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

6 x 2266 x 1 1711 y 2266 x 281 175 4433

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Unformatted text preview: − x = (1) (1711 − y ) + (226.6 − x ) + 281 + 17.5 443.3 − y − x . . y Fractional purity of iC 4 in the distillate = = 0.95 (2) 2.2 + y + x Combining (1) and (2) to eliminate x, and optimization of P with respect to y gives: P = 0.828 or 82.8 mol% nC4 in the bottoms, x = 6.8 lbmol/h, y = 171.1 lbmol/h Therefore, 100% recovery of iC4 in the distillate maximizes the purity of nC4 in the bottoms. P= Exercise 1.15 Subject: Sequence of two distillation columns, C1- C2, for the separation of alcohols. Given: 500 kmol/h feed of 40% methanol (M), 35% ethanol (E), 15% isopropanol (IP), and 10% normal propanol (NP), all in mol%. Distillate from column C1 is 98 mol% M, with a 96% recovery. Distillate from column C2 is 92 mol% E, with a recovery of 95% based on the feed to column C1. Find: (a) Component flow rates in the feed, distillates and bottoms. (b) Mol% purity of combined IP and NP in the bottoms from column C2. (c) Maximum achievable purity of E in the distillate from column C2 for 95% recovery of E from the feed to column C1. (d) Maximum recovery of E from the feed to column C1 for a 92 mol% purity of E in the distillate from column C2. Assumptions: Because of the sharp separation in column C1, neglect the presence of propanols in the distillate from column C1. Neglect the presence of M in the bottoms from column C2. The distillate from C2 does not contain normal propanol. Analysis: (a) M in distillate from C1 = (0.96)(500)(0.40) = 192 kmol/h Total distillate from C1 = 192/0.98 = 195.92 kmol/h E in distillate from C1 = 195.92 - 192 = 3.92 kmol/h E in feed to C2 = (500)(0.35) - 3.92 = 171.08 kmol/h M in feed to C2 = M in distillate from C2 = (500)(0.40) - 192 = 8 kmol/h E in distillate from C2 = (500)(0.35)(0.95) = 166.25 kmol/h Total distillate from C2 = 166.25/0.92 = 180.71 kmol/h IP in distillate from C2 = 180.71 - 166.25 - 8 = 6.46 kmol/h Block flow diagram: Exercise 1.15 (continued) Analysis: (a) continued Material balance table (all flow rates in kmol/h): Component Stream 1 2 3 M 200 192.00 8.00 E 175 3.92 171.08 IP 75 0.00 75.00 0.00 50.00 NP 50 Total 500 195.92 304.08 4 8.00 166.25 6.46 0.00 180.71 5 0.00 4.83 68.54 50.00 123.37 The assumption of negligible NP in stream 4 is questionable and should be corrected when designing the column. (b) Mol% purity of (IP + NP) in bottoms of C2 = (68.54 + 50.00)/123.37 or 96.08% (c) If the overall recovery of E in the distillate from C2 is fixed at 95%, the maximum purity of E in that distillate occurs when no propanols appear in that distillate. Then, mol% purity of E = 100% x 166.25/(166.25 + 8.0) = 95.41% (d) The maximum recovery of E in the distillate from C2 occurs when E does not appear in the bottoms from C2. Thus, that maximum is 100% x (171.08/175) = 97.76% Exercise 1.16 Subject: Pervaporation for the partial separation of ethanol and benzene Given: 8,000 kg/h of 23 wt% ethanol and 77 wt% benzene. Polymer membrane is selective for ethanol. Permeate is 60 wt% ethanol. Retentate is 90 wt% benzene. F...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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