ERT 313 Bioseparation Engineering LIQUID-LIQUID EXTRACTION...
Transcript of ERT 313 Bioseparation Engineering LIQUID-LIQUID EXTRACTION...
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.
e.g. Phase separation in
separating funnel
Liquid solution (Feed)
Solvent
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
i. The solute originally present in the aqueous phase gets distributed in both phases
ii. If solute has preferential solubility in the organic solvent, more solute would be present in the organic phase at equilibrium
iii. The extraction is said to be more efficient iv. Extract= the layer of solvent + extracted solute v. Raffinate= the layer from which solute has been removed vi. The distribution of solute between two phases is express
quantitatively by distribution coefficient, KD
vii. Higher value of KD indicates higher extraction efficiency
phaseraffinateinionconcentratsolute
phaseextractinionconcentratsoluteKD
BASIC PRINCIPLES OF EXTRACTION
EQ. 1
PRINCIPLES OF EXTRACTION
Solute (Examples) Organic solvent (Examples)
KD (mol/L) at 250C
Amino acids Glycine Alanine 2-aminobutyric acid Lysine Glutamic acid
n-butanol n-butanol n-butanol n-butanol n-butanol
0.01 0.02 0.02 0.20 0.07
Antibiotics Erythromycin Novobiocin Penicillin F Penicillin K
Amyl acetate Butyl acetate
Amyl acetate
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
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:
Single stage extraction Solvent
Feed
Extract
Raffinate
First stage
Solvent
Feed
Extract
Raffinate Second stage
Solvent
Extract
Final Raffinate
Combined Extract
(a)
(b)
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:
Where:
E = extraction factor
KD = distribution coefficient
V = volume of solvent
L = volume of aqueous
EQ. 2
Refer to EQ. 1
CALCULATION METHODS Extraction of Dilute Solution ii. For a single-stage extraction with pure solvent: The fraction of solute remaining: The fraction recovered:
EQ. 3
E1
1
E
E
1
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)
Single stage extraction Solvent (V, y0)
Feed (L, x0)
Extract (V, y1)
Raffinate (L, x1) (a)
Material balance: L(x0) + V(y0) = L (x1) + V(y1) 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)
Since KD = y1/x1, y1=KDx1 Refer to EQ. 1 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
0
)1(
)1(
xphaseraffinateinionconcentratsolute
yphaseextractinionconcentratsoluteDK
CALCULATION METHODS Solution 1 (a) (Draw the material balance diagrams)
Single stage extraction Solvent (V, y0)
Feed (L, x0)
Extract (V, y1)
Raffinate (L, x1) (a)
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)
First stage
Solvent (V, y0)
Feed (L, x0)
Extract (V, y1)
Raffinate (L, x1) Second stage
Solvent (V, y0)
Extract (V, y2)
Final Raffinate (L, x2)
Combined Extract
(b)
Material balance: L(x1) + V(y0) = L (x2) + V(y2) 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)
0
CALCULATION METHODS
Material balance: Since KD = y2/x2, y2=KDx2 Refer to EQ. 1 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%
)2(
)2(
xphaseraffinateinionconcentratsolute
yphaseextractinionconcentratsoluteDK
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. Equilibrium relationship are more complicated-3 or more components present in each phase.
ii. Equilibrium data are often presented on a triangular diagram such as Fig 23.7 and 23.8.
Triangular diagram
i. Consider Fig 23.7
ii. Line ACE shows extract phase
iii. Line BDE shows raffinate phase
iv. 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. Consider Fig 23.8
ii. Line AD shows extract phase
iii. Line BC shows raffinate phase
iv. 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)
TRIANGULAR DIAGRAM
• 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
Coordinate Scale
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:
Single-Stage Extraction
• 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
MULTISTAGE CROSSCURRENT EXTRACTION
• 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:
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
Multistage Crosscurrent Extraction
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.
MULTISTAGE CROSSCURRENT EXTRACTION
Water layer (Raffinate phase)
No Acetic acid
(%) Acetic acid (C)
(wt. Fraction, x) Water
(%) Water (A)
(wt. fraction) Isopropyl ether (%)
Isopropyl ether (B) (wt. Fraction)
1 0.69 0.0069 98.1 0.981 1.2 0.012 2 1.41 0.0141 97.1 0.971 1.5 0.015 3 2.89 0.0289 95.5 0.955 1.6 0.016 4 6.42 0.0642 91.7 0.917 1.9 0.019 5 13.3 0.133 84.4 0.844 2.3 0.023 6 25.5 0.255 71.1 0.711 3.4 0.034 7 36.7 0.367 58.9 0.589 4.4 0.044 8 44.3 0.443 45.1 0.451 10.6 0.106 9 46.4 0.464 37.1 0.371 16.5 0.165
Isopropyl ether layer (Extract phase)
Acetic acid (%)
Acetic acid (C) (wt. Fraction, y)
Water (%)
Water (A) (wt. fraction)
Isopropyl ether (%)
Isopropyl ether (B) (wt. fraction)
1 0.18 0.0018 0.5 0.005 99.3 0.993 2 0.37 0.0037 0.7 0.007 98.9 0.989 3 0.79 0.0079 0.8 0.008 98.4 0.984 4 1.93 0.0193 1 0.01 97.1 0.971 5 4.82 0.0482 1.9 0.019 93.3 0.933 6 11.4 0.114 3.9 0.039 84.7 0.847 7 21.6 0.216 6.9 0.069 71.5 0.715 8 31.1 0.311 10.8 0.108 58.1 0.581 9 36.2 0.362 15.1 0.151 48.7 0.487
0.012, 0.0069 0.015, 0.0141 0.016, 0.0289
0.019, 0.0642
0.023, 0.133
0.034, 0.255
0.044, 0.367
0.106, 0.443 0.165, 0.464
0.993, 0.0018 0.989, 0.0037 0.984, 0.0079 0.971, 0.0193 0.933, 0.0482
0.847, 0.114
0.715, 0.216
0.581, 0.311
0.487, 0.362
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id (
C)
(x, y
)
wt. fraction of isopropyl ether (B)
Rectangular Coordinates
Where, x= wt. fraction of acetic acid in Raffinate Phase y= wt. fraction of acetic acid in Extract Phase
Figure 10.9
Equilibrium points at isopropyl ether layer (Extract phase)
Equilibrium points at water layer (Raffinate phase)
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id (
C)
(x, y
)
wt. fraction of isopropyl ether (B)
Rectangular Coordinates
Distribution curve
Where, x= wt. fraction of acetic acid in Raffinate Phase y= wt. fraction of acetic acid in Extract Phase
Figure 10.9
0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id (
C)
(x, y
)
wt. fraction of isopropyl ether (B)
Rectangular Coordinates
Distribution curve
Where, x= wt. fraction of acetic acid in Raffinate Phase y= wt. fraction of acetic acid in Extract Phase
Figure 10.9(a)
Tie line
0
0.05
0.1
0.15
0.2
0.25
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id (
C)
(x, y
)
wt. fraction of isopropyl ether (B)
Rectangular Coordinates
Distribution curve
Figure 10.9(b)
Tie line
For acetic concentrations up to x = 0.30 (0riginally 30% in Feed)
First stage
Solvent, S1 (ys1)
Feed, F (xF)
Extract, E1 (y1)
Raffinate R1 (x1)
Second stage
Solvent, S2 (ys2)
Extract, E2 (y2)
Raffinate R2(x2)
Third stage
Solvent, S3 (ys3)
Extract, E3 (y3)
Final Raffinate
R3 (x3)
Solution (Draw the material balance diagrams)
First stage
Solvent, S1= 40 kg (ys1 = 0)
Feed, F =100 kg (xF = 0.30)
Extract, E1 (y1)
Raffinate, R1 (x1)
Solution at 1st Stage (Draw the material balance diagram)
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
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19
0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id (
C)
wt. fraction of isopropyl ether (B)
Rectangular Coordinates
1. First draw the tie lines. 2. Then plot the initial point of wt. frac of acetic acid (C) in Feed, F (0, 0.3) and in Solvent, S1 (1, 0). 3. Draw the line FS1 on the Rectangular Coordinates by joining the two points.
FS1 line
S1(1,0)
F (0, 0.3)
Tie line
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19
0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id (
x, y
)
wt. fraction of isopropyl ether (B)
Rectangular Coordinates
4. Plot the point M1 (?, 0.214) where it is located on the line FS1. 5. 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.
M1 (?, 0.214)
XM1 = 0.214
X1 = 0.258
y1 = 0.117
Tie line R1E1
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
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
Solution at 2nd Stage (Draw the material balance diagram)
First stage
Solvent, S1 (ys1=0)
Feed, F (xF =0.3)
Extract, E1 (y1=0.117)
Raffinate R1=96.4 kg (x1=0.258)
Second stage
Solvent, S2 = 40 Kg (ys2=0)
Extract, E2 (y2)
Raffinate R2(x2)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19
0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id
wt. fraction of isopropyl ether
Rectangular Coordinates
4. Plot the point M2 (?, 0.1823) where it is located on the line R1S2. 5. 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.
X2 = 0.227
XM2 = 0.1823
y2 = 0.095
R1S2 line
Tie line R2E2
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
First stage
Solvent, S1 (ys1)
Feed, F (xF)
Extract, E1 (y1)
Raffinate R1 (x1)
Second stage
Solvent, S2 (ys2)
Extract, E2 (y2)
Raffinate R2 =90.2kg
(x2=0.227)
Third stage
Solvent, S3 =40 kg (ys3=0)
Extract, E3 (y3)
Final Raffinate
R3 (x3)
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
Solution at 3rd Stage (Draw the material balance diagram)
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19
0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id
wt. fraction of isopropyl ether
Rectangular Coordinates
4. Plot the point M3 (?, 0.1573) where it is located on the line R2S3. 5. 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.
X3 = 0.2
XM3 = 0.1573
y3 = 0.078
R2S3 line
Tie line R3E3
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.
0 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09
0.1 0.11 0.12 0.13 0.14 0.15 0.16 0.17 0.18 0.19
0.2 0.21 0.22 0.23 0.24 0.25 0.26 0.27 0.28 0.29
0.3
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1
wt.
fra
ctio
n o
f ac
eti
c ac
id
wt. fraction of isopropyl ether
Rectangular Coordinates
F (0,0.3) XF = 0.3
S1,S2,S3 (1,0) ys 1, ys 2 , ys 3 = 0
R1 (?,0.258) X1 = 0.258
R2 (?,0.227) X2 = 0.227
R3 (?,0.2) X3 = 0.2
E1 (?,0.117) y1 = 0.258 E2 (?,0.095) y2 = 0.095
E3 (?,0.078) y3 = 0.078
M1 (?,0.214) xM1 = 0.214
M2 (?,0.1823) xM2 = 0.1823
M3 (?,0.1573) xM3 = 0.1573
Figure 10.15
• 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.
• 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
Aqueous Two Phase Extraction
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.
Single Stage
Extraction
Feed
Solvent
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.
Mixer-settlers
→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
Packed Extraction Towers
→ 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.
Agitated Tower Extractors