Angela Foudray's Thermodynamics Slides

7/25/2019 Angela Foudray's Thermodynamics Slides

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Lecture 1

Systems and Measurement

Sections 1.1 1.4


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Why Thermodynamics?And where these techniques applied?


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infrastructure

transportation

ecosystems


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boundary

To learn about each of these, we cant treat everything all at once!

System

surroundings/

environment


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IsolatedClosed

(control mass)

Open

(control volume)

Types of systems

piston in cylinder engineinsulated bath


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Example

Are these systems isolated, closed or open?

electric car sliding on frictionless

ice, assume air drag is negligible,

boundary at surface of car

candle in a dark roomgreenhouse, windows closed,

boundary just outside structure

dog, boundary at surface


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How do you choose the boundary?

Depends on:

What you know

What you want to know

air intake

exhaust gas

-> external wheel speed

conventional car,

sliding on surface

with k = 0.1


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What are we looking at, when looking at the system?

Large scale features

Temperature

Pressure

Volume

System of particles

Molecular, atomic, quantum

energy levels

Macroscopic Microscopic

2 Regimes


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Ways we describe a system

Properties

independent of history of the

system

have numerical value

can be measured or

computed by looking at thesystem macroscopically

Examples:

pressure, temperature, mass, volume

A property is a property if, and only if, its change in value between two states is

independent of the process.

A test if a descriptor of a system is a property


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State

the set of particular values the

properties have at a particular time

P = 1.203 x 105 Pa

T = 385 K

V = 0.3 m3


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Process

how a system changes from one

state to another over time

steady state properties arent changing with time

equilibriumno net energy is transferred between system and surroundings


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Watch the video on

Equilibrium vs steady state

http://tll.mit.edu/help/equilibrium-vs-steady-state

from 1:30 7:30

(Of course, you can watch the rest if you like. It does describe some

engineering thought experiments, measurement and analysis but some

of the concepts are beyond the scope of this lecture)


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depends on the amount

not location dependent

can vary with time

MassLength

Volume

Shape

Energy

Temperature

Color

Pressure

Density

does not depend on the amount

can be location or time dependent

Intensive Properties


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Quantities, Units, and their Relationships

primary

secondary

(derived)

=

1 N = 1 kg m/s2SI

english 1 lbf = 1 lb * 32.1740 ft/s2


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Example (P1.13)

At a certain elevation, the pilot of a balloon has a mass of 120 lb and a weight

of 119 lbf. What is the local acceleration of gravity, in ft/s2, at that elevation?

If the balloon drifts to another elevation where g = 32.05 ft/s2,what is her

weight in lbf, and mass in lb?


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Work through For Example problemsin section 1.4

then

Work on homeworkproblem 1.10, 1.17


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Lecture 3

Measurements and Units

Sections 1.5 1.8


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Density () and Specific Volume (v)

From a macroscopic perspective, description

of matteris simplified by considering it to be

distributed continuously throughout a region.

When substances are treated as continua, it

is possible to speak of their intensivethermodynamic properties at a point.

At any instant the density (r ) at a point is

defined as

Vm

VVlim

'

r (Eq. 1.6)

where V ' is the smallest volume for which a definite

value of the ratio exists.


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Density is mass per unit volume.Density is an intensive property that may

vary from point to point.

SI units are (kg/m3).

English units are (lb/ft3).


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Specific volume is the reciprocal of

density: v = 1/r .

Specific volume is volume per unit mass.

Specific volume is an intensive propertythat may vary from point to point.

SI units are (m3/kg).

English units are (ft3/lb).

Specific volume is usually preferred for

thermodynamic analysis when working with

gases that typically have small density values.


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Molar Quantities of Matter (n)

We can quantify how much matter we have on a

molar basis instead:

M

mn (Eq. 1.8)

where M is molecular weight and

n (kmol) m (kg) M (kg/kmol)

n (lbmol) m (lb) M (lb/lbmol)n (mol, gmol) m (g) M (g/mol)

n: mol, kmol, lbmol

NA = 6.022 x 10^23 molecules/mol


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Example:

A) Is Avogadros number the same when written in terms of

# of molecules/lbmol? (1 lb = 453.6 g)

B) To the right is a list of molecular weights. If the table is in

g/mol, what would the values be in kg/kmol and lb/lbmol?


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Notation: (for some variable x)

x: x in terms of mol, kmol, lbmol

m

V

Vm

MM

n

VV

n

1

(multiply both sides by M)

M

mn

M (Eq. 1.9)


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Pressure (p)

Consider a small area A passing through a point

in a fluid at rest.The fluid on one side of the area exerts a

compressive force that is normal to the area, Fnormal.

An equal but oppositely directed force is exerted on

the area by the fluid on the other side.

The pressure (p) at the specified point is defined

as the limit

Anormal

'AA

lim Fp (Eq. 1.10)

whereA' is the area at the point in the same limiting sense

as used in the definition of density.


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Pressure Units

SI unit of pressure is the pascal:1 pascal = 1 N/m2

Multiples of the pascal are frequently used:

1 kPa = 103 N/m21 bar = 105 N/m2

1 MPa = 106 N/m2

English units for pressure are:pounds force per square foot, lbf/ft2

pounds force per square inch, lbf/in.2


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Absolute pressure: Pressure with respect tothe zero pressure of a complete vacuum.

Absolute pressure mustbe used in

thermodynamic relations.

Pressure-measuring devices often indicate

the difference between the absolute pressure of

a system and the absolute pressure of the

atmosphere outside the measuring device.

Absolute Pressure


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When system pressure is greater thanatmospheric pressure, the term gage

pressure is used.

p(gage) =p(absolute)patm(absolute)

(Eq. 1.14)

When atmospheric pressure is

greater than system pressure, the term

vacuum pressure is used.

p(vacuum) =patm(absolute) p(absolute)

(Eq. 1.15)

Gage and Vacuum Pressure


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Work through For Example problemin section 1.6

then

Work on homework problems

1.26, 1.40


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Temperature (T)

If two blocks (one warmer than the other) are

brought into contact and isolated from their

surroundings, they would interact thermally with

changes in observable properties.

When all changes in observable properties cease,the two blocks are in thermal equilibrium.

Temperature is a physical property that

determines whether the two objects are in thermal

equilibrium.


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Any object with at least one measurable property

that changes as its temperature changes can be

used as a thermometer.

Such a property is called a thermometric

property.The substance that exhibits changes in the

thermometric property is known as a thermometric

substance.

Thermometers


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Example: Liquid-in-glass thermometer

Consists of glass capillary tube connected to a bulb filledwith liquid and sealed at the other end. Space above liquid

is occupied by vapor of liquid or an inert gas.

As temperature increases, liquid expands in volume and

rises in the capillary. The length (L ) of the liquid in thecapillary depends on the temperature.

The liquid is the thermometric substance.

L is the thermometric property.

Other types of thermometers:Thermocouples

Thermistors

Radiation thermometers and optical pyrometers

Thermometers


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Kelvin scale: An absolute thermodynamic temperature

scale whose unit of temperature is the kelvin (K); an SI base

unit for temperature.

Rankine scale: An absolute thermodynamic temperature

scale with absolute zero that coincides with the absolute zeroof the Kelvin scale; an English base unit for temperature.

Temperature Scales

T(oR) = 1.8T(K) (Eq. 1.16)

Celsius scale (o

C):T(oC) = T(K)273.15 (Eq. 1.17)

Fahrenheit scale (oF):

T(

o

F) = T(

o

R)

459.67 (Eq. 1.18)


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Work through For Example problemin section 1.7

then


1.52, 1.56


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Problem-Solving Methodology

Known: Read the problem, think about it, and

identify what is known.

Find: State what is to be determined.

Schematic and Given Data: Draw a sketch of

system and label with all relevant information/data.

Engineering Model: List all simplifying

assumptions and idealizations made.

Analysis: Reduce appropriate governingequations and relationships to forms that will

produce the desired results.


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Lecture 4

Energy and Work

Sections 2.1 2.3

(not 2.2.6, 2.2.7, 2.2.8)


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Closed System Energy Balance

Energy is an extensive property thatincludes the kinetic and gravitational potential

energy of engineering mechanics.

For closed systems, energy is transferred inand out, across the system boundary, by two

means only: by work and by heat.

Energy is conserved

1st law of thermodynamics


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Lets look deeper into this energy balance,

including what we mean by energy change

and energy transfer.

The closed system energy balance states:

Net amount of energy

transferred in and out

across the system boundaryby heat and work during

the time interval

The change in the

amount of energy

contained withina closed system

during some time

interval


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Change in Energy of a System

In engineering thermodynamics the changein energy of a system has three parts:

Kinetic energy

Gravitational potential energy

Internal energy


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Change in Kinetic Energy

The change in kinetic energy is associated withthe motion of the system as a whole relative to

an external coordinate frame such as the surface

of the earth.

For a system of mass m the change in kinetic

energy from state 1 to state 2 is

KE = KE2 KE1 = ( )

2

1

2

2

VV

2

1m (Eq. 2.5)

whereV1 and V2 are the initial and final velocity magnitudes.

is the final value minus initial value.


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Change in Gravitational Potential Energy

The change in gravitational potential energy isassociated with the position of the system in the

earths gravitational field.

For a system of mass m the change in potential

energy from state 1 to state 2 is

PE = PE2 PE1 = mg(z2z1) (Eq. 2.10)

wherez1 andz2 are the initial and final elevations relative to

the surface of the earth, respectively.

g is acceleration due to gravity.


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Change in Internal Energy

The change in internal energy is associated with the

makeup of the system, including its chemical

composition.

There is no simple expression like Eqs. 2.5 and 2.10 for

finding internal energy change for a wide range of

applications. Usually, we will be able to use the datafrom tables in appendices of the textbook.

Like kinetic and gravitational potential energy, internal

energy is an extensive property.

Internal energy is represented by U.The specific internal energy on a mass basis is u.

The specific internal energy on a molar basis is .u


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Change in Energy of a System

So, the total change in energy of a system from

state 1 to state 2 is

E2E1 = (U2U1) + (KE2 KE1) + (PE2 PE1)

(Eq. 2.27a)

E= U+ KE + PE (Eq. 2.27b)

Since an arbitrary valueE1 can be given for the

energy of a system at state 1, no particularsignificance can be attached to the value of

energy at state 1 or any other state. Only

changes in the energy of a system between

states have significance.


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When a spring is compressed,

energy is transferred to the spring bywork.

When a gas in a closed vessel is

stirred, energy is transferred to the

gas by work.

When a battery is charged

electrically, energy is transferred to

the battery contents by work.

Illustrations of Work

The first two examples of work are familiar from

mechanics. The third example is an example of this

broader interpretation of work.


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Energy Transfer by Work

The symbol Wdenotes an amount of energy

transferred across the boundary of a system by work.

Since engineering thermodynamics is often

concerned with internal combustion engines,

turbines, and electric generators whose purpose is todo work, convention is that the work done by a

system as positive.

W> 0: work done by the system

W< 0: work done on the system

The same sign convention is used for the rate of

energy transfer by work called power, given by .W


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Modeling Expansion and Compression Work

We will see, a useful example to study is a gas (or

liquid) undergoing an expansion (or compression)process while confined in a piston-cylinder

assembly.

During the process, the gas exerts a normal force

on the piston, F=pA , wherep is the pressure at the

interface between the gas and piston and A is the

area of the piston face.


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Modeling Expansion and Compression Work

From mechanics, the work done by the gas as the

piston face moves fromx1 tox2 is given by

== dxpFdxW A

Since the product Adx = dV , whereVis the volumeof the gas, this becomes

(Eq. 2.17)

= pdVW

V1

V2

For compression, dVis negative and so is the

value of the integral, which keeps with the sign

convention for work.


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Quasiequilibrium Processes

In a quasiequilibriumexpansion, the gas moves

along a pressure-volume

curve, or path, as shown.

Using Eq. 2.17, the work

done by the gas on the

piston is given by the area

under the curve of pressureversus volume.


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Polytropic Indices


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Polytropic Indices

no mass transfer

no heat transferadiabatic (closed system)

n = 0 pV0 = p = const.

(isobaric)

n < 1 systems with high T, high thermal energyinput

n = 1 pV = ZnRT = const.

(isothermal)

1 < n < k quasi-adiabatic

n = k adiabatic, reversible

n = pV, V dominates -> V=const.(isochoric)

k: specific heat ratio


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Work through For Example problems

in sections 2.1 and 2.2,and Example Problem 2.1

then

Work on homework problems2.7, 2.14 (answer in SI),

2.23(a&b only use MATLAB for b)


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Lecture 5

Heat and Cycles

Sections 2.4 & 2.6


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Energy Transfer by Heat

The symbol Qdenotes an amount of energy

transferredacross the boundary of a system by heat.

Since engineering thermodynamics is often

concerned with steam engines, internal combustion

engines and turbines, and that heat will need to beadded to the system for it to function, convention is

that the heat transfer into a system as positive.

Q> 0: heat t ransfer to the system

Q< 0: heat t ransfer f rom the systemThe same sign convention is used for the rate of

energy transfer by heat, given by

Note: this is the opposite convention from work!

.Q


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Modes of Heat Transfer

For any particular application, energy

transfer by heatcan occur by one or more of

three modes:

conduction

radiation

convection


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Conduction

Conductionis the transfer of energyfrom more energetic particles of a

substance to less energetic adjacent

particles due to interactions between

them.The time rate of energy transfer by

conduction is quantified by Fouriers

law.

An application of Fouriers law to a

plane wall at steady state is shown at

right.


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wherekis a proportionality constant, a property of the wall

material called the thermal conductivity.

The minus signis a consequence of energy transfer in

the direction of decreasing temperature.

Conduction

By Fouriers law, the rate of heat transfer across any

plane normal to the xdirection, , is proportional to thewall area, A, and the temperature gradient in the x

direction, dT/dx,

xQ

dx

dTQx Ak (Eq. 2.31)

In this case, temperature varies linearly with x, and thus

L

TTQx

12Akand Eq. 2.31gives)0(12

L

TT

dx

dT


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Thermal Radiation

Thermal radiationis energy transported byelectromagnetic waves (or photons). Unlike

conduction, thermal radiation requires no

intervening medium and can take place in a

vacuum.The time rate of energy transfer by radiation is

quantified by expressions developed from the

Stefan-Boltzmanlaw.


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Net energy is transferred in the direction of the arrowand quantified by

whereAis the areaof the smaller surface,

is a property of the surface called its emissivity,

sis the Stefan-Boltzman constant.

Thermal Radiation

An application involving net

radiation exchange between asurface at temperature Tband a

much larger surface at Ts(< Tb)

is shown at right.

]A[ 4s4be TTQ es

(Eq. 2.33)


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Convection

Convectionis energy transfer between a solidsurface and an adjacent gas or liquid by the

combined effects of conduction and bulk flow

within the gas or liquid.

The rate of energy transfer by convection isquantified by Newtons law of cooling.


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Energy is transferred in the direction of the arrow and

quantified by

whereAis the areaof the transistors surface and

his an empirical parameter called the convection heat

transfer coefficient.

Convection

An application involving

energy transfer byconvection from a transistor

to air passing over it is

shown at right.

]hA[ fbc TTQ (Eq. 2.34)


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Work through For Example problem

in section 2.4.2, andExample Problems 2.4, 2.5

then

Work on homework problem

2.51


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Thermodynamic Cycles

A thermodynamic cycleis a sequence of

processes that begins and ends at the same

state.

Examples of thermodynamic cycles include

Power cyclesthat develop a net energy transfer byworkusing an energy input by heat transfer from hot

combustion gases.

Refrigeration cyclesthat provide coolingfor a

refrigerated space using an energy input by work inthe form of electricity.

Heat pump cyclesthat provide heatingto a dwelling

using an energy input by work in the form of electricity.


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The energy transfers by heat and work

shown on the figure are each positive in the

direction of the accompanying arrow. This

convention is commonly used for analysisof thermodynamic cycles.

Power Cycle

A system undergoing a power cycle is

shown at right.

Wcycleis the net energy transfer by workfrom the system

per cycle of operationin the form of electricity, typically.

Qinis the heat transfer of energy to the systemper cycle

from the hot bodydrawn from hot gases of combustion orsolar radiation, for instance.

Qoutis the heat transfer of energy from the systemper

cycle to the cold bodydischarged to the surrounding

atmosphere or nearby lake or river, for example.

C


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Power Cycle

Applying the closed system energy balance to each

cycle of operation,

(Eq. 2.39)DEcycle= QcycleWcycle

Since the system returns to its initial state after eachcycle, there is no net change in its energy: Ecycle= 0,

and the energy balance reduces to give

(Eq. 2.41)Wcycle= QinQout

In words, the netenergy transfer by work from the

systemequals the netenergy transfer by heat to the

system, each per cycle of operation.


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Power Cycle


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Power CycleUsing the second law of thermodynamics (Chapter 5), we will

show that the value of thermal efficiency must be less thanunity: < 1 (< 100%). That is, only a portion of the energy

added by heat,Qin,can be obtained as work. The remainder,

Qout, is discharged.

Example: A system undergoes a power cycle while receiving


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p y g p y g

1000 kJ by heat transfer from hot combustion gases at a

temperature of 500 K and discharging 600 kJ by heat transfer

to the atmosphere at 300 K. Taking the combustion gases and

atmosphere as the hot and cold bodies, respectively, determine

for the cycle, the net work developed, in kJ, and the thermal

efficiency.

Substituting into Eq. 2.41,

Wcycle= 1000 kJ600 kJ =400 kJ.

Then, with Eq. 2.42,= 400 kJ/1000 kJ = 0.4 (40%).

Note the thermal efficiency is commonly reported on a

percent basis.

Wcycle= QinQout

in

cycle

Q

W


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Heat Pump Cycle

As for the refrigeration cycle, the energy balancereads

(Eq. 2.44)Wcycle= QoutQin

As before,Wcycleis the netenergy transfer by workto

the system per cycle,

usually provided in the form

of electricity.

Heat Pump Cycle


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Heat Pump Cycle

The performance of a system undergoing a heat

pump cycleis evaluated on an energy basis as theratio of energy provided to the hot body, Qout, to the

net work required to accomplish this effect, Wcycle:

(heat pump cycle) (Eq. 2.47)

called the coefficient of performancefor the heat pump

cycle.

cycleout

W

Q

We can also write this as:

inout

out

QQ

Q

(heat pump cycle) (Eq. 2.48)

Example: A system undergoes a heat pump cycle


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p y g p p y

while discharging 900 kJ by heat transfer to a dwelling

at 20oC and receiving 600 kJ by heat transfer from the

outside air at 5oC. Taking the dwelling and outside airas the hot and cold bodies, respectively, determine for

the cycle, the net work input, in kJ, and the coefficient

of performance.

Substituting into Eq. 2.44,

Wcycle= 900 kJ600 kJ =300 kJ.

Then, with Eq. 2.47,= 900 kJ/300 kJ = 3.0.

cycle

out

W

Q

Wcycle= QoutQin


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in section 2.6

then

Work on homework problems2.71, 2.74


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Lect re 7


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Lecture 7

Using Data Tables

Sections 3.4, 3.5, 3.6, 3.8, 3.8.1, 3.9, 3.10.1


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Si l Ph R i


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Single-Phase Regions

Since pressure and temperatureare independent properties in

the single-phase liquid and

vapor regions, they can be used

to fix the state in these regions.Tables A-4/A-4E (Superheated

Water Vapor) andA-5/A-5E

(Compressed Liquid Water)

provide several properties asfunctions of pressure and

temperature, as considered

next.

Table A-4/A-4E

Table A-5/A-5E

Single-Phase Regions


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Temperature (T)Pressure (p)

Specific volume (v)

Specific internal energy (u)

Specific enthalpy (h), which is a

sum of terms that often appears in

thermodynamic analysis:

h = u +pv

Single Phase Regions

Table A-4/A-4E

Table A-5/A-5E

Specific entropy (s), an intensive property developed in

Chapter 6

Properties tabulated in Tables A-4 andA-5 include

Enthalpy is a property because it is defined in terms of

properties; physical significance is associated with it in Chapter 4.

Example: Single-Phase Regions of Tables

Wh t th ifi l ifi th l d ifi


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What are the specific volume, specific enthalpy and specific

internal energy for superheated water vapor at 10 MPa and

400oC?


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Linear Interpolation


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When a state does not fall exactly on the grid of values provided

by property tables, linear interpolation between adjacent entries

is used.

Example: Specific volume (v) associated with superheated

water vaporat 10 barand 215oC is found by linear interpolation

between adjacent entries in Table A-4.

ToC

v

m3/kgu

kJ/kgh

kJ/kgs

kJ/kgK

p= 10 bar = 1.0 MPa(Tsat= 179.91oC)

Sat. 0.1944 2583.6 2778.1 6.5865

200 0.2060 2621.9 2827.9 6.6940

240 0.2275 2692.9 2920.4 6.8817

Table A-4

(0.2275 0.2060) m3/kg (v 0.2060) m3/kg

(240 200)oC (215 200)oCslope = = v = 0.2141 m3/kg

Two-Phase Liquid-Vapor Region


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Two Phase Liquid Vapor RegionTables A-2/A-2E

(Temperature Table) andA-3/A-3E (Pressure Table)

provide

saturated liquid (f) data

saturated vapor (g) data

Specific Volume

m3/kgInternal Energy

kJ/kgEnthalpy

kJ/kgEntropykJ/kgK

TempoC

Press.bar

Sat.Liquid

vf103

Sat.Vapor

vg

Sat.Liquid

uf

Sat.Vapor

ug

Sat.Liquid

hf

Evap.

hfg

Sat.Vapor

hg

Sat.Liquid

sf

Sat.Vapor

sg

TempoC

.01 0.00611 1.0002 206.136 0.00 2375.3 0.01 2501.3 2501.4 0.0000 9.1562 .01

4 0.00813 1.0001 157.232 16.77 2380.9 16.78 2491.9 2508.7 0.0610 9.0514 4

5 0.00872 1.0001 147.120 20.97 2382.3 20.98 2489.6 2510.6 0.0761 9.0257 5

6 0.00935 1.0001 137.734 25.19 2383.6 25.20 2487.2 2512.4 0.0912 9.0003 6

8 0.01072 1.0002 120.917 33.59 2386.4 33.60 2482.5 2516.1 0.1212 8.9501 8

Table A-2

Table note: For saturated liquid specific volume, the table heading is vf103.

At 8oC, vf 103 = 1.002 vf= 1.002/10

3 = 1.002 103.



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The specific volume of a two-phase liquid-

vapor mixture can be determined by using

the saturation tables and quality, x.

The total volume of the mixture is the sum

of the volumes of the liquid and vapor

phases:

q p g

V= Vliq + Vvap

Dividing by the total mass of the mixture,m, an average

specific volume for the mixture is:

m

V

m

V

m

V vapliq+==v

With Vliq =mliqvf , Vvap =mvapvg ,mvap/m =x , andmliq/m = 1 x :

v = (1 x)vf+xvg = vf+x(vg vf) (Eq. 3.2)

m

V

m

V

m

V vapliq +==v Vliq =mliqvf , Vvap =mvapvg ,mvap/m =x ,mliq/m = 1 x


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v = (1 x)vf+xvg = vf+x(vg vf)



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Two Phase Liquid Vapor Region

Since pressure and temperature are NOT

independent properties in the two-phase liquid-vapor region, they cannot be used to fix the state

in this region.

The property, quality (x), defined only in the two-phase liquid-vapor region, and either temperature

or pressure can be used to fix the state in this

region.

v = (1 x)vf+xvg = vf+x(vg vf) (Eq. 3.2)

u = (1 x)uf+xug = uf+x(ug uf) (Eq. 3.6)

h = (1 x)hf+xhg =hf+x(hg hf) (Eq. 3.7)



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Two Phase Liquid Vapor Region

Example: A system consists of a two-phase liquid-vapor

mixture of waterat 6oC and a quality of 0.4. Determine thespecific volume, in m3/kg, of the mixture.

Solution: Apply Eq. 3.2, v = vf+x(vg vf)

Specific Volume

m3/kgInternal Energy

kJ/kgEnthalpy

kJ/kgEntropykJ/kgK

TempoC

Press.bar

Sat.Liquid

vf103

Sat.Vapor

vg

Sat.Liquid

uf

Sat.Vapor

ug

Sat.Liquid

hf

Evap.

hfg

Sat.Vapor

hg

Sat.Liquid

sf

Sat.Vapor

sg

TempoC

.01 0.00611 1.0002 206.136 0.00 2375.3 0.01 2501.3 2501.4 0.0000 9.1562 .01

4 0.00813 1.0001 157.232 16.77 2380.9 16.78 2491.9 2508.7 0.0610 9.0514 4

5 0.00872 1.0001 147.120 20.97 2382.3 20.98 2489.6 2510.6 0.0761 9.0257 5

6 0.00935 1.0001 137.734 25.19 2383.6 25.20 2487.2 2512.4 0.0912 9.0003 6

8 0.01072 1.0002 120.917 33.59 2386.4 33.60 2482.5 2516.1 0.1212 8.9501 8

Table A-2

Substituting values from Table 2: vf= 1.001103 m3/kg and

vg = 137.734 m3/kg:

v = 1.001103 m3/kg + 0.4(137.734 1.001103) m3/kg

v = 55.094 m3/kg

Property Data Use in the

Cl d S t E B l


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Example: A piston-cylinder assembly contains 2 kg of

water vapor at 100oC and 1 bar. The water vapor iscompressed to a saturated vaporstate where the

pressure is 2.5 bar. During compression, there is a heat

transfer of energy from the vapor to its surroundings

having a magnitude of 250 kJ. Neglecting changes inkinetic energy and potential energy, determine the work,

in kJ, for the process of the water vapor.

State 1

2 kg

of waterT1 = 100

oC

p1 = 1 bar

State 2

Saturated vapor

p2 = 2.5 bar

Q = 250 kJ

2

T

v

p1 = 1 bar

1

p2 = 2.5 bar

T1 = 100oC

Property Data Use in the

Cl d S t E B l


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Closed System Energy BalanceSolution: An energy balance for the closed system is

KE + PE +U= Q W0 0

where the kinetic and potential energy changes are neglected.

Thus W=Q m(u2 u1)

State 1 is in the superheated vaporregion and is fixed byp1 = 1 bar and T1 = 100

oC. From Table A-4, u1 = 2506.7 kJ/kg.

State 2 is saturated vaporatp2 = 2.5 bar. From Table A-3,

u2 = ug = 2537.2 kJ/kg.W= 250 kJ (2 kg)(2537.2 2506.7) kJ/kg = 311 kJ

The negative sign indicates work is done on the system as

expected for a compression process.

A Small Primer on Partial Derivatives


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A Small Primer on Partial Derivatives

A dependent variable can be a function of many independent

variables, for instance:

We can then label which variable(s) we held constant when we

were taking the partial derivative:

EDyzyzCxBxyzyxA +++= 222),,(

00)2()1(2 +++=

xCyzBy

x

A

The partial derivative is the derivative of the function with respect

to one variable as if all the other variables were constants:

yx

A

CxyzBy 22 +=

partial derivative of A with respect to x,

holding y constant


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Property Approximations for Liquids


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Approximate values for v, u, andh at liquid states can be

obtained using saturated liquid data.

Since the values of v and u for liquidschange very little with pressure at a fixed

temperature, Eqs. 3.11 and 3.12 can be used

to approximate their values.

v(T,p) vf(T)u(T,p) uf(T)

(Eq. 3.11)(Eq. 3.12)

An approximate value forh at liquid states can be obtained usingEqs. 3.11 and 3.12 in the definitionh = u + pv: h(T, p) uf(T) + pvf(T)

or alternatively h(T,p) hf(T) + vf(T)[p psat(T)] (Eq. 3.13)

wherepsat denotes the saturation pressure at the given temperature

When the underlined term in Eq. 3.13 is small

h(T,p) hf(T) (Eq. 3.14)

Saturated

liquid


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in sections 3.5, 3.6, ,and Example Problems

3.1, 3.2, 3.3, 3.4

then


3.42, 3.54 (answer in SI)

Lecture 8


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Lecture 8

Using Data Tables

Sections 3.5 & 3.6

Example: What is the specific volume of water at a

state where p = 10 bar and T = 215oC?


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state where p = 10 bar and T = 215oC?

Quality

isobaric lines(constant pressure)

Tempera

ture

Specific Volume

liquid

supercriticalfluid

vapor

saturated region

(vapor dome)


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p = 10 bar and T = 215oC


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Quality

isobaric lines

(constant pressure)

Temperature

Specific Volume

liquid

supercriticalfluid

vapor

saturated region

(vapor dome)



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(x1, y1)

(x2, y2)

(x3, y3)

(y2 - y1)

(x2 - x1)

(y3 - y1)

(x3 - x1)

(y2 - y1) (y3 - y1)


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y3 is our unknown spec. vol.

(x2 - x1) (x3 - x1)

multiply both sides by (x3 - x1)

add y1 to both sides



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Make sure you understand all for example


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Make sure you understand all for example ,

Sample and in-class problems to date, which

includes: Which data table corresponds to which region of

a T-v diagram and/or which approximations hold

there.

What phase(s) exist in a region of the T-vdiagram.

How p changes in different regions of the T-v

diagram.

Lecture 9


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Ideal Gas Law

Compressibility Factor

Sections 3.11, 3.12

Incompressibility of Liquids


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Approximate values for v, u, and h at liquid states can be

obtained using saturated liquid data.

Since the values of v and u for liquids

change very little with pressure at a fixed

temperature, Eqs. 3.11 and 3.12 can be used

to approximate their values.

v(T, p) vf(T)

u(T, p) uf(T)

(Eq. 3.11)

(Eq. 3.12)h(T, p) hf(T) (Eq. 3.14)

Saturated

liquid

Generalized Compressibility Chart


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vThe p- -Trelation for 10 common gases is

shown in the generalized compressibility chart.



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TR

pZ

v

In this chart, the compressibility factor, Z, is plotted versus

the reduced pressure, pR, and reduced temperature TR,

wherep

R= p/pc TR= T/Tc

(Eq. 3.27) (Eq. 3.28)(Eq. 3.23)

The symbols pc

and Tc

denote the

temperature and pressure at the

critical point for the particular gas

under consideration. These values

are obtained from Tables A-1 and

A-1E.

R

8.314 kJ/kmolK

1.986 Btu/lbmoloR

1545 ftlbf/lbmoloR

(Eq. 3.22)

is the universal gas constant

R



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When p, pc, T, T

c, v , and are used in consistent units, Z,

pR, and T

Rare numerical values without units.

Example: For air at 200 K, 132 bar, TR = 200 K/133 K= 1.5,pR = 132 bar/37.7 bar= 3.5 where Tc and pc for air are from

Table A-1. With these TR, p

Rvalues, the generalized

compressibility chart gives Z= 0.8.

R

With v = Mv,

an alternativeform of the compressibility

factor is

Z=pv

RT

(Eq. 3.24)

where

R= R

M(Eq. 3.25)

Studying the Generalized Compressibility Chart


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The solid lines labeled with TRvalues represent best fits to

experimental data. For the 10 different gases represented

there is little scatter in data about these lines.

At the lowest reduced

temperature value

shown, TR = 1.0, thecompressibility factor

varies between 0.2

and 1.0. Less

variation is observedas T

Rtakes higher

values.



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For each specified value of TR, the compressibility factor

approaches a value of 1.0 as pR

approaches zero.



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Low values of pR, where Z 1, do not necessarily

correspond to a range of low absolute pressures.For instance, if pR= 0.05, then p= 0.05pc. With pc values

from Table A-1

These pressure values range from 1.9 to 11 bar, which in

engineering practice are not normally considered as low

pressures.

Water vapor pc

= 220.9 bar p= 11 bar

Ammonia pc = 112.8 bar p= 5.6 bar

Carbon dioxide pc = 73.9 bar p= 3.7 bar

Air pc = 37.7 bar p= 1.9 bar

Introducing the Ideal Gas Model


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To recap, the generalized

compressibility chart shows that

at states where the pressure p

is small relative to the critical

pressure pc

(where pR

is small),

the compressibility factor Z is

approximately 1.

At such states, it can be assumed with reasonable

accuracy that Z = 1. Then

pv = RT (Eq. 3.32)



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g

Three alternative forms of Eq. 3.32 can be derived

as follows:

With v = V/m, Eq. 3.32 gives

pV= mRT (Eq. 3.33)

With v = v/M and R= R/M, Eq. 3.32 gives

(Eq. 3.34)TRp v

Finally, with v = V/n, Eq. 3.34 gives

pV= nRT (Eq. 3.35)



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While the ideal gas model does not provide an

acceptable approximation generally, in thelimiting case considered in the discussion of the

compressibility chart, it is justified for use, and

indeed commonly applied in engineeringthermodynamics at such states.

Appropriateness of the ideal gas model can be

checked by locating states under consideration

on one of the generalized compressibility charts

provided by appendix figures Figs. A-1 through

A-3.

Work Example Problems


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Work Example Problems

3.7, 3.8

then


3.103, 3.113

Lecture 11


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Control Volumes

Mass Rate Balance

Sections 4.1 4.3

Mass Rate Balance


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time rate of changeof

mass contained within the

control volume at time t

time rate of f lowof

mass inacross

inlet i at time t

time rate of f low

of mass outacross

exit e at time t

ei mmdt

dm cv (Eq. 4.1)

Mass Rate Balance


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e

e

i

i mm

dt

dm

cv(Eq. 4.2)

Eq. 4.2 is the mass rate balance for control

volumes with several inlets and exits.

In practice there may be several locations on theboundary through which mass enters or exits.

Multiple inlets and exits are accounted for by

summing over all entrances and exits:

Example: (Problem 4.2)Liquid propane enters and initially empty cylindrical storage tank at a mass flow rate

of 10 kg/s. Flow continues until the tank is filled with propane at 20oC, 9 bar. The

tank is 25 m long and has a 4 m diameter Determine the time in minutes to fill the


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tank is 25 m long and has a 4 m diameter. Determine the time, in minutes, to fill the

tank.


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Example: (Problem 4.4)Data are provided for the crude oil

storage tank. Determine:

a) The mass of oil in the tank in kg


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a) The mass of oil in the tank, in kg,

after 24 hours, and

b) The volume of oil in the tank, in

m3, at 24 hours


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Mass Rate Balance(Steady State Form)


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(Steady-State Form)

Steady-state: all properties are unchangingin time.

For steady-state control volume, dmcv/dt= 0.

e

e

i

i mm (Eq. 4.6)

(mass rate in) (mass rate out)

e

e

i

i mm

dt

dm

cv

Work through


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Work through

For Example problems in section 4.2

Example Problems 4.1 and 4.2

then


4.9, 4.16

Lecture 12


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Energy Rate Balance

Sections 4.4 4.5

Mass Rate Balance


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timerate of changeof

mass contained within the

control volumeattime t

timerate of flowof

mass inacross

inlet i attime t

timerate of flow

of massoutacross

exit e attime t

ei mmdt

dm =cv (Eq. 4.1)

Mass Rate Balance


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=e

e

i

i mmdt

dm

cv(Eq. 4.2)

Eq.4.2is the mass rate balancefor control

volumes with several inlets and exits.

In practice there may be several locations on theboundary through which mass enters or exits.

Multiple inletsand exitsare accounted for by

summing over all entrances and exits:

Energy Rate Balance


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)2

V()

2

V(

22cv

ee

eeii

ii gzumgzumWQdt

dE+++++= (Eq. 4.9)

timerate of changeof the energy

contained within

the control volume

attime t

netrateat whichenergy is being

transferred in

by heat transfer

attime t

netrateat whichenergy is being

transferred out

by workat

time t

netrateof energytransfer intothe

control volume

accompanying

mass flow

inlets and exitsinteraction with surroundings

control volume

Evaluating Work for a Control Volume


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)()(cv iiieee vpmvpmWW += (Eq. 4.12)

The expression for work is

cvW accounts for work associated with rotating

shafts, displacement of the boundary, and electrical

effects.

where

)( eee vpm is the flow workat exit e.

)( iii vpm is the flow workat inlet i.

Control Volume Energy Rate Balance(One-Dimensional Flow Form)


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)2

V()

2

V(

22

cvcvcv

ee

eeeeii

iiii gzvpumgzvpumWQdt

dE+++++++=

(Eq. 4.13)

For convenience substitute enthalpy,h= u+pv

)2

V()

2

V(

22

cvcvcv

ee

eeii

ii gzhmgzhmWQdt

dE+++++=

(Eq 4 14)

Using Eq. 4.12 in Eq. 4.9

control volume

interaction with surroundings with closed system defs.inlets and exits

Control Volume Energy Rate Balance(One-Dimensional Flow Form)


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(One Dimensional Flow Form)

++ +++=e

ee

eei

ii

ii gzhmgzhmWQdt

dE)

2

V()

2

V(

22

cvcvcv

(Eq. 4.15)

Eq. 4.15is the accounting balancefor the

energy of the control volume.

In practice there may be several locationson the boundary through which mass enters

or exits. Multiple inletsand exitsare

accounted for by introducing summations:

Control Volume Energy Rate Balance(Steady-State Form)


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(Steady State Form)

++ +++=e

ee

eei

ii

ii gzhmgzhmWQ )2

V()

2

V(0

22

cvcv

(Eq. 4.18)

Steady-state: all properties are unchangingin time.

For steady-state control volume,dEcv/dt= 0.

Control Volume Energy Rate Balance(Steady-State Form, One-Inlet, One-Exit)


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Many important applications involve one-inlet,one-exitcontrol volumes at steady state.

The mass rate balance reduces to .mmm == 21

+

++= )(

2

)V(V)(0 21

22

21

21cvcv zzghhmWQ

Eq.

4.20a

or dividing by mass flow rate

)(2

)V(V)(0 21

22

21

21cvcv zzghhm

W

m

Q+

++=

Eq.

4.20b

Example: (Problem 4.26)Air enters a horizontal, constant-diameter heating duct operating at steady state at

290 K, 1 bar, with a volumetric flow rate of 0.25 m3/s, and exits at 325 K, 0.95 bar.

The flow area is 0.04 m2. Assuming the ideal gas model with k = 1.4 for the air,


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determine:

a) The mass flow rate, in kg/s, b) the velocity at the inlet and exit, and

c) the rate of heat transfer in kW.


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Example: (Problem 4.28)At steady state, air at 200 kPa, 52oC, and a mass flow rate of 0.5 kg/s enters a

horizontal insulated duct having differing inlet and exit cross-sectional areas. At the

duct exit, the pressure of the air is 100 kPa, the velocity is 255 m/s, and the cross-3 2


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sectional area is 2 x 10-3m2. Assuming the ideal gas model, determine: a) the

temperature of the air at the exit, in C,

b) The velocity of the air at the inlet in m/s, and c) the inlet cross-sectional area, in m2.


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4.23, 4.24

L 13


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Lecture 13

Nozzles, Diffusers,Turbines, Compressor

and Pumps

Nozzles and Diffusers


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Nozzle: a flow passage of varying crosectional area in which the velocity of

or liquid increases in the direction of fl

Diff fl f i

2

2

221

2

112

1

2

1ghVPghVP Berno

N l d Diff M d


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If the change in potential energy from inlet

negligible, g(z1z2) drops out.

If the heat transfer with surroundings is neg

drops out.

(

2

)V(V)(0 1

22

21

21cvcv zzghhmWQ

Nozzle and Diffuser Mode

.0cvW

22

cvQ

Example: (Problem 4.31)

Steam enters a nozzle operating at steady state at 20 bar, 280oC, with


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m/s. The exit pressure and temperature are 7 bar and 180oC, respectiv

mass flow rate is 1.5 kg/s. Neglecting heat transfer and potential energa) The exit velocity in m/s, and b) the inlet and exit flow areas, in c


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T bi


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Turbines

Turbine: a device in which power is

developed as a result of a gas or liqu

passing through a set of blades attac

T bi M d li


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(

2

)V(V)(0 1

22

21

21cvcv zzghhmWQ

Turbine Modeling

If the change in kinetic energy of flowing m

negligible, (V12 V2

2) drops out.

If the change in potential energy of flowing



d tQ

Compressors and P


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Compressors and P

Compressors and Pum

devices in which work is

on the substance flowingthrough them to change t

state of the substance, ty

to increase the pressure elevation.

Compressor : substance

C d P M d


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(

2

)V(V)(0 1

22

21

21cvcv zzghhmWQ

Compressorand Pump Mod

If the change in kinetic energy of flowing m

negligible, (V12 V2

2) drops out.

If the change in potential energy of flowing



d tQ

Example: (Problem 4.41)

Steam enters a well-insulated turbine operating at steady state at 4 MPa

enthalpy of 3015 4 kJ/kg and a velocity of 10 m/s The steam expands t


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enthalpy of 3015.4 kJ/kg and a velocity of 10 m/s. The steam expands t

where the pressure is 0.07 MPa, specific enthalpy is 2431.7 kJ/kg, and t

m/s. The mass flow rate is 11.95 kg/s. Neglecting potential energy effe

the power developed by the turbine, in kW.


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Work through Example Proble4.3, 4.4, 4.5 and 4.6

then

Work on homework problem

4 33 4 43 4 53

Lecture 14


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Lecture 14

Heat Exchangers and

Throttling Devices

Sections 4.9 and 4.10

Heat Exchangers


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Direct contact:A mixing chamber in which hot and

cold streams are mixed directly.

Tube-within-a-tube counterflow:A gas or liquid

stream is separatedfrom another gas or liquid by awall through which energy is conducted. Heat

transfer occurs from the hot stream to the cold

stream as the streams flow in opposite directions.

ei VV22

Heat Exchanger Modeling


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If the kinetic energies of the flowing streams are

negligible, (Vi2/2) and (Ve2/2) drop out.If the potential energies of the flowing streams are

negligible, gziand gzedrop out.

If the heat transfer with surroundings is negligible,drops out.

++ +++=e

ee

ee

i

ii

ii gzhmgzhmWQ )

2

V()

2

V(0 cvcv

(Eq. 4.18)

cvQ

.0cv =W

ee

eii

i hmhm = 0

im

em

im em

=e

ei

i mm

for each connected flow volume

1

2

3

14


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2 3

Problem 4.76 Steam enters a heat exchanger operating at steady state at 250 kPa and

a quality of 90% and exits as a saturated liquid at the same pressure. A separate stream

of oil with a mass flow rate of 29 kg/s enters at 20oC, and exits at 100oC with no

significant change in pressure. The specific heat of the oil is c = 2.0 kJ/kgK. Kinetic and

potential energy effects are negligible. If heat transfer from the heat exchanger to its

di i 10% f h i d i h f h il


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surroundings is 10% of the energy required to increase the temperature of the oil,

determine the steam mass flow rate, in kg/s.


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Pause the video


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Pause the video

Work through examples 4.7 and 4.8

then

Do homework problem 4.74

Throttling Devices


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Throttling Device:a device that achieves

a significant reduction in pressureby

introducing a restriction into a line through

which a gas or liquid flows. Means to

introduce the restriction include a partially

opened valve or a porous plug.


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Pause the video


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Work through example 4.9

then

Try suggested problem 4.90(well work on this in class)


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spontaneous heat transfer


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falling mass

spontaneous expansion

-> observed direction of these

processes

Aspects of theSecond Law of Thermodynamics

F ti f d


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From conservation of mass and energy

principles, mass andenergy cannot be

created or destroyed.

For a process, conservation of mass and

energyprinciples indicate the disposition ofmass and energy but do not infer whether the

process can actually occur.

The second law of thermodynamicsprovides the guiding principle for whether a

process can occur.


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Aspects of theSecond Law of Thermodynamics


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defininga temperature scaleindependent of the

properties of any thermometric substance.

developing means for evaluating properties

such as uand hin terms of properties that aremore readily obtained experimentally.

Scientists and engineers have found additional uses

of the second law and deductions from it. It alsohas been used in philosophy, economics, and other

disciplinesfar removed from engineering

thermodynamics.

Other aspectsof the second law include:

Second Law of ThermodynamicsAlternative Statements


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Clausius Statement

Kelvin-Planck StatementEntropy Statement

There is no simple statement that captures allaspects of the second law. Several

alternativeformulationsof the second laware

found in the technical literature. Threeprominent ones are:

Second Law of ThermodynamicsAlternative Statements


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The focus of Chapter 5is on the ClausiusandKelvin-Planck statements.

The Entropy statementis developed and applied

in Chapter 6.

Like every physical law, the basis of the secondlaw of thermodynamics is experimental

evidence. While the three forms given are not

directly demonstrable in the laboratory,

deductions from them can be verified

experimentally, and this infers the validity of the

second law statements.


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Entropy Statementof the Second Law


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Massand energyare familiar examples of

extensive propertiesused in thermodynamics.

Entropyis another important extensive property.

How entropy is evaluated and applied is detailed

in Chapter 6.Unlike mass and energy, which are conserved,

entropy is produced within systemswhenever

non-idealities such as friction are present.

The Entropy Statement is:

It is impossible for any system to operate in

a way that entropy is destroyed.

Irreversibilities

One of the important uses of the second law of


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One of the important uses of the second law of

thermodynamics in engineering is to determinethe best theoretical performanceof systems.

By comparing actual performancewith best

theoretical performance, insights often can be hadabout the potential for improved performance.

Best theoretical performance is evaluated in terms

of idealizedprocesses.

Actual processes are distinguishable from suchidealized processes by the presence of non-

idealities called irreversibilities.

Irreversibilities CommonlyEncountered in Engineering Practice


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Heat transferthrough a finite temperaturedifference

Unrestrained expansionof a gas or liquid to

a lower pressureSpontaneouschemical reaction

Spontaneous mixingof matter at different

compositions or statesFriction sliding friction as well as friction in

the flow of fluids


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Irreversible and ReversibleProcesses


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within the system, or

within its surroundings (usually the

immediate surroundings), or

within both the system and itssurroundings.

During a process of a system,irreversibilitiesmay be present:

Irreversible and ReversibleProcesses


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A process is irreversiblewhenirreversibilities are present within the system

and/or its surroundings.

All actual processes are irreversible.A process is reversiblewhen no

irreversibilities are present within the system

and its surroundings.This type of process is fully idealized.


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Example: Internally Reversible Process

Water contained within a piston-cylinder evaporates

from saturated liquid to saturated vapor at 100oC As


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from saturated liquid to saturated vapor at 100oC. As

the water evaporates, it passes through a sequence of

equilibrium stateswhile there is heat transfer to the

water from hot gases at 500oC.

Such spontaneous

heat transferis an

irreversibility in its

surroundings: an

externalirreversibility.

For a system enclosing the water there are nointernalirreversibilities, but

Analytical Form of theKelvin-Planck Statement

For any system undergoing a


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For any system undergoing a

thermodynamic cycle while

exchanging energy by heat

transfer with asinglethermal

reservoir, the net work,W

cycle,can be only negative or zero

never positive:

Wcycle 0 < 0: Internal irreversibilities present= 0: No internal irreversibilitiessingle

reservoir

(Eq. 5.3)

NO!


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Thermal Efficiency and Coefficient of Performance


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,

(refrigeration, heat pump)

(power cycle)

Applications to Power Cycles Interacting

with TwoThermal Reservoirs


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For a system undergoing apower cycle whilecommunicating thermally with twothermal

reservoirs, a hot reservoir and a cold reservoir,

(Eq. 5.4)H

C

H

cycle

1 Q

Q

Q

W==

the thermal efficiency of any such cycle is

(Eq. 2.43)


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Carnot Corollaries

2. The thermal efficiency of an irreversible power


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In other words, the reversible cycle between two reservoirstells us what the maximum efficiency of the cycle is. If there

are reversibilities, the efficiency can only be smaller than the

max.

cycleis always less thanthe thermal efficiency of areversible power cyclewhen each operates between

the same two thermal reservoirs.


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Applications to Power Cycles Interactingwith TwoThermal Reservoirs


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H

C

H

cycle1

Q

Q

Q

W==

1)(


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pump cycle whilecommunicating thermally with twothermal reservoirs, a hot reservoir and a cold

reservoir,

(Eq. 5.5)CH

C

cycle

C

QQ

Q

W

Q

==

the coefficient of performance for the

refrigeration cycle is

(Eq. 5.6)CH

H

cycle

H

QQ

Q

W

Q

==

and for the heat pump cycle is

(Eq. 2.45)

(Eq. 2.47)

Applications to Refrigeration and Heat Pump CyclesInteracting with TwoThermal Reservoirs


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By applying the Kelvin-Planck statementof thesecond law, Eq. 5.3, three conclusionscan be drawn:

1. For a refrigeration effect to occur a net work input

Wcycleis required. Accordingly, the coefficient ofperformance must be finite in value.

Two other conclusions are:

CH

C

cycle

C

QQ

Q

W

Q

==

== CH

H

cycle

H

QQ

Q

W

Q



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2. The coefficient of performance of an irreversiblerefrigeration cycleis always less thanthe coefficient

of performance of a reversible refrigeration cycle

when each operates between the same two thermal

reservoirs.

In other words, the reversible cycle between two reservoirs

tells us what the maximum coefficient of performance (COP)of the cycle is. If there are reversibilities, the COP can only be

smaller than the max.


3 All reversible refrigeration cycles operating


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In other words, once you have found one way to calculate thecoefficient of performance for a reversible cycle, you know what

the max COP is.

3. All reversible refrigeration cycles operating

between the same two thermal reservoirshave the

same coefficient of performance.

All three conclusions also apply to a system

undergoing a heat pump cycle between hot and cold

reservoirs.


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Lecture 17


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The Kelvin Temperature Scaleand Theoretical Limits

of Performance of

Cycles Between Two Reservoirs

Sections 5.8 5.9

Kelvin Temperature Scale

Consider systems undergoing apower cycle and a

refrigerationor heat pump cycle, each while


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g p p y ,

exchanging energy by heat transfer with hot and coldreservoirs:

(Eq. 5.7)H

C

cyclerevH

C

T

T

Q

Q=

The Kelvin temperature is defined so that


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Maximum Performance Measures for CyclesOperating between Two Thermal Reservoirs

Previous deductions from the Kelvin-Planck statement of the second law

include:


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1. The thermal efficiency of anirreversiblepowercycleis always less thanthe thermal efficiency of a

reversiblepower cyclewhen each operates between

the same two thermal reservoirs.

2. The coefficient of performance of anirreversible

refrigeration cycleis always less thanthe

coefficient of performance of areversible

refrigeration cyclewhen each operates between the

same two thermal reservoirs.

3. The coefficient of performance of anirreversibleheat pump cycleis always less thanthe coefficient

of performance of areversibleheat pump cycle

when each operates between the same two thermal

reservoirs.

The reversible

processdetermines

the maximum

theoretical

performance

for a cycle.

Maximum Performance Measures for CyclesOperating between Two Thermal Reservoirs

Using Eq. 5.7 in Eqs. 5.4, 5.5, and 5.6, we get respectively:

TQ


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(Eq. 5.9)H

Cmax 1

T

T=Power Cycle:

(Eq. 5.10)CH

Cmax

TT

T

=Refrigeration Cycle:

(Eq. 5.11)CH

Hmax

TT

T

=Heat Pump Cycle:

where THand T

Cmust be on theKelvinorRankine scale.

HC

cyclerevH

CT

T

Q

Q

=


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Example: Power Cycle Analysis

kJ600

Actual Performance: Calculate using the heat

transfers:


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4.0kJ1000

kJ60011

H

C ===

QQ

Maximum Theoretical Performance: Calculate

max

from Eq. 5.9 and compare to :

(a) 6.0K500

K20011

H

Cmax ===

T

T

(b) 4.0K500

K300

11H

Cmax ===

T

T

(c) 2.0K500

K40011

H

Cmax ===

T

T

Reversibly0.4 =0.4

Impossible0.4 >0.2

Irreversibly0.4


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Problem 5.59

At steady state, a refrigeration cycle operating between hot and cold reservoirs at 300 Kand 275 K, respectively, removes energy by heat transfer from the cold reservoir at a

rate of 600 kW.

a) If the cycles coefficient of performance is 4, determine the power input required, in

kW.

b) Determine the minimum theoretical power require, in kW, for any such cycle.


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) p q , , y y


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Lecture 18


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Carnot CycleClausius Inequality

Introducing Entropy

Sections 5.10, 5.11, 6.1, 6.2

Carnot Cycle


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The Carnot cycleprovides a specificexample of a reversible cycle that operatesbetween two thermal reservoirs. Otherexamples are provided in Chapter 9: the

Ericsson and Stirling cycles.In a Carnot cycle, the system executing the

cycle undergoes a series of four internallyreversible processes: two adiabaticprocessesalternated with two isothermalprocesses.

Carnot Power Cycles

The p-vdiagramand schematicof a gas in a piston-cylinder


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assembly executing a Carnot cycleare shown below:


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Clausius Inequality

The Clausius inequalityconsidered next


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q yprovides the basis for developing theentropy concept in Chapter 6.

The Clausius inequalityis applicable to any

cycle without regard for the body, or bodies,from which the system undergoing a cyclereceives energy by heat transferor towhich the system rejects energy by heattransfer. Such bodies need not be thermalreservoirs.

Work through example after Eq 5.14


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then

Work on homework

problems 5.85, 5.87


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Defining Entropy Change

Consider two cycles, each composed


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of two internally reversible processes,process A plus process Cand

process B plus process C, as shown

in the figure.

Applying Eq. 5.13to these cycles gives,

wherecycleis zero because the cycles arecomposed of internally reversible processes.


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Defining Entropy Change

Recalling (from Sec. 1.3.3) that a quantity is a property if,


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and only if, its change in value between two states isindependent of the process linking the two states, we

conclude that the integral represents the change in some

property of the system.

We call this property entropyand represent it by S. Thechange in entropy is

wherethe subscript int rev signals that the integral is

carried out for any internally reversible process linking

states 1and 2.

(Eq. 6.2a)


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Entropy FactsEntropyis an extensive property.

Just as mass and energy are accounted for by mass and

energy balances, entropy is accounted for by an entropy


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balance.Like mass and energy, entropy can be transferred across

the system boundary.

Like any other extensive property, the change in entropy

can be positive, negative, or zero:

By inspection of Eq. 6.2a, units for entropySare kJ/Kand

Btu/oR.

Units forspecificentropysare kJ/kgKand Btu/lboR.


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Lecture 19

C l l ti E t d


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Calculating Entropy, andEntropy Change in

Incompressible Substances,

Ideal Gases

Sections 6.3 6.5


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Entropy and Heat Transfer

By inspection of Eq. 6.2a, the defining equation for

entropy change on a differential basis is


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(Eq. 6.2b)

Equation 6.2bindicates that when a closed system

undergoing an internally reversible process receivesenergy by heat transfer, the system experiences an

increase in entropy. Conversely, when energy is removed

by heat transfer, the entropy of the system decreases.

From these considerations, we say that entropy transferaccompanies heat transfer. The direction of the entropy

transfer is the same as the heat transfer.


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Entropy and Heat Transfer

F thi it f ll th t


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From this it follows thatan energy transfer by

heat to a closed system

during an internally

reversible process isrepresented by an area

on a temperature-entropy

diagram:


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Calculating Entropy Change

As an application, consider a

change in phase from saturated

liquid to saturated vapor at


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T

dhds =

liquid to saturated vapor atconstant pressure.

Since pressure is constant, Eq.

6.10breduces to give

Then, because temperature is also constant during the

phase change

(Eq. 6.12)

This relationship is applied in property tables for

tabulating (sgsf) from known values of (hghf).

Calculating Entropy Change

For example, consider water vapor at 100oC

(373.15 K). From Table A-2, (hghf) =2257.1 kJ/kg.


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Next, the TdSequationsare applied to twoadditional cases: substances modeled as

incompressibleand gases modeled as ideal

gases.

which agrees with the value from Table A-2, as

expected.

(sgsf) =(hghf) /T = (2257.1 kJ/kg)/373.15 K

= 6.049 kJ/kgK

Then


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Calculating Entropy Change of an Ideal Gas

The ideal gas modelassumes pressure, specific volume

and temperature are related bypv=RT. Also, specific

internal energy and specific enthalpy each depend solely


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internal energy and specific enthalpy each depend solelyon temperature: u= u(T),h=h(T), givingdu=c

vdTand

dh=cpdT, respectively.

Using these relations and integrating, the TdSequations

give, respectively

(Eq. 6.17) (Eq. 6.18)

Sincecvandcpare functions of temperature for ideal gases,

such functional relations are required to perform the

integration of the first term on the right of Eqs 6 17 and 6 18



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integration of the first term on the right of Eqs. 6.17and 6.18.

For several gases modeled as ideal gases, including air,

CO2, CO, O2, N2, and water vapor, the evaluation of

entropy change can be reduced to aconvenient tabularapproach using the variablesodefined by

(Eq. 6.19)

whereT

' is an arbitrary reference temperature.


Usingso, the integral term of Eq. 6.18can be expressed as


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Accordingly, Eq. 6.18becomes

(Eq. 6.20a)

or on a per mole basisas

(Eq. 6.20b)

For air, Tables A-22andA-22Eprovideso

in units ofkJ/kgK and Btu/lboR, respectively. For the other gases

mentioned, Tables A-23andA-23Eprovide in units of

kJ/kmolK and Btu/lbmoloR, respectively.

os

(Eq. 6.18)


Example: Determine the change in specific entropy, in

kJ/kgK, of air as an ideal gas undergoing a process from

T1= 300 K,p

1= 1 barto T

2= 1420 K,p

2= 5 bar.


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Ideal Gas Pro pert ies of Ai r

T(K), hand u(kJ/ kg ), so(kJ/ kg K)

when s = 0 when s = 0

T h u so p r vr T h u so p r vr

250 250 .05 178 .28 1.51917 0.7329 979. 1400 1515.42 1113.52 3.36200 450.5 8.919

260 260.09 185.45 1.55848 0.8405 887.8 1420 1539.44 1131 .77 3.37901 478.0 8.526

270 270.11 192.60 1.59634 0.9590 808.0 1440 1563.51 1150 .13 3.39586 506.9 8.153

280 280.13 199.75 1.63279 1.0889 738.0 1460 1587.63 1168 .49 3.41247 537.1 7.801

285 285.14 203.33 1.65055 1.1584 706.1 1480 1611.79 1186 .95 3.42892 568.8 7.468

290 290.16 206.91 1.66802 1.2311 676.1 1500 1635.97 1205 .41 3.44516 601.9 7.152

295 295.17 210.49 1.68515 1.3068 647.9 1520 1660.23 1223 .87 3.46120 636.5 6.854

300 300.19 214.07 1.70203 1.3860 621.2 1540 1684.51 1242 .43 3.47712 672.8 6.569

305 305.22 217.67 1.71865 1.4686 596.0 1560 1708.82 1260 .99 3.49276 710.5 6.301

310 310.24 221.25 1.73498 1.5546 572.3 1580 1733.17 1279 .65 3.50829 750.0 6.046

From Table A-22, we getso1= 1.70203andso

2= 3.37901,

each in kJ/kgK. Substituting into Eq. 6.20a

1 1 2 2

Kkg

kJ215.1

bar1

bar5ln

Kkg

kJ

97.28

314.8

Kkg

kJ)70203.137901.3(12

=

= ss

Table A-22

Problem 6.15

One-tenth kmol of carbon monoxide (CO) is a piston-cylinder assembly undergoes aprocess from p1= 150 kPa, T1= 300 K to p2= 500 kPa, T2= 370 K. For the process,

W = -300 kJ. Employing the ideal gas model, determine

a) the heat transfer, in kJ, and

b) the change in entropy, in kJ/K.

c) Show the process on a T-s diagram.


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