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

9213500 12420 kgh since all of the water appears in

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Unformatted text preview: ontinued) Exercise 4.45 Subject: Comparison of solvents for single-equilibrium-stage liquid-liquid extraction. Given: Feed, F = 13,500 kg/h of 8 wt% acetic acid (B) in water (A) at 25oC. Four solvents (C), each with a different distribution coefficient, KC, in mass fractions, xB, for acetic acid, as given in the table below, according to Eq. (2-20), where (1) is the extract of flow rate E and (2) is the raffinate of flow rate R, where for a single equilibrium stage, the raffinate is to contain only 1 wt% B ( ( K DB = xBE ) / xBR ) (1) Assumptions: Water is insoluble in the solvent and the solvent is insoluble in water. Find: The kg/h, S, of each solvent required. ( ( Analysis: In Eq. (1), xBR ) = 0.01. Therefore, xBE ) = 0.01 K DB A total material balance gives: (2) F = 13,500 = E + R - S (3) An acetic acid material balance gives: ( ( ( ( xBF ) F = (0.08)(13,500) = 1,080 = xBE ) E + xBR ) R = xBE ) E + 0.01R (4) 92 wt% of the feed is water, or (0.92)(13,500) = 12,420 kg/h. Since all of the water appears in the raffinate, which is 99 wt% water, R = 12,420/0.99 = 12,546 kg/h. Eq. (3) becomes: S = E - 954 (5) and Eq. (4) becomes: ( xBE ) E = 954.5 (6) ( Eqs. (2), (5), (6) are three equations in three unknowns: S, E, and xBE ) . For each solvent, solve ( Eq. (2) for xBE ) . Solve (6) for E. Solve (5) for S. The results are: Solvent Methyl acetate Isopropyl ether Heptadecanol Chloroform KD 1.273 0.429 0.312 0.178 ( xBE ) 0.01273 0.00429 0.00312 0.00178 E, kg/h S, kg/h 74.984 222,500 305,940 536,260 74,030 221,500 305,000 535,300 Although methyl acetate is the best solvent, the solvent rates required are very large. To reduce the solvent rate, use a countercurrent, multiple-stage system. Exercise 4.46 Subject: Liquid-liquid extraction of ethylene glycol from water by furfural with one stage. Given: Feed, F = 45 kg, of 30 wt% ethylene glycol (B) and 70 wt% water (A). Phase equilibrium diagrams of Fig. 4.14a and 4.14e for 25oC. Find: (a) Minimum amount of solvent. (b) Maximum amount of solvent. (c) % glycol extraction and amounts of solvent-free extract and raffinate for 45 kg solvent. (d) Maximum possible glycol purity in extract. Maximum purity of water in raffinate. Analysis: For a single stage, all mixtures of feed, F, and solvent, S, lie on a straight line between these two points as shown in the following ternary diagram of Fig. 4.14a. (a) The minimum amount of solvent corresponds to the maximum solubility of the solvent in the feed. This is point M1 in the diagram below. By the inverse lever arm rule, S/F = 0.097. Therefore, Smin = 0.0972(45) = 4.4 kg. In this case, no extract is obtained. (b) The maximum amount of solvent corresponds to maximum solubility of the feed in the solvent. This is point M2 in the diagram below. By the inverse lever arm rule, S/F = 11.15. Thus, Smax = 11.2(45) = 504 kg. In this case, no raffinate is obtained. Exercise 4.46 (continued) Analysis: (continued) (c) With 45 kg of solvent, S/F = 1. Therefore, the mixing point, M3 is at the mid point between F and S in the diagram below. A tie line drawn through point M3 determine...
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