Ekstraksi-ppt

Lecture 7: Liquid-Liquid Ternary Single Stage1

Liquid-Liquid Ternary Single Equilibrium Stage

Ekstraksi Cair-Cair TernerDalam kasus ini:•Membuat dua fase cairan dengan menambahkan solvent (C) ke dalam campuran cair carrier (B) dan solute (A)•Solvent (C) dan carrier (B) tidak atau sedikit saling larut.

Liquid Feed

Cairan kaya solvent

Liquid-Liquid Extraction

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

Cairan kaya carrier

E, A, B, C

Solvent Feed

S, C

Raffinate adalah fase kaya carrier.Extract adalah fase kaya solvent.



Jika solvent dan carrier sedikit saling larut, maka raffinate mengandung sedikit solvent di dalamnya dan extract mengandung sedikit carrier.


Liquid Feed

Extract out


F, XAF, XB, T, P

R, XAR, XBR, XCR, T, P

Raffinate out

E, XAE, XBE, XCE, T, P

Solvent Feed

S, XCS,T, P



Jika solvent dan carrier tidak saling larut, maka raffinate tidak mengandung solvent dan extract tidak mengandung carrier.

Liquid Feed

Extract out


F, XAF, XB, T, P

R, XAR, XBR, T, P

Raffinate out

E, XBE, XCE, T, P

Solvent Feed

S, XCS,T, P

Semua carrier berada dalam raffinate

Semua solvent beradadalam extract



Mass and Mole Ratios

Konsentrasi biasa dinyatakan sebagai rasio massa atau rasio mol dari pada fraksi mass ataufraksi mol. Ini digunakan untuk penyederhanaan dalam analisis.

Rasi massa XB = rasio massa komponen B terhadap komponen lain.Rasio mol XB = rasio mol komponen B terhadap komponen lain.

Liquid Feed

Extract out

F, XAF, XB, T, P R, XAR, XBR, T, PRaffinate out

E, XBE, XCE, T, P

Solvent Feed

S, XCS,T, P

Mass Ratio Example:

ABFB FXF

SXEXE BECBEB

ABRABRB FXRXR

0SXS CBSB


Material Balances

Liquid Feed

Extract out

F, XA(F), XB R, XA (R), XB (R)Raffinate out

E, XB (E), XC (E)

Solvent Feed

S, XC(S)

Solute Material Balance:

FB XBF FA

EB XBE EC XB

E S

RB XBR RA XB

R FA

SB XBS SC 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.

FB SB RB EB

XBF FA XB

S SC XBR RA XB

E EC

XBF FA XB

R FA XBE S

Solute Material Balance:


Equilibrium Distribution

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

XBE KDB

' XBR

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

Liquid Feed

Extract out


E, XB (E), XC (E)S, XC(S)

Solvent Feed

Note that concentrations of exiting streams froman equilibrium stage are alwaysrelated by equilibrium.

B


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.

EB XBE EC XB

E S

RB XBR RA XB

R FA

B EBRB

XBE S

XBR FA

Combining this definition with the equilibrium relationship:

results in another expression for the extraction factor:

XBE KDB

' XBR

B KDB

' S

FA

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


Extraction Efficiency

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

XBF FA XB

E S XBR FA

XBF FA KDB

' XBR S XB

R FA

XBR

XBF

FAKDB

' S FA

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 extract and raffinate and less solvent.

XBR

XBF

1

KDB' S

FA1

1

B 1

We substitute in the K-value ratio:


Ternary Phase Diagrams

A B

CIt 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).



A B

CCompositions are read as follows:Draw three lines from the compositionpoint 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 fractionsor 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]


Construction of Ternary Phase Diagrams

G


Ternary Eutectic Phase Diagrams

G

L

G

L


Ternary Eutectic Phase Diagrams

A B

C

+

+

+

L

+ +L

+ +L

+ +L

+L +L

+L

T above T of the ternary eutectic, butbelow the binaryeutectics.

0 XB 1

+

L

+L

T



A B

C

L

+L


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:

66% EthGly7% Furfural 27 % Water A composition whereonly a single liquidexists.

17% EthGly27% Furfural 56 % Water A composition wheretwo liquid phases coexist.

50 % EthGly50% Furfural

100% Furfural

Solvent Carrier

Solute

Ethylene Glycol

Furfural

Water


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.

Ethylene Glycol

Furfural

Water



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

Liquid Feed

Extract out


E, XB (E), XC (E)S, XC(S)Solvent Feed

Ethylene Glycol

Furfural

Water



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



Step 2: Locate the mixing point M:

S300 Kg

XBF FA XB

S S

F S0.3200kg 0300kg

500kg0.12

Ethylene Glycol

Furfural

Water

M

F 60 kg EG 140 kg water



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)

S300 Kg


ER

Ethylene Glycol

Furfural

Water

M



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

R f XCM XC

E

XCR XC

E 0.63 0.820.08 0.82

0.257 Extract (4% water, 14%EG, 82% furfural)Raffinate (87% water, 5%EG, 8% furfural)

R0.257500kg128.4kg

E 1 0.257 500kg371.6kg

S300 Kg


ER

Ethylene Glycol

Furfural

Water

M



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)

S300 Kg


ER

Ethylene Glycol

Furfural

Water

M

H

XBR

XBF


Summary

This lecture covered:• Ternary Liquid-Liquid extractions.• Ternary phase diagrams.• A procedure to determine the product compositions andflow rates of a liquid-liquid extraction separation.

Next lecture will cover:• Leaching• Crystallization

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