This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 2.
(2) Reflux ratio = 2(1.092) = 2.18. From Eq. (7-7), the slope of the rectifying section operating line = L/V = R/(1 + R) = 2.18/3.18 = 0.686. To determine the bottoms composition, use a McCabe-Thiele diagram in terms of benzene, the more volatile component. The q-line and the rectifying section operating line are fixed and 4 trays are stepped off from the top, starting at the distillate mole fraction for benzene, xD , of 0.96. Then, the stripping section operating line is positioned by trial and error so that 3 more stages plus the reboiler stage are stepped off to arrive at the point where the assumed location of the stripping section operating line intersects the 45o line. The result is shown on the next page where it is seen that xB = 0.18. (3) The products are now computed by overall material balances: F = 100 = D + B and 50 = xDD + xBB = 0.96D + 0.18B. Solving these two equations, D = 41.0 mol/100 mol feed and B = 59.0 mol/100 mol feed. Analysis: (a) (continued) Exercise 7.17 (continued) (b) (1) F...
View Full Document