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

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516 Chapter 11 Thermodynamic Relations

In developing this result, the relationship between partial derivatives expressed by Eq. 11.15

has been used.

Several important conclusions about the specific heats c

and c

can be drawn from

Eq. 11.69.  for example. . . since the factor 

cannot be negative and  is positive

for all substances in all phases, the value of c

is always greater than, or equal to, c

The specific heats would be equal when   0, as occurs in the case of water at 1 atmos-

phere and 4C, where water is at its state of maximum density. The two specific heats also

become equal as the temperature approaches absolute zero. For some liquids and solids at

certain states, c

and c

differ only slightly. For this reason, tables often give the specific heat

of a liquid or solid without specifying whether it is c

or c

. The data reported are normally

values, since these are more easily determined for liquids and solids. 

EVALUATING c

兾c

. Next, let us obtain expressions for the ratio of specific heats. Em-

ploying Eq. 11.16, we can rewrite Eqs. 11.46 and 11.55, respectively, as

Forming the ratio of these equations gives

(11.70)

Since ( s p)

 1( p s)

and ( p T )

 1( T p)

, Eq. 11.70 can be expressed as

(11.71)

Finally, the chain rule from calculus allows us to write ( v p)

 ( v s)

( s p)

and

( p v)

 ( p T )

( T v)

, so Eq. 11.71 becomes

(11.72)

This can be expressed alternatively in terms of the isothermal and isentropic compressibilities as

(11.73)

Solving Eq. 11.72 for ( )

and substituting the resulting expression into Eq. 9.36b

gives the following relationship involving the velocity of sound c and the specific heat ratio k

(11.74)

Equation 11.74 can be used to determine c knowing the specific heat ratio and p–v–T data,

or to evaluate k knowing c and ( p v)

.  for example. . . in the special case of an

ideal gas, Eq. 11.74 reduces to

(9.37)

as can easily be verified. 

In the next example we illustrate the use of specific heat relations introduced above.

c  1kRT

1ideal gas2

c  2kv

10p



0v2



k 

 a

000000

 ca

dca

00000000



10v



0s2

10T



0v2

10p



0s2

10T



0p2

 a



1

10p



0s2

10T



0p2

 a



1

10v



0s2

10T



0v2

11.5 Other Thermodynamic Relations 517

EXAMPLE 11.6 Using Specific Heat Relations

For liquid water at 1 atm and 20C, estimate (a) the percent error in c

that would result if it were assumed that c

 c

(b) the velocity of sound, in m/s.

SOLUTION

Known: The system consists of a fixed amount of liquid water at 1 atm and 20C.

Find: Estimate the percent error in c

that would result if c

were approximated by c

, and the velocity of sound,

in m/s.

Analysis:

(a) Equation 11.69 gives the difference between c

and c

. Table 11.2 provides the required values for the volume expansiv-

ity , the isothermal compressibility , and the specific volume. Thus

Interpolating in Table A-19 at 20C gives c

 4.188 Thus, the value of c

Using these values, the percent error in approximating c

by c

(b) The velocity of sound at this state can be determined using Eq. 11.65. The required value for the isentropic compressibility

 is calculable in terms of the specific heat ratio k and the isothermal compressibility . With Eq. 11.73,   k. Inserting

this into Eq. 11.65 results in the following expression for the velocity of sound

The values of v and  required by this expression are the same as used in part (a). Also, with the values of c

and c

from

part (a), the specific heat ratio is k  1.006. Accordingly

Consistent with the discussion of Sec. 3.3.6, we take c

at 1 atm and 20C as the saturated liquid value at 20C.

The result of part (a) shows that for liquid water at the given state, c

and c

are closely equal.

For comparison, the velocity of sound in air at 1 atm, 20C is about 343 m/s, which can be checked using Eq. 9.37.

c 

11.0062110

2 bar

1998.21 kg/m

2 145.902

N/m

1 bar

1 kg

m/s

1 N

 1482 m/s

c 

 c

b 11002 a

0.027

4.161

b 11002 0.6%

 4.188  0.027  4.161 kJ/kg

kJ/kg

 0.027

 a272.96  10

6

bar

b `

N/m

1 bar

1 kJ

 a

998.21 kg/m

b 1293 K2 a

206.6  10

6

bar

45.90  10

6

 c

 v

❶

❷

❸

❶

❷

❸

 11.5.3 Joule–Thomson Coefficient

The value of the specific heat c

can be determined from p–v–T data and the Joule–Thomson

coefficient. The Joule–Thomson coefficient 

is defined as

(11.75)

Like other partial differential coefficients introduced in this section, the Joule–Thomson co-

efficient is defined in terms of thermodynamic properties only and thus is itself a property.

The units of 

are those of temperature divided by pressure.

A relationship between the specific heat c

and the Joule–Thomson coefficient 

can be

established by using Eq. 11.16 to write

The first factor in this expression is the Joule–Thomson coefficient and the third is c

. Thus

With ( )

 1()

from Eq. 11.15, this can be written as

(11.76)

The partial derivative ( )

, called the constant-temperature coefficient, can be eliminated

from Eq. 11.76 by use of Eq. 11.56. The following expression results:

(11.77)

Equation 11.77 allows the value of c

at a state to be determined using p–v–T data and the

value of the Joule–Thomson coefficient at that state. Let us consider next how the

Joule–Thomson coefficient can be found experimentally.

EXPERIMENTAL EVALUATION. The Joule–Thomson coefficient can be evaluated experi-

mentally using an apparatus like that pictured in Fig. 11.3. Consider first Fig. 11.3a, which

shows a porous plug through which a gas (or liquid) may pass. During operation at steady

state, the gas enters the apparatus at a specified temperature T

and pressure p

and expands

through the plug to a lower pressure p

, which is controlled by an outlet valve. The temper-

ature T

at the exit is measured. The apparatus is designed so that the gas undergoes a

throttling process (Sec. 4.3) as it expands from 1 to 2. Accordingly, the exit state fixed by

and T

has the same value for the specific enthalpy as at the inlet, h

 h

. By progres-

sively lowering the outlet pressure, a finite sequence of such exit states can be visited, as in-

dicated on Fig. 11.3b. A curve may be drawn through the set of data points. Such a curve is

called an isenthalpic (constant enthalpy) curve. An isenthalpic curve is the locus of all points

representing equilibrium states of the same specific enthalpy.

The slope of an isenthalpic curve at any state is the Joule–Thomson coefficient at that state.

The slope may be positive, negative, or zero in value. States where the coefficient has a zero

value are called inversion states. Notice that not all lines of constant h have an inversion state.

The uppermost curve of Fig. 11.3b, for example, always has a negative slope. Throttling a gas



cT a

 v d







0h0h





1

10p



0h2

1

 a

518 Chapter 11 Thermodynamic Relations

Joule–Thomson

coefficient

inversion states

11.5 Other Thermodynamic Relations 519

disastrous. Businesses such as

shopping centers also must

avoid costly service interrup-

tions. With microturbine-based

systems, the needed reliability

is provided by modular units

that can run virtually unat-

tended for years.

Microturbines are also well

suited for cogeneration, provid-

ing both power and heating for

buildings or other uses. They

also can be effectively inter-

grated with fuel cells and energy storage units. Self-contained

power systems are especially attractive in places such as de-

veloping nations where the power grid may be unreliable or

even nonexistent.

Small Power Plants Pack Punch

Thermodynamics in the News...

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With distributed power, consumers hope to avoid unpre-

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regions of the country.

At the heart of many distributed power systems are clev-

erly engineered gas turbine units, called microturbines, less

than one-fifth the size of utility gas turbines with fewer mov-

ing parts. They also emit less nitric oxide (NO

) pollution.

Although the cost per kilowatt hour may be higher with mi-

croturbines, some customers are willing to pay more to gain

control over their electric supply. Computer networks and hos-

pitals need high reliability, since even short disruptions can be

from an initial state on this curve would result in an increase in temperature. However, for

isenthalpic curves having an inversion state, the temperature at the exit of the apparatus may

be greater than, equal to, or less than the initial temperature, depending on the exit pressure

specified. For states to the right of an inversion state, the value of the Joule–Thomson coef-

ficient would be negative. For these states, the temperature would increase as the pressure at

the exit of the apparatus is reduced. At states to the left of an inversion state, the value of the

Joule–Thomson coefficient would be positive. For these states, the temperature would decrease

as the pressure at the exit of the device is reduced. This can be used to advantage in systems

designed to liquefy gases.

Inversion state

Inversion

state

Inversion

state

Critical

point

Triple

point

Vapor

Inlet state

, p

)

∂T

––

∂p

(

)

Liquid

Solid

(b)(a)

Inlet

, p

Porous plug

, p

Valve

 Figure 11.3

Joule–Thomson expansion. (a) Apparatus. (b) Isenthalpics on a T–p diagram.

520 Chapter 11 Thermodynamic Relations



11.6 Constructing Tables of

Thermodynamic Properties

The objective of this section is to utilize the thermodynamic relations introduced thus far to

describe how tables of thermodynamic properties can be constructed. The characteristics of

the tables under consideration are embodied in the tables for water and the refrigerants pre-

sented in the Appendix. The methods introduced in this section are extended in Chap. 13 for

the analysis of reactive systems, such as gas turbine and vapor power systems involving com-

bustion. The methods of this section also provide the basis for computer retrieval of ther-

modynamic property data.

Two different approaches for constructing property tables are considered:

 The presentation of Sec. 11.6.1 employs the methods introduced in Sec. 11.4 for assign-

ing specific enthalpy, specific internal energy, and specific entropy to states of pure,

simple compressible substances using p–v–T data, together with a limited amount of

specific heat data. The principal mathematical operation of this approach is integration.

 The approach of Sec. 11.6.2 utilizes the fundamental thermodynamic function concept

introduced in Sec. 11.3.3. Once such a function has been constructed, the principal

mathematical operation required to determine all other properties is differentiation.

 11.6.1 Developing Tables by Integration Using p–v–T and

Specific Heat Data

In principle, all properties of present interest can be determined using

(11.78)

In Eqs. 11.78, c

(T ) is the specific heat c

for the substance under consideration extrapo-

lated to zero pressure. This function might be determined from data obtained calorimetrically

or from spectroscopic data, using equations supplied by statistical mechanics. Specific heat

expressions for several gases are given in Tables A-21. The expressions p(v, T ) and v(p, T )

represent functions that describe the saturation pressure–temperature curves, as well as the

p–v–T relations for the single-phase regions. These functions may be tabular, graphical, or

analytical in character. Whatever their forms, however, the functions must not only represent

the p–v–T data accurately but also yield accurate values for derivatives such as ( )

and

(dpdT )

sat

Figure 11.4 shows eight states of a substance. Let us consider how values can be assigned

to specific enthalpy and specific entropy at these states. The same procedures can be used to

assign property values at other states of interest. Note that when h has been assigned to a

state, the specific internal energy at that state can be found from u  h  pv.

 Let the state denoted by 1 on Fig. 11.4 be selected as the datum state for enthalpy and

entropy. Any value can be assigned to h and s at this state, but a value of zero would be

usual. It should be noted that the use of an arbitrary datum state and arbitrary reference

values for specific enthalpy and specific entropy suffices only for evaluations involving

differences in property values between states of the same composition, for then datums

cancel.

 Once a value is assigned to enthalpy at state 1, the enthalpy at the saturated vapor state,

state 2, can be determined using the Clapeyron equation, Eq. 11.40

 h

 T

 v

sat



p  p1v, T 2,

v  v1p, T 2

 c

1T 2

11.6 Constructing Tables of Thermodynamic Properties 521

where the derivative (dpdT)

sat

and the specific volumes v

and v

are obtained from

appropriate representations of the p–v–T data for the substance under consideration.

The specific entropy at state 2 is found using Eq. 11.38 in the form

 Proceeding at constant temperature from state 2 to state 3, the entropy and enthalpy are

found by means of Eqs. 11.59 and 11.60, respectively. Since temperature is fixed, these

equations reduce to give

With the same procedure, s

and h

can be determined.

 The isobar (constant-pressure line) passing through state 4 is assumed to be at a low

enough pressure for the ideal gas model to be appropriate. Accordingly, to evaluate

s and h at states such as 5 on this isobar, the only required information would be c

(T )

and the temperatures at these states. Thus, since pressure if fixed, Eqs. 11.59 and 11.60

give, respectively

 Specific entropy and enthalpy values at states 6 and 7 are found from those at state 5

by the same procedure used in assigning values at states 3 and 4 from those at state 2.

Finally, s

and h

are obtained from the values at state 7 using the Clapeyron equation.

 11.6.2 Developing Tables by Differentiating a Fundamental

Thermodynamic Function

Property tables also can be developed using a fundamental thermodynamic function. It is con-

venient for this purpose to select the independent variables of the fundamental function from

among pressure, specific volume (density), and temperature. This indicates the use of the

Helmholtz function (T, v) or the Gibbs function g(T, p). The properties of water tabulated

in Tables A-2 through A-6 have been calculated using the Helmholtz function. Fundamental

functions also have been employed successfully to evaluate the properties of other substances.

The development of a fundamental function requires considerable mathematical manipu-

lation and numerical evaluation. Prior to the advent of high-speed computers, the evaluation

 s





and

 h





 s





and

 h





cv  T a

 s



 h

Arbitrary datum

state h

= s

= 0

234

765

Isobar–

reduced pressure

low enough

for the ideal gas

model to be

appropriate

 Figure 11.4 T–v diagram used to dis-

cuss how h and s can be assigned to liquid

and vapor states.

522 Chapter 11 Thermodynamic Relations

of properties by this means was not feasible, and the approach described in Sec. 11.6.1 was

used exclusively. The fundamental function approach involves three steps:

1. The first step is the selection of a functional form in terms of the appropriate pair of

independent properties and a set of adjustable coefficients, which may number 50 or more.

The functional form is specified on the basis of both theoretical and practical considerations.

2. Next, the coefficients in the fundamental function are determined by requiring that a set of

carefully selected property values and /or observed conditions be satisfied in a least-squares

sense. This generally involves the use of property data requiring the assumed functional

form to be differentiated one or more times, such as p–v–T and specific heat data.

3. When all coefficients have been evaluated, the function is carefully tested for accuracy

by using it to evaluate properties for which accepted values are known. These may

include properties requiring differentiation of the fundamental function two or more

times. For example, velocity of sound and Joule–Thomson data might be used.

This procedure for developing a fundamental function is not routine and can be accomplished

only with a computer. However, once a suitable fundamental function is established, extreme

accuracy in and consistency among the thermodynamic properties is possible.

The form of the Helmholtz function used in constructing the steam tables from which

Tables A-2 through A-6 have been extracted is

(11.79)

where 

and Q are given as the sums listed in Table 11.3. The independent variables are

density and temperature. The variable  denotes 1000T. Values for pressure, specific internal

c1r, T 2 c

1T 2 RT 3ln r  rQ1r, t24

TABLE 11.3 Fundamental Equation Used to Construct the Steam Tables

a,b

(1)

where

(2)

and

(3)

In (1), (2), and (3), T denotes temperature on the Kelvin scale,  denotes 1000T,  denotes

density in g/cm

, R  4.6151 or 0.46151  1.544912,

E  4.8, and

The coefficients for 

in J/g are given as follows:

 1857.065 C

 36.6649 C

 46.0

 3229.12 C

20.5516 C

1011.249

419.465 C

 4.85233

Values for the coefficients A

are listed in the original source.

J. H. Keenan, F. G. Keyes, P. G. Hill, and J. G. Moore, Steam Tables, Wiley, New York, 1969.

Also see L. Haar, J. S. Gallagher, and G. S. Kell, NBS/ NRC Steam Tables, Hemisphere, Washington, D.C.,

1984. The properties of water are determined in this reference using a different functional form for the

Helmholtz function than given by Eqs. (1)–(3).

 2.5 1 j 7 12  1.0 1j 7 12

 t

j  12

 0.634 1j  12

J/g

K, t

 1000



bar

Q  1t  t

j1

1t  t

j2

i1

1r  r

i1

 e

Er

i9

i9



i1



i1

 C

ln T  C

ln T



c  c

1T 2 RT 3ln r  rQ 1r, t24

11.6 Constructing Tables of Thermodynamic Properties 523

energy, and specific entropy can be determined by differentiation of Eq. 11.79. Values for

the specific enthalpy and Gibbs function are found from h  u  pv and g    pv, re-

spectively. The specific heat c

is evaluated by further differentiation, c

 ( u T )

. With

similar operations, other properties can be evaluated. Property values for water calculated

from Eq. 11.79 are in excellent agreement with experimental data over a wide range of con-

ditions. Example 11.7 illustrates this approach for developing tables.

EXAMPLE 11.7 Determining Properties Using a Fundamental Function

The following expression for the Helmholtz function has been used to determine the properties of water:

where  denotes density and  denotes 1000T. The functions 

and Q are sums involving the indicated independent vari-

ables and a number of adjustable constants (see Table 11.3). Obtain expressions for (a) pressure, (b) specific entropy, and

SOLUTION

Known: An expression for the Helmholtz function  is given.

Find: Determine the expressions for pressure, specific entropy, and specific internal energy that result from this fundamen-

tal function.

Analysis: The expressions developed below for p, s, and u require only the functions 

(T ) and Q(, ). Once these func-

tions are determined, p, s, and u can each be determined as a function of density and temperature using elementary mathe-

matical operations.

(a) When expressed in terms of density instead of specific volume, Eq. 11.28 becomes

as can easily be verified. When T is held constant  is also constant. Accordingly, the following is obtained on differentiation

of the given function:

Combining these equations gives an expression for pressure

(b) From Eq. 11.29

Differentiation of the given expression for  yields



 R cln r  rQ  rt



 cR1ln r  rQ2 RTr a

a

1000



 cR1ln r  rQ2 RTr

s a

p  rRT c1  rQ  r

 RT c

 Q

1r, t2 r a

p  r

c1r, T2 c

1T 2 RT 3ln r  rQ 1r, t24

524 Chapter 11 Thermodynamic Relations

Combining results gives

s from part (b) results in

This can be written more compactly by noting that

Thus,

Finally, the expression for u becomes

u 

d1c

 RTrt

 T

 c

 t



d1c

 T

a

1000

bt

 c

 T

 RTrt

u  3c

 RT 1ln r  rQ24 T e

 R cln r  rQ  rt a

s 

 R cln r  rQ  rt



11.7 Generalized Charts for

Enthalpy and Entropy

Generalized charts giving the compressibility factor Z in terms of the reduced properties p

, and v

are introduced in Sec. 3.4. With such charts, estimates of p–v–T data can be ob-

tained rapidly knowing only the critical pressure and critical temperature for the substance

of interest. The objective of the present section is to introduce generalized charts that allow

changes in enthalpy and entropy to be estimated.

GENERALIZED ENTHALPY DEPARTURE CHART

The change in specific enthalpy of a gas (or liquid) between two states fixed by temperature

and pressure can be evaluated using the identity

(11.80)

The term [h(T, p)  h*(T)] denotes the specific enthalpy of the substance relative to that of

its ideal gas model when both are at the same temperature. The superscript * is used in this

section to identify ideal gas property values. Thus, Eq. 11.80 indicates that the change in

specific enthalpy between the two states equals the enthalpy change determined using the

ideal gas model plus a correction that accounts for the departure from ideal gas behavior.

The correction is shown underlined in Eq. 11.80. The ideal gas term can be evaluated using

methods introduced in Chap. 3. Next, we show how the correction term is evaluated in terms

of the enthalpy departure.

 53h1T

, p

2 h*1T

24 3h1T

, p

2 h*1T

246

h1T

, p

2 h1T

, p

2 3h*1T

2 h*1T

11.7 Generalized Charts for Enthalpy and Entropy 525

DEVELOPING THE ENTHALPY DEPARTURE. The variation of enthalpy with pressure at

fixed temperature is given by Eq. 11.56 as

Integrating from pressure p to pressure p at fixed temperature T

This equation is not altered fundamentally by adding and subtracting h*(T ) on the left side.

That is

(11.81)

As pressure tends to zero at fixed temperature, the enthalpy of the substance approaches

that of its ideal gas model. Accordingly, as p tends to zero

In this limit, the following expression is obtained from Eq. 11.81 for the specific enthalpy

of a substance relative to that of its ideal gas model when both are at the same temperature:

(11.82)

This also can be thought of as the change in enthalpy as the pressure is increased from zero

to the given pressure while temperature is held constant. Using p–v–T data only, Eq. 11.82

can be evaluated at states 1 and 2 and thus the correction term of Eq. 11.80 evaluated. Let

us consider next how this procedure can be conducted in terms of compressibility factor data

and the reduced properties T

and p

The integral of Eq. 11.82 can be expressed in terms of the compressibility factor Z and

the reduced properties T

and p

as follows. Solving Z  pvRT gives

On differentiation

With the previous two expressions, the integrand of Eq. 11.82 becomes

(11.83)

Equation 11.83 can be written in terms of reduced properties as

Introducing this into Eq. 11.82 gives on rearrangement

h*1T 2 h1T, p2

 T



v  T a



v  T a



ZRT

 T



d





v 

ZRT

h1T, p2 h*1T 2



cv  T a

d dp

lim

p¿S0

3h1T, p¿ 2 h*1T 24 0

3h1T, p2 h*1T 24 3h1T, p¿2 h*1T 24



p¿

cv  T a

d dp

h1T, p2 h1T, p¿ 2



p¿

cv  T a

d dp

 v  T a