leaching

FKKKSA Chem. Eng. Dept

SOLID-LIQUID EXTRACTION/LEACHING

•Separation of solutes from solid using liquid solvent

•Eg. 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


EQUILIBRIUM 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 solid

Weight fraction of solute (A) in underflow’s solution L,yA:

Fresh solventOverflow (enriched solvent)

Rich solids Underflow (spent solids)

Concentration of inert solid (B) in solution of underflow, N:

C kg AkgBkgN+

=

C kg AkgAkgyA +

=

C kg AkgAkgxA +

=

Concentration of solution in overflow = concentration of solution in underflow


EQUILIBRIUM RELATIONS

Constant underflow concentration – straight & horizontal line in N vs yA

Varying underflow concentration – a curve in N vs yA


SINGLE-STAGE LEACHING

Fresh solventOverflow

Rich solids Underflow

V = overflow flowrate (kg/h)L = underflow’s 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 =MxAM

Balance on B:

L0N0 + 0 = L1N1 + 0 =MNM

L0

M

L1

V1

V2 xAM


Example 12.9-1

Total material balance:L0 + V2 = L1 + V1 = M

Balance on A:20(1) + 100(0) = L1xA1 + V1yA1 =120(xAM)

From the graph: yA1 = 0.167 xA1 = 0.167

L1,N1

yA1

B0 = 100(1-0.2) = 80kg

L0 = 100 – 80 = 20 kg

N0 = 80/20=4, yA0 = 1

V1 ,xA1 V2 = 100 kg

xA2=01 atm

293KComp.B = soy beanComp.C = hexane

Comp. A = soy oil

100 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 kg


COUNTERCURRENT MULTISTAGE LEACHING

Total material balance:L0 + VN+1 = LN + V1 = M

Balance on A:

L0yA0 + VN+1xAN+1 = LNyAN+ V1xA1 =MxAM

Difference flows ∆:

∆ = L0 - V1 = LN- VN+1 = …


Example 12.10-1

Balance on A: L0yA0 + VN+1xAN+1 = LNyAN + V1xA1 =MxAM

Feed: 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 = 2.36

VN+1 = 1310 + 20 = 1330 kg/hxN+1A = 20/ 1330 = 0.015

LN = 120 + CyNA = 120/(120 +C)

Solvent free calculation for leached solid:

NN = 2000/120+C

yNA = 120/120 = 1NN = 2000/120 = 16.67

850(0.941) + 1330(0.015) = LNyAN + V1xA1 =2180(xAM)

Total material balance: L0 + VN+1 = LN + V1 = M850 + 1330 = LN + V1 = 2180

xAM = 0.376


Example 12.10-1

1. Plot N vs yA,xA with underflow & overflow. Locate L0, VN+1 & LN

L0 = 850 kg/hy0A = 800/ 850 = 0.941N0 = 2000/850 = 2.36VN+1 = 1330 kg/hxN+1A = 20/ 1330 = 0.015

yNA = 120/120 = 1

M = 2180xAM = 0.376∴ N = 1.67

L0

VN+1

LN

Solvent free calculation for leached solid:

NN = 2000/120 = 16.67= slope = 0y0N

−−

16.6700.10N

0y0N =

−−=

−−

M

V1

2. Locate M on the line joining VN+1 & L0. A line from LN through M will give V1on the overflow.


Example 12.10-1

3. From V1 draw a vertical tie line to give L1. Locate ∆ from the intersection of the lines L0V1 & VN+1 LN

L0

VN+1

LN

M

V1

L1

∆


Example 12.10-1

4. 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.

L0

VN+1

LN

V1

L1

∆

V2

L2

V3

L4 L3

V4

No. of theoretical stages = 3.6

leaching

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