Separation Process Principles- 2n - Seader & Henley - Solutions Manual

Of stages 7 5 in column reboiler condenser no of

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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 liquid-phase 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: Mole-fraction 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 Acetone-Rich 26.6 Slop Cut 48.4 Benzene-Rich 25 0.967 0.016 0.017 0.398 0.043 0.559 0.001 0.901 0.098 Note, that a chloroform-rich 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 n-hexane (C6), nheptane (C7), and n-octane (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.

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