0.6340 0.6495r*

r0

T*

T0

P*

P0

cen84959_ch17.qxd 4/26/05 4:17 PM Page 835

17–4 ■ ISENTROPIC FLOW THROUGH NOZZLESConverging or converging–diverging nozzles are found in many engineeringapplications including steam and gas turbines, aircraft and spacecraftpropulsion systems, and even industrial blasting nozzles and torch nozzles.In this section we consider the effects of back pressure (i.e., the pressureapplied at the nozzle discharge region) on the exit velocity, the mass flowrate, and the pressure distribution along the nozzle.

Converging NozzlesConsider the subsonic flow through a converging nozzle as shown inFig. 17–20. The nozzle inlet is attached to a reservoir at pressure Pr andtemperature Tr. The reservoir is sufficiently large so that the nozzle inletvelocity is negligible. Since the fluid velocity in the reservoir is zero andthe flow through the nozzle is approximated as isentropic, the stagnationpressure and stagnation temperature of the fluid at any cross sectionthrough the nozzle are equal to the reservoir pressure and temperature,respectively.

836 | Thermodynamics

EXAMPLE 17–4 Critical Temperature and Pressure in Gas Flow

Calculate the critical pressure and temperature of carbon dioxide for the flowconditions described in Example 17–3 (Fig. 17–19).

Solution For the flow discussed in Example 17–3, the critical pressure andtemperature are to be calculated.Assumptions 1 The flow is steady, adiabatic, and one-dimensional. 2 Carbondioxide is an ideal gas with constant specific heats.Properties The specific heat ratio of carbon dioxide at room temperature isk � 1.289 (Table A–2a).Analysis The ratios of critical to stagnation temperature and pressure aredetermined to be

Noting that the stagnation temperature and pressure are, from Example17–3, T0 � 473 K and P0 � 1400 kPa, we see that the critical temperatureand pressure in this case are

Discussion Note that these values agree with those listed in Table 17–1, asexpected. Also, property values other than these at the throat would indicatethat the flow is not critical, and the Mach number is not unity.

P* � 0.5477P0 � 10.5477 2 11400 kPa 2 � 767 kPa

T* � 0.8737T0 � 10.8737 2 1473 K 2 � 413 K

P*

P0� a 2

k � 1b k>1k�12

� a 2

1.289 � 1b 1.289>11.289�12

� 0.5477

T*

T0�

2

k � 1�

2

1.289 � 1� 0.8737

T *P *

= 473 K

= 1.4 MPaCO2T0

P0

FIGURE 17–19Schematic for Example 17–4.

x

Lowest exitpressure

P/P0

ReservoirPe

x

P*

Vr = 0

0

1

P0

Pb = P*

Pb < P*

Pb = 05

4

3

2

1

Pb > P*

Pb = P0

Pb(Back

pressure)

Pr = P0Tr = T0

FIGURE 17–20The effect of back pressure on thepressure distribution along aconverging nozzle.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 836

Now we begin to reduce the back pressure and observe the resultingeffects on the pressure distribution along the length of the nozzle, as shownin Fig. 17–20. If the back pressure Pb is equal to P1, which is equal to Pr,there is no flow and the pressure distribution is uniform along the nozzle.When the back pressure is reduced to P2, the exit plane pressure Pe alsodrops to P2. This causes the pressure along the nozzle to decrease in theflow direction.

When the back pressure is reduced to P3 (� P*, which is the pressurerequired to increase the fluid velocity to the speed of sound at the exit planeor throat), the mass flow reaches a maximum value and the flow is said tobe choked. Further reduction of the back pressure to level P4 or below doesnot result in additional changes in the pressure distribution, or anything elsealong the nozzle length.

Under steady-flow conditions, the mass flow rate through the nozzle isconstant and can be expressed as

Solving for T from Eq. 17–18 and for P from Eq. 17–19 and substituting,

(17–24)

Thus the mass flow rate of a particular fluid through a nozzle is a functionof the stagnation properties of the fluid, the flow area, and the Mach num-ber. Equation 17–24 is valid at any cross section, and thus m

.can be evalu-

ated at any location along the length of the nozzle.For a specified flow area A and stagnation properties T0 and P0, the maxi-

mum mass flow rate can be determined by differentiating Eq. 17–24 withrespect to Ma and setting the result equal to zero. It yields Ma � 1. Sincethe only location in a nozzle where the Mach number can be unity is thelocation of minimum flow area (the throat), the mass flow rate through anozzle is a maximum when Ma � 1 at the throat. Denoting this area by A*,we obtain an expression for the maximum mass flow rate by substitutingMa � 1 in Eq. 17–24:

(17–25)

Thus, for a particular ideal gas, the maximum mass flow rate through anozzle with a given throat area is fixed by the stagnation pressure and tem-perature of the inlet flow. The flow rate can be controlled by changingthe stagnation pressure or temperature, and thus a converging nozzle can beused as a flowmeter. The flow rate can also be controlled, of course, byvarying the throat area. This principle is vitally important for chemicalprocesses, medical devices, flowmeters, and anywhere the mass flux of agas must be known and controlled.

A plot of m.

versus Pb/P0 for a converging nozzle is shown in Fig. 17–21.Notice that the mass flow rate increases with decreasing Pb /P0, reaches amaximum at Pb � P*, and remains constant for Pb /P0 values less than this

m#

max � A*P0Bk

RT0a 2

k � 1b 1k�12> 321k�124

m#

�AMaP02k> 1RT0 2

31 � 1k � 1 2Ma2>2 4 1k�12> 321k�124

m#

� rAV � a P

RTbA 1Ma2kRT 2 � PAMaB

k

RT

Chapter 17 | 837

/P0

P*

0

1.0

P0

Pb

Pe

5 4 3

2

1

P0

1.0P*P0

mmax

Pb

5 4 3

2

1

P0

1.0P*P0

m

⋅

⋅

FIGURE 17–21The effect of back pressure Pb on themass flow rate and the exit pressurePe of a converging nozzle.

m#

cen84959_ch17.qxd 4/21/05 11:08 AM Page 837

critical ratio. Also illustrated on this figure is the effect of back pressure onthe nozzle exit pressure Pe. We observe that

To summarize, for all back pressures lower than the critical pressure P*,the pressure at the exit plane of the converging nozzle Pe is equal to P*, theMach number at the exit plane is unity, and the mass flow rate is the maxi-mum (or choked) flow rate. Because the velocity of the flow is sonic at thethroat for the maximum flow rate, a back pressure lower than the criticalpressure cannot be sensed in the nozzle upstream flow and does not affectthe flow rate.

The effects of the stagnation temperature T0 and stagnation pressure P0 onthe mass flow rate through a converging nozzle are illustrated in Fig. 17–22where the mass flow rate is plotted against the static-to-stagnation pressureratio at the throat Pt /P0. An increase in P0 (or a decrease in T0) will increasethe mass flow rate through the converging nozzle; a decrease in P0 (or anincrease in T0) will decrease it. We could also conclude this by carefullyobserving Eqs. 17–24 and 17–25.

A relation for the variation of flow area A through the nozzle relative tothroat area A* can be obtained by combining Eqs. 17–24 and 17–25 for thesame mass flow rate and stagnation properties of a particular fluid. Thisyields

(17–26)

Table A–32 gives values of A/A* as a function of the Mach number for air(k � 1.4). There is one value of A/A* for each value of the Mach number,but there are two possible values of the Mach number for each value ofA/A*—one for subsonic flow and another for supersonic flow.

A

A*�

1

Mac a 2

k � 1b a1 �

k � 1

2 Ma2 b d 1k�12> 321k�124

Pe � ePb for Pb � P*

P* for Pb 6 P*

838 | Thermodynamics

0Pt

Decrease in ,

1.0P*P0

m

P 0

P0

P0, T0

T0,increase inor both

Increase in ,P0

T0,decrease inor both

Mat = 1 Mat < 1

⋅

FIGURE 17–22The variation of the mass flow ratethrough a nozzle with inlet stagnationproperties.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 838

Another parameter sometimes used in the analysis of one-dimensionalisentropic flow of ideal gases is Ma*, which is the ratio of the local velocityto the speed of sound at the throat:

(17–27)

It can also be expressed as

where Ma is the local Mach number, T is the local temperature, and T* isthe critical temperature. Solving for T from Eq. 17–18 and for T* fromEq. 17–21 and substituting, we get

(17–28)

Values of Ma* are also listed in Table A–32 versus the Mach number fork � 1.4 (Fig. 17–23). Note that the parameter Ma* differs from the Machnumber Ma in that Ma* is the local velocity nondimensionalized withrespect to the sonic velocity at the throat, whereas Ma is the local velocitynondimensionalized with respect to the local sonic velocity. (Recall that thesonic velocity in a nozzle varies with temperature and thus with location.)

Ma* � MaBk � 1

2 � 1k � 1 2Ma2

Ma* �Vc

c

c*�

Mac

c*�

Ma2kRT

2kRT*� MaB

T

T*

Ma* �V

c*

Chapter 17 | 839

Ma*0 T0

0.90

A*T

P r0

P rAMa

...

...

...............

...............

1.001.10

0.91461.00001.0812

..

...

.

1.00891.00001.0079

..

...

.

0.59130.52830.4684

..

...

. ..

FIGURE 17–23Various property ratios for isentropicflow through nozzles and diffusers arelisted in Table A–32 for k � 1.4 forconvenience.

Pb

At = 50 cm2

Pi = 1 MPa

Vi = 150 m/sTi = 600°C

AIR

Convergingnozzle

FIGURE 17–24Schematic for Example 17–5.

EXAMPLE 17–5 Effect of Back Pressure on Mass Flow Rate

Air at 1 MPa and 600°C enters a converging nozzle, shown in Fig. 17–24,with a velocity of 150 m/s. Determine the mass flow rate through the nozzlefor a nozzle throat area of 50 cm2 when the back pressure is (a) 0.7 MPaand (b) 0.4 MPa.

Solution Air enters a converging nozzle. The mass flow rate of air throughthe nozzle is to be determined for different back pressures.Assumptions 1 Air is an ideal gas with constant specific heats at roomtemperature. 2 Flow through the nozzle is steady, one-dimensional, andisentropic.Properties The constant-pressure specific heat and the specific heat ratio ofair are cp � 1.005 kJ/kg � K and k � 1.4, respectively (Table A–2a). Analysis We use the subscripts i and t to represent the properties at thenozzle inlet and the throat, respectively. The stagnation temperature andpressure at the nozzle inlet are determined from Eqs. 17–4 and 17–5:

These stagnation temperature and pressure values remain constant through-out the nozzle since the flow is assumed to be isentropic. That is,

T0 � T0i � 884 K and P0 � P0i � 1.045 MPa

P0i � Pi a T0i

Ti

b k>1k�12� 11 MPa 2 a 884 K

873 Kb 1.4>11.4�12

� 1.045 MPa

T0i � Ti �V 2

i

2cp

� 873 K �1150 m>s 2 2

2 11.005 kJ>kg # K 2 a1 kJ>kg

1000 m2>s2 b � 884 K

cen84959_ch17.qxd 4/21/05 11:08 AM Page 839

840 | Thermodynamics

The critical-pressure ratio is determined from Table 17–2 (or Eq. 17–22) tobe P*/P0 � 0.5283.

(a) The back pressure ratio for this case is

which is greater than the critical-pressure ratio, 0.5283. Thus the exit planepressure (or throat pressure Pt) is equal to the back pressure in this case. Thatis, Pt � Pb � 0.7 MPa, and Pt /P0 � 0.670. Therefore, the flow is not choked.From Table A–32 at Pt /P0 � 0.670, we read Mat � 0.778 and Tt /T0 � 0.892.

The mass flow rate through the nozzle can be calculated from Eq. 17–24.But it can also be determined in a step-by-step manner as follows:

Thus,

(b) The back pressure ratio for this case is

which is less than the critical-pressure ratio, 0.5283. Therefore, sonic condi-tions exist at the exit plane (throat) of the nozzle, and Ma � 1. The flow ischoked in this case, and the mass flow rate through the nozzle can be calcu-lated from Eq. 17–25:

since .Discussion This is the maximum mass flow rate through the nozzle for thespecified inlet conditions and nozzle throat area.

kPa # m2>2kJ>kg � 21000 kg>s � 7.10 kg/s

� 150 � 10�4 m2 2 11045 kPa 2� B1.4

10.287 kJ>kg # K 2 1884 K 2 a2

1.4 � 1b 2.4>0.8

m#

� A*P0Bk

RT0a 2

k � 1b 1k�12>321k�124

Pb

P0�

0.4 MPa

1.045 MPa� 0.383

m#

� rt AtVt � 13.093 kg>m3 2 150 � 10�4 m2 2 1437.9 m>s 2 � 6.77 kg/s

� 437.9 m>s � 10.778 2B 11.4 2 10.287 kJ>kg # K 2 1788.5 K 2 a 1000 m2>s2

1 kJ>kgb

Vt � Mat ct � Mat2kRTt

rt �Pt

RTt

�700 kPa

10.287 kPa # m3>kg # K 2 1788.5 K 2 � 3.093 kg>m3

Tt � 0.892T0 � 0.892 1884 K 2 � 788.5 K

Pb

P0�

0.7 MPa

1.045 MPa� 0.670

EXAMPLE 17–6 Gas Flow through a Converging Nozzle

Nitrogen enters a duct with varying flow area at T1 � 400 K, P1 � 100 kPa,and Ma1 � 0.3. Assuming steady isentropic flow, determine T2, P2, and Ma2at a location where the flow area has been reduced by 20 percent.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 840

Converging–Diverging NozzlesWhen we think of nozzles, we ordinarily think of flow passages whosecross-sectional area decreases in the flow direction. However, the highestvelocity to which a fluid can be accelerated in a converging nozzle is limitedto the sonic velocity (Ma � 1), which occurs at the exit plane (throat) of thenozzle. Accelerating a fluid to supersonic velocities (Ma � 1) can be accom-plished only by attaching a diverging flow section to the subsonic nozzle atthe throat. The resulting combined flow section is a converging–divergingnozzle, which is standard equipment in supersonic aircraft and rocket propul-sion (Fig. 17–26).

Forcing a fluid through a converging–diverging nozzle is no guaranteethat the fluid will be accelerated to a supersonic velocity. In fact, the fluidmay find itself decelerating in the diverging section instead of acceleratingif the back pressure is not in the right range. The state of the nozzle flow isdetermined by the overall pressure ratio Pb/P0. Therefore, for given inletconditions, the flow through a converging–diverging nozzle is governed bythe back pressure Pb, as will be explained.

Chapter 17 | 841

Solution Nitrogen gas enters a converging nozzle. The properties at thenozzle exit are to be determined.Assumptions 1 Nitrogen is an ideal gas with k � 1.4. 2 Flow through thenozzle is steady, one-dimensional, and isentropic.Analysis The schematic of the duct is shown in Fig. 17–25. For isentropicflow through a duct, the area ratio A/A* (the flow area over the area of thethroat where Ma � 1) is also listed in Table A–32. At the initial Machnumber of Ma1 � 0.3, we read

With a 20 percent reduction in flow area, A2 � 0.8A1, and

For this value of A2/A* from Table A–32, we read

Here we chose the subsonic Mach number for the calculated A2/A* insteadof the supersonic one because the duct is converging in the flow directionand the initial flow is subsonic. Since the stagnation properties are constantfor isentropic flow, we can write

which are the temperature and pressure at the desired location.Discussion Note that the temperature and pressure drop as the fluid accel-erates in a converging nozzle.

P2

P1�

P2>P0

P1>P0 S P2 � P1 a P2>P0

P1>P0b � 1100 kPa 2 a 0.8993

0.9395b � 95.7 kPa

T2

T1�

T2>T0

T1>T0 S T2 � T1 a T2>T0

T1>T0b � 1400 K 2 a 0.9701

0.9823b � 395 K

T2

T0� 0.9701

P2

P0� 0.8993 Ma2 � 0.391

A2

A*�

A2

A1 A1

A*� 10.8 2 12.0351 2 � 1.6281

A1

A*� 2.0351

T1

T0� 0.9823

P1

P0� 0.9395

A1

P1 = 100 kPaT1 = 400 K

Ma1 = 0.3

P2

T2

Ma2

N2Nozzle

A2 = 0.8A1

FIGURE 17–25Schematic for Example 17–6 (not toscale).

cen84959_ch17.qxd 4/21/05 11:08 AM Page 841

Consider the converging–diverging nozzle shown in Fig. 17–27. A fluidenters the nozzle with a low velocity at stagnation pressure P0. When Pb �P0 (case A), there will be no flow through the nozzle. This is expected sincethe flow in a nozzle is driven by the pressure difference between the nozzleinlet and the exit. Now let us examine what happens as the back pressure islowered.

1. When P0 � Pb � PC, the flow remains subsonic throughout the nozzle,and the mass flow is less than that for choked flow. The fluid velocityincreases in the first (converging) section and reaches a maximum atthe throat (but Ma � 1). However, most of the gain in velocity is lostin the second (diverging) section of the nozzle, which acts as a dif-fuser. The pressure decreases in the converging section, reaches aminimum at the throat, and increases at the expense of velocity in thediverging section.

2. When Pb � PC, the throat pressure becomes P* and the fluid achievessonic velocity at the throat. But the diverging section of the nozzle stillacts as a diffuser, slowing the fluid to subsonic velocities. The massflow rate that was increasing with decreasing Pb also reaches its maxi-mum value.

Recall that P* is the lowest pressure that can be obtained at thethroat, and the sonic velocity is the highest velocity that can beachieved with a converging nozzle. Thus, lowering Pb further has noinfluence on the fluid flow in the converging part of the nozzle or the

842 | Thermodynamics

FIGURE 17–26Converging–diverging nozzles are commonly used in rocket engines to provide high thrust.

Courtesy of Pratt and Whitney, www.pratt-whitney.com/how.htm. Used by permission.

cen84959_ch17.qxd 4/26/05 4:17 PM Page 842

mass flow rate through the nozzle. However, it does influence the char-acter of the flow in the diverging section.

3. When PC � Pb � PE, the fluid that achieved a sonic velocity at thethroat continues accelerating to supersonic velocities in the divergingsection as the pressure decreases. This acceleration comes to a suddenstop, however, as a normal shock develops at a section between thethroat and the exit plane, which causes a sudden drop in velocity to sub-sonic levels and a sudden increase in pressure. The fluid then continuesto decelerate further in the remaining part of the converging–divergingnozzle. Flow through the shock is highly irreversible, and thus it cannotbe approximated as isentropic. The normal shock moves downstreamaway from the throat as Pb is decreased, and it approaches the nozzleexit plane as Pb approaches PE.

When Pb � PE, the normal shock forms at the exit plane of the noz-zle. The flow is supersonic through the entire diverging section in thiscase, and it can be approximated as isentropic. However, the fluidvelocity drops to subsonic levels just before leaving the nozzle as it

Chapter 17 | 843

0x

Subsonic flowat nozzle exit(shock in nozzle)

Exit

P*

Supersonic flowat nozzle exit(no shock in nozzle)

PA

A

Ma

ThroatInlet

B

C

D

}}}

Subsonic flowat nozzle exit(no shock)

Pe

x

Vi ≅ 0Pb

P0

E, F, GShockin nozzle

Sonic flowat throat

Throat ExitInlet0

1

Sonic flowat throat

Shockin nozzle

E, F, G

x

Subsonic flowat nozzle exit(shock in nozzle)

Supersonic flowat nozzle exit(no shock in nozzle)

AB

C

D

}}

Subsonic flowat nozzle exit(no shock)

P0

P

Throat

}PB

PC

PD

PE

PG

PF

Pb

FIGURE 17–27The effects of back pressure on theflow through a converging–divergingnozzle.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 843

crosses the normal shock. Normal shock waves are discussed inSection 17–5.

4. When PE � Pb � 0, the flow in the diverging section is supersonic,and the fluid expands to PF at the nozzle exit with no normal shockforming within the nozzle. Thus, the flow through the nozzle can beapproximated as isentropic. When Pb � PF, no shocks occur withinor outside the nozzle. When Pb � PF, irreversible mixing and expan-sion waves occur downstream of the exit plane of the nozzle. WhenPb � PF, however, the pressure of the fluid increases from PF to Pbirreversibly in the wake of the nozzle exit, creating what are calledoblique shocks.

844 | Thermodynamics

EXAMPLE 17–7 Airflow through a Converging–Diverging Nozzle

Air enters a converging–diverging nozzle, shown in Fig. 17–28, at 1.0 MPaand 800 K with a negligible velocity. The flow is steady, one-dimensional, andisentropic with k � 1.4. For an exit Mach number of Ma � 2 and a throatarea of 20 cm2, determine (a) the throat conditions, (b) the exit plane condi-tions, including the exit area, and (c) the mass flow rate through the nozzle.

Solution Air flows through a converging–diverging nozzle. The throat andthe exit conditions and the mass flow rate are to be determined.Assumptions 1 Air is an ideal gas with constant specific heats at roomtemperature. 2 Flow through the nozzle is steady, one-dimensional, andisentropic.Properties The specific heat ratio of air is given to be k � 1.4. The gasconstant of air is 0.287 kJ/kg � K.Analysis The exit Mach number is given to be 2. Therefore, the flow mustbe sonic at the throat and supersonic in the diverging section of thenozzle. Since the inlet velocity is negligible, the stagnation pressure andstagnation temperature are the same as the inlet temperature and pressure,P0 � 1.0 MPa and T0 � 800 K. The stagnation density is

(a) At the throat of the nozzle Ma � 1, and from Table A–32 we read

Thus,

Also,

� 517.5 m/s

V* � c* � 2kRT* � B 11.4 2 10.287 kJ>kg # K 2 1666.6 K 2 a 1000 m2>s2

1 kJ>kgb

r* � 0.6339r0 � 10.6339 2 14.355 kg>m3 2 � 2.761 kg/m3

T* � 0.8333T0 � 10.8333 2 1800 K 2 � 666.6 K

P* � 0.5283P0 � 10.5283 2 11.0 MPa 2 � 0.5283 MPa

P*

P0� 0.5283

T*

T0� 0.8333

r*

r0� 0.6339

r0 �P0

RT0�

1000 kPa

10.287 kPa # m3>kg # K 2 1800 K 2 � 4.355 kg>m3

At = 20 cm2

T0 = 800 K

Mae = 2

P0 = 1.0 MPa

Vi ≅ 0

FIGURE 17–28Schematic for Example 17–7.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 844

17–5 ■ SHOCK WAVES AND EXPANSION WAVESWe have seen that sound waves are caused by infinitesimally small pressuredisturbances, and they travel through a medium at the speed of sound. Wehave also seen that for some back pressure values, abrupt changes in fluidproperties occur in a very thin section of a converging–diverging nozzleunder supersonic flow conditions, creating a shock wave. It is of interest tostudy the conditions under which shock waves develop and how they affectthe flow.

Normal ShocksFirst we consider shock waves that occur in a plane normal to the directionof flow, called normal shock waves. The flow process through the shockwave is highly irreversible and cannot be approximated as being isentropic.

Next we follow the footsteps of Pierre Lapace (1749–1827), G. F. Bern-hard Riemann (1826–1866), William Rankine (1820–1872), Pierre HenryHugoniot (1851–1887), Lord Rayleigh (1842–1919), and G. I. Taylor

Chapter 17 | 845

(b) Since the flow is isentropic, the properties at the exit plane can also becalculated by using data from Table A–32. For Ma � 2 we read

Thus,

and

The nozzle exit velocity could also be determined from Ve � Maece, where ceis the speed of sound at the exit conditions:

(c) Since the flow is steady, the mass flow rate of the fluid is the same at allsections of the nozzle. Thus it may be calculated by using properties at anycross section of the nozzle. Using the properties at the throat, we find thatthe mass flow rate is

Discussion Note that this is the highest possible mass flow rate that canflow through this nozzle for the specified inlet conditions.

m#

� r*A*V* � 12.761 kg>m3 2 120 � 10�4 m2 2 1517.5 m>s 2 � 2.86 kg/s

� 845.2 m>s Ve � Maece � Mae2kRTe � 2B 11.4 2 10.287 kJ>kg # K 2 1444.5 K 2 a 1000 m2>s2

1 kJ>kgb

Ve � Mae*c* � 11.6330 2 1517.5 m>s 2 � 845.1 m/s

Ae � 1.6875A* � 11.6875 2 120 cm2 2 � 33.75 cm2

re � 0.2300r0 � 10.2300 2 14.355 kg>m3 2 � 1.002 kg/m3

Te � 0.5556T0 � 10.5556 2 1800 K 2 � 444.5 K

Pe � 0.1278P0 � 10.1278 2 110 MPa 2 � 0.1278 MPa

Pe

P0� 0.1278

Te

T0� 0.5556

re

r0� 0.2300 Mat* � 1.6330

Ae

A*� 1.6875

cen84959_ch17.qxd 4/21/05 11:08 AM Page 845

(1886–1975) and develop relationships for the flow properties before andafter the shock. We do this by applying the conservation of mass, momen-tum, and energy relations as well as some property relations to a stationarycontrol volume that contains the shock, as shown in Fig. 17–29. The normalshock waves are extremely thin, so the entrance and exit flow areas for thecontrol volume are approximately equal (Fig 17–30).

We assume steady flow with no heat and work interactions and nopotential energy changes. Denoting the properties upstream of the shockby the subscript 1 and those downstream of the shock by 2, we have thefollowing:

Conservation of mass: (17–29)

or

Conservation of energy: (17–30)

or

(17–31)

Conservation of momentum: Rearranging Eq. 17–14 and integrating yield

(17–32)

Increase of entropy: (17–33)

We can combine the conservation of mass and energy relations into a sin-gle equation and plot it on an h-s diagram, using property relations. Theresultant curve is called the Fanno line, and it is the locus of states thathave the same value of stagnation enthalpy and mass flux (mass flow perunit flow area). Likewise, combining the conservation of mass and momen-tum equations into a single equation and plotting it on the h-s diagram yielda curve called the Rayleigh line. Both these lines are shown on the h-s dia-gram in Fig. 17–31. As proved later in Example 17–8, the points of maxi-mum entropy on these lines (points a and b) correspond to Ma � 1. Thestate on the upper part of each curve is subsonic and on the lower partsupersonic.

The Fanno and Rayleigh lines intersect at two points (points 1 and 2),which represent the two states at which all three conservation equations aresatisfied. One of these (state 1) corresponds to the state before the shock,and the other (state 2) corresponds to the state after the shock. Note thatthe flow is supersonic before the shock and subsonic afterward. Thereforethe flow must change from supersonic to subsonic if a shock is to occur.The larger the Mach number before the shock, the stronger the shock willbe. In the limiting case of Ma � 1, the shock wave simply becomes a soundwave. Notice from Fig. 17–31 that s2 � s1. This is expected since the flowthrough the shock is adiabatic but irreversible.

s2 � s1 � 0

A 1P1 � P2 2 � m# 1V2 � V1 2

h01 � h02

h1 �V 2

1

2� h2 �

V 22

2

r1V1 � r2V2

r1AV1 � r2AV2

846 | Thermodynamics

Controlvolume

Flow

Ma1 > 1 PV1

s

Shock wave

PV2

1 2

1 2

1 2

1 2

hh

sr r

Ma2 < 1

FIGURE 17–29Control volume for flow across anormal shock wave.

FIGURE 17–30Schlieren image of a normal shock ina Laval nozzle. The Mach number inthe nozzle just upstream (to the left) ofthe shock wave is about 1.3. Boundarylayers distort the shape of the normalshock near the walls and lead to flowseparation beneath the shock.

Photo by G. S. Settles, Penn State University. Usedby permission.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 846

The conservation of energy principle (Eq. 17–31) requires that the stagna-tion enthalpy remain constant across the shock; h01 � h02. For ideal gasesh � h(T ), and thus

(17–34)

That is, the stagnation temperature of an ideal gas also remains constantacross the shock. Note, however, that the stagnation pressure decreasesacross the shock because of the irreversibilities, while the thermodynamictemperature rises drastically because of the conversion of kinetic energyinto enthalpy due to a large drop in fluid velocity (see Fig. 17–32).

We now develop relations between various properties before and after theshock for an ideal gas with constant specific heats. A relation for the ratio ofthe thermodynamic temperatures T2/T1 is obtained by applying Eq. 17–18twice:

Dividing the first equation by the second one and noting that T01 � T02, wehave

(17–35)

From the ideal-gas equation of state,

Substituting these into the conservation of mass relation r1V1 � r2V2 andnoting that Ma � V/c and , we have

(17–36)T2

T1�

P2V2

P1V1�

P2Ma2c2

P1Ma1c1�

P2Ma22T2

P1Ma12T1

� a P2

P1b 2 aMa2

Ma1b 2

c � 1kRT

r1 �P1

RT1and r2 �

P2

RT2

T2

T1�

1 � Ma21 1k � 1 2 >2

1 � Ma22 1k � 1 2 >2

T01

T1� 1 � a k � 1

2bMa2

1 andT02

T2� 1 � a k � 1

2bMa2

2

T01 � T02

Chapter 17 | 847

Ma = 1

0

s

SHO

CK

WAV

E

Subsonic flow

h

h

a

h01

1

1

2

2

= h02

h02h01

P 02

P 01

Ma = 1b

V

2

22

2

(Ma < 1)

Supersonic flow(Ma > 1)h 1

s s

V 2

1

2

Fanno lin

e

Rayleigh line

FIGURE 17–31The h-s diagram for flow across anormal shock.

Normalshock

PP0V

MaTT0r

s

increasesdecreasesdecreasesdecreasesincreasesremains constantincreasesincreases

FIGURE 17–32Variation of flow properties across anormal shock.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 847

Combining Eqs. 17–35 and 17–36 gives the pressure ratio across the shock:

(17–37)

Equation 17–37 is a combination of the conservation of mass and energyequations; thus, it is also the equation of the Fanno line for an ideal gas withconstant specific heats. A similar relation for the Rayleigh line can beobtained by combining the conservation of mass and momentum equations.From Eq. 17–32,

However,

Thus,

or

(17–38)

Combining Eqs. 17–37 and 17–38 yields

(17–39)

This represents the intersections of the Fanno and Rayleigh lines and relatesthe Mach number upstream of the shock to that downstream of the shock.

The occurrence of shock waves is not limited to supersonic nozzles only.This phenomenon is also observed at the engine inlet of a supersonic air-craft, where the air passes through a shock and decelerates to subsonicvelocities before entering the diffuser of the engine. Explosions also pro-duce powerful expanding spherical normal shocks, which can be verydestructive (Fig. 17–33).

Various flow property ratios across the shock are listed in Table A–33 foran ideal gas with k � 1.4. Inspection of this table reveals that Ma2 (theMach number after the shock) is always less than 1 and that the larger thesupersonic Mach number before the shock, the smaller the subsonic Machnumber after the shock. Also, we see that the static pressure, temperature,and density all increase after the shock while the stagnation pressuredecreases.

The entropy change across the shock is obtained by applying the entropy-change equation for an ideal gas across the shock:

(17–40)

which can be expressed in terms of k, R, and Ma1 by using the relationsdeveloped earlier in this section. A plot of nondimensional entropy change

s2 � s1 � cp ln T2

T1� R ln

P2

P1

Ma22 �

Ma21 � 2> 1k � 1 2

2Ma21k> 1k � 1 2 � 1

P2

P1�

1 � kMa21

1 � kMa22

P1 11 � kMa21 2 � P2 11 � kMa2

2 2

rV 2 � a P

RTb 1Mac 2 2 � a P

RTb 1Ma2kRT 2 2 � PkMa2

P1 � P2 �m#

A 1V2 � V1 2 � r2V

22 � r1V

21

P2

P1�

Ma121 � Ma21 1k � 1 2 >2

Ma221 � Ma22 1k � 1 2 >2

848 | Thermodynamics

cen84959_ch17.qxd 4/21/05 11:08 AM Page 848

across the normal shock (s2 � s1)/R versus Ma1 is shown in Fig. 17–34.Since the flow across the shock is adiabatic and irreversible, the second lawrequires that the entropy increase across the shock wave. Thus, a shockwave cannot exist for values of Ma1 less than unity where the entropychange would be negative. For adiabatic flows, shock waves can exist onlyfor supersonic flows, Ma1 � 1.

Chapter 17 | 849

FIGURE 17–33Schlieren image of the blast wave(expanding spherical normal shock)produced by the explosion of afirecracker detonated inside a metal canthat sat on a stool. The shock expandedradially outward in all directions at asupersonic speed that decreased withradius from the center of the explosion.The microphone at the lower rightsensed the sudden change in pressureof the passing shock wave andtriggered the microsecond flashlampthat exposed the photograph.

Photo by G. S. Settles, Penn State University. Usedby permission.

0

Impossible

Subsonic flowbefore shock

Ma1 Ma1= 1 Supersonic flowbefore shock

s2 – s1 < 0

s2 – s1 > 0

(s2 � s1)/R

FIGURE 17–34Entropy change across the normalshock.

EXAMPLE 17–8 The Point of Maximum Entropy on the Fanno Line

Show that the point of maximum entropy on the Fanno line (point b of Fig.17–31) for the adiabatic steady flow of a fluid in a duct corresponds to thesonic velocity, Ma � 1.

Solution It is to be shown that the point of maximum entropy on the Fannoline for steady adiabatic flow corresponds to sonic velocity.Assumptions The flow is steady, adiabatic, and one-dimensional.Analysis In the absence of any heat and work interactions and potentialenergy changes, the steady-flow energy equation reduces to

Differentiating yields

For a very thin shock with negligible change of duct area across the shock, thesteady-flow continuity (conservation of mass) equation can be expressed as

rV � constant

dh � V dV � 0

h �V 2

2� constant

cen84959_ch17.qxd 4/21/05 11:08 AM Page 849

850 | Thermodynamics

Differentiating, we have

Solving for dV gives

Combining this with the energy equation, we have

which is the equation for the Fanno line in differential form. At point a(the point of maximum entropy) ds � 0. Then from the second T ds relation(T ds � dh � v dP) we have dh � v dP � dP/r. Substituting yields

Solving for V, we have

which is the relation for the speed of sound, Eq. 17–9. Thus the proof iscomplete.

V � a 0P

0rb 1>2

s

dPr

� V2 dr

r� 0 at s � constant

dh � V2 dr

r� 0

dV � �V dr

r

r dV � V dr � 0

EXAMPLE 17–9 Shock Wave in a Converging–Diverging Nozzle

If the air flowing through the converging–diverging nozzle of Example 17–7experiences a normal shock wave at the nozzle exit plane (Fig. 17–35), deter-mine the following after the shock: (a) the stagnation pressure, static pres-sure, static temperature, and static density; (b) the entropy change across theshock; (c) the exit velocity; and (d ) the mass flow rate through the nozzle.Assume steady, one-dimensional, and isentropic flow with k � 1.4 from thenozzle inlet to the shock location.

Solution Air flowing through a converging–diverging nozzle experiences anormal shock at the exit. The effect of the shock wave on various propertiesis to be determined.Assumptions 1 Air is an ideal gas with constant specific heats at roomtemperature. 2 Flow through the nozzle is steady, one-dimensional, andisentropic before the shock occurs. 3 The shock wave occurs at the exitplane.Properties The constant-pressure specific heat and the specific heat ratio ofair are cp � 1.005 kJ/kg · K and k � 1.4. The gas constant of air is 0.287kJ/kg � K (Table A–2a).Analysis (a) The fluid properties at the exit of the nozzle just before theshock (denoted by subscript 1) are those evaluated in Example 17–7 at thenozzle exit to be

P01 � 1.0 MPa P1 � 0.1278 MPa T1 � 444.5 K r1 � 1.002 kg>m3

T = 444.5 K

Ma1 = 2P01

1

1

1

= 1.0 MPa

P = 0.1278 MPa

= 1.002 kg/mr 3

Shock wave

1 2m = 2.86 kg/s·

FIGURE 17–35Schematic for Example 17–9.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 850

Example 17–9 illustrates that the stagnation pressure and velocitydecrease while the static pressure, temperature, density, and entropyincrease across the shock. The rise in the temperature of the fluid down-stream of a shock wave is of major concern to the aerospace engineerbecause it creates heat transfer problems on the leading edges of wings andnose cones of space reentry vehicles and the recently proposed hypersonicspace planes. Overheating, in fact, led to the tragic loss of the space shuttleColumbia in February of 2003 as it was reentering earth’s atmosphere.

Chapter 17 | 851

The fluid properties after the shock (denoted by subscript 2) are relatedto those before the shock through the functions listed in Table A–33. ForMa1 � 2.0, we read

Then the stagnation pressure P02, static pressure P2, static temperature T2,and static density r2 after the shock are

(b) The entropy change across the shock is

Thus, the entropy of the air increases as it experiences a normal shock,which is highly irreversible.

(c) The air velocity after the shock can be determined from V2 � Ma2c2,where c2 is the speed of sound at the exit conditions after the shock:

(d) The mass flow rate through a converging–diverging nozzle with sonicconditions at the throat is not affected by the presence of shock waves inthe nozzle. Therefore, the mass flow rate in this case is the same as thatdetermined in Example 17–7:

Discussion This result can easily be verified by using property values at thenozzle exit after the shock at all Mach numbers significantly greater than unity.

m#

� 2.86 kg/s

� 317 m/s

� 10.5774 2B 11.4 2 10.287 kJ>kg # K 2 1750 K 2 a 1000 m2>s2

1 kJ>kgb

V2 � Ma2c2 � Ma22kRT2

� 0.0942 kJ/kg # K

� 11.005 kJ>kg # K 2 ln 11.6875 2 � 10.287 kJ>kg # K 2 ln 14.5000 2 s2 � s1 � cp ln

T2

T1� R ln

P2

P1

r2 � 2.6667r1 � 12.6667 2 11.002 kg>m3 2 � 2.67 kg/m3

T2 � 1.6875T1 � 11.6875 2 1444.5 K 2 � 750 K

P2 � 4.5000P1 � 14.5000 2 10.1278 MPa 2 � 0.575 MPa

P02 � 0.7209P01 � 10.7209 2 11.0 MPa 2 � 0.721 MPa

Ma2 � 0.5774P02

P01� 0.7209

P2

P1� 4.5000

T2

T1� 1.6875

r2

r1� 2.6667

cen84959_ch17.qxd 4/21/05 11:08 AM Page 851

Oblique ShocksNot all shock waves are normal shocks (perpendicular to the flow direc-tion). For example, when the space shuttle travels at supersonic speedsthrough the atmosphere, it produces a complicated shock pattern consistingof inclined shock waves called oblique shocks (Fig. 17–36). As you cansee, some portions of an oblique shock are curved, while other portions arestraight.

First, we consider straight oblique shocks, like that produced when a uni-form supersonic flow (Ma1 � 1) impinges on a slender, two-dimensionalwedge of half-angle d (Fig. 17–37). Since information about the wedgecannot travel upstream in a supersonic flow, the fluid “knows” nothingabout the wedge until it hits the nose. At that point, since the fluid cannotflow through the wedge, it turns suddenly through an angle called theturning angle or deflection angle u. The result is a straight oblique shockwave, aligned at shock angle or wave angle b, measured relative to theoncoming flow (Fig. 17–38). To conserve mass, b must obviously begreater than d. Since the Reynolds number of supersonic flows is typicallylarge, the boundary layer growing along the wedge is very thin, and weignore its effects. The flow therefore turns by the same angle as the wedge;namely, deflection angle u is equal to wedge half-angle d. If we take intoaccount the displacement thickness effect of the boundary layer, the deflec-tion angle u of the oblique shock turns out to be slightly greater thanwedge half-angle d.

Like normal shocks, the Mach number decreases across an oblique shock,and oblique shocks are possible only if the upstream flow is supersonic.However, unlike normal shocks, in which the downstream Mach number isalways subsonic, Ma2 downstream of an oblique shock can be subsonic,sonic, or supersonic, depending on the upstream Mach number Ma1 and theturning angle.

852 | Thermodynamics

FIGURE 17–36Schlieren image of a small model ofthe space shuttle Orbiter being testedat Mach 3 in the supersonic windtunnel of the Penn State GasDynamics Lab. Several oblique shocksare seen in the air surrounding thespacecraft.

Photo by G. S. Settles, Penn State University. Usedby permission.

db

uMa1

Ma1

Ma2

Obliqueshock

FIGURE 17–37An oblique shock of shock angle bformed by a slender, two-dimensionalwedge of half-angle d. The flow isturned by deflection angle udownstream of the shock, and theMach number decreases.

cen84959_ch17.qxd 4/27/05 11:00 AM Page 852

We analyze a straight oblique shock in Fig. 17–38 by decomposing thevelocity vectors upstream and downstream of the shock into normal and tan-gential components, and considering a small control volume around theshock. Upstream of the shock, all fluid properties (velocity, density, pres-sure, etc.) along the lower left face of the control volume are identical tothose along the upper right face. The same is true downstream of the shock.Therefore, the mass flow rates entering and leaving those two faces canceleach other out, and conservation of mass reduces to

(17–41)

where A is the area of the control surface that is parallel to the shock. SinceA is the same on either side of the shock, it has dropped out of Eq. 17–41.

As you might expect, the tangential component of velocity (parallel to theoblique shock) does not change across the shock (i.e., V1,t � V2,t). This iseasily proven by applying the tangential momentum equation to the controlvolume.

When we apply conservation of momentum in the direction normal to theoblique shock, the only forces are pressure forces, and we get

(17–42)

Finally, since there is no work done by the control volume and no heattransfer into or out of the control volume, stagnation enthalpy does notchange across an oblique shock, and conservation of energy yields

But since V1,t � V2,t, this equation reduces to

(17–43)

Careful comparison reveals that the equations for conservation of mass,momentum, and energy (Eqs. 17–41 through 17–43) across an obliqueshock are identical to those across a normal shock, except that they are writ-ten in terms of the normal velocity component only. Therefore, the normalshock relations derived previously apply to oblique shocks as well, but mustbe written in terms of Mach numbers Ma1,n and Ma2,n normal to the obliqueshock. This is most easily visualized by rotating the velocity vectors inFig. 17–38 by angle p/2 � b, so that the oblique shock appears to be verti-cal (Fig. 17–39). Trigonometry yields

(17–44)

where Ma1,n � V1,n /c1 and Ma2,n � V2,n /c2. From the point of view shownin Fig. 17–40, we see what looks like a normal shock, but with some super-posed tangential flow “coming along for the ride.” Thus,

All the equations, shock tables, etc., for normal shocks apply to oblique shocksas well, provided that we use only the normal components of the Mach number.

In fact, you may think of normal shocks as special oblique shocks inwhich shock angle b � p/2, or 90°. We recognize immediately that anoblique shock can exist only if Ma1,n � 1, and Ma2,n � 1. The normal shock

Ma1,n � Ma1 sin b and Ma2,n � Ma2 sin 1b � u 2

h1 �1

2 V 2

1,n � h2 �1

2 V 2

2,n

h01 � h02 � h0 S h1 �1

2V 2

1,n �1

2V 2

1,t � h2 �1

2V 2

2,n �1

2V 2

2,t

P1 A � P2A � rV2,n AV2,n � rV1,n AV1,n S P1 � P2 � r2V22,n � r1V

21,n

r1V1,n A � r2V2,n A S r1V1,n � r2V2,n

Chapter 17 | 853

→

V2,n

Obliqueshock

Controlvolume

V1,n

V1

→V2

V1,tP1 P2

V2,t

u

b

FIGURE 17–38Velocity vectors through an obliqueshock of shock angle b and deflectionangle u.

V1,n

P1 P2

P1 P2

V1

V1,t

Ma1,n � 1Ma2,n � 1

Obliqueshock

V2,n

V2,t

u

b

b� u

→

V2→

FIGURE 17–39The same velocity vectors of Fig.17–38, but rotated by angle p/2 – b,so that the oblique shock is vertical.Normal Mach numbers Ma1,n andMa2,n are also defined.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 853

equations appropriate for oblique shocks in an ideal gas are summarized inFig. 17–40 in terms of Ma1,n.

For known shock angle b and known upstream Mach number Ma1, we usethe first part of Eq. 17–44 to calculate Ma1,n, and then use the normal shocktables (or their corresponding equations) to obtain Ma2,n. If we also knew thedeflection angle u, we could calculate Ma2 from the second part of Eq. 17–44.But, in a typical application, we know either b or u, but not both. Fortunately,a bit more algebra provides us with a relationship between u, b, and Ma1. Webegin by noting that tan b � V1,n/V1,t and tan(b � u) � V2,n /V2,t (Fig. 17–39).But since V1,t � V2,t, we combine these two expressions to yield

(17–45)

where we have also used Eq. 17–44 and the fourth equation of Fig. 17–40.We apply trigonometric identities for cos 2b and tan(b � u), namely,

After some algebra, Eq. 17–45 reduces to

The u-b-Ma relationship: (17–46)

Equation 17–46 provides deflection angle u as a unique function ofshock angle b, specific heat ratio k, and upstream Mach number Ma1. Forair (k � 1.4), we plot u versus b for several values of Ma1 in Fig. 17–41.We note that this plot is often presented with the axes reversed (b versus u)in compressible flow textbooks, since, physically, shock angle b is deter-mined by deflection angle u.

Much can be learned by studying Fig. 17–41, and we list some observa-tions here:

• Figure 17–41 displays the full range of possible shock waves at a givenfree-stream Mach number, from the weakest to the strongest. For any valueof Mach number Ma1 greater than 1, the possible values of u range from u� 0° at some value of b between 0 and 90°, to a maximum value u� umaxat an intermediate value of b, and then back to u� 0° at b� 90°. Straightoblique shocks for u or b outside of this range cannot and do not exist. AtMa1 � 1.5, for example, straight oblique shocks cannot exist in air withshock angle b less than about 42°, nor with deflection angle u greater thanabout 12°. If the wedge half-angle is greater than umax, the shock becomescurved and detaches from the nose of the wedge, forming what is called adetached oblique shock or a bow wave (Fig. 17–42). The shock angle bof the detached shock is 90° at the nose, but b decreases as the shock curvesdownstream. Detached shocks are much more complicated than simplestraight oblique shocks to analyze. In fact, no simple solutions exist, andprediction of detached shocks requires computational methods.

• Similar oblique shock behavior is observed in axisymmetric flow overcones, as in Fig. 17–43, although the u-b-Ma relationship foraxisymmetric flows differs from that of Eq. 17–46.

tan u �2 cot b 1Ma2

1 sin2 b � 1 2Ma2

1 1k � cos 2b 2 � 2

cos 2b � cos2 b � sin2 b and tan 1b � u 2 �tan b � tan u

1 � tan b tan u

V2,n

V1,n�

tan 1b � u 2tan b

�2 � 1k � 1 2Ma2

1,n

1k � 1 2Ma21,n

�2 � 1k � 1 2Ma2

1 sin2 b

1k � 1 2Ma21 sin2 b

854 | Thermodynamics

P0202

P0101� c (k � 1)Ma1)Ma 1,1, n

2

2 � (k �1)Ma1)Ma 1,1, n2dk /(/(k�1)1) c (k � 1)1)

2k Ma Ma 21,1, n� k � 1

d1/1/(k�1)1)

T2

T1� [2[2 � (k � 1)Ma1)Ma1, 1, n

2 ] ] 2k Ma Ma 1, 1, n

2 � k � 1

(k � 1)1)2MaMa 1, 1, n2

r2

r1�

V1, 1, n

V2, 2, n�

(k � 1)Ma1)Ma 1, 1, n2

2 � (k � 1)Ma1)Ma 1, 1, n2

P2

P1�

2k Ma Ma 1, 1, n2 � k � 1

k � 1

Ma Ma 2, 2, n �B(k � 1)Ma1)Ma 1, 1, n

2 � 2

2k Ma Ma 1, 1, n2 � k � 1

h0101 � h0202 → T0101 � T0202

FIGURE 17–40Relationships across an obliqueshock for an ideal gas in terms of thenormal component of upstream Machnumber Ma1,n.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 854

• When supersonic flow impinges on a blunt body—a body without asharply pointed nose, the wedge half-angle d at the nose is 90°, and anattached oblique shock cannot exist, regardless of Mach number. In fact,a detached oblique shock occurs in front of all such blunt-nosed bodies,whether two-dimensional, axisymmetric, or fully three-dimensional. Forexample, a detached oblique shock is seen in front of the space shuttlemodel in Fig. 17–36 and in front of a sphere in Fig. 17–44.

• While u is a unique function of Ma1 and b for a given value of k, there aretwo possible values of b for u � umax. The dashed black line in Fig. 17–41passes through the locus of umax values, dividing the shocks into weakoblique shocks (the smaller value of b) and strong oblique shocks (thelarger value of b). At a given value of u, the weak shock is more commonand is “preferred” by the flow unless the downstream pressure conditionsare high enough for the formation of a strong shock.

• For a given upstream Mach number Ma1, there is a unique value of u forwhich the downstream Mach number Ma2 is exactly 1. The dashed grayline in Fig. 17–41 passes through the locus of values where Ma2 � 1.To the left of this line, Ma2 � 1, and to the right of this line, Ma2 � 1.Downstream sonic conditions occur on the weak shock side of the plot,with u very close to umax. Thus, the flow downstream of a strong obliqueshock is always subsonic (Ma2 � 1). The flow downstream of a weakoblique shock remains supersonic, except for a narrow range of u justbelow umax, where it is subsonic, although it is still called a weakoblique shock.

• As the upstream Mach number approaches infinity, straight obliqueshocks become possible for any b between 0 and 90°, but the maximumpossible turning angle for k � 1.4 (air) is umax � 45.6°, which occurs at b� 67.8°. Straight oblique shocks with turning angles above this value ofumax are not possible, regardless of the Mach number.

• For a given value of upstream Mach number, there are two shock angleswhere there is no turning of the flow (u � 0°): the strong case, b � 90°,

Chapter 17 | 855

0 10 20 30 40

b, degrees

u, d

egre

es

Ma2 � 1

Ma2 � 1

→Ma1 �

u � umax

Weak

50

1.21.52310 5

60 70 80 900

10

20

30

40

50

Strong

Ma2 � 1

FIGURE 17–41The dependence of straight obliqueshock deflection angle u on shockangle b for several values of upstreamMach number Ma1. Calculations arefor an ideal gas with k � 1.4. Thedashed black line connects points ofmaximum deflection angle (u � umax).Weak oblique shocks are to the left ofthis line, while strong oblique shocksare to the right of this line. The dashedgray line connects points where thedownstream Mach number is sonic(Ma2 � 1). Supersonic downstreamflow (Ma2 � 1) is to the left of thisline, while subsonic downstream flow(Ma2 � 1) is to the right of this line.

Ma1

Detachedobliqueshock

d � umax

FIGURE 17–42A detached oblique shock occursupstream of a two-dimensional wedgeof half-angle d when d is greater thanthe maximum possible deflectionangle u. A shock of this kind is calleda bow wave because of itsresemblance to the water wave thatforms at the bow of a ship.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 855

corresponds to a normal shock, and the weak case, b� bmin, representsthe weakest possible oblique shock at that Mach number, which is calleda Mach wave. Mach waves are caused, for example, by very smallnonuniformities on the walls of a supersonic wind tunnel (several can beseen in Figs. 17–36 and 17–43). Mach waves have no effect on the flow,since the shock is vanishingly weak. In fact, in the limit, Mach waves areisentropic. The shock angle for Mach waves is a unique function of theMach number and is given the symbol m, not to be confused with thecoefficient of viscosity. Angle m is called the Mach angle and is found bysetting u equal to zero in Eq. 17–46, solving for b� m, and taking thesmaller root. We get

Mach angle: (17–47)

Since the specific heat ratio appears only in the denominator of Eq.17–46, m is independent of k. Thus, we can estimate the Mach number ofany supersonic flow simply by measuring the Mach angle and applyingEq. 17–47.

Prandtl–Meyer Expansion WavesWe now address situations where supersonic flow is turned in the oppositedirection, such as in the upper portion of a two-dimensional wedge at anangle of attack greater than its half-angle d (Fig. 17–45). We refer to thistype of flow as an expanding flow, whereas a flow that produces an obliqueshock may be called a compressing flow. As previously, the flow changesdirection to conserve mass. However, unlike a compressing flow, an expand-ing flow does not result in a shock wave. Rather, a continuous expandingregion called an expansion fan appears, composed of an infinite number ofMach waves called Prandtl–Meyer expansion waves. In other words, theflow does not turn suddenly, as through a shock, but gradually—each suc-cessive Mach wave turns the flow by an infinitesimal amount. Since eachindividual expansion wave is isentropic, the flow across the entire expansionfan is also isentropic. The Mach number downstream of the expansionincreases (Ma2 � Ma1), while pressure, density, and temperature decrease,just as they do in the supersonic (expanding) portion of a converging–diverging nozzle.

Prandtl–Meyer expansion waves are inclined at the local Mach angle m,as sketched in Fig. 17–45. The Mach angle of the first expansion wave iseasily determined as m1 � sin�1(1/Ma1). Similarly, m2 � sin�1(1/Ma2),

m � sin�1 11>Ma1 2

856 | Thermodynamics

Ma1

d

FIGURE 17–43Still frames from schlierenvideography illustrating thedetachment of an oblique shock from acone with increasing cone half-angle din air at Mach 3. At (a) d� 20 and(b) d� 40, the oblique shock remainsattached, but by (c) d� 60, theoblique shock has detached, forming a bow wave.

Photos by G. S. Settles, Penn State University. Used by permission.

FIGURE 17–44Shadowgram of a one-half-in diametersphere in free flight through air at Ma� 1.53. The flow is subsonic behindthe part of the bow wave that is aheadof the sphere and over its surface backto about 45. At about 90 the laminarboundary layer separates through anoblique shock wave and quicklybecomes turbulent. The fluctuatingwake generates a system of weakdisturbances that merge into thesecond “recompression” shock wave.

Photo by A. C. Charters, Army Ballistic ResearchLaboratory.

(a) (b) (c)

cen84959_ch17.qxd 4/21/05 11:08 AM Page 856

where we must be careful to measure the angle relative to the new directionof flow downstream of the expansion, namely, parallel to the upper wall ofthe wedge in Fig. 17–45 if we neglect the influence of the boundary layeralong the wall. But how do we determine Ma2? It turns out that the turningangle u across the expansion fan can be calculated by integration, makinguse of the isentropic flow relationships. For an ideal gas, the result is(Anderson, 2003),

Turning angle across an expansion fan: (17–48)

where n(Ma) is an angle called the Prandtl–Meyer function (not to be con-fused with the kinematic viscosity),

(17–49)

Note that n(Ma) is an angle, and can be calculated in either degrees or radi-ans. Physically, n(Ma) is the angle through which the flow must expand,starting with n � 0 at Ma � 1, in order to reach a supersonic Mach number,Ma � 1.

To find Ma2 for known values of Ma1, k, and u, we calculate n(Ma1) fromEq. 17–49, n(Ma2) from Eq. 17–48, and then Ma2 from Eq. 17–49, notingthat the last step involves solving an implicit equation for Ma2. Since thereis no heat transfer or work, and the flow is isentropic through the expansion,T0 and P0 remain constant, and we use the isentropic flow relations derivedpreviously to calculate other flow properties downstream of the expansion,such as T2, r2, and P2.

Prandtl–Meyer expansion fans also occur in axisymmetric supersonicflows, as in the corners and trailing edges of a cone-cylinder (Fig. 17–46).Some very complex and, to some of us, beautiful interactions involvingboth shock waves and expansion waves occur in the supersonic jet pro-duced by an “overexpanded” nozzle, as in Fig. 17–47. Analysis of suchflows is beyond the scope of the present text; interested readers are referredto compressible flow textbooks such as Thompson (1972) and Anderson(2003).

n 1Ma 2 � Bk � 1

k � 1 tan�1 cB

k � 1

k � 11Ma2 � 1 2 d � tan�1 a2Ma2 � 1 b

u � n 1Ma2 2 � n 1Ma1 2

Chapter 17 | 857

FIGURE 17–46A cone-cylinder of 12.5 half-angle ina Mach number 1.84 flow. Theboundary layer becomes turbulentshortly downstream of the nose,generating Mach waves that are visiblein this shadowgraph. Expansion wavesare seen at the corners and at thetrailing edge of the cone.

Photo by A. C. Charters, Army Ballistic ResearchLaboratory.

d

uMa1 � 1

m1m2

Ma2

Expansionwaves

Obliqueshock

FIGURE 17–45An expansion fan in the upper portion of the flow formed by a two-dimensional wedge at the angle ofattack in a supersonic flow. The flowis turned by angle u, and the Machnumber increases across the expansionfan. Mach angles upstream anddownstream of the expansion fan areindicated. Only three expansion wavesare shown for simplicity, but in fact,there are an infinite number of them.(An oblique shock is present in thebottom portion of this flow.)

cen84959_ch17.qxd 4/21/05 11:08 AM Page 857

858 | Thermodynamics

FIGURE 17–47The complex interactions betweenshock waves and expansion waves inan “overexpanded” supersonic jet.The flow is visualized by a schlieren-like differential interferogram.

Photo by H. Oertel sen. Reproduced by courtesy ofthe French-German Research Institute of Saint-Louis, ISL. Used with permission.

EXAMPLE 17–10 Estimation of the Mach Number from Mach Lines

Estimate the Mach number of the free-stream flow upstream of the spaceshuttle in Fig. 17–36 from the figure alone. Compare with the known valueof Mach number provided in the figure caption.

Solution We are to estimate the Mach number from a figure and compareit to the known value.Analysis Using a protractor, we measure the angle of the Mach lines in thefree-stream flow: m � 19°. The Mach number is obtained from Eq. 17–47,

Our estimated Mach number agrees with the experimental value of 3.0 � 0.1.Discussion The result is independent of the fluid properties.

m � sin�1 a 1

Ma1b S Ma

1�

1

sin 19° S Ma1 � 3.07

EXAMPLE 17–11 Oblique Shock Calculations

Supersonic air at Ma1 � 2.0 and 75.0 kPa impinges on a two-dimensionalwedge of half-angle d � 10° (Fig. 17–48). Calculate the two possibleoblique shock angles, bweak and bstrong, that could be formed by this wedge.For each case, calculate the pressure and Mach number downstream of theoblique shock, compare, and discuss.

Solution We are to calculate the shock angle, Mach number, and pressuredownstream of the weak and strong oblique shocks formed by a two-dimensional wedge.Assumptions 1 The flow is steady. 2 The boundary layer on the wedge isvery thin.Properties The fluid is air with k � 1.4.

Ma1

Strongshock

d � 10°

bstrong

(a)

(b)

Ma1

Weakshock

d � 10°bweak

FIGURE 17–48Two possible oblique shock angles,(a) bweak and (b) bstrong, formed by atwo-dimensional wedge of half-angled � 10�.

cen84959_ch17.qxd 4/27/05 5:57 PM Page 858

Chapter 17 | 859

Analysis Because of assumption 2, we approximate the oblique shockdeflection angle as equal to the wedge half-angle, i.e., u � d � 10°. WithMa1 � 2.0 and u � 10°, we solve Eq. 17–46 for the two possible values ofoblique shock angle b: Bweak � 39.3° and Bstrong � 83.7°. From these values,we use the first part of Eq. 17–44 to calculate the upstream normal Machnumber Ma1,n,

Weak shock:

Strong shock:

We substitute these values of Ma1,n into the second equation of Fig. 17–40to calculate the downstream normal Mach number Ma2,n. For the weakshock, Ma2,n � 0.8032, and for the strong shock, Ma2,n � 0.5794. We alsocalculate the downstream pressure for each case, using the third equation ofFig. 17–40, which gives

Weak shock:

Strong shock:

Finally, we use the second part of Eq. 17–44 to calculate the downstreamMach number,

Weak shock:

Strong shock:

The changes in Mach number and pressure across the strong shock aremuch greater than the changes across the weak shock, as expected.Discussion Since Eq. 17–46 is implicit in b, we solve it by an iterativeapproach or with an equation solver such as EES. For both the weak andstrong oblique shock cases, Ma1,n is supersonic and Ma2,n is subsonic. How-ever, Ma2 is supersonic across the weak oblique shock, but subsonic acrossthe strong oblique shock. We could also use the normal shock tables inplace of the equations, but with loss of precision.

Ma2 �Ma2,n

sin 1b � u 2 �0.5794

sin 183.7° � 10° 2 � 0.604

Ma2 �Ma2,n

sin 1b � u 2 �0.8032

sin 139.3° � 10° 2 � 1.64

P2

P1�

2kMa21,n � k � 1

k � 1S P2 � 175.0 kPa 2 2 11.4 2 11.988 2 2 � 1.4 � 1

1.4 � 1� 333 kPa

P2

P1�

2kMa21,n � k � 1

k � 1S P2 � 175.0 kPa 2 2 11.4 2 11.267 2 2 � 1.4 � 1

1.4 � 1� 128 kPa

Ma1,n � Ma1 sin b S Ma1,n � 2.0 sin 83.7° � 1.988

Ma1,n � Ma1 sin b S Ma1,n � 2.0 sin 39.3° � 1.267

EXAMPLE 17–12 Prandtl–Meyer Expansion Wave Calculations

Supersonic air at Ma1 � 2.0 and 230 kPa flows parallel to a flat wall thatsuddenly expands by d � 10° (Fig. 17–49). Ignoring any effects caused bythe boundary layer along the wall, calculate downstream Mach number Ma2and pressure P2.

Ma1 � 2.0

Ma2

d � 10°

u

FIGURE 17–49An expansion fan caused by the suddenexpansion of a wall with d� 10.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 859

17–6 ■ DUCT FLOW WITH HEAT TRANSFER ANDNEGLIGIBLE FRICTION (RAYLEIGH FLOW)

So far we have limited our consideration mostly to isentropic flow, alsocalled reversible adiabatic flow since it involves no heat transfer and noirreversibilities such as friction. Many compressible flow problems encoun-tered in practice involve chemical reactions such as combustion, nuclearreactions, evaporation, and condensation as well as heat gain or heat lossthrough the duct wall. Such problems are difficult to analyze exactly sincethey may involve significant changes in chemical composition during flow,and the conversion of latent, chemical, and nuclear energies to thermalenergy (Fig. 17–50).

The essential features of such complex flows can still be captured by asimple analysis by modeling the generation or absorption of thermal energy

860 | Thermodynamics

Solution We are to calculate the Mach number and pressure downstream ofa sudden expansion along a wall.Assumptions 1 The flow is steady. 2 The boundary layer on the wall isvery thin.Properties The fluid is air with k � 1.4.Analysis Because of assumption 2, we approximate the total deflectionangle as equal to the wall expansion angle (i.e., u � d � 10°). With Ma1 �2.0, we solve Eq. 17–49 for the upstream Prandtl–Meyer function,

Next, we use Eq. 17–48 to calculate the downstream Prandtl–Meyer function,

Ma2 is found by solving Eq. 17–49, which is implicit—an equation solver ishelpful. We get Ma2 � 2.385. There are also compressible flow calculatorson the Internet that solve these implicit equations, along with both normaland oblique shock equations; e.g., see www.aoe.vt.edu/~devenpor/aoe3114/calc.html.

We use the isentropic relations to calculate the downstream pressure,

Since this is an expansion, Mach number increases and pressure decreases,as expected.Discussion We could also solve for downstream temperature, density, etc.,using the appropriate isentropic relations.

P2 �P2>P0

P1>P0 P1 �

c1 � a k � 1

2bMa2

2 d�k>1k�12

c1 � a k � 1

2bMa2

1 d�k>1k�12 1230 kPa 2 � 126 kPa

u � n 1Ma2 2 � n 1Ma1 2 S n 1Ma2 2 � u � n 1Ma1 2 � 10° � 26.38° � 36.38°

� B1.4 � 1

1.4 � 1 tan�1 cB

1.4 � 1

1.4 � 112.02 �1 2 d � tan�1 a22.02 �1 b � 26.38°

n 1Ma 2 � Bk � 1

k � 1 tan�1 cB

k � 1

k � 11Ma2 �1 2 d � tan�1 a2Ma2�1 b

Fuel nozzles or spray bars

Flame holders

Air inlet

FIGURE 17–50Many practical compressible flowproblems involve combustion, whichmay be modeled as heat gain throughthe duct wall.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 860

as heat transfer through the duct wall at the same rate and disregarding anychanges in chemical composition. This simplified problem is still too com-plicated for an elementary treatment of the topic since the flow may involvefriction, variations in duct area, and multidimensional effects. In this sec-tion, we limit our consideration to one-dimensional flow in a duct of con-stant cross-sectional area with negligible frictional effects.

Consider steady one-dimensional flow of an ideal gas with constant spe-cific heats through a constant-area duct with heat transfer, but with negligiblefriction. Such flows are referred to as Rayleigh flows after Lord Rayleigh(1842–1919). The conservation of mass, momentum, and energy equationsfor the control volume shown in Fig. 17–51 can be written as follows:

Mass equation Noting that the duct cross-sectional area A is constant, therelation m

.1 � m

.2 or r1A1V1 � r2A2V2 reduces to

(17–50)

x-Momentum equation Noting that the frictional effects are negligibleand thus there are no shear forces, and assuming there are no external

and body forces, the momentum equation

in the flow (or x-) direction becomes a balance between static pressureforces and momentum transfer. Noting that the flows are high speedand turbulent, the momentum flux correction factor is approximately 1(b � 1) and thus can be neglected. Then,

or

(17–51)

Energy equation The control volume involves no shear, shaft, or otherforms of work, and the potential energy change is negligible. If the rateof heat transfer is Q

.and the heat transfer per unit mass of fluid is q �

Q./m

., the steady-flow energy balance E

.in � E

.out becomes

(17–52)

For an ideal gas with constant specific heats, �h � cp �T, and thus

(17–53)

or

(17–54)

Therefore, the stagnation enthalpy h0 and stagnation temperature T0change during Rayleigh flow (both increase when heat is transferred tothe fluid and thus q is positive, and both decrease when heat is trans-ferred from the fluid and thus q is negative).

Entropy change In the absence of any irreversibilities such as friction,the entropy of a system changes by heat transfer only: it increases withheat gain, and decreases with heat loss. Entropy is a property and thus

q � h02 � h01 � cp 1T02 � T01 2

q � cp 1T2 � T1 2 �V2

2 � V21

2

Q#

� m# ah1 �

V21

2b � m

# ah2 �V2

2

2b S q � h1 �

V 21

2� h2 �

V 22

2

P1 � r1V21 � P2 � r2V

22

P1A1 � P2A2 � m#V2 � m

#V1 S P1 � P2 � 1r2V2 2V2 � 1r1V1 2V1

a F!�a

outbm

#V

!�a

inbm

#V

!

r1V1 � r2V2

Chapter 17 | 861

P1, T1, r1 P2, T2, r2

V1

Controlvolume

Q.

V2

FIGURE 17–51Control volume for flow in a constant-area duct with heat transfer andnegligible friction.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 861

a state function, and the entropy change of an ideal gas with constantspecific heats during a change of state from 1 to 2 is given by

(17–55)

The entropy of a fluid may increase or decrease during Rayleigh flow,depending on the direction of heat transfer.

Equation of state Noting that P � rRT, the properties P, r, and T of anideal gas at states 1 and 2 are related to each other by

(17–56)

Consider a gas with known properties R, k, and cp. For a specified inletstate 1, the inlet properties P1, T1, r1, V1, and s1 are known. The five exitproperties P2, T2, r2, V2, and s2 can be determined from the five equations17–50, 17–51, 17–53, 17–55, and 17–56 for any specified value of heattransfer q. When the velocity and temperature are known, the Mach number can be determined from .

Obviously there is an infinite number of possible downstream states 2corresponding to a given upstream state 1. A practical way of determiningthese downstream states is to assume various values of T2, and calculate allother properties as well as the heat transfer q for each assumed T2 from theEqs. 17–50 through 17–56. Plotting the results on a T-s diagram gives acurve passing through the specified inlet state, as shown in Fig. 17–52. Theplot of Rayleigh flow on a T-s diagram is called the Rayleigh line, and sev-eral important observations can be made from this plot and the results of thecalculations:

1. All the states that satisfy the conservation of mass, momentum, andenergy equations as well as the property relations are on the Rayleighline. Therefore, for a given initial state, the fluid cannot exist at anydownstream state outside the Rayleigh line on a T-s diagram. In fact,the Rayleigh line is the locus of all physically attainable downstreamstates corresponding to an initial state.

2. Entropy increases with heat gain, and thus we proceed to the right onthe Rayleigh line as heat is transferred to the fluid. The Mach number isMa � 1 at point a, which is the point of maximum entropy (see Exam-ple 17–13 for proof). The states on the upper arm of the Rayleigh lineabove point a are subsonic, and the states on the lower arm below pointa are supersonic. Therefore, a process proceeds to the right on the Ray-leigh line with heat addition and to the left with heat rejection regard-less of the initial value of the Mach number.

3. Heating increases the Mach number for subsonic flow, but decreases itfor supersonic flow. The flow Mach number approaches Ma � 1 in bothcases (from 0 in subsonic flow and from ∞ in supersonic flow) duringheating.

4. It is clear from the energy balance q � cp(T02 � T01) that heatingincreases the stagnation temperature T0 for both subsonic and super-sonic flows, and cooling decreases it. (The maximum value of T0 occursat Ma � 1.) This is also the case for the thermodynamic temperature T

Ma � V>c � V>1kRT

P1

r1T1�

P2

r2T2

s2 � s1 � cp ln T2

T1� R ln

P2

P1

862 | Thermodynamics

Mab = 1/ k

Maa = 1

a

a

bMa < 1

Ma > 1

Cooling (Ma S 0)

Cooling (Ma S �)

Heating (Ma S 1)

Heating (Ma S 1)

Tmax

smax

s

T

FIGURE 17–52T-s diagram for flow in a constant-areaduct with heat transfer and negligiblefriction (Rayleigh flow).

cen84959_ch17.qxd 4/21/05 11:08 AM Page 862

except for the narrow Mach number range of � Ma � 1 in sub-sonic flow (see Example 17–13). Both temperature and the Mach num-ber increase with heating in subsonic flow, but T reaches a maximumTmax at Ma � (which is 0.845 for air), and then decreases. It mayseem peculiar that the temperature of a fluid drops as heat is transferredto it. But this is no more peculiar than the fluid velocity increasing inthe diverging section of a converging–diverging nozzle. The coolingeffect in this region is due to the large increase in the fluid velocity andthe accompanying drop in temperature in accordance with the relation T0� T � V2/2cp. Note also that heat rejection in the region � Ma �1 causes the fluid temperature to increase (Fig. 17–53).

5. The momentum equation P � KV � constant, where K � rV � con-stant (from the conservation of mass equation), reveals that velocity andstatic pressure have opposite trends. Therefore, static pressuredecreases with heat gain in subsonic flow (since velocity and theMach number increase), but increases with heat gain in supersonic flow(since velocity and the Mach number decrease).

6. The continuity equation rV � constant indicates that density and veloc-ity are inversely proportional. Therefore, density decreases with heattransfer to the fluid in subsonic flow (since velocity and the Mach num-ber increase), but increases with heat gain in supersonic flow (sincevelocity and the Mach number decrease).

7. On the left half of Fig. 17–52, the lower arm of the Rayleigh line issteeper (in terms of s as a function of T), which indicates that theentropy change corresponding to a specified temperature change (andthus a given amount of heat transfer) is larger in supersonic flow.

The effects of heating and cooling on the properties of Rayleigh flow arelisted in Table 17–3. Note that heating or cooling has opposite effects on mostproperties. Also, the stagnation pressure decreases during heating and increasesduring cooling regardless of whether the flow is subsonic or supersonic.

1>1k

1>1k

1>1k

Chapter 17 | 863

T01

Supersonicflow

Heating

T02 � T01

T1 T2 � T1

T01

Subsonicflow

Heating

T02 � T01

T1

T2 � T1 orT2 � T1

FIGURE 17–53During heating, fluid temperaturealways increases if the Rayleigh flowis supersonic, but the temperature mayactually drop if the flow is subsonic.

TABLE 17–3The effects of heating and cooling on the properties of Rayleigh flow

Heating Cooling

Property Subsonic Supersonic Subsonic Supersonic

Velocity, V Increase Decrease Decrease IncreaseMach number, Ma Increase Decrease Decrease IncreaseStagnation temperature, T0 Increase Increase Decrease DecreaseTemperature, T Increase for Ma � 1/k1/2 Increase Decrease for Ma � 1/k1/2 Decrease

Decrease for Ma � 1/k1/2 Increase for Ma � 1/k1/2

Density, r Decrease Increase Increase DecreaseStagnation pressure, P0 Decrease Decrease Increase IncreasePressure, P Decrease Increase Increase DecreaseEntropy, s Increase Increase Decrease Decrease

cen84959_ch17.qxd 4/21/05 11:08 AM Page 863

864 | Thermodynamics

EXAMPLE 17–13 Extrema of Rayleigh Line

Consider the T-s diagram of Rayleigh flow, as shown in Fig. 17–54. Usingthe differential forms of the conservation equations and property relations,show that the Mach number is Maa � 1 at the point of maximum entropy(point a), and Mab � 1/ at the point of maximum temperature (point b).

Solution It is to be shown that Maa � 1 at the point of maximum entropyand Mab � 1/ at the point of maximum temperature on the Rayleigh line.Assumptions The assumptions associated with Rayleigh flow (i.e., steadyone-dimensional flow of an ideal gas with constant properties through a con-stant cross-sectional-area duct with negligible frictional effects) are valid.Analysis The differential forms of the mass (rV � constant), momentum[rearranged as P + (rV)V � constant], ideal gas (P � rRT), and enthalpychange (�h � cp �T) equations can be expressed as

(1)

(2)

(3)

The differential form of the entropy change relation (Eq. 17–40) of anideal gas with constant specific heats is

(4)

Substituting Eq. 3 into Eq. 4 gives

(5)

sincecp � R � cv S kcv � R � cv S cv � R/(k � 1)

Dividing both sides of Eq. 5 by dT and combining with Eq. 1,

(6)

Dividing Eq. 3 by dV and combining it with Eqs. 1 and 2 give, after rear-ranging,

(7)

Substituting Eq. 7 into Eq. 6 and rearranging,

(8)

Setting ds /dT � 0 and solving the resulting equation R2(kRT � V 2) � 0 forV give the velocity at point a to be

(9)Va � 2kRTa and Maa �Va

ca�2kRTa

2kRTa

� 1

ds

dT�

R

T 1k � 1 2 �R

T � V 2>R �R2 1kRT � V 2 2

T 1k � 1 2 1RT � V 2 2

dT

dV�

T

V�

V

R

ds

dT�

R

T 1k � 1 2 �R

V

dV

dT

ds � cp dT

T� R adT

T�

dr

rb � 1cp � R 2 dT

T� R

dr

r�

R

k � 1

dT

T� R

dr

r

ds � cp dT

T� R

dP

P

P � rRT S dP � rR dT � RT dr S dP

P�

dT

T�

dr

r

P � 1rV 2V � constant S dP � 1rV 2 dV � 0 S dP

dV� �rV

rV � constant S r dV � V dr � 0 S dr

r� �

dV

V

1k

1k

smax

Tmax

Ma � 1� 0

Ma � 1

� �

T

aa

b

bdsdT

s

� 0� �dTds

FIGURE 17–54The T-s diagram of Rayleigh flowconsidered in Example 17–13.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 864

Chapter 17 | 865

Therefore, sonic conditions exist at point a, and thus the Mach number is 1.Setting dT/ds � (ds/dT )�1 � 0 and solving the resulting equation

T(k � 1)(RT � V 2) � 0 for velocity at point b give

(10)

Therefore, the Mach number at point b is Mab � 1/ . For air, k � 1.4 andthus Mab � 0.845.Discussion Note that in Rayleigh flow, sonic conditions are reached as theentropy reaches its maximum value, and maximum temperature occurs dur-ing subsonic flow.

1k

Vb � 2RTb and Mab �Vb

cb�2RTb

2kRTb

�1

2k

EXAMPLE 17–14 Effect of Heat Transfer on Flow Velocity

Starting with the differential form of the energy equation, show that the flowvelocity increases with heat addition in subsonic Rayleigh flow, but decreasesin supersonic Rayleigh flow.

Solution It is to be shown that flow velocity increases with heat addition insubsonic Rayleigh flow and that the opposite occurs in supersonic flow.Assumptions 1 The assumptions associated with Rayleigh flow are valid.2 There are no work interactions and potential energy changes are negligible.Analysis Consider heat transfer to the fluid in the differential amount of dq.The differential form of the energy equations can be expressed as

(1)

Dividing by cpT and factoring out dV/V give

(2)

where we also used cp � kR/(k � 1). Noting that Ma2 � V 2/c2 � V 2/kRT andusing Eq. 7 for dT/dV from Example 17–13 give

(3)

Canceling the two middle terms in Eq. 3 since V 2/TR � kMa2 and rearrang-ing give the desired relation,

(4)

In subsonic flow, 1 � Ma2 � 0 and thus heat transfer and velocity changehave the same sign. As a result, heating the fluid (dq � 0) increases theflow velocity while cooling decreases it. In supersonic flow, however, 1 �Ma2 � 0 and heat transfer and velocity change have opposite signs. As aresult, heating the fluid (dq � 0) decreases the flow velocity while coolingincreases it (Fig. 17–55).Discussion Note that heating the fluid has the opposite effect on flow veloc-ity in subsonic and supersonic Rayleigh flows.

dV

V�dq

cpT

1

11 � Ma2 2

dq

cpT�

dV

Vc V

Ta T

V�

V

Rb � 1k � 1 2Ma2 d �

dV

Va1 �

V 2

TR� kMa2 � Ma2 b

dq

cpT�

dT

T�

V dV

cpT�

dV

Vc V

dV dT

T�1k � 1 2V 2

kRTd

dq � dh0 � d ah �V 2

2b � cp dT � V dV

Supersonicflow

V1 V2 � V1

Subsonicflow

dq

dq

V1 V2 � V1

FIGURE 17–55Heating increases the flow velocity insubsonic flow, but decreases it insupersonic flow.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 865

Property Relations for Rayleigh FlowIt is often desirable to express the variations in properties in terms of the Machnumber Ma. Noting that and thus ,

(17–57)

since P � rRT. Substituting into the momentum equation (Eq. 17–51) givesP1 � kP1Ma1

2 � P2 � kP2Ma22, which can be rearranged as

(17–58)

Again utilizing V � Ma , the continuity equation r1V1 � r2V2 can beexpressed as

(17–59)

Then the ideal-gas relation (Eq. 17–56) becomes

(17–60)

Solving Eq. 17–60 for the temperature ratio T2/T1 gives

(17–61)

Substituting this relation into Eq. 17–59 gives the density or velocity ratio as

(17–62)

Flow properties at sonic conditions are usually easy to determine, and thusthe critical state corresponding to Ma � 1 serves as a convenient referencepoint in compressible flow. Taking state 2 to be the sonic state (Ma2 � 1, andsuperscript * is used) and state 1 to be any state (no subscript), the propertyrelations in Eqs. 17–58, 17–61, and 17–62 reduce to (Fig. 17–56)

(17–63)

Similar relations can be obtained for dimensionless stagnation tempera-ture and stagnation pressure as follows:

(17–64)

which simplifies to

(17–65)T0

T*0�1k � 1 2Ma2 32 � 1k � 1 2Ma2 4

11 � kMa2 2 2

T0

T*0�

T0

T

T

T* T *

T*0� a1 �

k � 1

2 Ma2 b cMa 11 � k 2

1 � kMa2 d2 a1 �

k � 1

2b�1

P

P*�

1 � k

1 � kMa2 T

T*� cMa 11 � k 2

1 � kMa2 d2

andV

V*�r*

r�11 � k 2Ma2

1 � kMa2

r2

r1�

V1

V2�

Ma21 11 � kMa2

2 2Ma2

2 11 � kMa21 2

T2

T1� cMa2 11 � kMa2

1 2Ma1 11 � kMa2

2 2 d2

T2

T1�

P2

P1

r1

r2� a 1 � kMa2

1

1 � kMa22

b a Ma21T2

Ma11T1

b

r1

r2�

V2

V1�

Ma22kRT2

Ma12kRT1

�Ma22T2

Ma12T1

1kRT

P2

P1�

1 � kMa21

1 � kMa22

rV 2 � rkRTMa2 � kPMa2

V � Ma1kRTMa � V>c � V>1kRT

866 | Thermodynamics

V

V*

r*

r��

(1(1 � k)M)Ma2

1 � kMaMa2

P

P*�

1 � k

1 � kMaMa2

T

T*� aMa(Ma(1 � k)

1 � kMaMa2b 2

P0

P0*�

k � 1

1 � kMaMa2a2 � (k � 1)Ma1)Ma2

k � 1b k/(/(k�1)1)

T0

T0*�

(k � 1)Ma1)Ma2[2[2 � (k � 1)Ma1)Ma2]

(1(1 � kMaMa2)2

FIGURE 17–56Summary of relations for Rayleighflow.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 866

Also,

(17–66)

which simplifies to

(17–67)

The five relations in Eqs. 17–63, 17–65, and 17–67 enable us to calculatethe dimensionless pressure, temperature, density, velocity, stagnation tem-perature, and stagnation pressure for Rayleigh flow of an ideal gas with aspecified k for any given Mach number. Representative results are given intabular form in Table A–34 for k � 1.4.

Choked Rayleigh FlowIt is clear from the earlier discussions that subsonic Rayleigh flow in a ductmay accelerate to sonic velocity (Ma � 1) with heating. What happens ifwe continue to heat the fluid? Does the fluid continue to accelerate to super-sonic velocities? An examination of the Rayleigh line indicates that the fluidat the critical state of Ma � 1 cannot be accelerated to supersonic velocitiesby heating. Therefore, the flow is choked. This is analogous to not beingable to accelerate a fluid to supersonic velocities in a converging nozzle bysimply extending the converging flow section. If we keep heating the fluid,we will simply move the critical state further downstream and reduce theflow rate since fluid density at the critical state will now be lower. There-fore, for a given inlet state, the corresponding critical state fixes the maxi-mum possible heat transfer for steady flow (Fig. 17–57). That is,

(17–68)

Further heat transfer causes choking and thus the inlet state to change (e.g.,inlet velocity will decrease), and the flow no longer follows the same Rayleighline. Cooling the subsonic Rayleigh flow reduces the velocity, and the Machnumber approaches zero as the temperature approaches absolute zero. Notethat the stagnation temperature T0 is maximum at the critical state of Ma � 1.

In supersonic Rayleigh flow, heating decreases the flow velocity. Furtherheating simply increases the temperature and moves the critical state furtherdownstream, resulting in a reduction in the mass flow rate of the fluid.It may seem like supersonic Rayleigh flow can be cooled indefinitely, but itturns out that there is a limit. Taking the limit of Eq. 17–65 as the Machnumber approaches infinity gives

(17–69)

which yields T0/T*0 � 0.49 for k � 1.4. Therefore, if the critical stagnationtemperature is 1000 K, air cannot be cooled below 490 K in Rayleigh flow.Physically this means that the flow velocity reaches infinity by the time thetemperature reaches 490 K—a physical impossibility. When supersonic flowcannot be sustained, the flow undergoes a normal shock wave and becomessubsonic.

LimMaS� T0

T*0� 1 �

1

k 2

qmax � h*0 � h01 � cp 1T*0 � T01 2

P0

P*0�

k � 1

1 � kMa2 c 2 � 1k � 1 2Ma2

k � 1d k>1k�12

P0

P*0�

P0

P

P

P* P*

P*0� a1 �

k � 1

2 Ma2 b k>1k�12 a 1 � k

1 � kMa2 b a1 �k � 1

2b�k>1k�12

Chapter 17 | 867

T01

Rayleighflow

Chokedflow

qmax

T02 � T01

T1 T2 � T*

*

FIGURE 17–57For a given inlet state, the maximumpossible heat transfer occurs whensonic conditions are reached at the exitstate.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 867

868 | Thermodynamics

EXAMPLE 17–15 Rayleigh Flow in a Tubular Combustor

A combustion chamber consists of tubular combustors of 15-cm diameter.Compressed air enters the tubes at 550 K, 480 kPa, and 80 m/s (Fig.17–58). Fuel with a heating value of 42,000 kJ/kg is injected into the airand is burned with an air–fuel mass ratio of 40. Approximating combustionas a heat transfer process to air, determine the temperature, pressure, veloc-ity, and Mach number at the exit of the combustion chamber.

Solution Fuel is burned in a tubular combustion chamber with compressedair. The exit temperature, pressure, velocity, and Mach number are to bedetermined.Assumptions 1 The assumptions associated with Rayleigh flow (i.e., steadyone-dimensional flow of an ideal gas with constant properties through a con-stant cross-sectional-area duct with negligible frictional effects) are valid.2 Combustion is complete, and it is treated as a heat transfer process, withno change in the chemical composition of the flow. 3 The increase in massflow rate due to fuel injection is disregarded.Properties We take the properties of air to be k � 1.4, cp � 1.005 kJ/kg · K,and R � 0.287 kJ/kg · K (Table A–2a).Analysis The inlet density and mass flow rate of air are

The mass flow rate of fuel and the rate of heat transfer are

The stagnation temperature and Mach number at the inlet are

The exit stagnation temperature is, from the energy equation q � cp(T02 � T01),

T02 � T01 �q

cp� 553.2 K �

1050 kJ/ kg

1.005 kJ/ kg # K� 1598 K

Ma1 �V1

c1�

80 m/s

470.1 m/s� 0.1702

c1 � 2kRT1 � B 11.4 2 10.287 kJ>kg # K 2 1550 K 2 a 1000 m2>s2

1 kJ>kgb � 470.1 m>s

T01 � T1 �V 2

1

2cp

� 550 K �180 m>s 2 2

2 11.005 kJ>kg # K 2 a1 kJ>kg

1000 m2>s2 b � 553.2 K

q �Q#

m#

air�

4515 kJ>s4.299 kg>s � 1050 kJ>kg

Q#

� m#

fuel HV � 10.1075 kg>s 2 142,000 kJ>kg 2 � 4515 kW

m#

fuel �m#

air

AF�

4.299 kg>s40

� 0.1075 kg>s

m#

air � r1A1V1 � 13.041 kg>m3 2 3p 10.15 m 2 2>4 4 180 m>s 2 � 4.299 kg>s r1 �

P1

RT1�

480 kPa

10.287 kJ>kg # K 2 1550 K 2 � 3.041 kg>m3

Combustortube

P1 � 480 kPaP2, T2, V2T1 � 550 K

V1 � 80 m/s

Q.

FIGURE 17–58Schematic of the combustor tubeanalyzed in Example 17–15.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 868

Chapter 17 | 869

Discussion Note that the temperature and velocity increase and pressuredecreases during this subsonic Rayleigh flow with heating, as expected. Thisproblem can also be solved using appropriate relations instead of tabulatedvalues, which can likewise be coded for convenient computer solutions.

17–7 ■ STEAM NOZZLESWe have seen in Chapter 3 that water vapor at moderate or high pressuresdeviates considerably from ideal-gas behavior, and thus most of the rela-tions developed in this chapter are not applicable to the flow of steamthrough the nozzles or blade passages encountered in steam turbines. Giventhat the steam properties such as enthalpy are functions of pressure as wellas temperature and that no simple property relations exist, an accurate analy-sis of steam flow through the nozzles is no easy matter. Often it becomesnecessary to use steam tables, an h-s diagram, or a computer program forthe properties of steam.

A further complication in the expansion of steam through nozzles occursas the steam expands into the saturation region, as shown in Fig. 17–59. Asthe steam expands in the nozzle, its pressure and temperature drop, and

The maximum value of stagnation temperature T*0 occurs at Ma � 1, and itsvalue can be determined from Table A–34 or from Eq. 17–65. At Ma1 �0.1702 we read T0/T*0 � 0.1291. Therefore,

The stagnation temperature ratio at the exit state and the Mach number cor-responding to it are, from Table A–34,

The Rayleigh flow relations corresponding to the inlet and exit Mach num-bers are (Table A–34):

Ma2 � 0.3142:T2

T*� 0.4389

P2

P*� 2.1086

V2

V*� 0.2082

Ma1 � 0.1702:T1

T*� 0.1541

P1

P*� 2.3065

V1

V*� 0.0668

T02

T*0�

1598 K

4285 K� 0.3729 S Ma2 � 0.3142

T*0 �T01

0.1291�

553.2 K

0.1291� 4285 K

Then the exit temperature, pressure, and velocity are determined to be

V2

V1�

V2>V*

V1>V*�

0.2082

0.0668� 3.117 S V2 � 3.117V1 � 3.117 180 m>s 2 � 249 m/s

P2

P1�

P2/P*

P1/P*�

2.1086

2.3065� 0.9142 S P2 � 0.9142P1 � 0.9142 1480 kPa 2 � 439 kPa

T2

T1�

T2>T*

T1>T*�

0.4389

0.1541� 2.848 S T2 � 2.848T1 � 2.848 1550 K 2 � 1566 K

cen84959_ch17.qxd 4/21/05 11:08 AM Page 869

ordinarily one would expect the steam to start condensing when it strikesthe saturation line. However, this is not always the case. Owing to the highspeeds, the residence time of the steam in the nozzle is small, and there maynot be sufficient time for the necessary heat transfer and the formation ofliquid droplets. Consequently, the condensation of the steam may bedelayed for a little while. This phenomenon is known as supersaturation,and the steam that exists in the wet region without containing any liquid iscalled supersaturated steam. Supersaturation states are nonequilibrium (ormetastable) states.

During the expansion process, the steam reaches a temperature lower thanthat normally required for the condensation process to begin. Once the tem-perature drops a sufficient amount below the saturation temperature corre-sponding to the local pressure, groups of steam moisture droplets ofsufficient size are formed, and condensation occurs rapidly. The locus ofpoints where condensation takes place regardless of the initial temperatureand pressure at the nozzle entrance is called the Wilson line. The Wilsonline lies between the 4 and 5 percent moisture curves in the saturationregion on the h-s diagram for steam, and it is often approximated by the4 percent moisture line. Therefore, steam flowing through a high-velocitynozzle is assumed to begin condensation when the 4 percent moisture line iscrossed.

The critical-pressure ratio P*/P0 for steam depends on the nozzle inlet stateas well as on whether the steam is superheated or saturated at the nozzle inlet.However, the ideal-gas relation for the critical-pressure ratio, Eq. 17–22, givesreasonably good results over a wide range of inlet states. As indicated inTable 17–2, the specific heat ratio of superheated steam is approximated ask � 1.3. Then the critical-pressure ratio becomes

When steam enters the nozzle as a saturated vapor instead of superheatedvapor (a common occurrence in the lower stages of a steam turbine), thecritical-pressure ratio is taken to be 0.576, which corresponds to a specificheat ratio of k � 1.14.

P*

P0� a 2

k � 1b k>1k�12

� 0.546

870 | Thermodynamics

s

h

P 1

1

P 2

Saturationline

2

Wilson line (x = 0.96)

FIGURE 17–59The h-s diagram for the isentropicexpansion of steam in a nozzle.

EXAMPLE 17–16 Steam Flow through a Converging–Diverging Nozzle

Steam enters a converging–diverging nozzle at 2 MPa and 400°C with a neg-ligible velocity and a mass flow rate of 2.5 kg/s, and it exits at a pressure of300 kPa. The flow is isentropic between the nozzle entrance and throat, andthe overall nozzle efficiency is 93 percent. Determine (a) the throat and exitareas and (b) the Mach number at the throat and the nozzle exit.

Solution Steam enters a converging–diverging nozzle with a low velocity.The throat and exit areas and the Mach number are to be determined.Assumptions 1 Flow through the nozzle is one-dimensional. 2 The flow isisentropic between the inlet and the throat, and is adiabatic and irreversiblebetween the throat and the exit. 3 The inlet velocity is negligible.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 870

Chapter 17 | 871

Analysis We denote the entrance, throat, and exit states by 1, t, and 2,respectively, as shown in Fig. 17–60.(a) Since the inlet velocity is negligible, the inlet stagnation and static statesare identical. The ratio of the exit-to-inlet stagnation pressure is

It is much smaller than the critical-pressure ratio, which is taken to beP*/P01 � 0.546 since the steam is superheated at the nozzle inlet. There-fore, the flow surely is supersonic at the exit. Then the velocity at the throatis the sonic velocity, and the throat pressure is

At the inlet,

Also, at the throat,

Then the throat velocity is determined from Eq. 17–3 to be

The flow area at the throat is determined from the mass flow rate relation:

At state 2s,

The enthalpy of the steam at the actual exit state is (see Chap. 7)

Therefore,

Then the exit velocity and the exit area become

A2 �m#v2

V2�12.5 kg>s 2 10.67723 m3>kg 2

929.8 m>s � 18.21 � 10�4 m2 � 18.21 cm2

V2 �22 1h01 � h2 2 �B 32 13248.4 � 2816.1 2 kJ>kg 4 a1000 m2>s2

1 kJ>kgb � 929.8 m>s

P2 � 300 kPa

h2 � 2816.1 kJ>kgf v2 � 0.67723 m3>kg

s2 � 7.2019 kJ>kg # K

0.93 �3248.4 � h2

3248.4 � 2783.6 ¡ h2 � 2816.1 kJ>kg

hN �h01 � h2

h01 � h2s

P2s � P2 � 300 kPa

s2s � s1 � 7.1292 kJ>kg # Kf h2s � 2783.6 kJ>kg

At �m#vt

Vt

�12.5 kg/s 2 10.2420 m3/kg 2

585.8 m/s� 10.33 � 10�4 m2 � 10.33 cm2

Vt �22 1h01 � ht 2 � B 32 13248.4 � 3076.8 2 kJ/kg 4 a1000 m2>s2

1 kJ>kgb � 585.8 m/s

Pt � 1.09 MPa

st � 7.1292 kJ>kg # Kf ht � 3076.8 kJ>kg

vt � 0.24196 m3>kg

P1 � P01 � 2 MPa

T1 � T01 � 400°Cf h1 � h01 � 3248.4 kJ>kg

s1 � st � s2s � 7.1292 kJ>kg # K

Pt � 0.546P01 � 10.546 2 12 MPa 2 � 1.09 MPa

P2

P01�

300 kPa

2000 kPa� 0.15

s

h

1

P 2 = 300 kPa

22s

tP t

P 1 =

P 01 = 2 M

Pa

T1 = 400°C

P1 = 2 MPa

V1 ≅ 0

STEAM

Throat

m = 2.5 kg/s·

hN = 93%

FIGURE 17–60Schematic and h-s diagram forExample 17–16.

cen84959_ch17.qxd 4/21/05 11:08 AM Page 871

872 | Thermodynamics

(b) The velocity of sound and the Mach numbers at the throat and the exitof the nozzle are determined by replacing differential quantities withdifferences,

The velocity of sound at the throat is determined by evaluating the specificvolume at st � 7.1292 kJ/kg · K and at pressures of 1.115 and 1.065 MPa(Pt 25 kPa):

The Mach number at the throat is determined from Eq. 17–12 to be

Thus, the flow at the throat is sonic, as expected. The slight deviation ofthe Mach number from unity is due to replacing the derivatives bydifferences.

The velocity of sound and the Mach number at the nozzle exit are deter-mined by evaluating the specific volume at s2 � 7.2019 kJ/kg · K and atpressures of 325 and 275 kPa (P2 25 kPa):

and

Thus the flow of steam at the nozzle exit is supersonic.

Ma �Vc

�929.8 m>s515.4 m>s � 1.804

c � B1325 � 275 2 kPa

11>0.63596 � 1>0.72245 2 kg>m3 a 1000 m2>s2

1 kPa # m3>kgb � 515.4 m>s

Ma �Vc

�585.8 m>s584.6 m>s � 1.002

c � B11115 � 1065 2 kPa

11>0.23776 � 1>0.24633 2 kg>m3 a 1000 m2>s2

1 kPa # m3>kgb � 584.6 m>s

c � a 0P

0rb 1>2

s

� c ¢P

¢ 11>v 2 d1>2

s

SUMMARY

In this chapter the effects of compressibility on gas flow areexamined. When dealing with compressible flow, it is conve-nient to combine the enthalpy and the kinetic energy of thefluid into a single term called stagnation (or total) enthalpyh0, defined as

The properties of a fluid at the stagnation state are calledstagnation properties and are indicated by the subscript zero.The stagnation temperature of an ideal gas with constant spe-cific heats is

which represents the temperature an ideal gas would attain ifit is brought to rest adiabatically. The stagnation properties ofan ideal gas are related to the static properties of the fluid by

T0 � T �V 2

2cp

h0 � h �V 2

2

The speed at which an infinitesimally small pressure wavetravels through a medium is the speed of sound. For an idealgas it is expressed as

The Mach number is the ratio of the actual velocity of thefluid to the speed of sound at the same state:

The flow is called sonic when Ma � 1, subsonic when Ma � 1,supersonic when Ma � 1, hypersonic when Ma �� 1, andtransonic when Ma � 1.

Ma �Vc

c � B a0P

0rb

s

� 2kRT

P0

P� a T0

Tb k>1k�12

andr0

r� a T0

Tb 1>1k�12

cen84959_ch17.qxd 4/21/05 11:08 AM Page 872

Chapter 17 | 873

Nozzles whose flow area decreases in the flow directionare called converging nozzles. Nozzles whose flow area firstdecreases and then increases are called converging–divergingnozzles. The location of the smallest flow area of a nozzle iscalled the throat. The highest velocity to which a fluid can beaccelerated in a converging nozzle is the sonic velocity.Accelerating a fluid to supersonic velocities is possible onlyin converging–diverging nozzles. In all supersonic converging–diverging nozzles, the flow velocity at the throat is the speedof sound.

The ratios of the stagnation to static properties for idealgases with constant specific heats can be expressed in termsof the Mach number as

and

When Ma � 1, the resulting static-to-stagnation propertyratios for the temperature, pressure, and density are calledcritical ratios and are denoted by the superscript asterisk:

and

The pressure outside the exit plane of a nozzle is called theback pressure. For all back pressures lower than P*, the pres-

r*

r0� a 2

k � 1b 1>1k�12

T*

T0�

2

k � 1

P*

P0� a 2

k � 1b k>1k�12

r0

r� c1 � a k � 1

2bMa2 d 1>1k�12

P0

P� c1 � a k � 1

2bMa2 d k>1k�12

T0

T� 1 � a k � 1

2bMa2

sure at the exit plane of the converging nozzle is equal to P*,the Mach number at the exit plane is unity, and the mass flowrate is the maximum (or choked) flow rate.

In some range of back pressure, the fluid that achieved asonic velocity at the throat of a converging–diverging nozzleand is accelerating to supersonic velocities in the divergingsection experiences a normal shock, which causes a suddenrise in pressure and temperature and a sudden drop in veloc-ity to subsonic levels. Flow through the shock is highlyirreversible, and thus it cannot be approximated as isen-tropic. The properties of an ideal gas with constant specificheats before (subscript 1) and after (subscript 2) a shock arerelated by

and

These equations also hold across an oblique shock, providedthat the component of the Mach number normal to theoblique shock is used in place of the Mach number.

Steady one-dimensional flow of an ideal gas with constantspecific heats through a constant-area duct with heat transferand negligible friction is referred to as Rayleigh flow. Theproperty relations and curves for Rayleigh flow are given inTable A–34. Heat transfer during Rayleigh flow can be deter-mined from

q � cp 1T02 � T01 2 � cp 1T2 � T1 2 �V 2

2 � V 21

2

P2

P1�

1 � kMa21

1 � kMa22

�2kMa2

1 � k � 1

k � 1

T2

T1�

2 � Ma21 1k � 1 2

2 � Ma22 1k � 1 2

T01 � T02 Ma2 � B1k � 1 2Ma2

1 � 2

2kMa21 � k � 1

1. J. D. Anderson. Modern Compressible Flow with HistoricalPerspective. 3rd ed. New York: McGraw-Hill, 2003.

2. Y. A. Çengel and J. M. Cimbala. Fluid Mechanics:Fundamentals and Applications. New York: McGraw-Hill, 2006.

3. H. Cohen, G. F. C. Rogers, and H. I. H. Saravanamuttoo.Gas Turbine Theory. 3rd ed. New York: Wiley, 1987.

4. W. J. Devenport. Compressible Aerodynamic Calculator,http://www.aoe.vt.edu/~devenpor/aoe3114/calc.html.

5. R. W. Fox and A. T. McDonald. Introduction to FluidMechanics. 5th ed. New York: Wiley, 1999.

6. H. Liepmann and A. Roshko. Elements of Gas Dynamics.Dover Publications, Mineola, NY, 2001.

7. C. E. Mackey, responsible NACA officer and curator.Equations, Tables, and Charts for Compressible Flow.

NACA Report 1135, http://naca.larc.nasa.gov/reports/1953/naca-report-1135/.

8. A. H. Shapiro. The Dynamics and Thermodynamics ofCompressible Fluid Flow. vol. 1. New York: Ronald PressCompany, 1953.

9. P. A. Thompson. Compressible-Fluid Dynamics. NewYork: McGraw-Hill, 1972.

10. United Technologies Corporation. The Aircraft GasTurbine and Its Operation. 1982.

11. Van Dyke, 1982.

12. F. M. White. Fluid Mechanics. 5th ed. New York:McGraw-Hill, 2003.

REFERENCES AND SUGGESTED READINGS

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Stagnation Properties

17–1C A high-speed aircraft is cruising in still air. Howwill the temperature of air at the nose of the aircraft differfrom the temperature of air at some distance from theaircraft?

17–2C How and why is the stagnation enthalpy h0 defined?How does it differ from ordinary (static) enthalpy?

17–3C What is dynamic temperature?

17–4C In air-conditioning applications, the temperature ofair is measured by inserting a probe into the flow stream.Thus, the probe actually measures the stagnation temperature.Does this cause any significant error?

17–5 Determine the stagnation temperature and stagnationpressure of air that is flowing at 44 kPa, 245.9 K, and 470m/s. Answers: 355.8 K, 160.3 kPa

17–6 Air at 300 K is flowing in a duct at a velocity of (a) 1,(b) 10, (c) 100, and (d) 1000 m/s. Determine the temperaturethat a stationary probe inserted into the duct will read foreach case.

17–7 Calculate the stagnation temperature and pressure forthe following substances flowing through a duct: (a) helium at0.25 MPa, 50°C, and 240 m/s; (b) nitrogen at 0.15 MPa, 50°C,and 300 m/s; and (c) steam at 0.1 MPa, 350°C, and 480 m/s.

17–8 Air enters a compressor with a stagnation pressure of100 kPa and a stagnation temperature of 27°C, and it is com-pressed to a stagnation pressure of 900 kPa. Assuming thecompression process to be isentropic, determine the powerinput to the compressor for a mass flow rate of 0.02 kg/s.Answer: 5.27 kW

17–9E Steam flows through a device with a stagnationpressure of 120 psia, a stagnation temperature of 700°F, and avelocity of 900 ft/s. Assuming ideal-gas behavior, determinethe static pressure and temperature of the steam at this state.

17–10 Products of combustion enter a gas turbine with astagnation pressure of 1.0 MPa and a stagnation temperatureof 750°C, and they expand to a stagnation pressure of 100kPa. Taking k � 1.33 and R � 0.287 kJ/kg · K for the prod-ucts of combustion, and assuming the expansion process to

PROBLEMS*

be isentropic, determine the power output of the turbine perunit mass flow.

17–11 Air flows through a device such that the stagnationpressure is 0.6 MPa, the stagnation temperature is 400°C, andthe velocity is 570 m/s. Determine the static pressure and tem-perature of the air at this state. Answers: 518.6 K, 0.23 MPa

Speed of Sound and Mach Number

17–12C What is sound? How is it generated? How does ittravel? Can sound waves travel in a vacuum?

17–13C Is it realistic to assume that the propagation ofsound waves is an isentropic process? Explain.

17–14C Is the sonic velocity in a specified medium a fixedquantity, or does it change as the properties of the mediumchange? Explain.

17–15C In which medium does a sound wave travel faster:in cool air or in warm air?

17–16C In which medium will sound travel fastest for agiven temperature: air, helium, or argon?

17–17C In which medium does a sound wave travel faster:in air at 20°C and 1 atm or in air at 20°C and 5 atm?

17–18C Does the Mach number of a gas flowing at a con-stant velocity remain constant? Explain.

17–19 Determine the speed of sound in air at (a) 300 K and(b) 1000 K. Also determine the Mach number of an aircraftmoving in air at a velocity of 280 m/s for both cases.

17–20 Carbon dioxide enters an adiabatic nozzle at 1200 Kwith a velocity of 50 m/s and leaves at 400 K. Assuming con-stant specific heats at room temperature, determine the Machnumber (a) at the inlet and (b) at the exit of the nozzle.Assess the accuracy of the constant specific heat assumption.Answers: (a) 0.0925, (b) 3.73

17–21 Nitrogen enters a steady-flow heat exchanger at 150kPa, 10°C, and 100 m/s, and it receives heat in the amount of120 kJ/kg as it flows through it. Nitrogen leaves the heatexchanger at 100 kPa with a velocity of 200 m/s. Determinethe Mach number of the nitrogen at the inlet and the exit ofthe heat exchanger.

17–22 Assuming ideal-gas behavior, determine the speed ofsound in refrigerant-134a at 0.1 MPa and 60°C.

17–23 The Airbus A-340 passenger plane has a maximumtakeoff weight of about 260,000 kg, a length of 64 m, a wingspan of 60 m, a maximum cruising speed of 945 km/h, aseating capacity of 271 passengers, maximum cruising alti-tude of 14,000 m, and a maximum range of 12,000 km. Theair temperature at the crusing altitude is about �60°C. Deter-mine the Mach number of this plane for the stated limitingconditions.

*Problems designated by a “C” are concept questions, and studentsare encouraged to answer them all. Problems designated by an “E”are in English units, and the SI users can ignore them. Problemswith a CD-EES icon are solved using EES, and complete solutionstogether with parametric studies are included on the enclosed DVD.Problems with a computer-EES icon are comprehensive in nature,and are intended to be solved with a computer, preferably using theEES software that accompanies this text.

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17–24E Steam flows through a device with a pressure of120 psia, a temperature of 700°F, and a velocity of 900 ft/s.Determine the Mach number of the steam at this state byassuming ideal-gas behavior with k � 1.3. Answer: 0.441

17–25E Reconsider Prob. 17–24E. Using EES (orother) software, compare the Mach number of

steam flow over the temperature range 350 to 700°F. Plot theMach number as a function of temperature.

17–26 The isentropic process for an ideal gas is expressedas Pv k � constant. Using this process equation and the defi-nition of the speed of sound (Eq. 17–9), obtain the expressionfor the speed of sound for an ideal gas (Eq. 17–11).

17–27 Air expands isentropically from 1.5 MPa and 60°Cto 0.4 MPa. Calculate the ratio of the initial to final speed ofsound. Answer: 1.21

17–28 Repeat Prob. 17–27 for helium gas.

17–29E Air expands isentropically from 170 psia and200°F to 60 psia. Calculate the ratio of the initial to finalspeed of sound. Answer: 1.16

One-Dimensional Isentropic Flow

17–30C Consider a converging nozzle with sonic velocityat the exit plane. Now the nozzle exit area is reduced whilethe nozzle inlet conditions are maintained constant. What willhappen to (a) the exit velocity and (b) the mass flow ratethrough the nozzle?

17–31C A gas initially at a supersonic velocity enters anadiabatic converging duct. Discuss how this affects (a) thevelocity, (b) the temperature, (c) the pressure, and (d) thedensity of the fluid.

17–32C A gas initially at a supersonic velocity enters anadiabatic diverging duct. Discuss how this affects (a) thevelocity, (b) the temperature, (c) the pressure, and (d) thedensity of the fluid.

17–33C A gas initially at a supersonic velocity enters anadiabatic converging duct. Discuss how this affects (a) thevelocity, (b) the temperature, (c) the pressure, and (d) thedensity of the fluid.

17–34C A gas initially at a subsonic velocity enters an adi-abatic diverging duct. Discuss how this affects (a) the veloc-ity, (b) the temperature, (c) the pressure, and (d) the densityof the fluid.

17–35C A gas at a specified stagnation temperature andpressure is accelerated to Ma � 2 in a converging–divergingnozzle and to Ma � 3 in another nozzle. What can you sayabout the pressures at the throats of these two nozzles?

17–36C Is it possible to accelerate a gas to a supersonicvelocity in a converging nozzle?

17–37 Air enters a converging–diverging nozzle at a pres-sure of 1.2 MPa with negligible velocity. What is the lowest

pressure that can be obtained at the throat of the nozzle?Answer: 634 kPa

17–38 Helium enters a converging–diverging nozzle at 0.7MPa, 800 K, and 100 m/s. What are the lowest temperatureand pressure that can be obtained at the throat of the nozzle?

17–39 Calculate the critical temperature, pressure, and den-sity of (a) air at 200 kPa, 100°C, and 250 m/s, and (b) heliumat 200 kPa, 40°C, and 300 m/s.

17–40 Quiescent carbon dioxide at 600 kPa and 400 K isaccelerated isentropically to a Mach number of 0.5. Deter-mine the temperature and pressure of the carbon dioxide afteracceleration. Answers: 388 K, 514 kPa

17–41 Air at 200 kPa, 100°C, and Mach number Ma � 0.8flows through a duct. Find the velocity and the stagnationpressure, temperature, and density of the air.

17–42 Reconsider Prob. 17–41. Using EES (or other)software, study the effect of Mach numbers in

the range 0.1 to 2 on the velocity, stagnation pressure, tem-perature, and density of air. Plot each parameter as a functionof the Mach number.

17–43E Air at 30 psia, 212°F, and Mach number Ma � 0.8flows through a duct. Calculate the velocity and the stagna-tion pressure, temperature, and density of air.Answers: 1016 ft/s, 45.7 psia, 758 R, 0.163 lbm/ft3

17–44 An aircraft is designed to cruise at Mach numberMa � 1.2 at 8000 m where the atmospheric temperature is236.15 K. Determine the stagnation temperature on the lead-ing edge of the wing.

Isentropic Flow through Nozzles

17–45C Consider subsonic flow in a converging nozzlewith fixed inlet conditions. What is the effect of dropping theback pressure to the critical pressure on (a) the exit velocity,(b) the exit pressure, and (c) the mass flow rate through thenozzle?

17–46C Consider subsonic flow in a converging nozzlewith specified conditions at the nozzle inlet and critical pres-sure at the nozzle exit. What is the effect of dropping theback pressure well below the critical pressure on (a) the exitvelocity, (b) the exit pressure, and (c) the mass flow ratethrough the nozzle?

17–47C Consider a converging nozzle and a converging–diverging nozzle having the same throat areas. For the sameinlet conditions, how would you compare the mass flow ratesthrough these two nozzles?

17–48C Consider gas flow through a converging nozzlewith specified inlet conditions. We know that the highestvelocity the fluid can have at the nozzle exit is the sonicvelocity, at which point the mass flow rate through the nozzleis a maximum. If it were possible to achieve hypersonic

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velocities at the nozzle exit, how would it affect the massflow rate through the nozzle?

17–49C How does the parameter Ma* differ from the Machnumber Ma?

17–50C What would happen if we attempted to deceleratea supersonic fluid with a diverging diffuser?

17–51C What would happen if we tried to further acceler-ate a supersonic fluid with a diverging diffuser?

17–52C Consider the isentropic flow of a fluid through aconverging–diverging nozzle with a subsonic velocity at thethroat. How does the diverging section affect (a) the velocity,(b) the pressure, and (c) the mass flow rate of the fluid?

17–53C Is it possible to accelerate a fluid to supersonicvelocities with a velocity other than the sonic velocity at thethroat? Explain.

17–54 Explain why the maximum flow rate per unit areafor a given gas depends only on P0/ . For an ideal gaswith k � 1.4 and R � 0.287 kJ/kg · K, find the constant asuch that m

./A* � aP0/ .

17–55 For an ideal gas obtain an expression for the ratio ofthe velocity of sound where Ma � 1 to the speed of soundbased on the stagnation temperature, c*/c0.

17–56 An ideal gas flows through a passage that first con-verges and then diverges during an adiabatic, reversible,steady-flow process. For subsonic flow at the inlet, sketch thevariation of pressure, velocity, and Mach number along thelength of the nozzle when the Mach number at the minimumflow area is equal to unity.

17–57 Repeat Prob. 17–56 for supersonic flow at the inlet.

17–58 Air enters a nozzle at 0.2 MPa, 350 K, and a veloc-ity of 150 m/s. Assuming isentropic flow, determine the pres-sure and temperature of air at a location where the airvelocity equals the speed of sound. What is the ratio of thearea at this location to the entrance area?Answers: 0.118 MPa, 301 K, 0.629

17–59 Repeat Prob. 17–58 assuming the entrance velocityis negligible.

17–60E Air enters a nozzle at 30 psia, 630 R, and a veloc-ity of 450 ft/s. Assuming isentropic flow, determine the pres-sure and temperature of air at a location where the airvelocity equals the speed of sound. What is the ratio of thearea at this location to the entrance area?Answers: 17.4 psia, 539 R, 0.574

17–61 Air enters a converging–diverging nozzle at 0.5 MPawith a negligible velocity. Assuming the flow to be isen-tropic, determine the back pressure that will result in an exitMach number of 1.8. Answer: 0.087 MPa

17–62 Nitrogen enters a converging–diverging nozzle at 700kPa and 450 K with a negligible velocity. Determine the criti-cal velocity, pressure, temperature, and density in the nozzle.

1T0

1T0

876 | Thermodynamics

17–63 An ideal gas with k � 1.4 is flowing through a noz-zle such that the Mach number is 2.4 where the flow area is25 cm2. Assuming the flow to be isentropic, determine theflow area at the location where the Mach number is 1.2.

17–64 Repeat Prob. 17–63 for an ideal gas with k � 1.33.

17–65 Air at 900 kPa and 400 K enters a convergingnozzle with a negligible velocity. The throat area

of the nozzle is 10 cm2. Assuming isentropic flow, calculateand plot the exit pressure, the exit velocity, and the mass flowrate versus the back pressure Pb for 0.9 � Pb � 0.1 MPa.

17–66 Reconsider Prob. 17–65. Using EES (or other)software, solve the problem for the inlet condi-

tions of 1 MPa and 1000 K.

17–67E Air enters a converging–diverging nozzle of asupersonic wind tunnel at 150 psia and 100°F with a lowvelocity. The flow area of the test section is equal to the exitarea of the nozzle, which is 5 ft2. Calculate the pressure, tem-perature, velocity, and mass flow rate in the test section for aMach number Ma � 2. Explain why the air must be very dryfor this application. Answers: 19.1 psia, 311 R, 1729 ft/s,1435 lbm/s

Shock Waves and Expansion Waves17–68C Can a shock wave develop in the converging sec-tion of a converging–diverging nozzle? Explain.

17–69C What do the states on the Fanno line and theRayleigh line represent? What do the intersection points ofthese two curves represent?

17–70C Can the Mach number of a fluid be greater than 1after a shock wave? Explain.

17–71C How does the normal shock affect (a) the fluidvelocity, (b) the static temperature, (c) the stagnation temper-ature, (d) the static pressure, and (e) the stagnation pressure?

17–72C How do oblique shocks occur? How do obliqueshocks differ from normal shocks?

17–73C For an oblique shock to occur, does the upstreamflow have to be supersonic? Does the flow downstream of anoblique shock have to be subsonic?

17–74C It is claimed that an oblique shock can be analyzedlike a normal shock provided that the normal component ofvelocity (normal to the shock surface) is used in the analysis.Do you agree with this claim?

17–75C Consider supersonic airflow approaching the noseof a two-dimensional wedge and experiencing an obliqueshock. Under what conditions does an oblique shock detachfrom the nose of the wedge and form a bow wave? What isthe numerical value of the shock angle of the detached shockat the nose?

17–76C Consider supersonic flow impinging on the roundednose of an aircraft. Will the oblique shock that forms in frontof the nose be an attached or detached shock? Explain.

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17–77C Are the isentropic relations of ideal gases applica-ble for flows across (a) normal shock waves, (b) obliqueshock waves, and (c) Prandtl–Meyer expansion waves?

17–78 For an ideal gas flowing through a normal shock,develop a relation for V2/V1 in terms of k, Ma1, and Ma2.

17–79 Air enters a converging–diverging nozzle of a super-sonic wind tunnel at 1.5 MPa and 350 K with a low velocity.If a normal shock wave occurs at the exit plane of the nozzleat Ma � 2, determine the pressure, temperature, Mach num-ber, velocity, and stagnation pressure after the shock wave.Answers: 0.863 MPa, 328 K, 0.577, 210 m/s, 1.081 MPa

17–80 Air enters a converging–diverging nozzle with lowvelocity at 2.0 MPa and 100°C. If the exit area of the nozzleis 3.5 times the throat area, what must the back pressure be toproduce a normal shock at the exit plane of the nozzle?Answer: 0.661 MPa

17–81 What must the back pressure be in Prob. 17–80 for anormal shock to occur at a location where the cross-sectionalarea is twice the throat area?

17–82 Air flowing steadily in a nozzle experiences a normalshock at a Mach number of Ma � 2.5. If the pressure andtemperature of air are 61.64 kPa and 262.15 K, respectively,upstream of the shock, calculate the pressure, temperature,velocity, Mach number, and stagnation pressure downstreamof the shock. Compare these results to those for helium under-going a normal shock under the same conditions.

17–83 Calculate the entropy change of air across the nor-mal shock wave in Prob. 17–82.

17–84E Air flowing steadily in a nozzle experiences anormal shock at a Mach number of Ma �

2.5. If the pressure and temperature of air are 10.0 psia and440.5 R, respectively, upstream of the shock, calculate thepressure, temperature, velocity, Mach number, and stagnationpressure downstream of the shock. Compare these results tothose for helium undergoing a normal shock under the sameconditions.

17–85E Reconsider Prob. 17–84E. Using EES (orother) software, study the effects of both air

and helium flowing steadily in a nozzle when there is a nor-mal shock at a Mach number in the range 2 � Ma1 � 3.5. Inaddition to the required information, calculate the entropychange of the air and helium across the normal shock. Tabu-late the results in a parametric table.

17–86 Air enters a normal shock at 22.6 kPa, 217 K, and680 m/s. Calculate the stagnation pressure and Mach numberupstream of the shock, as well as pressure, temperature,velocity, Mach number, and stagnation pressure downstreamof the shock.

17–87 Calculate the entropy change of air across the nor-mal shock wave in Prob. 17–86. Answer: 0.155 kJ/kg · K

17–88 Using EES (or other) software, calculate andplot the entropy change of air across the nor-

mal shock for upstream Mach numbers between 0.5 and 1.5in increments of 0.1. Explain why normal shock waves canoccur only for upstream Mach numbers greater than Ma � 1.

17–89 Consider supersonic airflow approaching the nose ofa two-dimensional wedge at a Mach number of 5. Using Fig.17–41, determine the minimum shock angle and the maxi-mum deflection angle a straight oblique shock can have.

17–90 Air flowing at 60 kPa, 240 K, and a Mach number of3.4 impinges on a two-dimensional wedge of half-angle 12°.Determine the two possible oblique shock angles, bweak andbstrong, that could be formed by this wedge. For each case,calculate the pressure, temperature, and Mach number down-stream of the oblique shock.

17–91 Consider the supersonic flow of air at upstream con-ditions of 70 kPa and 260 K and a Mach number of 2.4 overa two-dimensional wedge of half-angle 10°. If the axis of thewedge is tilted 25° with respect to the upstream airflow,determine the downstream Mach number, pressure, and tem-perature above the wedge. Answers: 3.105, 23.8 kPa, 191 K

Ma1 � 2.4

Ma2

25°10°

FIGURE P17–91

17–92 Reconsider Prob. 17–91. Determine the downstreamMach number, pressure, and temperature below the wedge fora strong oblique shock for an upstream Mach number of 5.

17–93E Air at 8 psia, 20°F, and a Mach number of 2.0 isforced to turn upward by a ramp that makes an 8° angle offthe flow direction. As a result, a weak oblique shock forms.Determine the wave angle, Mach number, pressure, and tem-perature after the shock.

17–94 Air flowing at P1 � 40 kPa, T1 � 280 K, and Ma1 �3.6 is forced to undergo an expansion turn of 15°. Determinethe Mach number, pressure, and temperature of air after theexpansion. Answers: 4.81, 8.31 kPa, 179 K

17–95E Air flowing at P1 � 6 psia, T1 � 480 R, andMa1 � 2.0 is forced to undergo a compression turn of 15°.Determine the Mach number, pressure, and temperature of airafter the compression.

Duct Flow with Heat Transfer and Negligible Friction(Rayleigh Flow)

17–96C What is the characteristic aspect of Rayleigh flow?What are the main assumptions associated with Rayleighflow?

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17–97C On a T-s diagram of Rayleigh flow, what do thepoints on the Rayleigh line represent?

17–98C What is the effect of heat gain and heat loss on theentropy of the fluid during Rayleigh flow?

17–99C Consider subsonic Rayleigh flow of air with aMach number of 0.92. Heat is now transferred to the fluidand the Mach number increases to 0.95. Will the temperatureT of the fluid increase, decrease, or remain constant duringthis process? How about the stagnation temperature T0?

17–100C What is the effect of heating the fluid on the flowvelocity in subsonic Rayleigh flow? Answer the same ques-tions for supersonic Rayleigh flow.

17–101C Consider subsonic Rayleigh flow that is acceler-ated to sonic velocity (Ma � 1) at the duct exit by heating. Ifthe fluid continues to be heated, will the flow at duct exit besupersonic, subsonic, or remain sonic?

17–102 Consider a 12-cm-diameter tubular combustionchamber. Air enters the tube at 500 K, 400 kPa, and 70 m/s.Fuel with a heating value of 39,000 kJ/kg is burned by spray-ing it into the air. If the exit Mach number is 0.8, determinethe rate at which the fuel is burned and the exit temperature.Assume complete combustion and disregard the increase inthe mass flow rate due to the fuel mass.

878 | Thermodynamics

17–103 Air enters a rectangular duct at T1 � 300 K, P1 �420 kPa, and Ma1 � 2. Heat is transferred to the air in theamount of 55 kJ/kg as it flows through the duct. Disregardingfrictional losses, determine the temperature and Mach num-ber at the duct exit. Answers: 386 K, 1.64

Combustortube

Fuel

P1 � 400 kPaT1 � 500 K

V1 � 70 m/s

Ma2 � 0.8

FIGURE P17–102

Air

55 kJ/kg

P1 � 420 kPa

T1 � 300 K

Ma1 � 2

FIGURE P17–103

flow is observed to be choked, and the velocity and the staticpressure are measured to be 620 m/s and 270 kPa. Disregard-ing frictional losses, determine the velocity, static tempera-ture, and static pressure at the duct inlet.

17–106E Air flows with negligible friction through a 4-in-diameter duct at a rate of 5 lbm/s. The temperature and pres-sure at the inlet are T1 � 800 R and P1 � 30 psia, and theMach number at the exit is Ma2 � 1. Determine the rate ofheat transfer and the pressure drop for this section of theduct.

17–107 Air enters a frictionless duct with V1 � 70m/s, T1 � 600 K, and P1 � 350 kPa. Letting

the exit temperature T2 vary from 600 to 5000 K, evaluate theentropy change at intervals of 200 K, and plot the Rayleighline on a T-s diagram.

17–108E Air is heated as it flows through a 4 in � 4 insquare duct with negligible friction. At the inlet, air is at T1 �700 R, P1 � 80 psia, and V1 � 260 ft/s. Determine the rate atwhich heat must be transferred to the air to choke the flow atthe duct exit, and the entropy change of air during thisprocess.

17–109 Compressed air from the compressor of a gas tur-bine enters the combustion chamber at T1 � 550 K, P1 � 600kPa, and Ma1 � 0.2 at a rate of 0.3 kg/s. Via combustion,heat is transferred to the air at a rate of 200 kJ/s as it flowsthrough the duct with negligible friction. Determine the Machnumber at the duct exit and the drop in stagnation pressureP01 � P02 during this process. Answers: 0.319, 21.8 kPa

17–110 Repeat Prob. 17–109 for a heat transfer rate of 300kJ/s.

17–111 Argon gas enters a constant cross-sectional-areaduct at Ma1 � 0.2, P1 � 320 kPa, and T1 � 400 K at a rateof 0.8 kg/s. Disregarding frictional losses, determine thehighest rate of heat transfer to the argon without reducing themass flow rate.

17–112 Consider supersonic flow of air through a 6-cm-diameter duct with negligible friction. Air enters the duct atMa1 � 1.8, P01 � 210 kPa, and T01 � 600 K, and it is decel-erated by heating. Determine the highest temperature that aircan be heated by heat addition while the mass flow rateremains constant.

Steam Nozzles

17–113C What is supersaturation? Under what conditionsdoes it occur?

17–114 Steam enters a converging nozzle at 3.0 MPa and500°C with a negligible velocity, and it exits at 1.8 MPa. Fora nozzle exit area of 32 cm2, determine the exit velocity,mass flow rate, and exit Mach number if the nozzle (a) isisentropic and (b) has an efficiency of 90 percent. Answers:(a) 580 m/s, 10.7 kg/s, 0.918, (b) 551 m/s, 10.1 kg/s, 0.865

17–104 Repeat Prob. 17–103 assuming air is cooled in theamount of 55 kJ/kg.

17–105 Air is heated as it flows subsonically through aduct. When the amount of heat transfer reaches 60 kJ/kg, the

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17–115E Steam enters a converging nozzle at 450 psia and900°F with a negligible velocity, and it exits at 275 psia. Fora nozzle exit area of 3.75 in2, determine the exit velocity,mass flow rate, and exit Mach number if the nozzle (a) isisentropic and (b) has an efficiency of 90 percent. Answers:(a) 1847 ft/s, 18.7 lbm/s, 0.900, (b) 1752 ft/s, 17.5 lbm/s, 0.849

17–116 Steam enters a converging–diverging nozzle at 1 MPaand 500°C with a negligible velocity at a mass flow rate of 2.5kg/s, and it exits at a pressure of 200 kPa. Assuming the flowthrough the nozzle to be isentropic, determine the exit area andthe exit Mach number. Answers: 31.5 cm2, 1.738

17–117 Repeat Prob. 17–116 for a nozzle efficiency of 95percent.

Review Problems

17–118 Air in an automobile tire is maintained at a pres-sure of 220 kPa (gauge) in an environment where the atmo-spheric pressure is 94 kPa. The air in the tire is at theambient temperature of 25°C. Now a 4-mm-diameter leakdevelops in the tire as a result of an accident. Assuming isen-tropic flow, determine the initial mass flow rate of air throughthe leak. Answer: 0.554 kg/min

17–119 The thrust developed by the engine of a Boeing 777is about 380 kN. Assuming choked flow in the nozzles, deter-mine the mass flow rate of air through the nozzle. Take theambient conditions to be 265 K and 85 kPa.

17–120 A stationary temperature probe inserted into a ductwhere air is flowing at 250 m/s reads 85°C. What is theactual temperature of air? Answer: 53.9°C

17–121 Nitrogen enters a steady-flow heat exchanger at150 kPa, 10°C, and 100 m/s, and it receives heat in theamount of 125 kJ/kg as it flows through it. The nitrogenleaves the heat exchanger at 100 kPa with a velocity of 180m/s. Determine the stagnation pressure and temperature ofthe nitrogen at the inlet and exit states.

17–122 Derive an expression for the speed of sound basedon van der Waals’ equation of state P � RT(v � b) � a/v2.Using this relation, determine the speed of sound in carbondioxide at 50°C and 200 kPa, and compare your result to thatobtained by assuming ideal-gas behavior. The van der Waalsconstants for carbon dioxide are a � 364.3 kPa · m6/kmol2

and b � 0.0427 m3/kmol.

17–123 Obtain Eq. 17–10 by starting with Eq. 17–9 andusing the cyclic rule and the thermodynamic property relations

17–124 For ideal gases undergoing isentropic flows, obtainexpressions for P/P*, T/T*, and r/r* as functions of k and Ma.

17–125 Using Eqs. 17–4, 17–13, and 17–14, verify that forthe steady flow of ideal gases dT0 /T � dA/A � (1 � Ma2)

cp

T� a 0s

0Tb

P

and cv

T� a 0s

0Tb

v.

dV/V. Explain the effect of heating and area changes on thevelocity of an ideal gas in steady flow for (a) subsonic flowand (b) supersonic flow.

17–126 A subsonic airplane is flying at a 3000-m altitudewhere the atmospheric conditions are 70.109 kPa and 268.65 K.A Pitot static probe measures the difference between the staticand stagnation pressures to be 35 kPa. Calculate the speed ofthe airplane and the flight Mach number. Answers: 257 m/s,0.783

17–127 Plot the mass flow parameter m.

/(AP0) versusthe Mach number for k � 1.2, 1.4, and 1.6 in the range of 0 �Ma � 1.

17–128 Helium enters a nozzle at 0.8 MPa, 500 K, anda velocity of 120 m/s. Assuming isentropic flow, deter-mine the pressure and temperature of helium at a locationwhere the velocity equals the speed of sound. What is theratio of the area at this location to the entrance area?

17–129 Repeat Prob. 17–128 assuming the entrance veloc-ity is negligible.

17–130 Air at 0.9 MPa and 400 K enters a convergingnozzle with a velocity of 180 m/s. The throat

area is 10 cm2. Assuming isentropic flow, calculate and plotthe mass flow rate through the nozzle, the exit velocity, theexit Mach number, and the exit pressure–stagnation pressureratio versus the back pressure–stagnation pressure ratio for aback pressure range of 0.9 � Pb � 0.1 MPa.

17–131 Steam at 6.0 MPa and 700 K enters a con-verging nozzle with a negligible velocity. The

nozzle throat area is 8 cm2. Assuming isentropic flow, plotthe exit pressure, the exit velocity, and the mass flow ratethrough the nozzle versus the back pressure Pb for 6.0 � Pb� 3.0 MPa. Treat the steam as an ideal gas with k � 1.3, cp� 1.872 kJ/kg · K, and R � 0.462 kJ/kg · K.

17–132 Find the expression for the ratio of the stagnationpressure after a shock wave to the static pressure before theshock wave as a function of k and the Mach number upstreamof the shock wave Ma1.

17–133 Nitrogen enters a converging–diverging nozzle at 700kPa and 300 K with a negligible velocity, and it experiences anormal shock at a location where the Mach number is Ma �3.0. Calculate the pressure, temperature, velocity, Mach num-ber, and stagnation pressure downstream of the shock. Com-pare these results to those of air undergoing a normal shockat the same conditions.

17–134 An aircraft flies with a Mach number Ma1 � 0.8 atan altitude of 7000 m where the pressure is 41.1 kPa and thetemperature is 242.7 K. The diffuser at the engine inlet hasan exit Mach number of Ma2 � 0.3. For a mass flow rate of65 kg/s, determine the static pressure rise across the diffuserand the exit area.

1RT0

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17–147 Repeat Prob. 17–146 for helium.

17–148 Air is accelerated as it is heated in a duct with neg-ligible friction. Air enters at V1 � 100 m/s, T1 � 400 K, andP1 � 35 kPa and then exits at a Mach number of Ma2 � 0.8.Determine the heat transfer to the air, in kJ/kg. Also deter-mine the maximum amount of heat transfer without reducingthe mass flow rate of air.

17–149 Air at sonic conditions and static temperature andpressure of 500 K and 420 kPa, respectively, is to be acceler-ated to a Mach number of 1.6 by cooling it as it flows througha channel with constant cross-sectional area. Disregardingfrictional effects, determine the required heat transfer from theair, in kJ/kg. Answer: 69.8 kJ/kg

17–150 Saturated steam enters a converging–diverging noz-zle at 3.0 MPa, 5 percent moisture, and negligible velocity,and it exits at 1.2 MPa. For a nozzle exit area of 16 cm2,determine the throat area, exit velocity, mass flow rate, andexit Mach number if the nozzle (a) is isentropic and (b) hasan efficiency of 90 percent.

Fundamentals of Engineering (FE) Exam Problems

17–151 An aircraft is cruising in still air at 5°C at a velocityof 400 m/s. The air temperature at the nose of the aircraftwhere stagnation occurs is

(a) 5°C (b) 25°C (c) 55°C (d) 80°C (e) 85°C

17–152 Air is flowing in a wind tunnel at 15°C, 80 kPa,and 200 m/s. The stagnation pressure at the probe insertedinto the flow section is

(a) 82 kPa (b) 91 kPa (c) 96 kPa(d) 101 kPa (e) 114 kPa

17–153 An aircraft is reported to be cruising in still air at�20°C and 40 kPa at a Mach number of 0.86. The velocityof the aircraft is

(a) 91 m/s (b) 220 m/s (c) 186 m/s(d) 280 m/s (e) 378 m/s

17–135 Helium expands in a nozzle from 1 MPa, 500 K,and negligible velocity to 0.1 MPa. Calculate the throat andexit areas for a mass flow rate of 0.25 kg/s, assuming thenozzle is isentropic. Why must this nozzle be converging–diverging? Answers: 3.51 cm2, 5.84 cm2

17–136E Helium expands in a nozzle from 150 psia, 900R, and negligible velocity to 15 psia. Calculate the throat andexit areas for a mass flow rate of 0.2 lbm/s, assuming thenozzle is isentropic. Why must this nozzle be converging–diverging?

17–137 Using the EES software and the relations inTable A–32, calculate the one-dimensional

compressible flow functions for an ideal gas with k � 1.667,and present your results by duplicating Table A–32.

17–138 Using the EES software and the relations inTable A–33, calculate the one-dimensional

normal shock functions for an ideal gas with k � 1.667, andpresent your results by duplicating Table A–33.

17–139 Consider an equimolar mixture of oxygen andnitrogen. Determine the critical temperature, pressure, anddensity for stagnation temperature and pressure of 800 Kand 500 kPa.

17–140 Using EES (or other) software, determine theshape of a converging–diverging nozzle for air

for a mass flow rate of 3 kg/s and inlet stagnation conditionsof 1400 kPa and 200°C. Assume the flow is isentropic.Repeat the calculations for 50-kPa increments of pressuredrops to an exit pressure of 100 kPa. Plot the nozzle to scale.Also, calculate and plot the Mach number along the nozzle.

17–141 Using EES (or other) software and the rela-tions given in Table A–32, calculate the one-

dimensional isentropic compressible-flow functions byvarying the upstream Mach number from 1 to 10 in incre-ments of 0.5 for air with k � 1.4.

17–142 Repeat Prob. 17–141 for methane with k �1.3.

17–143 Using EES (or other) software and the rela-tions given in Table A–33, generate the one-

dimensional normal shock functions by varying the upstreamMach number from 1 to 10 in increments of 0.5 for air with k � 1.4.

17–144 Repeat Prob. 17–143 for methane with k �1.3.

17–145 Air is cooled as it flows through a 20-cm-diameterduct. The inlet conditions are Ma1 � 1.2, T01 � 350 K, and P01� 240 kPa and the exit Mach number is Ma2 � 2.0. Disregard-ing frictional effects, determine the rate of cooling of air.

17–146 Air is heated as it flows subsonically through a10 cm � 10 cm square duct. The properties of air at the inletare maintained at Ma1 � 0.4, P1 � 400 kPa, and T1 � 360 K

880 | Thermodynamics

at all times. Disregarding frictional losses, determine thehighest rate of heat transfer to the air in the duct withoutaffecting the inlet conditions. Answer: 1958 kW

P1 � 400 kPa

T1 � 360 K

Ma1 � 0.4

Qmax

FIGURE P17–146

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Chapter 17 | 881

(d) There will be no flow through the nozzle if the backpressure equals the stagnation pressure.

(e) The fluid velocity decreases, the entropy increases, andstagnation enthalpy remains constant during flowthrough a normal shock.

17–160 Combustion gases with k � 1.33 enter a convergingnozzle at stagnation temperature and pressure of 400°C and800 kPa, and are discharged into the atmospheric air at 20°Cand 100 kPa. The lowest pressure that will occur within thenozzle is

(a) 26 kPa (b) 100 kPa (c) 321 kPa(d) 432 kPa (e) 272 kPa

Design and Essay Problems

17–161 Find out if there is a supersonic wind tunnel onyour campus. If there is, obtain the dimensions of the windtunnel and the temperatures and pressures as well as theMach number at several locations during operation. For whattypical experiments is the wind tunnel used?

17–162 Assuming you have a thermometer and a device tomeasure the speed of sound in a gas, explain how you candetermine the mole fraction of helium in a mixture of heliumgas and air.

17–163 Design a 1-m-long cylindrical wind tunnel whosediameter is 25 cm operating at a Mach number of 1.8.Atmospheric air enters the wind tunnel through a converging–diverging nozzle where it is accelerated to supersonic veloci-ties. Air leaves the tunnel through a converging–diverging diffuser where it is decelerated to a very low velocity beforeentering the fan section. Disregard any irreversibilities. Specifythe temperatures and pressures at several locations as well asthe mass flow rate of air at steady-flow conditions. Why is itoften necessary to dehumidify the air before it enters the windtunnel?

17–154 Air is flowing in a wind tunnel at 12°C and 66 kPaat a velocity of 230 m/s. The Mach number of the flow is

(a) 0.54 m/s (b) 0.87 m/s (c) 3.3 m/s(d) 0.36 m/s (e) 0.68 m/s

17–155 Consider a converging nozzle with a low velocity atthe inlet and sonic velocity at the exit plane. Now the nozzleexit diameter is reduced by half while the nozzle inlet tem-perature and pressure are maintained the same. The nozzleexit velocity will

(a) remain the same (b) double (c) quadruple(d) go down by half (e) go down to one-fourth

17–156 Air is approaching a converging–diverging nozzlewith a low velocity at 20°C and 300 kPa, and it leaves thenozzle at a supersonic velocity. The velocity of air at thethroat of the nozzle is

(a) 290 m/s (b) 98 m/s (c) 313 m/s(d) 343 m/s (e) 412 m/s

17–157 Argon gas is approaching a converging–divergingnozzle with a low velocity at 20°C and 120 kPa, and it leavesthe nozzle at a supersonic velocity. If the cross-sectional areaof the throat is 0.015 m2, the mass flow rate of argon throughthe nozzle is

(a) 0.41 kg/s (b) 3.4 kg/s (c) 5.3 kg/s(d) 17 kg/s (e) 22 kg/s

17–158 Carbon dioxide enters a converging–diverging noz-zle at 60 m/s, 310°C, and 300 kPa, and it leaves the nozzle ata supersonic velocity. The velocity of carbon dioxide at thethroat of the nozzle is

(a) 125 m/s (b) 225 m/s (c) 312 m/s(d) 353 m/s (e) 377 m/s

17–159 Consider gas flow through a converging–divergingnozzle. Of the five following statements, select the one that isincorrect:

(a) The fluid velocity at the throat can never exceed thespeed of sound.

(b) If the fluid velocity at the throat is below the speed ofsound, the diversion section will act like a diffuser.

(c) If the fluid enters the diverging section with a Machnumber greater than one, the flow at the nozzle exit willbe supersonic.

P0Ma � 1.8 D � 25 cm

T0

FIGURE P17–163

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Errata Sheet Y. A. Çengel and M. A. Boles, Thermodynamics: An Engineering Approach, 5th ed, McGraw-Hill, 2006.

(Last update: Dec. 29, 2005) Page Location Error Correction

42 Prob. 1-50E, 2nd line 29.1 mm Hg 29.1 in Hg 45 Prob. 1-76, Answer 5.0 1.34 50 Prob. 1-128 It is a repeat of Prob. 1-127 Delete it.

177 Example 4-6, Assumptions Assumption #4 Delete it (Temp doesn’t have to be const) 327 Prob. 6-138, 2nd line Bobbler (spelling error) bubbler 461 Last line 4665 5072 462 1st calculation 4665, 0.923, 92.3% 5072, 0.849, 84.9% 462 1st line after 1st calculation 7.7 15.1 462 2nd calculation 4665-4306=359 5072-4306=766 462 1st line after 2nd calculation 359 766 619 Fig. 11-9 Light-gray and light-color lines They should be switched in both figures. 916 Table A-20, the line T=760 K 20,135 30,135

Note: Changes that will be made in the next edition are designated by “6th ed”.