HW_4_Solution

# HW_4_Solution - For the two component mixture 1 and 2 We...

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v1 v2 v1 v2 2 v2 v1 v2 v1 1 v1 1 1 v1 v2 v1 2 v2 v1 v2 1 v2 1 v1 1 2 1 2 2 1 v2 2 v1 1 v2 v1 P P P - P * P P y similarly P P P - P * P P y therefore P P x y relation the From P P P - P x P P P - P x )P x - (1 P x P equation, s Raoult' in for x ng Substituti x - 1 x 1 x x P x P x P equation, s Raoult' From P and P P, know We 2 and 1 mixture component two For the - = - = = - = - = + = = = + + =

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Problem 12-6 0.500 0.400 0.405 Component From Fig 2-8 1 2,2-dimethylbutane 3 0.035 86.18 0.665 9.2 0.8622 0.8140 0.8164 0.4437 38.2353 70.3530 2 2,2,4-trimethylpentane 2 0.018 114.23 0.335 1.65 0.1661 0.1847 0.1836 0.5564 63.5528 20.9724 Total 0.052 1.0282 0.9987 1.0000 1.0000 101.7881 91.3255 P 5 psia 0.595 0.031 lbmol 0.021 lbmol 2.158 2.842 m 5.000 n L bar n L bar n L bar m i , lb n i , lbmol M i , lb/lbmol z i P vi y i y i y i x i M L , lb/lbmol M G , lb/lbmol n i =m i /M i n i /n y i =z i /(1-n L bar *(P/P vi -1)) x i =y i *P/P vi n G bar ( = 1 - n L bar ) n G ( = n G bar * n ) n L ( = n L bar * n) m L ( = n L * M L ) lb mass m G ( = n G * M G ) lb mass lb mass
Problem 12-14 0.6 0.5 0.53 Component from App A Propane 0.65 1.43 0.7931 0.7650 0.7730 0.5405 n-Butane 0.30 0.55 0.2012 0.2129 0.2094 0.3807 n-Pentane 0.05 0.23 0.0163 0.0184 0.0177 0.0787 1.00 1.0106 0.9963 1.0000 1.0000 Gas Composition: Liquid Composition: 0.7730 0.5405 0.2094 0.3807 0.0177 0.0787 Iterating for ∑y i =1 using the equation and the trail values of n

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## This note was uploaded on 04/07/2008 for the course PGE 312 taught by Professor Peters during the Fall '08 term at University of Texas at Austin.

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HW_4_Solution - For the two component mixture 1 and 2 We...

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