# 25 l v from eq 7 28 min 0815 002 346 025

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: - 0.02 (VB )min = ( L /V ) 1 min -1 = 1 = 0.407 3.46 - 1 Analysis: Base Case (continued) Exercise 7.37 (continued) For column operation, VB = 1.15(VB)min = 1.15(0.407) = 0.468 From Eq. (7-12), the slope of the stripping section operating line is, V + 1 0.468 + 1 L /V = B = = 314 . VB 0.468 As shown in the McCabe-Thiele diagram, below, a line of this slope is drawn through the point x = y = 0.02 until it intersects the q-line. Equilibrium stages are stepped off, starting from the distillate point at y = x = 0.98. The optimal feed stage location is located as shown at stage 5 from the top. The total number of stages required is just less than 8, with one of those stages being the partial reboiler. Exercise 7.37 (continued) Analysis: Base Case (continued) Exercise 7.37 (continued) Analysis: (continued) Alternative unit with Interreboiler: For column operation with an interreboiler that provides 50% of the reboiler duty, the operating boilup ratio between the reboiler and the interreboiler is 50% of 0.468 or 0.234. From Eq. (7-12), the slope of the operating line in this r...
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