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)
In the metal industry - leaching of copper salts from ground ores
using sulfuric acid or ammoniacal solutions
Leaching of toxic materials into groundwater is a major health concern
FKKKSA
Chem. Eng. Dept
EQUILIBRIUM RELATIONS & SINGLE-STAGE LEACHING
Fresh solvent
Overflow (enriched solvent)
Rich solids
Underflow (spent solids)
Assumptions:
solute-free solid (B) is insoluble in the solvent (C)
sufficient time for equilibrium
no adsorption of solute (A) back into solid
Concentration of solution in overflow = concentration of solution in underflow
Concentration of inert solid (B) in solution of underflow, N:
kg B
N=
kg A + kg C
Weight fraction of solute (A) in overflow V,xA:
x A = kg A
kg A + kg C
kg A
Weight fraction of solute (A) in underflows solution L,yA: y A = kg A + kg C
FKKKSA
Chem. Eng. Dept
EQUILIBRIUM RELATIONS
Constant underflow concentration straight & horizontal line in N vs yA
Varying underflow concentration a curve in N vs yA
FKKKSA
Chem. Eng. Dept
SINGLE-STAGE LEACHING
Overflow
Fresh solvent
Rich solids
Underflow
V = overflow flowrate (kg/h)
L = underflows solid-free flowrate (kg/h)
L1 L0
M = imaginary total flowrate (kg/h)
Total material balance:
L0 + V2 = L1 + V1 = M
Balance on A:
L0yA0 + V2xA2 = L1yA1 + V1xA1 =MxAM
Balance on B:
V1
L0N0 + 0 = L1N1 + 0 =MNM
FKKKSA
Chem. Eng. Dept
V2
xAM
Example 12.9-1
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
V1 ,xA1
retained.
V2 = 100 kg
1 atm
Comp. A = soy oil
xA2=0
L1,N1
B0 = 100(1-0.2) = 80kg
Comp.B = soy bean
293K
yA1
L0 = 100 80 = 20 kg
Comp.C = hexane
Total material balance: N0 = 80/20=4, yA0 = 1
L0 + V2 = L1 + V1 = M
Balance on A:
20 + 100 = L1 + V1 = 120
20(1) + 100(0) = L1xA1 + V1yA1 =120(xAM)
(xAM) =0.167
From the graph:
yA1 = 0.167 xA1 = 0.167
Solving: L1 = 53.3 kg V1 = 66.7 kg
FKKKSA
Chem. Eng. Dept
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 =
FKKKSA
Chem. Eng. Dept
Example 12.10-1
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/h
y0A = 800/ 850 = 0.941
N0 = 2000/850 = 2.36
VN+1 = 1310 + 20 = 1330 kg/h
xN+1A = 20/ 1330 = 0.015
Solvent free calculation for leached solid:
LN = 120 + C
yNA = 120/(120 +C)
NN = 2000/120+C
yNA = 120/120 = 1
NN = 2000/120 = 16.67
Total material balance:
L0 + VN+1 = LN + V1 = M
850 + 1330 = LN + V1 = 2180
Balance on A: L y + V x
0 A0
N+1 AN+1 = LNyAN + V1xA1 =MxAM
850(0.941) + 1330(0.015) = LNyAN + V1xA1 =2180(xAM)
xAM = 0.376
FKKKSA
Chem. Eng. Dept
Example 12.10-1
1. Plot N vs yA,xA with underflow & overflow. Locate L0, VN+1 & LN
L0 = 850 kg/h
L0
LN
y0A = 800/ 850 = 0.941
N0 = 2000/850 = 2.36
VN+1
VN+1 = 1330 kg/h
V1
xN+1A = 20/ 1330 = 0.015
Solvent free calculation for leached solid:
yNA = 120/120 = 1
N 0
NN = 2000/120 = 16.67= slope = y 0
M = 2180
N 0 = N 0 = 16.67
xAM = 0.376
N = 1.67
y 0 0.1 0
2. Locate M on the line joining VN+1 & L0. A line from LN through M will give V1
on the overflow.
FKKKSA
Chem. Eng. Dept
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
L1
LN
M
VN+1
V1
FKKKSA
Chem. Eng. Dept
L0
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.
L4LN
VN+1V4
L3
L2
L1
V3
V2
V1
L0
No. of theoretical stages = 3.6
FKKKSA
Chem. Eng. Dept
