# From the equilibrium data this intersection is at y

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Unformatted text preview: ion operating line, (L/V)min is (0.95 - 0.72)/(0.95 - 0.50) = 0.511. From a rearrangement of Eq. (7-7), Rmin = (L/V)min /[1 - (L/V)min ] = 0.511/(1 0.511) = 1.045. (b) The McCabe-Thiele plot for minimum stages at total reflux is shown below. The operating lines are coincident with the 45o line. Equilibrium stages are stepped off starting from xB = 0.05 to xD =0.95. It is seen that just less than 7 equilibrium stages are needed. Call it Nt = 7. From Eq. (6-21), for an overall plate efficiency of 65%, i.e. Eo = 0.65, the actual minimum number of plates = Na = Nt / Eo = 7/0.65 = 10.8. (c) Operating reflux ratio = R = 1.5 Rmin = 1.5(1.045) = 1.57. From Eq. (7-7), the slope of the operating line for the rectifying section = L/V = R/(1 + R) = 1.57(1 + 1.57) = 0.611. On the McCabe-Thiele diagram on the next page, the rectifying section operating line has this slope and passes through the point, y=0.95, x=0.95. the stripping section operating line passes through the point, y=0.05, x=0.05 and intersects the ver...
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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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