SOLUTIONS to Problem Set I - EEE 2010

1 eq 354 eq 353 calculate z given eq 352

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Unformatted text preview: .15K; Tr =T/Tc; Tr = 1.058 Pc = 48.72 bar; P =15 bar; Pr = P/Pc; Pr = 0.308 ω = 0.100 (i) From truncated virial equation correlation B = -156.7 cm3/mol; C = 9650 cm6/mol2; 3 V = 1791cm /mol Given: PV / RT =1 ; V = Find (V); Z = PV / RT; Z = 0.907 Ans. V = RT/P; V = 1625 cm3/mol; (ii) From generalized Pitzer correlation Equation B0 = -0.302; B1 = 3.517×10-3 V = ZRT/P; Z = 1 + (B0 + ωB1). Pr/Tr; Z = 0.912; (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 = 1634cm3/mol Ans. Guess Z = 0.9 1 , ; . Z = 0.906; , , . V = ZRT / P; , , . . V = 1622.7 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 = 0.907; , q (Tr) = V = ZRT / P; Eq. (3.54); Calculate Z; Given Eq. (3.52) , , . . V = 1624.8 cm3/mol Ans. Ψ = 0.457245 Table 3.1 / 0.26992 ; Table 3.1 . 1 Guess Z=0.9 1 , ; i. , . (v) The Peng/Robinson Equation 1 √2 ; Ω = 0.07779; σ = 1 +√2; , 1 0.37464 1.54226 Eq. (3.53) Z = 0.896; β (Tr, Pr) = . , . V = ZRT / P; Eq. (3.53) , , , . . V = 1605.5 cm3/mol Ans. Solution to problem 3.34 (SVNA) Tc =318.7K; Pc =37.6 bar; ω = 0.286 (i) T =348.15K; Tr =T/Tc; P =15 bar; Pr = P/Pc; From truncated virial equation correlation Tr = 1.092 Pr = 0.399 B = -194 cm3/mol; C = 15300 cm6/mol2; V = RT/P; V = 1930cm3/mol Given: PV / RT =1 ; V = Find (V); V = 1722 cm3/mol; Z = PV / RT; Z = 0.893 Ans. (ii) From generalized Pitzer correlation Equation B0 = -0.283; B1 = 0.02 Z = 1 + (B0 + ωB1). Pr/Tr; Z = 0.899 V = ZRT/P; (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 = 1734cm3/mol Ans. Guess Z = 0.9 1 , ; . Z = 0.888; , , . V = ZRT / P; , , . . V = 1714.1 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 = 0.895; , q (Tr) = V = ZRT / P; Eq. (3.54); Calculate Z; Given Eq. (3.52) , , . . V = 1726.9 cm3/mol Ans. Ψ = 0.45724 Table 3.1 / 0.26992 . 1 ; Table 3.1 Guess Z=0.9 1 , ; j. , . (v) The Peng/Robinson Equation σ = 1 +√2; 1 √2 ; Ω = 0.07779; , 1 0.37464 1.54226 Eq. (3.53) β (Tr, Pr) = . Z = 0.882; , . V = ZRT / P; Eq. (3.53) , , , . . V = 1701.5 cm3/mol Ans. Solution to Problem 3.35(SVNA) T =523.15K; (i) P =1800kPa From truncated virial equation correlation B = -152.5 cm3/mol; C = -5800 cm6/mol2; Given: PV / RT =1 ; V = Find (V); Z = PV / RT; Z = 0.931 Ans. V = RT/P; V = 2250 cm3/mol; (ii) From generalized Pitzer correlation Equation B0 = -0.51; B1 = -0.281 Z = 1 + (B0 + ωB1). Pr/Tr; Z = 0.939; V = ZRT/P; (iii) From Steam Tables...
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