Then the feed contains 10 kmolh alcohol and 90 kmolh

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Unformatted text preview: f 98 mol%, alcohol, distillate contains 9.8 kmol/h of alcohol. For an alcohol purity of 67.5 mol%, the distillate rate = D = 9.8/0.675 = 14.52 kmol/h. Water in the distillate = 14.52 - 9.8 = 4.72 kmol/h. Alcohol in the bottoms = 10 - 9.8 = 0.2 kmol/h. (a) With a partial reboiler, no other water enters the system. Therefore, water in the bottoms = 90 - 4.72 = 85.28 kmol/h. Total bottoms rate = B = 85.28 + 0.2 = 85.48 kmol/h. Mole fraction of alcohol in bottoms = 0.2/85.48 = 0.0023. The minimum reflux is determined from the McCabe-Thiele diagram on the next page, where the equilibrium curve is drawn from the given data and the q-line is vertical, passing through x = 0.10. The rectifying section operating line for minimum reflux usually is a straight line that connects the distillate mole fraction on the 45o line to the intersection of the equilibrium curve and the q-line as shown by the dashed line on the diagram. However, in this case the line mistakenly crosses over the equilibrium curve. Ther...
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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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