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

Therefore can not compute hog from eq 6 89 lt hognog

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Unformatted text preview: quid mole fraction, as given by y* = Kxb . Solving Eq. (4) for the driving vapor and liquid phase driving forces, Adding Eqs. (5) and (6) to eliminate yi and solving for r and equating to the last term in Eq. (4), r= yb − y* = 1 K + kG acG k L acL Solving Eq. (7) for KGa, KG a = yb − y * 1 K G acG 1 KcG 1 + kG a k L acL The total gas concentration is obtained from the ideal gas law, (7) (8) Analysis: (d) continued) xercise 6.26 (continued) cG = P 2(101.3) = = 81.8 mol/m3 −3 RT 8.314 × 10 (298) ρ L 106 g / m3 = = 55,500 mol/m3 ML 18.02 1 1 KG a = = = 0.22 s-1 1 150(81.8) 1.25 + 3.29 + 0.80 0.067(55,500) This is with a concentration driving force. cL = The liquid phase controls with a relative resistance of 3.29 compared to 1.25 for the gas phase. To obtain KGa with a partial pressure driving force, we write the rate of mass transfer as, r = K G acG ( yb − y* ) = ( K G ) p aP ( yb − y* ) (9) where KGa is with the concentration driving force = 0.22 s-1 and (KG)pa is with the partial pressure driving force. Solving Eq. (9), and applying the ideal gas law in the form, cG=P/RT, (K G ) p a = K G a cG K G a 0.22 = = = 0.089 mol/kPa-m3-s P RT (8.314 × 10−3 )(298) (e) From Table 6.7, with a partial pressure driving force, H OG = G (K G ) p aPS However, the cross sectional area of the tower, S, is not known. Therefore, can not compute HOG From Eq. (6-89), lT = HOGNOG . Therefore, can not compute the height because HOG is unknown. However, we can compute the packed volume. Packed volume = lTS = NOGHOGS = N OG G (149)(454) = (14.2) = 14.8 m3 (K G ) p aP 0.089(3, 600)(2)(101.3) Exercise 6.27 Subject: Absorption of GeCl4 from air into dilute caustic solution in an existing packed column. Given: Column operates at 25oC (77oF) and 1 atm. Gas enters at 23,850 kg/day containing 288 kg/day of GeCl4 and 540 kg/day of Cl2. Dissolved GeCl4 and Cl2 react with the caustic so that neither has a vapor pressure. Packed tower is 2-ft diameter with 10 ft of 1/2-inch ceramic Raschig ring packing of given characteristics. Liquid rate is to give 75% of flooding. Equation is given for estimating Kya. Assumptions: No stripping of water. No absorption of air. Find: (a) Entering dilute caustic flow rate. (b) Required packed height. Which controls, GeCl4 or Cl2? Is 10-ft height adequate? (c) % absorption of GeCl4 and of Cl2 for 10 ft of packing. If necessary, select an alternative packing. Analysis: (a) Determine entering dilute caustic flow rate from the 75% of flooding specification, using the flooding curve of Fig. 6.36(a), with the correction factors of Figs. 6.36(b,c). The air rate in the entering gas = 23,850 - 288 - 540 = 23,022 kg/day. MW of Cl2 = 71. MW of GeCl4 = 214.6. MW of air = 29. 23,022 288 540 Molar gas rate = V = + + = 803 kmol/day or 73.7 lbmol/h or 2189 lb/h 29 214.6 71 Average molecular weight of gas = 23,850/803 = 29.7 Tower cross sectional area for 2-ft diameter = 3.14(2)2/4 = 3.14 ft2 The continuity equation for flow through the tower based on the gas s...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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