More Thermodynamic Relations : Helmholtz and Gibbs Functions.

More Thermodynamic Relations, Helmholtz and Gibbs Functions.
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More Thermodynamic Relations 11.1 KYNNING (INTRODUCTION) We have a problem. It turns out that no meters, gauges, or instruments of any kind can be used to directly measure the internal energy or the enthalpy or the entropy of a system. How, then, do you get numerical values for thermodynamic properties that are not directly measurable? In Chapter 3, we discuss this subject briefly and find that numerical values of properties that are not directly measurable (e.g., u, h,and s)cansometimesbe calculated from the numerical values of properties that are measurable (e.g., p, v, and T). For elementary materials, such as incompressible solids (or liquids) and ideal gases, we have relatively simple equations of state that provide the necessary relations. For example, the specific internal energy, specific enthalpy, and specific entropy of an incompressible material are related to its temperature, pressure, and specific volume by ðu −u Þ = cðT −T Þ,ðh −h Þ = cðT −T Þ+vðp −p Þ, and ðs −s Þ = clnðT /T Þ 2 1 2 1 2 1 2 1 2 1 2 1 2 1 incomp incomp incomp where c is the specific heat of the material. And, in the case of an ideal gas, these properties are related by ðu −u Þ = c ðT −T Þ,ðh −h Þ = c ðT −T Þ, and ðs −s Þ = c lnðT /T Þ−Rlnðp /p Þ 2 1 v 2 1 2 1 p 2 1 2 1 p 2 1 2 1 ideal gas ideal gas ideal gas Complex materials require more sophisticated equations of state plus a knowledge of various mathematical property interrelationships to be able to evaluate their unmeasurable thermodynamic properties. In this chapter, we build on the equations introduced in Chapter 3 to formulate new property relations that can be used to compute numerical values for u, h, and s for complex real materials. 361362 CHAPTER 11: More Thermodynamic Relations We begin this chapter by rounding out our list of useful thermodynamic properties by defining two new properties, the Helmholtz and Gibbs functions. We then move on to develop a series of general mathematical results, called the Maxwell equations, that relate a number of thermodynamic properties. We end this chapter by using the principle of corresponding states to develop a set of generalized thermodynamic property charts that are valid for many realgases. 11.2 TWO NEW PROPERTIES: HELMHOLTZ AND GIBBS FUNCTIONS If we consider a stationary closed system containing a pure substance subjected only to a moving boundary mechanical work mode, then the combined energy and entropy balance is given by Eq. (7.30) as du =Tds−pdv (11.1) Since any two independent properties fix the thermodynamic state of a pure substance subjected to only one work mode (see Chapter 4), we can take the two independent properties here to be s and v,or u =usðÞ ,v The total differential of this composite function, then, has the form   ∂u ∂u du = ds+ dv (11.2) ∂s v ∂v s Comparing Eqs. (11.1) and (11.2) we see that  ∂u T = ∂s v and  ∂u p = − ∂v s For this system, we can also write, from Eq. (7.31), dh =Tds+vdp (11.3) and, in this case, we take the two independent properties to be s and p, so that h =hsðÞ ,p whose total differential is   ∂h ∂h dh = ds+ dp (11.4) ∂s p ∂p s On comparing Eqs. (11.3) and (11.4), we see that  ∂h T = ∂s p and  ∂h v = ∂p s EXAMPLE 11.1 To illustrate the relation between the constant volume and constant pressure specific heats and entropy, begin with Eqs. (11.1) and (11.3) and show that the constant volume and constant pressure specific heats are related to specific entropy by:  ∂s c = T v ∂T v and  ∂s c = T p ∂T P11.2 Two New Properties: Helmholtz and Gibbs Functions 363 Solution The constant volume specific heat is defined by Eq. (3.15) as  ∂u c = v ∂T v and the constant pressure specific heat is defined by Eq. (3.19) as  ∂h c = p ∂T p Equation (11.1) is du = Tds – pdv, and if we it divide through by dT, we get du ds dv = T −p dT dT dT If we now require the specific volume v to be constant during this operation, then this equation becomes    du ds dv = T −p (a)    dT v dT v dT v Now, a total derivative restrained with a constant parameter is just a partial derivative, or   du ∂u = = c  v dT v ∂T v   ds ∂s  = dT v ∂T v    dv ∂v  = = 0 ðsince v is to be held constant hereÞ dT v ∂T v Then, substituting these results into Eq. (a) gives one of the desired relations:   ∂u ∂s = c = T v ∂T ∂T v v Similarly, beginning with Eq. (11.3), we have dh =Tds +vdp, and dividing this through by dT gives dp dh ds = T + v (b) dT dT dT Again, imposing the condition that p must be constant during this operation, we get   dh ∂h = = c  p dT p ∂T p   ds ∂s =  dT p ∂T p   dp ∂p   = = 0 ðsince p is to be held constant hereÞ dT p ∂T p Then, substituting these results into Eq. (b) gives the other desired relation:   ∂h ∂s = c = T p ∂T ∂T p p The following exercises are designed to strengthen your understanding of the thermodynamics and the mathematics of the material presented in this part of the chapter. Exercises 1. Use the results of Example 11.1 to show that a material that has an entropy function of the form s(T, v) = A + B(ln T) + C(ln v), where A, B, and C are constants, has a constant volume specific heat given by c = B. Hint: Substitute the given v function into the relation c = T(∂s/∂T) . v v 2. If the specific internal energy of a material is found to depend on its specific entropy and specific volume according to the 2 relation u(s, v) = A + Bs + Cv + Ds/v, where A, B, C, and D are all constants, then determine an expression for p(s, v) for 2 this material. Answer: p(s, v) =–(2Cv– Ds/v ). 2 3 3. If the specific enthalpy ofa material dependsonspecific entropy and pressure according to h(s, p) = A + Bs + Cp + Ds p, 2 where A, B, C,and D are all constants,thendeterminean expression for T(s, p) for this material.Answer: T(s, p) = B +3Ds p. We now introduce two new thermodynamic properties. The first is the total Helmholtz function F, named after the German physicist and physiologist Hermann Ludwig Ferdinand von Helmholtz (1821–1894), defined as F = U−TS Dividing by the system mass gives the specific Helmholtz function f as f = u−Ts364 CHAPTER 11: More Thermodynamic Relations Differentiating this equation gives df = du−Tds−sdT but from Eq. (11.1) we have du−Tds =−pdv so that df =−pdv−sdT (11.5) If we presume the existence of a functional relation of the form f =fvðÞ ,T then its total differential is   ∂f ∂f df = dv+ dT (11.6) ∂v ∂T T v and, on comparing Eqs. (11.5) and (11.6), we see that  ∂f p =− ∂v T and  ∂f s = − ∂T v The second new thermodynamic function is the total Gibbs function G, named after the American physicist Josiah Willard Gibbs (1839–1903), defined as G = H−TS Dividing by the system mass gives the specific Gibbs function g as g = h−Ts (11.7) Differentiating Eq. (11.7) gives dg = dh−Tds−sdT but from Eq. (11.3), we have dh =Tds+vdp so that dg =vdp−sdT (11.8) If we presume a functional relation of the form g = gðp,TÞ then its total differential is   ∂g ∂g dg = dp+ dT (11.9) ∂p ∂T T p and comparing Eqs. (11.8) and (11.9) gives  ∂g v = ∂p T and  ∂g s =− ∂T p11.2 Two New Properties: Helmholtz and Gibbs Functions 365 Table 11.1 Summary of Thermodynamic Property Relations New Property Relations     ∂u ∂h T = = ∂s ∂s v p     ∂g ∂h v = = ∂p ∂p s T    ∂f ∂u p = − =− ∂v ∂v s T     ∂f ∂g s = − = − ∂T ∂T v p Table 11.1 summarizes these results. The importance of this set of partial differential equations lies in the fact that they relate easily measurable properties (p, v, T) to nonmeasurable properties (u, h, s, f,and g). Therefore, accurate p, v, T data on any pure substance can be used to generate information about u, h, s, f,and g for that substance. However, they do not provide a direct method for calculating u, h,or s from p, v, and T information. We must look for additional information to complete this task. But, first, we take a short diversion into phase change processes for which we can determine important results based on what we already know. EXAMPLE 11.2 The design of a new Happy Food fast-food processing system requires the values of the specific Helmholtz and Gibbs func- tions for superheated water vapor at 200. psia and 400.°F. Since most thermodynamic tables do not list these properties directly, you are asked to calculate them for these conditions. Superheated water vapor Solution at 200. psia and 400.°F First, draw a sketch of the system (Figure 11.1). The unknowns are the values of the specific Helmholtz and Gibbs Happy food functions for superheated water vapor at 200. psia and 400.°F. New Happy Food The specific Helmholtz and Gibbs functions are defined in the fast−food processing text as system Specific Helmholtz function: f = u−Ts FIGURE 11.1 and Example 11.1. Specific Gibbs function: g = h−Ts Therefore, we need u, h, and s information at the state defined by p = 200. psia and T = 400.°F = 860.R. From Table C.3a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that, at this state, uðp = 200:psia,T = 400:°FÞ = 1123:5Btu/lbm hðp = 200:psia,T = 400:°FÞ = 1210:8Btu/lbm, and . sðp = 200:psia,T = 400:°FÞ = 1:5602Btu/lbm R Then, . f = 1123:5Btu/lbm− ð400:+459:67RÞð1:5602Btu/lbm RÞ = −218Btu/lbm and . g = 1210:8Btu/lbm−ð400:+459:67RÞð1:5602Btu/lbm RÞ = −131Btu/lbm The following exercises consider different pressures and temperatures and explore why the specific Helmholtz and Gibbs functions were negative in Example 11.2. Exercises 4. Determine the value of the specific Helmholtz function of the superheated water vapor in Example 11.2 if the pressure is maintained at 200. psia but the temperature is increased to 1000.°F. Answer: f =−1320 Btu/lbm. 5. Determine the specific Helmholtz function of saturated liquid water at 200. psia. Answer: f =−103 Btu/lbm. 6. The Helmholtz and Gibbs functions calculated in Example 11.2 are both negative. Though this has no particular significance at this point, use Table C.3a to determine the temperature and pressure at which the Helmholtz function is zero for water. Answer: f = 0 for saturated liquid water at 0.0887 psia and 32.018°F (the triple point of water).366 CHAPTER 11: More Thermodynamic Relations 11.3 GIBBS PHASE EQUILIBRIUM CONDITION During a phase change process, the system pressure and temperature are not independent properties. This means that, if we hold one of them constant during a phase change, the other must also remain constant. Under the condition of constant pressure and temperature, dp = dT = 0, and Eq. (11.9) gives dg = 0. Since g = g + xg ,we f fg can then write dg = dg +xdg +g dx = 0 f fg fg Again, Eq. (11.8) can be used to evaluate dg = dg = 0. Since x can vary during the phase equilibrium, dx f fg cannot be zero. Therefore, we are forced to conclude from the preceding equation that g = 0 at phase equilibrium, fg or g = g . Using the definitionoftheGibbs function, Eq. (11.7), wesee that,at phase equilibrium, f g g = g = h −ðT Þs = h −ðT Þs f g f sat f g sat g or h −h = h =ðT Þðs −s Þ =ðT Þs g f fg sat g f sat fg or s = h /T (11.10) sat fg fg This gives us an important relation between the entropy and the enthalpy of a phase change, but we need much more information to complete the process of determining the nonmeasurable properties u, h,and s from the measurable properties p, v, and T. The following example illustrates the use of Eq. (11.10). EXAMPLE 11.3 Use Eq. (11.10) to calculate the phase change entropy s for water at exactly 1.00 MPa and compare the result with the fg value for s at exactly 1.00 MPa listed in Table C.2b in Thermodynamic Tables to accompany Modern Engineering fg Thermodynamics. 1 Solution The unknown is the phase change entropy of water. From (Eq. 11.10), we have h fg s = fg T sat and from Table C.2b at p = 1.00 MPa, we find that h = 2015:3kJ/kg fg and T = 179:90°C sat then, Eq. (11.10) gives h 2015:3kJ/kg fg . s = = = 4:4482kJ/kg K fg T 179:90+273:15K sat Comparing this with the value for s listed in Table C.2b at p = 1.00 MPa, we find that it is exactly the same. fg The following exercises illustrate some of the many uses of Eq. (11.10). Exercises 7. Use Eq. (11.10) to compute the values of s for water at 0.0100 MPa and compare the result with the values listed in fg Table C.2b. Find h and T at 0.01 MPa from Table C.2b. Answer:(s ) = 7.5021 kJ/kg·K. fg sat fg calc 8. Use Eq. (11.10) to compute the value of h for water at 100.°F and compare the result with the value listed in Table fg C.1a. Find values for s and T at 100.°F from Table C.1a. Answer:(h ) = 1036.96 Btu/lbm. fg sat fg calc 9. Use the values for h and s found in Table C.2b for water at 10.0 MPa and Eq. (11.10) to calculate the value of T at fg fg sat this state. Answer:(T ) = 584.2 K = 311.06°C. sat calc 1 To achieve the desired result, we need to carry a lot more significant figures than usual.11.4 Maxwell Equations 367 11.4 MAXWELL EQUATIONS Two sets of equations are named after the Scottish physicist James Clerk Maxwell (1831−1879): the electromagnetic field equations and the thermodynamic property equations. The thermodynamic Maxwell equations allow additional numerical information to be obtained about the nonmeasurable properties u, h, and s from accurately measured p, v, and T data. Consider an arbitrarily continuous function of the form z =zxðÞ ,y Then, we can write its total differential as   ∂z ∂z dz = dx+ dy = Mdx+Ndy (11.11) ∂x ∂y y x where we set  ∂z M = ∂x y and  ∂z N = ∂y x If we now differentiate M with respect to y while holding x constant and differentiate N with respect to x while holding y constant, we get  2 ∂M ∂ z = ∂y ∂y∂x x and  2 ∂N ∂ z = ∂x y ∂x∂y Since we require z(x, y) to be a continuous function, it follows that 2 2 ∂ z ∂ z = ∂y∂x ∂x∂y or that   ∂M ∂N  ≡ (11.12) ∂y ∂x y x Recall that the thermodynamic state of any pure substance is fixed by any pair of independent intensive thermo- dynamic properties of that substance. That is, any property of a pure substance can be written as a function of any other two independent properties of that substance. Consequently, if x and y are such independent proper- ties, then z also is a property, provided that Eq. (11.12) is satisfied. EXAMPLE 11.4 Suppose we make a series of measurements in the laboratory and think we discovered a new thermodynamic property, call 2 it z. Our experimental data provide an empirical equation of the form: dz =pdv + v dp.Is z a new property? Solution The unknown is whether or not z is a new thermodynamic property. Equation (11.11) here has the form 2 dz = Mdx+Ndy =pdv+v dp 2 so M = p, N = v , x = v, and y = p. The cross differentials in Eq. (11.12) are   ∂p ∂M = = 1 ∂y ∂p x v (Continued)368 CHAPTER 11: More Thermodynamic Relations EXAMPLE 11.4 (Continued) and    2 ∂ðv Þ ∂N ∂M = = 2v≠ ∂x y ∂v ∂y p x Since Eq. (11.12) is not satisfied here, then z cannot be a thermodynamic property. Exercises 2 2 10. Suppose the expression experimentally discovered in Example 11.4 is dz = p dv + v dp, would z be the thermodynamic property? Answer: No, because 2p≠ 2v. 11. IftheexpressionexperimentallydiscoveredinExample11.4isdz=pdv–vdp,wouldzbeathermodynamicproperty?Answer:No,1≠–1. 12. Ifthe expressionreported in Example 11.4 is dz=pdv +vdp,would z be athermodynamicproperty?Answer:Yes, dz= d(pv). If we now look at our four basic property relationships as differential equations of the form of Eq. (11.11), du =Tds−pdv (11.1) dh =Tds +vdp (11.3) df = −pdv−sdT (11.5) dg =vdp−sdT (11.8) then, Eq. (11.12) must be valid for these equations, since we already know that all of these functions are ther- modynamic properties. Applying Eq. (11.12) to each of these equations yields a new set of equations, known as the Maxwell thermodynamic equations: Maxwell thermodynamic equations   ∂p ∂T = − (11.13) ∂v s ∂s v   ∂T ∂v = (11.14) ∂p ∂s p s   ∂p ∂s = (11.15) ∂T ∂v T v   ∂v ∂s = − (11.16) ∂T p ∂p T While the Maxwell thermodynamic equations provide additional information about u, h,and s in terms of p, v, and T, they cannot be solved to produce the direct functional relations between these properties that we seek. However, these relations are used a little later in this chapter in conjunction with other material to provide the desired u, h, and s relations from experimental p, v, T data. EXAMPLE 11.5 Suppose wehave the idealgas equationof state, pv = RT, but knownothingabout the entropy ofthis typeof gas. Use the appro- priate Maxwell equations todetermine a mathematicalrelationfor the entropy ofanidealgasduringan isothermalprocess. Solution The ideal gas equation of state is pv = RT, and so we know a p, v, T relation for our material. Perusing the Maxwell equa- tions, we see two, Eqs. (11.15) and (11.16), that involve only p, v, T variables on one side of the equation. We can choose either of these equations to satisfy the problem statement, so we select Eq. (11.16):   ∂v ∂s = − ∂T p ∂p T Solving for v from the ideal gas equation of state gives v = RT/p, so the partial derivative we need is  ∂v = R/p ∂T p11.4 Maxwell Equations 369 then,  ∂s = −R/p ∂p T = –R(dp/p) , where the subscript T is used to indicate that the temperature is to be held constant. This can be so that ds T T integrated for the constant temperature condition to give Z 2 ðs −s Þ = −R ðdp/pÞ +funcðTÞ = −Rlnðp /p Þ+funcðTÞ 2 1 2 1 T T 1 where func(T) is an arbitrary function of integration. The function of integration here depends on the temperature T, and for an isothermal process, it is treated as a constant. Since we happen to know that the entropy relation for an ideal gas is in fact s – s = c ln(T /T ) – R ln(p /p ), it is easy to see that the function of integration here is simply c ln(T /T ). 2 1 p 2 1 2 1 p 2 1 The following exercises reinforce the concepts presented in Example 11.5. Exercises 13. Use the other Maxwell equation (Eq. 11.15) available for the solution of Example 11.5 to find a different ideal gas entropy relation. Answer:(s – s ) = R ln(v /v ) + func(T). 2 1 T 2 1 14. Show that (∂s/∂) = (∂p/∂T). Answer: Invert Eq. (11.14). p s Z 2 15. Show that ðs −s Þ = − ð∂T/∂vÞ dp+funcðvÞ. Answer: Use Eq. (11.13). 2 1 y s 1 Before we continue with our search for the illusive u, h, s equations in terms of p, v, T variables, the following example shows that the form taken by the Maxwell equations depends on the type of reversible work mode pre- sent in the system. EXAMPLE 11.6 The equation of state for a nonlinear rubber band is given by 2 F = KTðÞ L/L −1 o where F is the stretching force, L is the stretched length, L is the initial length, K is the elastic constant, and T is the absolute o temperature of the material. Then, a. Determine the Maxwell equations for this material. b. Show that the internal energy of this material is a function of temperature only. c. Determine the heat transfer required when the rubber band is stretched isothermally and reversibly from L = 0.0700 m o to L = 0.200 m at T = 20.0°C when K = 0.150 N/K. Solution The unknowns here are the Maxwell equations for this material, showing that the internal energy of this material is a func- tion of temperature only and the heat transfer required when the rubber band is stretched isothermally and reversibly between two states. a. Since the reversible work mode involved in the stretching process is ðdWÞ = −FdL rev the Maxwell equations for this material can be easily obtained from those derived in the text by replacing p with –F and v with L/m = ℓ, the specific length of the material. Then Eqs. (11.13) to (11.16) become   ∂T ∂F = ∂ℓ ∂s s ℓ   ∂T ∂ℓ = − ∂F s ∂s F   ∂F ∂s = − ∂T ℓ ∂ℓ T and   ∂ℓ ∂s = ∂T F ∂F T (Continued)370 CHAPTER 11: More Thermodynamic Relations EXAMPLE 11.6 (Continued) b. The combined energy and entropy balance for this material is du =Tds+Fdℓ so that   ∂u ∂s = T +F ∂ℓ ∂ℓ T T From the third Maxwell equation for this substance listed in part a and the given equation of state, we have   ∂s ∂F =− ∂ℓ ∂T T ℓ 2 =−KLðÞ /L −1 o and  ∂s 2 T = −KTðÞ L/L −1 = −F o ∂ℓ T Therefore,  ∂u = −F+F = 0 ∂ℓ T If we now set u = u(T, ℓ) and differentiate it, we get   ∂u ∂u du = dT+ dℓ ∂T ℓ ∂ℓ T  ∂u = c dT+ dℓ ℓ ∂ℓ T where c is the constant length specific heat. Now since (∂u/∂ℓ) = 0 here, then this equation reduces to du = c dT,so u ℓ T ℓ is only a function of T. c. A closed system energy balance applied to this material for an isothermal process with (u – u ) = 0 gives 2 1 T Z L Q = W = − FdL 1 2 1 2 L o Z L 2 =−KT ðÞ L/L −1 dL o Lo 3 =−KTLðÞ L/L −1 /3 o o . 3 0:200 =−ðÞ 0:150N/KðÞ 293KðÞ 0:0700m −1 3 0:0700 . Q =−6:57N m 1 2 Consequently, there is a heat transfer out of the system equal in magnitude to the work input. Exercises 16. Determine the heat transfer and work required to stretch the rubber band in Example 11.6 if the elastic constant of the rubber is increased from 0.150 N/K to 10.0 N/K. Answer: Q = W = –438 N·m. 1 2 1 2 17. If the temperature of the rubber band in Example 11.6 is increased from 20.0°Cto60.0°C, determine the heat transfer and Q = W =–7.47 N·m. work required to stretch the rubber band assuming all the other variables remain unchanged.Answer: 1 2 1 2 18. How much heat transfer and work is required to stretch the rubber band in Example 11.6 twice as far, to L = 0.400 m instead of 0.200 m, if everything else remains constant? Answer: Q = W = –107 N·m. 1 2 1 2 11.5 THE CLAPEYRON EQUATION Benoit Pierre Emile Clapeyron (1799–1864) was a French mining engineer and a contemporary of Carnot who, in the, 1830s, took an interest in studying the physical behavior of gases and vapors. He was able to derive a relation for the enthalpy change of the liquid to vapor phase transition (h ) in terms of pressure, temperature, fg and specific volume, thus providing one of the first equations for calculating a property that is not directly measurable in terms of properties that are directly measurable. Today, this relation is most easily derived from one of the Maxwell equations, Eq. (11.15). For an isothermal phase change from a saturated liquid to a satu- rated vapor, the pressure and temperature are independent of volume. Then, Eq. (11.15) becomes   s −s ∂p dp g f = = = s /v fg fg ∂T dT v −v g f v sat11.5 The Clapeyron Equation 371 2 and, using Eq. (11.10), we obtain the Clapeyron equation as   dp = h / T v (11.17) fg sat fg dT sat For most substances, v ≫ v, so we can approximate v ≈ v . Also, for vapors at very low pressures, the saturated g f fg g vapor curve can be accurately approximated by the ideal gas equation of state, so we can write v = RT /p . g sat sat Then Eq. (11.17) becomes   dp 2 = p h / RT sat fg sat dT sat or   dp 2 = h / RT dT (11.18) fg sat sat p sat This equation is often called the Clapeyron-Clausius equation. For small pressure and temperature changes, h can fg be assumed to be constant and Eq. (11.18) can be integrated from a reference state to any other state to give  T −T sat 0 lnðÞ p/p = h /R 0 fg sat T T sat 0 or   T −T sat 0 p = p exp h /R sat 0 fg T T sat 0 where p and T are reference state values. An exponential relation between p and T fits experimental data 0 0 sat sat quite well for most substances at low pressure. EXAMPLE 11.7 In 1849, William Rankine proposed the following relation between the saturation pressure and saturation temperature of water: 6289:78 913,998:92 lnp = 14:05− − sat 2 T T sat sat where p is in psia, and T is the temperature in °F + 461.2 (at that time –461.2°F was Rankine’s best estimate of absolute sat sat zero temperature). Determine h at 212.0°F from the Rankine equation and compare the result with that listed in the steam fg tables in Thermodynamic Tables to accompany Modern Engineering Thermodynamics. Solution Differentiating Rankine’s equation, we obtain  dp 1 6289:78 1,827,997:8 = + 2 3 p dT T T sat sat sat then, using Eq. (11.18), we get   2 dp RT h = = RðÞ 6289:78+1,827,997:8/T fg sat p dT sat (Continued) 2 The Clapeyron equation is valid for any type of phase change in a simple substance. For example, if we let the i subscript denote the solid phase, then for melting we can write   dp = h / T v if sat if solid dT liquid saturation and, for sublimation,   dp = h / T v ig sat ig solid dT vapor saturation372 CHAPTER 11: More Thermodynamic Relations EXAMPLE 11.7 (Continued) From Table C.13a in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find R = 85.78 ft·lbf/(lbm·R) = 0.1102Btu/(lbm·R). Then, at212.0°F, 2 hðÞ 212:0°F =½ 6289:78R+ðÞ 1,827,997:8R /4ðÞ 61:2+212:0R fg ×½ 0:1102Btu/ðÞ lbm⋅R = 992:37Btu/lbm Table C.1a gives h ð212:0°FÞ = 970:4Btu/lbm: Thus, the value obtained from Rankine’s equation is in error by only +2.26%. fg Exercises 19. Determine p fromthe Rankine equation given in Example 11.7 when T =212.0°F and compareitwiththe valueof T sat sat sat givenin Table C.1a at212.0°F.Answer:(p ) = 14.73 psia, andfromTable C.1a, p (212.0°F)=14.696 psia. sat calc sat 20. Using the relations given in Example 11.7, find the value of T for water when h =1037 Btu/lbm and compare your sat fg result with the value given in Table C.1a. Answer:(T ) = 124.6°F, and from Table C.1a, T =100.0°F. sat calc sat 21. Determine h in Example 11.7 if the temperature is increased from 212.0°F to 500.0°F and compare your result with the fg value given in Table C.1a. Answer:(h ) = 902.7 Btu/lbm, and from Table C.1a, h (500.0°F) = 714.8 Btu/lbm. fg calc fg 11.6 DETERMINING u, h, AND s FROM p, v, AND T We are now ready to combine the previous results to produce u, h,and s relations from p, v,and T data. For a simple substance, any two independent intensive properties fix its thermodynamic state. Consider the specific internal energy described by a function of temperature and specific volume. We can write this as u = u(T, v). Differentiating this function, we get   ∂u ∂u du = dT+ dv ∂T v ∂v T From Eq. (11.1), we can write   ∂u ∂s = T −p ∂v T ∂v T and using the Maxwell Eq. (11.15), this becomes   ∂p ∂u = T −p ∂v ∂T T v In Chapter 3, we introduce the constant volume specific heat c as v  ∂u c = (3.15) v ∂T v and our equation for the total differential du then becomes   ∂p du = c dT+ T −p dv (11.19) v ∂T v Therefore, the change in specific internal energy for any simple substance can be determined by integrating Eq. (11.19): Z Z  T2 v2 ∂p u −u = c dT+ T −p dv (11.20) 2 1 v ∂T T v 1 1 v Here, we achieved what we set out to do. Equation (11.20) has u cast completely in terms of the measurable quantities p, v, T, and c . v Similarly, we can consider the specific enthalpy to be given by a continuous function of temperature and pres- sure, h = h(T, p). Then, its total differential is   ∂h ∂h dh = dT+ dp ∂T ∂p p T In Chapter 3, we introduce the constant pressure specific heat c as p  ∂h c = (3.19) p ∂T p11.6 Determining u, h, and s from p, v, and T 373 Introducing the definition of specific enthalpy into Eq. (11.1) gives du = dh−pdv−vdp =Tds−pdv or dh =Tds+vdp (11.21) and, from this equation, we can deduce that   ∂h ∂s = T +v ∂p ∂p T T Using the Maxwell Eq. (11.16), we get   ∂h ∂v = −T +v ∂p ∂T p T and our total differential dh becomes   ∂v dh = c dT+ v−T dp (11.22) p ∂T p The change in specific enthalpy for any simple substance is then given by  Z Z T p  2 2 ∂v h −h = c dT+ v−T dp (11.23) 2 1 p ∂T p T p 1 1 Again, we are successful. Equation (11.23) has h cast completely in terms of the measurable quantities, p, v, . Also note that Eqs. (11.20) and (11.23) are related by the fact that T, and c p h −h = u −u +p v −p v 2 1 2 1 2 2 1 1 Finally, we can carry out the same type of analysis for the specific entropy of a simple substance. If we let s = s (T, v), then   ∂s ∂s ds = dT+ dv ∂T v ∂v T From Eqs. (11.1) and (3.15), we can deduce that   ∂s 1 ∂u c v = = ∂T v T ∂T v T and, using the Maxwell Eq. (11.15), we can write the total differential ds as   ∂p c v ds = dT+ dv (11.24) T ∂T v Integrating this gives a relation for the change in specific entropy of a pure substance based completely on measurable quantities:  Z Z T v 2 2 c ∂p v s −s = dT+ dv (11.25) 2 1 T ∂T T v 1 1 v By assuming s = s(T, p), we can also show that (see Problem 27 at the end of this chapter)  c p ∂v ds = dT− dp (11.26) T ∂T p and Z Z T p 2 2 c p ∂v s −s = dT− dp (11.27) 2 1 T ∂T p T p 1 1 This completes the process of discovering relations for the unmeasurable u, h,and s properties in terms of the measurable properties p, v, and T. The following example illustrates the use of these results.374 CHAPTER 11: More Thermodynamic Relations EXAMPLE 11.8 In Chapter 3, an equation of state developed in 1903 by Pierre Berthelot (1827–1907) was briefly discussed. Using this equation of state, develop equations based on measurable properties for the changes in (a) specific internal energy, (b) spe- cific enthalpy, and (c) specific entropy for an isothermal process. Solution The Berthelot equation is given in Eq. (3.46) as  2 pvðÞ −b = RT−avðÞ −b / Tv where a and b are constants. Solving this equation for p gives  2 p = RT/ðÞ v−b −a/ Tv a. The change in specific internal energy is given by Eq. (11.20), for which we need   ∂p 2 2 = R/ðÞ v−b +a/ T v ∂T v Then, for an isothermal process (T = T ), Eq. (11.20) gives 1 2 Z v 2   2 2 ðÞ u −u = RT/ðÞ v−b +a/ Tv −RT/ðv−bÞ+a/ Tv dv 2 1 T v 1 =−ðÞ 2a/TðÞ 1/v −1/v = 2avðÞ −v /ðÞ Tv v 2 1 2 1 1 2 b. To find the change in specific enthalpy, we could use Eq. (11.23). However, to evaluate this equation, we need to be able to determine the relationðÞ ∂v/∂T : Since the Berthelot equation is not readily solvable for v = v (T, p), we choose p instead to use the simpler approach, utilizing only the definition of specific enthalpy, h = u + pv. Then, 2aðv −v Þ 2 1 ðh −h Þ =ðu −u Þ +p v −p v = +p v −p v 2 1 2 1 2 2 1 1 2 2 1 1 T T Tv v 2 1  3aðv −v Þ 2 1 v v 2 1 = +RT − Tv v v −b v −b 1 2 2 1 c. Finally,sincewealreadyevaluatedtherelationðÞ ∂p/∂T , wechoosetouseEq.(11.25)fortheisothermalspecificentropyrelation: v Z  v2 ∂p ðÞ s −s = dv 2 1 T ∂T v 1 v Z v 2 2 2 = R/ðv−bÞ+a/ðT v Þ dv v 1 2 = R ln½ ðv −bÞ/ðv −bÞ +aðv −v Þ/ðT v v Þ 2 1 2 1 1 2 Exercises 22. Setting a = 0 in the Berthelot equation of state used in Example 11.8 produces the Clausius equation of state, p(v– b)= RT (see Eq. (3.43)). Determine equations for the change in specific internal energy, specific enthalpy, and specific entropy for a Clausiusgasundergoinganisothermalprocess.Answer:(u –u ) =0,(h –h ) =RT(v /(v –b)–v /(v –b),and(s –s ) =R 2 1 T 2 1 T 2 2 1 1 2 1 T ln(v –b)/(v –b). 2 1 4 2 23. EvaluatethechangeinspecificinternalenergyofwatervaporwhenitismodeledasaBerthelotgas,witha=4.30MN·m ·K/kg –3 3 3 andb= 4.50× 10 m /kg, and undergoesan isothermalcompression from a specific volume 40.0 m /kg to a specific volume 3 of 5.00 m /kg at a constant temperature of 100.°C.Answer:(u –u ) =–4.03 kN·m/kg=–4.03 kJ/kg. 2 1 T 4 2 24. EvaluatethechangeinspecificenthalpyofwatervaporwhenitismodeledasaBerthelotgas,witha=4.30MN·m ·K/kg and –3 3 –3 3 b=4.50×10 m /kg,andundergoesanisothermalexpansionfromaspecificvolumeof10.0×10 m /kgtoaspecificvolume 3 of1.00m /kgataconstanttemperatureof500.°C.UseR =461N·m/(kg·K).Answer:(h –h ) =261kN·m/kg=261kJ/kg. water 2 1 T Note that, for an ideal gas undergoing an isothermal process, ðu −u Þ =ðh −h Þ = 0 and ðs −s Þ = R lnðÞ v /v 2 1 2 1 2 1 2 1 T T T Therefore, the equations developed in Example 11.8 can be considered to be Berthelot corrections to ideal gas behavior. Equation (11.24) has the same Mdx + Ndy form as Eq. (11.11), so that we can utilize Eq. (11.12) to produce the property relation    ∂ðc /TÞ ∂p v ∂ = ∂v ∂T ∂T T v v11.6 Determining u, h, and s from p, v, and T 375 or   2 ∂ p ∂c v = T (11.28) 2 ∂v T ∂T v Similarly, Eq. (11.26) has the same Mdx + Ndy form and application of Eq. (11.12), so it gives "    ∂ c /T p ∂ ∂v = − ∂p ∂T ∂T p p T or   2 ∂c p ∂ v = −T (11.29) 2 ∂p ∂T T p Both Eqs. (11.28) and (11.29) give specific heat information from measurable p, v, and T properties. EXAMPLE 11.9 Using the Berthelot equation of state given in Example 11.8, determine an equation for the isothermal variation in the constant volume specific heat with volume change. Solution From Eq. (11.28) we have what we seek,   2 ∂ p ∂c v = T 2 ∂v T ∂T v and from Example 11.8, the Berthelot equation of state can be written as RT a p = − 2 v−b Tv so that  ∂p R a = + 2 2 ∂T v−b T v v and  2 ∂ p 2a = − 2 3 2 ∂T T v v then,  ∂c v 2a = − 2 2 ∂v T T v and, to find an explicit c = c (T, v) equation, the preceding equation can be integrated from a reference state specific v v volume v to give o Z v  2a dv 2 c = − = 2avðÞ −v / T v v + fðTÞ v 0 0 2 2 T v v0 where f(T) is a function of integration. Note that c is independent of v only in the case where a = 0 in the Berthelot equa- v tion of state. Exercises 25. The Clausius equation of state, p(v – b) = RT (see Eq. (3.43)), can be obtained by setting a = 0 in the Berthelot equation of state. Rework Example 11.9 to determine how the constant volume specific heat of a Clausius gas undergoing an isothermal process depends on the specific volume of the gas. Answer: For a Clausius gas undergoing an isothermal process, c does not depend on the specific volume of the gas. υ 26. For the Berthelot equation of state used in Example 11.9, determine an expression for the mixed partial derivative   ∂c ∂ v = ? ∂T ∂v T v 3 2 Answer:4a/(T v ). 4 2 27. EvaluatetheconstantvolumespecificheatrelationdevelopedinExample11.9foramaterialinwhicha=2.30MN·m ·K/kg . 3 3 Usev =0.100m /kgandv=0.0200m /kg.Answer:c =1.321kN·m/kg=1.321kJ/kg. 0 v376 CHAPTER 11: More Thermodynamic Relations Finally, if Eqs. (11.24) and (11.26) are set equal to each other,     ∂p c c p v ∂v dT+ dv = dT− dp T ∂T T ∂T p v and, if we solve for dT,    ∂p T ∂v T dT = dp+ dv c c −c −c p v p v ∂T ∂T p v Then, writing the general relation T = T(p, v) and differentiating it, we get   ∂T ∂T dT = dp+ dv ∂p ∂v p v and, by comparing coefficients of dp and dv in these two equations, it is clear that   ∂T T ∂v = c −c p v ∂p ∂T p v and   ∂p ∂T T = c −c p v ∂v ∂T p v or     ∂p ∂p ∂v ∂v c −c = T = T p v ∂T p ∂T ∂T ∂T p v v Using Eq. (3.3), we can write    ∂p ∂p ∂v = − ∂T ∂T ∂v p v T which, when substituted into the previous equation, yields   2 ∂p ∂v c −c = −T p v ∂T p ∂v T In Chapter 3, we define the isobaric coefficient of volume expansion β as  1 ∂v β = (3.5) v ∂T p and the isothermal coefficient of compressibility κ as  1 ∂v κ = − (3.6) v ∂p T Substituting these two relations into the previous equation gives the final result: 2 c −c = Tβ v/k (11.30) p v This equation reveals several important results. First of all, c = c for all simple substances at absolute zero p v 2 temperature. Second, since β /κ = 0 for incompressible materials, then c = c for all incompressible materials. In p v = c = c for all incompressible materials. this case, the p and v subscripts are normally dropped and we write c p v Finally, since T, β,v, and κ are always positive, then c ≥ c for all simple substances. p v EXAMPLE 11.10 Using the data in Table 3.2, determine the difference between c and c for saturated liquid water at 20.0°C. p v Solution From Table 3.2, for water, we find that −6 −1 β = 0:207×10 K −11 2 κ = 45:9×10 m /N11.6 Determining u, h, and s from p, v, and T 377 and, from Table C.1b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics, we find that 3 v = v ð20:0°CÞ = 0:001002m /kg f Then, from Eq. (11.30), we have 2 2 −6 −1 3 ð293KÞð0:207×10 K Þ ð0:001002m /kgÞ Tβ v c −c = = p v −11 2 κ 45:9×10 m⋅s /kg −5 −8 = 2:74×10 J/ðkg⋅KÞ = 2:74×10 kJ/ðkg⋅KÞ In most applications, this difference is clearly negligible, since the value of c for liquid water at standard temperature and p pressure is 4.18 kJ/(kg·K). Exercises 28. Show that, for an ideal gas, defined by pv = RT, the isobaric coefficient of volume expansion isβ = 1/T and the isothermal coefficient of compressibility is κ = 1/p. Then, show that, for an ideal gas, Eq. (11.30) gives c – c = R. p v 29. Using the methods of Example 11.10, determine the difference between the constant pressure specific heat c and p –5 3 for liquid mercury at 20.0°C. For liquid mercury, v = 7.4 × 10 m /kg. Answer: the constant volume specific heat c v –5 (c – c ) = 1.79 × 10 J/(kg·K). p v mercury –3 3 30. Rework Example 11.10 for liquid benzene at 20°C. For liquid benzene, v = 1.15 × 10 m /kg. Answer:(c – c ) = p v benzene –4 5.46 × 10 J/(kg·K). Now we need to develop a strategy for the integration of Eqs. (11.19), (11.22), (11.24), and (11.26) for arbitrary states. Since u, h, and s are point functions, the integration results are independent of the actual integra- tion path, so we should pick a path that is easy to evaluate. In addition, since u and h lack well-defined absolute zero values, the integration path must begin at an arbitrary reference state. Since all equations of state should reduce to the ideal gas equation of state at low pressures, we can postulate that they must all be reducible to the following form: pv = RT+ fvðÞ ,T where the function f(v, T) must be on the order of 1/v so that it vanishes as p→ 0 and v→∞. Consequently, the reference state is usually taken to be at some arbitrary reference temperature T and at essentially zero pressure p =0, 0 0 and zero density or infinite specific volume, v =∞. To generate a numerical value from Eqs. (11.19), (11.22), (11.24), and (11.26) for u, h, and s, we start the inte- gration process at this reference state. Now, the choice of the easiest integration path from the reference state to the actual state depends on the form of the arguments in the integrals. For example, in evaluating Eq. (11.19) for the specific internal energy, the easiest path to follow is a constant specific volume line for the first integral (since c is defined for a constant volume process) and to follow a constant temperature line for the second inte- v 0 gral. Since the first integration line is at zero pressure, the constant volume specific heat along this path is c v (the superscript zero indicates a zero pressure constant volume specific heat). Note that the integration path must be only piecewise continuous; therefore, it can have “kinks.” Since all of our equations of state must have the form pv = RT + f(v, T), fðv,TÞ RT p = + v v   ∂p ∂f R 1 + = v v ∂T ∂T v v and   ∂p ∂f T T T = R + v v ∂T ∂T v v then,    ∂p ∂f f ∂f f RT T RT T T −p = + − + = − v v v v v v ∂T ∂T ∂T v v v so that Eq. (11.19) now gives Z Z  Z Z  T v T v ∂p ∂f f T 0 0 u−u = c dT+ T −p dv = c dT+ − dv 0 v v v v ∂T ∂T T v =∞ T v =∞ 0 0 v 0 0 v378 CHAPTER 11: More Thermodynamic Relations Note that following a path of constant T for the second integral in this equation is very logical since f depends on only v and T and we are integrating over v. 2 For example, if we had an equation of state of the form pv = RT + αT /v, then 2 RT αT p = + 2 v v  ∂p R 2αT = + 2 v ∂T v v and  2 ∂p RT 2αT T = + 2 v ∂T v v then,   2 2 2 ∂p RT 2αT RT αT αT T −p = + − + = 2 2 2 v v ∂T v v v v Equation (11.19) now gives Z Z  T v 1 1 2 αT 0 −u = c dT+ u dv 1 0 v 2 v T v =∞ 0 0 Z T  1 v 1 1  0 2 = c dT+αT −  v v ∞ T 0 Z T 1 2 αT 0 1 = c dT+ v v T 1 0 0 and if the zero pressure specific heat c is constant over the temperature range T to T , this equation reduces to 0 1 v 2 αT 0 1 u −u = c ðT −T Þ− 1 0 1 0 v v 1 A similar equation can be easily developed for a second state, and we can then combine them to produce an equation for the change in specific internal energy between these two states for this material as  2 2 T T 0 2 1 u −u = c ðT −T Þ−α − 2 1 2 1 v v v 2 1 and the reference state values have completely cancelled out. 11.7 CONSTRUCTING TABLES AND CHARTS We are now able to use Eqs. (11.19), (11.22), (11.24), and (11.26) to construct thermodynamic tables and charts. The construction of thermodynamic tables and charts like the ones in Thermodynamic Tables to accompany Modern Engineering Thermodynamics require, first ofall, that a great deal of accurate experimental p, v, T,and c (or c)databe v p obtained. These data are reduced to mathematical equations through curve-fitting techniques. The resultant mathe- matical equations are used to derive equations for u, h,and s using the thermodynamic property relations discussed previously. One of thesimplest methods for generating saturation and superheat tables is carriedout asfollows. A. The following four data sets must be developed from appropriate experiments: Data set 1. Saturation temperature and saturation pressure (T , p ). sat sat Data set 2. Pressure, specific volume, and temperature in the superheated vapor region and along the saturated vapor curve (p, v, T). Data set 3. Saturated liquid specific volume (or density) and saturation temperature (v, T ). f sat 0 Data set 4. Low- (or zero) pressure constant volume specific heat, c and temperature T in the superheated v 0 vapor region and along the saturated vapor curve. The superscript 0 is used to denote the fact that the c v values are measured at essentially zero pressure. B. Once these four data sets have been obtained, a mathematical equation is curve fit to each of them to obtain four mathematical equations of the form Curvefit1: p = p ðT Þ (11.31a) sat sat sat11.7 Constructing Tables and Charts 379 Curvefit2: p = pðv,TÞðforsuperheatedandsaturatedvaporÞ (11.31b) Curvefit3: v = v ðT Þ (11.31c) f f sat 0 0 Curvefit4: c = c ðTÞðforsuperheatedandsaturatedvaporÞ (11.31d) v v 0 C. If a very low-pressure reference state (p , v , T ) is chosen such that ðc Þ = c , then Eqs. (11.19), (3.17), and v 0 0 0 0 v (11.24) are used to calculate values for u, h, and s relative to this reference state as   Z Z T v ∂p 0 u = u + c dT+ T −p dv (11.32) 0 v ∂T T v 0 0 v h = u+pv (3.17) and   Z Z T 0 v c ∂p v s = s + dT+ dv (11.33) 0 T ∂T T v 0 0 v where u and s are the internal energy and entropy values of the reference state. Note that these reference 0 0 state properties always cancel out in a typical internal energy change (u – u ) or entropy change (s – s ) 2 1 2 1 calculation, so their values can be arbitrarily chosen and need not be made known to the user of the table or chart. Typically u and s are chosen so as to make h and s zero at the reference temperature T , and T is 0 0 f f 0 0 often takento bethe triple point temperature (see, forexample, thefirst row ofvalues for water in Table C.1a), because the triple pointis a well-defined and easily reproducible referencestate. Therefore, u and s are seldom 0 0 chosentobezerothemselves. Thegeneration ofthe tables can now be carriedout asfollows. 11.7.1 Saturation tables A temperature entry saturation table can be constructed as follows: 1. A temperature T = T is chosen at which the properties are to be determined. sat 2. Next, p is calculated from Eq. (11.31a). sat 3. Equation (11.31b), which must be valid for saturated vapor as well as superheated vapor, is used to calculate v at these p and T values. g sat sat 4. Chose a reference temperature T and assign arbitrary values to u and s . o o o 3 5. The expression for (dp/dT) is determined by differentiating Eq. (11.31a). The values of u , h , and s are sat g g g then calculated from Eqs. (11.32), (3.17), and (11.33) by setting (∂p/∂T) = (dp/dT) . v sat 6. Equation (11.31c) is used to calculate v = v – v at the T value. fg g f sat 7. The remaining saturated liquid properties are determined from the Clapeyron and Gibbs Eqs. (11.17) and (11.10) as follows:   dp h = h −h = h −T v f g fg g sat fg dT sat u = u −u = u −ðh −p v Þ f g fg g fg sat fg and s = s −s = s −h /T f g fg g fg sat This sequence of operations is repeated for a variety of T values, and the compilation of all these results sat gives a temperature entry saturation table like Table C.1a or C.1b in Thermodynamic Tables to accompany Modern Engineering Thermodynamics. Beginning the calculation sequence with a p = p value and calculating the corresponding T value sat sat from Eq. (11.31a) and continuing as just described produces a pressure entry saturation table like Table C.2a or C.2b. 3 Note that Eq. (11.31b) should yield the same values of p and T as Eq (11.31a). However, they both are both empirical equations sat sat and may not yield the same values of (dp/dT) . In this case, Eq. (11.31a) is preferable. sat380 CHAPTER 11: More Thermodynamic Relations 11.7.2 Superheated vapor tables Superheated vapor tables are somewhat easier to construct. 1. Begin by choosing a pair of pressure-temperature (p, T) values and calculate the corresponding specific volume v from Eq. (11.31b). 2. Values are then calculated for u, h, and s from Eqs. (11.32), (3.17), and (11.33) utilizing (∂p/∂T) v determined from Eq. (11.31b). The compilation of v, u, h,and s for each set of p and T values chosen forms a superheated vapor table like Table C.3a or C.3b. Many tables do not list both u and h values, since these properties are simply related to each other through h = u + pv. Therefore, if only one is listed in a table (it is usually h), the other can be easily calculated. 11.8 THERMODYNAMIC CHARTS When accurate values for p, T, v, u, h,and s have been determined for the construction of saturated and superheated property tables, it is a relatively simple task to plot these values to form thermodynamic charts. Two-dimensional plots allow only two independent properties to be plotted, and the remaining properties have to be added to the plot as parametric families of lines representing constant property values (isotherms, isobars, etc.). For example, the Mollier diagram (see Figure 7.15a) has h and s as coordinate axes. This means that all the remaining property information must be displayed as families of lines of constant p, constant T, constant v, and so forth. Because of the large number of variables to choose from and the lack of any standard thermodynamic chart format, the charts found in Thermodynamic Tables to accompany Modern Engineering Thermodynamics have many coordinate axes (h-s, T-s, p-h, p-v, v-u, etc.). Although all thermodynamic tables and charts up to about 1950 were generated from manual calculations, the use of a modern digital computer can substantially reduce the amount of human labor involved. Most of the required software programming is straightforward, simply by following the steps outlined previously. However, oneaspectofthisprocessthatisnotsoobvious involves solving Eq. (11.31b) for v when p and T are known. These equations are often so algebraically complex that v cannot be determined explicitly in terms of p and T. EXAMPLE 11.11 A new substance has the following equations of state corresponding to Eqs. (11.31a) through (11.31d). Here, we just letter the equations (a) through (d) to avoid any confusion. That is, Eq. (11.31a) is just called (a) here.  A 2 p = exp A − (a) sat 1 T sat hi B RT T 2 p = − exp B − (b) 1 2 v v T 1 v = (c) f C +C T 1 2 sat 0 c = D = constant (d) 1 v where A , A , B , B , C , C,and D are all empirical constants. Determine the equations for u , u, h , h, s,and s for this 1 2 1 2 1 2 1 g f g f g f material in the saturated region. Solution For the saturation tables, let A = A – A /T and B = B – B /T, then p = expA and 1 2 sat 1 2 sat  RT T p = − exp½B 2 v v then,   dp A 2 = exp½A 2 dT T sat sat

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