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Lecture 7: Liquid-Liquid Ternary Single Stage1

Liquid-Liquid Extractions

This lecture will cover: Ternary Liquid -Liquid extractions. Ternary phase diagrams. A procedure to determine the product compositions and flow rates of a liquid-liquid extraction separation.

Last lecture covered: Using a simple spreadsheet to apply the Rachford RiceProcedure Bubble point and Dew Point temperature and pressure calculations An example of a dew point temperature calculation

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Liquid-Liquid Ternary Single Equilibrium Stage

We last covered the Flash Calculations where a liquid phase and a vapor phase werein equilibrium. For that system we needed equilibrium data, for example, from aDepriester chart to determine the distribution of components between the two phases.

Now well analyze the following system:

Ternary Liquid-Liquid Extraction In this case: We create two liquid phases by introducing a solvent (C) (MSA) to a liquid mixture of a carrier (A)

and a solute (B) Solvent (C) and carrier (A) have very little solubility in each other

Liquid Feed

Solvent Rich Liquid Out

Liquid-Liquid Extraction

F, A, B R, A, B, C

Carrier Rich Liquid out E, A, B, C

Solvent Feed

S, C

Define the raffinate as the exiting phase rich in carrier .Define the extract as the exiting phase rich in solvent .

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If the solvent and carrier have some solubility in each other, then the raffinate will havea small amount of solvent in it and the extract will have a small concentration of carrier:

The raffinate is the exiting phase rich in carrier .The extract is the exiting phase rich in solvent .

Liquid Feed

Extract out


F, X A(F), X B, T, P R, X A (R), X B (R), X C (R), T, P

Raffinate out

E, X A (E), X B (E), X C (E), T, P

Solvent Feed

S, X C(S),T, P

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If the solvent and carrier have no solubility in each other, then the raffinate willhave no solvent in it and the extract will have no carrier in it:

Liquid Feed

Extract out


F, X A(F), X B, T, P R, X A (R), X B (R), T, P

Raffinate out

E, X B (E), X C (E), T, P

Solvent Feed S, X C(S),T, P

The raffinate is the exiting phase rich in carrier .The extract is the exiting phase rich in solvent .

All of carrier exitsin the raffinate

All of solvent exitsin the extract

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Mass and Mole Ratios

Often the concentrations are as mass or mole ratios , rather than mass or mole fractions.This is generally done to simplify the expressions used in the analysis.

Mass ratio X B: The ratio of mass of component B to another component of the stream.Mole ratio X B: The ratio of moles of component B to another component of the stream.

Note that the basis (choice of component) for the mass or mole ratio must be chosen.

Liquid Feed

Extract out

F, X A(F), X B, T, P R, X A (R), X B (R), T, PRaffinate out

E, X B (E), X C (E), T, P

Solvent Feed S, X C(S),T, P

Mass Ratio Example :

F B X B F F A

E B X B E E C X B

E S

R B X B R R A X B

R F A

S B X BS

S C 0

Rate of B in the feed is the ratio of B to A, times feed rate of A.

Rate of B in the extract is the ratio of B to C, times rate of C.

Rate of B in the raffinate is the ratio of B to A, times rate of A.

Rate of B in the solvent is the ratio of B to C, times feed rate of C.

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Material Balances

Liquid Feed

Extract out

F, X A(F), X B R, X A (R), X B (R)Raffinate out

E, X B (E), X C (E)

Solvent Feed

S, X C(S)

Solute Material Balance :

F B X B F F A

E B X B E E C X B E S

R B X B R R A X B

R F A

S B X BS S C 0

Rate of B in the feed is the ratio of B to A, times feed rate of A.

Rate of B in the extract is the ratio of B to C, times rate of C.

Rate of B in the raffinate is the ratio of B to A, times rate of A.

Rate of B in the solvent is the ratio of B to C, times feed rate of C.

F B S B R B E B

X B F F A X B

S S C X B R R A X B

E E C

X B

F F A

X B

R F A

X B

E S

Solute Material Balance :

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Equilibrium Distribution

The way the solute will distribute itself between the extract and raffinate atequilibrium is given by the K-Value:

X B E K D B

' X B

R

Note that the K-value is primed to signify that thisis a ratio of mass or moleratios, not a ratio of molefractions.

Liquid Feed

Extract out

F, X A(F), X B R, X A (R), X B (R)Raffinate out

E, X B (E), X C (E)S, X C(S)

Solvent Feed

Note that concentrationsof exiting streams from

an equilibrium stage are alwaysrelated by equilibrium.

B

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The Extraction Factor

The degree of separation of the solute between the exiting streams is expressed as theextraction factor :

Extraction Factor : The ratio of solute flow inthe extract to solute flow in the raffinate.

E B X B E E C X B

E S

R B X B R

R A X B R

F A

BE B R B

X B E S

X B

R F A

Combining this definition with the equilibrium relationship:

results in another expression for the extraction factor :

B

E K D B

' X

B

R

B K D B

'S

F A

The larger the equilibriumdriving force to separate B, andthe larger the ratio of solvent tofeed, the larger the extraction

factor.

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Extraction Efficiency

We can determine the amount not extracted starting with the material balance of the solute:

X B F

F A X B E

S X B R

F A

B F F A K D B

' X B

R S X B R F A

X B R

X B F

F A K D B

'S F A

And simplify:

This ratio gives the amountof solute left in the raffinateto the amount originally inthe feed stream,

The amount not extracted increaseswith the feed rate, and smaller ratio distribution between extractand raffinate and less solvent.

X B R

X B F

1 K D B

'S

F A1

1 B 1

We substitute in the K-value ratio:

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Ternary Phase Diagrams

A B

CIt is convenient to construct ternary phase diagrams on aGibbs Triangle (shown at right). Note that the variables

for these diagrams are only composition and that pressure and temperature are held constant (that is thatthese diagrams are slices through a four dimensionalspace with constant T and P).

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A B

CCompositions are read as follows:Draw three lines from the composition

point parallel to the composition lines.Read the compositions off of the threeaxes.

Note: Only two mole fractions areneeded (use the third as a check).Compositions can be mole fractions

or mass fractions.

[94% C, 3% B, 3% A]

[20% C, 20% B, 60% A]

[100% A]

[33% C, 33% B, 33% A]

[30% C, 70% B]

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Construction of Ternary Phase Diagrams

G

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Ternary Eutectic Phase Diagrams

G

L

G

L

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Ternary Eutectic Phase Diagrams

A B

C

+

+

+

L

+ +L

+ +L

+ +L

+L +L

+L

T above T of theternary eutectic, but

below the binaryeutectics.

0 XB 1

+

L

+L

T

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A B

C

L+L

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Partially Soluble Ternary Systems

If the two phases both have a partial solubility of the other component, then theanalysis is somewhat more complicated:

The difficulty is that now equilibrium data must be obtained for the ternarywhich relates the partial solubilities. Equilibrium data can be obtained graphically, or from tables. The ternary phase diagram is a typical way of representing the equilibriumcompositions of the two phases:

66% EthGly7% Furfural27 % Water A composition whereonly a single liquidexists.

17% EthGly27% Furfural56 % Water A composition wheretwo liquid phasescoexist.

50 % EthGly50% Furfural

100% Furfural

Solvent Carrier

Solute

Ethylene Glycol

Furfural

Water

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For two phase equilibrium (either complete insolubility, or partially solubility): the equilibrium is between two liquids phases ( = 2) three components (ternary) distribute between the two phases (C = 3)

For the static equilibrium case we can specify 3 variables:

If we specify T and P we are left with one additional variable:

Thus if we specify the concentration of one component in either of the phasesthis completely defines the state of the system.

Specification of Liquid-Liquid Equilibrium

Tie-Lines:Show the compositionsof the equilibrium phases.

Ethyl ene Glycol

Furfural

Water

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Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Add 300kg of purefurfural solvent.

Liquid Feed

Extract out

F, X A(F), X B R, X A (R), X B (R)Raffinate out E, X B (E), X C (E)

S, X C(S)

Solvent Feed

Ethylene Glycol

Furfural

Water

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Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Add 300kg of solventwhich is pure furfural.

Step 1: Locate the Solvent and Feed points

S300 Kg

F60 kg EG140 kg water

Ethylene Glycol

Furfural

Water

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Step 2: Locate the mixing point M:

S 300 Kg

F 60 kg EG140 kg water

X B F F A X B

E S

F S

0.3 200kg 0 300kg

500kg 0. 12

Ethyl ene Glycol

Furfural

Water

M

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Step 3: Use the tie-line to get the raffinate and extract compositions.

Extract (4% water, 14%EG, 82% furfural)

Raffinate (87% water, 5%EG, 8% furfural)

S 300 Kg


E R

Ethylene Glycol

Furfural

Water

M

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Step 4: Determine the amount of extract and raffinate (can use lever rule)

R f

X C M X C

E

X C R X C E 0.63 0. 82

0.08 0.820.257 Extract (4% water, 14%EG, 82% furfural)

Raffinate (87% water, 5%EG, 8% furfural)

R 0.257 500kg 128 .4kg

E 1 0.257 500kg 371 .6kg

S 300 Kg


E R

Ethylene Glycol

Furfural

Water

M

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Step 5: Determine the solvent free extract: Mixtures of E and S. Extend line from S throughE to solvent free point at H.

Solvent free extract H (20% water, 80% EG)

S 300 Kg


E R

Ethylene Glycol

Furfural

Water

M

H

X B

R

X B F

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L 7 Li id Li id T Si l S24

Summary

This lecture covered: Ternary Liquid -Liquid extractions.

Ternary phase diagrams. A procedure to determine the product compositions and flow rates of a liquid-liquid extraction separation.

Next lecture will cover: Leaching Crystallization

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