EP325/BEP3043 Separation Process
Lesson 7
LIQUID-LIQUID EXTRACTION
EXTRACTION
•Removal of one or more components (solutes) from solids or liquids
using liquid solvent
LIQUID-LIQUID EXTRACTION/SOLVENT EXTRACTION
•Separation of two miscible liquids using another liquid (solvent)
•Eg. Vitamin A and D (solute) from fish oil (inert liquid) using liquid
propene (solvent)
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)
LIQUID-LIQUID EXTRACTION
• Takes advantage of the relative solubility of solutes in immiscible or nearly
immiscible liquids
• Solute dissolves more readily in the solvent in which it has a higher solubility
• Distribution Coefficient, K, determines the ratio of the concentration of the
solute in each liquid.
WHY TO CHOOSE LIQUID-LIQUID
EXTRACTION METHOD ???
• Separation by distillation is ineffective or difficult
• Boiling points of mixtures are close
• Flexibility in operation conditions choice is desired
• More than two components are present
• The material is heat sensitive
COUNTER CURRENT EXTRACTION
• Two immiscible fluids, usually one
light and one heavy fluid, flowing
continuously in opposite directions are
brought together and allowed to
separate
• Lighter liquid flows upward while the
heavier liquid flows downward
• Extract is the exit solvent rich stream
containing the desired extracted solute
• Raffinate is the exit residual stream
containing little solute
In solvent extraction, a raffinate is a liquid stream that remains
after the extraction with the immiscible liquid to remove solutes
EQUILIBRIUM RELATIONS IN EXTRACTION
component A = solute
component B = inert/carrier liquid
component C = liquid solvent
TERNARY DIAGRAM PLOTTING
TERNARY DIAGRAM PLOTTING
Acetic acid
100
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
30
25
20
15
10
5
Isopropyl 0 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
TERNARY DIAGRAM PLOTTING
Acetic acid
100
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
Step 1: Plotting liquid-liquid
equilibrium line for 3 components
using data above.
45
40
35
Step 2: Create equilibrium tie-line
by connecting a line between
exact and raffinate points.
30
25
20
15
10
5
0
Isopropyl
ether
0
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.1
Acetic acid
100
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
30
25
20
15
M
10
5
0
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.1
Acetic acid
100
95
90
85
To determine composition on A, B, C for extract & raffinate:
Extend the line (equilibrium tie-line) pass M point, read the
values (A, B, C) when at intersect between tie-line & extract
and raffinate.
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
Extract 30
Xc = 0.94
25
XA = 0.04
20
XB = 0.02
15
M
Raffinate
Xc = 0.02
XA = 0.13
XB = 0.85
x
10
5
x
0
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
SINGLE-STAGE EQUILIBRIUM EXTRACTION
Acetic acid
100
The composition of the two equilibrium layer in Example 7.1 are for the
extract layer (E) 𝑦𝐴 = 0.04, 𝑦𝐵 = 0.02, 𝑦𝐶 = 0.94, and for the raffinate layer
(R) 𝑥𝐴 = 0.12, 𝑥𝐵 = 0.86, and 𝑥𝐶 = 0.02. The original mixture contained 100
kg and 𝑥𝐴𝑀 = 0.10. Compute the amounts of E and R.
95
90
85
80
E
75
70
65
wt% Acetic acid
60
R
55
50
Overall mass balance:
E+R=M
45
40
35
30
A balance:
E𝑦𝐴 + 𝑅𝑥𝐴 = 𝑀𝑥𝐴𝑀
25
20
15
M
𝑥𝐴𝑀 10
5
x
x
0
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
SINGLE-STAGE EQUILIBRIUM EXTRACTION
Acetic acid
100
The composition of the two equilibrium layer in Example 7.1 are for the
extract layer (E) 𝑦𝐴 = 0.04, 𝑦𝐵 = 0.02, 𝑦𝐶 = 0.94, and for the raffinate layer
(R) 𝑥𝐴 = 0.12, 𝑥𝐵 = 0.86, and 𝑥𝐶 = 0.02. The original mixture contained 100
kg and 𝑥𝐴𝑀 = 0.10. Compute the amounts of E and R.
95
90
85
80
E
75
70
65
wt% Acetic acid
60
R
55
50
45
Overall mass balance:
V + L = M =100
40
35
30
A balance:
𝑉(0.04) + 𝐿 0.12
25
20
15
M
𝑥𝐴𝑀 10
5
Solving,
E = 75.0; R=25.0
x
x
0
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
= 100(0.10)
90
95
100
Water
Level-Arm Rule
SINGLE-STAGE EQUILIBRIUM EXTRACTION
Determine E & R, using LEVER-ARM RULE
Acetic acid
100
The composition of the two equilibrium layer in Example 7.1 are for the
extract layer (E) 𝑦𝐴 = 0.04, 𝑦𝐵 = 0.02, 𝑦𝐶 = 0.94, and for the raffinate layer
(R) 𝑥𝐴 = 0.12, 𝑥𝐵 = 0.86, and 𝑥𝐶 = 0.02. The original mixture contained 100
kg and 𝑥𝐴𝑀 = 0.10. Compute the amounts of E and R.
95
90
85
80
E
75
70
65
wt% Acetic acid
60
RR
55
50
45
Lever-arm rule:
𝐸 𝑅𝑀
Raffinate
=
Layer, R 𝑅
𝐸𝑀
40
Extract
35
Layer, E
30
25
20
15
M
𝑥𝐴𝑀 10
5
x
x
0
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
(𝑉𝑀) ̅ wt% water
𝑉𝑀=12cm
60
65
70
75
80
85
90
95
100
Water
SINGLE-STAGE EQUILIBRIUM EXTRACTION
Acetic acid
100
The composition of the two equilibrium layer in Example 7.1 are for the
extract layer (E) 𝑦𝐴 = 0.04, 𝑦𝐵 = 0.02, 𝑦𝐶 = 0.94, and for the raffinate layer
(R) 𝑥𝐴 = 0.12, 𝑥𝐵 = 0.86, and 𝑥𝐶 = 0.02. The original mixture contained 100
kg and 𝑥𝐴𝑀 = 0.10. Compute the amounts of E and R.
95
90
85
80
E
75
70
65
wt% Acetic acid
60
R
55
50
Lever-arm rule:
𝐸 𝑅𝑀
=
Raffinate
𝑅 𝐸𝑀
Layer, R
45
40
Extract
35
Layer, E
30
25
20
15
M
𝑥𝐴𝑀 10
5
x
x
0
Isopropyl 0
ether
5
10
15
20
25
30
35
𝑉𝑀
40
45 50 55
(𝑉𝑀) ̅ wt% water
60
65
70
75
80
85
𝐿𝑀
𝐿𝑀=5.2cm
90
95
100
Water
SINGLE-STAGE EQUILIBRIUM EXTRACTION
Acetic acid
100
95
The composition of the two equilibrium layer in Example 7.1 are for the
extract layer (E) 𝑦𝐴 = 0.04, 𝑦𝐵 = 0.02, 𝑦𝐶 = 0.94, and for the raffinate layer
(R) 𝑥𝐴 = 0.12, 𝑥𝐵 = 0.86, and 𝑥𝐶 = 0.02. The original mixture contained 100
kg and 𝑥𝐴𝑀 = 0.10. Compute the amounts of E and R.
90
85
80
75
E
70
65
R
wt% Acetic acid
60
55
50
Lever-arm rule:
𝐸 𝑅𝑀
=
Raffinate
𝑅 𝐸𝑀
Layer, R
45
40
Extract
35
Layer, E
30
25
𝐸 𝑅𝑀
=
𝑀 𝐸𝑅
12
E=
100
17.2
= 69.77
20
15
M
𝑥𝐴𝑀 10
5
x
x
0
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.2
Pure solvent isopropyl ether at the rate of 600 kg/h is being used to extract an
aqueous solution of 200 kg/h containing 30 wt% acetic acid concentration by
counter-current multistage extraction. The desired exit acetic acid concentration in
the aqueous phase is 4%. Use the equilibrium data in A.3-24, compute the
compositions and amounts of extract and raffinate.
Example 7.2 (solution)
1. Write down the known points:
Feed (F)
Solvent (S)
200 kg/h
600 kg/h
Acetic acid
30%
-
Water
70%
-
𝑥𝑤
-
100%
𝑥𝐼𝐸
Isopropyl ether
Raffinate
(R)
Extract
(E)
4%
𝑥𝐴𝐴
Mixture
(M)
Example 7.2 (solution)
2.Locate the F, R, and S point on the ternary diagram.
Acetic acid
100
95
3.Using the Lever rule to locate mixing point (M):
90
𝐹𝑆 = 𝐹𝑀 + 𝑀𝑆 = 10.7cm
85
80
𝑆 𝐹𝑀
=
𝐹 𝑀𝑆
75
70
65
wt% Acetic acid
60
600 10.7 − 𝑀𝑆
=
200
𝑀𝑆
55
50
45
𝑀𝑆 =
40
35
x
30
F
10.7
= 2.675 𝑐𝑚
4
25
20
15
M
10
x
5
S0x
Isopropyl 0
ether
x
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
R
𝑦𝐴𝐴 = 0.04
𝑦𝑤 = 0.94
𝑦𝐼𝐸 = 0.02
100
Water
Example 7.2 (solution)
4. Draw a straight line connecting point R to point M and
extended to Extract layer.
Acetic acid
100
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
15
10
M
Ex
x
5
S0x
Isopropyl 0
ether
x
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
R
100
Water
Example 7.2 (solution)
5. Read off 𝑥𝐴𝐴 = 0.08 at point E.
Acetic acid
100
6. Mass balance:
95
90
Overall balance: F + S = E + R
800 = E + R
85
80
75
Acetic acid balance: (0.3)(200) + 0 = 0.08E + 0.04R
R=100 kg/h
E= 700 kg/h
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
15
10
M
Ex
x
5
S0x
Isopropyl 0
ether
x
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
R
100
Water
Example 7.2 (solution)
5. Using the lever rule to find R and E:
Acetic acid
100
𝑅 𝐸𝑀
2.1
=
=
= 0.1935
𝐸 𝑀𝑅 10.85
95
90
85
R=0.1935E
80
Overall balance: F+S=R+E
800=0.1935E+E
E=670.297 kg/h
R=0.1935(670.297)=129.702 kg/h
75
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
15
10
M
Ex
x
5
S0x
Isopropyl 0
ether
x
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
R
100
Water
Number of Stages in Counter-current Extraction
Example 7.3
Pure isopropyl ether of 450 kg/h is being used to extract an
aqueous solution of 150 kg/h with 30 wt% acetic acid by
countercurrent multistage extraction. The exit acid concentration
in the aqueous phase is 10 wt%. Compute the number of stages
required, the amount of 𝑬𝟏 and 𝑹𝑵 .
Example 7.3 (solution)
1. Write down the known points:
Feed (F)
Solvent (S)
150 kg/h
450 kg/h
Acetic acid
30%
-
Water
70%
-
𝑥𝑤
-
100%
𝑥𝐼𝐸
Isopropyl ether
Raffinate
(𝑹𝑵 )
Extract
(𝑬𝟏 )
10%
𝑥𝐴𝐴
Mixture
(M)
Example 7.3 (solution)
2.Locate the F, 𝑅𝑁 , and S point on the ternary diagram.
Acetic acid
100
95
3.Using the Lever rule to locate mixing point (M):
90
𝐹𝑆 = 𝐹𝑀 + 𝑀𝑆 = 10.7cm
85
80
𝑆 𝐹𝑀
=
𝐹 𝑀𝑆
75
70
65
wt% Acetic acid
60
450 10.7 − 𝑀𝑆
=
150
𝑀𝑆
55
50
45
𝑀𝑆 =
40
35
x
30
F
𝑹𝑵
𝑦𝐴𝐴 = 0.100
𝑦𝑤 = 0.875
𝑦𝐼𝐸 = 0.025
25
20
15
M
10
10.7
= 2.675 𝑐𝑚
4
x
x
5
S0 x
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.3 (solution)
Acetic acid
100
4.Draw a straight line connecting point 𝑅𝑁 and M
extending to extract layer 𝐸1
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
15
M
10
𝑬𝟏
5
x
x
x
𝑹𝑵
S0 x
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.3 (solution)
5.Locate the operating point Δ.
Acetic acid
100
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
15
x
10
𝑬𝟏
Δ
x
5
𝑹𝑵
x
S0 x
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.2 (solution)
Acetic acid
5. Draw a tie line from 𝐸1 to raffinate layer 𝑅1
100
95
6. Draw a line to connect 𝑅1 to Δ.
90
7. Draw a tie line from 𝐸2 to raffinate layer 𝑅2
85
80
8. Draw a line to connect 𝑅2 to Δ.
75
70
9. Draw a tie line from 𝐸3 to raffinate
layer 𝑅3 (Exceeded 𝑅𝑁 ).
65
wt% Acetic acid
60
55
10. Number of stages
50
= number of tie lines ≅ 𝟐. 𝟑
45
40
35
x
30
F
25
20
x
𝑹𝟏
𝑹𝟐 𝑹
15
Δ
x
𝑬𝟏
𝑬𝟐10 x
𝑬𝟑 5 x
S0 x
Isopropyl 0
ether
xx
𝑵
𝑹𝟑
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Water
Example 7.3 (solution)
𝑥𝐴𝐴 = 0.07
Mass balance:
Overall balance: F + S = E1 + RN
150 + 450 = E1 + RN
Acetic acid balance: (0.3)(200) + 0 = 0.07E1 + 0.10RN
RN=100 kg/h
E1= 500 kg/h
MINIMUM SOLVENT RATE
An aqueous feed solution of 150 kg/h with 30 wt%
acetic acid is extracted using pure isopropyl ether.
The exit acid concentration in the aqueous phase is
2.5 wt%. Compute the minimum solvent rate
required to achieve this separation.
Acetic acid
100
95
90
85
80
1. Draw a straight line connecting S and F.
75
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
15
10
5
x
S0 x
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
𝑹𝑵
100
Water
MINIMUM SOLVENT RATE
An aqueous feed solution of 150 kg/h with 30
wt% acetic acid is extracted using pure isopropyl
ether. The exit acid concentration in the aqueous
phase is 2.5 wt%. Compute the minimum solvent
rate required to achieve this separation.
Acetic acid
100
95
90
85
80
2. Draw a tie line connecting F to
intercepting the extract layer
75
70
65
wt% Acetic acid
60
55
50
45
40
35
x
30
F
25
20
𝑬𝒎𝒂𝒙
M
15
10
𝑹𝑵
5
x
S0 x
Isopropyl 0
ether
5
10
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100
Wate
3. Using lever rule:
𝐹𝑆 = 𝐹𝑀 + 𝑀𝑆𝑚𝑖𝑛 =9.4 cm
𝑆𝑚𝑖𝑛
𝐹𝑀
=
𝐹
𝑀𝑆𝑚𝑖𝑛
𝑆𝑚𝑖𝑛 =
9.4 − 3.8
× 150 = 𝟐𝟐𝟏. 𝟎𝟓𝟑 𝒌𝒈/𝒉
3.8
THE END
By:
Dr. Ng Ching Yin
ngcy@ucsiuniversity.edu.my
100
95
90
85
80
75
70
65
wt% Acetic acid
60
55
50
45
40
35
30
25
20
15
10
5
0
0
5
10
wt% Isopropyl ether
15
20
25
30
35
40
45 50 55
wt% water
60
65
70
75
80
85
90
95
100