ERT 313 Bioseparation Engineering LIQUID-LIQUID EXTRACTION ...portal.unimap.edu.my/portal/page/portal30/Lecturer Notes... · An inlet water solution of 100 kg/h containing 0.010

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


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)


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

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.


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

)



Distribution curve


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

)



Distribution curve


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

)



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)



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

)



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


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



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



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.


• 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.

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

Documents

ERT 313 Bioseparation Engineering LIQUID-LIQUID EXTRACTION ...portal.unimap.edu.my/portal/page/portal30/Lecturer Notes... · An inlet water solution of 100 kg/h containing 0.010