# Ch07

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Unformatted text preview: a boilup ratio = 1.2 times minimum. Analysis: From a rearrangement of the equilibrium equation, Eq. (7-3), y y x= = (1) y + (1 - y ) 6 - 5 y (a) For minimum stages, have total reflux, so that y = x for passing streams. Begin calculations from the top. yD = y1 = 0.35. From Eq. (1), x1 = 0.35/[6 - 5(0.35)] = 0.0824. Therefore, y2 = x1 = 0.0824. From Eq. (1), x2 = 0.0824/[6-5(0.0824)] = 0.0147. Therefore, y3 = x2 = 0.0147. From Eq. (1), x3 = 0.0147/[6-5(0.0147)] = 0.0025. This is close to but not quite equal to the desired value of 0.002. Thus, we need just slightly more than 3 minimum equilibrium stages. (b) For minimum boilup ratio, the stripping section operating line connects the two points for {y, x} of {0.002, 0.002} and {y in equilibrium with x = 0.05}. From a rearrangement of Eq. (1), the y in equilibrium with x = 0.05 is: y = x/[1 + x( - 1)] = 6(0.05)/[1 + 0.05(6 - 1)] = 0.24. The slope of the operating line = ( L / V ) = (0.24 0.002)/(0.05 - 0.002) = 4.96. From a rearrangement of Eq. (7-12), (VB)...
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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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