Extraction

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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:

FS M ER

Fx F  Sy S  Mx M  Ey E  Rx R

xm 

Fx F  Sy s
FS

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 n1

3) Draw the stages
on the graph

THE END

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