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CHAPTER 8 LEACHING OPERATION(LIQUID-SOLID SEPARATION)

Mohd Fadhil MajnisDepartment of Chem. & Pet. Eng.FETBE, UCSI University1

1TOPIC LEARNING OUTCOMEIntroduction of leaching. Analyze the equilibrium relations and single stage leaching. Analyze the countercurrent multistage leaching.2SOLID-LIQUID EXTRACTION/LEACHINGSeparation of solutes from solid using liquid solventEg. Soya milk (solute) from soya bean (inert solid) using water (liquid solvent) Leaching of toxic materials into groundwater is a major health concern In the metal industry - leaching of copper salts from ground ores using sulfuric acid or ammoniacal solutions

33EQUILIBRIUM RELATIONS & SINGLE-STAGE LEACHING solute-free solid (B) is insoluble in the solvent (C)Weight fraction of solute (A) in overflow V,xA:Assumptions: sufficient time for equilibrium no adsorption of solute (A) back into solidWeight fraction of solute (A) in underflows solution L,yA:

Fresh solventOverflow (enriched solvent)Rich solidsUnderflow (spent solids)Concentration of inert solid (B) in solution of underflow, N:

Concentration of solution in overflow = concentration of solution in underflow44EQUILIBRIUM RELATIONSConstant underflow concentration straight & horizontal line in N vs yAVarying underflow concentration a curve in N vs yA

55SINGLE-STAGE LEACHING

Fresh solventOverflowRich solidsUnderflowV = overflow flowrate (kg/h)L = underflows solid-free flowrate (kg/h)M = imaginary total flowrate (kg/h)

Total material balance:L0 + V2 = L1 + V1 = MBalance on A:L0yA0 + V2xA2 = L1yA1 + V1xA1 =MxAMBalance on B:L0N0 + 0 = L1N1 + 0 =MNML0ML1V1V2xAM66Example 12.9-1Total material balance:L0 + V2 = L1 + V1 = MBalance on A:20(1) + 100(0) = L1xA1 + V1yA1 =120(xAM)From the graph:yA1 = 0.167 xA1 = 0.167

L1,N1yA1B0 = 100(1-0.2) = 80kgL0 = 100 80 = 20 kgN0 = 80/20=4, yA0 = 1V1 ,xA1V2 = 100 kgxA2=01 atm293KComp.B = soy beanComp.C = hexaneComp. A = soy oil100 kg soy beans containing 20 wt. % oil is leached with 100 kg of fresh hexane. N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained.20 + 100 = L1 + V1 = 120(xAM) =0.167

Solving: L1 = 53.3 kg V1 = 66.7 kg7

7COUNTERCURRENT MULTISTAGE LEACHINGTotal material balance:L0 + VN+1 = LN + V1 = MBalance on A:L0yA0 + VN+1xAN+1 = LNyAN+ V1xA1 =MxAMDifference flows : = L0 - V1 = LN- VN+1 =

88Example 12.10-1Balance on A:L0yA0 + VN+1xAN+1 = LNyAN + V1xA1 =MxAMFeed: B = 2000 kg/h meal A = 800 kg oil C = 50 kg benzene Fresh solvent: C = 1310 kg benzene A = 20 kg oil Leached solids: A = 120 kg oil.L0 = 800 + 50 = 850 kg/hy0A = 800/ 850 = 0.941N0 = 2000/850 = 2.36VN+1 = 1310 + 20 = 1330 kg/hxN+1A = 20/ 1330 = 0.015LN = 120 + CyNA = 120/(120 +C)Solvent free calculation for leached solid:NN = 2000/120+CyNA = 120/120 = 1NN = 2000/120 = 16.67850(0.941) + 1330(0.015) = LNyAN + V1xA1 =2180(xAM)Total material balance:L0 + VN+1 = LN + V1 = M850 + 1330 = LN + V1 = 2180xAM = 0.3769

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Example 12.10-11. Plot N vs yA,xA with underflow & overflow. Locate L0, VN+1 & LNL0 = 850 kg/hy0A = 800/ 850 = 0.941N0 = 2000/850 = 2.36VN+1 = 1330 kg/hxN+1A = 20/ 1330 = 0.015yNA = 120/120 = 1M = 2180xAM = 0.376 N = 1.67L0VN+1LNSolvent free calculation for leached solid:NN = 2000/120 = 16.67= slope =

MV12. Locate M on the line joining VN+1 & L0. A line from LN through M will give V1 on the overflow.1010Example 12.10-13. From V1 draw a vertical tie line to give L1. Locate from the intersection of the lines L0V1 & VN+1 LN

L0VN+1LNMV1L11111Example 12.10-14. From L1 draw a line to to giveV2 on the overflow. From V2 draw a vertical tie line to give L2. Repeat until LN or exceed LN.

L0VN+1LNV1L1V2L2V3L4L3V4No. of theoretical stages = 3.61212