Liquid-liquid extraction (LLE)

IntroductionLLE vs DistillationEquipment for LLE

Principle of LLE

Introduction Liquid-liquid extraction (LLE) is a separation technique in which a solute of a liquid solution is separated out of the solution by contact with another liquid.

The constituent we want to extract is called solute or component A.

Before the contact, the solute originally

exists as a liquid solution with an

original solvent B or diluent

When this solution (containing A in B) is

mixed with a different liquid C, a portion of A

will cross to form a new solution in C.

This separation technique is called extraction since C has

extracted (pulled away) some of A from B.

C is known as an extracting solvent.

AB

BBB

B

BB

BA A

A

A

A

A

A

AB

BBB

B

BB

BA A

A

A

A

A

A

A

A

A

A

AA

B

BB

B BB

B

Extracting solvent (C)

Shake/stir to allow molecules

to partition

A = soluteB = original solventC = extracting solvent

Original solvent (B)

containing A

A portion of A will cross to form a new

solution in C.

• A partially miscible solvent will extract mostly the desired component A and some of component B.

• The resulting solution is a ternary solution.

Ternary solutionA solution that

consists of more than two

components

Consists of 3 components

Solute A Original solvent B

Extracting solvent C

example

ComparisonLiquid-liquid extraction Distillation

Depends on solubility1. The constituents of liquid solution

have relatively volatility near unity2. The constituents of a liquid solution

have close boiling points3. The constituents of liquid solution

form an azeotropic mixture

Depends on relative volatility between the components in the solution

No heat input required for the process1. The component involved is

sensitive to heat.Required heat upon distillation process

Driven by chemical differences Vapor pressure differences

Equipment for Liquid-Liquid ExtractionIntroduction and Equipment Types

As in the separation processes of distillation, the two phases in liquid-liquid extraction must be brought into intimate contact with a high degree of turbulence in order to obtain high mass-transfer rates.

Distillation: Rapid and easy because of the large difference in

density (Vapor-Liquid).

Liquid extraction: Density difference between the two phases is not

large and separation is more difficult.

Liquid extraction equipment

Mixing by mechanical agitation

Mixing by fluid flow themselves

6

Mixer-Settles for Extraction

Separate mixer-settler Combined mixer-settler

7

Plate and Agitated Tower Contactors for Extraction

Perforated plate tower Agitated extraction tower

8

Packed and Spray Extraction Towers

Spray-type extraction tower Packed extraction tower

9

Principle of liquid-liquid extraction

• As the solution (A+B) is mixed and agitated with the solvent C, an intermediate solution will form at which a lighter liquid bubbles through a heavy liquid.

• The intermediate will settle into two distinct layers:– Solvent-rich (extract) layer : mainly C, extracted A and

any dissolved B– Diluent-rich (raffinate) layer : rich with B, leftover A

and tiny fraction of dissolved C

Solvent selection

• y/x at equilibrium; large values preferable. (The distribution coefficient (m) or partition coefficient for a component (A) is defined as the ratio of concentration of a A in extract phase to that in raffinate phase.

Distribution Coefficients

• can be defined as the ability of the solvent to pick up the desired component in the feed as compared to other components

Selectivity

• solvent should not be soluble in carrier liquid

Insolubility

• consider constraints (i.e. azeotropes)

Recoverability

• if too high, liquids will be difficult to mix

Interfacial Tension

•must be different so that phases can be separated by settling

Density

• solvent should be inert and stable

Chemical Reactivity

• low values make storage easier

Viscosity, Vapor Pressure, Freezing Point

• toxicity, flammability

Safety

Cost

Ternary phase equilibrium diagram

• Equilibrium relationship can be presented in triangular diagram

• Two types of triangular diagrams1. Right-angle triangular diagram2. Equilateral triangular diagram

Ternary equilateral triangular Phase Diagram

Acetone

TCE Water

Ternary Phase DiagramSolvent: TCESolute: AcetoneCarrier: Water

Acetone

TCE Water

Ternary Single-Stage

Plait PointP

Tie-lines

Extract

Raffinate

Solvent

Carrier

Solute

Mixing point E

R

Feed

S

F

A point where the composition of the extract is the same as the composition of the raffinate

Right-angle triangular diagram

Single-stage LLE• In single-stage LLE, the solvent and solution are in contact with each other

only ONCE and thus the raffinate and extract are in equilibrium only once.

• Normally the solution is a binary solution containing solute A dissolved in an initial solvent B (diluent)

solution

• The extracting solvent can be either • Pure solvent C or• Recycled solvent that may

contain a little A

solvent

• Resulting liquid phase formed at equilibrium that contains:• Mostly B• Residual A and• Little C that may have dissolved

raffinate

• Resulting liquid phase formed at equilibrium consists of:• Mainly C• Extracted A (molecules A that has

transferred from B)• Little B that have dissolved

extract

• In most single-stage extraction, we are interested to determine the equilibrium composition (y of A in E and x of A in R) and masses of raffinate (R) and extract (E) phases.

• These can be easily determined by means or ternary phase diagram and simple material balances.

How to determine the equilibrium compositions and masses of a typical single-stage extraction

Single-stage LLE

F, Feed solution

S, extracting solvent

M, Intermediate

E, Extract phase

R, Raffinate phase

xF

yS

yE

xM

xR

Where F : mass of feed solution in kg or lbmS : mass of extracting solvent in kg or lbmE : mass of extract phase in kg or lbmR : mass of raffinate phase in kg or lbmM : mass of an intermediate in kg or lbmxF : mass fraction of solute A in solution with ByS : mass fraction of solute A in solution with extracting solvent CxM : mass fraction of solute A in intermediate phasexR : equilibrium mass fraction of solute A in raffinate phaseyE : equilibrium mass fraction of solute A in extract phase

Step 1• Calculate the mass

of intermediate M using total material balance

• F + S = M

Step 2• Do component

material balance to find xM

• xF(F) + yS(S) = xM(M)

Step 3• On ternary diagram:

• Locate point F and S

• Draw straight line from F to S

• Using the calculated value of xM, locate point M on the FS line

• Draw a new tie line that pass through point M

• From the new tie line, locate point E and point R and hence can determine composition of raffinate and extract in equilibrium

Step 4• Determine the

masses of E and R using material balance

• Total material balance:• F + S = R + E

• Material balance for solute A• xF(F) + yS(S) = xR(R)

+ yE(E)

Example 1

• 100 kg of a solution containing 0.4 mass fraction ethylene glycol (EG) in water is to be extracted with equal mass of furfural at 25oC and 101 kPa. Using the ternary phase equilibrium diagram method, determine:i) The composition of raffinate and extract phasesii) The mass of extract and raffinateiii) The percent glycol extracted

Answer: i) 0.075, 0.26; ii) 135.3 kg and 64.9 kg; iii) 86.7%

Example 2

A mixture containing 33 wt% acetic acid, 2 wt% isopropyl ether and 65 wt% water is treated with equal weight of recycled solvent which contains 3 wt% acetic acid in a simple one stage batch extraction.

a) Determine the compositions and mass fraction of the raffinate and extract, xR and yE.

b) Calculate the percent extraction of acetic acid by isopropyl ether

Multi-stage continuously countercurrent LLE

In multi-stage continuously countercurrent LLE, the solvent and solution come into contact MORE THAN ONCE and thus the raffinate and extract reach equilibrium more than once.

F kg/h feed solution

xF A

E kg/h final extract

yE A

xR AyS A

S kg/h extracting solventR kg/h final raffinate

Stage 1Stage 2Stage 3Stage 4

Stage n-2Stage n-1

Stage nStage n+1

Stage n+2

example

Minimum solvent flow rate

• Minimum solvent flow rate is the lowest rate/amount at which solvent could be theoretically used for a specified extraction.

• Point M is dependant upon the solvent flow rate.– The larger the rate, the closer is point M to point S

on FS line.– The smaller the flow rate, the closer in point M to

point F on FS line

Step by step guidelines to determine minimum solvent flow rate (Smin)

Step 1

• Find the positions of point F and S on a triangular ternary diagram

Step 2

• Draw the best tie line that originate from F.

• The intersection of this line with extract half-dome is point Emin (minimum extract flow rate)

Step 3

• Draw a straight line from point Emin to point R.

• The intersection of this line with FS gives point Mmin

• From point Mmin can read the value xmin

Step 4

• Using the xmin, use material balance to calculate Smin.

Overall efficiency of a multi-stage extraction column

% Overall efficiency of extraction column = Number of ideal stages x 100 number of real stages

Example 3

5000 kg/h of a solution containing 25 wt% ethylene glycol in water is to be reduced to 5% (solvent-free) by continuous extraction in a countercurrent column using furfural as the extracting solvent.a) Determine the minimum solvent flow rate for the

extraction.b) If the solvent enters at twice the minimum solvent

rate, how many ideal stages are required?c) Determine the number of real stages if the overall

efficiency of the column is 60%

solution