# HW_4_key - HOMEWORK# 4 SOLUTION KEY 12-6. Mass of 2,2...

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HOMEWORK# 4 SOLUTION KEY 12-6. Mass of 2,2 dimethylbutane = 3 lbs mass of 2,2,4 trimethylpentane = 2lbs pressure = 5 psia temperature = 100 ° F Component Mass, lbs, mi Molecular wt, lb/lbmol, Mi Number of moles, ni (=m/M) Mole fraction, zi Vapor pressure, psia P v.pi /P 2,2 dimethylbutane 3 86 0.0349 0.6654 9 1.8 2,2,4 trimethylpentane 2 114 0.0175 0.3346 1.7 0.34 n = 0.0349+0.0175 = 0.0524 lb mole Because this is a binary system, the flash calculation can be done directly without trial and error. Will trial and error work? Yes, but why waste time with trial and error if you can solve the problem directly. Let A = 2,2 dimethylbutane and B = 2,24 trimethylpentane k A = P vA /P = 9/5 = 1.80 k B = P vB /P = 1.7/5 = 0.34 x A = (1-k B )/(k A -k B ) = (1-0.34)/(1.80-0.34) = 0.4521 x B = (k A -1)/(k A -k B ) = (1.8-1)/(1.80-0.34) = 0.5479 y A = k A x A = (1.8)(0.4521) = 0.8138 y B = k B x B = (0.34)(0.5479) = 0.1863 g n = n g /n = (z A -x A )/(y A -x A ) = (0.6654-0.4521)/(0.8138-0.4521) = 0.5897 ( 29 ( 29 g g n n n = =(0.5897)(0.0524) = 0.0309 lb mole of gas L n = n L /n = (y A -z A )/(y A -x A ) = (0.8138-0.6654)/(0.8138-0.4521) = 0.4103 ( 29 ( 29 L L n n n = = (0.4103)(0.0524) = 0.0215 lb mole of liquid M g

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## This note was uploaded on 02/13/2012 for the course PGE 312 taught by Professor Peters during the Spring '08 term at University of Texas.

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HW_4_key - HOMEWORK# 4 SOLUTION KEY 12-6. Mass of 2,2...

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