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

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Unformatted text preview: on has properties of water. Initial estimate of column diameter is 30 inches. Try 2-inch Intalox saddles packing. Find: (a) (b) (c) (d) (e) (f) (g) Minimum caustic solution-to-air molar flow-rate ratio. Maximum possible concentration of CO2 in caustic solution. Nt at L/V = 1.4 times minimum. Caustic solution rate. Pressure drop per ft of column packed height. NOG Packed height for KGa = 2.5 lbmol/h-ft3-atm. Analysis: Gas flow rate = 5,000(60)/379 = 792 lbmol/h. Therefore, air in entering gas = V' = 0.97(792) = 768 lbmol/h The CO2 in the entering gas = 0.03(792) = 24 lbmol/h The CO2 in the exiting gas = 0.03(24) = 0.72 lbmol/h The CO2 absorbed in the leaving liquid = 24 - 0.72 = 23.28 lbmol/h The mole ratios are Yin = 24/768 = 0.03125, Yout = 0.72/768 = 0.0009375, Xin = 0.0 (a) From Eq. (6-11), L'min = V'K(fraction absorbed) = 768(1.75)(0.97) = 1,304 lbmol/h (b) The maximum possible CO2 concentration in the caustic solution occurs at infinite stages with the minimum liquid rate. The leaving liquid is in equilibrium with the entering gas. Therefore, Xout = Yin/K = 0.03125/1.75 = 0.0179 mol CO2 / mol caustic solution. (c) Let L/V = L'/V' . Then, L = 1.4 (1,304) = 1,826 lbmol/h and L/G = 1,826/768 = 2.38. For this liquid flow rate, a material balance for CO2 is, 768(0.03125 - 0.0009375) = 23.28 = 1,826(Xout - 0.0). Solving, Xout = 0.01275 In the Y-X plot on the next page, the equilibrium line, Y = 1.75X, and the straight operating line, passing through the column Y-X end points {Yin = 0.03125, Xout = 0.01275} and {Yout = 0.0009375, Xin = 0.0} are shown, with the equilibrium stages stepped off, giving Nt = 7.3. (d) From Part (c), caustic rate = 1,826 lbmol/h. (e) Assume the use of 2-inch of ceramic Intalox saddles. From Table 6.8, FP = 40 ft2/ft3. Use Fig. 6.36(a). Assume for the dilute caustic solution, f{ρL} = 1.0 and f{µL} = 1.0. Take MV = 29, ML = 18, ρL = 62.4 lb/ft3, ρV = 29/379 = 0.0765 lb/ft3 Using conditions at the bottom of the column, Exercise 6.31 (conditions) Analysis: (d) and (e) (continued) LM L ρV X= VM V ρ L 1/ 2 (1,849)(18) 0.0765 = (792)(29) 62.4 1/ 2 = 0.051 From Fig. 6.36(a), at flooding, Y = 0.18 Now compute Y for the suggested column diameter of 30 inches = 2.5 ft. From the continuity equation, uo = m/SρV = [(792)(29)/3600]/[3.14(2.5)2/4](0.0765) = 17 ft/s From Table 6.8, the packing factor, FP = 40 ft2/ft3. 2 uo FP ρV 17 2 (40) 0.0765 Y= = = 0.440 , which is much greater than 0.18 g ρH 2O 32.2 62.4 Therefore, a diameter of 30 inches places the operation badly into the flooding region and a pressure drop calculation is meaningless. A larger diameter is necessary. Exercise 6.31 (continued) (f) From Eq. (6-95), with A = L/KV = (1,826)/1.75(768) = 1.36, NOG = Nt = A ln(1/A)/(1- A) = 7.3(1.36) ln (1/1.36)/(1 - 1.36) = 8.5 (g) From Table 6.7, HOG = V/KGaPS, with V = 792 lbmol/h, KGa = 2.5 lbmol/h-ft3-atm, P = 1 atm For column cross sectional area, S, assume 50% of flooding. At flooding, with Y = 0.18, by ratio with the Y of 0.440 for a 2.5 ft...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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