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

024 x 100 234 mol co2 and 9766 mol air a a plot of

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Unformatted text preview: rom air at 25oC by 5-N aqueous triethanolamine Given: Feed gas containing 10 mol% CO2 and 90 mol% air. Absorbent of 5N aqueous triethanolamine containing 0.04 moles of CO2 per mole of amine solution. Column with 6 equilibrium stages. Exit liquid to contain 78.4% of the CO2 in the feed gas. Therefore, exit gas contains 21.6% of the CO2 in the entering gas. Equilibrium data for CO2 at 25oC in terms of mole ratios. Assumptions: Negligible absorption of air and stripping of amine and water. Find: (a) Moles of amine solution required per mole of feed gas. (b) Exit gas composition. Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Therefore, for CO2, X0 = 0.04 mol CO2/mol amine solution YN+1 = Y7 = 10/90 = 0.1111 mol CO2/mol air Y1 = 0.216(10)/90 = 0.024 mol CO2/mol air (b) Therefore, the exit gas composition is 0.024 mol CO2/mol air or 0.024/(1 + 0.024) x 100% = 2.34 mol% CO2 and 97.66 mol% air. (a) A plot of the equilibrium data as Y vs. X is given below. The operating point (X0, Y1) at the top of the column is included. A straight operating line through this point is found by trial and error to give 6 equilibrium stages, when using Y7 = 0.1111. The resulting XN = X6 = 0.085. From Eq. (6-3), the slope of the operating line = L'/V' = (0.1111 - 0.024)/(0.085 - 0.04) = 1.936 mol triethanolamine solution/mol air. The feed gas contains 9 mol air/10 mol feed gas. Therefore, mols of amine solution/mol feed gas = 1.936(0.9) = 1.74. See plot on next page. Exercise 6.7 (continued) Exercise 6.8 Subject: Absorption of acetone from air by water at 20oC and 101 kPa (760 torr) in a valve-tray column. Given: 100 kmol/h of feed gas containing 85 mol% air and 15 mol% acetone. Pure water is the absorbent. Overall tray efficiency is 50%. Absorb 95% of the acetone. Equilibrium p-x data for acetone are given as listed below. Assumptions: Negligible absorption of air and stripping of water. Find: (a) Minimum ratio, L'/V' of moles of water/mole of air. (b) Number of equilibrium stages for L'/V' = 1.25 times minimum. (c) Concentration of acetone in the exit water. Analysis: Use the nomenclature and type of plot shown in Fig. 6.11(a). Then, the operating line will be straight. For acetone, X0 = 0.0 mol acetone/mol entering water YN+1 = 0.15/0.85 = 0.1765 mol acetone/mol air in entering gas Flow rate of acetone in exit gas = (1 - 0.95)(15) = 0.75 kmol/h. With 85 kmol/h of air, Y1 = 0.75/85 = 0.00882 mol acetone/mol air Convert the p-x equilibrium data to mole ratio, Y-X data, using y = p/P, Y = y/(1- y), X = x/(1-x) p, torr x y X Y 30.0 0.033 0.0395 0.0341 0.0411 62.8 0.072 0.0826 0.0776 0.0901 85.4 0.117 0.1124 0.1325 0.1266 103.0 0.171 0.1355 0.2063 0.1568 (a) With the type of curvature in the Y-X equilibrium curve, shown below, the minimum absorbent rate is determined by a straight operating line that passes through the point (Y1 , X0 ) and is drawn tangent to the equilibrium curve, as shown. From Eq. (6-3), the slope of the operating line = L'/V' = 1.06 mol water/mol of air on an acetone-free basis = minimum ratio. (b) For 1.25 times minimum, L'/V' = 1.25(1.06) = 1.325. Now a stra...
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