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

75 55 1275 composition wt chloroform 35 0 115 745

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Unformatted text preview: th 22.75 kg of water (S) in a single-stage extraction. (b) Compositions and weights of raffinate and extract if the raffinate from part (a) is extracted again with half its weight of water. (c) Composition of the raffinate from part (b) if all the water is removed from it. Use a right-triangle diagram of the equilibrium data, which is easily produced Analysis: with a spreadsheet. (a) In the diagram on the next page, the feed and solvent are represented by points F and S, respectively. The mixing point is M1, which is the sum of the feed and solvent (67.75 kg) with an overall composition of 33.58 wt% S, 23.25 wt% C, and 43.17 wt% A. A tie line passing through point M1 locates, on the equilibrium curve, the extract E1 and the raffinate R1. If the inverse lever arm rule is used to obtain E1 and R1, and their compositions are read from the diagram, the following results are obtained: Feed Solvent Extract Raffinate F S E1 R1 Amount, kg 45 22.75 55 12.75 Composition, wt%: Chloroform 35 0 11.5 74.5 Acetic acid 65 0 48.0 22.0 Water 0 100 40.5 3.5 (b) If R1 is mixed with half its weight of solvent (6.375 kg), the mixing point is M2, shown in the diagram on the next page, with an overall composition of 14.6 wt% A, 49.7 wt% C, and 35.7 wt% S. . A tie line passing through point M2 locates, on the equilibrium curve, the extract E2 and the raffinate R2. If the inverse lever arm rule is used to obtain E2 and R2, and their compositions are read from the diagram, the following results are obtained: Feed Solvent Extract Raffinate Water-free (R1) S E2 R2 Raffinate Amount, kg 12.75 6.375 8.895 10.23 10.10 Composition, wt%: Chloroform 74.5 0 1.0 91.0 93.2 Acetic acid 22.0 0 24.0 6.7 6.8 Water 3.5 100 74.9 1.3 0.0 (c) The water-free raffinate is included in the above table. Exercise 4.48 (continued) Exercise 4.49 Subject: L-L extraction of acetic acid from water by isopropyl ether at 25oC and 1 atm. Given: Equilibrium data for the ternary mixture Assumptions: Equilibrium stages. Find: (a) Compositions and weights of raffinate and extract when 100 kg of a 30 wt% acetic acid (A) and 70 wt% water (W) feed mixture (F) is extracted with 120 kg of isopropyl ether (E) in a single-stage extraction. Weight % of A in the extract if E is removed. (b) Compositions and weights of raffinate and extract if 52 kg of A and 48 kg of W are contacted with 40 kg of E Analysis: Use a right-triangle diagram of the equilibrium data, which is easily produced with a spreadsheet. (a) In the diagram on the next page, the feed and solvent are represented by points F1 and S, respectively. The mixing point is M1, which is the sum of the feed and solvent (220 kg) with an overall composition of 13.6 wt% A, 31.8 wt% W, and 54.6 wt% E. A tie line passing through point M1 locates, on the equilibrium curve, the extract E1 and the raffinate R1. If the inverse lever arm rule is used to obtain E1 and R1, and their compositions are read from the diagram, the following results are obtained, including the amount and composi...
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