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

Y procedure 1 solve with material balances and eq 1

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Unformatted text preview: mol% benzene (xF = 0.7) . Procedures are: 1. No column. Just a partial condenser on top of a partial reboiler. Feed is to the reboiler. Reflux ratio, L/D = 0.5. Vapor distillate is totally condensed. 2. Same as 1 except that one equilibrium stage sits between the condenser and reboiler. 3. Same as 1 except two equilibrium stages between the condenser and reboiler. 4. Same as 3 except that reflux bypasses the top equilibrium stage. 5. Same as 2 except that feed is sent to the stage between the condenser and the reboiler. Assumptions: Constant molar overflow. Constant relative volatility = αΑ,Β = 2.5. Find: For each procedure, determine: (a) Moles of distillate per 100 moles of feed. (b) Moles of total vapor generated per mole of distillate. (c) Mole percent of benzene in the bottoms (residue). (d) y-x diagram, indicating compositions of distillate, reflux, and residue. Also, (e) For maximization of benzene recovery (in the distillate), which procedure is preferred. Analysis: For each procedure, the partial condenser and the partial reboiler are equilibrium stages. Benzene is the more volatile component, so the y-x diagram is based on benzene. Because the relative volatility = constant = 2.5, the equilibrium relationship is given by Eq. (7-3), αx 2.5x = (1) 1 + x (α − 1) 1 + 15x . Take as a basis, 100 mol/s of feed. Therefore, the feed contains 70 mol/s of A and 30 mol/s of B. From the reflux ratio, L = 0.5D, V = L + D = 1.5D. Therefore, D/V = 2/3 and L/V = 1/3. Use a subscript of C for streams leaving the condenser, R for streams leaving the reboiler, 1 for the top stage when used, and 2 for the second stage when used. y= Procedure 1: Solve with material balances and Eq. (1). The liquid leaving the partial condenser is in equilibrium with the vapor distillate of yC = yD = 0.8. Solving Eq. (1), yC 0.8 = = 0.615 yC + α (1 − yC ) 0.8 + 2.5(1 − 0.8) Benzene material balance around condenser, xC = (2) y RV = yC D + xC L or y R = yC D L 2 1 + xC = 0.80 + 0.615 = 0.738 V V 3 3 Exercise 7.11 (continued) Analysis: Procedure 1 (continued) The vapor from the reboiler is in equilibrium with the liquid bottoms (residue). From the lefthand part of Eq. (2), yR 0.738 xR = = = 0.530 yR + α (1 − yR ) 0.738 + 2.5(1 − 0.738) Overall total material balance, F = 100 = D + B (3) Overall benzene material balance, xFF = yCD + xRB or 70 = 0.8D + 0.530B (4) Solving Eqs. (3) and (4), D = 62.9 mol/s or 62.9 mol/100 mol feed, and B = 37.1 mol/s. Therefore, vapor generated = V = 1.5D = 1.5(62.9) = 94.4 mol/s. The operating line for the y-x diagram passes through the (y, x) point (0.8, 0.8) with a slope, L/V = 1/3, as shown in the diagram below. Exercise 7.11 (continued) Analysis: (continued) Procedure 2: The slope and top point of the operating line are the same as for Procedure 1. We just have to step off one more stage. Therefore from the results above, we have: yC = 0.80 xC = 0.615 y1 = 0.738 x1 = 0.530 Benzene material balance around Stage 1, y RV + xC L = y1V + x1 L (5) L 1 = 0.738 + (0.530 − 0.615) = 0.710 V 3 The...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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