This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Exercise 13.1 Subject: Evaporation of a small amount of nheptane (C7) from a drum of toluene (T). Given: A drum of toluene containing 2 mol% C7. Simple batch differential (Rayleigh) distillation at 1 atm. Vaporliquid equilibrium data for the C7toluene system at 1 atm. Assumptions: Perfect mixing in the still. Exiting vapor in equilibrium with the liquid. Find: (a) Composition of the residue in the pot as a function of wt% distilled. Composition of residue and the cumulative distillate at 50 wt% distilled. (b) Composition of the cumulative distillate and the residue when 50% of the C7 has been distilled. Weight percent of the original charge distilled. Analysis: The vaporliquid equilibrium data indicate that C7 is more volatile than T. Therefore, make the calculations in terms of C7. Because the initial concentration of C7 in the charge is low (2 mol%), assume that the distillation takes place at constant relative volatility, = C7T . Alternatively, the y  x data in the low concentration region could be fitted to a quadratic equation and used with Eq. (132). The lowest concentration in the data table is 2.5 mol% C7, where x = 0.025 and y = 0.048. Therefore, for the binary system, using Eqs. (219) and (221), K C7 = y/x = 0.048/0.025 = 1.92 K T = (1 y )/(1 x ) = (1  0.048)/(1  0.025) = 0.976 C7T = K C7 / K T = 1.92/0.976 = 1.97 A similar calculation at the next point in the data table, x = 0.062 and y = 0.107, gives C7T = 1.81. Therefore the relative volatility increases as x drops to zero. Take C7T = 2 for the region of interest. From Eq. (135), ln ln ln ln . ln . W W x x x x x x 1 1 1 1 0 025 2 1 1 0 025 & = & + & = & + & (1) Taking a basis of W = 1 lbmole, this equation can be solved with a spreadsheet for W for a series of values of x from 0.020 to 0.0. For each value of x , a material balance on C7 gives Eq. (136) for the mole fraction of C7 in the cumulative distillate, y W x Wx W W D avg = (2) With values of W , x , and y D avg , together with the molecular weights of 100.21 for C7 and 92.14 for T, amounts in mass units can be computed, using Eqs. (1) and (2). The spreadsheet is given on the following page, from which the following results are obtained. (a) Initial residue = 92.29 lb. At 50 wt% distilled, residue = 46.145 lb. From the spreadsheet, the mole fraction of C7 in the residue = 0.0102, the mass fraction of C7 in the residue = 0.0110 . The mass fraction of C7 in the cumulative distillate = 0.0324 . (b) When 50% of the C7 has distilled, the mole fraction of C7 in the residue = 0.0142, the mass fraction of C7 in the cumulative distillate = 0.0365, and the wt% of charge distilled = 29.9%. Exercise 13.1 (continued) x ln( W o/ W ) W mols C7 lb C7 mols T lb T Total lb wt% C7 mass lb cum. C7 mass in in in in residue distilled fraction distillate fraction residue residue...
View
Full
Document
This note was uploaded on 04/03/2008 for the course CHE 312 taught by Professor Ofoli during the Spring '08 term at Michigan State University.
 Spring '08
 Ofoli
 Equilibrium, Distillation

Click to edit the document details