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

No absorption of air find a b c d e f minimum

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Unformatted text preview: 5 = 1.16 lb1/2-s-1-ft-1/2 or 2.83 kg1/2-s-1-m-1/2 and uL = [(526)(17.9)/3,600)]/(2.35)(62.4) = 0.018 ft/s = 0.00547 m3/m2-s Substitution of these values into Eq. (3) gives, kGa = 43.4(2.83)0.7(0.00547)0.45 = 8.6 s-1 Note that this kGa is in concentration units for the driving force. Therefore, from Table 6.7 by analogy to the liquid phase case, HOG = HG = VMV/kGaSρV = (100/3,600)(24.2)/(8.6)(2.35)(0.0618) = 0.54 ft Now, we must correct this value to 1.5-inch metal Pall rings. From Table 6.8, using interpolation when necessary, we have the following characteristics: Packing 1.5-inch Pall rings 2.0-inch Pall rings a, m2/m3 139.4 112.6 Ch 0.644 0.784 ε 0.965 0.951 CV 0.373 0.410 From Eqs. (6-136) to (6-140), aPh/a is proportional to ε1/2/a . Therefore, aPh / a for 1.5-inch Rings 0.965 = aPh / a for 2.0-inch Rings 0.951 1/ 2 112.6 = 0.814 139.4 From Eq. (6-133), if we ignore holdup in the term (ε - hL), then, 1ε a H G is proportional to 1.25 CV a aPh 1 0.965 H G for 1.5-inch Rings 0.341 139.41.25 1 Therefore, = = 1.15 1 0.951 H G for 2.0-inch Rings 0.814 0.410 112.61.25 This ratio should be about the same for HOG . Therefore, for the 1.5-inch rings, HOG = 0.54(1.15) = 0.62 ft. From Eq. (2), column height = 3.46(0.62) = 2.1 ft. Exercise 6.37 Subject: Absorption of acetone from air with water in a packed column. Given: Column operates at 60oF and 1 atm. Entering air is 50 ft3/min at 60oF and 1 atm, containing 3 mol% acetone. 97% of the acetone is to be absorbed. Maximum allowable gas superficial velocity is 2.4 ft/s. Equilibrium equation is Y = 1.75 X (1) Assumptions: No stripping of water. No absorption of air. Find: (a) (b) (c) (d) (e) (f) Minimum water-to-air ratio. Maximum acetone concentration possible in the water. Nt for absorbent flow rate equal to 1.4 times the minimum. Number of overall gas transfer units. Height of packing for Kya = 12.0 lbmol/h-ft3-mole ratio difference. Height of packing as a function of molar flow ratio for constant V and HTU. Analysis: First compute material balance. Total gas in = 50(60)/379 = 7.916 lbmol/h Acetone in entering gas = 0.03(7.916) = 0.237 lbmol/h Air in entering gas = 7.916 - 0.237 = 7.679 lbmol/h Acetone in exiting gas = 0.03(0.237) = 0.007 lbmol/h Acetone in entering liquid = 0.0 lbmol/h Acetone in exiting liquid = 0.237 - 0.007 = 0.230 lbmol/h Xin = 0.0 Yin = 0.03/0.97 = 0.03093 Yout = 0.007/7.679 = 0.000912 (a) For minimum water rate, the exiting liquid will be in equilibrium with the entering air. Therefore, using the equilibrium equation, Eq. (1), Xout = Yin /1.75 = 0.03093/1.75 = 0.01767 Therefore, water rate out and in = 0.230/0.01767 = 13.02 lbmol/h Minimum water-to-air molar ratio = 13.02/7.679 = 1.70 (b) Maximum possible acetone concentration in the liquid is at min. water rate, X = 0.01767 This corresponds to an acetone mole fraction of 0.01767/(1 + 0.01767) = 0.01736. (c) Use Lin = L' = 1.4 Lmin = 1.4(13.02) = 18.23 lbmol/h Using mole ratios with no acetone in entering liquid, and K' = constant...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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