Liquid-Liquid Extraction

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Liquid-Liquid Extraction

 CHOICE OF SOLVENT

Choosing the best solvent is the most critical aspect of developing a liquid-liquid extraction process. The solvent should have a high selectivity for the extracted solute. The selectivity  of a solvent is similar to relative volatility and is given by

R R

S S

x x

2 1

/

 /



x₁^S = weight fraction of component 1 (solute) in the solvent phase x₂^S = weight fraction of component 2 in the solvent phase

x₁^R = weight fraction of component 1 (solute) in the raffinate phase x₂^R = weight fraction of component 2 in the raffinate phase

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Liquid-Liquid Extraction

 SINGLE-STAGE CALCULATIONS

 In one-stage liquid-liquid extractor as shown in the next Figure, Component 1 is the solute, component 2 is the other component in the feed that we are trying to separate from component 1, and component 3 is the solvent.

Feed F, z₁, z₂, z₃

Fresh solvent S_o, x₁^So, x₂^So, x₃^So

solvent phase S₁, x₁^S1, x₂^S1, x₃^S1

Raffinate phase R₁, x₁^R1, x₂^R1, x₃^R1 Stage

 In the normal situation, the process feed contains no solvent, so that z₃ = 0. We

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Liquid-Liquid Extraction

 The total mass balance at this stage is

 A component balance on the jth component yields

 Now we define the parameter

Where the point M must lie on the straight line joining F and S_o. It also must lie on a straight line joining S₁ and R₁

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(5)

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Liquid-Liquid Extraction

 Since the two liquid phases leaving the system are in phase equilibrium, the points R₁ and S₁ must be connected by an LLE tie-line.

1. Calculate M from equation (7).

2. Calculate x₁^M and x₃^M from equations (8) and (9).

(7) (8) (9)

 the procedure for determining the compositions and flow rates of the two liquid streams leaving the system is as follows:

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6. Calculate the flow rates S₁ and R₁ by solving equations (4) and (5) simultaneously:

Liquid-Liquid Extraction

5. The points at the two ends of the LLE tie-line give the compositions of the two phases leaving the system: the raffinate-rich phase with composition x₁^R1 and x₃^R1 and the solvent-rich phase with composition x₁^S1 and x₃^S1

Splitting the mixture of F and So into raffinate-rich phase R₁ and solvent-rich phase S₁ at ends of a tie-line.

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Example: An organic stream, with composition 30 weight percent acetone and 70 weight percent methyl isobutyl ketone and flow rate 10,000 kg/h, is mixed with a pure water solvent with flow rate 5,000 kg/h. What are the compositions and flow rates of the two liquid phases leaving a single- stage liquid-liquid extractor operating at 25^oC.

Solution:

x₁^So = 0 x₂^So= 0 x₃^So= 1.0 z₁ = 0.3

z₂ =0.7 z₃= 0

Feed= 10000 z₁=0.3 z₂= 0.7, z₃=0

Fresh solvent=5000 x₁^So,=0 x₂^So=0, x₃^So=1.0

solvent phase S₁, x₁^S1, x₂^S1, x₃^S1

Raffinate phase R₁, x₁^R1, x₂^R1, x₃^R1 Stage

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Liquid-Liquid Extraction

M x S x Fz

S M

0

1 0 1

1

 

M x S x Fz

S M

0

3 0 3

3

 

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 The point M is thus located at (0.333,0.20), as shown in Figure

 Note that M lies on the straight line connecting the points F and S_o

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 The LLE conjugate line is used to determine the other end of the LLE tie-line on the solvent phase solubility curve. If the tie-line goes through the point M, the compositions of the water solvent phase and organic raffinate phase have been found. We have:

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Liquid-Liquid Extraction

 MULTIPLE STAGES WITH CROSSFLOW OF SOLVENT

 If the process liquid stream from the first stages fed into a second extractor and mixed with more fresh solvent, as shown in the Figure, we have what is called cross-flow extraction.

 The process can be described the same as for a single stage .it is simply repeated again for each stage, using the raffinate phase from the upstream stage as the feed to each stage.

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Liquid-Liquid Extraction

Mix points for cross-flow extractor.

2 1 0 1 1

1

0 1

2

0 2

M x S x R

S R M

X S

M 







1 3 0 3

3

1 1 0 1

1

0 1

M x S x Fz

S F M

S M

 





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 MULTISTAGE COUNTERCURRENT EXTRACTION

Multistage countercurrent extraction is the most commonly encountered liquid-liquid extraction process. The raffinate and solvent streams travel countercurrent to each other through N stages. The flow rate of the raffinate leaving the last stage (tray N) is R.

 Mass and component balances around the entire cascade give (10)

1

0 R S

S

F   _N 

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Liquid-Liquid Extraction

 We next define a pseudo flow rate M and pseudo compositions x₁^M and x₃^M : (13)

S0

F M  

(14)

0 1 1

1

0

S x F

z M

x

^M

 

^S

(15)

0 3 3

3

0

S x F

z M

x

^M

 

^S

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 If the fresh solvent flow rate S_o and composition are given along with the feed flow rate F and composition, we can locate the point M on the straight line connecting the points F and S_o.

 From equations (10) through (12), it is clear that M must also lie on the straight line

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Liquid-Liquid Extraction

The point R_N will be given, i.e., the concentration of the raffinate phase leaving the final stage will be specified so as to recover the desired amount of the solute from the feed.

 In the typical design problem, you will be given:

Note: in any real system we cannot recover all of the solute from the feed

 Since the points R_N and M are known, a straight line can be drawn to the solubility curve to determine the composition of the ^x₁^S1^and

x₃^S1 of the S₁ stream.

 Equations (10) and (11) can be used to solve for the flow rates R_N and S₁.

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Liquid-Liquid Extraction

 Since S₁ and R₁ are in phase equilibrium, we can use an LLE tie- line to determine the point R₁ on the solubility curve.

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 Component and mass balances around the first stage can then be used to calculate the flow rate S₂ and compositions x₁^S2 and x₃^S2 of the solvent entering stage 1 from stage 2:

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1 1

2 R S

S

F   

(17)

1 1 1

1 2

1 1

2

S x R x S

x F

z 

^S



^R



^S

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1 3 1

3 2

3 3

1 1

2

S x R x S

x F

z 

^S



^R



^S

 Once S₂ is known, R₂ can be found at the other end of the LLE tie-line.

 This computational procedure can be repeated from stage to stage until enough stages have been used to produce a process stream that meets or exceeds the specifications on R_N (the final raffinate phase leaving the unit).

Liquid-Liquid Extraction

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 Graphical procedure

First, define another pseudo flow rate Δ and pseudo compositions x₁^Δ and x₃^Δ This fictitious Δ stream serves the same purpose as the operating lines did in binary distillation. We can calculate the compositions of "passing" streams in the column. It provides a graphical way to solve the three mass balances that describe this ternary system simultaneously:

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S0

R_N 





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0 1 1

1

0S x R

x

x^  ^R^N _N  ^S

S

R 





Liquid-Liquid Extraction

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 The pseudo composition x_j^Δ is entirely fictitious and, therefore, can be less than zero or greater than unity.

 Using equations (10) and (19), we see that

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₁

0

F S

S

R

_N

  





 Therefore, the Δ point must lie on two straight lines, one through the points R_N and S₀ and the other through the points F and S₁.

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 Δ can lie either to the left or to the right of the phase diagram

Liquid-Liquid Extraction

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Liquid-Liquid Equilibrium

 The mass balances for the first stage (equations (16) through (18), the definition of Δ (equations (19) through (21), and equation(22) can be combined to give

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₁ ₁ ₂

0 F S R S

S

R_N     





 The two streams R₁ and S₂ that pass each other between the first and second stages are related to each other by the Δ point. Hence, if we know R₁ we can determine S₂ by using the straight line that connects R₁ and Δ.

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Liquid-Liquid Extraction

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Example: A liquid-liquid ternary phase diagram for isopropyl alcohol (IPA), toluene, and water at 25^oC. Feed flow rate is 100 kg/hr, and feed compositions 40 weight percent IPA and 60 weight percent toluene. Fresh solvent is pure water at a flow rate of 100 kghr. Determine the number of equilibrium stages required to produce a raffinate stream that contains 3 weight percent IPA.