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

975 which is the solvent free composition of the

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Unformatted text preview: rmine the extract E1, followed by drawing lines from E! to RN and from S to F, with the intersection of these two lines determining the mixing point, M. The wt% A at the mixing point is approximately 17 wt% From Eq. (8-10), S min ( xA ) F − ( xA ) M 0.27 − 0.17 = 0.59 = = F ( xA )M − ( xA )S 0.17 − 0.0 (b) The composition of the extract at the minimum solvent rate is: 34 wt% acetone, 1.5 wt% water, and 64.5 wt% solvent (c) If the pinch point is at the feed end of the system as assumed above, then the extract leaving Stage 2 is essentially the same composition as the final extract. Analysis: (continued) Exercise 8.24 (continued) Exercise 8.25 Subject: Liquid-liquid extraction of methylcyclohexane (A) from a mixture with n-heptane (C) by aniline (S) as the solvent, using extract reflux. Given: Feed of 50 wt% C and 50 wt% A. Extract to contain 92.5 wt% A in the extract, and 7.5 wt% A in the raffinate, both on a solvent-free basis. Equilibrium data at 25oC and 1 atm. Assumptions: Perfect solvent separator. Find: Using the graphical method of Maloney and Schubert: (a) Minimum stages. (b) Minimum extract reflux ratio. (c) Theoretical stages for an extract reflux ratio = 7. Analysis: The cascade is like that shown in Fig. 8.26b or 8.32, where extract reflux is used. The Maloney-Schubert method for the use of extract reflux is conducted on a Janecke diagram like that in Figs. 8.31 and 8.33. For the system here, the diagram is shown on the next page, where the composition on the ordinate, Y, is mass of solvent (S) per mass of solvent-free extract (upper curve), and per mass of solvent-free raffinate (lower curve). The abscissa composition, X, is mass of solute (A) per mass of solvent-free extract or raffinate. The dashed lines are the liquidliquid equilibrium tie lines. (a) For the minimum number of equilibrium stages, the construction is that is Fig. 8.31a, where all operating lines on the Y - X diagram are vertical lines. As shown on the following page, the number of stages stepped off by alternating between equilibrium tie lines (many of which are added by interpolation) and vertical operating lines starting from the raffinate, L1, and ending at the extract, VN , is just less than 10. Call the minimum number of stages 10. (b) For the minimum extract reflux ratio, the construction is that shown in Figs. 8.31a and b and is given here on the third page of this exercise. A tie line through the feed point, F, establishes the point P' on a vertical line through X = 0.975, which is the solvent-free composition of the solute in the extract. No other line drawn through a tie line to the right of the feed tie line or drawn from P'' points (established by tie lines to the right of the feed tie line) through the feed point, intersects the vertical line through X = 0.975 at a higher P' point. Therefore, as shown, P'min = 20.5. Using Eq. (8-18) with measurements from the diagram, the minimum reflux ratio = line P'VN / line VNLR = LR/D = 2.5 (c) With an extract reflux r...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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