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Unformatted text preview: ich is the slope of the operating line. Therefore, L = 0.5(85.5) = 42.75 mol/s. Therefore, the distillate rate = 85.5  42.75 = 42.75 mol/s. Passing to the reboiler is a liquid rate of 42.75 + 100 = 142.75 mol/s. The bottoms rate = 142.75  85.5 = 57.25 mol/s. The slope of the stripping section operating line is L / V = 142.75/85.5 = 1.67. The qline is a vertical line because the feed is a saturated liquid. To solve for the compositions of the distillate and bottoms on a McCabeThiele diagram, we must locate the operating lines to obtain three equilibrium stages that satisfy an overall benzene material balance given by,
xFF = 54.4 = xDD + xBB = 42.75xD + 57.25xB
Solving Eq. (1), xB = 0.952  0.7467 xD (1) (2) Therefore, an approach to solving this exercise is to assume a value of xD and then compute the value of xB from Eq. (2). Then construct the McCabeThiele diagram with the above operating lines and qline to see if three stages are required with the feed to the second plate. See plot below...
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 Spring '06
 selebi
 Distillation, pH, Mass Transfer

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