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

1 ky 005y y 0050111 0007 0480 kmolh ft2 mole

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Unformatted text preview: diameter, need uo at flooding = 17(0.18/0.44)0.5 = 10.9 ft/s At 50% of flooding, u = 5.45 ft/s. By ratio, column diameter = 2.5(17/5.45)0.5 = 4.4 ft Column cross sectional area = S = 3.14(4.4)2/4 = 15.2 ft2. Therefore, HOG = (792)/(2.5)(1)(15.2) = 20.8 ft, which is very high. From Eq. (6-89), lT = HOG NOG = 20.8(8.5) = 177 ft. Given value of KGa seems very low. Exercise 6.32 Subject: Absorption of NH3 from air into water with a packed column. Given: Conditions at a point in the column where P = 101.3 kPa, T = 20oC, 10 mol% NH3 in bulk gas, 1 wt NH3 in bulk liquid, NH3 partial pressure of 2.26 kPa at the interface, and ammonia absorption rate of 0.05 kmol/h-m2. Find: (a) X, Y, Yi , Xi , X*, Y*, KY, KX, kY, and kX in Fig. 6.49. (b) % of mass transfer resistance in each phase. 1 1 H′ = + (c) Verify that KY kY k X Analysis: (a) At the point, as shown in Fig. 6.49, the mole ratios in the bulk are X = (1/17)/(99/18) = 0.0107 and Y = 0.1/0.9 = 0.111. At the interface, Yi = 2.26/(101.3 - 2.26) = 0.0229. Then, using the equilibrium curve in Fig. 6.49, Xi = 0.028 The values of the mole fractions in equilibrium with the bulk values are obtained from the equilibrium curve in Fig. 6.49. Thus, in equilibrium with Y = 0.111 is X* = 0.114. In equilibrium with X = 0.0107 is Y* = 0.007 The given NH3 mass-transfer flux can be written with the following combinations of masstransfer coefficients and driving forces, using Table 6.7 as a guide: N = 0.05 = KY Y − Y * = K X X * − X = kY Y − Yi = k X X i − X (1) From Eq. (1), KY = 0.05/(Y - Y*) = 0.05/(0.111 - 0.007) = 0.480 kmol/h-ft2-mole ratio From Eq. (1), KX = 0.05/(X* - X) = 0.05/(0.114 - 0.0107) = 0.484 kmol/h-ft2-mole ratio From Eq. (1), kY = 0.05/(Y - Yi) = 0.05/(0.111 - 0.0229) = 0.568 kmol/h-ft2-mole ratio From Eq. (1), kX = 0.05/(Xi - X) = 0.05/(0.028 - 0.0107) = 2.89 kmol/h-ft2-mole ratio (b) To determine the relative mass-transfer resistances in each phase, Eq. (1) can be rearranged as in Eq. (6-78): kX (Y − Yi ) 2.89 =− =− = −5.08 kY ( X − Xi ) 0.568 Thus, the lowest coefficient and the highest driving force is that associated with the gas phase. From the ratio of 5.08, the largest mass-transfer resistance is in the gas phase. That resistance is about five times the resistance in the liquid phase, with 84% in the gas phase, 16% in the liquid. (c) From a mole ratio form of Eq. (6-80), 1 1 1 Yi − Y * = + (2) KY kY k X X i − X Therefore, Therefore, H′ = Yi − Y * 0.0229 − 0.007 = = 0.919 0.028 − 0.0107 Xi − X 1 H′ 1 0.919 1 1 + = + = 2.08 compared to = = 2.08 kY k X 0.568 2.89 KY 0.480 Exercise 6.33 Subject: Stripping of ammonia from an aqueous solution with air in a packed column. Given: Aqueous solution of 20 wt% NH3. Exit liquid to contain no more than 1 wt% NH3. Exit gas to be 1,000 ft3/h of a 10 mol% NH3 in air. Equilibrium data in Fig. 6.49. Temperature = 20oC = 68oF. Assumptions: No stripping of water. No absorption of air. Operation at 1 atm. Find: Volume of packing for KGa = 4 lbmol/h-ft3-atm. Analysi...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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