# 16754 00021 with isopropanol as the most volatile

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: given below, where the equilibrium curve is obtained from the data in Exercise 7.33 and the q-line is vertical through x = 0.10. The stripping section operating line passes through the point {x=0.0021, y=0.0021}with a slope = L/V = F/V =100/24.6 = 4.065. It also passes through the point {x=0.1, y=0.4}. From the plot, the number of equilibrium stages = just less than 3. Call it 2 equilibrium stages in the column + partial reboiler. (2) The open steam rate = V = 24.6 kmol/h. The liquid rate = L = 100 kmol/h. Therefore, the slope of the stripping section operating line is the same as for part (1), i.e. L./V = 100/24.6 = 4.065. Now the mole fraction of alcohol in the bottoms = xB = 0.16/100 = 0.0016. Thus, as shown in the McCabe-Thiele diagram below, the operating line passes through the points {x=0.0016, y=0}and {x=0.10, y=0.40}, with the slope of 4.065. Now, The number (1) of equilibrium stages is equal to 3, all of them in the column. Exercise 7.34 (continued) Analysis: Partial Reboiler Case: Exercise 7.34 (continued) Analysis:...
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