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

# Given 1000 kmolh of rich gas at 70of with in kmolh 250

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Unformatted text preview: 086 0.0385 0.1282 0.4098 1.3160 S 0.0044 φA 0.9914 0.9615 0.8718 0.5916 0.0541 φS 0.9956 υ1 l6 lbmol/h 14.0 6.5 12.3 21.2 22.7 497.8 574.5 υ1 l3 lbmol/h 30.0 12.0 22.6 33.2 22.9 498.9 619.6 lbmol/h 1,646.0 161.5 83.7 30.8 1.3 2.2 1,925.5 N = 3 and P = 150 psia: Component K-value A C1 C2 C3 nC 4 nC 5 nC10 Total 14.0 3.5 1.05 0.33 0.105 0.00055 0.0179 0.0714 0.2381 0.7576 2.3810 S 0.0022 φA 0.9821 0.9284 0.7644 0.3615 0.0443 φS 0.9978 lbmol/h 1,630.0 156.0 73.4 18.8 1.1 1.1 1,880.4 Exercise 6.13 (continued) Analysis: (continued) (c) N = 6 and P = 150 psia: Component C1 C2 C3 nC 4 nC 5 nC10 Total K-value 14.0 3.5 1.05 0.33 0.105 0.00055 A 0.0179 0.0714 0.2381 0.7576 2.3810 S 0.0022 φA 0.9821 0.9284 0.7619 0.2829 0.0032 φS 0.9978 υ1 lbmol/h 1,630.0 156.0 73.1 14.7 0.1 1.1 1,875.0 l6 lbmol/h 30.0 12.0 22.9 37.3 23.9 498.9 625.0 Exercise 6.14 Subject: Absorption of a light hydrocarbon gas by nC10. Given: 1,000 kmol/h of rich gas at 70oF with, in kmol/h, 250 C1, 150 C2, 200 C3, 200 nC4, and 150 nC5. 500 kmol/h of nC10 at 90oF. Absorber operates at 4 atm. K-value of nC10 at 80oF and 4 atm = 0.0014. Assumptions: Using the Kremser method, use V = 1,000 kmol/h and L = 500 kmol/h, with an average temperature of 80oF for K-values from Fig. 2.8 or other source. Find: Percent absorption of each component for 4, 10, and 30 equilibrium stages. Analysis: The K-values and absorption factors, from Eq. (6-15), A = L/KV, are: Component C1 C2 C3 nC4 nC5 K-value 38 7.6 2.25 0.64 0.195 A 0.0132 0.0658 0.2222 0.7813 2.564 For each component in the feed gas, the percent absorption is given by Eq. (6-13) x 100%, A N +1 − A Percent absorbed = N +1 × 100% A −1 (1) Using Eq. (1) with the values of A from the above table, the results are: Component C1 C2 C3 nC 4 nC 5 Percent absorbed: for N = 4 for 1.32 6.58 22.03 65.14 96.30 N = 10 1.32 6.58 22.22 76.58 99.995 for N = 30 1.32 6.58 22.22 78.12 100.00 The number of stages affects only the heavier hydrocarbons that are absorbed to the greatest extent. Exercise 6.15 Subject: Back calculation of tray efficiency from performance data on absorption of propane. Given: Large commercial absorber with Na = 16 actual plates. From Example 6.3, average total liquid rate, L, is 446.7 lbmol/h, average total gas rate, V, is 866.5 lbmol/h. Propane in gas feed = 213.8 lbmol/h. Propane in exit liquid = 43 lbmol/h. P = 93.7 psia. Average T = 111oF. Find: Tray efficiency from performance data. Estimated tray efficiency from DrickamerBradford and O'Connell correlations. Analysis: Fraction of propane absorbed = 43/213.8 = 0.20 From Fig. 2.8, for propane, K-value = 2.0 From Eq. (6-15), A = L/KV = 446.7/(2.0)(866.5) = 0.258 From Eq. (6-13), A N +1 − A 0.258 N +1 − 0.258 Fraction absorbed = 0.20 = N +1 = A −1 0.258 N +1 − 1 (1) Solving Eq. (1), N = Nt = 0.93 From Eq. (6-1), Eo = Nt /Na = 0.93/16 = 0.058 or 5.8% From Drickamer-Bradford correlation, Eq. (6-22), with µ L = 1.4 cP from Example 6.3, Eo , % = 19.2 - 57.8 log µL = 19.2 - 57.8 log(1.4) = 10.8 % Fr...
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