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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
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
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
0.00178 E, kg/h S, kg/h 74.984
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)
(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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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.
- Spring '11
- The Land