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

94 044 4544 1999 by interpolation for wx 24 moles x

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Unformatted text preview: he residue = 0.667(250) = 166.7 lb. Benzene is the more volatile component. Therefore, base the calculations on B. Use Eq. (13-3), with x0 = 0.7 and W0 = 3.037 lbmol. Thus, Eq. (13-3) becomes: x 0.7 dx dx W 3.037 (1) = ln or = ln x 0.7 y − x 3.037 y−x W where x is the mole fraction of B in the final residue and y is in equilibrium with x. Equilibrium data at 1 atm are conveniently obtained from a simulation program. For example, data from the Chemcad program are included below for the range of interest, based on Raoult's law (ideal K values). Using a spreadsheet with the equilibrium y-x data, integrate Eq. (1) using the trapezoidal rule with a ∆x increment of 0.02 starting from x = 0.70 with decreasing values of x. As an example, using this method for the first four increments, the integral is equal to ∆x[ 0.5fx=0.40 + fx=0.39 + fx=0.38 + fx=0.37 + 0.5fx=0.36 ], where f = 1/(y-x). At each step , W is computed and then the mass in pounds until 166.7 lb of residue is reached. The complete spreadsheet is given below From it, the following result is obtained: By interpolation, mole fraction of B in the residue = 0.630, corresponding to W = 2.005 lbmol. Using these values, a material balance on B gives for Eq. (13-6), 3.037(0.7) − 2.005(0.63) = 0.836 3.037 − 2.005 y f =1/(y -x) Value of W, lbmol Integral ( yD )avg = x 0.70 0.68 0.66 0.64 62 0.8550 0.8428 0.8302 0.8172 0.8037 6.452 6.143 5.875 5.643 5.444 0.1259 0.2461 0.3613 0.4722 3.307 2.678 2.374 2.116 1.894 W, lb 250.0 221.2 196.8 176.0 158.0 Exercise 13.5 Subject: Simple differential batch (Rayleigh) distillation of a mixture of benzene and toluene. Given: Mixture of 60 mol% benzene (B) and 40 mol% toluene (T). Distillation at 1 atm at the conditions below. Assumptions: Relative volatility = α = αB-T = 2.43, a constant. Perfect mixing in the still. Exiting vapor in equilibrium with the liquid. Find: (a) (b) for: (1) (2) (3) Number of moles in the distillate for 100 moles of feed Compositions of distillate and residue. A (cumulative) distillate of 70 mol% B. 40 mol% of the feed distilled. 60 percent of B distilled. Analysis: Calculations are made in terms of benzene, the more volatile component. Because the relative volatility is assumed constant, Eq. (13-5) applies, which for x0 = 0.60, W0 = 100 moles, and α = 2.43 is: ln 100 0.60 1− x = 0.6993 ln + 2.43 ln W x 0.40 (1) The composition of the cumulative distillate is given by Eq. (13-6), which becomes: yD Case 1: avg From Eq. (2), y D = avg 60 − Wx 100 − W = 0.70 = (2) 60 − Wx 100 − W or W = 10 0.7 − x (3) Substituting Eq. (3) into Eq. (1) and solving the resulting nonlinear equation gives x = 0.33. Then from Eq. (3), W = 27.0 moles. The distillate = W0 - W = 100 - 27 = 73 moles. In lieu of a nonlinear equation solver, a spreadsheet can be used by assuming a value of W and solving Eq. (3) for x. Then substitute this value into Eq. (1) and solve for W. Repeat until the assumed value of W equals the calculated value. This is...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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