SOLUTIONS to Problem Set I - EEE 2010

# 5 1227 1336 1489 1583 1704 1871 1532 1642 1791 2014

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Unformatted text preview: .7cm3/mol, we take V1 = 30713.44cm3/mol Similarly for quadratic in V2 gives no real roots so we use next approach ; By iteration, starting with V1ig = 31023.7 cm3/mol, we get V1 = 30780 1 cm3/mol Similarly for V2, starting with V2ig = 564.07cm3/mol, we iterate to get V2 = 241.35 cm3/mol ; After integration and plugging in the values we get W = 13.29kJ/mol d. Solution to Problem 3.38 (SVNA) Units are cm3/mol a R/K Liq. 108.1 R/K Vap. 1499.2 Rackett 94.2 Pitzer 1537.8 b c d e f g h i j k l m n o p q r s t 114.5 122.7 133.6 148.9 158.3 170.4 187.1 153.2 164.2 179.1 201.4 61.7 64.1 66.9 70.3 64.4 67.4 70.8 74.8 1174.7 920.3 717.0 1516.2 1216.1 971.1 768.8 1330.3 1057.9 835.3 645.8 1252.5 1006.9 814.5 661.2 1318.7 1046.6 835.6 669.5 98.1 102.8 109.0 125.4 130.7 137.4 146.4 133.9 140.3 148.6 160.6 53.5 55.1 57.0 59.1 54.6 56.3 58.3 60.6 1228.7 990.4 805.0 1577.0 1296.8 1074.0 896.0 1405.7 1154.3 955.4 795.8 1276.9 1038.5 853.4 707.8 1319.0 1057.2 856.4 700.5 e. The derivation can be referred to from the text book. f. Solution to problem 3.30 (SVNA) B = -242.5cm3/mol; C = 25200cm6/mol2; T = 373.15K P1 = 1bar; P2 = 55bar B’= B/RT; B’ = -7817×10-3 bar-1 C’ = (C-B2) / R2T2; C’ = -3.492 ×10-5 bar-2 Part (a): Solve virial equation for initial V Guess: V1 = RT/P1 Given (P1V1 / RT) =1 ; V1 = Find (V1); V1 = 30780 cm3/mol Solve virial equation for final V Guess V2 = RT/P2 Given (P2V2 / RT) =1 ; V2 = Find (V2); V2 = 241.33 cm3/mol Eliminate P from Eq. (1.3) by the virial equation 1 ; Work = 12.62 kJ/mol Ans. Part (b) Eliminate dV from Eqn. 1.3 by the virial equation in P: ; W = 12.597kJ/mol ; g. Solution to problem 3.32 (SVNA) Tc =282.3K; Pc =50.4bar; ω = 0.087 T =298.15K; Tr =T/Tc; P =12 bar; Pr = P/Pc; Tr = 1.056 Pr = 0.238 (i) From truncated virial equation correlation B = -140 cm3/mol; C = 7200 cm6/mol2; V = RT/P; V = 2066cm3/mol Given: PV / RT =1 ; V = Find (V); V = 1919 cm3/mol; Z = PV / RT; Z = 0.929 Ans. (ii) From generalized Pitzer correlation Equation B0 = -0.304; B1 = 2.262×10-3 V = ZRT/P; Z = 1 + (B0 + ωB1). Pr/Tr; Z = 0.932; (iii) From Redlich/Kwong EOS σ = 1; ε = 0; Ω = 0.08664; -0.5 Table 3.1; α (Tr) = Tr β (Tr, Pr) = (Table 3.1) Eq. (3.54) Eq. (3.53) Calculate Z; Given Eq. (3.52) Ψ = 0.42748 q (Tr) = V = 1924cm3/mol Ans. Guess Z = 0.9 1 , ; . Z = 0.928; , , . V = ZRT / P; , , . . V = 1916.5 cm3/mol Ans. (iv) From Soave/Redlich/Kwong Equation σ = 1; ε = 0; Ω = 0.08664; Ψ = 0.42748 (Table 3.1) / , 1 0.480 1.574 0.176 . 1 (Table 3.1) q (Tr) = Eq. (3.54); Calculate Z; Given Eq. (3.52) Guess Z=0.9 1 , ; β (Tr, Pr) = . Z = 0.928; , q (Tr) = V = ZRT / P; Eq. (3.54); Calculate Z; Given Eq. (3.52) , . (v) The Peng/Robinson Equation 1 √2 ; Ω = 0.07779; σ = 1 +√2; , 1 0.37464 1.54226 Eq. (3.53) , , . . V = 1918 cm3/mol Ans. Ψ = 0.457245 Table 3.1 / 0.26992 ; Table 3.1 . 1 Guess Z=0.9 1 , ; Z = 0.92; β (Tr, Pr) = . , . V = ZRT / P; Eq. (3.53) , , , . . V = 1900.6 cm3/mol Ans. h. Solution to Problem 3.33 (SVNA) (Type out the problem) Tc =305.3K; T = 323...
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## This homework help was uploaded on 04/03/2014 for the course CHME 2020 taught by Professor None during the Spring '08 term at Rensselaer Polytechnic Institute.

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