# 4 125 kmolh the total feed rate of feed 2 5006 833

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Unformatted text preview: a recovery of 95% of A in the distillate, the flow rate of A in the distillate = 0.95(100) = 95 kmol/h. With a mole fraction of 0.95 for A in the distillate, the total flow rate of the distillate = 95/0.95 = 100 kmol/h. From Eq. (7-3) for = 3, the equilibrium mole fractions of A are related by, 3x x (1) y= = 1 + ( - 1) x 1 + 2 x Equation (1) is plotted in the McCabe-Thiele diagram on the next page. Because Feed 2 is richer in A than Feed 1, Feed 2 enters the column above Feed 1. At minimum reflux, the pinch condition will occur at either Feed 1 or Feed 2. Assume that the pinch occurs at Feed 2. For a saturated liquid feed, using Eq. (1), the upper section operating line will intersect the equilibrium curve for xF = 0.6 at y = 3(0.6)/[1 + 2(0.6)] = 0.818. Therefore, the slope of this operating line is, (L/V)min = (0.95 - 0.818)/(0.950 - 0.6) = 0.377 Correspondingly, using Eq. (7-17), R = L/D = (L/V)min/[1 - (L/V)min] = 0.377/(1 - 0.377) = 0.605 and Lmin = 0.605(100) = 60.5 kmol/h. Now check the m...
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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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