CHAPTER 4:

EXTRACTION PROCES

What separation method do all the following

processes have in common?

Decaffeination Wastewater Treatment

Separation of Aromatics

from Hydrocarbons

Manufacture

of Penicillin

LIQUID-LIQUID EXTRACTION

Introduction to Liquid-Liquid Extraction

• Objectives

Understand concept of LLE

Equilibrium Relations in

extraction

Single stage equilibrium

extraction

Understand equipment for LLE

Calculate 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

solutionRICH

containing little solute.

Extract (solute _________)

Feed (containing solute)

Solvent

POOR

Raffinate

(solute________)

Advantages and Disadvantages of

LLE

Advantages:

• Effective at low pH and low

temperature

• Separation by distillation is not feasible

•

•

•

•

•

Disadvantages:

• Incomplete

separation, which

contaminates the

product.

• Multiple processes

are required

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

Baird, M. “Handbook of Solvent Extraction”

Solute concentration low

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

Solvent is the key to a successful separation

by liquid-liquid extraction.

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

SELECTIVITY

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

feed

extraction phase

1 .0

• Good selectivity =

feed phase

• 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

•

DISTRIBUTION

COEFFICIENT

ratio (at equilibrium) of the concentration of solute

in the extract and raffinate phases.

K y/x

Distribution coefficient , less solvent is required for a

given degree of extraction

RECOVERABILITY

No azeotrope formed between solvent and

solute

Mixtures should have a high relative volatility

Solvent should have a small latent heat of

vaporization.

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 solutio

otherwise the separation is not "clean".

DENSITY

• A large difference in density between extract and raffinat

phases permits high capacities in equipment.

OTHERS

• Non-flammable

• Non-toxic

• Low cost

Types of Extractors

Mixer-Settles for Extraction

To provide efficient mass transfer, a mechanical

mixer is often used to provide intimate contact of 2

liquid phases.

Separate mixer-settler

Combined mixer-settler

16

Types of Extractors

Plate and Agitated Tower Contactors for

Extraction

The

rising droplets of the

A serial of paddle

light solvent liquid are

dispersed. The dispersed

droplets combine below

each tray and then re

formed on each tray by

passing through the

perforations.

Perforated plate (sieve tray)

tower

agitatorss mounted on a

central rotating shaft

provides the agitation for

the 2 phases.

Agitated extraction tower

17

Types of Extractors

Packed and Spray Extraction Towers

Packed & spray tower extractors give differential

contacts, where mixing and settling proceed

continuously and simultaneously.

Packed extraction tower

Spray-type extraction tower

18

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

•

Equilateral triangular coordinates are often used to represent the

equilibrium data of a three component system (3 axes)

xA + xB + xC = 1

Equilibrium Relations in extraction

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

xA + xB + xC = 0.4

represents

xB.+ 0.2 + 0.4 = 1

Coordinates for a triangular

diagram

(A and B are partially miscible.)

xB = 1.0 - xA - xC

yB = 1.0 - yA - yC

Reading ternary phase diagrams

Consider the point M:

water content (xA) is ?

0.19

ethylene glycol content (xB) is

?

0.20

0.61

furfural content (xC) is ?

•

check: xA + xB + xC = 1

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.

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

region of partial miscibility A-C

M.

The compositions of E and R converge at the plait point, P (i.e., no

separation).

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

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% 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.

23

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

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).

Right-triangle phase diagrams

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

25

4.6 Single-Stage Equilibrium

Extraction

Total material balance:

Solute

balance:

Combine both

eqn:

FS M ER

Fx F Sy S Mx M Ey E Rx R

xm

Fx F Sy s

FS

4.2

4.3

Fx F Sy S ( F S ) x M

Rx R Ey E ( R E ) x M

4.1

F xM y s

S xF xM

4.4

E xM xR

R y E xM

4.5

26

EXAMPL

1000 kg of an aqueous solution containing 50%

acetone is contacted with E800 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.

•

•

Mass of feed, F =1000k g

mass fraction acetone (C ) in the feed, x

•

•

•

•

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

0.995

Total material F S M 1000 800 1800kg

balance:

xcm

From figure:

Fx F Sy s 1000(0.5) 800(0.005)

0.28

F S

1800

xc R 0.236

E R 1800kg

E 0.28 0.236

R 0.302 0.28

yc E 0.302

E 1200kg

R 600kg

CF

= 0.5

BS

=

Single stage extraction calculation ilustrated;

(a) Right triangular coordinate and

(b)x-y diagram

Continuous multistage countercurrent extraction

Countercurrent process and overall balance

An overall mass balance:

A balance on C:

L0 V N 1 L N V1 M

4.12

L0 xC 0 V N 1 y C N 1 L N xC N V1 y C1 Mx C M

Combining 4.12 and 4.13

x CM

Balance on component A gives

x AM

L0 xC 0 V N 1 y CN 1

L0 V N 1

L N xC N V1 y C1

4.13

4.14

LN V1

L0 x A0 V N 1 y AN 1 L N x AN V1 y A1

L0 V N 1

LN V1

30

4.15

Continuous multistage countercurrent extraction

Countercurrent 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 V 1 and

the aqueous raffinate LN. Use equilibrium data from the table.

VN+1 =

600kg/h,

yAN+1 = 0,

yCN+1 = 1.0,

L0 =200kg/h,

xA0 = 0.30,

xB0 = 0.70, xC0 = 0,

xAN = 0.04.

Solution: For the mixture point M,

xCM

x AM

L0 xC 0 VN 1 yC N 1

L0 VN 1

substituting into eqs. below,

200(0) 600(1.0)

0.75

200 600

L0 x A0 VN 1 y AN 1 200(0.30) 600(0)

0.075

L0 VN 1

200 600

32

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 L 0 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 L N

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

Distillation

Extraction

Solute diffuses through the gas

phase to the interface between the

phases

-Low boiler diffuses through the

liquid phase to the interface

- away from interface to the vapor

Solute diffuses through the raffinate

phase to interface and then into the

extract phase

SOLID-LIQUID EXTRACTIO

Solid –Liquid Extraction

• Objectives

Understand concept of leaching

Carry out mass balance for leaching

Calculate 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.

Applications in

Industry

• 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

• Food Industry

• sugar industry when

soluble sucrose is

removed by water

extraction from

sugar cane or beet.

• Pharmaceutical

• Herbal and oil

extraction

Before extraction

solven

t

Inert

After extraction

Solve

nt +

solut

e

Inert

Solute (transition

component)

• 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)

Particle

size

Agitation

of fluid

Factors

Temperature

Solvent

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

(d)rate

Agitation

of the solvent

of extraction.

- 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

i. Solute= A

ii. Solvent = C

iii. An inert solid= B

Solute

completely

dissolves

Ratio of solid to liquid in

the underflow is a

constant

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.

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 solid

yN : fresh solvent entering system

y1 : 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.

Number of ideal stages by using

Right-angled Triangular Diagram

•

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 abscissa

The proportion of solvent S is plotted as the

ordinate

The 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

•

Mass fractions of the solute, the solvent, and

the inerts(underflow) are calculated from:

For the solute (A):

xA

yAK

1 K

For the solvent (S):

K (1 y A )

xS

1 K

xB

1

K 1

yA = mass of solute / lb of

solution in the overflow

xA= mass frac. of solute

xs-= mass frac of solvent

xB= mass fraction of inert

K= mass of solution

removed in the underflow

per unit mass of solids

Overflow:

y

component

solvent solute

component

yx ssolvent

yA 1

solute carrier

ys 1 y A

y= fraction in

overflow

Underflow:

x= fraction in

underflow

component

x

solvent solute carrier

xs x A xB 1

xs 1 ( xA xB )

S

General formula:

overflow

Mass fraction A

Underflow

A

Overflow:

y s solvent /( solvent solute )

y A solute /( solvent solute )

ys

yA

solvent / solute

Underflow:

xs solvent /( solvent solute carrier )

x A solute /( solvent solute carrier )

xs

xA

solvent / solute

Mass of A

Total mass

Construct graph by right

triagular Diagram Method

To determine the number of stages

for countercurrent leaching

Underflow line ( xs vs

xA)

1.Plot line for all underflow( xs

vs xA)

2. Material balance to

find outgoing overflow

,y1

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.

x1

5. Find x1 by drawing tie

line from origo to y1.

6. Find y2 by connecting F

’,x1 and overflow line.

7. Step off until x1< xn.

Example

Seeds 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

Solute

:

xA

y Ak

1 k

Solvent

s:

xS

k (1 y A )

1 k

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

n+1

yn+1

Vn+1

Ln

Xn

n

yn

Vn

Ln-1

Xn-1

2

1

ya

Va

La

Xa

In the underflow feed:

The seeds contain 25% oil and 75% inert,

xS 1 0

.

x A1 0.25

In the overflow feed:

Pure solvent is used,

y An 1 0

y sn 1 1.0

This point is marked as yn

on the graph

+ 1

In the overflow product:

The oil concentration is 50%

y S 1 0.50

and

y A1 0.50

y1

This point lies on the hypotenuse and is marked

on the graph

the underflow product:

% of the oil is recovered, leaving 25(1 – 0.90) = 2.5 kg assoc

.

th 75 kg inerts;

atio (oil/inerts) = (2.5/75) = 0.033 = kyA

k (0.7 0.5 y A 3 y A2 )

ky A (0.7 y A 0.5 y 3 y )

2

A

3

A

0.033 (0.7 y A 0.5 y A2 3 y 3A )

y A 0.045

Multiply by yA

ky A 0.033

k (0.045) 0.033

0.033

k

0.733

0.045

x An 1

x Sn 1

y Ak

(0.045)( 0.733)

0.019

k 1

0.733 1

(1 y A ) k (1 0.045)( 0.733)

0.404

k 1

0.733 1

This point is drawnx

n 1on the graph.

as

y n 1 . x n 1

The pole point

is obtained where

extended meet.

y1 .x1

and

Construct graph :

xA

1) Plot x s against

on the graph.

2) Marking point

x1 , y1

and

x n 1 y n1

3) Draw the stages

on the graph

THE END

EXTRACTION PROCES

What separation method do all the following

processes have in common?

Decaffeination Wastewater Treatment

Separation of Aromatics

from Hydrocarbons

Manufacture

of Penicillin

LIQUID-LIQUID EXTRACTION

Introduction to Liquid-Liquid Extraction

• Objectives

Understand concept of LLE

Equilibrium Relations in

extraction

Single stage equilibrium

extraction

Understand equipment for LLE

Calculate 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

solutionRICH

containing little solute.

Extract (solute _________)

Feed (containing solute)

Solvent

POOR

Raffinate

(solute________)

Advantages and Disadvantages of

LLE

Advantages:

• Effective at low pH and low

temperature

• Separation by distillation is not feasible

•

•

•

•

•

Disadvantages:

• Incomplete

separation, which

contaminates the

product.

• Multiple processes

are required

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

Baird, M. “Handbook of Solvent Extraction”

Solute concentration low

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

Solvent is the key to a successful separation

by liquid-liquid extraction.

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

SELECTIVITY

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

feed

extraction phase

1 .0

• Good selectivity =

feed phase

• 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

•

DISTRIBUTION

COEFFICIENT

ratio (at equilibrium) of the concentration of solute

in the extract and raffinate phases.

K y/x

Distribution coefficient , less solvent is required for a

given degree of extraction

RECOVERABILITY

No azeotrope formed between solvent and

solute

Mixtures should have a high relative volatility

Solvent should have a small latent heat of

vaporization.

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 solutio

otherwise the separation is not "clean".

DENSITY

• A large difference in density between extract and raffinat

phases permits high capacities in equipment.

OTHERS

• Non-flammable

• Non-toxic

• Low cost

Types of Extractors

Mixer-Settles for Extraction

To provide efficient mass transfer, a mechanical

mixer is often used to provide intimate contact of 2

liquid phases.

Separate mixer-settler

Combined mixer-settler

16

Types of Extractors

Plate and Agitated Tower Contactors for

Extraction

The

rising droplets of the

A serial of paddle

light solvent liquid are

dispersed. The dispersed

droplets combine below

each tray and then re

formed on each tray by

passing through the

perforations.

Perforated plate (sieve tray)

tower

agitatorss mounted on a

central rotating shaft

provides the agitation for

the 2 phases.

Agitated extraction tower

17

Types of Extractors

Packed and Spray Extraction Towers

Packed & spray tower extractors give differential

contacts, where mixing and settling proceed

continuously and simultaneously.

Packed extraction tower

Spray-type extraction tower

18

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

•

Equilateral triangular coordinates are often used to represent the

equilibrium data of a three component system (3 axes)

xA + xB + xC = 1

Equilibrium Relations in extraction

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

xA + xB + xC = 0.4

represents

xB.+ 0.2 + 0.4 = 1

Coordinates for a triangular

diagram

(A and B are partially miscible.)

xB = 1.0 - xA - xC

yB = 1.0 - yA - yC

Reading ternary phase diagrams

Consider the point M:

water content (xA) is ?

0.19

ethylene glycol content (xB) is

?

0.20

0.61

furfural content (xC) is ?

•

check: xA + xB + xC = 1

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.

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

region of partial miscibility A-C

M.

The compositions of E and R converge at the plait point, P (i.e., no

separation).

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

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% 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.

23

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

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).

Right-triangle phase diagrams

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

25

4.6 Single-Stage Equilibrium

Extraction

Total material balance:

Solute

balance:

Combine both

eqn:

FS M ER

Fx F Sy S Mx M Ey E Rx R

xm

Fx F Sy s

FS

4.2

4.3

Fx F Sy S ( F S ) x M

Rx R Ey E ( R E ) x M

4.1

F xM y s

S xF xM

4.4

E xM xR

R y E xM

4.5

26

EXAMPL

1000 kg of an aqueous solution containing 50%

acetone is contacted with E800 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.

•

•

Mass of feed, F =1000k g

mass fraction acetone (C ) in the feed, x

•

•

•

•

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

0.995

Total material F S M 1000 800 1800kg

balance:

xcm

From figure:

Fx F Sy s 1000(0.5) 800(0.005)

0.28

F S

1800

xc R 0.236

E R 1800kg

E 0.28 0.236

R 0.302 0.28

yc E 0.302

E 1200kg

R 600kg

CF

= 0.5

BS

=

Single stage extraction calculation ilustrated;

(a) Right triangular coordinate and

(b)x-y diagram

Continuous multistage countercurrent extraction

Countercurrent process and overall balance

An overall mass balance:

A balance on C:

L0 V N 1 L N V1 M

4.12

L0 xC 0 V N 1 y C N 1 L N xC N V1 y C1 Mx C M

Combining 4.12 and 4.13

x CM

Balance on component A gives

x AM

L0 xC 0 V N 1 y CN 1

L0 V N 1

L N xC N V1 y C1

4.13

4.14

LN V1

L0 x A0 V N 1 y AN 1 L N x AN V1 y A1

L0 V N 1

LN V1

30

4.15

Continuous multistage countercurrent extraction

Countercurrent 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 V 1 and

the aqueous raffinate LN. Use equilibrium data from the table.

VN+1 =

600kg/h,

yAN+1 = 0,

yCN+1 = 1.0,

L0 =200kg/h,

xA0 = 0.30,

xB0 = 0.70, xC0 = 0,

xAN = 0.04.

Solution: For the mixture point M,

xCM

x AM

L0 xC 0 VN 1 yC N 1

L0 VN 1

substituting into eqs. below,

200(0) 600(1.0)

0.75

200 600

L0 x A0 VN 1 y AN 1 200(0.30) 600(0)

0.075

L0 VN 1

200 600

32

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 L 0 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 L N

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

Distillation

Extraction

Solute diffuses through the gas

phase to the interface between the

phases

-Low boiler diffuses through the

liquid phase to the interface

- away from interface to the vapor

Solute diffuses through the raffinate

phase to interface and then into the

extract phase

SOLID-LIQUID EXTRACTIO

Solid –Liquid Extraction

• Objectives

Understand concept of leaching

Carry out mass balance for leaching

Calculate 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.

Applications in

Industry

• 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

• Food Industry

• sugar industry when

soluble sucrose is

removed by water

extraction from

sugar cane or beet.

• Pharmaceutical

• Herbal and oil

extraction

Before extraction

solven

t

Inert

After extraction

Solve

nt +

solut

e

Inert

Solute (transition

component)

• 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)

Particle

size

Agitation

of fluid

Factors

Temperature

Solvent

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

(d)rate

Agitation

of the solvent

of extraction.

- 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

i. Solute= A

ii. Solvent = C

iii. An inert solid= B

Solute

completely

dissolves

Ratio of solid to liquid in

the underflow is a

constant

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.

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 solid

yN : fresh solvent entering system

y1 : 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.

Number of ideal stages by using

Right-angled Triangular Diagram

•

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 abscissa

The proportion of solvent S is plotted as the

ordinate

The 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

•

Mass fractions of the solute, the solvent, and

the inerts(underflow) are calculated from:

For the solute (A):

xA

yAK

1 K

For the solvent (S):

K (1 y A )

xS

1 K

xB

1

K 1

yA = mass of solute / lb of

solution in the overflow

xA= mass frac. of solute

xs-= mass frac of solvent

xB= mass fraction of inert

K= mass of solution

removed in the underflow

per unit mass of solids

Overflow:

y

component

solvent solute

component

yx ssolvent

yA 1

solute carrier

ys 1 y A

y= fraction in

overflow

Underflow:

x= fraction in

underflow

component

x

solvent solute carrier

xs x A xB 1

xs 1 ( xA xB )

S

General formula:

overflow

Mass fraction A

Underflow

A

Overflow:

y s solvent /( solvent solute )

y A solute /( solvent solute )

ys

yA

solvent / solute

Underflow:

xs solvent /( solvent solute carrier )

x A solute /( solvent solute carrier )

xs

xA

solvent / solute

Mass of A

Total mass

Construct graph by right

triagular Diagram Method

To determine the number of stages

for countercurrent leaching

Underflow line ( xs vs

xA)

1.Plot line for all underflow( xs

vs xA)

2. Material balance to

find outgoing overflow

,y1

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.

x1

5. Find x1 by drawing tie

line from origo to y1.

6. Find y2 by connecting F

’,x1 and overflow line.

7. Step off until x1< xn.

Example

Seeds 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

Solute

:

xA

y Ak

1 k

Solvent

s:

xS

k (1 y A )

1 k

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

n+1

yn+1

Vn+1

Ln

Xn

n

yn

Vn

Ln-1

Xn-1

2

1

ya

Va

La

Xa

In the underflow feed:

The seeds contain 25% oil and 75% inert,

xS 1 0

.

x A1 0.25

In the overflow feed:

Pure solvent is used,

y An 1 0

y sn 1 1.0

This point is marked as yn

on the graph

+ 1

In the overflow product:

The oil concentration is 50%

y S 1 0.50

and

y A1 0.50

y1

This point lies on the hypotenuse and is marked

on the graph

the underflow product:

% of the oil is recovered, leaving 25(1 – 0.90) = 2.5 kg assoc

.

th 75 kg inerts;

atio (oil/inerts) = (2.5/75) = 0.033 = kyA

k (0.7 0.5 y A 3 y A2 )

ky A (0.7 y A 0.5 y 3 y )

2

A

3

A

0.033 (0.7 y A 0.5 y A2 3 y 3A )

y A 0.045

Multiply by yA

ky A 0.033

k (0.045) 0.033

0.033

k

0.733

0.045

x An 1

x Sn 1

y Ak

(0.045)( 0.733)

0.019

k 1

0.733 1

(1 y A ) k (1 0.045)( 0.733)

0.404

k 1

0.733 1

This point is drawnx

n 1on the graph.

as

y n 1 . x n 1

The pole point

is obtained where

extended meet.

y1 .x1

and

Construct graph :

xA

1) Plot x s against

on the graph.

2) Marking point

x1 , y1

and

x n 1 y n1

3) Draw the stages

on the graph

THE END