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

Given plate column absorber and stripper feed gas

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Unformatted text preview: ight operating line that passes through the point (Y1 = 0.00882 and X0 = 0.0) is drawn as shown below, where equilibrium stages are stepped off, as in Figs. 6.11(a) and 6.12, until the steps reach or exceed the required value of YN+1 = 0.1765. As seen, in the figure below, the determined number of equilibrium stages is approximately 9. (c) The concentration of acetone in the exit liquid, XN is the point on the operating line for Part (b), where YN+1 = 0.1765. From Eq. (6-3), for the operating line for an absorber, Y −Y 0.1765 − 0.00882 X N = N +1 1 + X 0 = + 0.0 = 0.127 mol acetone/mol water L′ 1.325 V′ or an exit mole fraction of acetone in the water of 0.127/(1.127) = 0.113 Analysis: (continued) Exercise 6.8 (continued) Exercise 6.9 Subject: Absorption-stripping system for benzene (B) removal from gas by oil and subsequent stripping of benzene from oil by steam. Given: Plate column absorber and stripper. Feed gas containing 0.06 mol B/mol B-free gas. Absorbent oil containing 0.01 mol B/mol B-free oil. Liquid leaving absorber contains 0.19 mol B/mol B-free oil. 90% of B in the gas stream is absorbed. Liquid leaving stripper contains 0.01 mol B/mol oil. Flow ratio of B-free oil-to-B-free steam = 2.0. Molecular weights are 200, 78, and 32 for oil, benzene, and gas, respectively. Equilibrium X-Y data given for absorber (25oC) and stripper (110oC). Assumptions: Gas is not soluble in oil and oil is not volatile. Find: (a) Molar flow rate ratio of B-free oil to B-free gas in the absorber. (b) Number of equilibrium stages in the absorber. (c) Minimum steam flow rate in the stripper per mol of B-free oil. Analysis: A flow diagram of the absorber-stripper system is shown in Fig. 5.10(a), except that there is no makeup absorbent oil because the oil is assumed to be non-volatile. (a) Consider just the absorber. Take a basis of V ' = 1 mol/h of B-free gas entering the absorber. Because YN+1 = 0.06 mol B/mol B-free gas, we have 0.06 mol/h of B in entering gas. With 90% absorption, 0.06(0.9) = 0.054 mol/h of B absorbed, leaving 0.006 mol/h of B in the exiting gas. In the entering absorbent oil, X0 = 0.01 mol B/mol B-free oil. In the liquid leaving the absorber, XN = 0.19 mol B/mol B-free oil. A material balance around the absorber on the benzene gives: YN +1V ′ + X 0 L′ = YV ′ + X N L′ Therefore, 1 0.06(1.0)+0.01L′ = 0.006 + 0.19 L′ (1) Solving Eq. (1), L' = 0.30 mol/h of B-free oil, giving L'/V’' = 0.30 mol B-free oil to B-free gas. (b) Use a graphical method to determine the number of equilibrium stages in the absorber, using Y-X coordinates to give a straight operating line. Data for the equilibrium curve of the absorber are given and the operating line is obtained from terminal values of Y and X. The results are given in the plot below. As seen, between 9 and 10 equilibrium stages are needed. (c) The equilibrium curve for the stripper is shown in the plot below. With the type of curvature in the Y-X equilibrium curve the minimum stripping gas rate is determined by...
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