Ch. 12 a - 1 PGE 312 Physical and Chemical Behavior of...

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1 PGE 312 Physical and Chemical Behavior of Fluids Home Work 4 Solutions Problem 12-6 Given: Pressure = 5psia Temperature = 100°F Compound A- 3lb of 2,2-dimethylbutane Compound B - 2lb of 2,2,4-trimethylpentane Solution: Molecular weight of A = 86lb/lbmol Molecular weight of B = 114lb/lbmol No. of moles of A = 3/86 = 0.0348 moles No. of moles of B = 2/114= 0.0175 moles Total no. of moles in the mixture = 0.0348 + 0.0175 = 0.0523 moles Mole fraction of compound A = 0.0348/0.0523 = 0.665 Mole fraction of compound B = 0.0175/0.0523 = 0.335 ± ² = 1 - ± ³ = 1- 0.4053 = 0.595 n g (lbmol)= ± ² *n = 0.595*0.0523 = 0.03111 moles n L (lbmol)= ± ³ *n = 0.4053*0.0523 = 0.02118 moles m L (lbs) = n L *M L = 0.02118*101.78 = 2.158 lb m G (lbs) = n G *M G = 0.03111*91.3271 = 2.8412 lb Component z i p vi y i = z i / [1+( ´ µ · *(p/p vi -1))] x i M L , lb/lbmol M G , lb/lbmol n i /n y i at different ´ µ L x i = y i *(p/p vi ) 0.5 0.45 0.4053 A 0.6654 9.2 0.8622 0.8374 0.8164 0.4437 38.2375 70.3570 B 0.3346 1.65 0.1661 0.1749 0.1836 0.5563 63.5457 20.9701 Total 1.0282 1.0123 1.0000 1.0000 101.7832 91.3271
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2 Problem 12-8 We know that the bubble point pressure is given by p b = Σ z j p vj ------- (1) Therefore, we iterate for bubble pt. temperature (T B ) using the values of P vi from Fig 2-8. The mole fractions of compounds A and B are 0.665 and 0.335 respectively. The calculations are shown in the table: 1st iteration 2nd iteration 3rd iteration Component z T B 90 T B 85 T B 86 1 0.67 p V1 7.5 p V1 6.7 p V1 6.90 2 0.33 p V2 1.7 p V2 1.4 p V2 1.25 1 ∑z i p vi = 5.59 ∑z i p vi = 4.95 ∑z i p vi = 5.04 The bubble point temperature of the mixture is around 86°F. We know that the dew point pressure of the mixture is given by p d = 1/ Σ (z j /p vj ) ------ (2) Therefore, we iterate for T D once again using values of p vi from Fig. 2-8The mole fractions of compounds A and B are 0.665 and 0.335 respectively. The calculations are shown in the table. 1st iteration 2nd iteration 3rd iteration Component z T D 150 T D 110 T D 115 1 0.67 p V1 23 p V1 11 p V1 12 2 0.33 p V2 5 p V2 2.1 p V2 2.35 1 1/(∑z i p vi ) = 10.51 1/(∑z i p vi ) = 4.59 1/(∑z i p vi ) = 5.10 The dew point temperature is around 115°F .
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3 Problem 12-17 Given: Non-ideal Solution Propane - 25 lb n-butane - 15 lb n-pentane - 20 lb n-hexane - 10 lb Temperature = 160°F Solution: Molecular weight of propane = 44lb/lbmol Molecular weight of n-butane = 58lb/lbmol Molecular weight of n-pentane = 72lb/lbmol
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Ch. 12 a - 1 PGE 312 Physical and Chemical Behavior of...

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