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

# K values in fig28 find percent absorption for each

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Unformatted text preview: ride equation: By material balances on A and W, B/D = 0.753/0.247 = 3.05 NR = NS zW, F xA, B zA, F xW, D 2 0.206 B D = 0.75 0.25 0.02 0.05 0.206 2 3.05 = 1.08 (c) McCabe-Thiele method: From the McCabe-Thiele graph of Exercise 9.6, shown on the next page, where the total number of minimum stages = 5. N R 4.2 = = 5.25 N S 0.8 The McCabe-Thiele method is the most accurate. The Fenske equation is better than the Kirkbride equation because the Kirkbride equation does not take into account the drastic change in relative volatility from the distillate to the bottoms. Exercise 9.20 (continued) Analysis: McCabe-Thiele method (continued) Exercise 9.21 Subject: Derivation of Eq. (9-37) for the effective stripping factor Given: Equations (5-46) and (5-47) Analysis: By analogy to Eqs. (5-46) and (5-47), the fraction of species in the entering feed liquid that is not stripped is given for just 2 stages by: φS = 1 1 =2 S N S1 + S N + 1 Se + Se + 1 Therefore, Se2 + Se + 1 = S N S1 + S N + 1 : Solving for Se , Se2 + Se − S N S1 + 1 = 0 Se = −1 + 1 + 4 S N S1 + 1 Which is Eq. (9-37). 2 = 0.25 + S N S1 + 1 − 0.5 Exercise 9.22 Subject: Effect of number of stages and pressure on multicomponent absorption Given: Feed gas of 2,000 lbmol/h of light paraffin hydrocarbons at 60oF, and 500 lbmol/h of nC10 absorbent at 90oF. K-values in Fig. 2.8. Find: Separation at following conditions: (a) 6 stages and 75 psia. (b) 3 stages and 150 psia. (c) 6 stages and 150 psia Analysis: Apply Kremser's method, with absorption factors, A = L/KV, computed for L = entering absorbent rate = 500 lbmol/h, V = entering gas rate = 2,000 lbmol/h, and K-values from Fig. 2.8 at the average of the two entering temperatures, (90 + 60)/2 = 75oF and each of the two pressures. 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: Feed, lbmol/h 1660 168 96 52 24 2000 K at 75 psia 29 5.8 1.68 0.48 0.145 A at 75 psia 0.00862 0.0431 0.149 0.521 1.724 K at 150 psia 15 3.1 0.93 0.27 0.08 A at 150 psia 0.0167 0.0806 0.269 0.926 3.13 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 l N = 1 − φ A f A −1 The results for all three cases are as follows: Component C1 C2 C3 nC 4 nC 5 Total: Case (a): Case (b): φA lN, lbmol/h φA lN, lbmol/h 0.991 14.9 0.983 28.2 0.957 7.2 0.919 13.6 0.851 14.3 0.731 25.8 0.484 26.8 0.280 37.4 0.0164 23.6 0.0224 23.5 86.8 128.5 Case (c): φA lN, lbmol/h 0.983 28.2 0.919 13.6 0.731 25.8 0.178 42.7 0.00072 23.98 134.28 The results show that doubling the pressure is much more effective than doubling the number of equilibrium stages. Even Case (c) results in little additional absorption over Case (b) when the stages are doubled for the same pressure. Exercise 9.23 Subject: Effect of the number of stages on the absorption of a light normal paraffin hydrocarbon gas by the Kremser method. Given: 1000 kmol/h of gas at 70oF of composition below, and 500 km...
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## This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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