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

Case 3 same as case 2 except replace reflux with the

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Unformatted text preview: h a reflux ratio of 3.5, from Eq. (7-7), the slope of the rectifying section operating line is, L/V = R/(1 + R) = 3.5/4.5 = 0.778 The q-line is a horizontal line at y = 0.44. For 14 plates with 50% efficiency, the column has the equivalent of 7 equilibrium stages + 1 for the partial reboiler. The McCabe-Thiele construction is shown on the next page, where it is seen that it is possible to obtain the desired distillate composition. (b) and (c) From the McCabe-Thiele diagram, the mole fraction of benzene in the bottoms is xB = 0.24. As a weight fraction, this corresponds to, 0.24(78.11) = 0.211 weight fraction or 21.1 wt% benzene 0.24(7811) + 0.76(92.13) . Compute the distillate rate by overall molar material balances. Overall total mass balance: 158.22 = D + B (1) Overall benzene mass balance: 69.65 = 0.974D + 0.240B (2) Solving Eqs. (1) and (2): D = 43.16 kmol/h B = 115.06 kmol/h By weight, D = 43.16[0.974(78.11) + 0.026(92.13)] = 3,387 kg/h Analysis: (a) (continued) Exercise 7.15 (continued) McCabe-Thiele Diagram Exercise 7.16 Subject: Effect on the separation of A from B by distillation when 3 of 7 theoretical plates rust and drop to the bottom of the column. Given: . Column has 7 theoretical plates + partial reboiler. Saturated liquid feed of 100 kmol/h of 50 mol% A is sent to plate 5 from the top. Distillate contains 90 mol% A. The L/V = 0.75 in the rectifying section. Vapor-liquid equilibrium data. Assumptions: Constant molar overflow. Total condenser. Find: Case 1: Column before the 3 plates rust and drop. (a) Composition of the bottoms product. (b) The L/V in the stripping section. (c) The kmol/h of bottoms product. Case 2: If plates 5, 6, and 7 counted down from the top are lost: (a) Composition of bottoms product. Case 3: Same as Case 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 utility rates are such that L/V and L / V are the same as in Case 1. Then, on the McCabe-Thiele diagram, the val...
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