Moran M.J., Shapiro H.N. Fundamentals of Engineering Thermodynamics

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The SI unit for entropy is J/K. However, in this book it is convenient to work in

terms of kJ/K. A commonly employed English unit for entropy is Btu/8R. Units in SI

for specific entropy are kJ/kg

K for s and kJ/kmol

K for s. Commonly used English

units for specific entropy are Btu/lb

8R and Btu/lbmol

8R.

It should be clear that entropy is defined and evaluated in terms of a particular

expression (Eq. 6.2a) for which no accompanying physical picture is given. We encoun-

tered this previously with the property enthalpy. Enthalpy is introduced without

physical motivation in Sec. 3.6.1. Then, in Chap. 4, we learned how enthalpy is used

for thermodynamic analysis of control volumes. As for the case of enthalpy, to gain

an appreciation for entropy you need to understand how it is used and what it is used

for. This is the aim of the rest of this chapter.

6.1.2

Evaluating Entropy

Since entropy is a property, the change in entropy of a system in going from one state

to another is the same for all processes, both internally reversible and irreversible,

between these two states. Thus, Eq. 6.2a allows the determination of the change in

entropy, and once it has been evaluated, this is the magnitude of the entropy change

for all processes of the system between the two states.

The defining equation for entropy change, Eq. 6.2a, serves as the basis for evaluat-

ing entropy relative to a reference value at a reference state. Both the reference value

and the reference state can be selected arbitrarily. The value of entropy at any state

y relative to the value at the reference state x is obtained in principle from

5 S

int

rev

(6.3)

where S

is the reference value for entropy at the specified reference state.

The use of entropy values determined relative to an arbitrary reference state is satis-

factory as long as they are used in calculations involving entropy differences, for then the

reference value cancels. This approach suffices for applications where composition remains

constant. When chemical reactions occur, it is necessary to work in terms of absolute

values of entropy determined using the third law of thermodynamics (Chap. 13).

6.1.3

Entropy and Probability

The presentation of engineering thermodynamics provided in this book takes a mac-

roscopic view as it deals mainly with the gross, or overall, behavior of matter. The

macroscopic concepts of engineering thermodynamics introduced thus far, including

energy and entropy, rest on operational definitions whose validity is shown directly

or indirectly through experimentation. Still, insights concerning energy and entropy

can result from considering the microstructure of matter. This brings in the use of

probability and the notion of disorder. Further discussion of entropy, probability, and

disorder is provided in Sec. 6.8.2.

units for entropy

6.2 Retrieving Entropy Data

In Chap. 3, we introduced means for retrieving property data, including tables, graphs,

equations, and the software available with this text. The emphasis there is on evaluat-

ing the properties p, y, T, u, and h required for application of the conservation of

mass and energy principles. For application of the second law, entropy values are

usually required. In this section, means for retrieving entropy data for water and

several refrigerants are considered.

6.2 Retrieving Entropy Data 283

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284 Chapter 6

Using Entropy

Tables of thermodynamic data are introduced in Secs. 3.5 and 3.6 (Tables A-2

through A-18). Specific entropy is tabulated in the same way as considered there for

the properties y, u, and h, and entropy values are retrieved similarly. The specific

entropy values given in Tables A-2 through A-18 are relative to the following reference

states and values. For water, the entropy of saturated liquid at 0.018C (32.028F) is set

to zero. For the refrigerants, the entropy of the saturated liquid at 2408C (2408F) is

assigned a value of zero.

6.2.1

Vapor Data

In the superheat regions of the tables for water and the refrigerants, specific entropy

is tabulated along with y, u, and h versus temperature and pressure.

consider two states of water. At state 1 the pressure is 3 MPa

and the temperature is 5008C. At state 2, the pressure is 0.3 MPa and the specific

entropy is the same as at state 1, s

5 s

. The object is to determine the temperature

at state 2. Using T

and p

, we find the specific entropy at state 1 from Table A-4 as

5 7.2338 kJ/kg

K. State 2 is fixed by the pressure, p

5 0.3 MPa, and the specific

entropy, s

5 7.2338 kJ/kg

K. Returing to Table A-4 at 0.3 MPa and interpolating

with s

between 160 and 2008C results in T

5 1838C. b b b b b

6.2.2

Saturation Data

For saturation states, the values of s

and s

are tabulated as a function of either

saturation pressure or saturation temperature. The specific entropy of a two-phase

liquid–vapor mixture is calculated using the quality

s 5

1 2 x

1 xs

(6.4)

5 s

1 x

2 s

These relations are identical in form to those for y, u, and h (Secs. 3.5 and 3.6).

let us determine the specific entropy of Refrigerant 134a at a

state where the temperature is 08C and the specific internal energy is 138.43 kJ/kg.

Referring to Table A-10, we see that the given value for u falls between u

and u

08C, so the system is a two-phase liquid–vapor mixture. The quality of the mixture

can be determined from the known specific internal energy

u 2 u

2 u

138

227.06 2 49.79

5 0.5

Then with values from Table A-10, Eq. 6.4 gives

s 5

1 2 x

1 xs

0.5

0.1970

0.5

0.9190

5 0.5580 kJ

kg ? K b b b b b

6.2.3

Liquid Data

Compressed liquid data are presented for water in Tables A-5. In these tables s, y, u, and

h are tabulated versus temperature and pressure as in the superheat tables, and the tables

are used similarly. In the absence of compressed liquid data, the value of the specific

entropy can be estimated in the same way as estimates for y and u are obtained for

liquid states (Sec. 3.10.1), by using the saturated liquid value at the given temperature

s1T, p2< s

1T2 (6.5)

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suppose the value of specific entropy is required for water at

25 bar, 2008C. The specific entropy is obtained directly from Table A-5 as s 5 2.3294 kJ/

K. Using the saturated liquid value for specific entropy at 2008C from Table A-2,

the specific entropy is approximated with Eq. 6.5 as s 5 2.3309 kJ/kg

K, which

agrees closely with the previous value. b b b b b

6.2.4

Computer Retrieval

The software available with this text, Interactive Thermodynamics: IT, provides data

for the substances considered in this section. Entropy data are retrieved by simple

call statements placed in the workspace of the program.

consider a two-phase liquid–vapor mixture of H

O at p 5 1 bar,

y 5 0.8475 m

/kg. The following illustrates how specific entropy and quality x are

obtained using IT

p 5 1 // bar

v 5 0.8475 // m

/kg

v 5 vsat_Px(“Water/Steam”,p,x)

s 5 ssat_Px(“Water/Steam”,p,x)

The software returns values of x 5 0.5 and s 5 4.331 kJ/kg

K, which can be checked

using data from Table A-3. Note that quality x is implicit in the expression for specific

volume, and it is not necessary to solve explicitly for x. As another example, consider

superheated ammonia vapor at p 5 1.5 bar, T 5 88C. Specific entropy is obtained

from IT as follows:

p 5 1.5 // bar

T 5 8 // 8C

s 5 s_PT (“Ammonia”, p,T)

The software returns s 5 5.981 kJ/kg

K, which agrees closely with the value obtained

by interpolation in Table A-15. b b b b b

6.2.5

Using Graphical Entropy Data

The use of property diagrams as an adjunct to problem solving is emphasized through-

out this book. When applying the second law, it is frequently helpful to locate states

and plot processes on diagrams having entropy as a coordinate. Two commonly used

figures having entropy as one of the coordinates are the temperature–entropy dia-

gram and the enthalpy–entropy diagram.

Temperature–Entropy Diagram

The main features of a temperature–entropy diagram are shown in Fig. 6.2. For detailed

figures for water in SI and English units, see Figs. A-7. Observe that lines of constant

enthalpy are shown on these figures. Also note that in the superheated vapor region

constant specific volume lines have a steeper slope than constant-pressure lines. Lines

of constant quality are shown in the two-phase liquid–vapor region. On some figures,

lines of constant quality are marked as percent moisture lines. The percent moisture

is defined as the ratio of the mass of liquid to the total mass.

In the superheated vapor region of the T–s diagram, constant specific enthalpy

lines become nearly horizontal as pressure is reduced. These superheated vapor states

are shown as the shaded area on Fig. 6.2. For states in this region of the diagram, the

enthalpy is determined primarily by the temperature: h(T, p)

h(T). This is the

T–s diagram

TAKE NOTE...

Note that IT does not

provide compressed liquid

data for any substance. IT

returns liquid entropy data

using the approximation of

Eq. 6.5. Similarly, Eqs. 3.11,

3.12, and 3.14 are used

to return liquid values for

y, u, and h, respectively.

6.2 Retrieving Entropy Data 285

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286 Chapter 6

Using Entropy

region of the diagram where the ideal gas model provides a reasonable approxima-

tion. For superheated vapor states outside the shaded area, both temperature and

pressure are required to evaluate enthalpy, and the ideal gas model is not suitable.

Enthalpy–Entropy Diagram

The essential features of an enthalpy–entropy diagram, commonly known as a Mollier

diagram

, are shown in Fig. 6.3. For detailed figures for water in SI and English units,

see Figs. A-8. Note the location of the critical point and the appearance of lines of

constant temperature and constant pressure. Lines of constant quality are shown in

the two-phase liquid–vapor region (some figures give lines of constant percent mois-

ture). The figure is intended for evaluating properties at superheated vapor states and

for two-phase liquid–vapor mixtures. Liquid data are seldom shown. In the super-

heated vapor region, constant-temperature lines become nearly horizontal as pressure

is reduced. These superheated vapor states are shown, approximately, as the shaded

area on Fig. 6.3. This area corresponds to the shaded area on the temperature–entropy

diagram of Fig. 6.2, where the ideal gas model provides a reasonable approximation.

to illustrate the use of the Mollier diagram in SI units, consider

two states of water. At state 1, T

5 240°C, p

5 0.10 MPa. The specific enthalpy and

quality are required at state 2, where p

= 0.01 MPa and s

5 s

. Turning to Fig. A-8,

state 1 is located in the superheated vapor region. Dropping a vertical line into the

two-phase liquid–vapor region, state 2 is located. The quality and specific enthalpy at

state 2 read from the figure agree closely with values obtained using Tables A-3 and

A-4: x

5 0.98 and h

5 2537 kJ/kg. b b b b b

Mollier diagram

6.3 Introducing the T dS Equations

Although the change in entropy between two states can be determined in principle

by using Eq. 6.2a, such evaluations are generally conducted using the T dS equations

developed in this section. The T dS equations allow entropy changes to be evaluated

from other more readily determined property data. The use of the T dS equations to

Fig. 6.2

Temperature–entropy diagram.

Critical point

p = constant

v = constant

x = 0.2 x

= 0.9

h = constant

v = constant

p = constant

Fig. 6.3 Enthalpy–entropy diagram.

p = constant

T = constant

p = constant

Critical point

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evaluate entropy changes for an incompressible substance is illustrated in Sec. 6.4

and for ideal gases in Sec. 6.5. The importance of the T dS equations is greater than

their role in assigning entropy values, however. In Chap. 11 they are used as a point

of departure for deriving many important property relations for pure, simple com-

pressible systems, including means for constructing the property tables giving u, h,

and s.

The T dS equations are developed by considering a pure, simple compressible

system undergoing an internally reversible process. In the absence of overall system

motion and the effects of gravity, an energy balance in differential form is

int

5 dU 1

int

(6.6)

By definition of simple compressible system (Sec. 3.1.2), the work is

1dW2

int

rev

5 p dV (6.7a)

On rearrangement of Eq. 6.2b, the heat transfer is

1dQ2

int

rev

5 T dS (6.7b)

Substituting Eqs. 6.7 into Eq. 6.6, the first T dS equation results

T ds 5 dU 1 p dV (6.8)

The second T dS equation is obtained from Eq. 6.8 using H 5 U 1 pV. Forming

the differential

dH 5 dU 1 d1pV25 dU 1 p dV 1 V dp

On rearrangement

dU 1 p dV 5 dH 2 V dp

Substituting this into Eq. 6.8 gives the second T dS equation

T dS 5 dH 2 V dp (6.9)

The T dS equations can be written on a unit mass basis as

T ds 5 du 1 p dy (6.10a)

T ds 5 dh 2 y dp (6.10b)

or on a per mole basis as

T ds 5 du 1 p

dy (6.11a)

T ds 5 dh 2

ydp

(6.11b)

Although the T dS equations are derived by considering an internally reversible

process, an entropy change obtained by integrating these equations is the change for

any process, reversible or irreversible, between two equilibrium states of a system.

Because entropy is a property, the change in entropy between two states is indepen-

dent of the details of the process linking the states.

To show the use of the T dS equations, consider a change in phase from saturated

liquid to saturated vapor at constant temperature and pressure. Since pressure is

constant, Eq. 6.10b reduces to give

ds 5

6.3 Introducing the T dS Equations 287

first T dS equation

second T dS equation

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288 Chapter 6

Using Entropy

Then, because temperature is also constant during the phase change

2 s

2 h

(6.12)

This relationship shows how s

2 s

is calculated for tabulation in property tables.

consider Refrigerant 134a at 08C. From Table A-10, h

2 h

197.21 kJ/kg, so with Eq. 6.12

2 s

197.21

273.15 K

5 0.7220

kg ? K

which is the value calculated using s

and s

from the table. To give a similar exam-

ple in English units, consider Refrigerant 134a at 08F. From Table A-10E, h

2 h

90.12 Btu/lb, so

2 s

90.12 Btu

459.678R

5 0.1961

Btu

lb ? 8R

which agrees with the value calculated using s

and s

from the table. b b b b b

6.4 Entropy Change of an Incompressible

Substance

In this section, Eq. 6.10a of Sec. 6.3 is used to evaluate the entropy change between

two states of an incompressible substance. The incompressible substance model intro-

duced in Sec. 3.10.2 assumes that the specific volume (density) is constant and the

specific internal energy depends solely on temperature. Thus, du 5 c(T)dT, where c

denotes the specific heat of the substance, and Eq. 6.10a reduces to give

ds 5

c1T2d

c1T

On integration, the change in specific entropy is

2 s

c1T2

When the specific heat is constant, this becomes

2 s

5 c ln



1incompressible, constant c2

(6.13)

Equation 6.13, along with Eqs. 3.20 giving Du and Dh, respectively, are applicable to

liquids and solids modeled as incompressible. Specific heats of some common liquids

and solids are given in Table A-19.

consider a system consisting of liquid water initially at T

5 300 K,

5 2 bar undergoing a process to a final state at T

5 323 K, p

5 1 bar. There are

two ways to evaluate the change in specific entropy in this case. The first approach is to

use Eq. 6.5 together with saturated liquid data from Table A-2. That is, s

) 5

0.3954 KJ/kg ? K and s

) 5 0.7038 KJ/kg ? K, giving s

2 s

5 0.308 KJ/kg

The second approach is to use the incompressible model. That is, with Eq. 6.13 and

c 5 4.18 KJ/kg ? K from Table A-19, we get

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2 s

5 c ln

5 a4.18

kg ? K

b lna

323 K

300 K

b5 0.309 KJ

kg ? K

Comparing the values obtained for the change in specific entropy using the two

approaches considered here, we see they are in agreement. b b b b b

6.5 Entropy Change of an Ideal Gas

In this section, the T dS equations of Sec. 6.3, Eqs. 6.10, are used to evaluate the

entropy change between two states of an ideal gas. For a quick review of ideal gas

model relations, see Table 6.1.

It is convenient to begin with Eqs. 6.10 expressed as

ds 5

(6.14)

(6.15)

For an ideal gas, du 5 c

(T) dT, dh 5 c

(T) dT, and py 5 RT. With these relations,

Eqs. 6.14 and 6.15 become, respectively

ds 5 c

1T2

1 R



and



ds 5 c

1T2

2 R

(6.16)

On integration, Eqs. 6.16 give, respectively

s1T

, y

22 s1T

, y

1T2

1 R ln

(6.17)

s1T

, p

22 s1T

, p

1T2

2 R ln

(6.18)

Since R is a constant, the last terms of Eqs. 6.16 can be integrated directly. However,

because c

and c

are functions of temperature for ideal gases, it is necessary to have

information about the functional relationships before the integration of the first term

in these equations can be performed. Since the two specific heats are related by

1T25 c

1T21 R (3.44)

where R is the gas constant, knowledge of either specific function suffices.

6.5.1

Using Ideal Gas Tables

As for internal energy and enthalpy changes of ideal gases, the evaluation of entropy

changes for ideal gases can be reduced to a convenient tabular approach. We begin

by introducing a new variable s8(T) as

s81T25

T¿

1T2

(6.19)

where T9 is an arbitrary reference temperature.

6.5 Entropy Change of an Ideal Gas 289

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290 Chapter 6

Using Entropy

The integral of Eq. 6.18 can be expressed in terms of s° as follows

T ¿

5 s81T

22 s81T

Thus, Eq. 6.18 can be written as

s1T

, p

22 s1T

, p

25 s81T

22 s81T

22 R ln

(6.20a)

or on a per mole basis as

s1T

, p

22 s1T

, p

25 s81T

22 s81T

22 R ln

(6.20b)

Because s8 depends only on temperature, it can be tabulated versus temperature,

like h and u. For air as an ideal gas, s8 with units of kJ/kg ? K or Btu/lb ? 8R is given

in Table A-22 and A-22E, respectively. Values of s8 for several other common gases are

given in Tables A-23 with units of kJ/kmol ? K or Btu/lbmol ? 8R. In passing, we note the

arbitrary reference temperature T9 of Eq. 6.19 is specified differently in Tables A-22 than

in Tables A-23. As discussed in Sec. 13.5.1, Tables A-23 give absolute entropy values.

Using Eqs. 6.20 and the tabulated values for s8 or s8, as appropriate, entropy

changes can be determined that account explicitly for the variation of specific heat

with temperature.

let us evaluate the change in specific entropy, in kJ/kg ? K, of air

modeled as an ideal gas from a state where T

5 300 K and p

= 1 bar to a state

where T

5 1000 K and p

5 3 bar. Using Eq. 6.20a and data from Table A-22

2 s

5 s81T

22 s81T

22 R ln

5 12.96770 2 1.702032

kg ? K

8.314

28.97

kg ? K

3 bar

1 bar

5 0.9504 kJ

kg ? K b b b b b

Ideal Gas Model Review

Equations of state:

py 5 RT (3.32)

pV 5 mRT (3.33)

Changes in u and h:

u(T

) 2 u(T

)

1T2dT

(3.40)

h(T

) 2 h(T

)

1T2dT

(3.43)

Constant Specific Heats Variable Specific Heats

TABLE 6.1

u(T ) and h(T ) are evaluated from Tables A-22

for air (mass basis) and Tables A-23 for several

other gases (molar basis).

u(T

) 2 u(T

) 5 c

2 T

) (3.50)

h(T

) 2 h(T

) 5 c

2 T

) (3.51)

See Tables A-20, 21 for c

and c

data.

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If a table giving s8 (or s8) is not available for a particular gas of interest, the inte-

grals of Eqs. 6.17 and 6.18 can be performed analytically or numerically using specific

heat data such as provided in Tables A-20 and A-21.

6.5.2

Assuming Constant Specific Heats

When the specific heats c

and c

are taken as constants, Eqs. 6.17 and 6.18 reduce,

respectively, to

s1T

, y

22 s1T

, y

5 c

1 R ln

(6.21)

s1T

, p

22 s1T

, p

25 c

2 R ln

(6.22)

These equations, along with Eqs. 3.50 and 3.51 giving ¢u and ¢h, respectively, are

applicable when assuming the ideal gas model with constant specific heats.

let us determine the change in specific entropy, in KJ/kg

K, of

air as an ideal gas undergoing a process from T

5 300 K, p

5 1 bar to T

5 400 K,

5 5 bar. Because of the relatively small temperature range, we assume a constant value

of c

evaluated at 350 K. Using Eq. 6.22 and c

5 1.008 KJ/kg

K from Table A-20

¢s 5 c

2 R ln

5 a1.008

kg ? K

b lna

400 K

300 K

b2 a

8.314

28.97

kg ? K

b lna

5 bar

1 bar

520.1719 kJ

kg ? K b b b b b

6.5.3

Computer Retrieval

For gases modeled as ideal gases, IT directly returns s(T, p) using the following special

form of Eq. 6.18:

s1T, p22 s1T

ref

, p

ref

1T2

dT 2 R ln

ref

and the following choice of reference state and reference value: T

ref

5 0 K (08R),

ref

5 1 atm, and s(T

ref

, p

ref

) 5 0, giving

s1T, p25

1T2

dT 2 R ln

ref

(a)

Such reference state and reference value choices equip IT for use in combustion

applications. See the discussion of absolute entropy in Sec. 13.5.1.

Changes in specific entropy evaluated using IT agree with entropy changes evalu-

ated using ideal gas tables.

consider a process of air as an ideal gas from T

5 300 K, p

1 bar to T

5 1000 K, p

5 3 bar. The change in specific entropy, denoted as dels, is

determined in SI units using IT as follows:

p1 5 1//bar

T1 5 300//K

p2 5 3

T2 5 1000

s1 5 s_TP(“Air”,T1,p1)

s2 5 s_TP(“Air”,T2,p2)

dels 5 s2 2 s1

6.5 Entropy Change of an Ideal Gas 291

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292 Chapter 6 Using Entropy

The software returns values of s

5 1.706, s

5 2.656, and dels 5 0.9501, all in units

of kJ/kg ? K. This value for ¢s agrees with the value obtained using Table A-22: 0.9504

kJ/kg ? K, as shown in the concluding example of Sec. 6.5.1. b b b b b

Note again that IT returns specific entropy directly using Eq. (a) above. IT does

not use the special function s8.

6.6 Entropy Change in Internally Reversible

Processes of Closed Systems

In this section the relationship between entropy change and heat transfer for internally

reversible processes is considered. The concepts introduced have important applications

in subsequent sections of the book. The present discussion is limited to the case of

closed systems. Similar considerations for control volumes are presented in Sec. 6.13.

As a closed system undergoes an internally reversible process, its entropy can

increase, decrease, or remain constant. This can be brought out using

dS 5 a

int

rev

(6.2b)

which indicates that when a closed system undergoing an internally reversible process

receives energy by heat transfer, the system experiences an increase in entropy. Con-

versely, when energy is removed from the system by heat transfer, the entropy of the

system decreases. This can be interpreted to mean that an entropy transfer accompa-

nies heat transfer. The direction of the entropy transfer is the same as that of the heat

transfer. In an adiabatic internally reversible process, entropy remains constant. A

constant-entropy process is called an isentropic process.

6.6.1

Area Representation of Heat Transfer

On rearrangement, Eq. 6.2b gives

1dQ2

int

rev

5 T dS

Integrating from an initial state 1 to a final state 2

int

rev

T dS

(6.23)

From Eq. 6.23 it can be concluded that an energy transfer by heat to a closed system

during an internally reversible process can be represented as an area on a temperature–

entropy diagram. Figure 6.4 illustrates the area representation of heat transfer for an

arbitrary internally reversible process in which temperature varies. Carefully note that

temperature must be in kelvins or degrees Rankine, and the area is the entire area

under the curve (shown shaded). Also note that the area representation of heat trans-

fer is not valid for irreversible processes, as will be demonstrated later.

6.6.2

Carnot Cycle Application

To provide an example illustrating both the entropy change that accompanies heat

transfer and the area representation of heat transfer, consider Fig. 6.5a, which shows a

Carnot power cycle (Sec. 5.10.1). The cycle consists of four internally reversible processes

in series: two isothermal processes alternated with two adiabatic processes. In Process 2–3,

heat transfer to the system occurs while the temperature of the system remains constant

entropy transfer

Carnot cycle

isentropic process

Fig. 6.4 Area representation

of heat transfer for an

internally reversible process

of a closed system.

(δQ) = T dS

Q =

T dS

int

rev

int

rev

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