# Case 3 same as case 2 except replace reflux with the

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Unformatted text preview: e 2, except replace reflux with the same molar flow rate of product containing 80 mol% A: (a) Composition of distillate. (b) Composition of bottoms. Analysis: Case 1: Apply the McCabe-Thiele method in terms of component A, which is more volatile than B. The rectifying section operating line passes through [0.90, 0.90] with a slope of 0.75. The q-line is vertical through x = 0.50. Step off 4 stages in the rectifying section. Then, by trial and error, find an xB with a corresponding stripping section operating line that gives 4 equilibrium stages in the stripping section. The result is shown on the following page, where: (a) Bottoms contains 7 mol% A and 93 mol% B. (b) The slope of the stripping section operating line from the coordinates of the line is: L / V = {[0.90 - 0.75(0.90 - 0.50] - 0.07}/(0.50 - 0.07) = 1.23 (c) By material balances, F = D + B and FxF = 50 = 0.9D + 0.07B. Solving these two equations, distillate flow rate = 51.8 kmol/h and bottoms flow rate = 48.2 kmol/h Case 2: We now have 4 equilibrium stages and a partial reboiler, with the feed being sent to the reboiler. Assume that uti...
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## This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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