Liquid-Liquid Extractions
Last lecture covered:
• Using a simple spreadsheet to apply the Rachford Rice
Procedure
• Bubble point and Dew Point temperature and pressure
calculations
• An example of a dew point temperature calculation
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.
Lecture 7: Liquid-Liquid Ternary Single Stage
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Liquid-Liquid Ternary Single Equilibrium Stage
We last covered the Flash Calculations where a liquid phase and a vapor phase were
in equilibrium. For that system we needed equilibrium data, for example, from a
Depriester chart to determine the distribution of components between the two phases.
Now we’ll 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-Liquid Extraction
Solvent Feed
S, C
Solvent Rich Liquid Out
E, A, B, C
Liquid Feed
F, A, B
Carrier Rich Liquid out
R, A, B, C
Define the raffinate as the exiting phase rich in carrier.
Define the extract as the exiting phase rich in solvent.
Lecture 7: Liquid-Liquid Ternary Single Stage
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Liquid-Liquid Ternary Single Equilibrium Stage
If the solvent and carrier have some solubility in each other, then the raffinate will have
a small amount of solvent in it and the extract will have a small concentration of carrier:
Liquid-Liquid Extraction
Solvent Feed
Extract out
S, XC(S),T, P
E, XA (E), XB (E), XC (E), T, P
Liquid Feed
F, XA(F), XB, T, P
Raffinate out
R, XA (R), XB (R), XC (R), T, P
The raffinate is the exiting phase rich in carrier.
The extract is the exiting phase rich in solvent.
Lecture 7: Liquid-Liquid Ternary Single Stage
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Liquid-Liquid Ternary Single Equilibrium Stage
If the solvent and carrier have no solubility in each other, then the raffinate will
have no solvent in it and the extract will have no carrier in it:
All of solvent exits
in the extract
Liquid-Liquid Extraction
Solvent Feed
S, XC(S),T, P
Extract out
E, XB (E), XC (E), T, P
Liquid Feed
F, XA(F), XB, T, P
Raffinate out
R, XA (R), XB (R), T, P
All of carrier exits
in the raffinate
The raffinate is the exiting phase rich in carrier.
The extract is the exiting phase rich in solvent.
Lecture 7: Liquid-Liquid Ternary Single Stage
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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 XB: The ratio of mass of component B to another component of the stream.
Mole ratio XB: 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.
Mass Ratio Example:
Solvent Feed
Extract out
S, XC(S),T, P
E, XB (E), XC (E), T, P
Liquid Feed
Raffinate out
F, XA(F), XB, T, P
R, XA (R), XB (R), T, P
F
FB XB FA
E
E
EB X B EC XB S
R
R
RB XB RA XB FA
S
S B X B SC 0
Lecture 7: Liquid-Liquid Ternary Single Stage
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
Solute Material Balance:
Solvent Feed
Extract out
S, XC(S)
E, XB (E), XC (E)
Liquid Feed
Raffinate out
F, XA(F), XB
R, XA (R), XB (R)
F
FB XB FA
Rate of B in the feed is the ratio of B to A, times feed rate of A.
E
E
EB X B EC XB S
R
R
RB XB RA XB FA
S
S B X B SC 0
Solute Material Balance:
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.
FB SB RB EB
F
S
R
E
X B FA XB SC XB RA X B EC
F
R
E
X B FA XB FA X B S
Lecture 7: Liquid-Liquid Ternary Single Stage
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Equilibrium Distribution
The way the solute will distribute itself between the extract and raffinate at
equilibrium is given by the K-Value:
E
R
X B K'DB XB
Solvent Feed
Extract out
S, XC(S)
Liquid Feed
F, XA(F), XB
Note that the K-value is
primed to signify that this
is a ratio of mass or mole
ratios, not a ratio of mole
fractions.
B
E, XB (E), XC (E)
Raffinate out
R, XA (R), XB (R)
Note that concentrations
of exiting streams from
an equilibrium stage are always
related by equilibrium.
Lecture 7: Liquid-Liquid Ternary Single Stage
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The Extraction Factor
The degree of separation of the solute between the exiting streams is expressed as the
extraction factor:
Extraction Factor: The ratio of solute flow in
the extract to solute flow in the raffinate.
E
E
EB X B EC XB S
R
R
RB XB RA XB FA
XB E S
EB
B
R
RB X FA
B
Combining this definition with the equilibrium relationship:
E
R
X B K'DB XB
results in another expression for the extraction factor:
B
Lecture 7: Liquid-Liquid Ternary Single Stage
K'DB S
FA
The larger the equilibrium
driving force to separate B, and
the larger the ratio of solvent to
feed, 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:
F
E
R
X B FA XB S X B FA
We substitute in the K-value ratio:
F
R
R
X B FA K'DB XB S XB FA
And simplify:
XB R
This ratio gives the amount
of solute left in the raffinate
to the amount originally in
the feed stream,
FA
F K' S F
XB
DB
A
XBR
1
F K' S
XB
DB
FA
Lecture 7: Liquid-Liquid Ternary Single Stage
1
1
B 1
The amount not extracted increases
with the feed rate, and smaller
ratio distribution between extract
and raffinate and less solvent.
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Ternary Phase Diagrams
It is convenient to construct ternary phase diagrams on a
Gibbs Triangle (shown at right). Note that the variables
for these diagrams are only composition and that
pressure and temperature are held constant (that is that
these diagrams are slices through a four dimensional
space with constant T and P).
A
Lecture 7: Liquid-Liquid Ternary Single Stage
C
B
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Ternary Phase Diagrams
Compositions are read as follows:
Draw three lines from the composition
point parallel to the composition lines.
Read the compositions off of the three
axes.
Note: Only two mole fractions are
needed (use the third as a check).
Compositions can be mole fractions
or mass fractions.
[20% C, 20% B, 60% A]
C
[94% C, 3% B, 3% A]
[33% C, 33% B, 33% A]
[30% C, 70% B]
[100% A]
A
Lecture 7: Liquid-Liquid Ternary Single Stage
B
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Partially Soluble Ternary Systems
If the two phases both have a partial solubility of the other component, then the
analysis is somewhat more complicated:
The difficulty is that now equilibrium data must be obtained for the ternary
which relates the partial solubilities. Equilibrium data can be obtained graphically, or
from tables. The ternary phase diagram is a typical way of representing the equilibrium
compositions of the two phases:
Solute
Ethylene Glycol
Water
50 % EthGly
50% Furfural
17% EthGly
27% Furfural
56 % Water
A composition where
two liquid phases
coexist.
100% Furfural
Solvent
Lecture 7: Liquid-Liquid Ternary Single Stage
66% EthGly
7% Furfural
27 % Water
A composition where
only a single liquid
exists.
Furfural
Carrier
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Specification of Liquid-Liquid Equilibrium
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 phases
this completely defines the state of the system.
Ethylene Glycol
Water
Tie-Lines:
Show the compositions
of the equilibrium phases.
Lecture 7: Liquid-Liquid Ternary Single Stage
Furfural
17
Partially Soluble Ternary Systems
Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Add 300kg of pure
furfural solvent.
Solvent Feed
Extract out
S, XC(S)
E, XB (E), XC (E)
Liquid Feed
Raffinate out
F, XA(F), XB
R, XA (R), XB (R)
Ethylene Glycol
Water
Furfural
Lecture 7: Liquid-Liquid Ternary Single Stage
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Partially Soluble Ternary Systems
Example: Consider a feed of 200 kg of 30% ethylene glycol in water. Add 300kg of solvent
which is pure furfural.
Step 1: Locate the Solvent and Feed points
Ethylene Glycol
S
300 Kg
Lecture 7: Liquid-Liquid Ternary Single Stage
Water
F
60 kg EG
140 kg water
Furfural
19
Partially Soluble Ternary Systems
Step 2: Locate the mixing point M:
X BF FA XBS S 0.3 200kg 0 300kg
0.12
FS
500kg
Ethylene Glycol
Water
F 60 kg EG
140 kg water
M
S
300 Kg
Lecture 7: Liquid-Liquid Ternary Single Stage
Furfural
20
Partially Soluble Ternary Systems
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)
Ethylene Glycol
Water
F 60 kg EG
140 kg water
M
E
R
S
300 Kg
Lecture 7: Liquid-Liquid Ternary Single Stage
Furfural
21
Partially Soluble Ternary Systems
Step 4: Determine the amount of extract and raffinate (can use lever rule)
M
E
XC
0.63 0.82
R f R
E 0.08 0.82 0.257
XC XC
XC
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
Ethylene Glycol
Water
F 60 kg EG
140 kg water
M
E
R
S
300 Kg
Lecture 7: Liquid-Liquid Ternary Single Stage
Furfural
22
Partially Soluble Ternary Systems
Step 5: Determine the solvent free extract: Mixtures of E and S. Extend line from S through
E to solvent free point at H.
Solvent free extract H (20% water, 80% EG)
R
XB
XB
H
Ethylene Glycol
Water
F 60 kg EG
140 kg water
M
E
F
R
S
300 Kg
Lecture 7: Liquid-Liquid Ternary Single Stage
Furfural
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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
Lecture 7: Liquid-Liquid Ternary Single Stage
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