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Unformatted text preview: − x
=
(1)
(1711 − y ) + (226.6 − x ) + 281 + 17.5 443.3 − y − x
.
.
y
Fractional purity of iC 4 in the distillate =
= 0.95 (2)
2.2 + y + x
Combining (1) and (2) to eliminate x, and optimization of P with respect to y gives:
P = 0.828 or 82.8 mol% nC4 in the bottoms,
x = 6.8 lbmol/h, y = 171.1 lbmol/h
Therefore, 100% recovery of iC4 in the distillate maximizes the purity of nC4 in the
bottoms.
P= Exercise 1.15
Subject: Sequence of two distillation columns, C1 C2, for the separation of alcohols.
Given: 500 kmol/h feed of 40% methanol (M), 35% ethanol (E), 15% isopropanol (IP),
and 10% normal propanol (NP), all in mol%. Distillate from column C1 is 98 mol% M,
with a 96% recovery. Distillate from column C2 is 92 mol% E, with a recovery of 95%
based on the feed to column C1.
Find: (a) Component flow rates in the feed, distillates and bottoms.
(b) Mol% purity of combined IP and NP in the bottoms from column C2.
(c) Maximum achievable purity of E in the distillate from column C2 for 95%
recovery of E from the feed to column C1.
(d) Maximum recovery of E from the feed to column C1 for a 92 mol% purity of
E in the distillate from column C2.
Assumptions: Because of the sharp separation in column C1, neglect the presence of
propanols in the distillate from column C1. Neglect the presence of M in the bottoms
from column C2. The distillate from C2 does not contain normal propanol.
Analysis: (a) M in distillate from C1 = (0.96)(500)(0.40) = 192 kmol/h
Total distillate from C1 = 192/0.98 = 195.92 kmol/h
E in distillate from C1 = 195.92  192 = 3.92 kmol/h
E in feed to C2 = (500)(0.35)  3.92 = 171.08 kmol/h
M in feed to C2 = M in distillate from C2 = (500)(0.40)  192 = 8 kmol/h
E in distillate from C2 = (500)(0.35)(0.95) = 166.25 kmol/h
Total distillate from C2 = 166.25/0.92 = 180.71 kmol/h
IP in distillate from C2 = 180.71  166.25  8 = 6.46 kmol/h
Block flow diagram: Exercise 1.15 (continued)
Analysis: (a) continued
Material balance table (all flow rates in kmol/h):
Component
Stream 1
2
3
M
200
192.00
8.00
E
175
3.92
171.08
IP
75
0.00
75.00
0.00
50.00
NP
50
Total
500
195.92
304.08 4
8.00
166.25
6.46
0.00
180.71 5
0.00
4.83
68.54
50.00
123.37 The assumption of negligible NP in stream 4 is questionable and should be corrected
when designing the column.
(b) Mol% purity of (IP + NP) in bottoms of C2 = (68.54 + 50.00)/123.37 or
96.08%
(c) If the overall recovery of E in the distillate from C2 is fixed at 95%, the
maximum purity of E in that distillate occurs when no propanols appear in that distillate.
Then,
mol% purity of E = 100% x 166.25/(166.25 + 8.0) = 95.41%
(d) The maximum recovery of E in the distillate from C2 occurs when E does not
appear in the bottoms from C2. Thus, that maximum is 100% x (171.08/175) = 97.76% Exercise 1.16
Subject: Pervaporation for the partial separation of ethanol and benzene
Given: 8,000 kg/h of 23 wt% ethanol and 77 wt% benzene. Polymer membrane is
selective for ethanol. Permeate is 60 wt% ethanol. Retentate is 90 wt% benzene.
F...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
 Spring '11
 Levicky
 The Land

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