ERT 313
Bioseparation Engineering
LIQUID-LIQUID EXTRACTION
(LLE)
Zulkarnain Bin Mohamed Idris
e-mail: [email protected]
COURSE OUTCOME (CO)
Ability to APPLY principles of extraction, ANALYZE
extraction equipment and extraction operating
modes and DEVELOP basic calculations of extraction!
OUTLINES
1. Introduction to extraction
2. Principles of extraction
3. Operating modes of extraction (batch extraction, continuous
extraction and aqueous two phase extraction)
4. Basic calculations of extraction
5. Equipment for extraction
INTRODUCTION
Definition of Liquid-Liquid Extraction:
 is a mass transfer operation in which a liquid solution (the feed) is
contacted with an immiscible or nearly immiscible liquid (solvent) that
exhibits preferential affinity or selectivity towards one or more of the
components in the feed.
Solvent
Liquid
solution
(Feed)
e.g. Phase
separation in
separating funnel
INTRODUCTION
Purpose of Liquid-Liquid Extraction:
i.
To separate closed-boiling point mixture (acetic
acid, b.p 118 °C and water, b.p 100 °C)
ii. Mixture that cannot withstand high temperature or
heat sensitive components (such as antibiotics)
Example:
i.
Recovery of penicillin F (antibiotic) from
fermentation broth (feed) using butyl acetate
(solvent)
ii. Recovery of acetic acid from dilute aqueous
solutions (feed) using ethyl-acetate (solvent)
BASIC PRINCIPLES OF
EXTRACTION
BASIC PRINCIPLES OF
EXTRACTION
i.
ii.
iii.
iv.
v.
vi.
The solute originally present in the aqueous phase gets
distributed in both phases
If solute has preferential solubility in the organic solvent, more
solute would be present in the organic phase at equilibrium
The extraction is said to be more efficient
Extract= the layer of solvent + extracted solute
Raffinate= the layer from which solute has been removed
The distribution of solute between two phases is express
quantitatively by distribution coefficient, KD
KD 
vii.
solute concentration in extract phase
EQ. 1
solute concentration in raffinate phase
Higher value of KD indicates higher extraction efficiency
PRINCIPLES OF
EXTRACTION
Organic solvent
(Examples)
KD (mol/L) at 250C
n-butanol
n-butanol
n-butanol
n-butanol
n-butanol
0.01
0.02
0.02
0.20
0.07
Antibiotics
Erythromycin
Novobiocin
Amyl acetate
Butyl acetate
Penicillin F
Amyl acetate
Penicillin K
Amyl acetate
120
100 at pH 7.0
0.01 at pH 10.5
32 at pH 4.0
0.06 at pH 6.0
12 at pH 4.0
0.1 at pH 6.0
Solute (Examples)
Amino acids
Glycine
Alanine
2-aminobutyric acid
Lysine
Glutamic acid
OPERATING MODES OF
EXTRACTION
i.
Batch Extraction:
Single stage or Multiple stage
ii. Continuous Extraction:
Co-current or Countercurrent extraction
Batch Extraction:
i. The aqueous feed is mixed with the organic solvent
ii. After equilibration, the extract phase containing
the desired solute is separated out for further
processing
iii. Is routinely utilized in laboratory procedures
iv. This can be carried out for example in separating
funnel or in agitated vessel
OPERATING MODES OF
EXTRACTION
Batch Extraction (Single & Multiple Stages):
Schematic representations of (a) single & (b) multiple
stages (crosscurrent) batch operation:
(a)
Feed
Solvent
Raffinate
Single stage
extraction
Solvent
Extract
Solvent
(b)
Feed
First
stage
Extract
Raffinate
Second
stage
Final
Raffinate
Extract
Combined
Extract
OPERATING MODES OF
EXTRACTION
Continuous Extraction(Co-current & Counter-current):
Schematic representations of (a) co-current & (b)
countercurrent operations:
(a) Co-current extraction
(b) Counter-current extraction
CALCULATION METHODS
1. Extraction of Dilute Solution
i. Extraction factor is defined as:
EQ. 2
Where:
E = extraction factor
KD = distribution coefficient
V = volume of solvent
L = volume of aqueous
Refer to
EQ. 1
CALCULATION METHODS
Extraction of Dilute Solution
ii.
For a single-stage extraction with pure
solvent:
The fraction of solute remaining:
1
1 E
EQ. 3
The fraction recovered:
E
1 E
EQ. 4
CALCULATION METHODS
Example 1
Penicillin F is recovered from a dilute aqueous fermentation
broth by extraction with amyl acetate, using 6 volumes of
solvent (V) per 100 volumes of the aqueous phase (L). At pH
3.2 the distribution coefficient KD is 80.
(a) What fraction of the penicillin would be recovered in a
single ideal stage?
(b) What would be the recovery with two-stage
extraction using fresh solvent in both
stages?
CALCULATION METHODS
Solution 1 (a) (Draw the material balance diagrams)
(a)
Feed (L, x0)
Solvent (V, y0)
Single stage
extraction
Raffinate (L, x1)
Extract (V, y1)
Material balance:
L(x0) + V(y0) = L (x1) + V(y1) 0
L(x0) – L(x1) = V(y1) – V(y0)
Since y0=0 (at initial no penicillin in solvent phase)
So, L(x0)-L(x1) = V(y1)
L(x0-x1)= V(y1)
solute concentration in extract phase( y1)
Since KD = y1/x1, y1=KDx1  Refer to EQ. 1 K D 
solute concentration in raffinate phase( x1)
So, L(x0-x1)=V(KDx1)
x1[(VKD/L )+ 1)]= x0, where VKD/L = E  Refer to EQ. 2
E = E= (6)(80)/100 = 4.8
CALCULATION METHODS
Solution 1 (a) (Draw the material balance diagrams)
(a)
Feed (L, x0)
Solvent (V, y0)
Single stage
extraction
Raffinate (L, x1)
Extract (V, y1)
Material balance (continued):
x1/x0 = 1/ (1+E)  Refer to EQ. 3 (frac. of penicillin in raffinate phase = frac.
remaining)
= 1/ (1+ 4.8)
= 0.1724
Fraction of penicillin recovered = Fraction of penicillin in extract phase
= 1- 0.1724
= 0.828
= 82.8%
Or calculated using EQ. 4, E/(1+E)= 4.8/ (1+4.8) =0.828; 82.8% recovery
CALCULATION METHODS
Solution 1 (b) (Draw the material balance diagrams)
(b)
Solvent (V, y0)
Feed (L, x0)
First
stage
Solvent (V, y0)
Raffinate (L, x1)
Extract (V, y1)
Second
stage
Final
Raffinate (L, x2)
Extract (V, y2)
Combined
Extract
Material balance:
L(x1) + V(y0) = L (x2) + V(y2) 0
L(x1) – L(x2) = V(y1) – V(y0)
Since y0=0 (at initial no penicillin in solvent phase)
So, L(x1)-L(x2) = V(y2)
L(x1-x2)= V(y2)
CALCULATION METHODS
Material balance:
solute concentration in extract phase( y 2)
K

Since KD = y2/x2, y2=KDx2  Refer to EQ. 1 D
solute concentration in raffinate phase( x2)
So, L(x1-x2)=V(KDx2)
x2[(VKD/L )+ 1)]= x1, where VKD/L = E  Refer to EQ. 2
E= (6)(80)/100 = 4.8
x2/x1 = 1/ (1+E)  Refer to EQ. 3 (frac. of penicillin in final raffinate phase from
raffinate phase in 1st stage) = frac. remaining)
= 1/ (1+ 4.8)
= 0.1724
x2/x0 = (x2/x1) * (x1/x0)
= (0.1724) * (0.1724)
= 0.0297 (frac. of penicillin in final raffinate phase from feed phase = frac.
remaining from )
Fraction of penicillin recovered = Fraction of penicillin in extract phase from
feed phase
= 1- 0.0297
= 0.9703
= 97.0%
CALCULATION METHODS
Example 2
An inlet water solution of 100 kg/h containing 0.010 wt
fraction nicotine in water is stripped with a kerosene
stream of 200 kg/h containing 0.0005 wt fraction nicotine
in a single stage extraction unit. It is desired to reduce the
concentration of the exit water to 0.0010 wt fraction
nicotine. Calculate the flow rate of the nicotine in both of
the exit streams.
CALCULATION METHODS
Solution 2
i.
Nicotine in the feed solution = 100 (0.01) = 1 kg/h nicotine
Water in feed = 100 (1 - 0.01) = 99 kg/h water
ii.
Nicotine in solvent = 200 (0.0005) = 0.1 kg/h nicotine
Kerosene = 200 (1 – 0.0005) = 199.9 kg/h kerosene
iii.
Exit stream of aqueous phase, L1
Water = 99 kg/h = (1 – 0.0010) L1
L1 = 99.099 kg/h (nicotine + water)
Nicotine = 99.099 – 99 = 0.099 kg/h nicotine in exit
stream
iv.
Exit stream of solvent phase, V1
Solvent = 199.9 kg/h
Nicotine in solvent = 0.1 + (1 – 0.099) = 1.001 kg/h
in exit stream
Solvent + Nicotine = 199.9 + 1.001 = 200.9 kg/h
CALCULATION METHODS
2. Extraction of Concentrated Solution
i.
ii.
Equilibrium relationship are more complicated-3 or
more components present in each phase.
Equilibrium data are often presented on a triangular
diagram such as Fig 23.7 and 23.8.
Triangular diagram
TRIANGULAR DIAGRAM
i.
ii.
iii.
iv.
Consider Fig 23.7
Line ACE shows extract phase
Line BDE shows raffinate phase
Point E is the plait point – the composition of extract &
raffinate phases approach each other
v. Tie line – a straight line joining the composition of extract &
raffinate phases.
vi. Tie line in Fig 23.7 slope up to the left – extract phase is
richer in acetone than the raffinate phase.
vii. This suggest that most of the acetone could be extract from
water phase using moderate amount of solvent.
TRIANGULAR DIAGRAM
• How to obtain the phase composition using the triangular
diagram?
- Example: if a mixture with 40 % acetone and 60 percent water
is contacted with equal mass of MIK, the overall mixture is
represented by point M in Figure 23.7:
Point M: 0.2 Acetone, 0.3 water, 0.5 MIK
- Draw a new tie line
- Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK
- Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK
- Ratio of acetone to water in the product = 0.232/0.043 =
5.4
- Ratio of acetone to water in the raffinate = 0.132/0.845 =0.156
Triangular diagram
TRIANGULAR DIAGRAM
i.
ii.
iii.
iv.
Consider Fig 23.8
Line AD shows extract phase
Line BC shows raffinate phase
Tie line in Fig 23.8 slope up to the right – extraction
would still be possible
v. But more solvent would have to use.
vi. The final extract would not be as rich in desired
component (MCH)
Coordinate Scale
• Refer to Treybal, Mass Transfer Operation, 3rd ed., McGraw
Hill
• The book use different triangular system
• The location of solvent (B) is on the right of the triangular
diagram (McCabe use on the left)
• Coordinate scales of equilateral triangles can be plotted as y
versus x as shown in Fig 10.9
• Y axis = wt fraction of component C (acetic acid)
• X axis = wt fraction of solvent B (ethyl acetate)
Coordinate Scale
Single-Stage Extraction
Single-Stage Extraction
• The triangular diagram in Fig 10.12 (Treybal) is
a bit different as compared to Fig. 23.7 (McCabe)
• Extract phase – on the left
• Raffinate phase - on the right
• Fig 10.12 shows that we want to extract
component C from A by using solvent B.
• Total material balance:
• Material balance on C:
• Amount of solvent to provide a given location for M1 on the line
FS:
• The quantities of extract and raffinate:
• Minimum amount of solvent is found by locating M1 at D
• Maximum amount of solvent is found by locating M1 at K
M ULTISTAGE C ROSSCURRENT
E XTRACTION
• Continuous or batch processes
• Refer to Fig 10.14
• Raffinate from the previous stage will be the feed for the next
stage
• The raffinate is contacted with fresh solvent
• The extract can be combined to provide the composited extract
• The total balance for any stage n:
• Material balance on C:
M ULTISTAGE C ROSSCURRENT
E XTRACTION
EXAMPLE
If 100 kg of a solution of acetic acid (C) and water (A) containing
30% acid is to be extracted three times with isopropyl ether (B)
at 20°C, using 40 kg of solvent in each stage, determine the
quantities and compositions of the various streams. How much
solvent would be required if the same final raffinate
concentration were to be obtained with one stage?
The equilibrium data at 20°C are listed below [Trans. AIChE, 36,
628 (1940), with permission].
Multistage Crosscurrent Extraction
M ULTISTAGE C ROSSCURRENT
E XTRACTION
SOLUTION
The horizontal rows give the concentrations in
equilibrium solutions. The system is of the type
shown in Fig. 10.9a, except that the tie lines slope
downward toward the B apex. The rectangular
coordinates of Fig. l0.9b will be used, but only for
acid concentrations up to x = 0.30. These are plotted
in Fig. 10.15.
No
1
2
3
4
5
6
7
8
9
Water layer
(Raffinate phase)
Acetic acid Acetic acid (C)
(%)
(wt. Fraction, x)
0.69
0.0069
1.41
0.0141
2.89
0.0289
6.42
0.0642
13.3
0.133
25.5
0.255
36.7
0.367
44.3
0.443
46.4
0.464
Water
(%)
98.1
97.1
95.5
91.7
84.4
71.1
58.9
45.1
37.1
Water (A)
(wt. fraction)
0.981
0.971
0.955
0.917
0.844
0.711
0.589
0.451
0.371
Isopropyl
ether (%)
1.2
1.5
1.6
1.9
2.3
3.4
4.4
10.6
16.5
Isopropyl ether (B)
(wt. Fraction)
0.012
0.015
0.016
0.019
0.023
0.034
0.044
0.106
0.165
1
2
3
4
5
6
7
8
9
Isopropyl ether layer
(Extract phase)
Acetic acid Acetic acid (C)
(%)
(wt. Fraction, y)
0.18
0.0018
0.37
0.0037
0.79
0.0079
1.93
0.0193
4.82
0.0482
11.4
0.114
21.6
0.216
31.1
0.311
36.2
0.362
Water
(%)
0.5
0.7
0.8
1
1.9
3.9
6.9
10.8
15.1
Water (A)
(wt. fraction)
0.005
0.007
0.008
0.01
0.019
0.039
0.069
0.108
0.151
Isopropyl
ether (%)
99.3
98.9
98.4
97.1
93.3
84.7
71.5
58.1
48.7
Isopropyl ether (B)
(wt. fraction)
0.993
0.989
0.984
0.971
0.933
0.847
0.715
0.581
0.487
Where,
x= wt. fraction of acetic acid in Raffinate Phase
y= wt. fraction of acetic
1 acid in Extract Phase
Rectangular Coordinates
wt. fraction of acetic acid (C) (x, y)
0.9
0.8
0.7
Equilibrium points at water
layer (Raffinate phase)
0.6
0.5
0.4
0.044, 0.367
0.487, 0.362
0.581, 0.311
0.3
0.034, 0.255
0.715, 0.216
0.2
0.023, 0.133
0.1
0.847, 0.114
0.019, 0.0642
0.016, 0.0289
0.015, 0.0069
0.0141
0.012,
0
0
Figure 10.9
Equilibrium points at
isopropyl ether
layer (Extract phase)
0.165, 0.464
0.106, 0.443
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
wt. fraction of isopropyl ether (B)
0.933, 0.0482
0.971,
0.0193
0.984,
0.989,
0.0037
0.993,0.0079
0.0018
1
Where,
x= wt. fraction of acetic acid in Raffinate Phase
y= wt. fraction of acetic
1 acid in Extract Phase
Rectangular Coordinates
wt. fraction of acetic acid (C) (x, y)
0.9
0.8
0.7
0.6
Distribution curve
0.5
0.4
0.3
0.2
0.1
0
0
Figure 10.9
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
wt. fraction of isopropyl ether (B)
1
Where,
x= wt. fraction of acetic acid in Raffinate Phase
y= wt. fraction of acetic
1 acid in Extract Phase
Rectangular Coordinates
wt. fraction of acetic acid (C) (x, y)
0.9
0.8
0.7
0.6
0.5
Distribution curve
Tie line
0.4
0.3
0.2
0.1
0
0
Figure 10.9(a)
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
wt. fraction of isopropyl ether (B)
1
wt. fraction of acetic acid (C) (x, y)
For acetic concentrations up to x = 0.30 (0riginally 30% in Feed)
Rectangular Coordinates
0.3
0.25
0.2
Distribution curve
0.15
Tie line
0.1
0.05
0
0
Figure 10.9(b)
0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9
wt. fraction of isopropyl ether (B)
1
Solution (Draw the material balance diagrams)
Solvent, S1 (ys1)
Feed, F
(xF)
First
stage
Extract, E1 (y1)
Solvent, S2 (ys2)
Raffinate
R1 (x1)
Second
stage
Extract, E2 (y2)
Solvent, S3 (ys3)
Raffinate
R2(x2)
Third
stage
Extract, E3 (y3)
Final
Raffinate
R3 (x3)
Solution at 1st Stage (Draw the material balance diagram)
Solvent, S1= 40 kg (ys1 = 0)
Feed, F =100 kg
(xF = 0.30)
First
stage
Raffinate, R1
(x1)
Extract, E1 (y1)
Material Balance:
Total Balance:
F + S1 = E1 + R1 = M1-----------------------------EQ 10.4
M1 = 1oo kg + 40 kg = 140kg
Material Balance on acetic acid (C):
F(xF) + S1(ys) = E1(y1) + R1(x1) = M1(xM1) ----------------------------EQ 10.5
F(xF) + S1(ys) = M1(xM1)
100 kg(0.30) + 40 kg (0)= (140 kg)(xM1)
Thus, xM1 = 30 kg/140 kg =0.214
Rectangular Coordinates
wt. fraction of acetic acid (C)
F (0, 0.3)
1.
2.
3.
First draw the tie lines.
Then plot the initial point of
wt. frac of acetic acid (C)
in Feed, F (0, 0.3) and in
Solvent, S1 (1, 0).
Draw the line FS1 on the
Rectangular Coordinates
by joining the two
points.
S1(1,0)
0
Tie line
FS1 line
0.3
0.29
0.28
0.27
0.26
0.25
0.24
0.23
0.22
0.21
0.2
0.19
0.18
0.17
0.16
0.15
0.14
0.13
0.12
0.11
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
wt. fraction of isopropyl ether (B)
1
wt. fraction of acetic acid (x, y)
Rectangular Coordinates
0.3
0.29
0.28
0.27
X1 = 0.258 0.26
0.25
0.24
0.23
XM1 = 0.214 0.22
0.21
0.2
0.19
0.18
0.17
0.16
0.15
0.14
0.13
y1 = 0.117 0.12
0.11
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
4.
0
Tie line R1E1
5.
M1 (?, 0.214)
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
wt. fraction of isopropyl ether (B)
1
Plot the point M1 (?, 0.214)
where it is located on
the line FS1.
With the help of a
distribution curve,
draw the tie line passing
through M1 is located as
shown, and x1 = 0.258,
y1 = 0.117 wt fraction
acetic acid were determine
at the intersection points
with the distribution curve.
The quantities of extract (E1)and raffinate (Y1)in Stage 1:
Total Balance:
E1 + R1 = M1-----------------------------EQ 10.5
M1 = 1oo kg + 40 kg = 140kg
Material Balance on acetic acid (C):
E1(y1) + R1(x1) = M1(xM1) -------------------EQ 10.7
Since E1 + R1 = M1-----EQ 10.5, thus R1 = M1-E1,
So , substitute R1=M1- E1 into EQ 10.7 and simplified, thus E1 will equal to:
E1 = [M1 (xM1 –x1)]/ (y1-x1)
= [(140 kg) (0.214- 0.258)]/(0.117-0.258)
= 43.6 kg
R1 =M1-E1
= 140 kg – 43.6 kg
= 96.4 kg
Solution at 2nd Stage (Draw the material balance diagram)
Solvent, S1 (ys1=0)
Solvent, S2 = 40 Kg (ys2=0)
First
stage
Second
stage
Feed, F
(xF =0.3)
Raffinate
R1=96.4 kg
(x1=0.258)
Extract, E1 (y1=0.117)
Material Balance:
Total Balance:
R1 + S2 = E2 + R2 = M2
M2 = 96.4 kg + 40 kg = 136.4kg
Material Balance on acetic acid (C):
R1(x1) + S2(ys2) = E2(y2) + R2(x2) = M2(xM2)
R1(x1) + S2(ys2) = M2(xM2)
96.4 kg(0.258) + 40 kg (0)= (136.4 kg)(xM2)
Thus, xM2 = 24.871 kg/136.4 kg =0.1823
Extract, E2 (y2)
Raffinate
R2(x2)
wt. fraction of acetic acid
Rectangular Coordinates
R1S2 line
0.3
0.29
0.28
0.27
0.26
0.25
0.24
X2 = 0.227 0.23
0.22
0.21
0.2
0.19
XM2 = 0.1823 0.18
0.17
0.16
0.15
0.14
0.13
0.12
0.11
0.1
y2 = 0.095 0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
4.
5.
0
Tie line R2E2
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
wt. fraction of isopropyl ether
0.9
1
Plot the point M2 (?, 0.1823)
where it is located on
the line R1S2.
With the help of a
distribution curve,
draw the tie line passing
through M2 is located as
shown, and x2 = 0.227,
y2 = 0.095 wt fraction
acetic acid were determine
at the intersection points
with the distribution curve.
The quantities of extract (E2)and raffinate (Y2)in Stage 2:
Total Balance:
E2 + R2 = M2
M2 = 136.4kg
Material Balance on acetic acid (C):
E2(y2) + R2(x2) = M2(xM2)----------------EQ 10.7(a)
Since E2 + R2 = M2, thus R2 = M2-E2,
So , substitute R2=M2- E2 into EQ 10.7(a) and simplified, thus E2 will equal to:
E2 = [M2 (xM2 –x2)]/ (y2-x2)
= [(136.4 kg) (0.1823- 0.227)]/(0.095-0.227)
= 46.2 kg
R2 =M2-E2
= 136.4 kg – 46.2 kg
= 90.2 kg
Solution at 3rd Stage (Draw the material balance diagram)
Solvent, S1 (ys1)
Feed, F
(xF)
First
stage
Solvent, S2 (ys2)
Raffinate
R1 (x1)
Extract, E1 (y1)
Second
stage
Extract, E2 (y2)
Material Balance:
Total Balance:
R2 + S3 = E3 + R3 = M3
M3 = 90.2 kg + 40 kg = 130.2kg
Material Balance on acetic acid (C):
R2(x2) + S3(ys3) = E3(y3) + R3(x3) = M3(xM3)
R2(x2) + S3(ys3) = M3(xM3)
90.2 kg(0.227) + 40 kg (0)= (130.2 kg)(xM3)
Thus, xM3 = 20.475 kg/130.2 kg =0.1573
Solvent, S3 =40 kg (ys3=0)
Raffinate
R2 =90.2kg
(x2=0.227)
Third
stage
Extract, E3 (y3)
Final
Raffinate
R3 (x3)
wt. fraction of acetic acid
Rectangular Coordinates
R2S3 line
0.3
0.29
0.28
0.27
0.26
0.25
0.24
0.23
0.22
0.21
0.2
X3 = 0.2
0.19
0.18
0.17
XM3 = 0.1573 0.16
0.15
0.14
0.13
0.12
0.11
0.1
0.09
y3 = 0.078 0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
4.
5.
0
Tie line R3E3
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
wt. fraction of isopropyl ether
0.9
1
Plot the point M3 (?, 0.1573)
where it is located on
the line R2S3.
With the help of a
distribution curve,
draw the tie line passing
through M3 is located as
shown, and x3 = 0.2,
y3 = 0.078 wt fraction
acetic acid were determine
at the intersection points
with the distribution curve.
The quantities of extract (E3)and raffinate (Y3)in Stage 3:
Total Balance:
E3 + R3 = M3
M2 = 130.2kg
Material Balance on acetic acid (C):
E3(y3) + R3(x3) = M3(xM3) ------------------EQ 10.7 (b)
Since E3 + R3 = M3, thus R3 = M3-E3,
So , substitute R3=M3- E3 into EQ 10.7 (b) and simplified, thus E3 will equal to:
E3 = [M3 (xM3 –x3)]/ (y3-x3)
= [(130.2kg) (0.1573- 0.2)]/(0.078-0.2)
= 45.6kg
R3 =M3-E3
= 130.2 kg – 45.6 kg
= 84.6 kg
So, the acetic acid content in the final raffinate:
= R3*x3
= 84.6 kg (0.2) =16.92 kg
The composited extract is:
E1 + E2 + E3 = 43.6 + 46.2 + 45.6 = 135.4 kg,
The acid content in the composited extract:
E1y1 + E2y2 + E3y3 = [(43.6 *0.117) + (46.2 * 0.095) + (45.6 *0.078) =13.05 kg.
Rectangular Coordinates
wt. fraction of acetic acid
F (0,0.3)
XF = 0.3
R1 (?,0.258)
X1 = 0.258
R2 (?,0.227)
X2 = 0.227
R3 (?,0.2)
X3 = 0.2
0.3
0.29
0.28
0.27
0.26
0.25
0.24
0.23
0.22
0.21
0.2
0.19
0.18
0.17
0.16
0.15
0.14
0.13
0.12
0.11
0.1
0.09
0.08
0.07
0.06
0.05
0.04
0.03
0.02
0.01
0
M1 (?,0.214)
xM1 = 0.214
M2 (?,0.1823)
xM2 = 0.1823
M3 (?,0.1573)
xM3 = 0.1573
E1 (?,0.117)
y1 = 0.258
E2 (?,0.095)
y2 = 0.095
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
wt. fraction of isopropyl ether
Figure 10.15
0.9
S1,S2,S3 (1,0)
1 ys 1, ys 2 , ys 3 = 0
E3 (?,0.078)
y3 = 0.078
• If an extraction to give the same final raffinate concentration,
x = 0.20.
• were to be done in one stage, the point M would be at the
intersection of tie line R3E3 and line FS of Figure 10.15.
• So, XM = 0.12.
• The solvent required would then be, by Eq. (10.6),
• S1 = 100(0.30 - 0.12)/(0.12 - 0) = 150 kg,
• Hence, 150 kg of solvent is required for single stage
extraction
• 120 kg of solvent is required in the three-stage
extraction.
Aqueous Two Phase Extraction
• Use widely in separation of proteins, enzymes, viruses,
cells and cell organels.
• not denature the biological entities as they might be in
organic solvents.
• The proteins are partitioning between two aqueous
phases which contains mutually incompatible polymers
or other solutes.
Aqueous Two Phase Extraction
• For example;
- light phase is water + 10% polyethylene glycol (PEG) and
0.5% dextran
- heavy phase is water + 1% glycol and 15% dextran
• Proteins are partitioned between phases with
distribution coefficient (KD) that depends on the pH.
• KD can vary from 0.01 to more than 100.
Aqueous Two Phase Extraction
• Factors that affect protein partitioning in Aqueous
Two Phase System:
1. Protein molecular weight
2. Protein charge, surface properties
3. Polymer(s) molecular weight
4. Phase composition, tie-line length
5. Salt effects
6. Affinity ligands attached to polymers
Extraction Equipment
• Extraction Equipments:
- Mixer settlers
- Packed extraction towers
- Perforated plate towers
- Baffle towers
- Agitated tower extractors
• Auxiliary equipment:
- stills, evaporators, heaters and condenser
Mixer-settlers
• For Batchwise Extraction:
→ The mixer and settler may be the same unit.
→ A tank containing a turbine or propeller agitator is most
common.
→ At the end of mixing cycle the agitator is shut off, the layers are
allowed to separate by gravity.
→ Extract and raffinate are drawn off to separate receivers through
a bottom drain line carrying a sight glass.
→ The mixing and settling times required for a given extraction can
be determined only by experiment.
(e.g: 5 min for mixing and 10 min for settling are typical)
- both shorter and much longer times are common.
MIXER-SETTLERS
Feed
Solvent
Single
Stage
Extraction
Raffinate
Extract
Schematic Diagram Representation of a Single
Stage Batch Extraction
Mixer-settlers
• For Continous Extraction:
→ The mixer and settler are usually separate pieces of equipment.
→ The mixer; small agitated tank provided with a drawoff line and
baffles to prevent short-circuiting, or it may be motionless mixer
or other flow mixer.
→The settler; is often a simple continuous gravity decanter.
→In common used; several contact stages are required, a train of
mixer-settlers is operated with countercurrent flow.
Mixer-settlers
Note: The raffinate from each settler becomes a feed to the next
mixer, where it meets intermediate extract or fresh solvent.
Packed Extraction Towers
→ Tower extractors give differential contacts, not stage contacts,
and mixing and settling proceed continuously and
simultaneously.
→ Extraction; can be carried out in an open tower, with drops of
heavy liquid falling through the rising light liquid or vice versa.
→ The tower is filled with packings such as rings or saddles, which
causes the drops to coalesce and reform, and tends to limit axial
dispersion.
→ In an extraction tower there is continuous transfer of material
between phases, and the composition of each phase changes as
it flows through the tower.
→ The design procedure ; is similar to packed absorption towers.
Packed Extraction Towers
Tower packings;
(a) Raschig rings,
(b) metal Pall ring,
(c) plastic Pall ring,
(d) Berl saddle, (e) ceramic
Intalox saddle, (f) plastic
Super Intalox saddle,
(g) metal Intalox saddle
Agitated Tower Extractors
→ It depends on gravity flow for mixing
and for separation.
→ Mechanical energy is provided by
internal turbines or other agitators,
mounted on a central rotating shaft.
→ Fig (a), flat disks disperse the liquids and
impel them outward toward the tower
wall, where stator rings create quite
zones in which the two phases can
separate.
→ In other designs, set of impellers are
separated by calming sections to give, in
effect, a stack of mixer-settlers one
above the other.
Agitated Tower Extractors
→ In the York-Scheibel extractor (Fig.
b), the region surrounding the
agitators are packed with wire mesh
to encounter coalescence and
separation of the phases.
→ Most of the extraction takes place in
the mixing sections, but some also
occurs in the calming sections.
→ The efficiency of each mixer-settler
unit is sometimes greater than 100
percent.