cen84959_ch12

cen84959_ch12 - cen84959_ch12.qxd 9/15/06 6:03 AM Page 669...

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: cen84959_ch12.qxd 9/15/06 6:03 AM Page 669 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 Joule-Thomson 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 Joule-Thomson coefficient. • Develop a method of evaluating the h, u, and s of real gases through the use of generalized enthalpy and entropy departure charts. | 669 cen84959_ch12.qxd 9/15/06 6:03 AM Page 670 670 | Thermodynamics 12–1 ■ A LITTLE MATH—PARTIAL 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 1 x, y 2 (x +Δx) Δf f(x) Δx Slope x x+Δx 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 12–1 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 ¢x S 0 FIGURE 12–1 The derivative of a function at a specified point represents the slope of the function at that point. lim ¢f ¢x ¢x S 0 lim f 1x ¢x 2 ¢x f 1x 2 (12–1) Therefore, the derivative of a function f(x) with respect to x represents the rate of change of f with x. EXAMPLE 12–1 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 A–17, and compare it to the value listed in Table A–2b. 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 A–2b 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. 12–2): cp 1 300 K 2 c dh 1 T 2 dT d c ¢h 1T 2 ¢T d h 1 305 K 2 1 305 295 2 K h 1 295 K 2 295.17 1 305.22 295 300 305 T, K T 1 305 295.17 2 kJ> kg 295 2 K 300 K T 300 K 1.005 kJ/kg # K FIGURE 12–2 Schematic for Example 12–1. Discussion Note that the calculated cp value is identical to the listed value. Therefore, differential quantities can be viewed as differences. They can cen84959_ch12.qxd 9/15/06 6:03 AM Page 671 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 ∂ ( ––xz )y ∂ | 671 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 ¢x S 0 y x lim a ¢z b ¢x y ¢x S 0 z 1x lim ¢ x, y 2 ¢x z 1 x, y 2 (12–2) FIGURE 12–3 Geometric representation of partial derivative ( z/ x)y. This is illustrated in Fig. 12–3. 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. 12–4. 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 1 x, y ¢y 2 z 1 x, y 2 z 1 x, y z 1 x, y z z(x, y) z(x + Δ x, y + Δy) x x + Δ x, y y x, y + Δy x + Δ x, y + Δy Adding and subtracting z(x, y ¢z ¢ x, y ¢y 2 ¢y 2 y), we get ¢y 2 ¢y 2 ¢y 2 z 1 x, y 2 z 1 x, y 2 or ¢z FIGURE 12–4 Geometric representation of total derivative dz for a function z(x, y). ¢y z 1x ¢ x, y ¢x z 1 x, y ¢y 2 ¢x ¢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 (12–3) Equation 12–3 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. cen84959_ch12.qxd 9/15/06 6:03 AM Page 672 672 | Thermodynamics EXAMPLE 12–2 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. 12–3, 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. 12–3 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 1 302 300 2 K 2K 0.01 m3> kg dv ¢v 1 0.87 0.86 2 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. 12–3 and using dP a 0P b dT 0T v a 0P b dv 0v T R dT v RT dv v2 1 0.287 kPa 0.664 kPa # m3> kg # K2 c 2K 0.865 m3> kg 1 301 K 2 1 0.01 m3> kg 2 1 0.865 m3> kg 2 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 2-K rise in temperature (Fig. 12–5). 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 0P 2 T 1 0P 2 v 1 0P 2 T 0.664 kPa 1.155 kPa 0.86 0.87 300 302 T, K v, m3/kg dP 1 0P 2 v 0.664 1.155 0.491 kPa FIGURE 12–5 Geometric representation of the disturbance discussed in Example 12–2. Discussion Of course, we could have solved this problem easily (and exactly) by evaluating the pressure from the ideal-gas 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. cen84959_ch12.qxd 9/15/06 6:03 AM Page 673 Chapter 12 | 673 Partial Differential Relations Now let us rewrite Eq. 12–3 as dz M dx N dy a 0z b 0y x (12–4) where M a 0z b and N 0x y 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 0N b and a 0x 0y 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 (12–5) 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 derivatives—the 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. 12–3, 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 (12–6) Eliminating dx by combining Eqs. 12–3 and 12–6, we have dz ca 0z 0x ba b 0x y 0y z 0z b d dy 0y x Rearranging, ca 0z 0x ba 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 (12–7) 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 ba b 0z y 0x y a 1S a 0x b 0z y 1 1 0 z> 0 x 2 y 1 (12–8) 0z 0x ba b 0x y 0y z 0y 0x 0x 0z b Sa ba b a b 0y x 0y z 0z x 0x y (12–9) cen84959_ch12.qxd 9/15/06 6:03 AM Page 674 674 | 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. 12–6). The second relation is called the cyclic relation, and it is frequently used in thermodynamics (Fig. 12–7). Function: z + 2xy – 3y2z = 0 2xy 1) z = —–––– 3y2 – 1 3y2z – z 2) x = —–––– 2y Thus, () 2y z –– = —–––– xy 3y2 – 1 2 EXAMPLE 12–3 Verification of Cyclic and Reciprocity Relations Using the ideal-gas equation of state, verify (a) the cyclic relation and (b) the reciprocity relation at constant P. 3y – 1 ( ––xz )y = —–––– 2y 1 –––––– x –– zy ( ––xz )y = () Solution The cyclic and reciprocity relations are to be verified for an ideal gas. Analysis The ideal-gas 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. 12–9 by P, v, and T, respectively, we can express the cyclic relation for an ideal gas as a 0P 0v 0T ba ba b 0v T 0T P 0P v RT 0P Sa b v 0v T RT 0v Sa b P 0T P 0T Pv Sa b R 0P v RT Pv 1 FIGURE 12–6 Demonstration of the reciprocity relation for the function z 2xy 3y2z 0. where P v T Substituting yields P 1 v, T 2 v 1 P, T 2 T 1 P, v 2 RT v2 R P v R a v RT R ba ba b P R v2 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 0 T> 0 v 2 P R P Performing the differentiations and substituting, we have R P FIGURE 12–7 Partial differentials are powerful tools that are supposed to make life easier, not harder. © Reprinted with special permission of King Features Syndicate. 1 R S P> R P Thus the proof is complete. 12–2 ■ 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. cen84959_ch12.qxd 9/15/06 6:03 AM Page 675 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 (12–10) (12–11) | 675 The other two Gibbs relations are based on two new combination properties—the Helmholtz function a and the Gibbs function g, defined as a g u h Ts Ts (12–12) (12–13) Differentiating, we get da dg du dh T ds T ds s dT s dT Simplifying the above relations by using Eqs. 12–10 and 12–11, we obtain the other two Gibbs relations for simple compressible systems: da dg s dT s dT P dv v dP (12–14) (12–15) A careful examination of the four Gibbs relations reveals that they are of the form dz M dx N dy (12–4) with a 0M b 0y x a 0N b 0x y (12–5) since u, h, a, and g are properties and thus have exact differentials. Applying Eq. 12–5 to each of them, we obtain a a 0T b 0v s 0T b 0P s a a 0P b 0s v (12–16) 0v b 0s P (12–17) P T ( ∂∂–– )s = – (∂–– )v v ∂s T ∂v ( ∂∂–– )s = ( ∂–– )P P s P ∂ ( ∂––s )T = (∂–– )v v ∂T ∂ v –– ( ∂––s )T = – ( ∂∂T )P P 0s ab 0v T a 0s b 0P T 0P ab 0T v a 0v b 0T P (12–18) (12–19) These are called the Maxwell relations (Fig. 12–8). 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 12–8 Maxwell relations are extremely valuable in thermodynamic analysis. cen84959_ch12.qxd 9/15/06 6:03 AM Page 676 676 | Thermodynamics EXAMPLE 12–4 Verification of the Maxwell Relations Verify the validity of the last Maxwell relation (Eq. 12–19) for steam at 250°C 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. 12–19 with corresponding finite quantities, using property values from the tables (Table A–6 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 250°C ? 300 kPa T 250°C ? v300°C v200°C 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. 12–19 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. 12–3 ■ THE CLAPEYRON EQUATION The Maxwell relations have far-reaching 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. 12–18: a 0P b 0T v a 0s b 0v T During a phase-change process, the pressure is the saturation pressure, which depends on the temperature only and is independent of the specific cen84959_ch12.qxd 9/15/06 6:03 AM Page 677 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 P-T diagram at a specified saturation state (Fig. 12–9). This slope is independent of the specific volume, and thus it can be treated as a constant during the integration of Eq. 12–18 between two saturation states at the same temperature. For an isothermal liquid–vapor phase-change process, for example, the integration yields sg sf dP a b 1v dT sat g sfg vfg vf 2 (12–20) P | 677 LIQUID SOLID P (∂–– ) ∂T VAPOR T = const. sat or dP a b dT sat (12–21) T During this process the pressure also remains constant. Therefore, from Eq. 12–11, dh T ds 0 → v dP S f g g FIGURE 12–9 The slope of the saturation curve on a P-T diagram is constant at a constant T or P. dh f T ds S h fg Tsfg Substituting this result into Eq. 12–21, we obtain a dP b dT sat hfg Tvfg (12–22) which is called the Clapeyron equation after the French engineer and physicist E. Clapeyron (1799–1864). 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 P-T diagram and the specific volume of saturated liquid and saturated vapor at the given temperature. The Clapeyron equation is applicable to any phase-change process that occurs at constant temperature and pressure. It can be expressed in a general form as a dP b dT sat h12 Tv12 (12–23) where the subscripts 1 and 2 indicate the two phases. EXAMPLE 12–5 Evaluating the h fg of a Substance from the P - v - T Data Using the Clapeyron equation, estimate the value of the enthalpy of vaporization of refrigerant-134a at 20°C, and compare it with the tabulated value. Solution The hfg of refrigerant-134a is to be determined using the Clapeyron equation. Analysis From Eq. 12–22, hfg Tvfg a dP b dT sat cen84959_ch12.qxd 9/15/06 6:03 AM Page 678 678 | Thermodynamics where, from Table A–11, vfg a dP b d T sat,20°C 1 vg a vf 2 @ 20°C 0.035969 Psat @ 24°C 24°C 0.0008161 Psat @ 16°C 16°C 0.035153 m3> kg ¢P b ¢ T sat,20°C 646.18 since T (°C) 504.58 kPa 8°C 17.70 kPa> K T (K). Substituting, we get h fg 1 293.15 K 2 1 0.035153 m3> kg 2 1 17.70 kPa> K 2 a 182.40 kJ/kg 1 kJ b 1 kPa # m3 The tabulated value of hfg at 20°C 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 20°C. The Clapeyron equation can be simplified for liquid–vapor and solid–vapor 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. 12–22, 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 (12–24) This equation is called the Clapeyron–Clausius equation, and it can be used to determine the variation of saturation pressure with temperature. It can also be used in the solid–vapor region by replacing hfg by hig (the enthalpy of sublimation) of the substance. EXAMPLE 12–6 Extrapolating Tabular Data with the Clapeyron Equation 50°F, using the Estimate the saturation pressure of refrigerant-134a at data available in the refrigerant tables. Solution The saturation pressure of refrigerant-134a is to be determined using other tabulated data. Analysis Table A–11E lists saturation data at temperatures 40°F and above. Therefore, we should either resort to other sources or use extrapolation cen84959_ch12.qxd 9/15/06 6:03 AM Page 679 Chapter 12 to obtain saturation data at lower temperatures. Equation 12–24 provides an intelligent way to extrapolate: | 679 ln a P2 b P1 sat hfg R a 1 T1 1 b T2 sat 40°F and T2 50°F. For refrigerant-134a, R 0.01946 In our case T1 Btu/lbm · R. Also from Table A–11E at 40°F, we read hfg 97.100 Btu/lbm and P1 Psat @ 40°F 7.432 psia. Substituting these values into Eq. 12–24 gives ln a P2 b 7.432 psia P2 0.01946 Btu> lbm 5.56 psia 97.100 Btu> lbm # 1 a R 420 R 1 b 410 R Therefore, according to Eq. 12–24, the saturation pressure of refrigerant-134a at 50°F is 5.56 psia. The actual value, obtained from another source, is 5.506 psia. Thus the value predicted by Eq. 12–24 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.) 12–4 ■ 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. 12–3): 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 (12–25) cen84959_ch12.qxd 9/15/06 6:03 AM Page 680 680 | 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) (12–26) 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 (12–27) Equating the coefficients of dT and dv in Eqs. 12–25 and 12–27 gives a a cv T Ta 0s b 0v T 0P b 0T v 0P b 0T v P (12–28) Using the third Maxwell relation (Eq. 12–18), we get a Ta P Substituting this into Eq. 12–25, we obtain the desired relation for du: du cv dT cTa P d dv (12–29) 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 (12–30) 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 (12–31) 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 (12–32) Substituting this into the T ds relation dh dh Ta 0s b dT 0T P T ds Ta v dP gives (12–33) 0s b d dP 0P T cen84959_ch12.qxd 9/15/06 6:03 AM Page 681 Chapter 12 Equating the coefficients of dT and dP in Eqs. 12–31 and 12–33, 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 (12–34) | 681 Using the fourth Maxwell relation (Eq. 12–19), we have a v Ta Ta Substituting this into Eq. 12–31, we obtain the desired relation for dh: dh cp d T cv (12–35) 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 (12–36) In reality, one needs only to determine either u2 u1 from Eq. 12–30 or h2 h1 from Eq. 12–36, 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 1 P2v2 P1v1 2 (12–37) 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. 12–26) by Eq. 12–28 and the second partial derivative by the third Maxwell relation (Eq. 12–18), yielding ds cv dT T T2 a 0P b dv 0T v v2 (12–38) and s2 s1 T1 cv dT T v1 a 0P b dv 0T v (12–39) The second relation is obtained by replacing the first partial derivative in the total differential of ds (Eq. 12–32) by Eq. 12–34, and the second partial derivative by the fourth Maxwell relation (Eq. 12–19), yielding ds cP dT T T2 a 0v b dP 0T P P2 (12–40) and s2 s1 cp T dT P1 T1 a 0v b dP 0T P (12–41) Either relation can be used to determine the entropy change. The proper choice depends on the available data. cen84959_ch12.qxd 9/15/06 6:03 AM Page 682 682 | 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 ideal-gas, specific heats (denoted cv 0 and cp 0), 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 P-v-T behavior of the substance. Such relations are obtained by applying the test of exactness (Eq. 12–5) on Eqs. 12–38 and 12–40, which yields a 0 cv b 0v T b Ta 0 2P b 0T 2 v 0 2v b 0T 2 P (12–42) and a 0 cp 0P T Ta (12–43) The deviation of cp from cp 0 with increasing pressure, for example, is determined by integrating Eq. 12–43 from zero pressure to any pressure P along an isothermal path: 1 cp cp0 2 T P T 0 a 0 2v b dP 0T 2 P (12–44) The integration on the right-hand side requires a knowledge of the P-v-T 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 P-v-T data of the substance. We start the development of such a relation by equating the two ds relations (Eqs. 12–38 and 12–40) and solving for dT: dT T 1 0 P> 0 T 2 v dv cp cv T 1 0 v> 0 T 2 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 ba b 0T P 0T v (12–45) cen84959_ch12.qxd 9/15/06 6:03 AM Page 683 Chapter 12 An alternative form of this relation is obtained by using the cyclic relation: a 0P 0T 0v ba ba b 0T v 0v P 0P T 1S a 0P b 0T v a 0v 0P ba b 0T P 0v T | 683 Substituting the result into Eq. 12–45 gives cp cv Ta 0v 2 0P ba b 0T P 0v T (12–46) 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. 12–10) b 1 0v ab v 0T P 1 0v ab v 0P T (12–47) 20°C 100 kPa 1 kg v –– ( ∂∂T ) P and a (12–48) 21°C 100 kPa 1 kg (a) A substance with a large β Substituting these two relations into Eq. 12–46, we obtain a third general relation for cp cv: cp cv vTb2 a (12–49) 20°C 100 kPa 1 kg v –– ( ∂∂T ) P It is called the Mayer relation in honor of the German physician and physicist J. R. Mayer (1814–1878). 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 4°C), 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 constant-pressure specific heat is always greater than or equal to the constant-volume specific heat: cp cv (12–50) 21°C 100 kPa 1 kg (b) A substance with a small β FIGURE 12–10 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 12–7 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. cen84959_ch12.qxd 9/15/06 6:03 AM Page 684 684 | Thermodynamics Analysis The change in internal energy of any simple compressible system in any phase during any process can be determined from Eq. 12–30: 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 1 c1 c2T 2 dT v1 a dv v2 u2 u1 c1 1 T2 T1 2 c2 2 1T 22 2 T12 aa 1 v1 1 b v2 which is the desired relation. EXAMPLE 12–8 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. 12–29 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. 12–42: a 0 cv b 0v T Ta 0 2P b 0T 2 v cen84959_ch12.qxd 9/15/06 6:03 AM Page 685 Chapter 12 For an ideal gas P RT/v. Then | 685 a Thus, 0P b 0T v R 0 2P and a 2 b v 0T v a 0 cv b 0v T c 0 1 R> v 2 0T d 0 v 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. 12–11). (b) For an incompressible substance, v from Eq. 12–49, cp cv c since a b Then Eq. 12–29 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. 12–43: u = u(T ) c = c(T ) LAKE a 0 cp 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 12–11 The internal energies and specific heats of ideal gases and incompressible substances depend on temperature only. EXAMPLE 12–9 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 right-hand side of Eq. 12–46 is equivalent to the gas constant R of the ideal gas: cp P v Substituting, cv Ta RT v2 a 0v 2 0P ba b 0T P 0v T P v RT 0P Sa b v 0v T RT 0v 2 Sa b P 0T P 0v 2 0P ba b 0T P 0v T cp cv R2 b P R2 ba P P b v Ta Therefore, Ta R R cen84959_ch12.qxd 9/15/06 6:03 AM Page 686 686 | T1 = 20°C Thermodynamics > T2 = 20°C < P2 = 200 kPa 12–5 ■ THE JOULE-THOMSON COEFFICIENT P1 = 800 kPa FIGURE 12–12 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. 12–12). The temperature behavior of a fluid during a throttling (h constant) process is described by the Joule-Thomson coefficient, defined as m a 0T b 0P h (12–51) Exit states 2 2 Thus the Joule-Thomson coefficient is a measure of the change in temperature with pressure during a constant-enthalpy process. Notice that if 6 0 temperature increases m JT • 0 temperature remains constant 7 0 temperature decreases 2 2 2 Inlet state h = constant line 1 P1 P FIGURE 12–13 The development of an h line on a P-T diagram. constant T Maximum inversion temperature μJT > 0 μJT < 0 h = const. Inversion line P FIGURE 12–14 Constant-enthalpy lines of a substance on a T-P diagram. during a throttling process. A careful look at its defining equation reveals that the Joule-Thomson coefficient represents the slope of h constant lines on a T-P 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 T-P diagram, as shown in Fig. 12–13. Repeating the experiment for different sets of inlet pressure and temperature and plotting the results, we can construct a T-P diagram for a substance with several h constant lines, as shown in Fig. 12–14. Some constant-enthalpy lines on the T-P diagram pass through a point of zero slope or zero Joule-Thomson coefficient. The line that passes through these points is called the inversion line, and the temperature at a point where a constant-enthalpy 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 constant-enthalpy 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 right-hand side of the inversion line. However, the fluid temperature decreases during a throttling process that takes place on the left-hand 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 cen84959_ch12.qxd 9/15/06 6:03 AM Page 687 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 68°C. 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 Joule-Thomson 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. 12–35) dh cp dT cv Ta 0v b d dP 0T P | 687 For an h constant process we have dh rearranged to give 1 cv cp Ta 0v bd 0T P a 0. Then this equation can be 0T b 0P h m JT (12–52) which is the desired relation. Thus, the Joule-Thomson coefficient can be determined from a knowledge of the constant-pressure specific heat and the P-v-T behavior of the substance. Of course, it is also possible to predict the constant-pressure specific heat of a substance by using the Joule-Thomson coefficient, which is relatively easy to determine, together with the P-v-T data for the substance. EXAMPLE 12–10 Joule-Thomson Coefficient of an Ideal Gas T h = constant line Show that the Joule-Thomson 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 0v b 0T P 1 cp cv R P R Td P 1 1v cp v2 P1 P2 P Substituting this into Eq. 12–52 yields mJT 1 cp cv 0v Ta b d 0T P 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. 12–15). FIGURE 12–15 The temperature of an ideal gas remains constant during a throttling process since h constant and T constant lines on a T-P diagram coincide. 12–6 ■ 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 ideal-gas behavior, and it becomes necessary to account for this deviation. cen84959_ch12.qxd 9/15/06 6:03 AM Page 688 688 | 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. 12–36) 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 12–16 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 constant-pressure 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. 12–36. Consider, for example, the process shown on a T-s diagram in Fig. 12–16. The enthalpy change during this process h2 h1 can be determined by performing the integrations in Eq. 12–36 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. 12–16. 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 ideal-gas state, we can express the enthalpy change of a real gas during process 1-2 as h2 P2 P 0 =0 P 2 h1 1 h2 h*2 2 1 h* 2 h* 2 1 P2 1 h* 1 h1 2 (12–53) where, from Eq. 12–36, h2 h* 2 h* 1 h* 2 h* 1 T1 * P1 0 P* 2 T2 cv 0 Ta 0v bd 0T P T T2 dP T2 P0 cv Ta 0v bd 0T P T dP (12–54) T2 cp dT cv T1 cp0 1 T 2 dT P1 (12–55) h1 0 P1 Ta 0v bd 0T P T dP T1 P0 cv Ta 0v bd 0T P T dP (12–56) 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 P-v-T behavior of the gas. In the absence of such data, we can use the relation Pv ZRT, where Z is the compressibility factor. Substituting cen84959_ch12.qxd 9/15/06 6:03 AM Page 689 Chapter 12 v ZRT/P and simplifying Eq. 12–56, we can write the enthalpy departure at any temperature T and pressure P as 1 h* h2T P | 689 RT 2 0 a 0 Z 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 1 h* R uTcr h2T PR 2 TR 0 a 0Z b d 1 ln PR 2 0 TR PR (12–57) 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. A–29. 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. 12–53 for the enthalpy change of a gas during a process 1-2 can be rewritten as h2 h1 1 h2 1 h2 h1 2 ideal h1 2 ideal RuTcr 1 Zh2 RTcr 1 Zh2 Zh1 2 Zh1 2 (12–58) or h2 h1 (12–59) where the values–of Zh–are determined from the generalized enthalpy deparh1)ideal is determined from the ideal-gas tables. Notice ture chart and (h2 that the last terms on the right-hand 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 1 h2 h1 2 R u 1 Z 2 T2 Z 1T1 2 (12–60) 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 ideal-gas entropy on pressure as well as the temperature. The general relation for ds was expressed as (Eq. 12–41) 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 cen84959_ch12.qxd 9/15/06 6:03 AM Page 690 690 | Thermodynamics finally along the T2 constant line to P2, as we did for the enthalpy. This approach is not suitable for entropy-change 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. 12–17. Then the entropy change can be expressed as s2 s1 1 s2 * sb 2 * 1 sb * s2 2 * 1 s2 * s1 2 * 1 s1 * sa 2 * 1 sa s1 2 (12–61) 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 entropy-change relations for ideal gases. The calculation of entropy change between an actual state and the corresponding imaginary ideal-gas 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. 12–17. The entropy change during this isothermal process can be expressed as 1 sP * sP 2 T 1 sP P 0 T1 s* 2 T 0 a 1s* 0 * sP 2 T 0 P 0v b dP 0T P a 0 v* b dP 0T P Alternative process path s where v ZRT/P and v * and rearranging, we obtain 1 sP * sP 2 T videal P RT/P. Performing the differentiations Z2R RT 0 Zr a b d dP P 0T P FIGURE 12–17 An alternative process path to evaluate the entropy changes of real gases during process 1-2. 0 c 11 P PcrPR and rearranging, the entropy By substituting T TcrTR and P departure can be expressed in a nondimensionalized form as Zs 1 s* Ru s 2 T,P 0 PR cZ 1 TR a 0Z b d d 1 ln PR 2 0 TR PR (12–62) – – 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. A–30. 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. 12–61 for the entropy change of a gas during a process 1-2 as s2 s1 1 s2 s 1 2 ideal Ru 1 Zs2 Zs1 2 (12–63) cen84959_ch12.qxd 9/15/06 6:03 AM Page 691 Chapter 12 or s2 s1 1 s2 s1 2 ideal R 1 Zs2 Zs1 2 (12–64) | 691 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 right-hand side are zero for an ideal gas. EXAMPLE 12–11 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 ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior. Solution Oxygen undergoes a process between two specified states. The enthalpy and entropy changes are to be determined by assuming ideal-gas behavior and by accounting for the deviation from ideal-gas behavior. Analysis The critical temperature and pressure of oxygen are Tcr 154.8 K and Pcr 5.08 MPa (Table A–1), 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 ideal-gas 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 ideal-gas table of O2 (Table A–19) at the specified temperatures: 1 h2 h 1 2 ideal h 2,ideal 1 8736 h 1,ideal 6404 2 kJ> kmol 2332 kJ/kmol The entropy depends on both temperature and pressure even for ideal gases. Under the ideal-gas assumption, the entropy change of oxygen is determined from 1s2 s 1 2 ideal s° 2 s° 1 R u ln P2 P1 1 205.213 196.171 2 kJ> kmol # K 1 8.314 kJ> kmol # K 2 ln 10 MPa 5 MPa 3 .28 kJ/kmol # K (b) The deviation from the ideal-gas 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 cen84959_ch12.qxd 9/15/06 6:03 AM Page 692 692 | 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. 12–58 and 12–63, h2 h1 2332 kJ> kmol 2396 kJ/kmol 1 h2 h 1 2 ideal R uTcr 1 Z h2 1 8.314 kJ> kmol Z h1 2 # K 2 3 154.8 K 1 0.48 0.53 2 4 and s2 s1 3.28 kJ> kmol 1 s2 s 1 2 ideal # # R u 1 Z s2 K K 1 8.314 kJ> kmol Z s1 2 # K 2 1 0.20 0.25 2 3.70 kJ/kmol Discussion Note that the ideal-gas 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 cen84959_ch12.qxd 9/15/06 6:03 AM Page 693 Chapter 12 For liquid–vapor and solid–vapor phase-change 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 | 693 where b is the volume expansivity and a is the isothermal compressibility, defined as b 1 0v a b and a v 0T P 1 0v ab 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 Joule-Thomson coefficient, defined as mJT a 0T b 0P h The Joule-Thomson 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 bd 0T P For specific heats, we have the following general relations: 0 cv b 0v T 0 cp 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 ideal-gas behavior by using the following relations: h2 u2 s2 h1 u1 s1 1 h2 1 h2 1 s2 h1 2 ideal h1 2 Ru 1 Z2T2 Z1T1 2 s 1 2 ideal Ru 1 Zs2 Zs1 2 RuTcr 1 Zh2 Zh1 2 cp0,T cv 0 0 2v a 2 b dP 0T P Ta 0v 2 0P ba b 0T P 0v T where the values of Zh and Zs are determined from the generalized charts. REFERENCES AND SUGGESTED READINGS 1. A. Bejan. Advanced Engineering Thermodynamics. 2nd ed. New York: Wiley, 1997. 2. K. Wark, Jr. Advanced Thermodynamics for Engineers. New York: McGraw-Hill, 1995. PROBLEMS* Partial Derivatives and Associated Relations 12–1C Consider the function z(x, y). Plot a differential surface on x-y-z coordinates and indicate x, dx, y, dy, ( z)x, ( z)y, and dz. 12–2C 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 the icon are solved using EES, and complete solutions together with parametric studies are included on the enclosed DVD. Problems with the icon are comprehensive in nature, and are intended to be solved with a computer, preferably using the EES software that accompanies this text. cen84959_ch12.qxd 9/15/06 6:03 AM Page 694 694 | Thermodynamics 12–17 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 12–3C 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? 12–4C Consider a function z(x, y) and its partial derivative ( z/ y)x. Can this partial derivative still be a function of x? 12–5C Consider a function f (x) and its derivative df/dx. Can this derivative be determined by evaluating dx/df and taking its inverse? 12–6 Consider air at 400 K and 0.90 m3/kg. Using Eq. 12–3, 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. 12–7 Repeat Problem 12–6 for helium. 12–8 Prove for an ideal gas that (a) the P constant lines on a T-v diagram are straight lines and (b) the high-pressure lines are steeper than the low-pressure lines. 12–9 Derive a relation for the slope of the v constant lines on a T-P diagram for a gas that obeys the van der Waals equation of state. Answer: (v b)/R 12–10 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 A–18, and compare them to the values listed in Table A–2b. 12–11E 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 A–18E, and compare them to the values listed in Table A–2Eb. Answers: 0.249 Btu/lbm · R, 0.178 Btu/lbm · R 12–18 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. 12–19 Using the Maxwell relations and the ideal-gas equation of state, determine a relation for ( s/ v)T for an ideal gas. Answer: R/v 0P k 0P a b. 12–20 Prove that a b 0T s k 1 0T v 12–21 Show how you would evaluate T, v, u, a, and g from the thermodynamic function h h(s, P). The Clapeyron Equation 12–22C What is the value of the Clapeyron equation in thermodynamics? 12–23C What approximations are involved in the ClapeyronClausius equation? 12–24 Using the Clapeyron equation, estimate the enthalpy of vaporization of steam at 300 kPa, and compare it to the tabulated value. 12–25 Calculate the hfg and sfg of steam at 120°C from the Clapeyron equation, and compare them to the tabulated values. 12–26E Determine the hfg of refrigerant-134a at 50°F on the basis of (a) the Clapeyron equation and (b) the Clapeyron-Clausius equation. Compare your results to the tabulated hfg value. 12–27 Plot the enthalpy of vaporization of steam as a function of temperature over the temperature range 10 to 200°C by using the Clapeyron equation and steam data in EES. 12–28 Two grams of a saturated liquid are converted to a saturated vapor by being heated in a weighted piston-cylinder device arranged to maintain the pressure at 200 kPa. During the phase conversion, the volume of the system increases by 1000 cm 3; 5 kJ of heat are required; and the temperature of the substance stays constant at 80 C. Estimate the boiling temperature of this substance when its pressure is 180 kPa. Answer: 352 K Weight 12–12 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. 12–3 and (b) the ideal-gas relation at each state. 12–13 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 12–14 Verify the validity of the last Maxwell relation (Eq. 12–19) for refrigerant-134a at 80°C and 1.2 MPa. 12–15 Reconsider Prob. 12–14. Using EES (or other) software, verify the validity of the last Maxwell relation for refrigerant-134a at the specified state. 12–16E Verify the validity of the last Maxwell relation (Eq. 12–19) for steam at 800°F and 400 psia. 200 kPa 80°C 2 grams sat. liquid Q FIGURE P12–28 cen84959_ch12.qxd 9/15/06 6:03 AM Page 695 Chapter 12 12–29 Estimate the saturation pressure Psat of the substance in Prob. 12–28 when its temperature is 100 C. 12-30 Estimate sfg of the substance in Prob. 12–28 at 80 C. Show that cp,g cp,f Ta 0 1 hfg> T 2 0T b vfg a 0P b. 0 T sat Answer: 7.08 kJ/kg # K | 695 12–43 Determine the change in the entropy of air, in kJ/kg # K, as it undergoes a change of state from 100 kPa and 20 C to 600 kPa and 300 C using the equation of state P(v a) RT where a 0.10 m 3/kg, and compare the result to the value obtained by using the ideal gas equation of state. 12–44 Determine the change in the internal energy of helium, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 C to 600 kPa and 300 C using the equation of state P(v a) RT where a 0.10 m 3/kg, and compare the result to the value obtained by using the ideal gas equation of state. Answers: 872 kJ/kg, 872 kJ/kg 12–45 Determine the change in the enthalpy of helium, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 C to 600 kPa and 300 C using the equation of state P(v a) RT where a 0.10 m 3/kg, and compare the result to the value obtained by using the ideal gas equation of state. 12–46 Determine the change in the entropy of helium, in kJ/kg # K, as it undergoes a change of state from 100 kPa and 20 C to 600 kPa and 300 C using the equation of state P(v a) RT where a 0.10 m 3/kg, and compare the result to the value obtained by using the ideal gas equation of state. Answers: 0.239 kJ/kg # K, 0.239 kJ/kg # K 12–47 Derive an expression for the volume expansivity of a substance whose equation of state is P(v a) RT. 12–48 The Helmholtz function of a substance has the form a RT ln v v0 cT0 a 1 T T0 T T ln b T0 T0 12–31 P 12–32E A table of properties for methyl chloride lists the saturation pressure as 116.7 psia at 100 F. At 100 F, this table also lists hfg 154.85 Btu/lbm, and vfg 0.86332 ft 3/lbm. Estimate the saturation pressure Psat of methyl chloride at 90 F and 110 F. 12–33 The saturation table for refrigerant-134a lists the following at 40 C: P 51.25 kPa, hfg 225.86 kJ/kg, and 0.36010 m 3/kg. Estimate the saturation pressure of vfg refrigerant-134a at 50 C and –30 C. General Relations for du, dh, ds, cv, and cp 12–34C Can the variation of specific heat cp with pressure at a given temperature be determined from a knowledge of P-v-T data alone? 12–35 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) 12–36 Derive expressions for ( u/ P)T and ( h/ v)T in terms of P, v, and T only. 12–37E Estimate the specific heat difference cp cv for liquid water at 1000 psia and 150°F. Answer: 0.057 Btu/lbm · R 12–38 Estimate the volume expansivity b and the isothermal compressibility a of refrigerant-134a at 200 kPa and 30°C. 0P 0v 12–39 Show that cp cv T a b a b . 0T v 0T P 12–40 Temperature may alternatively be defined as T a 0u b 0s v where T0 and v0 are the temperature and specific volume at a reference state. Show how to obtain P, h, s, cv, and cp from this expression. 12–49 Derive an expression for the volume expansivity of a substance whose equation of state is P RT v b a v 1 v b 2 T1>2 Prove that this definition reduces the net entropy change of two constant-volume systems filled with simple compressible substances to zero as the two systems approach thermal equilibrium. 12–41 Determine the change in the internal energy of air, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 C to 600 kPa and 300 C using the equation of state P(v a) RT where a 0.10 m 3/kg, and compare the result to the value obtained by using the ideal gas equation of state. 12–42 Determine the change in the enthalpy of air, in kJ/kg, as it undergoes a change of state from 100 kPa and 20 C to 600 kPa and 300 C using the equation of state P(v a) RT where a 0.10 m 3/kg, and compare the result to the value obtained by using the ideal gas equation of state. Answers: 335 kJ/kg, 285 kJ/kg where a and b are empirical constants. 12–50 Derive an expression for the specific heat difference of a substance whose equation of state is P RT v b a v 1 v b 2 T1>2 where a and b are empirical constants. 12–51 Derive an expression for the volume expansivity of a substance whose equation of state is P RT v b a v 2T where a and b are empirical constants. cen84959_ch12.qxd 9/15/06 6:03 AM Page 696 696 | Thermodynamics 12–69 The equation of state of a gas is given by v RT P a T b 12–52 Derive an expression for the isothermal compressibility of a substance whose equation of state is a P vb v 2T where a and b are empirical constants. 12–53 Show that b RT a 1 0 P> 0 T 2 v . cp va 12–54 Demonstrate that k . cv 1 0 v> 0 P 2 s 12–55 Derive an expression for the specific heat difference of a substance whose equation of state is a v 2T where a and b are empirical constants. P v b RT where a and b are constants. Use this equation of state to derive an equation for the Joule-Thomson coefficient inversion line. The dh, du, and ds of Real Gases 12–70C What is the enthalpy departure? 12–71C 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? 12–72C Why is the generalized enthalpy departure chart prepared by using PR and TR as the parameters instead of P and T ? 12–73 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% 12–74 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 ideal-gas behavior and (b) by accounting for the deviation from ideal-gas behavior through the use of generalized charts. 12–75 Methane is compressed adiabatically by a steady-flow compressor from 2 MPa and 10°C to 10 MPa and 110°C at a rate of 0.55 kg/s. Using the generalized charts, determine the required power input to the compressor. Answer: 133 kW 10 MPa 110°C CH4 The Joule-Thomson Coefficient 12–56C What does the Joule-Thomson coefficient represent? 12–57C Describe the inversion line and the maximum inversion temperature. 12–58C The pressure of a fluid always decreases during an adiabatic throttling process. Is this also the case for the temperature? 12–59C Does the Joule-Thomson coefficient of a substance change with temperature at a fixed pressure? 12–60C Will the temperature of helium change if it is throttled adiabatically from 300 K and 600 kPa to 150 kPa? 12–61E Estimate the Joule-Thomson 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. 12–62E Reconsider Prob. 12–61E. Using EES (or other) software, plot the Joule-Thomson coefficient for nitrogen over the pressure range 100 to 1500 psia at the enthalpy values 100, 175, and 225 Btu/lbm. Discuss the results. 12–63 Estimate the Joule-Thomson coefficient of refrigerant-134a at 0.7 MPa and 50°C. 12–64 Steam is throttled slightly from 1 MPa and 300°C. Will the temperature of the steam increase, decrease, or remain the same during this process? 12–65 Demonstrate that the Joule-Thomson coefficient is given by T 2 0 1 v> T 2 d m c cp 0T P. ˛ · m = 0.55 kg/s 2 MPa –10°C FIGURE P12–75 12–76E Water vapor at 3000 psia and 1500 F is expanded to 1000 psia and 1000 F. Calculate the change in the specific entropy and enthalpy of this water vapor using (a) the departure charts and (b) the property tables. 12–77 Water vapor at 1000 kPa and 600 C is expanded to 500 kPa and 400 C. Calculate the change in the specific entropy and enthalpy of this water vapor using the departure charts and the property tables. 12–78 Saturated water vapor at 300 C is expanded while its pressure is kept constant until its temperature is 700 C. Cal- 12–66 What is the most general equation of state for which the Joule-Thomson coefficient is always zero? 12–67E Estimate the Joule-Thomson coefficient of refrigerant-134a at 30 psia and 20 F. 12–68 Estimate the Joule-Thomson coefficient of refrigerant-134a at 200 kPa and 20 C. Answer: 0.0245 K/kPa cen84959_ch12.qxd 9/15/06 6:03 AM Page 697 Chapter 12 culate the change in the specific enthalpy and entropy using (a) the departure charts and (b) the property tables. Answers: (a) 973 kJ/kg, 1.295 kJ/kg # K, (b) 1129 kJ/kg, 1.541 kJ/kg # K | 697 12–79 Propane is to be adiabatically and reversibly compressed in a steady-flow device from 500 kPa and 100 C to 4000 kPa. Calculate the specific work required for this compression treating the propane as an ideal gas with temperature variable specific heats and using the departure charts. 4 MPa 12–84E Propane is compressed isothermally by a piston– cylinder device from 200°F 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 12–85 A 0.08-m3 well-insulated 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 paddle-wheel work done during this process. Answers: (a) 12,190 kPa, (b) 393 kJ 12–86 Carbon dioxide is contained in a constant-volume tank and is heated from 100°C and 1 MPa to 8 MPa. Determine the heat transfer and entropy change per unit mass of the carbon dioxide using (a) the ideal-gas assumption, (b) the generalized charts, and (c) real fluid data from EES or other sources. Propane Review Problems 500 kPa 100 C FIGURE P12–79 12–80E Oxygen is adiabatically and reversibly expanded in a nozzle from 200 psia and 600 F to 70 psia. Determine the velocity at which the oxygen leaves the nozzle, assuming that it enters with negligible velocity, treating the oxygen as an ideal gas with temperature variable specific heats and using the departure charts. Answers: 1738 ft/s, 1740 ft/s 12–87 For b 0, prove that at every point of a singlephase region of an h-s diagram, the slope of a constantpressure (P constant) line is greater than the slope of a constant-temperature (T constant) line, but less than the slope of a constant-volume (v constant) line. 12–88 Using the cyclic relation and the first Maxwell relation, derive the other three Maxwell relations. 12–89 Starting with the relation dh T ds + v dP, show that the slope of a constant-pressure line on an h-s diagram (a) is constant in the saturation region and (b) increases with temperature in the superheated region. 12–90 cv Show that Ta 0v 0P b a b and cp 0T s 0T v Ta 0P 0v ba b 0T s 0T P 200 psia 600 F 0 ft/s O2 70 psia 12–91 Temperature and pressure may be defined as T a 0u b and P 0s v a 0u b 0v s FIGURE P12–80E 12–81 Propane is compressed isothermally by a piston– cylinder device from 100°C and 1 MPa to 4 MPa. Using the generalized charts, determine the work done and the heat transfer per unit mass of propane. 12–82 Reconsider Prob. 12–81. Using EES (or other) software, extend the problem to compare the solutions based on the ideal-gas assumption, generalized chart data, and real fluid data. Also extend the solution to methane. 12–83 Determine the exergy destruction associated with the process described in Prob. 12–81. Assume T0 30°C. Using these definitions, prove that for a simple compressible substance a 0s b 0v u P T 12–92 For ideal gases, the development of the constantpressure specific heat yields a 0h b 0P T 0 Prove this by using the definitions of pressure and tempera( u/ v)s . ture, T ( u/ s)v and P cen84959_ch12.qxd 9/15/06 6:03 AM Page 698 698 | Thermodynamics 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. 12–93 For ideal gases, the development of the constantvolume specific heat yields a 0u b 0v T 0 Prove this by using the definitions of pressure and temperature, T ( u/ s)v and P ( u/ v)s . 12–94 Develop expressions for h, u, s , Pr, and vr for an ideal gas whose cp is given by e b>T 1 where ai , a0, n, and b are empirical constants. c° p a aiT in a0e b>T a b> T b N2 225 K 10 MPa 12–95 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 A–2b. 12–96 Steam is throttled from 4.5 MPa and 300°C to 2.5 MPa. Estimate the temperature change of the steam during this process and the average Joule-Thomson coefficient. Answers: 26.3°C, 13.1°C/MPa 0.2 m 3 Initially evacuated 12–97 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 25°C 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. v a T FIGURE P12–99 12–100 Propane at 500 kPa and 100 C is compressed in a steady-flow device to 4000 kPa and 500 C. Calculate the change in the specific entropy of the propane and the specific work required for this compression (a) treating the propane as an ideal gas with temperature variable specific heats and (b) using the departure charts. Answers: (a) 1121 kJ/kg, 1.587 kJ/kg # K, (b) 1113 kJ/kg, 1.583 kJ/kg # K u g 12–101 Determine the second-law efficiency of the compression process described in Prob.12–100. Take T0 25 C. 12–102E Methane is to be adiabatically and reversibly compressed from 50 psia and 100 F to 500 psia. Calculate the specific work required for this compression treating the methane as an ideal gas with variable specific heats and using the departure charts. 500 psia s h P FIGURE P12–97 12–98 Reconsider Prob. 12–97. 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. 12–99 An adiabatic 0.2-m3 storage tank that is initially evacuated is connected to a supply line that carries nitrogen Methane 50 psia 100 F FIGURE P12–102E cen84959_ch12.qxd 9/15/06 6:03 AM Page 699 Chapter 12 12–103 The volume expansivity of water at 20°C 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 10°C to 30°C at constant pressure. 12–104 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 Joule-Thomson coefficient inversion curve on the T-P plane is given by the equation ( Z/ T)P 0. 12–105 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 ab P 0v s | 699 (e) inversely proportional to the entropy change sfg at that temperature. 12–110 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) 33% (d ) 26% (e) 65% 12–111 Based on data from the refrigerant-134a tables, the Joule-Thompson coefficient of refrigerant-134a at 0.8 MPa and 100°C is approximately (a) 0 (d ) 8°C/MPa (b) 5°C/MPa (e) 26°C/MPa (c) 11°C/MPa b) 12–112 For a gas whose equation of state is P(v RT, 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 12–113 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. 12–114 There have been several attempts to represent the thermodynamic relations geometrically, the best known of these 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. v a T Also, show that the isentropic expansion exponent k reduces to the specific heat ratio cp /cv for an ideal gas. 12–106 Refrigerant-134a undergoes an isothermal process at 60°C from 3 to 0.1 MPa in a closed system. Determine the work done by the refrigerant-134a by using the tabular (EES) data and the generalized charts, in kJ/kg. 12–107 Methane is contained in a piston–cylinder device and is heated at constant pressure of 4 MPa from 100 to 350°C. Determine the heat transfer, work and entropy change per unit mass of the methane using (a) the ideal-gas assumption, (b) the generalized charts, and (c) real fluid data from EES or other sources. Fundamentals of Engineering (FE) Exam Problems 12–108 A substance whose Joule-Thomson 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. 12–109 Consider the liquid–vapor saturation curve of a pure substance on the P-T diagram. The magnitude of the slope of the tangent line to this curve at a temperature T (in Kelvin) is (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. u g s h P FIGURE P12–114 12–115 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 well-known 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. cen84959_ch12.qxd 9/15/06 6:03 AM Page 700 ...
View Full Document

Ask a homework question - tutors are online