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

0004573 00001676 000007272 000003556 recovery 8683

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Unformatted text preview: rmine percent recoveries (extraction) of carbonate for 5 stages and various S/Smin , apply Eq. (5-10) to determine the mass ratio of carbonate to solvent in the final underflow leaving the F 1 last Stage N: (1) XN = B S W N −1 where, from Eq. (5-5), W =S/RFA = S/Smin (2) Combining Eqs. (1) and (2), noting that FB = 1,350 kg/h, FA = 2,400 kg/h, R = 0.667, and N = 5, XN = FB 1 FA R W N or X 5 = 1,350 (2,400)(0.667) 1 S S min 5 = 0.84375 S S min 5 (3) With a solvent rate in the underflow of 1,600 kg/h, the carbonate leaving in the underflow is 1,600 X5 . Therefore, the carbonate leaving in the final overflow (extract) = FB - 1,600 X5 = 1,350 - 1,600 X5 . 1,350 − 1,600 X 5 The percent extraction = × 100% (4) 1,350 Exercise 5.5 (continued) Analysis: (continued) Solving Eqs. (3) and (4), the following results are obtained: S/Smin 1.5 2.5 3.5 4.5 5.5 6.5 7.5 X5 0.1111 0.00864 0.001607 0.0004573 0.0001676 0.00007272 0.00003556 % Recovery 86.83 98.98 99.81 99.95 99.98 99.99 99.996 Exercise 5.6 Subject: Recovery of Al2(SO4)3 from bauxite ore by reaction of Al2O3 with sulfuric acid followed by three-stage countercurrent washing with water, with the flowsheet below Given: 40,000 kg/day of bauxite ore containing 50 wt% Al2O3 and 50 wt% inert.Stoichiometric amount of 50 wt% aqueous H2SO4 to convert the Al2O3 to Al2(SO4) by the reaction: Al2O3 + 3 H2SO4 = Al2(SO4)3 + 3H2O (1) S = 240,000 kg/day of water to extract the sulfate in three countercurrent washing stages. Underflow from each washing stage contains 1 kg of water/kg of insoluble inert. Assumptions: Equilibrium washing stages where, for each stage, the weight fraction of sulfate in the overflow equals that in the underflow liquid. No solids in the overflows. Reaction of oxide with sulfuric acid is complete. Find: Flow rates in kg/day of sulfate, water, and inert solid in each product stream. Percent recovery of sulfate. Effect of adding one more stage. Analysis: In Eq. (1), the molecular weights from left to right are 101.96, 98.082, 342.158, and 18.016. Therefore, the yield of sulfate = 0.50(40,000)(342.158)/101.96 = 67,116 kg/day. Water leaving the reactor = water entering with the sulfuric acid + water from the reaction. Sulfuric acid entering the reactor = 0.50(40,000)(3)(98.082)/101.96 = 57,718 kg/day. Same amount of water, 57,718 kg/day, enters the reactor. Analysis: (continued) Exercise 5.6 (continued) Water from the reaction = 0.50(40,000)(3)(18.016)/101.96 = 10,602 kg/day. Total water leaving reactor and entering Stage 1 = 57,718 + 10,602 = 68,320 kg/day = L0 Inert solid leaving reactor = 0.5(40,000) = 20,000 kg/day. For the three-section washing cascade shown above, reactor effluent enters Stage 1 at a liquid rate of 68,320 + 67,116 = 135,436 kg/day and an inert solids rate of 20,000 kg/day. Water solvent enters Stage 3 at a rate, S, of 240,000 kg/day. Let L1 = L2 = L3 = water leaving in each underflow, which equals the inert solids rate or 20,000 kg/day. Let Vi = water rate in overflow from Stage i. Water material balances around Stages...
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