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

The constructions to obtain p1 and p2 are shown on

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Unformatted text preview: diagram. Analysis: Because the flow rates and compositions of both feeds and the solvent are given, the mixing point, M, can be computed, with the following results: kg/h: Component Feed, F Feed, F' Solvent, S Mixing point, M Acetone 3,750 1,875 0 5,625 Water 3,750 5,625 0 9,375 Trichloroethane 0 0 5,000 5,000 Total: 7,500 7,500 5,000 20,000 Thus, the composition of the mixing point is 28.125 wt% A, 46.875 wt% C, and 25.0 wt% S. This is shown below on the right-triangle equilibrium diagram. Also shown is the composition of the raffinate, R, on the equilibrium curve and corresponding to 10 wt% A. A straight line extending from R, through M, to its intersection with the equilibrium curve, establishes the extract composition at point E, as shown in the diagram below. The compositions of the raffinate and extract are thus determined from the diagram to be: Composition Acetone Water Trichloroethane Total: Raffinate, wt% 10.0 89.5 0.5 100.0 Extract, wt% 46.5 3.7 49.8 100.0 By component material balances or the inverse lever-arm rule, using the mixing point, the raffinate rate is 10,060 kg/h, the extract rate is 9,940 kg/h. Because the two feeds differ in composition, they should not be mixed before entering the cascade, but should enter separately at different stages. Note that feed F' has a composition closer to the final raffinate, leaving from stage N, than feed F. Therefore, feed F' should be fed to an intermediate stage, m, and feed F should be fed to stage 1, from which the final extract leaves. Feed F' should enter at that stage, m, where it matches most closely the intermediate raffinate entering stage m. In the diagram on the next page, two operating points are shown, P1 and P2. Analysis: (continued) Exercise 8.17 (continued) Point P1 is used to step off stages from m to N, based on a material balance for the entire cascade, because both F and F' move in the direction toward the final raffinate. Thus, the material balance is: F + F' + S = E1 + RN = M Rearranging for passing streams, F + F' - E1 = RN - S = P1 (1) Straight lines drawn through (F + F1) and E1, and through RN and S, intersect at the operating point P1. For stages 1 through m-1, the material balance is: F + Em = Rm-1 +E1 Rearranging for passing streams, F - E1 = Rm-1 - Em = P2 (2) Substituting Eq. (2) into Eq. (1), P2 + F' = RN - S = P1 (3) Exercise 8.17 (continued) Analysis: (continued) Therefore, operating point P2 is obtained by the intersection of straight lines drawn through F and E1, and through P1 and F'. The constructions to obtain P1 and P2 are shown on the next page. Note that for this system, P1 and P2 are on opposite sides of the triangular diagram. The equilibrium stages are stepped off starting with P2 from stage 1 at the end where the final extract, E1, is produced. By a tie line, E1 is in equilibrium with R1. An operating line connecting R1 with P2 gives E2. The stepping off, using P2, is continued until R2 is reached, which is very close to the composition of F'. Therefore, F' is fed to stage 2. The remaining stages are stepped off...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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