Tema 5 - Liquid-liquid Extraction (2014-15)

7/23/2019 Tema 5 - Liquid-liquid Extraction (2014-15)

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Liquid-liquid extraction (1)

Unit Operations of Chemical Engineering

LIQUID-LIQUID

EXTRACTION


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Introduction

Introduction

This subject deals with the separation of the components of a liquid mixture by treatment with a solvent in which one or more of

the desired components is preferentially soluble. After this, two immiscible phases of different density are obtained.

Basic Nomenclature:

Feed: solution to be treated.

Solvent: liquid which is added to the feed (it may contain some solute).

When the feed and solvent are mixed, two phases are obtained:

Extract (E): the richest phase in solvent.

Raffinate (R): the poorest phase in solvent.

Extracted product: is the remaining mixture after separating the solvent from the extract.

Raffinate product: is the remaining after separating the solvent from the raffinate.

It is fundamental an easy recovery of the solvent which has been added (by distillation, evaporation, etc.)

The addition of a solvent to a binary mixture of the solute in a solvent may lead to the formation of several types of mixture:

a. A homogeneous solution may be formed, then the solvent selected is unsuitable.

b. The solvent may be completely immiscible with the initial solvent

c. The solvent may be partially miscible wilt the original solvent, resulting in the formation of two or three pair of partially miscible

liquids.

d. The types b), and c) generate systems that may be useful.

Generally, it can be said that for carrying out a liquid-liquid extraction the following conditions are necessary:

1. Contact between the original solution with the solvent.

2. Separation of the obtained liquid phases (extract and raffinate).

3. Recovery of the solvent.

The first two previous sections constitute the stage or extraction unit, and it is called ideal or theoretic when the contact between

the solution and the solvent has been deep enough to consider that the separated phases have the concentrations correspondingwith equilibrium conditions.

Distillation is either uneconomic or

impracticable in the following

cases:1. When an excessive amount of

steam must be applied

2. When the relative volatility of the

two main components is near unity.

3. When azeotropes are formed.


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Equilibrium Conditions in Liquid-Liquid Extraction

The equilibrium condition for the distribution of one solute between two liquid phases may be considered in

terms of the distribution law. Thus, at equilibrium, the ratio of the concentrations of the solute in the two

phases is given by:

The Kivalue depends on the solvent, the mixture composition and the temperature.

In many cases the two solvents are partially miscible and the concentration of the solute influences the

solubility; in any case the experimental data are essential.

Sequence of studyIntroductory exercises

a.- Single Contact (In one stage)

a1- Solvent partially miscible in one of the feed components and completely miscible in the other

one.

a2- Immiscible solvents.

b.- Co-current Multiple Contact

b1- Co-current contact with partially miscible solventsb2- Immiscible solvents.

c.- Countercurrent Multiple Contact

c1- Solvent partially miscible in one of the feed components and completely miscible in the other

one.

c2-Immiscible solvents.

Concentration of the component inDistribution constant

Concentration of the component in

E

i

R

i E CK

i R C


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Three-component systems and their graphical representation

a) Equilateral triangular coordinates

Each of the three corners represents a pure component. On each side there are all the binary mixtures of the two components

placed on the corners of this side. The points placed inside represent the ternary mixtures, and the relative amount in relation

to the mixture of each component is proportional to the distance from the point to the opposite side (a, b, c). Because of the

triangle height is equal to the sum of the three distances, relative amounts can be found when dividing each distance by thesum of the three ones.

The segments a, b, c are proportional to the segments a, b, c (parallel to the triangle sides) and their sum is equal to the

triangle side value, and the following expressions can be assumed:

Properties:1.- The compositions of mixtures which are represented by points belonging

to a straight line which is parallel to a triangle side have the same

concentration corresponding to the component placed on the opposite

vertex.

2.- All the points belonging to the segment which links a vertex to a opposite

side point represent mixtures which have the same ratio between the

concentrations of the components placed on the other two vertexes.

3.- When a pure component or a mixture (represented by P2) are added to a

mixture (represented by P1) the point Pwhich represents the finalcomposition must be found of the segment P1P2which links the points P1and P2and determines the segments P1Pand P2P, which are inversely

proportional to the amounts of each point (lever rule), such that:

Some problems where the previous properties are applied will be stated

below.

Ternary Systems

' ' '% 100 % 100 % 100

triangle side triangle side triangle side

a b cA x B x C x

1 2

2 1

amount of

amount of

P P P

PP P


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LLE-1. In order to prepare 500 kg of a solution which must contain 35% of HNO3 by weight, 20% of H2SO4 by weight and 45%

of H2O by weight there are available sulphuric acid of 98% by weight, nitric acid of 68% by weight and water. Calculate the

necessary amounts of each component, using the triangular diagram.

Solution:

On the triangular diagram the points corresponding with H2SO4 of 98% (point S on the sideAB), with HNO3 of 68% (point Non

the side BC), with H2O (on the vertex B) and with the solution to be prepared (point P) are represented on the graph.The intersection of the straight lines BPand SNgives the point Mcorresponding with the mixture of S and N, in such a way

that on adding B it can be reached the point P(xM = 0,4865).

It is accomplished that:

kg,kg ; M =,B==M+B

B,M=

,,,

,,

MP

PB

B

M

73593140500

5642

564235048650

000350

kg,kg S =,N=,= MS+N

N,S=

,,,

,,

SM

MN

N

S

410232577359

39830

3983000048650

48650680

Ternary Systems (Problems)

C

HNO3

A

H2SO4

B

H2O


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LLE-2. 1000 kg of an acid mixture whose composition by weight is 30% of H2SO4, 20% of HNO3 and 50% of water must be

changed for another whose composition by weight is 30% of HNO3, 50% of H2SO4 and 20% of H2O by adding H2SO4 of 98%

and HNO3 of 90%. Calculate the necessary amounts of the two acids which must be added, using the triangular diagram.

Solution:

The values are placed on the triangular diagram and having into account the above statement.

and the following system must be solved:

The solution is: S = 1225 kg H2SO4 and N = 775 kg HNO3

2000300350

2003001000

,,

,-,=

MP

FPM=FMPMFPF

N,,

,,NSNMNSMS 581

350

350900

N,S

NS

581

2000

Ternary Systems (Problems)

C

HNO3

A

H2SO4

B

H2O


7/31Liquid-liquid extraction (7)

Ternary Systems

When the solvent is partially miscible with one component of the mixture and completely miscible with the

other component the ternary systems of the figure are obtained. With the aim of simplify the study of ternary

systems, the following nomenclature will be followed: on the vertex C is placed the solute to be extracted,

on the vertex A is placed the original solvent partially miscible with the solvent (added later), placed on the

vertex B. It may be considered the ternary system most frequently found.

When dealing with a mixture of A and B, represented by D, themixture will separate into two liquid phases, represented by E and

R, if the two solvents were totally immiscible E and R match B

and A.

If C (miscible in A and B) is added to the mixture D, Cwill

distribute between the two phases and conjugate ternary

solutions will be formed (E1,R1); (E2, R2),...the relative proportion

of the two phases is varying according to the relations D1R1/D1E1,

D2R2/D2E2, etc., until reaching the point Dwhere the two phaseswill disappear.

When starting from mixtures like F, G and H, and Cis added, the points F, Gi Hare obtained and they

determine the binodal solubility curve. All the

mixtures with overall compositions placed inside the

binodal solubility curve will split into two conjugate

phases with compositions placed on the binodal

solubility curve. The straight lines which link pairs of

points corresponding with conjugate phases arecalled tie lines.



Ternary Systems

The slope of the tie lines is function of the concentrations and the mixture components. According to the

below figures it may be stated the following:

Case a) The phase Eis richer in Cthan the phase R.

Case b) The phases Eand Rhave a similar concentration in C.

Case c) The phase Ris richer in Cthat the phase R.

If Cincreases the concentrations of the two phases come closer until reaching the point Pwhere the two

phases have the same concentration and they turn into one phase (critical point or plait point).

Also, there are more complex ternary systems as shown in the figures below:



Interpolation of tie linesAlders Method:

When the binodal solubility curve and the tie lines have been drawn from the experimental data, parallel

straight lines to the triangle sides are drawn as shown in the below figure. Points where the straight lines cut

make the curve ILMKHLM(conjugate line) which cuts the binodal curve at the plait point. To draw a tie line

passing through the point S, it must be drawn a parallel straight line to CB which will cut the conjugate lineat the point Rfrom where the parallel straight line toACis drawn, its intersection with the binodal curve

gives the point S, which is the conjugate phase of the point S.

Sherwood Method:

The conjugate line is found in a similar way that by applying the Alders

Method.

Ternary Systems



Calculations in Ternary Systems

All such processes require:

1.- Bringing the solvent and solution into intimate contact.

2.- Separation of the resulting two phases.

3.- Removal and recovery of the solvent from each phase.

Single-stage batch extraction

Solvent partially miscible with one of the feed components and completely miscible with the other

one.

This extraction system is drawn in the figure below:

The evolution of this kind of extraction has been drawn on a ternary diagram. From the figure it can be seen

that for having two phases it is necessary to be inside the binodal curve. If the overall composition M1 must

be obtained, the necessary amount of solvent can be calculated by means of a mass balance on the

component C:

1

1

1

1

1

that isF M

F

M

M

F x x FxB x

x F B



The point M1 may be graphically calculated taking into account that M1 must be on the straight line FB and it

must be accomplished:

E1 and R1 may be calculated applying a mass balance if their compositionsx1 and y1 are known and whichcan be read where the tie line which passes through M1 intersects the binodal curve.

The values are:

The above equations can be used to calculate Eand Rgives:

If the solvent is not a pure component (S) and the mixture is not

a binary composition (it may contain an amount of solvent used

in the extraction, F), the evolution of this case may be on the

ternary diagram. It can be inferred:

Also, the remainder quantities can be calculated in the same way.

1

1

amount of

amount of

B M F

F M B

11

111111

11

1111

)(

xy

xyMEBFR

xy

xxME

MM

'')(' EFRXYXxFE F

sM

MFsFM

xx

xxFS

SF

xSFxx

1

11

1

11

)(

Calculations in Ternary Systems

1



LLE-3.A mixture of acetic acid (30%) and chloroform weighting 100 kg is treated with water at 18C with the

aim of extracting the acetic acid. Calculate:

a) The amounts of water, minimum and maximum, to be added.

b) The maximum concentration reached by the acetic acid in the extracted product.

c) The amount of water to be added to obtain the maximum concentration of the extracted product.

Equilibrium data of the liquid phases at 18C corresponding to the tie line ends are shown on the belowtable.

Solution:

a) The minimum amount of solvent will be the needed

one to saturate the mixture. In other words, to reach

the two phases zone (punt D), where its composition

read on the ternary diagram isxD = 0.283. Applying a

mass balance on the component C (acetic acid):

Taking this minimum amount of B, the amount of

extract would be infinitesimal and the extract one

would be maximum.

The maximum amount of solvent to be used would be that one which gives an infinitesimal amount ofraffinate (maximum amount of extract), i.e. the necessary amount for reaching the point G on the binodal

curve, where the composition value isxG = 0.005.

Applying a mass balance on the component C(acetic acid) gives:

D mn DF

Fx = Dx = F + B x

100 0.30 100 0.283 6mn mn x = + B B = kg

GmxGF xF + B== GxFx

100 0.30 100 0.005 5900

mx mx

x = + B B = kg

Heavy phase (% by weight) Light phase (% by weight)

CHCl3 H2O CH3COOH CHCl3 H2O CH3COOH

99.01 0.99 0.00 0.84 99.16 0.00

91.85 1.38 6.77 1.21 73.69 25.10

80.00 2.28 17.72 7.30 48.58 44.12

70.13 4.12 25.75 15.11 34.71 50.18

67.15 5.20 27.65 18.33 31.11 50.56

59.99 7.93 32.08 25.20 25.39 49.41

55.81 9.53 34.61 28.85 23.28 47.87

Single stage batch extraction (Problems)

Si l t b t h t ti (P bl )



b) The straight line BEmust be tangent to the binodal curve and the point of tangency will give the

maximum composition of the extracted product. On the ternary diagram may be read:

Y= 0.955; i.e. 95.5% by weight of acetic acid y = 0.28

c) A tie line is drawn passing through the tangency point Tand its intersection with the straight line FB gives

the point Mwhere the composition read on the ternary diagram isxM= 0.185. The necessary amount of

water will be: 100 0.30 0.185

62.20.185

-B kg

Single stage batch extraction (Problems)

Si l t b t h t ti ith i i ibl l t



Single stage batch extraction with immiscible solvents

Immiscible solvents

If the componentsA i B are completely immiscible each other the calculations are easier. The

concentrations of each phase will be referred to the component whose amount does not change:

Applying a mass balance on the component Cgives:

Rearranging gives:

The above straight line equation (operating line) allowsto calculate the equilibrium concentrations on the point where

intersects the equilibrium curve.

If the equilibrium line is a straight line

Substituting the straight line equation in the operating line

gives:

In case of dealing with a pure solvent (above).

Dealing with a recycled solvent with concentration ys gives:

weight of mol of weight of mol of'( , ) '( , )

weight of mol of weight of mol of

C C C C x y

A A B B

)()( ''1''

1'1

'1

''sFsF yyBxxAByAxByAx

B

A

xx

yy

F

s

''

1

''

1

'

1

'

1 mxy

' ' '

( / )F F

Ax x x

A mB m B A

1

1

1

m

y

A

Bm

yxx SsF

''''

1

1

1

Single stage batch e traction ith immiscible sol ents (Problems)



LLE-4.Acetone must be separated from an aqueous solution whose concentration in acetone is 20% by

weight. The extraction must be carried out using pure benzene at 15C. Calculate:

a) The extract and raffinate compositions if the amount of benzene used is 2 kg per kg of mixture to be

extracted.

b) The amount of benzene needed when extracting 100 kg of mixture if the acetone concentration in

raffinate should not exceed 3% by weight.The equilibrium data for this system at 15C and corresponding to the ends of the tie lines are shown on the

below table.

Solution:

a) Calculating the values ofx(kg of acetone/kg of

water) and y(kg acetone/kg benzene) from the

equilibrium data gives:

Calculating the remainder values in the same way

gives:

The composition of the mixture to be extracted is:

The solvent is pure benzene:

The A/B ratio will be: A/B = 0.80/2 = 0.40

Aqueous phase Benzene phase

Benzene Acetone Water Benzene Acetone Water

0.1 5.0 94.9 95.2 4.7 0.1

0.1 10 89.9 89.0 10.8 0.2

0.3 20 79.7 73.4 26.1 0.5

0.7 30 69.3 55.2 43.0 1.8

1.4 40 58.6 39.1 56.5 4.4

3.2 50 46.8 27.6 63.9 8.5

5 4.70.0526 0.0494

94.9 95.2x' = y' =

10 10.80.1112 0.1213

89.9 89.0x' = y' =

x y

0.0526 0.0494

0.1112 0.1213

0.2509 0.3556

0.4329 0.7789

0.6826 1.4450

1.0684 2.3152

200.25

80

'

Fx

0'sy

Single stage batch extraction with immiscible solvents (Problems)




Usingx-y coordinates the following lines are drawn: the equilibrium curve and the straight line of slope

-0.40 passing through the point (0.25; 0). Their intersection gives the extract and raffinate concentrations:

b) The extract concentration in equilibrium with 3/97 = 0.031 is

calculated from the equilibrium curve or by interpolating in the

data table:

Applying the equation below:

Substituting the values gives:

The total amount of benzene needed to extract 100 kg of mixture, where there are 80 kg of water, will be:

B = 6.63x80 = 530.4 kg of benzene

1

1

0.070

0.070

'

'

x

y

1 0.033

'

y

B

A

xx

yy'F

'

's

'

1

1

1

1

0.25 0.0316.63kg of benzene kg of water

0.033

' 'F

'

B x - x /

A y


Co current multiple contact with partially miscible solvents


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Co-current Multiple ContactIt is the repetition of the single stage extraction when the raffinate leaving the stage is brought into contact

by adding new solvent in the following stage and so on. Finally, the extracts obtained are joined together.

Co-current contact with partially miscible solventsRepresentation on the ternary diagram is similar to that

obtained by a single stage. Applying a mass balance on the n-stage gives:

The raffinate is the corresponding to that leaving the last stage.The raffinate product is calculated by using the intersection

of the straight line which passes through B and

R3 (corresponding to the last raffinate) and its intersection

with the sideAC. The amount and composition of the extract E

is calculated by applying mass balances and the extracted product

is found on the intersection of the straight line BE and the side AC.

Co-current multiple contact with partially miscible solvents

nm

nnMnn

nM

nMnnn

nn

nn

n

nnnM xy

xxME

x

xxRB

BR

xR

M

xRx

)()( 11

1

1111

Co-current contact with partially miscible solvents (Problems)


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LLE-5. A mixture ofA and Cweighting 100 kg and containing 30% of Cis extracted in co-current contact

adding the solvent B . The extraction is carried out in three stages adding 50 kg of B to each one.

Calculate:

a) The amount and composition of the extract and raffinate in each stage.

b) The amount and composition of the extracted and raffinate product.

c) The maximum composition of Cwhich can be obtained when using this extraction method.The compositions of the conjugate phases expressed in percent by weight are shown on the table below.

A B C A B C

X N Y N

95.0 5.0 0.0 10.0 90.0 0.0 0.00 0.053 0.00 9.000

92.5 5.0 2.5 10.1 82.0 7.9 0.026 0.053 0.44 4.555

89.9 5.1 5.0 10.8 74.2 15.0 0.053 0.054 0.58 2.876

87.3 5.2 7.5 11.5 67.5 21.0 0.079 0.055 0.646 1.985

84.6 5.4 10.0 12.7 61.1 26.2 0.106 0.057 0.674 1.570

81.9 5.6 12.5 14.2 55.8 30.0 0.132 0.059 0.679 1.263

79.1 5.9 15.0 15.9 50.3 33.8 0.159 0.063 0.680 1.012

78.3 6.2 17.5 17.8 45.7 36.5 0.186 0.066 0.672 0.850

73.4 6.6 20.0 19.6 41.4 39.0 0.221 0.071 0.665 0.690

67.5 7.5 25.0 24.6 32.9 42.5 0.270 0.081 0.643 0.513

61.1 8.9 30.0 28.0 27.5 44.5 0.329 0.098 0.614 0.379

54.4 10.6 35.0 33.3 21.7 45.0 0.391 0.119 0.575 0.277

46.6 13.4 40.0 40.5 16.5 43.0 0.461 0.155 0.515 0.198

43.4 15.0 41.6 43.4 15.0 41.6 0.492 0.176 0.492 0.176

Raffinate Extract




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Solution:

The data of the previous table are drawn on a ternary diagram as below.

1st stage.- The straight line BF is drawn on the ternary diagram and the point M1 is found on this straight

line, considering that:

M = F + B = + = kg

1 1 100 50 150

1 11 1

0 150 1000.3 0 150F M M

Fx + B = x + B = x 1

1000.300.20

150M

x =

Co current contact with partially miscible solvents (Problems)

Look out!:There is a

mistake when drawing

the straight line



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The compositions of extract and raffinate are read on the points where the tie line that passes through the

point M1 intersects the binodal curve, their values are:

y1 = 0.28 x1 = 0.11

The amounts of extract and raffinate will be:

2nd Stage.- The straight line BR1 is drawn and on the point M2on this straight line:

The extract and raffinate compositions leaving this stage are read on the intersections of the tie line which

passes through the point M2and the binodal curve; their values are:

y2= 0.100 x2= 0.031

The amounts of raffinate and extract will be:

1 1 1

11 1

(0.20 0.11) (0.28 0.20) 79.41kg

70.59 kg150

R E E

RE R

2 1 2 70.59 50 120.59 kgM = R + B = + =

22 1 2

0.11 0M

M x = R + B

2

70.59 0 110.064

120.59M

x ,x =


2 2 2

22 2

(0.064 0,031) (0.10 0.064) 57.63 kg

62.96 kg120.59

R E E

RE R



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3th Stage.- Proceeding in the same way that in the previous stages:

y3 = 0.023 x3 = 0.008

b) The composition of the solvent read on the diagram (point R3) is 0.050; so, the amount of raffinate product

will be:

R= 45.19 (1-0.050) = 42.93 kg

The composition of raffinate product is read on the intersection of the straight line BR3 and the sideAC, the

value is:X= 0,009.

The total amount of extract will be: 79.41 + 57.63 + 67.77 = 204.81 kgThe amount of C in the extract will be:

79.41x0.28 + 57.63x0.100 + 67.77x0.023 = 29.55 kg

The amount ofA in the extracted product will be:

Amount of A in the feed amount of A in the raffinate product:

100(1 - 0,30) 42.93(1 - 0,009) = 27,45 kg

The total amount of extracted product will be:

E= 29,55 + 27.45 = 57.0 kgThe extract composition will be:

xE= 29.55/204.81 = 0.144

The composition of extracted product will be:Y= 29,55/57.0 = 0,518

c) The maximum concentration of Cwhich can be obtained in the extracted product is that is read on the point

of the sideACintercepted by the straight line BEtangent to the binodal curve, in this case the value will be

68% and it is reached when applying the first stage.

3 2 3 62.96 50 112.96M = R + B = + = kg

3

62.96 0.0310.017

112.96M

xx =


3 3 3

33 3

(0.017 0.008) (0.023 0.017) 67.77 kg

45.19 kg112.96

R E E

RE R

Co-current contact with immiscible solvents


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Co-current contact with immiscible solvents

In this case the solvent B is miscible with the component to be extracted Cand completely immiscible with

A. The equilibrium concentrations of the point (x1, y1) are found on the intersection of the straight line

calculated before:

and the equilibrium curve. If the raffinate R1 is extracted using new solvent B2, the new equilibrium

concentrations are found on the intersection of the straight line with slopeA/B2and the equilibrium curve

and so on for each stage until reaching the required raffinate concentration.

When the equilibrium line is a straight line y = mx, the solute which has not been extracted in the raffinate

leaving the last stage is:

where n is the number of stages.

B

A

xx

yy

F

s

''

1

''

1

A

Bm

x

x

nx

A

Bmx F

n

Fnn

1log

log

1

1 '

'

''

Co-current contact with immiscible solvents (Problems)


23/31


LLE-6.An aqueous solution weighting 1000 kg contains 0,3% of benzoic acid which must be extracted by

means of an co-current extraction process consisting in two stages at 20C using 1800 kg of benzene as

solvent in each stage. Calculate the percent of benzoic acid extracted. The equilibrium data at 20C are

shown on the below table.

Solution:

Considering that benzene is immiscible in water the diagramx(g of benzoic acid/kg of water), y(g of

benzoic acid/kg of benzene). Using the data of the table the equilibrium curve can be drawn. On the x-axis

the point corresponding to the feed is drawn:

The slope of the operating line passing through this point is:

-A/B1 = -997/1800 = -0.555

The intersection of this straight line and the equilibrium curve gives the extract and raffinate concentrations

leaving the first stage:

The feed of the second stage is the raffinate of the first stage which is extracted using 1800 kg

of benzene. Consequently, for this stage the operating line is drawn passing through the point 0.00096 on

the x-axis and whose slope is:

-A/B2 = -0.555

x (g of benzoic acid/kg of water) 0.915 1.135 1.525 2.04 2.56 3.99 5.23y(g of benzoic acid/kg of benzene) 1.025 1.60 2.91 5.33 7.94 20.1 36.0

33

1 0.003

'

F

g cidx =

- kg aigua

1 10.00096 benzoic acid/kg water 0.00113 benzoic acid/kgbenzene' 'x kg y kg

( )

'

1 0.00096x

Co-current contact with immiscible solvents (Problems)


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The intersections of this straight line with the equilibrium curve are:

Calculating the benzoic acid extracted gives:

Amount of benzoic acid extracted = 2991 g 418.74 g = 2572.26 g

And the percent of benzoic acid extracted will be: 2572.26/3000 = 0.857 (85.7%)

2 20.00042 0.00030

' 'x y

g of benzoic acidBefore starting the extration: 3 997 kg of water 2991g of benzoic acid

kg of water x =

g of benzenic acidAfter ending the extraction 0.42 997 kg of water 418.74 g of benzoic acid

kg of water: x =

Countercurrent multiple contact with partially miscible solvents


25/31


Countercurrent Multiple ContactFeed and solvent enter through the opposite ends of the extraction system. In the first stage the feed

comes into contact with the solvent which is rich in solute. In the following stages the raffinates are

extracted with solvents whose concentration in solute is decreasing. Thus, the concentration in solute of the

raffinate is decreasing from the first stage to the last one while the concentration in solute of the extract is

increasing from the last stage to the first one. This layout uses less solvent than the co-current extraction

method.

Countercurrent contact with partially miscible solvents

Applying a mass balance on the overall system:

The point Mmust be found on the ternary diagram.

It must be on the straight line FB and it must be accomplished:

Rearranging the balance equation gives

The above equation points out that the difference between the inlet and outlet flows which enter through theopposite ends is constant.

MREBF n 1

BF

FxxxBFFx

FMMF

PBREF n 1

Countercurrent multiple contact with partially miscible solvents


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Applying a mass balance on any stage it can be seen that the difference between the inlet and outlet flows

of this stage is the same that the corresponding to the ends of the extraction system, i.e. the difference

keeps constant in the whole system:

The equation points out that the straight lines which link Fwith E1

and Rn

with B must

have a common point P (Pole or operating common point). The straight lines PEintersect points on the

binodal curve which are the extract and the raffinate of two adjacent stages. Once the point Pis set the

number of stages are calculated graphically.

The points F, E1, Rn, Mand Pare set and the tie line passing through E1 is drawn giving R1. The straight

line PR1 is drawn and E2 is obtained. Taking the tie line which passes through E2gives R2and so on, until

reaching the value of Rcorresponding to Rn.

The composition of the extracted product will be the intersection of BE1

andAC, the composition of the raffinate product will be the intersectionof BRn andAC.

To calculate E1 i Rn:

PBRERREBR nn 2121

PBREFn

1

11

1

1

111

EMEBFR

xy

xxME

MxxEMyE

n

n

nM

Mn

Countercurrent multiple contact with partially miscible solvents (Problems)


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LLE-7. In order to reduce the content of diphenylhexane from 50% by weight in the mixture to less than

10% by weight in the raffinate product, a mixture of diphenylhexane-docosane is extracted in countercurrent

with furfurol at 45C. Calculate the number of theoretical stages required to achieve the concentration of

diphenylhexane indicated if 800kg furfurol/500kg feed are used.

The table below shows the concentrations which define the points of the binodal curve for the mixture at

45C.

The below table shows the equilibrium compositions which determine tie lines.

Solution:

Using the above tables the binodal curve and tie lines are drawn. The point Mis set on the straight line FB

considering:

Furfurol Diphenylhexane Docosane Furfurol Diphenylhexane Docosane

0.040 0.000 0.960 0.500 0.423 0.077

0.050 0.110 0.840 0.600 0.356 0.044

0.070 0.260 0.670 0.700 0.274 0.026

0.100 0.375 0.525 0.800 0.185 0.015

0.200 0.474 0.326 0.900 0.090 0.010

0.300 0.487 0.213 0.993 0.000 0.007

0.400 0.468 0.132

Furfurol Difenilhex Docos Furfurol Difenilhex Docos

0.048 0.100 0.852 0.891 0.098 0.011

0.065 0.245 0.690 0.736 0.242 0.022

0.133 0.426 0.439 0.523 0.409 0.068

500 0 50

0.192800 500M

x ,

x = +

Countercurrent multiple contact with partially miscible solvents (Problems)


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The composition of the raffinate leaving the last stage Rn, is read on the intersection of the straight line RB

and the binodal curve in the zone corresponding to the raffinate, its value isxn = 0.093.

The composition of the extract leaving the first stage y1, is found on the intersection of the straight line RnM

and the binodal curve in the zone corresponding to the extract, its value is y1 = 0.216.

The pole Pis set on the intersection of the straight lines FE1 and RB. The composition of the raffinate

leaving the first stage is set on the intersection of the binodal curve and the tie line which passes through

E1; its value isx1 = 0.215.

If the straight line R1Pis drawn the point E2 is located where y2= 0,050. Drawing the tie line which passes

through the point E2 the value of R2 is found:x2= 0,052. This value is lower than that of the raffinate to be

obtained. Consequently, two theoretical stages are required.

Countercurrent multiple contact with immiscible solvents


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Countercurrent contact with immiscible solventsIn this case the solvent B is immiscible with the solventA.

Applying a mass balance on the overall extraction process:

On thex-yaxis the operating line is a straight line whose slope isA/B and passing through the points

. . The number of theoretical stages is the number of steps traced between the

equilibrium curve and the operating line.

weight of C weight of C'weight of A weight of Bx' y

''

'0

'1'

1''

0'

nFnF

xx

yy

B

AByAxByAx

' ' ' '1 0, and ,F nx y x y

Countercurrent multiple contact with immiscible solvents (Problems)


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LLE-08. 100 kg of a 20% solution of acetone (by weight) in water are extracted at 25C in countercurrent

with monochlorobenzene which contains 0.4% of acetone with the aim of decreasing the content of

acetone. Considering that the monochlorobenzene and water are immiscible in the extraction conditions,

calculate:

a) The minimum amount of solvent required if the acetone concentration in the raffinate should not exceed

2%.

b) The number of theoretical stages required if the amount of solvent to be used is 25% higher than theminimum.

Equilibrium data for this ternary system at 25C are shown in the below table.

Solution:a) F= 100 kg xF= 0.20 water = 100( 1 0.20 ) = 80 kg

The operating line must pass through the point:

If the minimum amount of solvent is used, the number of stages must be infinite, this condition is

accomplished if the straight line intersects the equilibrium curve on the point M when ; at this point

the value of yis y= 0.224.

x(kg acetone/kg water) 0.0258 0.0739 0.1605 0.267

y(kgacetone/kg monochlorobenzene) 0.0258 0.0754 0.156 0.236

0.20

0.25kg acetone/kg water1 0.20

'

Fx

-

0.02

0 020 kg acetone/kg water1 0.02

'

nx = ,

-

0

0.0040.004 kg acetone kg monochlorobenzene

1 0.004

'y = /

-

00.0204 0.004

' '

nx y

'0.25

Fx

Countercurrent multiple contact with immiscible solvents (Problems)


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Applying the equation corresponding to this extraction process gives:

b) If the amount of solvent used is 25% higher than the minimum: B = 83.5x1.25 = 104.4 kg

The equilibrium stages are drawn on the graph and

approximately 5.5 theoretical stages are obtained.

0.224 0.0040.958

0.25 0.0204

8083.5

0.958 0.958

mn

mn

A

B

AB = kg

800.776

104.4

A

B

1 0 .766 0.25 0.0204 0.004 0.187

'

y )

Documents

Tema 5 - Liquid-liquid Extraction (2014-15)