Chapter14_A - Chapter 14 - Section A - Mathcad Solutions...

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Φ 2 PT , y 1 , y 2 , () exp B 22 P Psat 2 Py 1 2 ⋅δ 12 + RT := Φ 1 , y 1 , y 2 , exp B 11 P Psat 1 2 2 12 + := δ 12 2B 12 B 11 B 22 := B 12 52 cm 3 mol := B 22 1523 cm 3 mol := B 11 963 cm 3 mol := BUBL P calculations with virial coefficients: (b) y 1 x 1 0.808 = P bubl x 1 85.701kPa = x 1 0.75 := y 1 x 1 0.731 = P bubl x 1 80.357kPa = x 1 0.50 := y 1 x 1 0.562 = P bubl x 1 64.533kPa = x 1 0.25 := y 1 x 1 x 1 γ 1 x 1 Psat 1 P bubl x 1 := P bubl x 1 x 1 γ 1 x 1 Psat 1 1x 1 γ 2 x 1 Psat 2 + := BUBL P calculations based on Eq. (10.5): (a) Psat 2 37.31 kPa := Psat 1 82.37 kPa := γ 2 x 1 exp x 1 2 A 21 2A 12 A 21 1 + := γ 1 x 1 exp 1 x 1 2 A 12 21 A 12 x 1 + := Margules equations: T 55 273.15 + K := A 21 1.42 := A 12 0.59 := 14.1 Chapter 14 - Section A - Mathcad Solutions 539
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y 1 y 2 P kPa 0.812 0.188 85.14 = y 1 y 2 P Find y 1 y 2 , P , () := y 2 1y 1 = y 2 Φ 2 PT , y 1 , y 2 , P 1x 1 γ 2 x 1 Psat 2 = y 1 Φ 1 , y 1 , y 2 , P x 1 γ 1 x 1 Psat 1 = Given x 1 0.75 := y 1 y 2 P kPa 0.733 0.267 79.621 = y 1 y 2 P Find y 1 y 2 , P , := y 2 1 = y 2 Φ 2 , y 1 , y 2 , P 1 γ 2 x 1 Psat 2 = y 1 Φ 1 , y 1 , y 2 , P x 1 γ 1 x 1 Psat 1 = Given x 1 0.50 := y 1 y 2 P kPa 0.558 0.442 63.757 = y 1 y 2 P Find y 1 y 2 , P , := y 2 1 = y 2 Φ 2 , y 1 , y 2 , P 1 γ 2 x 1 Psat 2 = y 1 Φ 1 , y 1 , y 2 , P x 1 γ 1 x 1 Psat 1 = Given x 1 0.25 := y 2 1 := y 1 0.5 := P Psat 1 Psat 2 + 2 := Guess: 540
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Combining this with Eq. (12.10a) yields the required expression ln γ 1 A 12 = It follows immediately from Eq. (12.10a) that: (a) Psat 2 P 1 := x 2 1x 1 := i 2 rows P () .. := y 1 0.000 0.2716 0.4565 0.5934 0.6815 0.7440 0.8050 0.8639 := P 12.30 15.51 18.61 21.63 24.01 25.92 27.96 30.12 := x 1 0.000 0.0895 0.1981 0.3193 0.4232 0.5119 0.6096 0.7135 := Data: Pressures in kPa 14.4 Ans. x 1 0.118 = x 1 y 1 φ 1 P H 1 := Equate the liquid- and vapor-phase fugacities and solve for x1: φ 1 0.827 = φ 1 exp BP RT := By Eq. (11.36): fhat 1 v y 1 φ 1 P = fhat 1 l H 1 x 1 = Assume Henry's law applies to methane(1) in the liquid phase, and that the Lewis/Randall rule applies to the methane in the vapor: B 105 cm 3 mol := H 1 200 bar := y 1 0.95 := P 30 bar := T 200 K := 14.3 541
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Ans. A 12 A 21 H 1 0.348 0.178 51.337 = A 12 A 21 H 1 Find A 12 A 21 , H 1 , () := 0 i H 1 P i x 1 i γ 1 x 1 i x 2 i , A 12 , A 21 , H 1 exp A 12 x 2 i γ 2 x 1 i x 2 i , A 12 , A 21 , Psat 2 + ...
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This note was uploaded on 04/18/2011 for the course CEM 497 taught by Professor Ku during the Spring '11 term at Punjab Engineering College.

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Chapter14_A - Chapter 14 - Section A - Mathcad Solutions...

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