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

206 b d 075 025 002 005 0206 2 305 108 c

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Unformatted text preview: o of which are nonlinear, in 6 unknowns gives the following results: Component A B C zF 0.4 0.2 0.4 xD 0.662 0.307 0.031 xB 0.007 0.039 0.954 (b) Assuming that component A does not distribute, the Underwood Eq. (9-28) for Class 2 applies: α i , D zi , F i αi ,D − θ Analysis: (continued) = 1− q = 0 = 5(0.40) 3(0.20) 1(0.40) + + 5− θ 3− θ 1− θ (1) Exercise 9.19 (continued) There are two roots to Eq. (1). The one of interest is the one between 3 and 1.0, which from a spreadsheet calculation is θ = 1.425. If component A does not distribute, the composition of the distillate by material balances is as follows, if the total distillate rate of 60 kmol/h is maintained: kmol/h: xD xB Component Feed Distillate Bottoms A 40 40 0.0 0.6667 0.0000 B 20 18.3 1.7 0.3050 0.0425 C 40 1.7 38.3 0.0283 0.9575 Total: 100 60 40 1.0000 1.0000 Apply Underwood Eq. (9-29): α i , D xi , D i αi ,D − θ = 5(0.6667) 3(0.3050) 1(0.0283) + + = 1 + Rmin 5 − 1425 . 3 − 1.425 1 − 1.425 = 0.9324 + 0.5810 − 0.0666 = 1447 = 1 + Rmin . Therefore, Rmin = 1.447 - 1.000 = 0.447. Assume this is also the external reflux ratio. Assume constant molar overflow. Then, the reflux rate in the rectifying section = 0.447(60) = 26.8 kmol/h. The reflux rate in the stripping section = 100 + 26.8 = 126.8 kmol/h. The boilup rate in the stripping section = 126.8 - 40 = 86.8 kmol/h. The boilup ratio = 86.8/40 = 2.17. (c) Operating reflux ratio = 1.2(0.0.447) = 0.536. In the Gilliland equation, (9-34), X = (R - Rmin)/(R + 1) = (0.536 - 0.447)/0.536 + 1) = 0.058. Using Eq. (9.34), Y = 0.60 = (N - Nmin)/(N + 1). Solving, N = 14 stages. For feed stage location, apply the Kirkbride equation (9-36): From above, using the Fenske distribution with a material balance, D/B = 1.5. NR = NS zC , F xB, B zB , F xC , D 2 0.206 B D = 0.40 0.20 0.039 0.031 2 1 15 . 0.206 = 1166 . Therefore, of 14 equilibrium stages, (1.166/2.166)(14) = 7.5 stages are in the rectifying section. Assuming a total condenser, the feed stage is stage 7 or 8 from the top. Exercise 9.20 Subject: Comparison of the Kirkbride, Fenske, and McCabe-Thiele methods for determining feed-stage location. Given: A feed mixture of 25 mol% acetone (A) and 75 mol% water (W). Distillate to contain 95 mol% A, and bottoms to contain 2 mol% A. Infinite dilution activity coefficients from Exercise 9.6. Find: Ratio of rectifying to stripping stages by: (a) Fenske equation. (b) Kirkbride equation. (c) McCabe-Thiele method. Analysis: (a) Fenske equation: From Exercise 9.6: zF xD xB αi,W, D αi,W, B Component Acetone 0.25 0.95 0.02 1.48 27.7 Water 0.75 0.05 0.98 1.00 1.0 The relative volatility at feed conditions is also needed. Assume a bubble-point liquid feed. Also from vapor-liquid, y-x table of Exercise 9.6, αι,W = [(0.789/0.250)]/[(0.211/0.750)] = 11.2 From Eq. (9-35), 1/ 2 N R log ( 0.95 / 0.25 ) ( 0.75 / 0.05 ) log[(27.7)(11.2)] 1.756(1.246) = = = 2. 96 1/ 2 1.213(0.6097) N S log ( 0.25 / 0.02 )( 0.98 / 0.75 ) log[(1.48)(11.2)] (b) Kirkb...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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