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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...
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