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

Oil material balances around stages 1 2 and 3

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Unformatted text preview: L0 +V2y2 = L1x1+ V1y1 or 2,500 + 5,000y2 = 1,346x1 + 6,154y1 (1) Oil material balance around Stage 2: L1x1 = L2x2+ V2y2 or 1,346x1 = 1,346x2 + 5,000y2 (2) From equilibrium at each stage for the leaving liquids, y1 = x1 and y2 = x2 (3, 4) Solving the four linear equations, (1), (2), (3), and (4), we obtain: y1 = 0.3882 x1 = 0.3882 y2 = 0.08234 x2 = 0.08234 Analysis: (continued) Exercise 5. 4 (continued) Oil in the final extract = V1y1 = 6,154(0.3882) = 2,389 kg/h Percent extraction of oil = (2,389/2,500) x 100% = 95.6% This compares to 83.3% for one stage. (b) ) The three-section cascade is shown below. Soybean meal enters Stage 1 at 5,000 kg/h and the benzene solvent enters Stage 3 at 5,000 kg/h. Underflow liquids are L1 = L2 = L3 = 1,346 kg/h. Total liquid material balances around Stages 3, 2, and 1, respectively, similar to those in Part (a), give: V3 = V2 = 5,000 kg/h and V1 = 6,154 kg/h. Oil material balances around Stages 1, 2, and 3, respectively, similar to those in Part (a) give: 2,500 + 5,000y2 = 1,346x1 + 6,154y1 (5) 1,346x1 + 5,000y3 = 1,346x2 + 5,000y2 (6) 1,346x2 = 1,346x3 + 5,000y3 (7) From equilibrium at each stage for the leaving liquids, y2 = x2 y3 = x3 (8, 9, 10) y1 = x1 Solving the six linear equations, (5) through (10), we obtain: y1= x1 = 0.4015 y2 = x2 = 0.1023 y3 = x3 = 0.02169 Oil in the final extract = V1y1 = 6,154(0.4015) = 2,471 kg/h Percent extraction of oil = (2,471/2,500) x 100% = 98.8% This compares to 83.3% for one stage and 95.6% for two stages. (c) From Eq. (5-12), the minimum solvent rate corresponds to zero overflow, with all liquid going to the underflow. Liquid in the underflow = 1,346 kg/h. Can not apply the minimum solvent rate concept here because the liquid in the underflow is not stated as just solvent, but as total liquid including the oil. The total liquid in the underflow is less than the oil in the feed. However, note that for the given solvent rate, somewhat more than the desired 98% extraction is achieved with three stages. Exercise 5.5 Subject: Leaching of sodium carbonate from an insoluble oxide with water in a countercurrent-flow cascade. Given: From Example 5.1, entering flow rates are: Na2CO3 = FB = 1,350 kg/h, insoluble oxide = FA = 2,400 kg/h, and water = S = 4,000 kg/h. Underflow from each stage = 40 wt% on a solute (Na2CO3)-free basis. Assumptions: Equilibrium leaching stages where, for each stage, the weight fraction of carbonate in the overflow equals that in the underflow liquid. No solids in the overflows. Find: Minimum solvent water rate in lb/h. Ratio of solvent rate to minimum solvent rate. Percent recovery of carbonate with five stages and solvent rates from 1.5 to 7.5 times the minimum solvent rate. Analysis: R = mass of solvent/mass of insoluble solids = 40/60 = 0.667. Therefore solvent rate in the underflow = 2,400(0.667) = 1,600 kg/h. From Eq. (5-12), Smin = RFA = 0.667(2,400) = 1,600 kg/h or 3,630 lb/h Therefore, the ratio of solvent to minimum solvent = 2,400/1,600 = 2.5 To dete...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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