For a 90 recovery the distillate contains 0912 108

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Unformatted text preview: Since the distillate is 85 mol% ethanol, the total distillate rate = D = 10.8/0.85 = 12.7 kmol/h. The bottoms contains 12 - 10.8 = 1.2 kmol/h of ethanol. The distillate contains 12.7 - 10.8 = 1.9 kmol/h of water. The bottoms contains 88 - 1.9 + open steam = 89.9 + open steam in kmol/h. (a) For a reflux ratio of 3, L = 3D = 3(12.7) = 38.1 kmol/h. Overhead vapor rate = V = L + D = 38.1 + 12.7 = 50.8 kmol/h. Below the feed stage, L' = L + F = 38.1 + 100 = 138.1 kmol/h. Boilup rate = V' = V = 50.8 kmol/h = flow rate of open steam. (b) The bottoms rate = 138.1 kmol/h. The bottoms consists of 1.2 kmol/h of ethanol and 138.1 - 1.2 = 136.9 kmol/h of water. The mole fraction of ethanol in the bottoms = 1.2/138.1 = 0.0087. The McCabe-Thiele diagram is given on the next page, where the equilibrium curve is obtained from Exercise 7.29 and the q-line is vertical at x = 0.12. The rectifying section operating line passes through the point {0.85, 0.85}and has a slope, L/V = 38.1/50.8 = 0.75. The stripping section operat...
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