Redo Example 7.1, keeping everything the same but assuming a relative volatility of 1.7. and Toluene.EXAMPLE 7.1 Distillation of a Mixture of BenzeneFour hundred and fifty lbmol/h (204 kmol/h) of a mixture of 60 mol%liquid distillate and a liquid bottoms product of 95 mol% and 5 mol %benzene (LK) and 40 mol % toluene (HK) is to be separated into a, respectively. The feed enters the column with a molar per-vaporization equal to the distillate-to-feed ratio. Use theMcCabe-Thiele method to compute, at 1 atm (101.3 kPa): (a) min,(b) Rmin, and (c) number of equilibrium stages N, for R/Rmin = 1.3,and the optimal feed-stage location. Also, compare the results withbenzene,centthose from a process sSolutionsimulator.First calculate D and B. An overall material balance on benzenegivesFrom (7-26),0.60(450) = 0.95D +0.05B450D + BA total balance givesCombining (1) and (2) and solving, D = 275 lbmol/h, B = 175Ibmol/h, and D/F = 0.611. Thus, the molar vaporization of the feedis 61.1%.Calculate the slope of the q-line:VF/F = D/F = 0.611, and g for a partially vaporized feed isLF (F-VF)FFthe slope of the q-line is= 19q-1=VFF= 0.3890.3890.389-1= -0.637R1.59= 0.614R+1 1.59 + 1(1)(2)=(a) In Figure 7.13, where y and x refer to benzene, xp = 0.95 andx = 0.05, the minimum stages are stepped off between theequilibrium curve and the 45° line, giving Nmin = 6.7.(b) In Figure 7.14, a q-line is drawn that has a slope of -0.637 andpasses through the feed composition (z= 0.60) on the 45° line.For Rmin, an operating line for the rectifying section passesthrough the point x = xp = 0.95 on the 45° line and throughthe point of intersection of the q-line and the equilibrium curve(y=0.684, x = 0.465). The slope of this operating line is 0.55,which from (7-9) equals R/(R+ 1). Therefore, Rmin = 1.22.(c) The operating reflux ratio is 1.3 Rmin = 1.3 (1.22) = 1.59. From(7-9), the rectifying-section operating-line slope is=XBThe two operating lines and the q-line are shown in Figure7.15, where the stripping-section operating line is drawn topass through the point x = xp = 0.05 on the 45° line andthrough the intersection of the q-line with the rectifying-sectionoperating line. The equilibrium stages are stepped off, first,between the rectifying-section operating line and the equilibriumcurve, and then between the stripping-section operating line andthe equilibrium curve, starting from point A (distillate composi-tion) and finishing at point B (bottoms composition). For the opti-mal feed stage, the transfer from the rectifying-section operatingline to the stripping-section operating line takes place at pointP, giving N= 13.2 equilibrium stages, the feed going into stage7 from the top, and N/Nmin== 13.2/6.7 1.97. The bottomNCabe-Thiele Graphical Method for Trayed TowersMole fraction of benzene in the vapor, yMole fraction of benzene in the vapor, y1.00.80.60.4Mole fraction of benzene in the vapor, y0.201.00.80.60.40.200.40.60.8XD 1.0Mole fraction of benzene in the liquid, xFigure 7.13 Determination of minimum stages for Example 7.1.1.00.80.60.4Benzene-toluene at 1 atm14-6XB0XBLO0.2 1213,50.2Benzene-toluene at 1 atmXB0.6xD 1.0ZFMole fraction of benzene in the liquid, xFigure 7.14 Determination of minimum reflux for Example 7.1.Equilibrium Curve0.2114B10q-lineBenzene-toluene at 1 atmEquilibrium0.40.245° linecurve9Equilibrium curve650.445° line410.6ZF0.83269245° lineA0.8 x1.0Mole fraction of benzene in the liquid, xFigure 7.15 Determination of number of equilibrium stages andfeed-stage location for Example 7.1.

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