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HW_5_Solution

# HW_5_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 86.177 0.665 9.2 0.8622 0.8140 0.8164 0.4437 38.2353 70.3530 2 2,2,4-trimethylpentane 2 0.018 114.231 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 12-8 P (psia) = 5 1st iteration 2nd iteration 3rd iteration Component z 90 85 86 1 0.67 7.5 6.7 6.9 2 0.33 1.7 1.4 1.25 1 5.59 4.95 5.04 Therefore, the bubblepoint temper 86 1st iteration 2nd iteration 3rd iteration z 150 110 115 1 0.67 23 11 12 2 0.33 5 2.1 2.35 1 10.51 4.59 5.10 Therefore, the dewpoint temperatu 115 At bubble point conditions, ∑z i P vi = 5. Therefore, iterating for bubble pt. temperature (T B ) using the values of P vi from Fig 2-8 T B T B T B P V1 P V1 P V1 P V2 P V2 P V2 ∑z i P vi = ∑z i P vi = ∑z i P vi = o F At dew point conditions, [1 / (∑z i P vi )] = 5. Therefore, iterating for T

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HW_5_Solution - For the two component mixture 1 and 2 We...

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