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

# 0 030 02 034 04 038 06 042 analysis because the

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Unformatted text preview: il in the final extract, if all leaching takes place in the leaching stage. (b) The % recovery of oil in the final extract, if leaching takes 3 of the total of 4 stages, with the amount of leaching divided equally among the three stages. Analysis: Because the underflow basis is on total liquid (not just the hexane), use mass fractions to express the oil composition in the overflow and in the liquid in the underflow. Let: yi = mass fraction of oil in the overflow from Stage i xi = mass fraction of oil in the underflow from Stage i Vi = kg/h of liquid in the overflow from Stage i Li = kg/h of liquid in the underflow from Stage i S = kg/h of fresh n-hexane entering the last stage (a) This part can be solved from oil mass balances around each stage or by applying (16-8) of the McCabe-Thiele algebraic method. In either method: (1) the flowsheet of Figure 16.7 applies with a leaching stage, L, and N = 3 washing stages; and (2) the leaching stage is computed separately. Solve with (16-8). For both methods, 40,000 kg/h of insoluble solids appear in each underflow. Therefore, all L = 0.8( 40,000) = 32,000 kg/h. The total final underflow = 40,000 + 32,000 = 72,000 kg/h The total flow rate of flaked soybeans = 50,000 kg/h The flow rate of entering n-hexane solvent = 50,000 kg/h Therefore, by overall total mass balance, VL = 50,000 + 50,000 – 72,000 = 28,000 kg/h However, since all L = 32,000 kg/h, total mass balances around each washing stage give: V = 50,000 kg/h for all stages except the overflow leaving the leaching stage. Oil mass balance around the leaching stage: y1V1 + 10,000 = yLVL + xLLL or, since xL = yL, y1(50,000) + 10,000 = yL(28,000) + yL(32,000) = yL(60,000) Because there are two unknowns, we cannot compute yL. We need more equations. A total oil mass balance around all stages gives: 10,000 + 0 = x3(32,000) + yL(28,000) (1) (2) Exercise 16.5 (continued) From (16-8), log N= x3 − 0 yL − y1 or, 3 = 32, 000 log 50, 000 log or rearranging, x3 − y N +1 yL − y1 L log V x3 3log 0.64 = 10 ( ) = 0.262 yL − y1 (3) Equations (1) to (3) are three linear equations in three unknowns: y1, yL, and x3 Solving with Polymath, y1 = 0.1794, yL = 0.3162, x3 = 0.03585 Oil in the final extract = yLVL = 0.3162(28,000) = 8,854 kg/h The % recovery of oil = 8,854/10,000 = 0.8854 or 88.54% (b) We cannot use (16-8) for this part because the leaching is divided equally among the leaching stage and the first 2 washing stages. Until the oil is leached, it is part of the solid. However, because the underflow from each stage is given as 0.8 kg liquid/kg of oil-free soybeans, values of V and L remain the same as in Part (a) and xi = yi still applies. Write oil mass balances for each stage, using the nomenclature of Figure 16.7, noting that the following flow rates apply for unleached oil in the underflows: 10,000 – (1/3)(10,000) = 6, 667 kg/h to Stage 1 6,667 – (1/3)(10,000) = 3,333 kg/h to Stage 2 3,333 – (1/3)(10,000) = 0 kg/h to Stage 3 Stage L: y1V1 + 10,000 = yLVL + xLLL + 6,667 (4) or, y1(50,000) + 10,000 = yL(28,000) + yL(32,000) + 6,667 Stage 1: y2V2 + xLLL + 6,667 = y1V1 + x1L1 + 3,333 or, y2(50,000) + yL(32,000) + 6,667 = y1(50,000) + y1(32,000) + 3,333 (5) Stage 2: y3V3 + x1L1 + 3,333 = y2V2 + x2L2 or, y3(50,000) + y1(32,000) + 3,333 = y2(50,000) + y2(32,000) (6) Stage 3: 0 + x2L2 = y3V3 + x3L3 or, y2(32,000) = y3(50,000) + y3(32,000) (7) Solving linear Equations (4) to (7) with Polymath for the fou...
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