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Unformatted text preview: each with a reflux ratio of 12, the Chemcad input
data and results are as follows:
Batch column:
No. of stages = 7 (5 in column + reboiler + condenser)
No. of operation steps = 3 (no second slop cut)
Total condenser
Condenser pressure = 1 atm
Condenser holdup = 0
Stage holdup = 0
Method of calculation = Insideout
Operation parameters:
Step
Startup option
Tank no.
First specification
Second specification
Step size, hr
Record frequency
Stop criterion
Stop tolerance 1
Total reflux
3
Reflux ratio = 12
Boilup rate = 100
kmol/h
0.01
10
Mole fraction of C6 in
accumulator = 0.95
0.001 2
Column content
4
Reflux ratio = 12
Boilup rate = 100
kmol/h
0.01
5
Amount in the
accumulator = 6 kmol
0.001 3
Column content
5
Reflux ratio = 12
Boilup rate = 100
kmol/h
0.01
5
Mole fraction of C8 in
residue = 0.95
0.001 Results:
C6 product:
Slop cut 1:
C7 product:
C8 product: 49.15 kmols of 95 mol% C6 and 5 mol% C7
6.05 kmols of 38.6 mol% C6, 60.6 mol% of C7, and 0.8 mol% C8.
46.1 kmols of 90 mol% C7, 2 mol% C6, and 8 mol% C8.
48.7 kmols of 95 mol% C8 and 5 mol% C7. Exercise 13.32 (continued)
Analysis (continued):
(b) A number of changes can accomplish the goal of reducing the amount of the first slop
cut, which, from Example 13.11 for a reflux ratio of 8 for all 4 steps, is a total of 16.67 kmols out
of the 150 kmol charge. For three steps, each with a reflux ratio of 12, the Chemcad input data
and results are as follows, where the amount of the first slop cut is reduced from an amount of 6
kmols in part (a) to 5 kmols.
Batch column:
No. of stages = 7 (5 in column + reboiler + condenser)
No. of operation steps = 3 (no second slop cut)
Total condenser
Condenser pressure = 1 atm
Condenser holdup = 0
Stage holdup = 0
Method of calculation = Insideout
Operation parameters:
Step
Startup option
Tank no.
First specification
Second specification
Step size, hr
Record frequency
Stop criterion
Stop tolerance 1
Total reflux
3
Reflux ratio = 12
Boilup rate = 100
kmol/h
0.01
10
Mole fraction of C6 in
accumulator = 0.95
0.001 2
Column content
4
Reflux ratio = 12
Boilup rate = 100
kmol/h
0.01
5
Amount in the
accumulator = 5 kmol
0.001 3
Column content
5
Reflux ratio = 12
Boilup rate = 100
kmol/h
0.01
5
Mole fraction of C8 in
residue = 0.95
0.001 Results:
C6 product:
Slop cut 1:
C7 product:
C8 product: 49.15 kmols of 95 mol% C6 and 5 mol% C7
5.00 kmols of 41.7 mol% C6, 57.6 mol% of C7, and 0.7 mol% C8.
47.2 kmols of 89.6 mol% C7, 2.6 mol% C6, and 7.8 mol% C8.
48.7 kmols of 94.9 mol% C8 and 5.1 mol% C7. Exercise 14.1
Subject: Differences of membrane separations from some other separation operations.
Find: How membrane separations differ from:
(a) Absorption and stripping
(b) Distillation
(c) Liquidliquid extraction
(d) Extractive distillation
Analysis: In general, membrane separations differ from absorption, stripping, distillation,
liquidliquid extraction, and extractive distillation in the following respects:
1. The separating agent for membrane separations is a semipermeable membrane.
2....
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 Spring '11
 Levicky
 The Land

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