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Unformatted text preview: agram below, where the equilibrium curve is plotted from the data, the qline for Feed 1 is vertical through the point, x = 0.75, the qline for Feed 2 has a slope of 1 starting from x = 0.50, and the operating line for the upper section between Feed 1 and the condenser is drawn through the two points, {x=0.98, y=0.98} and the intersection of the equilibrium curve and the qline for Feed 1. From the plot, for the upper section, L/V = (0.980.828)/(0.980.75) = 0.661. From Eq. (727), R = L/D = (L/V)/[1  (L/V)] = 0.661/(1  0.661) = 1.95. Therefore, L = 1.95D = 1.95(123.66) = 241.1 kmol/h and V = L + D = 241.1 + 123.66 = 364.8 kmol/h. In the middle section, between the two feeds, L' = L + F1 = 241.1 + 100 = 341.1 kmol/h and V' = V = 364.8 kmol/h. Therefore, the slope of the middle section operating line = L'/V' = 341.1/364.8 = 0.935. As seen in the diagram below, this line does not cause a pinched region at Feed 2. Therefore, the assumption is correct and Rmin = 1.95. For an operating reflux ratio of 1.2 times mi...
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 Spring '06
 selebi
 Distillation, pH, Mass Transfer

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