Unformatted text preview: antaneous distillate
accumulator = 0.04
= 0.80
Stop tolerance
0.0001
0.0001
0.0001
Case 2: Plate column:
Chemcad input data same as for Case 1 except:
Condenser holdup = 1 lbmol
Stage holdup = 0.8 lbmol per stage (total of 8 lbmol for 10 stages)
Case 3: Packed column:
Chemcad input data same as for Case 1 except:
Condenser holdup = 1 lbmol
Stage holdup = 0.2 lbmol per stage (total of 2 lbmol for 10 stages) Exercise 13.26 (continued)
Analysis:
All 3 cases converged. The results are as follows for the contents of tanks 3, 4, and 5
after Steps 1, 2, and 3, respectively, and the final residue:
Case 1: No holdups:
Vessel
Tank 3 Tank 4
Tank 5 Final residue
Amount, lbmol
32.9987 7.6664 27.1655
32.1696
Mole fractions:
C6
0.9487 0.3537
0.0362
0.0000
C7
0.0513 0.6460
0.9231
0.1019
C8
0.0000 0.0003
0.0407
0.8981
Operation time, hr
3.96
0.92
3.26
8.14
Case 2: Plate column:
Vessel
Tank 3
Amount, lbmol
34.9987
Mole fractions:
C6
0.9478
C7
.0.0522
C8
0.0000
Operation time, hr
4.20
Case 3: Packed column:
Vessel
Tank 3
Amount, lbmol
33.4988
Mole fractions:
C6
0.9497
C7
.0.0503
C8
0.0000
Operation time, hr
4.02 Tank 4
3.4999 Tank 5
30.9987 Final residue
21.5030 0.2859
0.7136
0.0005
0.42 0.0267
0.9136
0.0417
3.72 0.0000
0.0244
0.9756
8.34 Tank 4
6.6664 Tank 5
28.1655 Final residue
28.6694 0.3334
0.6662
0.0004
0.80 0.0342
0.9255
0.0403
3.38 0.0000
0.0677
0.9323
8.20 The above results show a significant effect of holdup. If the 4 amounts are totaled, Case 1 gives
100.0002 lbmol (since no holdup), Case 2 gives 91.0003 lbmol (since 1+ 9 lbmol of holdup), and
Case 3 gives 97.0001 lbmol (since 1+2 lbmol of holdup). The plate column gives the most
product from Step 1, by far the least from Step 2, and the most from Step 3. The compositions
for Tank 3 (Step 1) are all close. The compositions for Tank 4 (Step 2) are all different, with the
best cumulative purity for C7 being the plate column. The compositions for Tank 5 (Step 3) are
fairly close, while the compositions of the final residue are different, with the best purity for C8
given by the plate column. Total operation times are close, but the time for Step 2 is much lower
for the plate column. This exercise shows the great difficulty in predicting the effect of holdup.
Composition plots for the no holdup case are shown on the following pages. Exercise 13.26 (continued)
Analysis: (continued) No holdup case Exercise 13.26 (continued)
Analysis: (continued) No holdup case Exercise 13.26 (continued)
Analysis: (continued) No holdup case Exercise 13.27
Subject: Calculation of batch rectification of a fourcomponent mixture by a method that
accounts for holdup.
Given: Charge of 100 lbmol of 10 mol% propane (C3), 30 mol% nbutane (C4), 10 mol% npentane (C5), and 50 mol% nhexane (C6). Batch rectification with 50 psia in condenser a
column pressure drop of 2 psia. Column with the equivalent of 8 theoretical stages, plus a partial
reboiler and a total condenser. Vapor boilup rate, V = 40 lbmol/h. Holdup of condenserreflux
drum = 1 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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