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Unformatted text preview: Chapter 12
THERMODYNAMIC PROPERTY RELATIONS
n the preceding chapters we made extensive use of the property tables. We tend to take the property tables for granted, but thermodynamic laws and principles are of little use to engineers without them. In this chapter, we focus our attention on how the property tables are prepared and how some unknown properties can be determined from limited available data. It will come as no surprise that some properties such as temperature, pressure, volume, and mass can be measured directly. Other properties such as density and specific volume can be determined from these using some simple relations. However, properties such as internal energy, enthalpy, and entropy are not so easy to determine because they cannot be measured directly or related to easily measurable properties through some simple relations. Therefore, it is essential that we develop some fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties. By the nature of the material, this chapter makes extensive use of partial derivatives. Therefore, we start by reviewing them. Then we develop the Maxwell relations, which form the basis for many thermodynamic relations. Next we discuss the Clapeyron equation, which enables us to determine the enthalpy of vaporization from P, v, and T measurements alone, and we develop general relations for cv, cp, du, dh, and ds that are valid for all pure substances under all conditions. Then we discuss the JouleThomson coefficient, which is a measure of the temperature change with pressure during a throttling process. Finally, we develop a method of evaluating the h, u, and s of real gases through the use of generalized enthalpy and entropy departure charts. I Objectives
The objectives of Chapter 12 are to: Develop fundamental relations between commonly encountered thermodynamic properties and express the properties that cannot be measured directly in terms of easily measurable properties. Develop the Maxwell relations, which form the basis for many thermodynamic relations. Develop the Clapeyron equation and determine the enthalpy of vaporization from P, v, and T measurements alone. Develop general relations for cv, cp, du, dh, and ds that are valid for all pure substances. Discuss the JouleThomson coefficient. Develop a method of evaluating the h, u, and s of real gases through the use of generalized enthalpy and entropy departure charts.  651 652  Thermodynamics 121 A LITTLE MATHPARTIAL DERIVATIVES AND ASSOCIATED RELATIONS Many of the expressions developed in this chapter are based on the state postulate, which expresses that the state of a simple, compressible substance is completely specified by any two independent, intensive properties. All other properties at that state can be expressed in terms of those two properties. Mathematically speaking,
f(x) z z 1x, y2 (x +x) f f(x) x Slope x x+x x where x and y are the two independent properties that fix the state and z represents any other property. Most basic thermodynamic relations involve differentials. Therefore, we start by reviewing the derivatives and various relations among derivatives to the extent necessary in this chapter. Consider a function f that depends on a single variable x, that is, f f (x). Figure 121 shows such a function that starts out flat but gets rather steep as x increases. The steepness of the curve is a measure of the degree of dependence of f on x. In our case, the function f depends on x more strongly at larger x values. The steepness of a curve at a point is measured by the slope of a line tangent to the curve at that point, and it is equivalent to the derivative of the function at that point defined as
df dx lim
xS0 FIGURE 121 The derivative of a function at a specified point represents the slope of the function at that point. f x lim
xS0 f 1x x2 x f 1x2 (121) Therefore, the derivative of a function f(x) with respect to x represents the rate of change of f with x. EXAMPLE 121 Approximating Differential Quantities by Differences The cp of ideal gases depends on temperature only, and it is expressed as cp(T ) dh(T )/dT. Determine the cp of air at 300 K, using the enthalpy data from Table A17, and compare it to the value listed in Table A2b.
h(T ), kJ/kg Slope = cp(T ) 305.22 Solution The cp value of air at a specified temperature is to be determined using enthalpy data. Analysis The cp value of air at 300 K is listed in Table A2b to be 1.005 kJ/kg K. This value could also be determined by differentiating the function h(T ) with respect to T and evaluating the result at T 300 K. However, the function h(T ) is not available. But, we can still determine the cp value approximately by replacing the differentials in the cp(T ) relation by differences in the neighborhood of the specified point (Fig. 122):
cp 1300 K 2 c dh 1T2 dT d c h 1T2 T d h 1305 K2 1305 295 2 K h 1295 K2 295.17 1305.22
295 300 305 T, K T 300 K T 300 K 1305 295.172 kJ>kg 295 2 K 1.005 kJ/kg # K FIGURE 122 Schematic for Example 121. Discussion Note that the calculated cp value is identical to the listed value. Therefore, differential quantities can be viewed as differences. They can Chapter 12
even be replaced by differences, whenever necessary, to obtain approximate results. The widely used finite difference numerical method is based on this simple principle.
z z ( )y x  653 Partial Differentials
Now consider a function that depends on two (or more) variables, such as z z(x, y). This time the value of z depends on both x and y. It is sometimes desirable to examine the dependence of z on only one of the variables. This is done by allowing one variable to change while holding the others constant and observing the change in the function. The variation of z(x, y) with x when y is held constant is called the partial derivative of z with respect to x, and it is expressed as
a 0z b 0x y
xS0 y x lim a z b x y lim
xS0 z 1x x, y2 x z 1x, y2 (122) FIGURE 123 Geometric representation of partial derivative ( z/ x)y. This is illustrated in Fig. 123. The symbol represents differential changes, just like the symbol d. They differ in that the symbol d represents the total differential change of a function and reflects the influence of all variables, whereas represents the partial differential change due to the variation of a single variable. Note that the changes indicated by d and are identical for independent variables, but not for dependent variables. For example, ( x)y dx but ( z)y dz. [In our case, dz ( z)x ( z)y.] Also note that the value of the partial derivative ( z/ x)y, in general, is different at different y values. To obtain a relation for the total differential change in z(x, y) for simultaneous changes in x and y, consider a small portion of the surface z(x, y) shown in Fig. 124. When the independent variables x and y change by x and y, respectively, the dependent variable z changes by z, which can be expressed as
z z 1x z 1x x, y z 1x, y y2 z 1x, y2 z 1x, y z 1x, y z z(x, y) z(x + x, y + y) x x + x, y y x, y + y x + x, y + y Adding and subtracting z(x, y
z x, y y2 y), we get
y2 y2 z 1x, y2 z 1x, y2 or
z FIGURE 124 Geometric representation of total derivative dz for a function z(x, y).
y z 1x x, y y2 x z 1x, y y2 x y2 y Taking the limits as x 0 and y 0 and using the definitions of partial derivatives, we obtain
dz a 0z b dx 0x y a 0z b dy 0y x
(123) Equation 123 is the fundamental relation for the total differential of a dependent variable in terms of its partial derivatives with respect to the independent variables. This relation can easily be extended to include more independent variables. 654  Thermodynamics
EXAMPLE 122 Total Differential versus Partial Differential Consider air at 300 K and 0.86 m3/kg. The state of air changes to 302 K and 0.87 m3/kg as a result of some disturbance. Using Eq. 123, estimate the change in the pressure of air. Solution The temperature and specific volume of air changes slightly during a process. The resulting change in pressure is to be determined. Assumptions Air is an ideal gas. Analysis Strictly speaking, Eq. 123 is valid for differential changes in variables. However, it can also be used with reasonable accuracy if these changes are small. The changes in T and v, respectively, can be expressed as dT
and T 1302 3002 K 2K 0.01 m3>kg dv v 10.87 0.862 m3>kg RT v An ideal gas obeys the relation Pv RT. Solving for P yields P
Note that R is a constant and P average values for T and v, P(T, v). Applying Eq. 123 and using dP a 0P b dT 0T v a 0P b dv 0v T R dT v RT dv v2 10.287 kPa 0.664 kPa # m3>kg # K2 c 2K 0.865 m3>kg 1301 K2 10.01 m3>kg 2 10.865 m3>kg2 2 d 1.155 kPa 0.491 kPa
Therefore, the pressure will decrease by 0.491 kPa as a result of this disturbance. Notice that if the temperature had remained constant (dT 0), the pressure would decrease by 1.155 kPa as a result of the 0.01 m3/kg increase in specific volume. However, if the specific volume had remained constant (dv 0), the pressure would increase by 0.664 kPa as a result of the 2K rise in temperature (Fig. 125). That is, P, kPa (P)v = 0.664 dP = 0.491 (P)T = 1.155 a a
and 0P b dT 0T v 0P b dv 0v T 1 0P2 T 10P2 v 10P2 T 0.664 kPa 1.155 kPa 0.86 0.87 300 302 T, K v, m3/kg dP 10P2 v 0.664 1.155 0.491 kPa FIGURE 125 Geometric representation of the disturbance discussed in Example 122. Discussion Of course, we could have solved this problem easily (and exactly) by evaluating the pressure from the idealgas relation P RT/v at the final state (302 K and 0.87 m3/kg) and the initial state (300 K and 0.86 m3/kg) and taking their difference. This yields 0.491 kPa, which is exactly the value obtained above. Thus the small finite quantities (2 K, 0.01 m3/kg) can be approximated as differential quantities with reasonable accuracy. Chapter 12  655 Partial Differential Relations
Now let us rewrite Eq. 123 as
dz M dx N dy a 0z b 0y x
(124) where
M a 0z b 0x y and N Taking the partial derivative of M with respect to y and of N with respect to x yields
a 0M b 0y x 0 2z 0x 0y and a 0N b 0x y 0 2z 0y 0x The order of differentiation is immaterial for properties since they are continuous point functions and have exact differentials. Therefore, the two relations above are identical:
a 0M b 0y x a 0N b 0x y
(125) This is an important relation for partial derivatives, and it is used in calculus to test whether a differential dz is exact or inexact. In thermodynamics, this relation forms the basis for the development of the Maxwell relations discussed in the next section. Finally, we develop two important relations for partial derivativesthe reciprocity and the cyclic relations. The function z z(x, y) can also be expressed as x x(y, z) if y and z are taken to be the independent variables. Then the total differential of x becomes, from Eq. 123,
dx a 0x b dy 0y z a a 0x b dz 0z y a 0x 0z b a b dz 0z y 0x y
(126) Eliminating dx by combining Eqs. 123 and 126, we have
dz ca 0z 0x b a b 0x y 0y z 0z b d dy 0y x Rearranging,
ca 0z 0x b a b 0x y 0y z a 0z b d dy 0y x c1 a 0x 0z b a b d dz 0z y 0x y
(127) The variables y and z are independent of each other and thus can be varied independently. For example, y can be held constant (dy 0), and z can be varied over a range of values (dz 0). Therefore, for this equation to be valid at all times, the terms in the brackets must equal zero, regardless of the values of y and z. Setting the terms in each bracket equal to zero gives
a a 0z 0x b a b 0z y 0x y a 1S a 0x b 0z y 1 10z>0x2 y 1
(128) 0z 0x b a b 0x y 0y z 0y 0x 0x 0z b S a b a b a b 0y x 0y z 0z x 0x y (129) 656  Thermodynamics
The first relation is called the reciprocity relation, and it shows that the inverse of a partial derivative is equal to its reciprocal (Fig. 126). The second relation is called the cyclic relation, and it is frequently used in thermodynamics (Fig. 127). Function: z + 2xy 3y2z = 0 2xy 1) z =  3y2 1 3y2z z 2) x =  2y Thus,  ( xz )y = 3y 2y 1
2 EXAMPLE 123 Verification of Cyclic and Reciprocity Relations Using the idealgas equation of state, verify (a) the cyclic relation and (b) the reciprocity relation at constant P. 3y 1 ( xz )y =  2y 1 x z y 2 Solution The cyclic and reciprocity relations are to be verified for an ideal gas.
Analysis The idealgas equation of state Pv RT involves the three variables P, v, and T. Any two of these can be taken as the independent variables, with the remaining one being the dependent variable. (a) Replacing x, y, and z in Eq. 129 by P, v, and T, respectively, we can express the cyclic relation for an ideal gas as ( xz )y = ( ) a 0P 0v 0T b a b a b 0v T 0T P 0P v RT 0P S a b v 0v T RT 0v S a b P 0T P 0T Pv S a b R 0P v RT Pv 1 FIGURE 126 Demonstration of the reciprocity relation for the function z 2xy 3y2z 0. where P v T
Substituting yields P 1v, T2 v 1P, T2 T 1P, v2 RT v2 R P v R a RT R v ba ba b 2 P R v 1 which is the desired result. (b) The reciprocity rule for an ideal gas at P constant can be expressed as a 0v b 0T P 1 1 0T> 0v2 P R P Performing the differentiations and substituting, we have R P 1 R S P>R P FIGURE 127 Partial differentials are powerful tools that are supposed to make life easier, not harder. Reprinted with special permission of King Features Syndicate. Thus the proof is complete. 122 THE MAXWELL RELATIONS The equations that relate the partial derivatives of properties P, v, T, and s of a simple compressible system to each other are called the Maxwell relations. They are obtained from the four Gibbs equations by exploiting the exactness of the differentials of thermodynamic properties. Chapter 12
Two of the Gibbs relations were derived in Chap. 7 and expressed as
du dh T ds T ds P dv v dP
(1210) (1211)  657 The other two Gibbs relations are based on two new combination propertiesthe Helmholtz function a and the Gibbs function g, defined as
a g u h Ts Ts
(1212) (1213) Differentiating, we get
da dg du dh T ds T ds s dT s dT Simplifying the above relations by using Eqs. 1210 and 1211, we obtain the other two Gibbs relations for simple compressible systems:
da dg s dT s dT P dv v dP
(1214) (1215) A careful examination of the four Gibbs relations reveals that they are of the form
dz M dx N dy
(124) with
a 0M b 0y x a 0N b 0x y
(125) since u, h, a, and g are properties and thus have exact differentials. Applying Eq. 125 to each of them, we obtain
a a 0T b 0v s 0T b 0P s a a 0P b 0s v
(1216) 0v b 0s P (1217) ( T )s = (P)v v s T ( P)s = ( sv )P s ( )T = (P)v v T s v ( P)T = ( T )P 0s a b 0v T a 0s b 0P T 0P a b 0T v a 0v b 0T P (1218) (1219) These are called the Maxwell relations (Fig. 128). They are extremely valuable in thermodynamics because they provide a means of determining the change in entropy, which cannot be measured directly, by simply measuring the changes in properties P, v, and T. Note that the Maxwell relations given above are limited to simple compressible systems. However, other similar relations can be written just as easily for nonsimple systems such as those involving electrical, magnetic, and other effects. FIGURE 128 Maxwell relations are extremely valuable in thermodynamic analysis. 658  Thermodynamics
EXAMPLE 124 Verification of the Maxwell Relations Verify the validity of the last Maxwell relation (Eq. 1219) for steam at 250C and 300 kPa. Solution The validity of the last Maxwell relation is to be verified for steam
at a specified state. Analysis The last Maxwell relation states that for a simple compressible substance, the change in entropy with pressure at constant temperature is equal to the negative of the change in specific volume with temperature at constant pressure. If we had explicit analytical relations for the entropy and specific volume of steam in terms of other properties, we could easily verify this by performing the indicated derivations. However, all we have for steam are tables of properties listed at certain intervals. Therefore, the only course we can take to solve this problem is to replace the differential quantities in Eq. 1219 with corresponding finite quantities, using property values from the tables (Table A6 in this case) at or about the specified state. s P s P s400 kPa s200 kPa (400 200) kPa ? T
? a a c 0v b 0T P 0v b 0T P T 250C
? 300 kPa T 250C
? v300C v200C d (300 200)C P 300 kPa (7.3804 7.7100) kJ kg # K (400 200) kPa 0.00165 m3/kg K (0.87535 0.71643) m3 kg (300 200)C 0.00159 m3/kg K since kJ kPa m3 and K C for temperature differences. The two values are within 4 percent of each other. This difference is due to replacing the differential quantities by relatively large finite quantities. Based on the close agreement between the two values, the steam seems to satisfy Eq. 1219 at the specified state. Discussion This example shows that the entropy change of a simple compressible system during an isothermal process can be determined from a knowledge of the easily measurable properties P, v, and T alone. 123 THE CLAPEYRON EQUATION The Maxwell relations have farreaching implications in thermodynamics and are frequently used to derive useful thermodynamic relations. The Clapeyron equation is one such relation, and it enables us to determine the enthalpy change associated with a phase change (such as the enthalpy of vaporization hfg) from a knowledge of P, v, and T data alone. Consider the third Maxwell relation, Eq. 1218:
a 0P b 0T v a 0s b 0v T During a phasechange process, the pressure is the saturation pressure, which depends on the temperature only and is independent of the specific Chapter 12
volume. That is, Psat f (Tsat). Therefore, the partial derivative ( P/ T )v can be expressed as a total derivative (dP/dT )sat, which is the slope of the saturation curve on a PT diagram at a specified saturation state (Fig. 129). This slope is independent of the specific volume, and thus it can be treated as a constant during the integration of Eq. 1218 between two saturation states at the same temperature. For an isothermal liquidvapor phasechange process, for example, the integration yields
sg sf dP a b 1v dT sat g sfg vfg vf 2
(1220) P  659 LIQUID SOLID (P) T VAPOR T = const. sat or
dP a b dT sat
(1221) T During this process the pressure also remains constant. Therefore, from Eq. 1211,
dh T ds 0 v dP S
f g g FIGURE 129 The slope of the saturation curve on a PT diagram is constant at a constant T or P. dh
f T ds S h fg Tsfg Substituting this result into Eq. 1221, we obtain
a dP b dT sat hfg Tvfg
(1222) which is called the Clapeyron equation after the French engineer and physicist E. Clapeyron (17991864). This is an important thermodynamic relation since it enables us to determine the enthalpy of vaporization hfg at a given temperature by simply measuring the slope of the saturation curve on a PT diagram and the specific volume of saturated liquid and saturated vapor at the given temperature. The Clapeyron equation is applicable to any phasechange process that occurs at constant temperature and pressure. It can be expressed in a general form as
a dP b dT sat h12 Tv12
(1223) where the subscripts 1 and 2 indicate the two phases. EXAMPLE 125 Evaluating the h fg of a Substance from the PvT Data Using the Clapeyron equation, estimate the value of the enthalpy of vaporization of refrigerant134a at 20C, and compare it with the tabulated value. Solution The hfg of refrigerant134a is to be determined using the Clapeyron
equation. Analysis From Eq. 1222, hfg Tvfg a dP b dT sat 660  Thermodynamics
where, from Table A11, vfg a dP b d T sat,20C 1vg a vf 2 @ 20C 0.035969 Psat @ 24C 24C 0.0008161 Psat @ 16C 16C 0.035153 m3>kg P b T sat,20C 646.18
since T(C) 504.58 kPa 8C 17.70 kPa>K T(K). Substituting, we get h fg 1293.15 K2 10.035153 m3>kg2 117.70 kPa>K2 a 182.40 kJ/kg 1 kJ b 1 kPa # m3 The tabulated value of hfg at 20C is 182.27 kJ/kg. The small difference between the two values is due to the approximation used in determining the slope of the saturation curve at 20C. The Clapeyron equation can be simplified for liquidvapor and solidvapor vf , phase changes by utilizing some approximations. At low pressures vg and thus vfg vg. By treating the vapor as an ideal gas, we have vg RT/P. Substituting these approximations into Eq. 1222, we find
a dP b dT sat h fg R Ph fg RT 2 a dT b T 2 sat or
a dP b P sat For small temperature intervals hfg can be treated as a constant at some average value. Then integrating this equation between two saturation states yields
ln a P2 b P1 sat h fg R a 1 T1 1 b T2 sat
(1224) This equation is called the ClapeyronClausius equation, and it can be used to determine the variation of saturation pressure with temperature. It can also be used in the solidvapor region by replacing hfg by hig (the enthalpy of sublimation) of the substance. EXAMPLE 126 Extrapolating Tabular Data with the Clapeyron Equation
50F, using the Estimate the saturation pressure of refrigerant134a at data available in the refrigerant tables. Solution The saturation pressure of refrigerant134a is to be determined
using other tabulated data. Analysis Table A11E lists saturation data at temperatures 40F and above. Therefore, we should either resort to other sources or use extrapolation Chapter 12
to obtain saturation data at lower temperatures. Equation 1224 provides an intelligent way to extrapolate:  661 ln a P2 b P1 sat hfg R a 1 T1 1 b T2 sat 40F and T2 50F. For refrigerant134a, R 0.01946 In our case T1 Btu/lbm R. Also from Table A11E at 40F, we read hfg 97.100 Btu/lbm and P1 Psat @ 40F 7.432 psia. Substituting these values into Eq. 1224 gives ln a P2 b 7.432 psia P2 97.100 Btu>lbm 0.01946 Btu>lbm 5.56 psia # 1 a R 420 R 1 b 410 R Therefore, according to Eq. 1224, the saturation pressure of refrigerant134a at 50F is 5.56 psia. The actual value, obtained from another source, is 5.506 psia. Thus the value predicted by Eq. 1224 is in error by about 1 percent, which is quite acceptable for most purposes. (If we had used linear extrapolation instead, we would have obtained 5.134 psia, which is in error by 7 percent.) 124 GENERAL RELATIONS FOR du, dh, ds, cv, AND cp The state postulate established that the state of a simple compressible system is completely specified by two independent, intensive properties. Therefore, at least theoretically, we should be able to calculate all the properties of a system at any state once two independent, intensive properties are available. This is certainly good news for properties that cannot be measured directly such as internal energy, enthalpy, and entropy. However, the calculation of these properties from measurable ones depends on the availability of simple and accurate relations between the two groups. In this section we develop general relations for changes in internal energy, enthalpy, and entropy in terms of pressure, specific volume, temperature, and specific heats alone. We also develop some general relations involving specific heats. The relations developed will enable us to determine the changes in these properties. The property values at specified states can be determined only after the selection of a reference state, the choice of which is quite arbitrary. Internal Energy Changes
We choose the internal energy to be a function of T and v; that is, u u(T, v) and take its total differential (Eq. 123):
du a 0u b dT 0T v a a 0u b dv 0v T Using the definition of cv, we have
du cv dT 0u b dv 0v T
(1225) 662  Thermodynamics
Now we choose the entropy to be a function of T and v; that is, s and take its total differential,
ds a 0s b dT 0T v a 0s b dv 0v T s(T, v) (1226) Substituting this into the T ds relation du
du Ta 0s b dT 0T v 0s b 0T v 0u b 0v T 0u b 0v T cTa T ds
0s b 0v T P dv yields
P d dv
(1227) Equating the coefficients of dT and dv in Eqs. 1225 and 1227 gives
a a cv T Ta 0s b 0v T 0P b 0T v 0P b 0T v P
(1228) Using the third Maxwell relation (Eq. 1218), we get
a Ta P Substituting this into Eq. 1225, we obtain the desired relation for du:
du cv dT cTa P d dv
(1229) The change in internal energy of a simple compressible system associated with a change of state from (T1, v1) to (T2, v2) is determined by integration:
T2 v2 u2 u1
T1 cv dT
v1 cTa 0P b 0T v P d dv (1230) Enthalpy Changes
The general relation for dh is determined in exactly the same manner. This time we choose the enthalpy to be a function of T and P, that is, h h(T, P), and take its total differential,
dh a 0h b dT 0T P a a 0h b dP 0P T Using the definition of cp, we have
dh cp dT 0h b dP 0P T
(1231) Now we choose the entropy to be a function of T and P; that is, we take s s(T, P) and take its total differential,
ds a 0s b dT 0T P cv a 0s b dP 0P T
(1232) Substituting this into the T ds relation dh
dh Ta 0s b dT 0T P T ds
Ta v dP gives
(1233) 0s b d dP 0P T Chapter 12
Equating the coefficients of dT and dP in Eqs. 1231 and 1233, we obtain
a a 0s b 0T P 0h b 0P T 0h b 0P T cp T v Ta 0s b 0P T 0v b 0T P 0v b d dP 0T P
(1234)  663 Using the fourth Maxwell relation (Eq. 1219), we have
a v Ta Ta Substituting this into Eq. 1231, we obtain the desired relation for dh:
dh cp d T cv
(1235) The change in enthalpy of a simple compressible system associated with a change of state from (T1, P1) to (T2, P2) is determined by integration:
T2 P2 h2 h1
T1 cp dT
P1 cv Ta 0v b d dP 0T P (1236) In reality, one needs only to determine either u2 u1 from Eq. 1230 or h2 h1 from Eq. 1236, depending on which is more suitable to the data at hand. The other can easily be determined by using the definition of enthalpy h u Pv:
h2 h1 u2 u1 1P2v2 P1v1 2
(1237) Entropy Changes
Below we develop two general relations for the entropy change of a simple compressible system. The first relation is obtained by replacing the first partial derivative in the total differential ds (Eq. 1226) by Eq. 1228 and the second partial derivative by the third Maxwell relation (Eq. 1218), yielding
ds cv dT T
T2 a 0P b dv 0T v
v2 (1238) and
s2 s1
T1 cv dT T v1 a 0P b dv 0T v (1239) The second relation is obtained by replacing the first partial derivative in the total differential of ds (Eq. 1232) by Eq. 1234, and the second partial derivative by the fourth Maxwell relation (Eq. 1219), yielding
ds cP dT T
T2 a 0v b dP 0T P
P2 (1240) and
s2 s1 cp T dT
P1 T1 a 0v b dP 0T P (1241) Either relation can be used to determine the entropy change. The proper choice depends on the available data. 664  Thermodynamics Specific Heats cv and cp
Recall that the specific heats of an ideal gas depend on temperature only. For a general pure substance, however, the specific heats depend on specific volume or pressure as well as the temperature. Below we develop some general relations to relate the specific heats of a substance to pressure, specific volume, and temperature. At low pressures gases behave as ideal gases, and their specific heats essentially depend on temperature only. These specific heats are called zero pressure, or idealgas, specific heats (denoted cv 0 and cp0), and they are relatively easier to determine. Thus it is desirable to have some general relations that enable us to calculate the specific heats at higher pressures (or lower specific volumes) from a knowledge of cv 0 or cp 0 and the PvT behavior of the substance. Such relations are obtained by applying the test of exactness (Eq. 125) on Eqs. 1238 and 1240, which yields
a 0cv b 0v T b Ta 0 2P b 0T 2 v 0 2v b 0T 2 P
(1242) and
a 0cp 0P
T Ta (1243) The deviation of cp from cp 0 with increasing pressure, for example, is determined by integrating Eq. 1243 from zero pressure to any pressure P along an isothermal path:
1cp cp0 2 T
P T
0 a 0 2v b dP 0T 2 P (1244) The integration on the righthand side requires a knowledge of the PvT behavior of the substance alone. The notation indicates that v should be differentiated twice with respect to T while P is held constant. The resulting expression should be integrated with respect to P while T is held constant. Another desirable general relation involving specific heats is one that relates the two specific heats cp and cv. The advantage of such a relation is obvious: We will need to determine only one specific heat (usually cp) and calculate the other one using that relation and the PvT data of the substance. We start the development of such a relation by equating the two ds relations (Eqs. 1238 and 1240) and solving for dT:
dT T 10P>0T2 v dv cp cv T 10v>0T2 P dP cp cv Choosing T T(v, P) and differentiating, we get
dT a 0T b dv 0v P a 0T b dP 0P v Equating the coefficient of either dv or dP of the above two equations gives the desired result:
cp cv Ta 0v 0P b a b 0T P 0T v
(1245) Chapter 12
An alternative form of this relation is obtained by using the cyclic relation:
a 0P 0T 0v b a b a b 0T v 0v P 0P T 1S a 0P b 0T v a 0v 0P b a b 0T P 0v T  665 Substituting the result into Eq. 1245 gives
cp cv Ta 0v 2 0P b a b 0T P 0v T
(1246) This relation can be expressed in terms of two other thermodynamic properties called the volume expansivity b and the isothermal compressibility a, which are defined as (Fig. 1210)
b 1 0v a b v 0T P 1 0v a b v 0P T
(1247) 20C 100 kPa 1 kg v ( T ) P and
a
(1248) 21C 100 kPa 1 kg (a) A substance with a large v ( T ) Substituting these two relations into Eq. 1246, we obtain a third general relation for cp cv:
cp cv vTb2 a
(1249) 20C 100 kPa 1 kg P It is called the Mayer relation in honor of the German physician and physicist J. R. Mayer (18141878). We can draw several conclusions from this equation: 1. The isothermal compressibility a is a positive quantity for all substances in all phases. The volume expansivity could be negative for some substances (such as liquid water below 4C), but its square is always positive or zero. The temperature T in this relation is thermodynamic temperature, which is also positive. Therefore we conclude that the constantpressure specific heat is always greater than or equal to the constantvolume specific heat:
cp cv
(1250) 21C 100 kPa 1 kg (b) A substance with a small FIGURE 1210 The volume expansivity (also called the coefficient of volumetric expansion) is a measure of the change in volume with temperature at constant pressure. 2. The difference between cp and cv approaches zero as the absolute temperature approaches zero. 3. The two specific heats are identical for truly incompressible substances since v constant. The difference between the two specific heats is very small and is usually disregarded for substances that are nearly incompressible, such as liquids and solids. EXAMPLE 127 Internal Energy Change of a van der Waals Gas Derive a relation for the internal energy change as a gas that obeys the van der Waals equation of state. Assume that in the range of interest cv varies according to the relation cv c1 c2T, where c1 and c2 are constants. Solution A relation is to be obtained for the internal energy change of a
van der Waals gas. 666  Thermodynamics
Analysis The change in internal energy of any simple compressible system in any phase during any process can be determined from Eq. 1230:
T2 v2 u2 u1
T1 cv dT
v1 cTa 0P b 0T v P d dv The van der Waals equation of state is P
Then RT v 0P b 0T v RT v b v b a v2 R v b RT b a v2
v2 a
Thus, Ta
Substituting gives 0P b 0T v P a v2 T2 u2
Integrating yields u1
T1 1c1 c2T2 dT
v1 a dv v2 u2 u1 c1 1T2 T1 2 c2 2 1T 2 2 2 T12 aa 1 v1 1 b v2 which is the desired relation. EXAMPLE 128 Internal Energy as a Function of Temperature Alone Show that the internal energy of (a) an ideal gas and (b) an incompressible substance is a function of temperature only, u u(T). Solution It is to be shown that u u(T) for ideal gases and incompressible substances. Analysis The differential change in the internal energy of a general simple compressible system is given by Eq. 1229 as
du
(a) For an ideal gas Pv cv dT
RT. Then cTa 0P b 0T v P d dv Ta
Thus, 0P b 0T v P Ta du R b v cv dT P P P 0 To complete the proof, we need to show that cv is not a function of v either. This is done with the help of Eq. 1242: a 0cv b 0v T Ta 0 2P b 0T 2 v Chapter 12
For an ideal gas P RT/v. Then  667 a
Thus, 0P b 0T v R v and a 0 2P b 0T 2 v c 0 1R>v 2 0T d 0
v a 0cv b 0v T 0 which states that cv does not change with specific volume. That is, cv is not a function of specific volume either. Therefore we conclude that the internal energy of an ideal gas is a function of temperature only (Fig. 1211). (b) For an incompressible substance, v from Eq. 1249, cp cv c since a b Then Eq. 1229 reduces to constant and thus dv 0. Also 0 for incompressible substances.
u = u(T ) cv = cv (T) cp = cp(T ) AIR du c dT Again we need to show that the specific heat c depends on temperature only and not on pressure or specific volume. This is done with the help of Eq. 1243: u = u(T ) c = c(T ) LAKE a 0cp 0P b T Ta 0 2v b 0T 2 P 0 since v constant. Therefore, we conclude that the internal energy of a truly incompressible substance depends on temperature only. FIGURE 1211 The internal energies and specific heats of ideal gases and incompressible substances depend on temperature only. EXAMPLE 129
Show that cp cv The Specific Heat Difference of an Ideal Gas
R for an ideal gas. Solution It is to be shown that the specific heat difference for an ideal gas
is equal to its gas constant. Analysis This relation is easily proved by showing that the righthand side of Eq. 1246 is equivalent to the gas constant R of the ideal gas: cp P v
Substituting, cv Ta RT v2 a 0v 2 0P b a b 0T P 0v T P v RT 0P S a b v 0v T RT 0v 2 S a b P 0T P 0v 2 0P b a b 0T P 0v T cp cv R 2 b P R 2 b a P P b v Ta
Therefore, Ta R R 668 
T1 = 20C Thermodynamics
> T2 = 20C < P2 = 200 kPa 125 THE JOULETHOMSON COEFFICIENT P1 = 800 kPa FIGURE 1212 The temperature of a fluid may increase, decrease, or remain constant during a throttling process.
T P2, T2 (varied) P1, T1 (fixed) When a fluid passes through a restriction such as a porous plug, a capillary tube, or an ordinary valve, its pressure decreases. As we have shown in Chap. 5, the enthalpy of the fluid remains approximately constant during such a throttling process. You will remember that a fluid may experience a large drop in its temperature as a result of throttling, which forms the basis of operation for refrigerators and air conditioners. This is not always the case, however. The temperature of the fluid may remain unchanged, or it may even increase during a throttling process (Fig. 1212). The temperature behavior of a fluid during a throttling (h constant) process is described by the JouleThomson coefficient, defined as
m a 0T b 0P h
(1251) Exit states 2 2 Thus the JouleThomson coefficient is a measure of the change in temperature with pressure during a constantenthalpy process. Notice that if
6 0 m JT 0 7 0 temperature increases temperature remains constant temperature decreases 2 2 2 Inlet state h = constant line 1 P1 P FIGURE 1213 The development of an h line on a PT diagram. constant T Maximum inversion temperature JT > 0 JT < 0
h = const. Inversion line P FIGURE 1214 Constantenthalpy lines of a substance on a TP diagram. during a throttling process. A careful look at its defining equation reveals that the JouleThomson coefficient represents the slope of h constant lines on a TP diagram. Such diagrams can be easily constructed from temperature and pressure measurements alone during throttling processes. A fluid at a fixed temperature and pressure T1 and P1 (thus fixed enthalpy) is forced to flow through a porous plug, and its temperature and pressure downstream (T2 and P2) are measured. The experiment is repeated for different sizes of porous plugs, each giving a different set of T2 and P2. Plotting the temperatures against the pressures gives us an h constant line on a TP diagram, as shown in Fig. 1213. Repeating the experiment for different sets of inlet pressure and temperature and plotting the results, we can construct a TP diagram for a substance with several h constant lines, as shown in Fig. 1214. Some constantenthalpy lines on the TP diagram pass through a point of zero slope or zero JouleThomson coefficient. The line that passes through these points is called the inversion line, and the temperature at a point where a constantenthalpy line intersects the inversion line is called the inversion temperature. The temperature at the intersection of the P 0 line (ordinate) and the upper part of the inversion line is called the maximum inversion temperature. Notice that the slopes of the h constant 0) at states to the right of the inversion line and lines are negative (mJT positive (mJT 0) to the left of the inversion line. A throttling process proceeds along a constantenthalpy line in the direction of decreasing pressure, that is, from right to left. Therefore, the temperature of a fluid increases during a throttling process that takes place on the righthand side of the inversion line. However, the fluid temperature decreases during a throttling process that takes place on the lefthand side of the inversion line. It is clear from this diagram that a cooling effect cannot be achieved by throttling unless the fluid is below its maximum inversion Chapter 12
temperature. This presents a problem for substances whose maximum inversion temperature is well below room temperature. For hydrogen, for example, the maximum inversion temperature is 68C. Thus hydrogen must be cooled below this temperature if any further cooling is to be achieved by throttling. Next we would like to develop a general relation for the JouleThomson coefficient in terms of the specific heats, pressure, specific volume, and temperature. This is easily accomplished by modifying the generalized relation for enthalpy change (Eq. 1235)
dh cp dT cv Ta 0v b d dP 0T P  669 For an h constant process we have dh rearranged to give
1 cv cp Ta 0v b d 0T P a 0. Then this equation can be
0T b 0P h m JT (1252) which is the desired relation. Thus, the JouleThomson coefficient can be determined from a knowledge of the constantpressure specific heat and the PvT behavior of the substance. Of course, it is also possible to predict the constantpressure specific heat of a substance by using the JouleThomson coefficient, which is relatively easy to determine, together with the PvT data for the substance.
EXAMPLE 1210 JouleThomson Coefficient of an Ideal Gas
T h = constant line Show that the JouleThomson coefficient of an ideal gas is zero. Solution It is to be shown that mJT
Analysis For an ideal gas v 0 for an ideal gas. RT/P, and thus a cv 0v Ta b d 0T P 0v b 0T P 1 cp cv R P R T d P 1 1v cp
P1 P2 P Substituting this into Eq. 1252 yields mJT 1 cp v2 0 Discussion This result is not surprising since the enthalpy of an ideal gas is a function of temperature only, h h(T), which requires that the temperature remain constant when the enthalpy remains constant. Therefore, a throttling process cannot be used to lower the temperature of an ideal gas (Fig. 1215). FIGURE 1215 The temperature of an ideal gas remains constant during a throttling process since h constant and T constant lines on a TP diagram coincide. 126 THE h, u, AND s OF REAL GASES We have mentioned many times that gases at low pressures behave as ideal gases and obey the relation Pv RT. The properties of ideal gases are relatively easy to evaluate since the properties u, h, cv, and cp depend on temperature only. At high pressures, however, gases deviate considerably from idealgas behavior, and it becomes necessary to account for this deviation. 670  Thermodynamics
In Chap. 3 we accounted for the deviation in properties P, v, and T by either using more complex equations of state or evaluating the compressibility factor Z from the compressibility charts. Now we extend the analysis to evaluate the changes in the enthalpy, internal energy, and entropy of nonideal (real) gases, using the general relations for du, dh, and ds developed earlier. Enthalpy Changes of Real Gases
The enthalpy of a real gas, in general, depends on the pressure as well as on the temperature. Thus the enthalpy change of a real gas during a process can be evaluated from the general relation for dh (Eq. 1236)
T2 P2 h2
T Actual process path T2 T1 1
P
1 h1
T1 cp dT
P1 cv Ta 0v b d dP 0T P 2 1* 2* Alternative process path s FIGURE 1216 An alternative process path to evaluate the enthalpy changes of real gases. where P1, T1 and P2, T2 are the pressures and temperatures of the gas at the initial and the final states, respectively. For an isothermal process dT 0, and the first term vanishes. For a constantpressure process, dP 0, and the second term vanishes. Properties are point functions, and thus the change in a property between two specified states is the same no matter which process path is followed. This fact can be exploited to greatly simplify the integration of Eq. 1236. Consider, for example, the process shown on a Ts diagram in Fig. 1216. The enthalpy change during this process h2 h1 can be determined by performing the integrations in Eq. 1236 along a path that consists of constant and T2 constant) lines and one isobaric two isothermal (T1 constant) line instead of the actual process path, as shown in (P0 Fig. 1216. Although this approach increases the number of integrations, it also simplifies them since one property remains constant now during each part of the process. The pressure P0 can be chosen to be very low or zero, so that constant process. the gas can be treated as an ideal gas during the P0 Using a superscript asterisk (*) to denote an idealgas state, we can express the enthalpy change of a real gas during process 12 as
h2
P2 P 0 =0 P 2 h1 1h 2 h* 2 2 1h* 2 h* 2 1
P2 1h* 1 h1 2 (1253) where, from Eq. 1236,
h2 h* 2 h* 1 h* 2 h* 1
T1 * P1 0
P* 2 T2 cv 0 Ta 0v b d 0T P T
T2 dP
T2 P0 cv Ta 0v b d 0T P T dP (1254)
T2 cp dT cv T1 cp0 1T 2 dT
P1 (1255) h1 0
P1 Ta 0v b d 0T P T dP
T1 P0 cv Ta 0v b d 0T P T dP (1256)
T1 The difference between h and h* is called the enthalpy departure, and it represents the variation of the enthalpy of a gas with pressure at a fixed temperature. The calculation of enthalpy departure requires a knowledge of the PvT behavior of the gas. In the absence of such data, we can use the relation Pv ZRT, where Z is the compressibility factor. Substituting Chapter 12
v ZRT/P and simplifying Eq. 1256, we can write the enthalpy departure at any temperature T and pressure P as
1h*
P  671 h2 T RT 2
0 a 0Z dP b 0T P P The above equation can be generalized by expressing it in terms of the reduced Pcr PR. After some manipulations, the coordinates, using T TcrTR and P enthalpy departure can be expressed in a nondimensionalized form as
Zh 1h* R uTcr h2T
PR 2 TR 0 a 0Z b d 1ln PR 2 0TR PR (1257) where Zh is called the enthalpy departure factor. The integral in the above equation can be performed graphically or numerically by employing data from the compressibility charts for various values of PR and TR. The values of Zh are presented in graphical form as a function of PR and TR in Fig. A29. This graph is called the generalized enthalpy departure chart, and it is used to determine the deviation of the enthalpy of a gas at a given P and T from the enthalpy of an ideal gas at the same T. By replacing h* by hideal for clarity, Eq. 1253 for the enthalpy change of a gas during a process 12 can be rewritten as
h2 h1 1h2 1h2 h1 2 ideal h1 2 ideal RuTcr 1Zh2 RTcr 1Zh2 Zh1 2 Zh1 2
(1258) or
h2 h1 (1259) where the valuesof Zhare determined from the generalized enthalpy deparh1)ideal is determined from the idealgas tables. Notice ture chart and (h2 that the last terms on the righthand side are zero for an ideal gas. Internal Energy Changes of Real Gases
The internal energy change of a real gas is determined by relating it to the u ZR T: enthalpy change through the definition h u Pv u
u2 u1 1h 2 h1 2 R u 1Z 2 T2 Z 1T1 2
(1260) Entropy Changes of Real Gases
The entropy change of a real gas is determined by following an approach similar to that used above for the enthalpy change. There is some difference in derivation, however, owing to the dependence of the idealgas entropy on pressure as well as the temperature. The general relation for ds was expressed as (Eq. 1241)
T2 s2 s1
T1 cp T P2 dT
P1 a 0v b dP 0T P where P1, T1 and P2, T2 are the pressures and temperatures of the gas at the initial and the final states, respectively. The thought that comes to mind at this point is to perform the integrations in the previous equation first along a constant line to zero pressure, then along the P 0 line to T2, and T1 672  Thermodynamics
finally along the T2 constant line to P2, as we did for the enthalpy. This approach is not suitable for entropychange calculations, however, since it involves the value of entropy at zero pressure, which is infinity. We can avoid this difficulty by choosing a different (but more complex) path between the two states, as shown in Fig. 1217. Then the entropy change can be expressed as
s2 s1 1s2 * sb 2 * 1sb * s2 2 * 1s2 * s1 2 * 1s1 * sa 2 * 1sa s1 2 (1261) T T2 Actual process path 2 1* P2 P1 1 a* P0
b* 2* * * States 1 and 1* are identical (T1 T1 and P1 P1 ) and so are states 2 and 2*. The gas is assumed to behave as an ideal gas at the imaginary states 1* and 2* as well as at the states between the two. Therefore, the entropy change during process 1*2* can be determined from the entropychange relations for ideal gases. The calculation of entropy change between an actual state and the corresponding imaginary idealgas state is more involved, however, and requires the use of generalized entropy departure charts, as explained below. Consider a gas at a pressure P and temperature T. To determine how much different the entropy of this gas would be if it were an ideal gas at the same temperature and pressure, we consider an isothermal process from the actual state P, T to zero (or close to zero) pressure and back to the imaginary idealgas state P*, T* (denoted by superscript *), as shown in Fig. 1217. The entropy change during this isothermal process can be expressed as
1sP * sP 2 T 1sP
P 0 T1 s* 2 T 0 a 1s * 0 * sP 2 T
0 P 0v b dP 0T P a 0v* b dP 0T P Alternative process path s where v ZRT/P and v * and rearranging, we obtain
1sP * sP 2 T videal
P RT/P. Performing the differentiations
Z2 R P RT 0Zr a b d dP P 0T P FIGURE 1217 An alternative process path to evaluate the entropy changes of real gases during process 12. 0 c 11 PcrPR and rearranging, the entropy By substituting T TcrTR and P departure can be expressed in a nondimensionalized form as
Zs 1 s* s 2 T,P
0 PR Ru cZ 1 TR a 0Z b d d 1ln PR 2 0TR PR (1262) The difference (s * s )T,P is called the entropy departure and Zs is called the entropy departure factor. The integral in the above equation can be performed by using data from the compressibility charts. The values of Zs are presented in graphical form as a function of PR and TR in Fig. A30. This graph is called the generalized entropy departure chart, and it is used to determine the deviation of the entropy of a gas at a given P and T from the entropy of an ideal gas at the same P and T. Replacing s* by sideal for clarity, we can rewrite Eq. 1261 for the entropy change of a gas during a process 12 as s2 s1 1 s2 s 1 2 ideal Ru 1Zs2 Zs1 2 (1263) Chapter 12
or
s2 s1 1s2 s1 2 ideal R 1Zs2 Zs1 2
(1264)  673 where the values of Zs are determined from the generalized entropy departure chart and the entropy change (s2 s1)ideal is determined from the idealgas relations for entropy change. Notice that the last terms on the righthand side are zero for an ideal gas.
EXAMPLE 1211 The h and s of Oxygen at High Pressures Determine the enthalpy change and the entropy change of oxygen per unit mole as it undergoes a change of state from 220 K and 5 MPa to 300 K and 10 MPa (a) by assuming idealgas behavior and (b) by accounting for the deviation from idealgas behavior. Solution Oxygen undergoes a process between two specified states. The
enthalpy and entropy changes are to be determined by assuming idealgas behavior and by accounting for the deviation from idealgas behavior. Analysis The critical temperature and pressure of oxygen are Tcr 154.8 K and Pcr 5.08 MPa (Table A1), respectively. The oxygen remains above its critical temperature; therefore, it is in the gas phase, but its pressure is quite high. Therefore, the oxygen will deviate from idealgas behavior and should be treated as a real gas. (a) If the O2 is assumed to behave as an ideal gas, its enthalpy will depend on temperature only, and the enthalpy values at the initial and the final temperatures can be determined from the idealgas table of O2 (Table A19) at the specified temperatures: 1h 2 h 1 2 ideal h 2,ideal 18736 h 1,ideal 6404 2 kJ>kmol 2332 kJ/kmol
The entropy depends on both temperature and pressure even for ideal gases. Under the idealgas assumption, the entropy change of oxygen is determined from
1s2 s 1 2 ideal s 2 s 1 R u ln P2 P1 1205.213 196.171 2 kJ>kmol # K 18.314 kJ>kmol # K 2 ln 10 MPa 5 MPa 3.28 kJ/kmol # K (b) The deviation from the idealgas behavior can be accounted for by determining the enthalpy and entropy departures from the generalized charts at each state: TR1 PR1 T1 Tcr P1 Pcr 220 K 1.42 154.8 K Z h1 5 MPa 0.98 5.08 MPa 0.53, Z s1 0.25 674  Thermodynamics
and TR2 PR2 T2 Tcr P2 Pcr 300 K 1.94 154.8 K Z h2 10 MPa 1.97 5.08 MPa 0.48, Z s2 0.20 Then the enthalpy and entropy changes of oxygen during this process are determined by substituting the values above into Eqs. 1258 and 1263, h2 h1 1h 2 h 1 2 ideal R uTcr 1Z h2 2332 kJ>kmol 2396 kJ/kmol
and 18.314 kJ>kmol Z h1 2 # K2 3 154.8 K 10.48 0.53 2 4 s2 s1 1 s2 s 1 2 ideal 3.28 kJ>kmol 3.70 kJ/kmol # # R u 1Z s2 K K 18.314 kJ>kmol Z s1 2 # K 2 10.20 0.25 2 Discussion Note that the idealgas assumption would underestimate the enthalpy change of the oxygen by 2.7 percent and the entropy change by 11.4 percent. SUMMARY
Some thermodynamic properties can be measured directly, but many others cannot. Therefore, it is necessary to develop some relations between these two groups so that the properties that cannot be measured directly can be evaluated. The derivations are based on the fact that properties are point functions, and the state of a simple, compressible system is completely specified by any two independent, intensive properties. The equations that relate the partial derivatives of properties P, v, T, and s of a simple compressible substance to each other are called the Maxwell relations. They are obtained from the four Gibbs equations, expressed as du dh da dg T ds T ds s dT s dT P dv v dP P dv v dP The Maxwell relations are a a a a 0T b 0v s 0T b 0P s 0s b 0v T 0s b 0P T a a a 0P b 0s v 0v b 0s P 0P b 0T v a 0v b 0T P The Clapeyron equation enables us to determine the enthalpy change associated with a phase change from a knowledge of P, v, and T data alone. It is expressed as a dP b dT sat hfg T vfg Chapter 12
For liquidvapor and solidvapor phasechange processes at low pressures, it can be approximated as ln a P2 b P1 sat h fg T2 T1 a b R T1T2 sat cp cv vTb2 a  675 where b is the volume expansivity and a is the isothermal compressibility, defined as b 1 0v a b v 0T P and a 1 0v a b v 0P T The changes in internal energy, enthalpy, and entropy of a simple compressible substance can be expressed in terms of pressure, specific volume, temperature, and specific heats alone as 0P du cv dT cTa b P d dv 0T v 0v dh cp dT c v T a b d dP 0T P cv 0P dT a b dv ds T 0T v or ds cp T a a cp,T cp dT a 0v b dP 0T P 0 2P b 0T 2 v 0 2v b 0T 2 P
P The difference cp cv is equal to R for ideal gases and to zero for incompressible substances. The temperature behavior of a fluid during a throttling (h constant) process is described by the JouleThomson coefficient, defined as mJT a 0T b 0P h The JouleThomson coefficient is a measure of the change in temperature of a substance with pressure during a constantenthalpy process, and it can also be expressed as mJT 1 cv cp Ta 0v b d 0T P For specific heats, we have the following general relations: 0cv b 0v T 0cp 0P b Ta T Ta T The enthalpy, internal energy, and entropy changes of real gases can be determined accurately by utilizing generalized enthalpy or entropy departure charts to account for the deviation from the idealgas behavior by using the following relations: h2 u2 s2 h1 u1 s1 1h2 1h2 1 s2 h1 2 ideal h1 2 Ru 1Z2T2 Z1T1 2 s 1 2 ideal Ru 1Zs2 Zs1 2 RuTcr 1Zh2 Zh1 2 cp0,T cv 0 0 2v a 2 b dP 0T P Ta 0v 2 0P b a b 0T P 0v T where the values of Zh and Zs are determined from the generalized charts. REFERENCES AND SUGGESTED READINGS
1. G. J. Van Wylen and R. E. Sonntag. Fundamentals of Classical Thermodynamics. 3rd ed. New York: John Wiley & Sons, 1985. 2. K. Wark and D. E. Richards. Thermodynamics. 6th ed. New York: McGrawHill, 1999. PROBLEMS*
Partial Derivatives and Associated Relations
121C Consider the function z(x, y). Plot a differential surface on xyz coordinates and indicate x, dx, y, dy, ( z)x, ( z)y, and dz. 122C What is the difference between partial differentials and ordinary differentials?
*Problems designated by a "C" are concept questions, and students are encouraged to answer them all. Problems designated by an "E" are in English units, and the SI users can ignore them. Problems with a CDEES icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with a computerEES icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. 676  Thermodynamics
1217 Reconsider Prob. 1216. Using EES (or other) software, verify the validity of the last Maxwell relation for refrigerant134a at the specified state. 123C Consider the function z(x, y), its partial derivatives ( z/ x)y and ( z/ y)x, and the total derivative dz/dx. (a) How do the magnitudes ( x)y and dx compare? (b) How do the magnitudes ( z)y and dz compare? (c) Is there any relation among dz, ( z)x, and ( z)y? 124C Consider a function z(x, y) and its partial derivative ( z/ y)x. Under what conditions is this partial derivative equal to the total derivative dz/dy? 125C Consider a function z(x, y) and its partial derivative ( z/ y)x. If this partial derivative is equal to zero for all values of x, what does it indicate? 126C Consider a function z(x, y) and its partial derivative ( z/ y)x. Can this partial derivative still be a function of x? 127C Consider a function f(x) and its derivative df/dx. Can this derivative be determined by evaluating dx/df and taking its inverse? 128 Consider air at 400 K and 0.90 m3/kg. Using Eq. 123, determine the change in pressure corresponding to an increase of (a) 1 percent in temperature at constant specific volume, (b) 1 percent in specific volume at constant temperature, and (c) 1 percent in both the temperature and specific volume. 129 Repeat Problem 128 for helium. 1210 Prove for an ideal gas that (a) the P constant lines on a Tv diagram are straight lines and (b) the highpressure lines are steeper than the lowpressure lines. 1211 Derive a relation for the slope of the v constant lines on a TP diagram for a gas that obeys the van der Waals equation of state. Answer: (v b)/R 1212 Nitrogen gas at 400 K and 300 kPa behaves as an ideal gas. Estimate the cp and cv of the nitrogen at this state, using enthalpy and internal energy data from Table A18, and compare them to the values listed in Table A2b. 1213E Nitrogen gas at 600 R and 30 psia behaves as an ideal gas. Estimate the cp and cv of the nitrogen at this state, using enthalpy and internal energy data from Table A18E, and compare them to the values listed in Table A2Eb.
Answers: 0.249 Btu/lbm R, 0.178 Btu/lbm R 1218E Verify the validity of the last Maxwell relation (Eq. 1219) for steam at 800F and 400 psia. 1219 Using the Maxwell relations, determine a relation for ( s/ P)T for a gas whose equation of state is P(v b) RT.
Answer: R/P 1220 Using the Maxwell relations, determine a relation a/v 2) for ( s/ v)T for a gas whose equation of state is (P (v b) RT. 1221 Using the Maxwell relations and the idealgas equation of state, determine a relation for ( s/ v)T for an ideal gas. Answer: R/v The Clapeyron Equation
1222C What is the value of the Clapeyron equation in thermodynamics? 1223C Does the Clapeyron equation involve any approximations, or is it exact? 1224C What approximations are involved in the ClapeyronClausius equation? 1225 Using the Clapeyron equation, estimate the enthalpy of vaporization of refrigerant134a at 40C, and compare it to the tabulated value. Reconsider Prob. 1225. Using EES (or other) software, plot the enthalpy of vaporization of refrigerant134a as a function of temperature over the temperature range 20 to 80C by using the Clapeyron equation and the refrigerant134a data in EES. Discuss your results. 1227 Using the Clapeyron equation, estimate the enthalpy of vaporization of steam at 300 kPa, and compare it to the tabulated value. 1228 Calculate the hfg and sfg of steam at 120C from the Clapeyron equation, and compare them to the tabulated values. Determine the hfg of refrigerant134a at 50F on the basis of (a) the Clapeyron equation and (b) the ClapeyronClausius equation. Compare your results to the tabulated hfg value. 1230 Plot the enthalpy of vaporization of steam as a function of temperature over the temperature range 10 to 200C by using the Clapeyron equation and steam data in EES. 1229E 1226 1214 Consider an ideal gas at 400 K and 100 kPa. As a result of some disturbance, the conditions of the gas change to 404 K and 96 kPa. Estimate the change in the specific volume of the gas using (a) Eq. 123 and (b) the idealgas relation at each state. 1215 Using the equation of state P(v a) RT, verify (a) the cyclic relation and (b) the reciprocity relation at constant v. The Maxwell Relations
1216 Verify the validity of the last Maxwell relation (Eq. 1219) for refrigerant134a at 80C and 1.2 MPa. 1231 Using the ClapeyronClausius equation and the triplepoint data of water, estimate the sublimation pressure of water at 30C and compare to the value in Table A8. Chapter 12
General Relations for du, dh, ds, cv, and cp
1232C Can the variation of specific heat cp with pressure at a given temperature be determined from a knowledge of PvT data alone? 1233 Show that the enthalpy of an ideal gas is a function of temperature only and that for an incompressible substance it also depends on pressure. 1234 Derive expressions for (a) u, (b) h, and (c) s for a gas that obeys the van der Waals equation of state for an isothermal process. 1235 Derive expressions for (a) u, (b) h, and (c) s for a gas whose equation of state is P(v a) RT for an isothermal process. Answers: (a) 0, (b) a(P2 P1), (c) R ln (P2/P1) 1236 Derive expressions for ( u/ P)T and ( h/ v)T in terms of P, v, and T only. 1237 Derive an expression for the specificheat difference cv for (a) an ideal gas, (b) a van der Waals gas, and cp (c) an incompressible substance. 1238 Estimate the specificheat difference cp cv for liquid water at 15 MPa and 80C. Answer: 0.32 kJ/kg K 1239E Estimate the specificheat difference cp cv for liquid water at 1000 psia and 150F. Answer: 0.057 Btu/lbm R 1240 Derive a relation for the volume expansivity b and the isothermal compressibility a (a) for an ideal gas and (b) for a gas whose equation of state is P(v a) RT. 1241 Estimate the volume expansivity b and the isothermal compressibility a of refrigerant134a at 200 kPa and 30C. 1250E  677 Estimate the JouleThomson coefficient of nitrogen at (a) 200 psia and 500 R and (b) 2000 psia and 400 R. Use nitrogen properties from EES or other source. 1251E Reconsider Prob. 1250E. Using EES (or other) software, plot the JouleThomson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 225 Btu/lbm. Discuss the results. 1252 Estimate the JouleThomson coefficient of refrigerant134a at 0.7 MPa and 50C. 1253 Steam is throttled slightly from 1 MPa and 300C. Will the temperature of the steam increase, decrease, or remain the same during this process? The dh, du, and ds of Real Gases
1254C What is the enthalpy departure? 1255C On the generalized enthalpy departure chart, the normalized enthalpy departure values seem to approach zero as the reduced pressure PR approaches zero. How do you explain this behavior? 1256C Why is the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T ? 1257 Determine the enthalpy of nitrogen, in kJ/kg, at 175 K and 8 MPa using (a) data from the idealgas nitrogen table and (b) the generalized enthalpy departure chart. Compare your results to the actual value of 125.5 kJ/kg. Answers:
(a) 181.5 kJ/kg, (b) 121.6 kJ/kg The JouleThomson Coefficient
1242C What does the JouleThomson coefficient represent? 1243C Describe the inversion line and the maximum inversion temperature. 1244C The pressure of a fluid always decreases during an adiabatic throttling process. Is this also the case for the temperature? 1245C Does the JouleThomson coefficient of a substance change with temperature at a fixed pressure? 1246C Will the temperature of helium change if it is throttled adiabatically from 300 K and 600 kPa to 150 kPa? 1247 Consider a gas whose equation of state is P(v a) RT, where a is a positive constant. Is it possible to cool this gas by throttling? 1248 Derive a relation for the JouleThomson coefficient and the inversion temperature for a gas whose equation of state is (P a/v2)v RT. 1249 Estimate the JouleThomson coefficient of steam at (a) 3 MPa and 300C and (b) 6 MPa and 500C. 1258E Determine the enthalpy of nitrogen, in Btu/lbm, at 400 R and 2000 psia using (a) data from the idealgas nitrogen table and (b) the generalized enthalpy chart. Compare your results to the actual value of 177.8 Btu/lbm. 1259 What is the error involved in the (a) enthalpy and (b) internal energy of CO2 at 350 K and 10 MPa if it is assumed to be an ideal gas? Answers: (a) 50%, (b) 49% 1260 Determine the enthalpy change and the entropy change of nitrogen per unit mole as it undergoes a change of state from 225 K and 6 MPa to 320 K and 12 MPa, (a) by assuming idealgas behavior and (b) by accounting for the deviation from idealgas behavior through the use of generalized charts. 1261 Determine the enthalpy change and the entropy change of CO2 per unit mass as it undergoes a change of state from 250 K and 7 MPa to 280 K and 12 MPa, (a) by assuming idealgas behavior and (b) by accounting for the deviation from idealgas behavior. 1262 Methane is compressed adiabatically by a steadyflow compressor from 2 MPa and 10C to 10 MPa and 110C at a rate of 0.55 kg/s. Using the generalized charts, determine the required power input to the compressor. Answer: 133 kW 678  Thermodynamics
10 MPa 110C constanttemperature (T constant) line, but less than the slope of a constantvolume (v constant) line. 1272 Using the cyclic relation and the first Maxwell relation, derive the other three Maxwell relations. 1273 Starting with the relation dh T ds + v dP, show that the slope of a constantpressure line on an hs diagram (a) is constant in the saturation region and (b) increases with temperature in the superheated region. 1274 Derive relations for (a) u, (b) h, and (c) s of a gas that obeys the equation of state (P + a/v 2)v RT for an isothermal process. 1275 cv Show that Ta 0v 0P b a b 0T s 0T v and cp Ta 0P 0v b a b 0T s 0T P m = 0.55 kg/s CH4 2 MPa 10C FIGURE P1262
1263 Propane is compressed isothermally by a piston cylinder device from 100C and 1 MPa to 4 MPa. Using the generalized charts, determine the work done and the heat transfer per unit mass of propane. 1264 Reconsider Prob. 1263. Using EES (or other) software, extend the problem to compare the solutions based on the idealgas assumption, generalized chart data, and real fluid data. Also extend the solution to methane. 1276 Estimate the cp of nitrogen at 300 kPa and 400 K, using (a) the relation in the above problem and (b) its definition. Compare your results to the value listed in Table A2b. 1277 Steam is throttled from 4.5 MPa and 300C to 2.5 MPa. Estimate the temperature change of the steam during this process and the average JouleThomson coefficient.
Answers: 26.3C, 13.1C/MPa 1265E Propane is compressed isothermally by a piston cylinder device from 200F and 200 psia to 800 psia. Using the generalized charts, determine the work done and the heat transfer per unit mass of the propane.
Answers: 45.3 Btu/lbm, 141 Btu/lbm 1266 Determine the exergy destruction associated with the process described in Prob. 1263. Assume T0 30C. 1267 Carbon dioxide enters an adiabatic nozzle at 8 MPa and 450 K with a low velocity and leaves at 2 MPa and 350 K. Using the generalized enthalpy departure chart, determine the exit velocity of the carbon dioxide. Answer: 384 m/s 1268 Reconsider Prob. 1267. Using EES (or other) software, compare the exit velocity to the nozzle assuming idealgas behavior, the generalized chart data, and EES data for carbon dioxide. 1278 A rigid tank contains 1.2 m3 of argon at 100C and 1 MPa. Heat is now transferred to argon until the temperature in the tank rises to 0C. Using the generalized charts, determine (a) the mass of the argon in the tank, (b) the final pressure, and (c) the heat transfer.
Answers: (a) 35.1 kg, (b) 1531 kPa, (c) 1251 kJ 1279 Argon gas enters a turbine at 7 MPa and 600 K with a velocity of 100 m/s and leaves at 1 MPa and 280 K with a velocity of 150 m/s at a rate of 5 kg/s. Heat is being lost to the surroundings at 25C at a rate of 60 kW. Using the generalized charts, determine (a) the power output of the turbine and (b) the exergy destruction associated with the process. 1269 A 0.08m3 wellinsulated rigid tank contains oxygen at 220 K and 10 MPa. A paddle wheel placed in the tank is turned on, and the temperature of the oxygen rises to 250 K. Using the generalized charts, determine (a) the final pressure in the tank and (b) the paddlewheel work done during this process. Answers: (a) 12,190 kPa, (b) 393 kJ 1270 Carbon dioxide is contained in a constantvolume tank and is heated from 100C and 1 MPa to 8 MPa. Determine the heat transfer and entropy change per unit mass of the carbon dioxide using (a) the idealgas assumption, (b) the generalized charts, and (c) real fluid data from EES or other sources. 7 MPa 600 K 100 m/s 60 kW W Ar m = 5 kg/s T0 = 25C 1 MPa 280 K 150 m/s Review Problems
1271 For b 0, prove that at every point of a singlephase region of an hs diagram, the slope of a constantpressure (P constant) line is greater than the slope of a FIGURE P1279 Chapter 12
1280 Reconsider Prob. 1279. Using EES (or other) software, solve the problem assuming steam is the working fluid by using the generalized chart method and EES data for steam. Plot the power output and the exergy destruction rate for these two calculation methods against the turbine exit pressure as it varies over the range 0.1 to 1 MPa when the turbine exit temperature is 455 K.  679 1285 The volume expansivity of water at 20C is b 0.207 10 6 K 1. Treating this value as a constant, determine the change in volume of 1 m3 of water as it is heated from 10C to 30C at constant pressure. 1286 The volume expansivity b values of copper at 300 K 10 6 K 1, and 500 K are 49.2 10 6 K 1 and 54.2 respectively, and b varies almost linearly in this temperature range. Determine the percent change in the volume of a copper block as it is heated from 300 K to 500 K at atmospheric pressure. 1287 Starting with mJT (1/cp) [T( v/ T )p v] and noting that Pv ZRT, where Z Z(P, T ) is the compressibility factor, show that the position of the JouleThomson coefficient inversion curve on the TP plane is given by the equation ( Z/ T)P 0. 1288 Consider an infinitesimal reversible adiabatic compression or expansion process. By taking s s(P, v) and using the Maxwell relations, show that for this process Pv k constant, where k is the isentropic expansion exponent defined as k v 0P a b P 0v s 1281E Argon gas enters a turbine at 1000 psia and 1000 R with a velocity of 300 ft/s and leaves at 150 psia and 500 R with a velocity of 450 ft/s at a rate of 12 lbm/s. Heat is being lost to the surroundings at 75F at a rate of 80 Btu/s. Using the generalized charts, determine (a) the power output of the turbine and (b) the exergy destruction associated with the process. Answers: (a) 922 hp, (b) 121.5 Btu/s 1282 An adiabatic 0.2m3 storage tank that is initially evacuated is connected to a supply line that carries nitrogen at 225 K and 10 MPa. A valve is opened, and nitrogen flows into the tank from the supply line. The valve is closed when the pressure in the tank reaches 10 MPa. Determine the final temperature in the tank (a) treating nitrogen as an ideal gas and (b) using generalized charts. Compare your results to the actual value of 293 K. N2 225 K 10 MPa Also, show that the isentropic expansion exponent k reduces to the specific heat ratio cp /cv for an ideal gas. 1289 Refrigerant134a undergoes an isothermal process at 60C from 3 to 0.1 MPa in a closed system. Determine the work done by the refrigerant134a by using the tabular (EES) data and the generalized charts, in kJ/kg. 0.2 m 3 Initially evacuated FIGURE P1282
1283 For a homogeneous (singlephase) simple pure substance, the pressure and temperature are independent properties, and any property can be expressed as a function of these two properties. Taking v v(P, T), show that the change in specific volume can be expressed in terms of the volume expansivity b and isothermal compressibility a as dv v b dT a dP 1290 Methane is contained in a pistoncylinder device and is heated at constant pressure of 4 MPa from 100 to 350C. Determine the heat transfer, work and entropy change per unit mass of the methane using (a) the idealgas assumption, (b) the generalized charts, and (c) real fluid data from EES or other sources. Fundamentals of Engineering (FE) Exam Problems
1291 A substance whose JouleThomson coefficient is negative is throttled to a lower pressure. During this process, (select the correct statement) (a) the temperature of the substance will increase. (b) the temperature of the substance will decrease. (c) the entropy of the substance will remain constant. (d) the entropy of the substance will decrease. (e) the enthalpy of the substance will decrease. 1292 Consider the liquidvapor saturation curve of a pure substance on the PT diagram. The magnitude of the slope of the tangent line to this curve at a temperature T (in Kelvin) is Also, assuming constant average values for b and a, obtain a relation for the ratio of the specific volumes v2/v1 as a homogeneous system undergoes a process from state 1 to state 2. 1284 Repeat Prob. 1283 for an isobaric process. 680  Thermodynamics
v
a T (a) proportional to the enthalpy of vaporization hfg at that temperature. (b) proportional to the temperature T. (c) proportional to the square of the temperature T. (d) proportional to the volume change vfg at that temperature. (e) inversely proportional to the entropy change sfg at that temperature. 1293 Based on the generalized charts, the error involved in the enthalpy of CO2 at 350 K and 8 MPa if it is assumed to be an ideal gas is (a) 0 (b) 20% (c) 35% (d) 26% (e) 65% 1294 Based on data from the refrigerant134a tables, the JouleThompson coefficient of refrigerant134a at 0.8 MPa and 100C is approximately (a) 0 (d ) 8C/MPa (b) 5C/MPa (e) 26C/MPa (c) 11C/MPa b) RT, u g s h P FIGURE P1297
being Koenig's thermodynamic square shown in the figure. There is a systematic way of obtaining the four Maxwell relations as well as the four relations for du, dh, dg, and da from this figure. By comparing these relations to Koenig's diagram, come up with the rules to obtain these eight thermodynamic relations from this diagram. 1298 Several attempts have been made to express the partial derivatives of the most common thermodynamic properties in a compact and systematic manner in terms of measurable properties. The work of P. W. Bridgman is perhaps the most fruitful of all, and it resulted in the wellknown Bridgman's table. The 28 entries in that table are sufficient to express the partial derivatives of the eight common properties P, T, v, s, u, h, f, and g in terms of the six properties P, v, T, cp, b, and a, which can be measured directly or indirectly with relative ease. Obtain a copy of Bridgman's table and explain, with examples, how it is used. 1295 For a gas whose equation of state is P(v the specified heat difference cp cv is equal to (a) R (b) R b (c) R b (d ) 0 (e) R(1 + v/b) Design and Essay Problems
1296 Consider the function z z(x, y). Write an essay on the physical interpretation of the ordinary derivative dz/dx and the partial derivative ( z/ x)y. Explain how these two derivatives are related to each other and when they become equivalent. 1297 There have been several attempts to represent the thermodynamic relations geometrically, the best known of these ...
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This homework help was uploaded on 04/18/2008 for the course MAE 204 taught by Professor Errington during the Spring '08 term at SUNY Buffalo.
 Spring '08
 Errington

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