hw4key - HOMEWORK# 4 SOLUTION KEY 12-6. Mass of 2,2...

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Unformatted text preview: 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) = (0....
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hw4key - HOMEWORK# 4 SOLUTION KEY 12-6. Mass of 2,2...

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