LiquidLiquid Extractions

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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.
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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 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-Liquid Extraction
Solvent Feed
Solvent Rich Liquid Out
S, C
E, A, B, C
Liquid Feed
Carrier Rich Liquid out
F, A, B
R, A, B, 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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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
Raffinate out
F, XA(F), XB, T, P
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.
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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 exitsin the extract
Liquid-Liquid Extraction
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
All of carrier exitsin the raffinate
The raffinate is the exiting phase rich in
carrier. The extract is the exiting phase rich in
solvent.
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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
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)
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.
Solute Material Balance
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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
Note that the K-value isprimed to signify that
thisis a ratio of mass or mole ratios, not a
ratio of molefractions.
Solvent Feed
Extract out
S, XC(S)
E, XB (E), XC (E)
B
Liquid Feed
Raffinate out
F, XA(F), XB
R, XA (R), XB (R)
Note that concentrations of exiting streams
froman equilibrium stage are alwaysrelated by
equilibrium.
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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 in
the extract to solute flow in the raffinate.
Combining this definition with the equilibrium
relationship results in another expression
for the extraction factor
The larger the equilibrium driving force to
separate B, andthe larger the ratio of solvent
tofeed, the larger the extractionfactor.
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Extraction Efficiency
We can determine the amount not extracted
starting with the material balance of the solute
We substitute in the K-value ratio
And simplifyThis 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
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 variablesfor these diagrams are
only composition and thatpressure and
temperature are held constant (that is thatthese
diagrams are slices through a four
dimensionalspace with constant T and P).
C
A
B
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Ternary Phase Diagrams
Compositions are read as follows Draw three
lines from the compositionpoint 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). Composition
s can be mole fractions or mass fractions.
C
94 C, 3 B, 3 A
33 C, 33 B, 33 A
20 C, 20 B, 60 A
30 C, 70 B
100 A
A
B
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Construction of Ternary Phase Diagrams
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Ternary Eutectic Phase Diagrams
?G
?G
L
L
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?
?
?
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Ternary Eutectic Phase Diagrams
C
T above T of the ternary eutectic, butbelow the
binaryeutectics.
?
T
L
?L
?
?
? ?
??
? ?
? ? L
? ? L
? L
XB
0
1
L
?L
?L
? ?L
?
?
? ?
A
B
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Ternary Phase Diagrams
C
?
?L
L
A
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 ternarywhich relates the partial
solubilities. Equilibrium data can be obtained
graphically, orfrom tables. The ternary phase
diagram is a typical way of representing the
equilibrium compositions of the two phases
Solute
66 EthGly 7 Furfural 27 Water A composition
whereonly a single liquidexists.
Ethylene Glycol
Water
50 EthGly 50 Furfural
17 EthGly 27 Furfural 56 Water A
composition wheretwo liquid phases coexist.
100 Furfural
Solvent
Carrier
Furfural
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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
phasesthis completely defines the state of the
system.
Ethylene Glycol
Water
Tie-Lines Show the compositionsof the
equilibrium phases.
Furfural
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Partially Soluble Ternary Systems
Example Consider a feed of 200 kg of 30
ethylene glycol in water. Add 300kg of
purefurfural 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
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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
Water
F 60 kg EG 140 kg water
Furfural
S 300 Kg
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Partially Soluble Ternary Systems
Step 2 Locate the mixing point M
Ethylene Glycol
Water
F 60 kg EG 140 kg water
M
S 300 Kg
Furfural
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Partially Soluble Ternary Systems
Step 3 Use the tie-line to get the raffinate and
extract compositions. Extract (4 water, 14EG,
82 furfural) Raffinate (87 water, 5EG, 8
furfural)
Ethylene Glycol
Water
F 60 kg EG 140 kg water
M
E
R
S 300 Kg
Furfural
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Partially Soluble Ternary Systems
Step 4 Determine the amount of extract and
raffinate (can use lever rule)
Extract (4 water, 14EG, 82 furfural) Raffinate
(87 water, 5EG, 8 furfural)
Ethylene Glycol
Water
F 60 kg EG 140 kg water
M
E
R
S 300 Kg
Furfural
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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)
H
Ethylene Glycol
Water
F 60 kg EG 140 kg water
M
E
R
S 300 Kg
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