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

# 28 by extrapolation k 0003 from eq 2 eo 035 therefore

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Unformatted text preview: 2 0.287 4.1+1 − 0.287 From Eq. (6-13, fraction of H2 absorbed = = = 0.286 N 0.287 4.1+1 − 1 AH 2+1 − 1 This exceeds the minimum 20% specified. Actually, this problem is overspecified. The absorber can be designed for 95 mol% purity of H2 in the exit gas or 20% absorption of H2. In this problem, we can not achieve both specifications, unless the plate efficiency for H2 is much less than that for CH4. Check this next. (d) From the O'Connell correlation, Eq. (6-23), with ML = 114, ρL = 44 lb/ft3, and µL= 0.47 cP from Perry's Handbook, log E o = 1587 − 0.199 log C − 0.0896 log C . where, for H2, C = 2 (2) KM L µ L (30)(114)(0.47) = = 36.5 ρL 44 Therefore, from Eq. (2), 2 log E o = 1587 − 0.199 log 36.5 − 0.0896 log 36.5 = 1057 . . which gives Eo = 11.4 % for H2. Similarly, Eo = 21.1 % for CH4, and Eo = 32.5 % for C2H6. (e) From Eq. (6-21), using CH4, Na = Nt /Eo = 4.1/0.211 = 19.4 or say 20 trays. (f) For the stripping of nC8, from Fig. 2.8 by extrapolation, K = 0.003. From Eq. (2), Eo = 0.35. Therefore, if 20 trays, have 20(0.35) = 7 equilibrium stages. From Eq. (6-16), S = KV/L =0.003(2,230)/19,200 = 0.00035. From Eq. (6-14) for 7 stages, S N +1 − S 0.000357 +1 − 0.00035 fraction stripped = N +1 = = 0.00035 0.000357 +1 − 1 S −1 Therefore, 0.00035(19,200) = 7 lbmol/h of nC8 leaves in the exit gas. For H2 , with 20 trays and Eo = 0.114, we have (0.114)(20) = 2.3 equilibrium stages. From Eq. N AH 2+1 − AH 2 0.287 2.3+1 − 0.287 (6-13), with A = 0.287, fraction H2 absorbed = = = 0.275 N AH 2+1 − 1 0.287 2.3+1 − 1 Therefore, we can not meet the specification of less than 20% absorbed. For C2H6, with 20 trays and Eo = 0.325, we have (0.325)(20) = 6.5 equilibrium stages. From Eq. (6-13), with A = (19,200)/(1.75)(2,230) = 4.92, A N +1 − A 4.92 6.5+1 − 4.92 fraction C2H6 absorbed = = = 0.999975 A N +1 − 1 4.92 6.5+1 − 1 Composition of exit gas in lbmol/min is 1,172 H2 , 68 CH4 , and 7 nC8 . H2 purity = 94% (g) Cost of lost oil = 7(114)(60)(7,900)(\$1.00)/(5.86 lb/gal) = 64.6 million \$/year. Exercise 6.17 Subject: Scale-up of absorber using Oldershaw column efficiency. Given: Absorption operation of Examples 6.1 and 6.4 with a column diameter of 3 ft. New column with 11.5 ft diameter. New tray with Oldershaw-column efficiency of 55% (EOV = 0.55). Liquid flow path length from Fig. 6.16. Value of u/DE = 6 ft-1. From Example 6.4, λ = 0.68. Assumptions: Straight operating line and equilibrium line as stated in Example 6.4 Find: Efficiencies EMV and Eo . Analysis: Consider three cases: (a) complete mixing, (b) plug flow, and (c) partial mixing. (a) From Eq. (6-31), EMV = EOV = 0.55. From Eq. (6-37), log [1 + EMV (λ − 1) ] log [1 + 0.55(0.68 − 1) ] = Eo = = 0.50 log λ log 0.68 (b) From Eq. (6-32), 1 1 EMV = eλEOV − 1 = ( e(0.68)(0.55) − 1) = 0.67 λ 0.68 ( ) From Eq. (6-37), log [1 + EMV (λ − 1) ] log [1 + 0.67(0.68 − 1) ] = Eo = = 0.63 log λ log 0.68 (c) To use Fig. 6.16, we need an estimate of the liquid flow ra...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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