# Then the feed contains 10 kmolh alcohol and 90 kmolh

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

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...
View Full Document

## 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 .

Ask a homework question - tutors are online