Separation Process Principles- 2n - Seader &amp; Henley - Solutions Manual

# It is seen that for benzene xd 090 and xb 028 exercise

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Unformatted text preview: chlorobenzene with specified number of equilibrium stages, boilup ratio, and reflux ratio. Given: Feed is a saturated liquid of 54.5 mol% benzene. Column contains two equilibrium plates with feed to the bottom plate. Column is equipped with total condenser and partial reboiler. Boilup is V/F = 0.855. Reflux ratio, L/V = 0.5. Vapor-liquid equilibrium data given. Find: Compositions of distillate and bottoms, assuming constant molar overflow. Analysis: Take as a basis, F = 100 mol/s. Therefore, vapor generated in reboiler = 0.855(100) = 85.5 mol/s. Since the feed is a saturated liquid, this vapor rate continues up the column to the condenser. L/V = 0.5, which 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, where benzene mole fractions are plotted because it is the more volatile component. It is seen that for benzene, xD = 0.90 and xB = 0.28. Exercise 7.14 (continued) McCabe-Thiele Diagram Exercise 7.15 Subject: Effect of loss of plates in a distillation column separating a benzene-toluene mixture. Given: Saturated vapor feed of 13,600 kg/h of 40 wt% benzene and 60 wt% toluene. Column with 14 plates above the feed location. Plate efficiency is 50%. Reflux ratio is 3.5. Previously, with 10 plates in the stripping section, column could achieve a distillate of 97 wt% benzene and a bottoms of 98 wt% toluene. Vapor-liquid equilibrium data in Exercise 7.13. Assumptions: Constant molar overflow. Total condenser and partial reboiler. Find: (a) If column with 10 inoperable plates can yield a distillate of 97 wt% benzene, assuming that we no longer can achieve the 98 wt% bottoms product. (b) The distillate flow rate. (c) The composition of the bottoms. Analysis: (a) First convert the feed to kmol/h and mole fractions, using molecular weights of 78.11 for benzene and 92.13. The result is: Component kmol/h Mole fraction Benzene 69.65 0.44 Toluene 88.57 0.56 Total: 158.22 1.00 For a distillate of 97 wt% benzene, the mole fraction for benzene, the more volatile of the two components, is, 97 7811 . xD = = 0.974 97 3 + 78.11 92.13 Wit...
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