This is the second time I am posting this. Please explain how to get alfa average (average volatility ) of 3.58 based on these given informations. Please explain

Exercise 2.7 Methanol (M) is to be separated from water (W) by atmospheric distillation. The feed contains 14.46kg/h methanol and 10.44kg/hr water. The distillate is 99mol% pure, the bottom product contains 5mol% of methanol. The feed is subcooled such that q=1.12. a. Determine the minimum number of stages and minimum reflux b. Determine the feed stage location and number of theoretical stages required for a reflux ratio of 1 . Vapor-liquid equilibrium data (1.0 atm, mole fraction methanol) Exercise 2.7 Given FM=14.46kg/hFW=10.44kg/hxD=0.98xB=0.05q=1.12P=1bar Find a. the minimum number of stages and minimum reflux. b. the feed stage location and number of theoretical stages for a reflux ratio R=1.0. Thoughts The minimum number of stages is obtained at maximum reflux, i.e. D=0 and B=0. Hence the slope of both operating lines equals unity. The minimum reflux is obtained at infinite number of stages, i.e. both operating lines intersect the feed line on the equilibrium line. Draw the operating line for the rectifying section for R=1.0 according to Eq. 2.23. Calculate V and L" as explained in Exercise 2.6b and draw the operating line for the stripping section. Solution a. The minimum number of stages is obtained at maximum reflux, slopes of operating lines amount to unity. The stair case is constructed between the diagonal of the yx diagram and the equilibrium curve. As shown in the first diagram below, the minimum number of stages amounts to 5.8. With av= 3.58 (Eq. 2.48), the Fenske expression (Eq. 2.47 ) would give Nmin=5.4 Note that in this case the reboiler does not contribute to the separation because the liquid entering the reboiler is entirely vaporized: its composition