Question

The feed to Column C1 is given in Figure 1. The separation is performed to produce a distillate of 95 mol% pure isobutane with a recovery in the distillate of 96%. Because of the sharp separation in Column C1 between isobutane and normal butane, asuume all propane goes to the distillate and all pentanes go to the bottoms. ( Propane = 2.2 lbmol/h, Isobutane = 171.1 lbmol/h, Normal butane = 226.6 lbmol/h, Isopentane = 28.1 lbmol/h, Normal pentane = 17.5 lbmol/h ). a) Compute the flow rates in lbmol/h of each component in each of the two product leaving Column 1. b) What is the percent purity of the normal butane bottoms product ? c) If distillate contained no normal butane, what would be the purity of normal butane at the bottoms ?

Answer 1

Feed consists of

A-propane

B-isobutane

C- normal butane

D- isopentane

Component	F(lbmol/h)	feed mole percent	D(lbmol/h)	distillate mole percent	B(lbmol/h)	bottoms mole percent
Propane	2.2	0.514	2.2	1.272	0	0
isobutane	171.1	39.97	0.96(171.1)= 164.256	95	6.844	2.682
Normal butane	226.6	52.94	6.445	3.728	220.155	86.302
isopentane	28.1	6.56	0	0	28.1	11.01
Total	428	100	172.901	100	255.098	100

F = 428 lbmol/h

Amount of isobutane in distillate = 171.1(0.96)=164.256 lbmol/h

D = 164.265/0.95 = 172.901 lbmol/h

F = D+B

B = F-D

B = 428-172.901 =255.098

B)

From above table

Percent purity of normal butane =86.302%

C)

If distillate contains no normal butane and all normal butane is in bottoms then

Total normal butane in bottoms = 220.155+6.445 = 226.6 lbmol/h

The total B = 255.098+6.445 = 261.543 lbmol/h

mole fraction of normal butane in bottoms =

(226.6/261.543)= 0.86639 = 86.639 %

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