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

# 2736 y1 02616 y2 01873 y1 00731 therefore recovery

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Unformatted text preview: 4 and insoluble residue called gangue. Given: 20,000 kg/h of decanted sludge, consisting of 5 wt% water, 10 wt% ZnSO4, and 85 wt% gangue. A leaching stage and a continuous, countercurrent washing system, using wash water to produce an extract (strong solution) of 10 wt% ZnSO4 in water with a ZnSO4 recovery of 98%. Assumptions: Each underflow consists of two parts of water (ZnSO4-free basis) per part gangue. No solids in the overflows. Liquid in the underflow and liquid in the overflow leaving a stage have the same composition. Find: The number of stages Analysis: Because the underflow liquid is given as a mass ratio of ZnSO4-free water to the solid gangue in the underflow, use ZnSO4 compositions in the liquid as mass ratios, X, to water, and liquid flow rates on a ZnSO4-free basis. Let: Yi = mass ratio of ZnSO4 to water in the overflow from Stage i Xi = mass ratio of ZnSO4 to water in the underflow from Stage i Vi = kg/h of water in the overflow from Stage i Li = kg/h of water in the underflow from Stage i S = kg/h of wash water entering the last stage Decanted sludge consists of: 0.05(20,000) = 1,000 kg/h of water 0.10(20,000) = 2,000 kg/h of ZnSO4 0.85(20,000) = 17,000 kg/h of solid gangue ZnSO4 in the extract = 0.98(2,000) = 1,960 kg/h Water in the extract of 90% water = (9/1)(1,960 = 17,640 kg/h Final underflow is: 17,000 kg/h of gangue 2(17,000) = 34,000 kg/h of water 2,000 – 1,960 = 40 kg/h of soluble ZnSO4 By overall water mass balance, flow rate of wash water = 34,000 + 17,640 – 1,000 = 50,640 kg/h Use (16-8) of the McCabe-Thiele algebraic method to determine the number of stages. That equation is in terms of mass fractions, but it also applies to the use of mass ratios. Thus, log N= X N − YN +1 YL − Y1 L log V (1) where the subscripts on the symbols refer to Figure 16.7. Thus, the countercurrent system consists of one leaching stage, L, and N washing stages for a total of N+1 stages. Exercise 16.4 (continued) From the calculations above, XN = 40/34,000 = 0.001176 Also, YN+1 = 0 YL = 10/90 = 0.1111 S = 17,000 kg/h leaving each stage Therefore, L = 2(17,000) = 34,000 kg/h, constant from each stage And, by mass balance for each washing stage, V = 50,640 kg/h, a constant Obtain Y1 by a ZnSO4 mass balance around just the leaching stage, L: Y1V1 + 2,000 = 1,960 + XLLL But, V1 = V, LL = L, and XL = YL , and therefore, Y1(50,640) + 2,000 = 1,960 + 0.1111(34,000) Solving, Y1 = 0.0738 Now apply (1) to calculate the number of washing stages: log N= 0.001176 − 0 0.1111 − 0.0738 = 8.7 34, 000 log 50, 640 Therefore, need 9 ideal washing stages plus a leaching stage or 10 stages total. Exercise 16.5 Subject: Leaching of oil from flaked soybeans with n-hexane in a continuous, countercurrent system consisting of a leaching stage and three washing stages. Given: 50,000 kg/h of flaked soybeans containing 20 wt% oil and 80 wt% insoluble solids. 50,000 kg/h of n-hexane solvent for washing. From experiments, all underflows contain 0.8 kg of liquid/kg of oil-free soybeans. Assumptions: No solids in the overflows. Liquid in the underflow and liquid in the overflow leaving a stage have the same composition. Find: (a) The % recovery of o...
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