# Take as a basis 100 mols of feed therefore the feed

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Unformatted text preview: 00 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 + (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 + (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) Solvi...
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## This note was uploaded on 09/08/2008 for the course CHE 244 taught by Professor Selebi during the Spring '06 term at Lehigh University .

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