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

Let na moles of a still in the gas at equilibrium at

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Unformatted text preview: equilibrium. Then by Dalton's law, s PPA nPA = [(8,000 − 67) + nPA ] (1) P s PPA (7,933) Solving for nPA , with P = 770 torr, nPA = 770 s (2) PPA 1− 770 and the percent recovery = (97 - nPA)/97 x 100% Solving Eqs. (2) and (3), Ps, torr nPA , lbmol/h % Recovery in gas of PA 0.7 7.2 89.3 0.4 4.1 93.9 0.1 1.03 98.5 (3) Exercise 4.66 Subject: Desublimation of anthraquinone (A) from nitrogen (N). Given: Gas containing 10 mol% A and 90 mol% N at 760 torr and 300oC is cooled to 200oC. Assumptions: Given vapor pressure data for solid A can be fitted to Antoine equation Find: % desublimation of A assuming no change in total pressure. Analysis: Using Polymath, the given vapor pressure data are fitted to the following Antoine eq.: PAs = 10 ss A o 32 .8653− 49 , 762 .2 1327 .24 + T (1) where P is in torr and T is in C. From Eq. (1), P = 1.915 torr at 200oC. Assume 100 moles of initial gas, containing 90 moles N and 10 moles A. Let nA = moles of A still in the gas at equilibrium at 200oC. Then by Dalton's law, where pA = PAs , s A pA Ps 1.915 ntotal = A 90 + nA = 90 + nA P P 760 1.915 (90) nA = 760 = 0.227 mole Solving for nA , 1.915 1− 770 and the percent recovery = (10 - 0.227)/10 x 100% = 97.73% nA = (2) (3) Exercise 4.67 Subject: Separation of a gas mixture of propylene (A) and propane (P) by adsorption. Given: T = 25oC, P = 101 kPa, gas containing 0.7 mole A and 1.3 mole P, and S = 0.1 kg silica gel. Equilibrium adsorption data in Fig. 4.30. Find: Equilibrium compositions and amounts of adsorbate and final gas. Analysis: Let y and x be the mole fractions of P in the vapor and adsorbate, respectively. Let G and W be the moles of equilibrium vapor and adsorbate, respectively. Then a propane material balance is: 1.3 = Wx + Gy . A total material balance is: 2.0 = W + G. Combining balances, y= 13 − Wx . 2 −W (1) To solve, assume W, plot the equation on Fig. 4.30a and find intersection with the equilibrium curve to find y and x. Read the mmole adsorbate/g adsorbent = moles adsorbate/kg adsorbent from Fig. 4.30b. Calculate kg adsorbent. If it is not 0.1 kg, repeat with a new guess of W. For W = 0.195 moles, G = 1.805 moles, y = 0.72 - 0.108x, y = 0.67, x = 0.41, S = 0.10 kg. Exercise 4.68 Subject: Separation of a gas mixture of propylene (A) and propane (P) by adsorption. Given: Gas containing 50 mol% A and 50 mol% P, and S = 1,000 lb of less of silica gel/lbmol of feed gas. Equilibrium adsorption data in Fig. 4.30 apply. Find: If one equilibrium stage can produce a vapor with 90 mol% P and an adsorbate of 75 mol% A. If not, what separation can be achieved. Analysis: With a basis of 1 lbmole of feed gas, let y and x be the mole fractions of P in the equilibrium vapor and adsorbate, respectively. Then we desire, y = 0.9 and x = 0.25. Let G and W be the lbmol of equilibrium vapor and adsorbate, respectively. Then a propane material balance is: 0.5 = Wx + Gy . A total material balance is: 1.0 = W + G. Combining balances, y= 0.5 − Wx 1−W (1) From the equilibr...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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