Lecture 5 - Energy Transfer

8/4/2019 Lecture 5 - Energy Transfer

1/50

ME2121 Engineering Thermodynamics

ENERGY TRANSFER

First Law Of Thermodynamics

A concise review

Chapter 3 of C&B

ASMRefer to text for details

Copyright 2005 Prof. Arun S. Mujumdar.


2/50

Energy Transfer 2

The First Law of Thermodynamics

Introduction - Outline The first law of thermodynamics is a statement of the

conservation of energy principle and it asserts thattotal energy is a thermodynamic property.

Energy transfer with heat and work is introduced,and the mechanisms of heat transfer as well asvarious work modes are discussed.

The first-law relation for closed systems is developed

in a step-by-step manner using an intuitive approach.

Specific heats are defined, and relations are obtainedfor the internal energy and enthalpy of ideal gases interms of specific heats and temperature.



3/50

Energy Transfer 3

This approach is also applied to solids and liquids -approximated as incompressible substances.

Energy can be neither created nor destroyed; it canonly change forms. Energy can cross the boundary ofa closed system in two distinct forms: heat and work.

Heat is defined as the form of energy that istransferred between two systems (or a system andits surroundings) by virtue of a temperaturedifference.

Heat is energy in transition. It is recognized only as itcrosses the boundary of a system. A process duringwhich there is no heat transfer is called an adiabaticprocess.



4/50

Energy Transfer 4

Work like heat, is an energy interaction between asystem and its surroundings. Energy can cross theboundary of a closed system in the form of heat or

work.

If the energy crossing the boundary of a closedsystem is not heat, it must be work.

Work is the energy transfer associated with a forceacting through a distance. A rising piston, a rotatingshaft, and an electric wire crossing the systemboundaries are all associated with work interactions.

Sign convention: heat transfer toa system and workdone bya system are positive; heat transfer from asystem and work done on a system are negative.



5/50

Energy Transfer 5

Heat and work are energy transfer mechanismsbetween a system and its surroundings, there aremany similarities between them:

Both are recognized at the boundaries of a system asthey cross them. Both heat and work are boundaryphenomena.

Systems possess energy, but not heat and work.

Both are associated with a process, not a state.Unlike thermodynamic properties, heat or work hasno meaning at a state.

Both are path functions(i.e. their magnitudes dependon the path followed during a process as well as the

end states).

These are basic concepts useful in further deliberations.Be clear about their precise meaning.



6/50

Energy Transfer 6

ENERGY TRANSFER BY WORKSome definitions/terminology

Work, like heat, is an energy interaction betweena system and its surroundings

If the energy crossing the boundary of a closed

system is not heat, it must be work Work is the energy transfer associated with a

force acting through a distance

Work done per unit mass of a system is denoted

by w and is expressed as

)kg/kJ(m

Ww



7/50

Energy Transfer 7

ENERGY TRANSFER BY WORK (Contd)

Work done per unit time is called power and is denotedW (Figure 3-8). The unit of power is kJ/s or kW

Heat and work are directional quantities; requires thespecification of both the magnitude and direction; signconvention

Heat transfer to a system and work done by a systemare positive; heat transfer from a system and work doneon a system are negative

Use this intuitive approach in this book as it eliminatesthe need to adopt a formal sign convention



8/50

Energy Transfer 8

Similarities between Heat andWork - Summary

Both heat and work are boundaryphenomena

Systems possess energy but not heat orwork

Both are associated with a process, not astateimportant to note!

Both are path functions (magnitudedepend on the path followed)



9/50

Energy Transfer 9

Path functions have inexact differentialsdesignated by the symbol , e.g. Q orW

Properties are point functions , i.e. theydepend on the state only. They have exact

differentials designated by the symbol d.



10/50

Energy Transfer 10

MECHANICAL FORMS OF WORK

The work done by a constant force F on a bodydisplaced a distance s in the direction of the forceis given by

W = F s (kJ) If the force F is not constant i.e. it is a function of

distance, s ,then

21

)kJ(dsFW



11/50

Energy Transfer 11

MECHANICAL FORMS OF WORK (contd)

Two requirements exist for a work interactionbetween a system and its surroundings There must be a force acting on the boundary The boundary must move

Displacement of the boundary without any force

to oppose or drive this motion (such as theexpansion of a gas into an evacuated space) isnot a workinteraction since no energy istransferred

Mechanical work is associated with themovement of the boundary of a systemor withthe movement of the entire systemas a whole



12/50

Energy Transfer 12

MOVING BOUNDARY WORK

Moving boundary work is the primary form ofwork involved in automobile engines

We analyze the moving boundary work for aquasi-equilibrium process, a process duringwhich the system remains in equilibrium at all

times Consider the gas enclosed in the piston-cylinder

device shown in Fig. 3-19, C &B. The initialpressure of the gas is P, the total volume is V

and the cross sectional area of the piston is A. Ifthe piston is allowed to move a distance ds, thedifferential work done during this process is

Wb = F ds = PA ds = P dV



13/50

Energy Transfer13

MOVING BOUNDARY WORK (contd)

Note P is the absolute pressure, which is alwayspositive. However, the volume change dV ispositive during an expansion process (volumeincreasing) and negative during a compressionprocess (volume decreasing).

Thus, the boundary work is positive during anexpansion process and negative during acompression process. Therefore the aboveequation can be viewed as an expression forboundary work output, Wb,out.A negative resultindicates boundary work input (compression)



14/50

Energy Transfer14


Moving boundary work is sometimes called theP dV work

The total boundary work done during the entire

process as the piston moves is obtained byadding all the differential works from the initialstate to the final state

21

b )kJ(dVPW



15/50

Energy Transfer 15


P = f(v) should be available. Note that P = f(v)is simply the equation of the process path on aP-V diagram

The total area A under the process curve 1-2 (to go from state 1 to state 2) is obtained byadding these differential areas:

21

2

1

dVPdAAArea



16/50

Energy Transfer 16


A gas can follow several different paths as it

(say) expands from state 1 to state 2. Ingeneral, each path will have a different areaunderneath it, and since this area represents themagnitude of work, the work done will bedifferent for each process (See Fig 3-21).

Note: work is a path function (i.e., it depends onthe path followed as well as the end states).

If work were not a path function, no cyclicdevices (car engines, power plants) couldoperate as work producing devices.



17/50

Energy Transfer 17

Boundary Work during a Constant-Volume Process (Example 3-5)

A rigid tank contains air at 500 kPa and 150C. Asa result of heat transfer to the surroundings,the temperature and pressure inside the tank

drop to 65C and 400 kPa, respectively.Determine the boundary work done during thisprocess. No volume change. Hence,

21

0b 0dVPW



18/50

Energy Transfer 18

Boundary Work for a Constant-PressureProcess (Example 3-6)

A frictionless piston-cylinder device contains 10lbm of water vapor at 60 psia and 320F. Heat isnow transferred to the steam until thetemperature reaches 400F. If the piston is notattached to a shaft and its mass is constant,determine the work done by the steam duringthis process.

Not in SI units but illustrates a point!Note this is a constant pressure process.



19/50

Energy Transfer 19

Boundary Work for a Constant-PressureProcess (Example 3-6)

21

2

11200b )VV(PdVPdVPW

)VV(mPW 120b Note:Work = mass x pressure x change of volume of

system



20/50

Energy Transfer 20

Isothermal Compression of anIdeal Gas (Example 3-7)

A piston-cylinder device initially contains0.4m3 of air at 100 kPa and 80C. The air

is now compressed to 0.1m

3

in such a waythat the temperatureinside the cylinderremains constant. Determine the workdone during this process.

Assume ideal gas.



21/50

Energy Transfer 21

Isothermal Compression of anIdeal Gas (Analysis - Example 3-7)

For an ideal gas at constant temperature T0,

V

CPorCmRTPV 0

1

211

1

22

1

2

1

2

1

b

V

VlnVP

V

VlnC

V

dVCdV

V

CdVPW

,Then

Insert values given in problem statements to find Wb



22/50

Energy Transfer 22

POLYTROPIC PROCESS

During actual expansion and compressionprocesses of gases, pressure and volume areoften related by PVn = C, where n and C are

constants. A process of this kind is called apolytropic process

P = CV-n

n1

VPVP

1n

VVCdVCVdVPW

,Then

11221n

11n

22

1

2

1

nb



23/50

Energy Transfer 23

Gravitational Work

Gravitational work is defined as the work done by or

against a gravitational force field.

F = mg

21

2

1

12g )kJ()zz(mgdZmgdZFW



24/50

Energy Transfer 24

Accelerational Work

The work associated with a change in velocity of asystem is called accelerational work. Theaccelerational work required to accelerate a body of

mass m from the definition of acceleration andNewtons Second Law:

dt

dmF

dt

da

maFV

V



25/50

Energy Transfer 25

dtds

dt

dsVV

Interesting Observation:The work done to accelerate a body is independent ofthe path followed and is equivalent to the change inthe kinetic energy of the body.

)VV(m2

1VdVm)dtV(

dt

VdmdsFW 2

1

2

2

2

1

2

1

2

a



26/50

Energy Transfer 26

Shaft Work:

Energy transmission with a rotating shaft is verycommon in engineering practice. Often the torque applied to the shaft is constant, which means thatthe force F applied is also constant.

F acting through a moment arm r generates a torque of

rFFr

This force acts through a distance s, which is related to

the radius r by

n)r2(2s Copyright 2005 Prof. Arun S. Mujumdar.


27/50

Energy Transfer 27

Then the shaft work is given by:

n2rn2rFsWshSpring Work:

To determine the spring work, we need a frelationship between force F and displacement x.For linear elastic springs, the displacement x isproportional to the force applied,or

F = kx



28/50

Energy Transfer 28

Here k is the spring constant and has the units kN/m.

The displacement x is measured from theundisturbed position of the spring (that is x = 0 whenF = 0). Substituting and integrating,

)kJ()xx(k2

1W 2122spring

Here x1 and x2 are the initial and the final

displacements of the spring, measured from theundisturbed position of the spring.



29/50

Energy Transfer 29

Non-mechanical Forms of Work

Some examples of non-mechanical work modes areelectrical work, where the generalized force is thevoltage (the electrical potential) and the generalized

displacement is the electrical charge, and magneticwork, where the generalized force is the magneticfield strength and the generalized displacement is thetotal magnetic dipole moment; and electricalpolarization work.

These forms of work will not be considered inME2121. However you should be aware of these.



30/50

Energy Transfer 30

Total EnergyAn Outcome of the First Law

A major consequence of the first law is the existenceand the definition of the property total energy E. thenet work is the same for all adiabatic processes of aclosed system between two specified states, thevalue of the net work must depend on the end statesof the system only, and thus it must correspond to achange in a property of the system.

This property is the total energy. The first law statesthat the change in the total energy during anadiabatic process must be equal to the net workdone.

Therefore, any convenient arbitrary value can beassigned to total energy at a specified state to serveas a reference point.



31/50


32/50

Energy Transfer 32

Energy Change of a System, Esystem

Energy change = Energy at final stateEnergy at initial state

Esystem = Efinal Einitial = E2 E1

Energy can exist in numerous forms such asinternal (sensible, latent, chemical andnuclear), kinetic, potential, electrical, andmagnetic and their sum constitutes the total

energy E of a system. In the absence ofelectric, magnetic and surface tension effects

E = U + KE PE



33/50

Energy Transfer 33

where

)zz(mgPE

)(m2

1KE

)uu(mU

12

21

22

12

VV

Systems that do not involve any changes in theirvelocity or elevation during a process are stationarysystems.Changes in kinetic and potential energies are zero

(that is KE = PE = 0), and E = U for stationarysystems.



34/50

Energy Transfer 34

First Law (Contd)

Mechanisms of Energy Transfer, Einand Eout Heat Transfer, Q

Work, W

Mass Flow, m

When mass enters an open system, the energy of the

system increases because mass carries energy withit.

The first law or energy balance relation for a closedsystembecomes

Q W = E



35/50

Energy Transfer 35

First Law: Statement

Heat input to a system minus work output by the

system is equal to the change in the energy of thesystem. A negative quantity for Q or W simply meansthat the assumed direction for that quantity is wrong.

Various forms of this traditional first law relation for

closed systems are given as

1. General Q W = E

2. Stationary systems Q W = U

3. Per unit mass q w = e

4. Differential form q - w = de



36/50

Energy Transfer 36

The specific heat is defined as the energy required toraise the temperature of a unit mass of a substance

by one degree. In thermodynamics, we are interested in two kinds of

specific heats: specific heat at constant volume Cvand specific heat at constant pressure Cp.

The specific heat at constant volume Cv is the energyrequired to raise the temperature of the unit mass ofa substance by one degree as the volume ismaintained constant.

TuCv Copyright 2005 Prof. Arun S. Mujumdar.


37/50

Energy Transfer 37

An expression for the specific heat at constantpressure Cp can be obtained by considering a

constant-pressure process (wb + u = h). It yields

The specific heats of a substance depend on thestate that, in general, is specified by twoindependent, intensive properties.

The energy required to raise the temperature of a

substance by one degree will be different at differenttemperatures and pressures. But this difference isusually not very large.

p

pT

hC



38/50

Energy Transfer 38

These equations are property relations and as suchare independent of the type of processes. They arevalid for any substance undergoing any process.

Cv is related to the changes in internal energy and Cpto the changes in enthalpy. Cv is a measure of thevariation of internal energy of a substance withtemperature, and Cp is a measure of the variation ofenthalpy of a substance with temperature.

Since u and h depend only on temperature for anideal gas, the specific heats Cv and Cp also depend ontemperature only. Therefore, at a given temperature,u, h, Cv and Cp of an ideal gas will have fixed valuesregardless of the specific volume or pressure. Forideal gases, thepartial derivatives can bereplaced by ordinary derivatives.



39/50

Energy Transfer 39

du = Cv(T) dT and

dh = Cp (T) dT

Hence,

2

1

v12 )kg/kJ(dT)T(Cuuu

21

p12 )kg/kJ(dT)T(Chhh

and



40/50

Energy Transfer 40

To carry out these integrations, we need to haverelations for Cv and Cp as functions of temperature.

The specific heats of gases with complex molecules(molecules with two or more atoms) are higher andincrease with temperature. The variation of specificheats with temperature is smooth and may beapproximated as linear over small temperatureintervals (a few hundred degrees or less).

Hence

u2 u1 = Cv,av(T2 T1) (kJ/kg)and

h2 h1 = Cp,av(T2 T1) (kJ/kg)



41/50

Energy Transfer 41

The ideal gas specific heats of monatomic gases suchas argon, neon and helium remain constant over theentire temperature range.

Specific-Heat Relations of Ideal Gases

A special relationship between Cp and Cp for ideal

gases can be obtained by differentiating the relationh = u + RT, which yields

dh = du + R dT

Replacing dh by Cp dT and du by Cp dT and dividingthe resulting expression by dT we obtain

Cp = Cp+ R [kJ/(kg . K)]



42/50

Energy Transfer 42

Another ideal-gas property called the specific heat ratiok, defined as

The specific heat ratio also varies mildly withtemperature. For monatomic gases, its value isessentially constant at 1.667. Many diatomic gases,

including air, have a specific heat ratio of about 1.4at room temperature.

v

p

C

Ck

)K.kmol/(kJ114.29C)K.kmol/(kJ314.8R

)K.kmol/(kJ80.20C

or

)K.kg/(kJ005.1C)K.kg/(kJ287.0R

)K.kg/(kJ718.0C

p

u

v

pv



43/50

Energy Transfer 43

Conservation of Mass PrincipleSection 3-5 C&B

Statement: Net mass transfer to or from a systemduring a process is equal to the net change oftotal mass of the system during that process.

Note: The change may be an increase or adecrease.

This is the same principle as conservation of

money in your bank account, viz.,Amount entering the systemAmount leaving the

system = Net change within the system


M B l f St d Fl


44/50

Energy Transfer 44

Mass Balance for Steady FlowProcesses (Figure 3-43 C&B)

In a steady process total mass in a control volume (CV)does not change with time. Hence mass entering a CVmust be equal to that leaving

There can be multiple entries and exits to a controlvolume

For steady flow, single stream

22211121 AVAVormm V is average velocity over cross section A.



45/50

Energy Transfer 45

Examples

For incompressible fluid (not flow) density is

constant, hence V1A1 = V2A2 Example 3-12 This is a simple problem

involving a constant flow rate through a gardenhose nozzle.

Example 3-13 This is an interesting example tocompute discharge time for water in a tank.Note that the discharge velocity is a function ofthe level of water in the tank which falls with

time as water is discharged. Clearly this is aproblem of mass conservation in rate form, i.e. itis a unsteady state problem. Pls review carefully.



46/50

Energy Transfer 46

Flow Work / Energy of a Flowing Fluid

Open systems or control volumes involve fluid

flowing into or out of the CV. This involves socalled flow work or flow energy.

Flow work is needed to maintain fluid flowthrough CV (See figure 3-48)

If P is fluid pressure and the cross sectional areafor fluid flow is A then work required to makethe fluid element move a distance L is

Wflow = FL = PAL = PV (kJ)

Flow work per unit mass is

wflow=Pv (kJ/kg)


Total Energy of a Flowing Fluid


47/50

Energy Transfer 47

Total Energy of a Flowing Fluid

It is sum of internal, kinetic, and potential energy of a

compressible system, i.e.

For a flowing system on a per unit mass basis we mustadd flow energy to obtain the total energy of a flowing

fluid, viz. = Pv + e = Pv + (u + ke + pe) or

= h + ke + pe where h = Pv + u

h is called specific enthalpy of the fluid. Using h instead ofu allows us to account for flow work in control volumeproblems.

)kg/kJ(gz2/Vupekeue 2


E T t b M


48/50

Energy Transfer 48

Energy Transport by Mass

In open systems (control volumes) note that

energy transfer can also occur due to mass flowin or out over control surfacesAmount of energy transported by mass, Emass=

m (kJ) This equation can also be written in rate form by

simply taking time derivative of the aboveequation, i.e. simply place a dot over E and m

Study Example 3-14 (estimation of mass and

energy leaving a pressure cooker in a given timeby application of the rate form of conservationof mass to a CV (figure 3-53))



49/50

Energy Transfer 49

Closing Remarks Chapter 3

This chapter defines various forms ofenergy and how energy transfer can occurin closed and open systems

Reviews definitions of work, heat, internal

energy, enthalpy, etc. In chapter 4 these concepts are applied

along with the First Law of

Thermodynamics to a number of simplepractical systems involving gases andliquids



50/50

Energy Transfer 50

Some Quotable Quotes

Thermodynamics is the only physicaltheory of universal content which, withinthe framework of the applicability of itsbasic concepts, I am convinced will neverbe overthrown .

Albert Einstein

Make it simple but not simpler.- Albert Einstein

Documents

Lecture 5 - Energy Transfer