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

Liquid in the underflow and liquid in the overflow

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Unformatted text preview: . Final Overflow 10 wt% Na2S 1 120,000 kg/d H2O 40,000 kg/d BaS solid Na2CO3 Washing Water 2 3 4 Each underflow is two parts water/part BaCO3. (b) Na2CO3 requirement and BaCO3 and Na2S produced by stoichiometry: Molecular Weight: BaS (aq) + Na2CO3 (s) = BaCO3 (s) + Na2S (aq) 169.42 106.00 197.37 78.05 5 Final Underflow Exercise 16.2 (continued) Flow rate of Na2CO3 required = 40,000(106.00/169.42) = 25,027 kg/d Flow rate of BaCO3 produced = 40,000(197.37/169.42) = 46,599 kg/d Flow rate of Na2S produced = 40,000(78.05/169.42) = 18,428 kg/d Determination of fresh water needed washing: Final underflow contains all of the solid BaCO3 = 46,599 kg/d Water in the final underflow = 2(46,599) = 93,198 kg/d Since the precipitation reaction is complete and under stoichiometric conditions, the final overflow contains only water and Na2S. Need to write mass balances for Na2S for each stage. Best to write these in terms of mass ratios of Na2S because underflow ratios are given as parts of H2O (rather than parts solution) per part of solid BaCO3. Yi = mass ratio of Na2S to water in the overflow from Stage i Xi = mass ratio of Na2S to water in the underflow from Stage i Vi = kg/d of water in the overflow from Stage i. Li = kg/d of water in the underflow from Stage i. S = kg/d of wash water entering Stage 5. Then: Y1 = 0.1/0.9 = 0.1111 = X1 and all Yi = Xi . Since the flow rate of water in the underflows is constant, L1 = L2 = L3 = L4 = L5 = 93,198 kg/d of water Overall water balance: 120,000 + S = V1 + 93,198 or S = V1 – 26,802 A water balance around each stage results in, V2 = V3 = V4 = V5 = V1 – 26,802 = S Mass balances for Na2S around each stage: Stage 1: 18,428 + Y2V2 = X1L1 + Y1V1 or 18,428 + Y2S = 0.1111(93,148) + 0.1111(S + 26,802) (1) Stage 2: X1L1 + Y3V3 = X2L2 + Y2V2 or 0.1111(93,198) + Y3S = Y2(93,198) + Y2S (2) Stage 3: X2L2 + Y4V4 = X3L3 + Y3V3 or Y2(93,198) + Y4S = Y3(93,198) + Y3S (3) Stage 4: X3L3 + Y5V5 = X4L4 + Y4V4 or Y3(93,198) + Y5S = Y4(93,198) + Y4S (4) Stage 5: X4L4 = X5L5 + Y5V5 or Y4(93,198) = Y5(93,198) + Y5S (5) We have five linear equations in five variables: S, Y2, Y3, Y4, and Y5 Solving with Polymath, S = 130,600 kg/d and Yi values from Stage 2 to 5 are 0.07207, 0.04421, 0.02433, and 0.01013. (c) Since Yi = Xi and Wt% Na2S in an underflow = [Xi/(1 + Xi)](100%) =[Yi/(1 + Yi)](100%), the Wt% values from Stage 2 to 5 are: 6.72%, 4.23%, 2.37%, and 1.00%. The final dried underflow contains 0.01013(93,198) = 944 kg/d of Na2S. Therefore, purity of the final dried BaCO3 is 46,599/(944 + 46,599) = 0.980 or 98.0%. Let: Exercise 16.3 Subject: Precipitation of CaCO3 from the reaction of solid CaO and an aqueous solution of Na2CO3, followed by two stages of washing. Given: A process consisting of a precipitation vessel followed by a two-stage, continuous, countercurrent washing system using water. Slurry of 100,000 lb/h from the precipitation tank is 5 wt% solid CaCO3, 0.1 wt% NaOH, and 94.9 wt% water. Wash is 20,000 lb/h of water. Underflow from each of the two washing stages is 20 wt% solids. Assumptions: No solids in the overflows. Liquid in the underflow and liquid in the overflow leaving a sta...
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This document was uploaded on 02/24/2014 for the course CBE 2124 at NYU Poly.

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