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SEPARATION PROCESSES

LIQUID-LIQUID EXTRACTION PROCESSES

Equilibrium Relations in LL Extraction

Transport Processes and Separation Process Principles CHRISTIE J.

GEANKOPLIS

Unit operations in chemical engineering by Mccabe, Smith and Harriot

PRINCIPLES AND MODERN APPLICATIONS OF MASS TRANSFER

OPERATIONS by Jaime Benitez

Separation process principles by Seader and Henley

Separation processes by King

Chemical Engineering by Coulson and Richardson

LIQUID-LIQUID EXTRACTION PROCESSES

removing one constituent from a liquid by means of an other liquid

solvent

When separation by distillation is ineffective or very difficult (close-boiling

mixtures or substances / or heat sensitive mixtures), liquid extraction is one

of the main alternatives to consider

recovery of penicillin from the fermentation broth by extraction with a

solvent such as butyl acetate

recovery acetic acid from dilute aqueous solutions

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Equilibrium Relations in LL Extraction

Equilateral triangular coordinates are often used to represent the equilibrium

data of a three-component system, since there are three axes

A, B, or C : a pure component

point M a mixture of A, B, and C

• the perpendicular distance from the

point M to

• the base AB : the mass fraction xc of

C in the mixture at M,

• to base CB the mass fraction xA of A,

• to base AC the mass fraction xB of B

xA + xB + xc = 0.40 + 0.20 + 0.40 = 1.0

Liquid-liquid phase diagram where components A and B are partially

miscible

• C dissolves completely in A or in B.

• A is only slightly soluble in B and B slightly soluble in A.

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

• An original mixture of composition M will separate into two phases a and b

which are on the equilibrium tie line through point M.

• Other tie lines are also shown. The two phases are identical at point P, the Plait

point.

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Phase diagram where the solvent pairs b—c and a-c are partially miscible.

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A: acetic acid, B: water, C: isopropyl ether solvent

The solvent pair B and C are partially miscible.

The concentration of the component C is plotted on the vertical axis and that

of A on the horizontal axis.

The concentration of component B is obtained by difference from

inside the envelope: two phase

Outside: one-phase region

A tie line g-i connecting the raffinate and extract layers

raffinate layer : water-rich layer

extract layer : ether-rich solvent layer g,

The raffinate composition is designated by x and the extract by y

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the mass fraction of C in the extract layer : yc

the mass fraction of C in the raffinate layer and as xc

To construct the tie line gi using the equilibrium yA — xA plot below the phase

diagram, vertical lines to g and i are drawn.

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Equilibrium: System Water-Chloroform-Acetone

Equilibrium extraction data for the system

water-chloroform-acetone at 298 K and 1 atm are given in Table. Plot these

data on a right-triangular diagram, including in the diagram a conjugate or

auxiliary line that will allow graphical construction of tie lines.

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Chloroform as solvent (B), water as diluent (A), and

acetone as solute (C).

This system exhibits a type I equilibrium behavior. The

graphical construction to generate the conjugate line

from the tie-line data given (e.g., line RE) is shown, as

well as the use of the conjugate line to generate other

tie lines not included in the original data. Line LRP is

the saturated raffinate line, while line PEK is the

saturated extract line.

Single-Stage Equilibrium Extraction

Derivation of lever-arm rule for graphical addition.

two streams, L kg and V kg, containing components A, B, and C, are mixed

(added) to give a resulting mixture stream M kg total mass.

overall mass balance and a balance on A,

Graphical addition and lever-arm rule : (a) process flow, (b) graphical addition

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An original mixture weighing 100 kg and containing isopropyl ether (C), acetic

acid (A), and water (B) is equilibrated and the equilibrium phases separated The

compositions of the two equilibrium layers are for the extract layer (V) yA = 0.04,

yB = 0.02, and yc = 0.94, and for the raffinate layer (L) xA =0.12, xB = 0.86, and

xc = 0.02. The original mixture contained 100 kg and xAM = 0.10. Determine the

amounts of V and L.

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M = 100 kg and x

AM

= 0.10

L = 75.0 and V = 25.0

L = 72.5 kg and V = 27.5 kg

Single-stage equilibrium extraction

separation of A from a mixture of A and B by a solvent C in a single equilibrium

stage.

the solvent, as stream V2 and the stream L0 enter

The streams are mixed and equilibrated and

the exit streams L1 and V1 leave in equilibrium with each other.

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Single-Stage Extraction

Pure chloroform is used to extract acetone from a feed containing 60 wt%

acetone and 40 wt% water in a co-current extractor. The feed rate is 50 kglh,

and the solvent rate is also 50 kg/h. Operation is at 298 K and 1 atm. Find the

extract and raffinate flow rates and compositions when one equilibrium stage

is used for the separation. (assume water and chloroform do not mix)

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V= C, Solvent, chloroform, ya1=0, yC1=1.0

L= Liquid mixture, A: acetone, B=water

xA1=0.60, xB1=0.40

A: acetone

C: chloroform

B: water

V1

L1

V2

L2

V

1

+L

1

=V

2

+L

2

=M

V

1

y

A1

+L

1

x

A1

=V

2

y

A2

+L

2

x

A2

=Mx

AM

V

1

y

B1

+L

1

x

B1

=V

2

y

B2

+L

2

x

B2

=Mx

BM

V

1

y

C1

+L

1

x

C1

=V

2

y

C2

+L

2

x

C2

=Mx

CM

L1

V1

yA2

xA2

V

1

+L

1

=V

2

+L

2

=M

V

1

y

A1

+L

1

x

A1

=V

2

y

A2

+L

2

x

A2

=Mx

AM

V= C, Solvent, chloroform, ya1=0, yC1=1.0

L= Liquid mixture, A: acetone, B=water

xA1=0.60, xB1=0.40

L1

V1

yA2

xA2

Tie line is constructed through by trial and error, using conjugate line extract

(V2, yA2) and rafinate (L2, xA2) locations are obtained. The concentrations

yA2=0.334, xA2=0.189

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L1

V1

yA2

xA2

Continuous Multistage Countercurrent Extraction

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Material Balance for Countercurrent Stage Process Pure solvent isopropyl

ether at the rate of VN+1= 600 kg/h is being used to extract an aqueous

solution of L0 = 200 kg/h containing 30 wt % acetic acid (A) by countercurrent

multistage extraction. The desired exit acetic acid concentration in the

aqueous phase is 4%. Calculate the compositions and amounts of the ether

extract V1 and the aqueous raffinate LN.

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Pure solvent isopropyl ether (C)

V

N+1

= 600 kg/h y

AN+1

=0 y

CN+1

=1.0

B: aqueous solution of L

0

= 200 kg/h containing 30 wt % acetic acid (A) exit

acetic acid concentration in the aqueous phase is 4%.

x

A0

= 0.30, x

B0

=0.70, x

C0

=0, x

AN

=0.04

L

0

x

A0

+V

N+1

y

AN+1

=L

N

x

AN

+V

1

y

A1

=Mx

AM

V

N+1

= 600 kg/h y

AN+1

=0 y

CN+1

=1.0

L

0

= 200 kg/h x

A0

= 0.30, x

B0

=0.70, x

C0

=0, x

AN

=0.04

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V

N+1

= 600 kg/h y

AN+1

=0 y

CN+1

=1.0

L

0

= 200 kg/h x

A0

= 0.30

x

B0

=0.70, x

C0

=0, x

AN

=0.04

x

CM

=0.75, x

AM

= 0.075

L

0

x

A0

+V

N+1

y

AN+1

=L

N

x

AN

+V

1

y

A1

=Mx

AM

V

N+1

and L

0

are plotted

L

N

is on phase boundary

It can be plotted at x

AN

=0.04

For the mixture point M

x

CM

=0.075, x

AM

=0.75

Draw a line from the point of

L

N

and M at equilibrium of

extract phase gives V

1

,

y

A1

=0.08 and

y

C1

=0.90

, y

CN+1

, y

AN+1

y

C1

, y

A1

x

CN

, x

AN

x

C0

, x

A0

V

N+1

= 600 kg/h y

AN+1

=0 y

CN+1

=1.0

L

0

= 200 kg/h x

A0

= 0.30

x

B0

=0.70, x

C0

=0, x

AN

=0.04

x

CM

=0.75, x

AM

= 0.075

y

A1

=0.08 and

y

C1

=0.90

L

N

= 136 kg/h

V

1

= 664 kg/h

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Stage-to-stage calculations for countercurrent extraction

stage by stage to determine the concentrations at each stage and the total

number of stages N needed to reach L

N

total balance on stage 1

total balance on stage n

Assumption: D (kg/h) is constant for all stages

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balance on component A, B, or C.

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Number of Stages in Countercurrent Extraction

Pure isopropyl ether (C) of 450 kg/h is being used to extract an aqueous solution of

150 kg/h with 30 wt % acetic acid (A) by countercurrent multistage extraction. The

exit acid concentration in the aqueous phase (B) is 10 wt %. Calculate the number

of stages required.

V

N+1

=450, y

AN+1

=0, y

CN+1

=1.0

L

0

=150, x

A0

=0.30, x

B0

=0.70, x

C0

=0, x

AN

=0.10

(150 0) (450 1.0)

0.75

150 450

x x +

= =

+

(150 0.30) (450 0)

0.075

150 450

× + ×

= =

+

V

N+1

=450, y

AN+1

=0, y

CN+1

=1.0

L

0

=150, x

A0

=0.30, x

B0

=0.70, x

C0

=0, x

AN

=0.10

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Countercurrent-Stage Extraction with Immiscible Liquids

If the solvent stream V

N+l

contains components A(solute) and C(solvent) and

the feed stream L

0

contains A (solute) and B(feed solvent) and components

B and C are relatively immiscible in each other, the stage calculations are

made more easily.

The solute A is relatively dilute and is being transferred from L

0

to V

N+1

where L’ = kg inert B/h, V’ = kg inert C/h, y = mass fraction A in V stream, and x = mass fraction A in L

stream

Slope of the operating line is L’ / V’ (if y and x are dilute)

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EXAMPLE Extraction of Nicotine with Immiscible Liquids. An inlet water solution of

100 kg/h containing 0.010 wt fraction nicotine (A) in water is stripped with a

kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a

countercurrent stage tower. The water and kerosene are essentially immiscible in

each other. It is desired to reduce the concentration of the exit water to 0.0010 wt

fraction nicotine. Determine the theoretical number of stages needed. The

equilibrium data are as follows (C5), with x the weight fraction of nicotine in the

water solution and y in the kerosene.

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L

0

=100 kg/h, x

0

=0.010

V

N+1

= 200 kg/h, y

N+1

= 0.0005, x

N

=0,0010

L’=L(1-x)=L

0

(1-x

0

)=100(1-0.010)=99.0 kg water/h

V’=V(1-y)=V

N+1

(1-y

N+1

)=200(1-0.0005)=199.9 kg kerosen/h

1

1

0.010 0.0005 0.0010

99.0 199.9 99.0 199.9

1 0.010 1 0.0005 1 0.0010 1

y

y

| |

| | | | | |

+ = +

| | | |

÷ ÷ ÷ ÷

\ . \ . \ .

\ .

y

1

=0.00497

N=3.8 theoretical plate

L

0

=100 kg/h, x

0

=0.010

V

N+1

= 200 kg/h, y

N+1

= 0.0005, x

N

=0,0010

y

1

=0.00497

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Minimum Solvent and Countercurrent Extraction of Acetone. An aqueous

feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is

being extracted in a countercurrent multistage extraction system using pure

methyl isobutyl ketone solvent at 298-299 K. The outlet water raffinate will

contain 2.5 wt % acetone. Use equilibrium data from Appendix A.3.

a) Calculate the minimum solvent that can be used. [Hint: In this case the tie

line through the feed L

0

represents the condition for minimum solvent flow rate.

This gives V

1 min

. Then draw lines L

N

V

1 min

and L

0

V

N+1

to give the mixture point

M

min

and the coordinate x

AMmin

, find V

N+ 1 min

the minimum value of the solvent

flow rate V

N+ 1

]

b) Using a solvent flow rate of 1.5 times the minimum, calculate the number of

theoretical stages.

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L

0

=1000 kg/h

x

A0

=0.235

x

C0

=0

V

N+1

=? kg/h

y

AN+

1=0

y

CN+1

=1.0

x

AN

=0.025

A:Acetone

C:MIK

Tie line on L0(xA0), V1(y1)min

xA

yA

x

A0

y

1min

L

0,

x

A0

V

N+1

A

C

L

N

,

x

AN

=0.025

V

1 min

y

1 min

A

min

M

min

Tie line

x

Amin

=0.175

0 0 1min 1 1min

min

0 1min 1min

1min

1000(0.235) (0)

0.175

1000

340 /

A N AN N

AM

N N

N

L x V y V

x

L V V

V kg h

+ + +

+ +

+

+ +

= =

+ +

=

V

N+1

=1.5(V

N+1 min

)=1.5(340)=510kg/k

L

0

=1000 kg/h

x

A0

=0.235

x

C0

=0

y

AN+1

=0

y

CN+1

=1.0

0 0 1 1

0 1

1000(0.235) 510(0)

1000 510

0.156

A N AN

AM

N

AM

L x V y

x

L V

x

+ +

+

+ +

= =

+ +

=

0 0 1 1

0 1

1000(0) 510(1.0)

1000 510

0.337

C N CN

CM

N

CM

L x V y

x

L V

x

+ +

+

+ +

= =

+ +

=

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M(x

AM

and x

CM

)

M(0.156, 0.337)

x

AN

=0.025 given

L

N

-M line gives V

1

, y

1

y

A1

=0.290, y

C1

=0.645

x

CN

=0.020 from graph

Lo+V

N+1

=L

N

+V

1

=M

1000+510=L

N

+V

1

V

1

=1510-L

N

| |

1 1

1

(0.025) (1510 )(0.290)

0.156

1510

N AN A

AM

N

N N

L x V y

x

L V

L L

+

=

+

+ ÷

=

L

N

=763, V

1

=747

To calculate A point

0 0 1 1

0 1

0 0 1 1

0 1

1000(0) 747(0.645)

1000 747

1.91

1000(0.235) 747(0.290)

1000 747

0.073

C C

C

C

A A

A

C

L x V y

x

L V

x

L x V y

x

L V

x

A

A

A

A

÷ ÷

= =

÷ ÷

= ÷

÷ ÷

= =

÷ ÷

=

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L

0,

x

A0

A

L

N

, x

AN

V

1 ,

y

1

M

V

N+1

C

xA

yA

x1

y

1

Draw tie line y

1

(V

1

), x

1

(L

1

)

Draw L

1

A,

Draw tie line y

2

(V

2

), x

2

(L

2

)

Draw L

2

A

Draw tie line y

3

(V

3

), x

3

(L

3

)

Repeat steps to L

N

N= 5 stage

V

2 ,

y

2

y

2

x2

L1

L2

V

3,

y

3

Countercurrent Extraction of Acetic Acid and Minimum Solvent. An

aqueous feed solution of 1000 kg/h of acetic acid-water solution contains 30.0 wt

% acetic acid and is to be extracted in a countercurrent multistage process with

pure isopropyl ether to reduce the acid concentration to 2.0 wt % acid in the final

raffinate. (a) Calculate the minimum solvent flow rate that can be used.

(b) If 2500 kg/h of ether solvent is used, determine the number of theoretical

stages required. (Note: It may be necessary to replot on an expanded scale the

concentrations at the dilute end.)

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Extraction with Immiscible Solvents. A water solution of 1000 kg/h containing

1.5 wt % nicotine in water is stripped with a kerosene stream of 2000 kg/h

containing 0.05 wt % nicotine in a countercurrent stage tower. The exit water is to

contain only 10% of the original nicotine, i.e., 90% is removed. Calculate the

number of theoretical stages needed.

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