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

5813929 1103 the resulting distillate and bottoms

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Unformatted text preview: nalysis: First, verify the separation for a feed rate of 100 kmol/h. This is shown in the McCabe-Thiele plot below in terms of mole fractions of A, the more volatile component. The equilibrium curve is computed from Eq. (7-3), 3x αx y= = 1 + x (α − 1) 1 + 2 x The given mole fractions are: xF = 0.50 xD = 0.90 xB = 0.20 The rectification section operating line has a slope of 0.75 and passes through point {0.9, 0.9}. The q-line is vertical at x = 0.5. The plot shows almost perfect agreement with the desired separation for a feed rate of 100 kmol/h. Analysis: (continued) Exercise 7.20 (continued) For the base case of F = 100 kmol/h, the material balance equations are F = D + B and xFF = 50 = xDD + xBB = 0.90D + 0.20B. Solving, these equations, along with V = L + D, V/L = 0.75, V = V , and L = L + F , gives the following results: Stream Feed Distillate Bottoms Reflux, L Overhead vapor, V Liquid to reboiler, L Vapor from reboiler, V Flow rate, kmol/h 100.00 42.86 57.14 128.58 171.44 228.58 171.44 Mol% A 50 90 20 Mol% B 50 10 80 When the feed rate is reduced to 25 kmol/h, the reflux rate, L, is maintained at 128.58 kmol/h and the boilup, V , is maintained at 171.44 kmol/h. Therefore by material balances, L = L + F = 128.58 + 25 = 153.58 and B = L - V = 153.58 - 171.44 = -17.86 kmol/h. This is impossible. Therefore, the column can not be operated with the same boilup rate. That rate would have to be reduced to achieve a desired bottoms rate, e. g. the 57.14 % of the feed, as in the base case or 0.5714(25) = 14.29 kmol/h. If this were done, we would now have, V = L - B = 153.58 - 14.29 = 139.29 kmol/h = V. Thus, in the rectifying section, L/V = 128.58/139.29 = 0.923 and in the stripping section, L / V = 153.58/139.29 = 1.103. The resulting distillate and bottoms compositions are determined by positioning the operating lines so that 3 stages + a reboiler can be stepped off. The result is shown below, where the mole fractions of A are 0.93 in the distillate and reflux, 0.18 in the bottoms, and 0.38 in the reboiler vapor. Analysis: (continued) Exercise 7.20 (continued) Exercise 7.21 Subject: Distillation of a saturated vapor of maleic anhydride (A) and benzoic acid (B) under vacuum at 13.3 kPa. Given: Feed contains 90 mol% A and 10 mol% B. Distillate to contain 99.5 mol% anhydride and bottoms to contain 0.5 mol% acid. Vapor pressure data. Assumptions: Raoult's law to compute K-values from vapor pressure data. Constant molar overflow. Find: Number of theoretical plates needed if a reflux ratio, L/D = 1.6 times minimum. Analysis: First compute an equilibrium y, x curve using Raoult's law with the vapor pressure data. Eq. (2-44 ) applies, as well as the sum of the mole fractions in the phases in equilibrium. Thus, Ps T Ps T y y KA = A = A , KB = B = B (1, 2) xA P xB P yA + y B = 1 , xA + x B = 1 (3, 4) Equations (1) to (4) can be reduced to the following equations for the mole fractions of maleic anhydride (A) in terms of the K-values: 1 − KB (5, 6) , y A...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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