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 q-line is a vertical line because the feed is a saturated liquid. To solve for the compositions of the distillate and bottoms on a McCabe-Thiele 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 McCabe-Thiele diagram with the above operating lines and q-line to see if three stages are required with the feed to the second plate. See plot below...
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