Take as a basis 100 mols of feed therefore the feed

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

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 .

Ask a homework question - tutors are online