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

In eq 1 for all three cases the new flow rates needed

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Unformatted text preview: the ratio of d to b for any component is given by a rearrangement of Eq. (5-66), d1 = b AF From Eq. (5-65), From Eq. (5-59), From just below Eq. (5-64), S B φ AX + 1 / φ SX (1) AC φ SE + 1 / φ AE AF = LF / K FVF (2) AC = LC / KCVC = LC / KC D (3) S B = K BVB / LB = K BVB / B (4) where values of K are assumed to be the same as for Example 5.4 because of the assumption of no change in stage temperatures. Thus, AF , AC , and SB change only by changes in the reflux rate, which changes stagewise vapor and liquid rates as given below. From Eqs. (5-48) and (5-50), A −1 A N +1 − 1 S −1 φ S = N +1 S −1 φA = where A = L/KV and S = KV/L. (5) (6) Analysis: (continued) Exercise 5.21 (continued) φAX and φSX pertain to the stripping (exhauster)section, φAE and φSE pertain to the rectifying (enricher) section, Because the temperatures change over the stripping and rectifying sections, average absorption and stripping factors are used. In Eq. (1), For all three cases, the new flow rates needed in the absorption and stripping factors are determined from the assumption of constant molar overflow, with the following results: Section Condenser Enricher Feed Exhauster Reboiler Flow rates, lbmol/h: L0 = 1,000 (a) L0 = 1,500 Vapor Liquid Vapor Liquid 530 1,000 530 1,500 1,530 1,000 2,030 1,500 1,530 1,000 2,030 1,500 730 1,000 1,230 1,500 730 270 1,230 270 (b) L0 = 2,000 (c) L0 = 2,500 Vapor Liquid Vapor Liquid 530 2,000 530 2,500 2,530 2,000 3,030 2,500 2,530 2,000 3,030 2,500 1,730 2,000 2,230 2,500 1,730 270 2,230 270 Assume that as with Example 5.4 and at higher reflux rates, components C1, C2, and nC5 will not distribute to any extent. Therefore, make distribution calculations only for the component of interest, propane. Using the above flow rates for Cases (a), (b), and (c) , the absorption and stripping factors for C3 with L0 = 1,000 lbmol/h become as follows, where, for example, AC for Case (a) is given from Eq. (3) by: AC Factor AC AE SE = 1/AE AF AX SX = 1/AX SB Case (a) = L0 = DKC L0 = 1,000 10.48 1.72 0.581 1.108 1.73 0.577 2.78 L0 D L0 AC D = Example 5.4 (a) L0 = 1,500 15.72 1.94 0.515 1.250 1.54 0.649 4.68 1,500 = 15.72 1,000 530 10.48(530) (b) L0 = 2,000 20.96 2.08 0.481 1.340 1.46 0.685 6.59 (c) L0 = 2,500 26.20 2.17 0.461 1.397 1.416 0.706 8.49 Analysis: (continued) Exercise 5.21 (continued) Using Eqs. (5) and (6), the following φ factors are determined for the enricher and exhauster from the above values of AE , SE , AX , and SX : Factor φAE φAX φSE φSX L0 = 1,000 0.0289 0.028 0.435 0.439 (a) L0 = 1,500 0.0180 0.0438 0.494 0.379 (b) L0 = 2,000 0.0135 0.0530 0.526 0.351 (c) L0 = 2,500 0.0113 0.0589 0.544 0.336 Using Eq. (1) with the above factors, the following d/b ratios for C3 are obtained: Case Example5.4 (a) (b) (c) d/b for C3 0.0114 0.00522 0.00322 0.00237 As the reflux rate increases, the separation is enhanced. Exercise 5.22 Subject: Effect of numbers of rectification and stripping equilibrium stages on keycomponent distribution in multicomponent distillation by the group method. Given: Result...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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