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

Analysis apply kremsers method with stripping factors

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Unformatted text preview: ol/h of nC10 absorbent at 90oF. Absorber operating at 4 atm. K-values in Fig.2.8. Find: Percent absorption for each component for (a) 4 stages, (b) 10 stages, and (c) 30 stages. Analysis: : Apply Kremser's method, with absorption factors, A = L/KV, computed for L = entering absorbent rate = 500 kmol/h, V = entering gas rate = 1,000 kmol/h, and K-values from Fig. 2.8 at the average of the two entering temperatures, (90 + 70)/2 = 80oF and a pressure of 4 atm = 59 psia. The feed gas composition and the resulting K-values and values of A are as follows: Component C1 C2 C3 nC 4 nC 5 Total: f, kmol/h 250 150 250 200 150 1000 K-value 35 7.7 2.2 0.65 0.20 A = L/KV 0.0142 0.065 0.227 0.769 2.50 The fraction of a gas component not absorbed, φA, is given by the Kremser Eq. (5-48), and by material balance the amount absorbed follows: A −1 φ A = N +1 and Percent absorbed = 100 1- φ A A −1 The results for all three cases are as follows: Component C1 C2 C3 nC 4 nC 5 N = 4 stages: φA 0.986 0.935 0.773 0.316 0.0155 % absorbed 1.4 6.5 22.7 68.4 98.4 N = 10 stages: N = 30 stages: % % φA φA absorbed absorbed 0.986 1.4 0.986 1.4 0.935 6.5 0.935 6.5 0.773 22.7 0.773 22.7 0.245 75.5 0.231 76.9 0.0001 99.99 0.0000 100.0 The above results show that as N increases, the percent absorption of the heavier components increases, but the absorption of the lighter components does not increase. However, beyond 10 stages, little change occurs. Exercise 9.24 Subject: Flow rate of stripping steam by the Kremser method Given: Feed of composition below enters a flash drum operating at 150oF and 2 atm. Equilibrium liquid is sent to a 5-equilibrium-stage stripper operating at 2 atm to give 0.5 kmol/h of nC5 in stripper bottoms. Find: Flow of steam needed in the stripper. Analysis: Use Chemcad with the SRK equation of state to flash the feed to determine the liquid feed to the stripper. The result is: Component C1 C2 C3 nC 4 nC 5 nC12 Total: Feed, kmol/h Distillate, kmol/h 13.7 13.6 101.3 98.2 146.9 134.9 23.9 19.1 5.7 3.4 196.7 0.4 488.2 269.6 Bottoms, kmol/h 0.1 3.1 12.0 4.8 2.3 196.3 218.6 The stripping steam enters the stripper at 2 atm and 300oF. For the stripper, the average temperature of the two feeds = (150 + 300) = 225oF. From Fig. 2.8, the K-value of nC5 at 2 atm and 225oF = 3.0. The average stripping factor for n-pentane can be taken as: S = KV/L = 3V/218.6 = 0.0137 V The fraction of nC5 not stripped = φS = 0.5/2.3 = 0.217 Use Kremser Eq. (5-50) to compute the value of S needed for N = 5 theoretical stages. By trial and error, the result from: φS = is S = 0.89. Therefore, S −1 = 0.217 S N +1 − 1 Flow rate of stripping steam = V = S/0.0137 = 0.89/0.0137 = 65 kmol/h Exercise 9.25 Subject: Stripping of a hydrocarbon liquid with superheated steam. Given: 1000 kmol/h of feed liquid at 250oF with composition below. 100 kmol/h of superheated steam at 300oF and 50 psia. Stripper operates at 50 psia and has 3 equilibrium stages. Assumptions: Negligible stripping of nC10 and negligible condensation of steam Find: 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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