cdb2013 liquid liquid extraction lecture 1

Separation Process ICDB2013

Dr Chew Thiam Leng

Chapter –Liquid-Liquid

Extraction

Lesson outline

• Introduction

• General design considerations

• Working principles of liquid-liquid processes

• Liquid-liquid equilibrium/equilibrium relation in extraction processes

• Single-stage equilibrium extraction

• Continuous multiple stage countercurrent extraction

Lesson outcome

At the end of the session, the students are able to:

1. Discuss the principles of liquid-liquid processes andbasic design considerations.

2. Able to estimate the exit stream amounts andcompositions of single stage extractor.

3. Able to conduct total and component material balanceon multistage liquid-liquid extraction equipment.

4. Able to estimate the number of stages required toachieve the desired separation in liquid-liquidextraction processes.

Recap of pervious lesson

Determination of multiple stages required for a desired

separation using

Material balance and graphical method

Analytical Equations (Kremser Equation)

• Countercurrent Multiple-Contact Stages

• General design considerations

• Working principles of absorption and desorption (stripping)

• Gas-liquid equilibrium

Introduction

• Liquid-liquid extraction (solvent extraction): is the separation of theconstituents of a liquid solution by contact with another insolubleliquid.

Important terms

• The solution which is to be extracted is called the feed,

• the liquid with which the feed is contacted is called the solvent.

• The solvent rich product of the operation is called the extract,the residual liquid from which the solute has been removed isthe raffinate.

• It is a mass transfer operation in which a liquid solution (feed) iscontacted with an immiscible or nearly immiscible liquid (solvent)that exhibits preferential affinity or selectivity towards one or moreof the components in the feed.

Working Principles of Liquid-liquid extraction

When Liquid-liquid extraction is carried out in a test tube or flaskthe two immiscible phases are shaken together to allow moleculesto partition (dissolve) into the preferred solvent phase.

7

CSOLVENT

A+BFEED

EXTRACT

B+A+C

C+A+B

RAFFINATE

• In liquid-liquid extraction, a soluble component (the

solute) moves from one liquid phase to another.

• The two liquid phases must be either immiscible, or

partially miscible.

usually isothermal and isobaric

can be done at low temperature

(good for thermally fragile solutes, such as large organic molecules or

biomolecules)

extracting solvent is usually recycled, often by

distillation.

can be single stage (mixer-settler) or multistage

(cascade)

• An example of extraction:

Acetic acid (A)

in Water (B) +Ethyl acetate

(C)

Extract (C+A+B)Organic layer contains most of

acetic acid in ethyl acetate with a small amount of water.

Raffinate (B+C+A)Aqueous layer contains a weak

acetic acid solution with a small amount of ethyl acetate.

• The amount of water in the extract and ethyl

acetate in the raffinate depends upon their solubility's in one another.

9

Laboratory separation funnel

Typical liquid-liquid extraction equipment

Mixer-settler for Extraction


12

b) Combined mixer-settlera) Separate mixer-settler



13



14

Advantages Disadvantages

Reliable scale up techniques. Large amounts of floor space required.

Can handle high flow rates. Only systems with a few stages are economical.

Ideal for processes with slow mass transfer.

Large amounts of valuable material, such as the solvent, are tied up during the process.

Intense mixing, resulting in small droplets for good mass transfer.

Poor solvent conservation.

Spray Extraction Tower



Vessel in which the mixingis done by the flow of thefluid themselves,

mixing and settlingproceed continuously andsimultaneously.




It has no internal parts except for the spray nozzles and hence eliminate the scale buildup and plugging problem.

Inefficient.

Low cost

The column consists of

• a packed bed,

• distributors for

the two liquid


Packed Extraction Tower


Packed Extraction Tower


Highly efficient when only a few stages are needed.

Suffer from solids plugging

The random packings in the tower cause the droplets to coalesce and redisperse at frequent intervals throughout the tower

The initial dispersion is a concern for packed columns, to avoid channeling

Reduced axial mixing Hard to scale up accurately

rotating-disk contactor

a. agitator; b. stator disk


Mechanically agitated extraction tower

VesselWalls

Shaft

Stators RotorsLightPhase In

HeavyPhase In

LightPhase Out

HeavyPhase Out

Drive Motor Gearbox

InterfaceControl

Interface

Mixer-settlers operate with a purely stage-wise contact.After every mixer there is a settler. Mixer-settlers can beoperated in a multistage, co- or countercurrent fashion.


Mechanically agitated extraction tower


Mechanical parts result in good dispersion.

Repair of internal mechanical parts can cause process delays.

Little axial mixing compared to non-agitated columns

General Design Considerations

Some factors that influence extraction processes are:

• Type of stage configurations

• Minimum solvent flow rate and actual flow as multiple ofminimum flow rate (or reflux for more than one stage system)

• Operating conditions

• Choice of liquid solvents

• Number of equilibrium stages

• Emulsification and scum formation tendencies

• Phase-density difference

• Interfacial tension

• Type of extractor

General Design Considerations

Some characteristics of an ideal solvent:

High selectivity for targeted solute to be extracted

High capacity for dissolving solute

Mininum or no corrosion problem.

Large density difference with carrier

High thermal and chemical stability

No or low toxicity

Low price

Equilibrium relations in extraction

• Extraction involves the use of systems of at least

three substances. The most important notation

scheme for extraction are:

1. Liquid B and C are pure, substantially insoluble

liquids, and A is the distributed solute.

2. Mixtures to be separated by extraction composed of A

and B where as C is the extracting solvent.

3. Equilateral triangular coordinates are often used to

represent the equilibrium data for a three component

systems as shown in the figure;

Fig.: Triangular diagram

Ternary relationship

xA + xB + xC = 0.40 + 0.20 + 0.40 = 1.0

3. Any point on the side of the triangle

represents a binary mixture

Point D for example contains a

fraction of 0.8 A and 0.2 B.

4. The perpendicular distance from any point such as M to the base ABrepresents the mass fraction of C in the mixture at M, the distance tothe base AC of B, and that to the base CB the mass fraction of A. Thus,at point M;

1. Each altitude of the trianglerepresents 100% pure componentand the distance to the three sidesrepresents the percentages or thefractions of the three components.

2. Point M represents the mixture of

the three components

Ternary relationship when A and B are partially miscible phase

• This is the most common type of systems in extraction

Example:

i. water (A)-chloroform(B)-acetone (C) and

ii. methyl isobutyl ketone (A)-water (B)- acetone (C)

iii. benzene (A)-water (B)-acetic acid (C)

26

• Liquid C dissolves completely in A and B, but Liquid A is only slightlysoluble in B and B slightly soluble in A.

• The more insoluble the liquid A and B , the nearer the apexes of the

triangle will points L and K be located.

• The two phase region is included inside below the curved envelope.

• Point M represents a mixture of the two phases.

Fig.: Liquid-liquid phase diagram where components A & B are partially miscible

• Any mixture outside the curve is ahomogenous mixture of one phase liquid.

Any ternary mixture underneath the curve, such as M, will form two

insoluble saturated liquid phase of equilibrium compositions indicated

by a (A-rich) and b (B-rich).

• An original mixture of composition M will separate into two phases a and b which are on the equilibrium tie line through point M.

• The two phases are identical at point P, the Plait point.

• Curve LPK is a bimodal solubility

curve indicating the change in solubility

of the A-and B-rich phases upon addition

of C.

Reading ternary phase diagram

28

• Example: Define the composition of point A, B, C, M, E, R, P and DEPRGin the ternary-mixture.

Define the composition of point A, B, C, M, E, R, P and DEPRG in the ternary-mixture.

Point A = 100% Water

Point B = 100% Ethylene Glycol

Point C = 100% Furfural

Point M = 30% Ethylene glycol, 40% water, 30% furfural

Point E = 43.8% glycol, 10% water, 46.2% furfural

Point R = 11% glycol, 81% water, 8% furfural

The miscibility limits for the furfural-water binary system are at point D and G.

Point P (Plait point), the two liquid phases have identical compositions.

DEPRG is saturation curve or bimodal curve

29

Example 5.1Any point on the side of the triangle represents a binary mixture

30

From the diagram below determine the concentrations (mole fractions) at

Point M, E and R?

Example 5.2

31

•

•

•

Consider the point M:

Furfural content (xC):

0.19

0.20

0.61

check: xA + xB + xC = 1

Solution

water content (xA) :

Ethylene glycol content (xB):

Do the same for point E, R and P.

Equilibrium data on rectangular coordinates

Since triangular coordinates have some disadvantage due to the special

coordinates, in order to be able to expand one concentration scale relative

to the other, a more useful method of plotting the three-component data is

to use rectangular coordinates.

The solvent pair B and C are partially miscible.

The two phase region is inside the envelop.

yB = 1.0 - yA - yC

The concentration of B is obtained by differences from equation below

xB = 1.0 - xA - xC

The tie line gi is connecting the water-rich layer i

called the raffinate, x layer and the ether-rich layer, g called the extract layer, y

To construct the tie line gi using the equilibrium

yA-xA plot below the phase diagram , vertical line to g and I are drawn

Raffinate compositions are represented by

coordinates: (xA, xC)

Extract compositions are represented by

coordinates: (yA, yC)

The system acetic acid (A) – water (B) – isopropyl ether solvent (C).

Liquid-liquid Equilibrium phase diagram

34

1. Find mixing point, h

2. Find tie-line through h (trial

and error);

3. Find g and i at either end

(co-linear with h).

4. Find flow rates of g and i.

Overview of solution using

Rectangular diagram:

An original mixture weighing 100 kg and containing 30 kg of

isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is

equilibrated and the equilibrium phases separated. What are the

compositions of the two equilibrium phases?

Solution:

Composition of original mixture is xc= 0.3, xA = 0.10, and xB = 0.60.

35

Example 5.3

36

1. Plot the composition of xC =

0.30, xA = 0.10 and xB on the

equilibrium diagram as at

point h.

2. The tie line gi is drawn

through point h by trial and

error.

3. The composition of the

extract (ether) layer at g is yA

= 0.04, yC = 0.94, and yB =

1.00 - 0.04 - 0.94 = 0.02

mass fraction.

4. The raffinate (water) layer

composition at i is xA = 0.12,

xC = 0.02, and xB = 1.00 –

0.12 – 0.02 = 0.86.

Solution

Equilibrium contact stages

Single stage equilibrium extraction

• As shown in Figure below the equation can be derived using rectangular coordinates.

• Two streams containing L kg and V kg containing component A, B, and C are mixed to give a resulting mixture M total mass.

• Writing an overall mass balance and balance on A and on C

VM

mixer

L

L + V = M

VyA + LxA = MxA,M = (V + L)xAM

VyC + LxC = MxCM = (V + L)xCM 5.4

5.3

5.2

AAM

AMA

xx

xy

V

L

• Similarly combining Eqs (5.2) and (5.4)

CMC

MCC

xx

xy

V

L

(5.5)

AAM

CMC

AMA

MCC

yx

yx

xx

xx

(5.6)

•Equating 5.5 and 5.6 and rearranging

• Combining Eqns (5.2) and (5.3)

V

M

mixerL

(5.7)

ML

MV

kgV

kgL

)(

)(

VL

MV

kgM

kgL

)(

)(

(5.8)

Lever arm’s rule

Eqn. 5.7 shows that points L, M, and V must lie on a straight line.

By using the properties of similar right triangles,

(5.9)

V•

M•

L•

similar triangles

The compositions of the two equilibrium layers in Example 5.1 are for the extract layer (V2) yA = 0.04, yB = 0.02, and yC = 0.94, and for the raffinate layer (L1) xA = 0.12, xB = 0.86, and xC = 0.02. The original mixture contained 100 kg and xAM = 0.10. Determine the amounts of V1 and L0.

Solution:

10012 MLV

)10.0(100)12.0()04.0( 12 LV

40

Substituting into eq. 5.3, where M = 100 kg and xAM = 0.10,

Substituting into eq. 5.2

V1V2

mixer

L0 L1

Example 5.4

Given:

extract layer (V2): yA = 0.04, yB = 0.02,

raffinate layer (L1) xA = 0.12, xB = 0.86,

M= 100 kg, xAM = 0.10

41

Solution:

)(10012 aMLV

)10.0(100)12.0()04.0( 12 LV

Substituting into eq. 5.3, where M = 100 kg and xAM = 0.10,

Substituting into eq. 5.2

V1V2

mixer

L0 L1

Given:

Extract layer (V2): yA = 0.04, yB = 0.02,

Raffinate layer (L1) xA = 0.12, xB = 0.86,

M= 100 kg, xAM = 0.10

Required: V1 and L0

)(10121.004.0 12 bLV

8.5

2.4

100

ig

ghL

M

L

Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the material-balance method.

•Solving eqns (a) and (b) simultaneously,

L1 = 75.0 and V2 = 25.0.

•Alternatively, using the lever-arm rule,

the distance hg in Figure below is

measured as 4.2 units and gi as 5.8 units. Then by eq. 5.8,

g•

h•

i•

similar triangles

Single-stage liquid-liquid extraction processes

Single-state equilibrium extraction

MVLVL 1120

AMAAAA MxyVxLyVxL 11112200

MCCCCC MxyVxLyVxL 11112200

We now study the separation of A from a mixture

of A and B by a solvent C in a single equilibrium

stage.

0.1 CBA xxx

An overall mass balance:

A balance on A:

A balance on C:

5.10

5.11

5.12

43

V1V2

L0 L1

To solve the three equations, the equilibrium-phase-diagram is used.

1. L0 and V2 are known.

2. We calculate M, xAM, and xCM by using

equation 5.10-5.12.

3. Plot L0, V2, M in the Figure.

4. Using trial and error a tie line is drawn

through the point M, which locates the

compositions of L1 and V1.

5. The amounts of L1 and V1 can be

determined by substitution in Equation

5.10-5.12 or by using lever-arm rule.

44

V1V2

L0 L1

Example 5.5

45

A mixture weighing 1000 kg contains 23.5 wt% acetic acid

(A) and 76.5 wt% water (B) and is to be extracted by 500

kg isopropyl ether (C) in a single-stage extraction.

Determine the amounts and compositions of the extract and

raffinate phases.

Given:

10,0500

0765.0,235.0,1000

2222

0

CBA

CoBoAo

yandyykgV

xandxxkgL

AMx)1500()0)(500()235.0)(1000(

kgMVL 1500500100020

157.0AMx

MCCC MxyVxL 2200

Similarly 0765.0235.00.11 000 BAc xxx

AMAA MxyVxL 2200

MCx)1500()1)(500()0)(1000(

33.0CMx

V1V2

L0 L1

Solution

M

V2 (0,1) = (yA2, yC2)

V1 (0.1,0.89) = (yA1, yC1)

L1(0.2,0.03) = (xA1, xC1)

L0(0.235,0) = (xA0, xC0)

M(0.157,0.33) = (xAM, xCM)

47

1. L0 and V2 are known.

2. We calculate M, xAM, and xCM

by using equation 5.10-5.12.

3. Plot L0, V2, M in the Figure.

4. Using trial and error a tie

line is drawn through the

point M, which locates the

compositions of L1 and V1.

5. The amounts of L1 and V1

can be determined by

substitution in Equation 5.10-

5.12 or

• Alternatively, by using lever-

arm rule.

(1)

AMAA MxyVxL 1111

MCCC MxyVxL 1111

)157.0)(1500()1.0()2.0( 11 VL

)33.0)(1500()89.0()03.0( 11 VL

From the graph: xC1 = 0.03 and yC1 = 0.89;

From the graph: xA1 = 0.2 and yA1 = 0.1;

(2)

5.177,15.0 11 VL

500,1667.29 11 VL

Solving eq(1) and eq(2) to get L1 and V1;

kgVandkgL 28.52586.914 11

89.003.0,1.0,2.0 1111 CCAA yandxyx Answer

48

49

A single-stage extraction is performed in which 400 kg of a solution

containing 35 wt% acetic acid in water is contacted with 400 kg of pure

isopropyl ether. Calculate the amounts and compositions of the extract

and raffinate layers. Solve for the amounts both algebraically and by

the lever-arm rule. What percent of the acetic acid is removed? Use

equilibrium data given below.

Exercise

50

Solution

AMx)800()0)(400()35.0)(400(

kgMVL 80040040020

0.165.0,35.0 200 ABA yandxx

Given:

175.0AMx

MCCC MxyVxL 2200

Given: 065.035.00.11 000 BAc xxx

AMAA MxyVxL 2200

MCx)800()1)(400()0)(400(

5.0CMx

Draw the equilibrium diagram and Locate point,

L0, V1 and M

51

(i)

AMAA MxyVxL 1111

)175.0)(800()12.0()22.0( 11 VL

• From the graph: xA1 = 0.22 and yA1 = 0.12;

36.63654.0 11 VL

V2 (0,1) = (yA2, yC2)

V1 (0.12,0.87) = (yA1, yC1)

L1(0.22,0.03) = (xA1, xC1)

L0(0.35,0) = (xA0, x

M(0.175,0.5) = (xAM, xCM)

M

• Using trial and error, the tieline is drawn through pointM that determines thecomposition of LN and V1

MCCC MxyVxL 1111

)5.0)(800()87.0()03.0( 11 VL

From the graph:

xC1 = 0.03 and

yC1 = 0.87;

(ii) 33.333,129 11 VL

Solving eq(i) and eq(ii) to get L1 and V1;

kgVandkgL 49.2412.623 11

88.003.0,12.0,23.0 1111 CCAA yandxyx

52

53

ML

MV

kgV

kgL

1

1

1

1

)(

)(

11

11

)(

)(

VL

MV

kgM

kgL

Lever arm’s rule

M

V1 (0.12,0.88) = (yA1, yC1)

L1(0.23,0.03) = (xA1, xC1)

M(0.175,0.5) = (xAM, xCM)

81.047.0

38.0

)(

)(

1

1 kgV

kgL

44.086.0

38.0

)(

)(1 kgM

kgL kg352kg80044.0)kg(1 L

kg57.43481.0/kg352)(1 kgV

The amount and composition of extract are

*57.4341 kgV

74.003.0,23.0 111 BCA xandxx

54

*3251 kgL

0

88.0

,12.0

1

1

1

B

C

A

y

andy

y

The amount and composition of raffinate are

*The correction of amount of extract and raffinate depends on the

identification of points on the graph (setting the pint). The value obtained from Lever arm’s rule is more reliable.

55

The percent of acetic acid removed is about

%𝑅𝑒𝑚𝑜𝑣𝑒𝑑

=𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 − 𝑎𝑚𝑜𝑢𝑛𝑡𝑜𝑓𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑𝑖𝑛 𝑡ℎ𝑒 𝑟𝑎𝑓𝑖𝑛𝑎𝑡

𝑎𝑚𝑜𝑢𝑛𝑡𝑜𝑓𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑𝑖𝑛 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑

%𝑅𝑒𝑚𝑜𝑣𝑒𝑑 =(400x0.35−325x0.23)x100%

400𝑥0.35= 𝟒𝟔. 𝟔%

56

3. A feed mixture weighing 200 kg of unknown composition

containing water, acetic acid, and isopropyl ether is contacted

in a single stage with 280 kg of a mixture containing 40 wt %

acetic acid, 10 wt % water, and 50 wt. % isopropyl ether. The

resulting raffinate layer weighs 230 kg and contains 29.5 wt. %

acetic acid,66.5 wt. % water, and 4 wt. % isopropyl ether.

Determine the original composition of the feed mixture and the

composition of the resulting extract layer. (Use Equilibrium

data Given below

57

TABLE : Acetic acid-water-isopropanol ether liquid-liquid

equilibrium data

Water layer (wt%) Isopropyl layer (wt%)

Acetic acid Water

Isopropyl ether Acetic acid Water

Isopropyl ether

0 98.8 1.2 0 0.6 99.4

0.69 98.1 1.2 0.18 0.5 99.3

1.41 97.1 1.5 0.37 0.7 98.9

2.89 95.5 1.6 0.79 0.8 98.4

6.42 91.7 1.9 1.93 1.0 97.1

13.30 84.4 2.3 4.82 1.9 93.3

25.50 71.1 3.4 11.4 3.9 84.7

36.70 58.9 4.4 21.6 6.9 71.5

44.30 45.1 10.6 31.10 10.8 58.1

46.40 37.1 16.5 36.20 15.1 48.7

cdb2013 liquid liquid extraction lecture 1

Documents

principles of liquid

liquid phases

residual liquid

liquid solution feed

immiscible liquid solventthat

example of extraction

single stage mixersettler

desired separation