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Unformatted text preview: 0.906
0.059 0.000
0.039
0.961 Exercise 13.31
Subject: Calculation of batch rectification of a ternary mixture with a process simulator
Given: Charge of 100 kmol of 45 mol% acetone, 30 mol% chloroform, and 25 mol% benzene.
Batch rectification at 101.3 kPa. Column with the equivalent of 10 theoretical stages, plus a
boiler and a total condenser. Total reflux condition is established by time zero. Use UNIFAC
for liquidphase activity coefficients. Two operation steps, each at a distillate flow rate of 2
kmol/h and a reflux ratio of 10, as follows:
Step 1: 13.3 h
Step 2: 24.2 h
Assumptions: No pressure drop and no holdup
Find: Molefraction compositions and amounts of each of the 3 products.
Analysis: Use the Batch distillation program of Chemcad to make the calculations. The
flowsheet below includes a batch column (1), a time switch (2), and 2 tanks (3 and 4). The following pages are printouts of the input data and the results of the calculations by the
Chemcad program. This is followed by a summary of the results for the three products. Analysis: (continued)
Process Simulator Input Data: Exercise 13.31 (continued) Analysis: (continued)
Process Simulator Results: Exercise 13.31 (continued) Analysis: (continued) Exercise 13.31 (continued) Analysis: (continued) Exercise 13.31 (continued) Summary of Results: Cut
Amount, kmol
Mole fractions:
Acetone
Benzene
Chloroform AcetoneRich
26.6 Slop Cut
48.4 BenzeneRich
25 0.967
0.016
0.017 0.398
0.043
0.559 0.001
0.901
0.098 Note, that a chloroformrich cut is not obtained. Exercise 13.32
Subject: Improving a batch distillation campaign to eliminate or reduce slop cuts.
Given: Conditions of Example 13.11. A charge of 150 kmol of equimolar nhexane (C6), nheptane (C7), and noctane (C8). Batch rectification at 1 atm. Column with the equivalent of 5
theoretical stages, plus a partial reboiler and a total condenser. Vapor boilup rate, V = 100
kmol/h. Initial campaign consists of four operation steps, all at a reflux ratio of 8, to produce
five products:
1. 95 mol% C6, 2. Slop cut one, 3. 90 mol% C7, 4. Slop cut two, 5. 95 mol% C8
Assumptions: No holdup and no pressure drop. Total reflux condition is established by time
zero. Reflux ratio is held constant during each operation step.
Find: (a) A reflux ratio above 8 that will eliminate the second slop cut.
(b) A revised termination specification for the second step that will reduce the amount of
the first slop cut, without failing to meet all three product specifications.
Analysis: Use the Batch distillation program of Chemcad to make the calculations. The
flowsheet for Chemcad is shown below. Note that it includes the batch column (1), a time switch
(2), and 4 tanks (3, 4, 5, and 6). Exercise 13.32 (continued)
Analysis (continued):
(a) A number of reflux ratios can accomplish the goal of eliminating the second slop cut,
which, from Example 13.11 for a reflux ratio of 8 for all 4 steps, is a total of only 4.38 kmols out
of the 150 kmol charge. For only three steps,...
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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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