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

The component material balance in lbh is as follows

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Unformatted text preview: corresponding flow rate of exiting solvent is SD = 15,893.6 - 1,986.7 = 13,906.9 lb/h The slope of the solvent-free operating line at the top of the cascade = LD/V'N = 1,644.2/1,986.7 = 0.828. However, this slope is not constant because as we move down from the top of the cascade, all three components undergo mass transfer between the two liquid phases in an overall non-equimolar fashion. At the bottom of the cascade, by overall solvent material balance, the entering solvent flow rate = SB = 13,906.9 + 67.4 - 140 = 13,834.3 lb/h Next, determine two Y - X points on the operating line above the feed entry by writing the following balances around a section of stages from Stage n above the feed to Stage N at the top, in terms of total stream mass flows, V and L, and mass fractions (including solvent), y and x: Total mass balance: Solute mass balance: Solvent mass balance: Vn-1 = Ln + 14,249.4 (yn-1)A Vn+1 - (xn)A Ln = 284.3 (yn-1)C Vn+1 - (xn)C Ln = 13,906.9 (5) (6) (7) Eqs. (5) to (7) are solved by trial and error for a selected value of Xn. To do this, combine Eq. (5) with (6) to eliminate Vn-1, and combine Eq. (5) with (7) to eliminate Vn-1. We now have two equations for Ln. The selected value of Xn fixes the mass fractions of A and C in the raffinate. Then find a value of Yn-1 (which fixes the mass fractions of A and C in the extract) that gives identical values of Ln. The results are as follows, with these two points for Y and X, together with Y = X = 0.83 establishing the upper operating line, as shown in the Y - X diagram on the second page of this exercise: Xn 0.621 0.405 Yn-1 0.670 0.520 Vn-1, lb/h 15,790 15,400 Ln, lb/h 1,541 1,151 Exercise 8.26 (continued) Analysis: (continued) Next, determine two Y - X points on the operating line below the feed entry by solving the following balances around a section of stages from Stage m below the feed to bottom Stage1: Total mass balance: Solute mass balance: Solvent mass balance: Vm = Lm+1 - 13,109.4 (ym)A Vm - (xm+1)A Lm+1 = -65.7 (ym)C Vm - (xm+1)C Lm+1 = 13,766.9 (8) (10) (9) . The results are as follows, with these two points for Y and X, together with Y = X = 0.10 establishing the lower operating line, as shown in the Y - X diagram on the second page of this exercise: Xm+1 0.287 0.166 Ym 0.390 0.205 Vm, lb/h 15,182 15,064 Lm+1,lb/h 2,073 1,955 From the Y - X plot on the second page of this exercise, 10 equilibrium stages are stepped off, with 2 above the feed stage and 7 below. The component material balance in lb/h is as follows: Component A B C Total Feed 350.0 650.0 140.0 1140.0 Solvent in 0.0 0.0 13834.3 13834.3 Extract 284.3 58.2 0.0 342.5 Raffinate 65.7 591.8 67.4 724.9 Solvent out 0.0 0.0 13906.9 13906.9 Reflux, LD 1364.7 279.5 0.0 1644.2 Exercise 8.27 Subject: Extraction of methylcyclohexane (MCH) from n-heptane (C) with aniline (S) at 25oC in a countercurrent-stage extractor. Given: Feed of 50 wt% MCH in C. On a solvent-free basis, extract contains 95 wt% MCH and raffinate contains 5 wt% MCH. Reflux at both ends as in...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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