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

14 1 permeability pm h 2 n h2 l m ph 2 avg 0105303

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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 = Inside-out 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 = Inside-out 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) Liquid-liquid extraction (d) Extractive distillation Analysis: In general, membrane separations differ from absorption, stripping, distillation, liquid-liquid extraction, and extractive distillation in the following respects: 1. The separating agent for membrane separations is a semi-permeable membrane. 2....
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