CHAPTER 4:EXTRACTION PROCESS

What separation method do all the following processes have in common?

Decaffeination

Manufacture of Penicillin

Separation of Aromatics from Hydrocarbons

Wastewater Treatment

LIQUID-LIQUID EXTRACTION

Introduction to Liquid-Liquid Extraction

• ObjectivesUnderstand concept of LLEEquilibrium Relations in extraction

Single stage equilibrium extraction

Understand equipment for LLECalculate the ideal stages required for LLE

What is LLE?• Mass transfer separation• Liquid solution (the feed) contacted with

immiscible liquid (the solvent)• The results of this contact are:• The extract: the solution containing the desired

extracted solute• The raffinate: the residual feed solution

containing little solute.

Feed (containing solute) Raffinate (solute________)

Extract (solute _________) SolventRICH

POOR

Advantages and Disadvantages of LLE

Advantages:

• Effective at low pH and low temperature• Separation by distillation is not feasible ie. expensive and complicated• Solute and solvent have similar volatility• Solute and solvent form azeotropic

mixture• The material is heat-sensitive ie.

penicillin• The material is non-volatile ie. mineral

salts• Solute concentration low

Disadvantages:

• Incomplete separation, which contaminates the product.

• Multiple processes are required

Baird, M. “Handbook of Solvent Extraction”www.modular-process.com/exraction_2.html

Applications of LLE• Wastewater treatment

– organic pollutants from a highly salted waste stream into another aqueous stream free of salts

• Separation of metals– Recovery of Zinc (and Copper) from Mine Waters

• Food industry – Decaffeination

• Lube oil extraction– Aromatics and unsaturated hydrocarbons ie. SO2 and benzene

extraction • Acetic acid extraction

– Manufacture of cellulose yields aqueous acetic acid • Pharmaceutical manufacturing

– Penicillin and vitamins extraction

From: Baird, Malcom “Handbook of Solvent Extraction”, 1982

SOLVENT SELECTION

Factors to be considered: Selectivity Distribution coefficient Insolubility of solvent Recoverability of solute from

solvent Density difference between liquid

phases Cost Viscosity, vapour pressure Flammability, toxicity

Solvent is the key to a successful separation by liquid-liquid extraction.

SELECTIVITY

• The ability of a solvent to extract the solute from the feed

• Good selectivity =

• For all useful extraction operation the selectivity must exceed unity. If the selectivity is unity, no separation is possible .

• Little or no miscibility with feed solution

• Depends on the nature of the solvent, pH, and residence time

From: www.cheresources.com/extraction.shtml

0.1phasefeed

phaseextraction

DISTRIBUTION COEFFICIENT• ratio (at equilibrium) of the concentration of solute

in the extract and raffinate phases.

Distribution coefficient , less solvent is required for a given degree of extraction

xyK /

No azeotrope formed between solvent and solute

Mixtures should have a high relative volatility Solvent should have a small latent heat of

vaporization.

RECOVERABILITY

CHEMICAL REACTIVITY• Solvent should be stable and inert

VISCOSITY, VAPOR PRESSURE, FREEZING POINT

• These should be low for ease in handling and storage, for example, a high viscosity leads to difficulties with

pumping , dispersion and mass-transfer rate.

INSOLUBILITY • The solvent should have low solubility in the feed solution,

otherwise the separation is not "clean".

DENSITY

• A large difference in density between extract and raffinate. phases permits high capacities in equipment.

OTHERS• Non-flammable• Non-toxic• Low cost

Mixer-Settles for Extraction

Separate mixer-settler Combined mixer-settler

Types of Extractors

16

To provide efficient mass transfer, a mechanical mixer is often used to provide intimate contact of 2 liquid phases.

Plate and Agitated Tower Contactors for Extraction

Perforated plate (sieve tray) tower Agitated extraction tower

Types of Extractors

17

The rising droplets of the light solvent liquid are dispersed. The dispersed droplets combine below each tray and then re formed on each tray by passing through the perforations.

A serial of paddle agitatorss mounted on a central rotating shaft provides the agitation for the 2 phases.

Packed and Spray Extraction Towers

Spray-type extraction towerPacked extraction tower

Types of Extractors

18

Packed & spray tower extractors give differential contacts, where mixing and settling proceed continuously and simultaneously.

Types of Extractors

Equilibrium Relations in Extraction• LLE system have 3 components, A ,B and C and 2 phases in

equilibrium.

• Total mass fraction = 1.0 xA + xB + xC = 1

• Equilateral triangular coordinates are often used to represent the equilibrium data of a three component system (3 axes)

Triangular coordinates and equilibrium data Each of the three corners

represents a pure component A, B, or C.Point M represents a mixture of A, B, and C.The perpendicular distance from the point M to the base AB represents the mass fraction xC. The distance to the base CB represents xA, and the distance to base AC represents xB.

Coordinates for a triangular diagram

(A and B are partially miscible.)

xA + xB + xC = 0.4 + 0.2 + 0.4 = 1

xB = 1.0 - xA - xC

yB = 1.0 - yA - yC

Equilibrium Relations in extraction

Consider the point M:water content (xA) is ?ethylene glycol content (xB) is ?furfural content (xC) is ?

0.190.200.61

Reading ternary phase diagrams

Read the mole/mass fraction of each component on the axis for that component, using the lines parallel to the edge opposite the corner corresponding to the pure component.

A 2-component mixture of furfural and water is partially miscible over the composition range from about 8 % furfural to 95 % furfural. Separation by extraction requires a furfural/water ratio in this range (otherwise – single phase).

The mixture M lies inside the miscibility boundary, and will spontaneously separate into two phases. Their compositions (E and R) are given by the tie-line through M. The compositions of E and R converge at the plait point, P (i.e., no

separation).

region of partial miscibility A-C

check: xA + xB + xC = 1

•

•

•

Point A = 100% Water

Point B = 100% Ethylene Glycol

Point C = 100% Furfural

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

Point E = 41.8% glycol, 10% water, 48.2% furfural

Point R = 11.5% glycol, 81.5% water, 7% 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.

Liquid-Liquid equilibrium, ethylene glycol-furfural-water, 25ºC,101 kPa.23

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

Right-triangle phase diagrams

• Since triangular diagram, have some disadvantages because of special coordinates

• Right triangle coordinates is more useful method plotting the 3 components data.

• Right triangle coordinates for system acetic acid (A) – water (B) – isopropyl ether solvent (C).

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

The solvent pair B and C are partially miscible.

vertical axis = comp. C horizontal axis.= comp. A

Equilibrium data(solubility curve) yA-xA is plot below phase diagram.

tie line gi is construct by connecting water rich layer i (raffinate layer) and the ether rich solvent layer g(extract layer) by using Equilibrium data yA-xA is plot below phase diagram.

xB = 1.0 - xA - xC

yB = 1.0 - yA - yC

Right-triangle phase diagrams

25

4.6 Single-Stage Equilibrium Extraction

REMSF Total material balance:

REMSF RxEyMxSyFx Solute balance:

4.1

4.2

4.3

26

SFSyFx

x sFm

Combine both

eqn:

MF

sMMSF xx

yxSFxSFSyFx

)(

ME

RMMER xy

xxxEREyRx

RE)(

4.4

4.5

1000 kg of an aqueous solution containing 50% acetone is contacted with 800 kg of chlorobenzene containing 0.5 mass % acetone in a mixer settler unit,followed by separation of the extract and the raffinate phase. Determine the composition extract & raffinate phases.

EXAMPLE

• Mass of feed, F =1000k g • mass fraction acetone (C ) in the feed, x CF = 0.5• mass fraction Chlorobenzene in the feed, x BF = 0• Solvent:• Mass of solvent, S =800k g • mass fraction acetone (C ) in the solvent, y CS = 0.005• mass fraction Chlorobenzene in the solvent, y BS =

0.995

kgMSF 18008001000 Total material balance:

28.01800

)005.0(800)5.0(1000

SFSyFx

xc sFm

From figure:

600kg R 1200kg E 28.0302.0

236.028.0RE

1800

kgRE

302.0 0.236xcR Eyc

Single stage extraction calculation ilustrated;(a)Right triangular coordinate and(b)x-y diagram

Continuous multistage countercurrent extractionCountercurrent process and overall balance

MVLVL NN 110

MCCNCNNCNC MxyVxLyVxL 111100

1

11

10

1100

VLyVxL

VLyVxL

xN

CNCN

N

NCNCMC

1

11

10

1100

VLyVxL

VLyVxL

xN

AANN

N

ANNAMA

An overall mass balance:

A balance on C:

Combining 4.12 and 4.13

Balance on component A gives

4.12

4.13

4.14

4.15

30

Continuous multistage countercurrent extractionCountercurrent process and overall balance

1. Usually, L0 and VN+1 are known and the desired exit composition xAN is set.2. Plot points L0, VN+1, and M as in the figure, a straight line must connect these three points. 3. LN, M, and V1 must lie on one line. Also, LN and V1 must also lie on the phase envelope.

31

Pure solvent isopropyl ether at the rate of VN+1 = 600 kg/h is being used to extract an aqueous solution of L0=200 kg/h containing 30 wt% acetic acid (A) by countercurrent multistage extraction. The desired exit acetic acid concentration in the aqueous phase is 4%. Calculate the compositions and amounts of the ether extract V1 and the aqueous raffinate LN. Use equilibrium data from the table.

Solution: For the mixture point M, substituting into eqs. below,

75.0600200

)0.1(600)0(200

10

1100

N

NCNCMC VL

yVxLx

075.0600200

)0(600)30.0(200

10

1100

N

ANNAMA VL

yVxLx32

L0 =200kg/h, xA0 = 0.30, xB0 = 0.70, xC0 = 0,

xAN = 0.04.

VN+1 = 600kg/h, yAN+1 = 0, yCN+1 = 1.0,

4.14

4.15

Using these coordinates, 1) In figure below, VN+1 and L0 are plotted. Also, since LN is on the

phase boundary, it can be plotted at xAN = 0.04.2) Point M is plotted in Figure below. 3) We locate V1 by drawing a line from LN through M and

extending it until it intersects the phase boundary. This gives yA1 = 0.08 and yC1 = 0.90.

4) For LN a value of xCN = 0.017 is obtained. By substituting into Eqs. 4.12 and 4.13 and solving, LN = 136 kg/h and V1 = 664 kg/h.

33

Stage-to-stage calculations for countercurrent extraction.

1. Δ is a point common to all streams passing each other, such as L0 and V1, Ln and Vn+1, Ln and Vn+1, LN and VN+1, and so on.

2. This coordinates to locate this Δ operating point are given for x cΔ and x AΔ in eqn 4.21. Since the end points VN+1, LN or V1, and L0 are known, xΔ can be calculated and point Δ located.

3. Alternatively, the Δ point is located graphically in the figure as the intersection of lines L0 V1 and LN VN+1.

4. In order to step off the number of stages using eqn. 4.22 we start at L0 and draw the line L0Δ, which locates V1 on the phase boundary.

5. Next a tie line through V1 locates L1, which is in equilibrium with V1.

6. Then line L1Δ is drawn giving V2. The tie line V2L2 is drawn. This stepwise procedure is repeated until the desired raffinate composition LN is reached. The number of stages N is obtained to perform the extraction.

34

Conclusion

• In all the operations, diffusion occurs in at least one phase OR both phases

Gas absorption Solute diffuses through the gas phase to the interface between the phases

Distillation -Low boiler diffuses through the liquid phase to the interface- away from interface to the vapor

Extraction Solute diffuses through the raffinate phase to interface and then into the extract phase

SOLID-LIQUID EXTRACTION

Solid –Liquid Extraction

• Objectives

Understand concept of leachingCarry out mass balance for leachingCalculate the ideal stages required

for leaching

• SLE or Leaching is an extraction of a soluble constituent from a solid by using a liquid solvent.

• In order to separate the desired solute constituent or remove an undesirable solute component from the solid phase, the solid is contacted with a liquid phase

• When two phases are in intimate contact and the solute can diffuse from the solid to the liquid phase, which causes separation of the components originally in the solid.

• Extraction of vegetable oils– extract oil from peanuts,

soybeans, castor beans by using Organic solvent (hexane, acetone, etc) .

– removal of nickel salts or gold from their natural solid beds with sulfuric acid solutions

Applications in Industry

• Food Industry• sugar industry when

soluble sucrose is removed by water extraction from sugar cane or beet.

• Pharmaceutical• Herbal and oil

extraction

solvent

Inert

Solute (transition component)

Before extraction After extraction

Solvent + solute

Inert

• Theory :

amount of soluble material removed is often greater than ordinary filtration.

properties of the solid may change considerably during the leaching process

coarse, hard or granular feed solids may disintegrate into pulp or mush when their content of soluble material is removed.

Factors influencing the solid liquid extraction(leaching)

Factors

Particlesize

SolventAgitationof fluid

Temperature

Factors influencing the extraction(a) Particle size- the smaller the particle size, the

interfacial area between the solid and liquid is greater.

- increase the rate of transfer of material and smaller the distance of the solute must diffuse within the solid.

(b) Solvent-have low viscosity easy for it to circulate freely.~ Low boiling point & non toxic.Easy to remove from product liquor by flash vaporization

(c) Temperature increase the temperature will increase

the solubility of material which is being extracted to give the higher rate of extraction.

the diffusion coefficient will increase with the rise of temp. and improve the rate of extraction.(d) Agitation of the solvent

- increase the eddy diffusion and increase the solid liquid mass transfer coeefficient of material from the surface of the particles to the bulk of the solution.

- prevents sedimentation.

Principles of Continuous Countercurrent Leaching

In leaching, the solvent is present to dissolve all the solute in the entering solid and no adsorption of solute by the solid.

Equilibrium is attained when the solute is completely dissolved and the concentration of the solution so formed is uniform.

Equilibrium Relationship in Leaching

General assumption;a) The system consists of three components:

i. Solute.ii. A solvent.iii. An inert solid.

b) The flowrate of the inerts from stage to stage is constant.

c) It is assumed that equilibrium is attained, thus the concentration of the solution leaving a stage is the same as the concentration of the solution adhering to the inerts. The equilibrium relationship is xe = ye

Ideal Leaching

Solute completely dissolves

Ratio of solid to liquid in the underflow is a constant

i. Solute= Aii. Solvent = Ciii. An inert solid= B

General arrangement in Continuous Countercurrent Leaching

The stages are numbered in the direction of flow of the solid.

V phase = Mass flow of solvent (it moves from stage N to stage n + 1).

L phase = Mass flow of solute (it moves from stage 1 to stage N). Underflow (L phase ) =Mass flow of soluble solid + residual

solvent. Overflow (V phase )= Mass flow of solvent+ mass flow of solute. Exhausted solids leave stage N. Concentrated solution overflows leave from stage 1.

Leaching The flow L and V may be expressed in mass per unit time. x1 : solution retained by entering solid

xN : solution retained by leaving solidyN : fresh solvent entering systemy1 : concentrated solution leaving system

Fraction in underflow

Fraction in Overflow

The overflow and underflow are brought into contact so that intimate mixing is achieved and the solution leaving in the overflow has the same composition as that associated with the solids in the underflow.

The solute free solid is assumed insoluble in the solvent and the flowrate of this solid is constant throughout the cascade.

• Principles of Continuous Countercurrent Leaching

The composition of 3 component mixture can be represented on a right-angled triangular diagram.

The proportion of solute A is plotted as the abscissaThe proportion of solvent S is plotted as the ordinateThe proportion of insoluble solid B is obtained by difference.

Number of ideal stages by using Right-angled Triangular Diagram

• Principles of Continuous Countercurrent Leaching

Figure 4.1b: Right-angled Triangular Diagram

Number of ideal stages by using Right-angled Triangular Diagram

Number of ideal stages by using Right-angled Triangular Diagram

• Mass fractions of the solute, the solvent, and the inerts(underflow) are calculated from:For the solute (A):

For the solvent (S):

11

K

xB

KKyx A

A

1

KyKx A

S

1

)1(

xA= mass frac. of solutexs-= mass frac of solventxB= mass fraction of inertK= mass of solution removed in the underflow per unit mass of solids

yA = mass of solute / lb of solution in the overflow

carriersolutesolventcomponentx

)(1 BA xxxs

1 Bxxx As

carriersolutesolventcomponentx

y= fraction in overflowx= fraction in underflow

Overflow:

solutesolventcomponenty

1 As yy

As yy 1

Underflow:

Soverflow

UnderflowA

Overflow:

solutesolventyy

solutesolventsoluteysolutesolventsolventy

A

s

A

s

/

)/()/(

Underflow:

solutesolventxx

carriersolutesolventsolutexcarriersolutesolventsolventx

A

s

A

s

/

)/()/(

massTotalAofMassAfractionMass

General formula:

Construct graph by right triagular Diagram Method

To determine the number of stages for countercurrent leaching

1.Plot line for all underflow( xs vs xA)

2. Material balance to find outgoing overflow ,y1

Underflow line ( xs vs xA)

3. The difference point, F ’ lies on a straight line through xo and y1.4. F ’ is constant and thus xn,yn+1 and F ’ also lie on common line. xn is on the line for all underflows.

5. Find x1 by drawing tie line from origo to y1.

x1

6. Find y2 by connecting F ’,x1 and overflow line.

7. Step off until x1< xn.

ExampleSeeds containing 25% by weight of oil, are

extracted in a countercurrent plant, and 90% of the oil is recovered in a solution containing 50% of oil. It has been found experimentally that the amount of the solution removed in the underflow in association with every kilogram of insoluble matter is given by the equation;

k = 0.7 + 0.5ya +3ya2

where yA is the concentration of the overflow solution (weight fraction of solute). If the seeds extracted with fresh solvent, how many ideal stages are required?

Basis: 100 kg underflow feed to the first stage

Make a table to get data underflow line x A and x S

kkyx A

A

1 kykx A

S

1

)1(Solute :

Solvents :

n+1 n 2 1

yn+1

Vn+1

yn

Vn

ya

Va

La

Xa

Ln

Xn

Ln-1

Xn-1

In the underflow feed:The seeds contain 25% oil and 75% inert,

01 Sx

25.01 Ax.

0.11 sny

01 Any

In the overflow feed:Pure solvent is used,

This point is marked as yn + 1 on the graph

In the overflow product:The oil concentration is 50%

50.01 Sy 50.01 Ayand

This point lies on the hypotenuse and is marked on the graph

1y

.

In the underflow product:90% of the oil is recovered, leaving 25(1 – 0.90) = 2.5 kg associated with 75 kg inerts;Ratio (oil/inerts) = (2.5/75) = 0.033 = kyA

)35.07.0(

)35.07.0(32

2

AAAA

AA

yyyky

yyk

045.0Ay)35.07.0(033.0 32

AAA yyy

Multiply by yA

033.0Aky

033.0)045.0( k

733.0045.0033.0

k

019.01733.0

)733.0)(045.0(11

k

kyx AAn

404.01733.0

)733.0)(045.01(1)1(

1

kky

x ASn

This point is drawn as on the graph.1nx

The pole point is obtained where and extended meet.

11. nn xy 11.xy

Construct graph : 1)Plot against

on the graph. 2)Marking point

and

3)Draw the stages on the graph

sx Ax

1x 1, y

1nx 1ny

THE END