ADVANCED THERMODYNAMICS ENGINEERING phần 5 potx

P = (RT/v) + (1/v 2 ) (bRT – a/T n ). (63) Therefore, Pv/RT = Z (v,T) = (1 + b/v) –a/(vRT (n+1) ) = 1 + B(T)/v, (64) Where B(T) = b – a/RT (n+1) . As P → 0, v → ∞, and Z → 1. Solving for v from Eq. (64), v = RT/2P (1 ± (1 + (4 P/RT)(b – a/RT n+1 )) 1/2 ). As P → 0, v ≈ RT/2P (1 ± (1 + 2 P/RT (b–a/RT n+1 ))), only positive values of which are acceptable. Therefore, v = RT/P + (b – a/RT n+1 ), Since RT/P = v 0 , v = v 0 + b – a/(RT n+1 ), for RK, VW and Berthelot (65) Therefore, at lower pressures, (v – v 0 ) P → 0 = b – a/RT n+1 , where n ≥ 0. For example, if n = 0, the volume deviation function has a value equal to (b – a/RT), and is a function of tem- perature. In this case, the real gas volume never approaches the ideal gas volume even when T → ∞. In dimensionless form ′v R – ′v oR, = a ∗ – b ∗ /T R (n+1) , for RK, VW and Berthelot fluids. 0 0.02 0.04 0.06 0.08 0.1 0.12 0.14 0.16 0.18 0.2 1 10 100 1000 T R V R ’-V 0R ’ B erthelot R K V W Figure 11: The deviation function for Berthelot, RK,and VW gases, P R →0 where n = 0, 1, 1/2, a ∗ = (27/64), (27/64), 0.4275, and b ∗ = 0.125, 0.125, 0.08664, respectively, for the Van der Waals, Berthelot and RK equations. The difference between ′v R and ′v oR, are illustrated with respect to T R for these equations as P R → 0 in Figure 11. If a = b = 0 in any of the real gas equations of state, these equations are identical to the ideal gas state equation. 8. Three Parameter Equations of State If v = v c (i.e., along the critical isochore), employing the Van der Waals equation, P = RT/(v c – b) – a/v c 2 , which indicates that the pressure is linearly dependent on the temperature along that isochore. Likewise, the RK equation also indicates a linear expression of the form P = RT/(v c – b) – a/(T 1/2 v c (v c + b)). However, experiments yield a different relation for most gases. Simple fluids, such as argon, krypton and xenon, are exceptions. The compressibility factors calculated from either the VW or RK equations (that are two parameter equations) are also not in favorable agreement with experiments. One solution is to increase the number of parameters. a. Critical Compressibility Factor (Z c ) Based Equations Clausius developed a three parameter equation of state which makes use of experi- mentally measured values of Z c to determine the three parameters, namely P = R T/( v – b ) – a /(T ( v + c ) 2 ). (66) where the constants can be obtained from two inflection conditions and experimentally known value of Z C , critical compressibility factor. b. Pitzer Factor The polarity of a molecule is a measure of the distribution of its charge. If the charge it carries is evenly or symmetrically distributed, the molecule is non–polar. However, for some chemical species, such as water, octane, toluene, and freon, the charge is separated across the Figure 12. Illustration of Pitzer factor estimation. molecule, making it uneven or polar. The compressibility factors for nonsymmetric or polar fluids are found to be different from those determined using two parameter equations of state. Therefore, a third factor, called the Pitzer or acentric factor ω has been added so that the em- pirical values correspond with those obtained from experiments. This factor was developed as a measure of the structural difference between the molecule and a spherically symmetric gas (e.g., a simple fluid, such as argon) for which the force–distance relation is uniform around the molecule. In case of the saturation pressure, all simple fluids exhibit universal relations for P R sat with respect to T R (as illustrated in Figure 14). In Chapter 7 we can derive such a relation using a two parameter equation of state. For instance, when T R = 0.7, all simple fluids yield P R sat ≈ 0.1, but polar fluids do not. The greater the polarity of a molecule, the larger will be its deviation from the behavior of simple fluids. Figure 14 could also be drawn for log 10 P r sat vs. 1/T R as illustrated in Figure 12. The acentric factor ω is defined as ω = –1.0 – log 10 (P R sat ) TR=0.7 = –1 – 0.4343 ln (P R sat ) TR=0.7 . (67) Table A-1 lists experimental values of ω” for various substances. In case they are not listed, it is possible to use Eq. (68). i. Comments The vapor pressure of a fluid at T R = 0.7, and its critical properties are required in or- der to calculate ω. For simple fluids ω = 0. For non-spherical or polar fluids, a correction method can be developed. If the com- pressibility factor for a simple fluid is Z (0) , for polar fluids Z ≠ Z (0) at the same values of T R and P R . We assume that the degree of polarity is proportional to ω. In general, the difference (Z – Z (0) ) at any specified T R and P R increases as ω becomes larger (as illustrated by the line SAB in Figure 13). With these observations, we are able to establish the following relation, namely. (Z (ω,T R ,P R )– Z (0) (P R , T R )) = ωZ (1) (T R , P R ). (68) Evaluation of Z(ω,T R ,P R ) requires a knowledge of Z (1) , w and Z (0) (P R , T R ). c. Evaluation of Pitzer factor, ω i. Saturation Pressure Correlations The function ln(P sat ) varies linearly with T –1 , i.e., ln P sat = A – B T –1 . (69) Using the condition T = T c , P = P c , if another boiling point T ref is known at a pressure P ref , then the two unknown parameters in Eq. (70) can be determined. Therefore, the saturation pressure at T = 0.7T c can be ascertained and used in Eq. (69) to determine ω. ii. Empirical Relations Empirical relations are also available, e.g., ω = (ln P R sat – 5.92714 + 6.0964/T R,BP +1.28862 ln T R,BP – 0.16935 T R,BP )/ (15.2578 – 15.6875/T R,NBP + 0.43577 T R,NBP ), (70) where P R denotes the reduced vapor pressure at normal boiling point (at P = 1 bar), and T R,NBP the reduced normal boiling point. An alternative expression involves the critical compressibility factor, i.e., ω = 3.6375 – 12.5 Z c . (71) Another such relation has the form ω = 0.78125/Z c – 2.6646. (72) 9. Other Three Parameter Equations of State Other forms of the equation of state are also available. a. One Parameter Approximate Virial Equation For values of v R > 2 (i.e., at low to moderate pressures), Z = 1 + B 1 (T R ) P R , (73) where B 1 (T R ) = B (0) (T R ) + ωB (1) (T R ), B (0) (T R ) = (0.083 T R –1 ) – 0.422 T R –2.6 , and B (1) (T R ) = 0.139 – 0.172 T R –5.2 . b. Redlich–Kwong–Soave (RKS) Equation Soave modified the RK equation into the form P = RT/(v–b) – a α (ω,T R )/(v(v+b)), (74) where a = 0.42748 R 2 T c 2 P c –1 , b = 0.08664 RT c P c –1 , and α (ω,T R ) = (1 + f(ω)(1 – T R 0.5 )) 2 , which is determined from vapor pressure correlations for pure hydrocarbons. Thus, f(ω) = (0.480 + 1.574 ω – 0.176 ω 2 ). c. Peng–Robinson (PR) Equation The Peng–Robinson equation of state has the form P = (RT/(v – b)) – (a α(ω,T R )/((v + b(1 + 2 0.5 ))(v + b(1 – 2 0.5 ))), (75) P R2 , T R2 Z ( 1 ) (P R , T R ) S A Z ref w ref P R1 , T R1 B Figure 13: An illustration of the variation in the compressibility factor with respect to the acentric factor. where a = 0.45724 R 2 T c 2 P c –1 , b = 0.07780 RT c P c –1 and α(ω,T R ) = (1 + f(ω) (1 – T R 0.5 )) 2 , f(ω) = 0.37464 + 1.54226 ω – 0.26992 ω 2 . Equation (75) can be employed to predict the variation of P sat with respect to T, and can be used to explicitly solve for T(P,v). 10. Generalized Equation of State Various equations of state (e.g., VW, RK, Berthelot, SRK, PR, and Clausius II) can be expressed in a general cubic form, namely, P = RT/(v–b) – aα (ω,T R )/(T n (v+c)(v+d)). (76) In terms of reduced variables this expression assumes the form P R = T R /( ′v R – b´) – a´α (ω,T R )/(T R n ( ′v R + c´) ( ′v R + d´ ) ), (77) where a´ = a/(P c ′v c 2 T c n ), b´ = b/ ′v c , c´ = c/ ′v c , and d´ = d/ ′v c . Tables are available for the pa- rameters a´ to d´. Using the relation Z = P R ′v R /T R , we can obtain a generalized expression for Z as a function of T R and P R , i.e., Z 3 + Z 2 ((c´ + d´ – b´) P R /T R – 1) + Z (a´α (ω,T R ) P R /T R 2+n – (1 + b´P R /T R ) (c´ + d´)P R /T R + c´d´ P R 2 /T R 2 ) – (a´α (ω,T R ) b´P R 2 /T R (3+n) + (1 + P R b´/T R ) (c´d´P R 2 /T R 2 )) = 0. (78) Writing this relation in terms of ′v R , ′v R 3 P R + ′v R 2 ((c´ + d´ – b´)(P R /T R ) –1) + ′v R ((c´d´ – b´c´ –b´d´)P R – (c´ + d´)T R + a×/T R n ) – P R b´c´d´ – a´α (ω,T R )b´/T R – T R c´d´ = 0. (79) Using this equation along with the relation T R = P R ′v R /Z, the compressibility factor can be obtained as a function of P R and ′v R , i.e., Z (3+n) (a´α(w, T R ) /( ′v R (2+n) P R (1+n) )) (1–b´/ ′v R ) + Z 3 (1+(c´+d´–b´)/ ′v R –(b´/ ′v R 2 )(c´+d´–d´/v R )) – Z 2 (1 + (c´ + d´)/ ′v R –c´d´P R / ′v R + d´/ ′v R 2 ) – Z(b´ c´/ ′v R ) – c´ = 0 (80) where T R in α (w, T R ) expression must be replaced by P R v R ´/Z. Table 2 tabulates values of α, n, a´, b´, c´, and d´ for various equations of state. Table 2: Constants for the generalized real gas equation of state. Berthelot Clausius II PR PR with w RK SRK VW a´ 0.421875 0.421875 0.45724 0.4572 0.42748 0.42748 0.421875 b´ 0.125 -0.02 0.0778 0.0778 0.08664 0.08664 0.125 c´ 0 0.145 0.187826 0.187826 0.08664 0.08664 0 d´ 0 0.375 -0.03223 -0.03223 0 0 0 n 11000.500 f(ω), H 2 O 0.873236 1.000629 Note that Z c is required for Clausius II while ω is required for RKS, PR, f(ω) for H 2 O with ω = 0.344 11. Empirical Equations Of State These equations accurately predict the properties of specified fluid; however, they are not suit- able for predicting the stability characteristics of a fluid (Chapter 10). a. Benedict–Webb–Rubin Equation The Benedict Webb Rubin (BWR) equation of state which was specifically devel- oped for gaseous hydrocarbons, has the form P = RT/v + (B 2 RT–A 2 –C 2 /T 2 )/v 2 + (B 3 RT–A 3 )/v 3 + A 3 C 6 /v 6 + (D 3 /(v 3 T 2 ))(1+E 2 /v 2 ) exp(–E 2 /v 2 ) (81) The eight constants in this relation are tabulated in the literature. This equation is not recom- mended for polar fluids. Table A-20A lists the constants. b. Beatie – Bridgemann (BB) Equation of State This equation is capable representing P-v-T data in the regions where VW and RK equations of state fail particularly when ρ < 0.8 ρ c . It has the form P v 2 = R T ( v + B 0 (1- ( b / v )) (1- c/( v T 3 ))- (A 0 / v 2 )(1-(a/ v )). Table A-20B contains several equations and constants. c. Modified BWR Equation The modified BWR equation is useful for halocarbon refrigerants and has the form P = n= ∑ 1 9 A n (T)/v n + exp(–v c 2 /v 2 ) n= ∑ 10 15 A n (T)/v (2n –17) . (82) Figure 14: Relation between pressure and volume for compres- sion/expansion of air (from A. Bejan, Advanced Engineering Ther- modynamics, John Wiley and Sons., 1988, p 281). d. Lee–Kesler Equation of State This is another modified form of the BWR equation which has 12 constants and is applicable for any substance. This relation is of the form P R = (T R / ′v R ) (1+A/ ′v R +B/ ′v R 2 +C/ ′v R 5 +(D/ ′v R )(β+γ/ ′v R 2 )exp(–γ/ ′v R 2 )), (83a) Z = P R ′v R /T R = 1+A/ ′v R +B/ ′v R 2 +C/ ′v R 5 +(D/ ′v R )(β+γ/ ′v R 2 )exp(–γ/ ′v R 2 ), (83b) where A = a 1 – a 2 /T R – a 3 /T R 2 – a 4 /T R 3 , B = b 1 – b 2 /T R + b 3 /T R 3 , C = c 1 + c 2 /T R , and D = d 1 /T R 3 . The constants are usually tabulated to determine Z (0) for all simple fluids and Z (ref) for a refer- ence fluid, that is usually octane (cf. Table A-21). Assuming that Z (ref) – Z (0) = ω Z (1) , (83c) Z (1) can be determined. A general procedure for specified values of P R and T R is as follows: solve for v R ´ from Eq. (83a) with constants for simple fluids and use in Eq. (83b) to obtain Z (0) . Then repeat the procedure for the same P R and T R with different constants for the reference fluid, obtain Z (ref) , and determine Z (1) from Eq.(83c). The procedure is then repeated for different sets of P R and T R . A plot of Z (0) is contained in the Appendix and tabulated in Table A–23A. The value of Z (1) so determined is assumed to be the same as for any other fluid. Tables A-23A and A-23B tabulate Z (0) and Z (1) as function of P R and T R . e. Martin–Hou The Martin–Hou equation is expressed as P = RT/(v – b) + j= ∑ 2 5 F j (T)/(v – b) j + F 6 (T)/e Bv , (84) where F i (T) = A i + B i T + C i exp (–KT R ), b, B and F j are constants (typically B 4 = 0, C 4 = 0 and F 6 (T) = 0). This relation is accurate within 1 % for densities up to 1.5 ρ c and temperatures up to 1.5T c . 12. State Equations for Liquids/Solids a. Generalized State Equation The volume v = v (P,T), and dv = (∂v/∂P) T dP + (∂v/∂T) P dT, i.e., dv = (∂v/∂P) T dP + (∂v/∂T) P dT. (85a) We define β P = (1/v)(∂v/∂T) P , (85b) β T = –(1/v) (∂v/∂P) T , (85c) κ T = 1/(β T P) = (–v/P) (∂P/∂T) T (85d) where β P , β T and κ T are, respectively, the isobaric expansivity, isothermal compressibility, and isothermal exponent. The isobaric expansivity is a measure of the volumetric change with re- spect to temperature at a specified pressure. We will show in Chapter 10 that β T >0 for stable fluids. Upon substituting these parameters in Eq. (86a), dv = vβ P dT – vβ T dP, or d(ln v) = β P dT – β T dP. If β P and β T are constant, the general state equation for liquids and solids can be written as ln(v/v ref ) = β P (T – T ref ) – β T (P – P ref ). (86) This relation is also referred to as the explicit form of the thermal equation of state. In terms of pressure, the relation P = P ref + (β P /β T )(T – T ref ) – ln (v/v ref )/( β T v ref ), (87) is an explicit, although approximate, state equation for liquids and solids. Both Eqs. (87) or (88) can be approximated as (v–v ref )/v ref = β P (T – T ref ) – β T (P – P ref ). (88) Solving the relation in terms of pressure P = P ref + (β P /β T ) (T – T ref ) – (v–v ref )/( β T v ref ), (89) which is an explicit, although approximate, state equation for liquids and solids. The pressure effect is often small compared to the temperature effect. Therefore, Eq. (89) can be approximated in the form ln(v/v ref ) ≈ β P (T – T ref ). (90) In case β P (T – T ref ) « 1, then v/v ref = (1 + β P (T – T ref )). (91) which is another explicit, although approximate, state equation for liquids and solids Copper has the following properties at 50ºC: v, β P , and β T are, respectively, 7.002×10 –3 m 3 kmole –1 , 11.5×10 –6 K –1 , and 10 –9 bar –1 . Therefore, heating 10 kmole of the sub- stance from 50 to 51ºC produces a volumetric change that can be determined from Eq.(87) as 7.002×10 –3 × 10 × 11.5×10 –6 = 805 cm 3 . If a copper bar containing 10 kmole of the substance is vertically oriented and a weight is placed on it such that the total pressure on the mass equals 2 bar, the volume of the copper will reduce by a value equal to –7.002×10 –3 × 10 × 0.712×10 –9 = –0.05 mm 3 . Therefore, changing the state of the 10 kmole copper mass from 50ºC and 1 bar to 51ºC and 2 bars, will result in a volumetric change that equals 805 – 0.05 = 804.95 mm 3 . For solids β P is related to the linear expansion coefficient α. The total volume V ∝ L 3 , and β P = 1/V(∂V/∂T) P = 1/L 3 ∂(L 3 )/∂T = (3/L) ∂L/∂T = 3α, (92) where α = (1/L) (∂L/∂T) P . g. Example 7 v = 0.00101 m 3 kg –1 , and c p = 4.178 kJ kg –1 K –1 . Solution Since β P = 44.8×10 –6 bar –1 and dv = –β P dP v, ln v 2 /v = –β T (P 2 – P 1 ) = – 0.00268, i.e., v 2 /v = 0.997. Now, v 2 = 0.997×0.00101 = 0.001007 m 3 kg –1 , so that v 2 – v = 0.001007 – 0.001010 = 0.000997 m 3 kg –1 . pressible substance, and assume that at 30ºC, β P = 2.7×10 –4 K –1 , β T = 44.8×10 –6 bar –1 , change in volume, and work required to compress the fluid. Treat water as a com- Water is compressed isentropically from 0.1 bar and 30ºC to 60 bar. Determine the δw = –vdP (for a reversible process in an open system). ∴ δw = – v (dP/dv) dv = (1/β T ) dv. Integrating this expression, w = (1/β T )(v 2 – v . ) = 100 kPa bar –1 ×(0.001007–0.00101)÷44.8×10 –6 = –6.76 kJ kg –1 . h. Example 8 diator? Solution Since d ln v = β P dT – β T dP and the volume is constant, dP/dT = β P /β T = 2.7×10 –4 K –1 /44.8×10 –6 bar –1 = 6.03 bar K –1 . Assuming that β T and β P are constants, ∆P = 6.03×65 = 391 bar. b. Murnaghan Equation of State If we assume that the isothermal bulk modulus B T (= 1/β T ) is a linear function of the pressure, then B T (T,P) = (1/β T ) = –v(∂P/∂v) T = B T (T,0) + αP (93) where α = (∂B T /∂P) T . Therefore, ∂P β T (1,0)/(1 + αP β T (1,0)) = –dv/v. (94) Integrating, and using the boundary condition that as P → 0, v → v 0 , we obtain the following relation v/v 0 = 1/(1 + (αPβ T (1,0)) (1/ α ) , i.e., (95) P(T,v) = ((v 0 /v) α – 1) (1/ αβ T (1,0)). (96) c. Racket Equation for Saturated Liquids The specific volume of saturated liquid follows the relation given by the Racket equation, namely, vvZ fcc T R = ( – ) . 1 0 2857 . (97) d. Relation for Densities of Saturated Liquids and Vapors. If ρ f denotes the saturated liquid density, and ρ g the saturated vapor density, then ρ Rf = ρ f /ρ c = 1 + (3/4)(1 – T R ) + (7/4)(1– T R ) 1/3 , and (98) ρ Rg = ρ g /ρ c = 1 + (3/4)(1 – T R ) – (7/4)(1– T R ) 1/3 . (99) These relations are based on curve fits to experimental data for Ne, Ar, Xe, O 2 , CO, and CH 4 . It is also seen that ρ Rf – ρ Rg = (7/2)(1 – T R ) 1/3 . (100) 90ºC. Assuming that the radiator is rigid, what is the final water pressure in the ra- provision for the reservoir. The radiator water temperature increases from 25ºC to A defective radiator does not have a pressure relief valve and there is no drainage At low pressures, ρ Rf ≈ (7/2)(1 – T R ) 1/3 since ρ Rf >> ρ Rg In thermodynamics, ρ Rf – ρ Rg is called order of parameter. If ρ Rg is known at low pressures (e.g., ideal gas law), then ρ Rf can be readily determined. Another empirical equation follows the relation ρ R,f = 1 + 0.85(1–T R ) + (1.6916 + 0.9846ψ)(1–T R ) 1/3 (101) where ψ ≈ ω. e. Lyderson Charts (For Liquids) Lyderson charts can be developed based on the following relation, i.e., ρ R = ρ/ρ c = v c /v. (102) The appendix contains charts for ρ R vs. P R with T R as a parameter. In case the density is known at specified conditions, the relation can be used to determine P c , T c and ρ c . Alternatively, if the density is not known at reference conditions, the following relation, namely, ρ/ρ ref = v ref /v = ρ R /ρ R,ref (103) can be used. f. Incompressible Approximation Recall that liquid molecules experience stronger attractive forces compared to gases due to the smaller intermolecular spacing. The molecules are at conditions close to the lowest potential energy where the maximum attractive forces occur. Therefore, any compression of liquids results in strong repulsive forces that produce an almost constant intermolecular dis- tance. This allows us to use the incompressible approximation, i.e., v = constant. D. SUMMARY This chapter describes how some properties can be determined for liquids, vapors, and gases at specified conditions, e.g., the volume at a given pressure and temperature. Com- pressibility charts can be constructed using the provided information and fluid characteristics, such as the Boyle temperature, can be determined. The relations can be used to determine the work done as the state of a gas is changed. Various methods to improve the predictive accu- racy are discussed, e.g., by introducing the Pitzer factor. State equations for liquids and solids are also discussed. E. APPENDIX 1. Cubic Equation One real and three imaginary solutions are obtained for Z when T R >1. However, when T R <1, we may obtain one to three real solutions. The following method is used in spreadsheet software to determine the compressibil- ity factor. Consider the relation Z 3 + a 2 Z 2 + a 1 Z + a 0 = 0. Furthermore, let α = a 2 2 /9 – a 1 /3, β= –a 2 3 /27 + a 1 a 2 /6 – a 0 /2, and γ = α 2 - β 3 . [...]... 3 85 K, and Pc = 41.2 bar Therefore a =208 .59 bar (m3 kmole–1) 2 K1/2, and b = 0.06731 m3 kmole–1 The molecular weight M = 120.92 kg kmole–1, and a = a /M2 = 208 .59 bar (m3 kmole–1) 2K1/2÷120.92 2(kg kmole–1)2 = 1.427 k Pa (m3 kg–1) 2 K1/2, and b = b /M = 0 .55 7×10–3 m3 kg–1 Since, R = 8.314 ÷ 120.92 = 0.06876 kJ kg–1 K–1, s2 – s1 = 0.06876 ln[(0.0126 – 0.00 055 7) ÷ (0.0194 – 0.00 055 7)] + (1/2){172 .5. .. 0.37780 .5/ tan (2×27.84) + 1/3 = 0.8392 + 0.333 = 1.17 25 For the third case, a× = 0.3204, b× =0. 056 93, a1 =.2602, a0 = -0.01824, and α = 0.02437, β = 0.002789, and γ = -6.78×10-6 Case II applies, and there are three roots to the equation cos φ = β/α1 .5 = 0.7329, i.e., φ = 42.87 Z1 = 2α0 .5 cos(φ/3) + 1/3 = 2 × 0.024370 .5 cos(42.87/3) + 1/3 = 0.3122 × 0.9691 + 1/3 = 0.30 25 + 0.333= 0.6 359 Z2 = 2α0 .5 cos(φ/3... + γ0 .5) 1/3 + (α – γ0 .5) 1/3 + 1/3 = (0.02481 + 0.00021 450 .5) 1/3 + (0.02481 – 0.00021 450 .5) 1/3 + 1/3 = 0.340 + 0.2166 + 0.333 = 0.8899 In the second case, a× = 2.7109, b× = 0.722, a1 =1.4668, a0 = -1. 956 9 so that α = -0.3778, β =0.7709, and γ = 0.6482 Hence, Case Ib is applicable tan φ = (-α)1 .5/ β = 0.3012, i.e., φ = 16.76 tan θ = (tan (φ/2))1/3 = 0.4366, i.e., θ = 27.84 Therefore, Z = 2 (–α)0 .5/ tan(2θ)... 0 .55 7×10–3 m3 kg–1 Since, R = 8.314 ÷ 120.92 = 0.06876 kJ kg–1 K–1, s2 – s1 = 0.06876 ln[(0.0126 – 0.00 055 7) ÷ (0.0194 – 0.00 055 7)] + (1/2){172 .5 ÷ (3331 .5 0.00 055 7)} ln [0.0126 (0.0194 + 0.00 055 7) ÷ {0.0194 × (0.0126+ 0.00 055 7)}] = – 0.06876 × 0.448 – 0.211 × 0.014 95 = –0.03396 kJ kg–1 K–1 4 Fourth Maxwell Relation By subtracting d(Ts) from both sides of Eq (16) and using the relation g = h – Ts, we obtain... α > 0 There is one real root for this case, i.e Z = Z = (α + γ 0 .5) 1/3 + ( α – γ 0 .5) 1/3 + 1/3 ii Case Ib: α < 0 Again, only one real root exists If tan ϕ = (–p)1 .5/ q, tan θ = (tan(ϕ/2))1/3 if ϕ > 0, and –(tan(–ϕ/2))1/3 if ϕ . SRK VW a´ 0.4218 75 0.4218 75 0. 457 24 0. 457 2 0.42748 0.42748 0.4218 75 b´ 0.1 25 -0.02 0.0778 0.0778 0.08664 0.08664 0.1 25 c´ 0 0.1 45 0.187826 0.187826 0.08664 0.08664 0 d´ 0 0.3 75 -0.03223 -0.03223. relations are also available, e.g., ω = (ln P R sat – 5. 92714 + 6.0964/T R,BP +1.28862 ln T R,BP – 0.169 35 T R,BP )/ ( 15. 257 8 – 15. 68 75/ T R,NBP + 0.4 357 7 T R,NBP ), (70) where P R denotes the reduced. α 3 = 0.00021 45. Since, γ > 0 and α > 0, Case Ia is applicable, and Z = (α + γ 0 .5 ) 1/3 + (α – γ 0 .5 ) 1/3 + 1/3 = (0.02481 + 0.00021 45 0 .5 ) 1/3 + (0.02481 – 0.00021 45 0 .5 ) 1/3 + 1/3 =