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Unformatted text preview: .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. Distilla...
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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 .
 Spring '06
 selebi
 Distillation, pH, Mass Transfer

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