01 Liqliq Extraction

Lecture 7: Liquid-Liquid Ternary Single Stage1

Liquid-Liquid Extractions

•Liquid-liquid extraction (also known as solvent extraction) involves the separation of the constituents (solutes) of a liquid solution by contact with another insoluble liquid.

•Solutes are separated based on their different solubilities in different liquids.

•Separation is achieved when the substances constituting the original solution is transferred from the original solution to the other liquid solution.


The Figure showed a feed liquid (the "first" liquid) containing the desirable compound to be separated together with other compounds. Then an immiscible extraction liquid (the "second" liquid) is added and mixed with the feed liquid through agitation. The species re-distribute themselves between the 2 liquid phases. Agitation of the 2 phases is continued until equilibrium, and then agitation is stopped and the liquids are allowed to settle until both phases are clear. The 2 phases can then be separated.that is

The simplest liquid-liquid extraction involves only a ternary (i.e. 3 components) system. The solution which is to be extracted is called the feed, and the liquid with which the feed is contacted is the solvent. The feed can be considered as comprising the solute A and the "carrier" liquid C. Solvent S is a pure liquid. During contact, mass transfer of A from the feed to the solvent S occurs, with little transfer of C to S.

http://www.separationprocesses.com/Extraction/SE_Chp01a.htm


Operasi Perpindahan Massa Ekstraksi Dalam Industri Pengolahan Minyak

• The purpose of solvent extraction is to prevent corrosion, protect catalyst in subsequent processes, and improve finished products by removing unsaturated, aromatic hydrocarbons from lubricant and grease stocks.

• The solvent extraction process separates aromatics, naphthenes, and impurities from the product stream by dissolving or precipitation. The feedstock is first dried and then treated using a continuous countercurrent solvent treatment operation.


Application of Liquid-Liquid ExtractionExtraction processes are well suited to the petroleum industry because of the need to separate heat-sensitive liquid feeds according to chemical type (e.g. aliphatic, aromatic, naphthenic) rather than by molecular weight or vapour pressure. Other major applications exist in the biochemical or pharmaceutical industry, where emphasis is on the separation of antibiotics and protein recovery. In the inorganic chemical industry, they are used to recover high-boiling components such as phosphoric acid, boric acid, and sodium hydroxide from aqueous solutions.Some examples are given below:

•Extraction of nitrobenzene after reaction of HNO3 with toluene in H2SO4

•Extraction of methylacrylate from organic solution with perchlorethylene

•Extraction of benzylalcohol from a salt solution with toluene

•Removing of H2S from LPG with MDEA

•Extraction of caprolactam from ammonium sulfate solution with benzene

•Removing residual alkalis from dichlorohydrazobenzene with water

•Extraction of methanol from LPG with water

•Extraction of chloroacetic acid from methylchloroacetate with water

•Extraction of acrylic acid from wastewater with butanol


Solvent extraction

• In one type of process, the feedstock is washed with a liquid in which the substances to be removed are more soluble than in the desired resultant product. In another process, selected solvents are added to cause impurities to precipitate out of the product. In the adsorption process, highly porous solid materials collect liquid molecules on their surfaces.

• The solvent is separated from the product stream by heating, evaporation, or fractionation, and residual trace amounts are subsequently removed from the raffinate by steam stripping or vacuum flashing.


• Electric precipitation may be used for separation of inorganic compounds.• The solvent is regenerated for reused in the process.• The most widely used extraction solvents are phenol, furfural, and cresylic acid.• Other solvents less frequently used are liquid sulfur dioxide, nitrobenzene, and 2,2' dichloroethyl ether.• The selection of specific processes and chemical agents depends on the nature of the feedstock being

treated, the contaminants present, and the finished product requirements.


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 Feed

Solvent Rich Liquid Out

Liquid-Liquid Extraction

F, A, B R, A, B, CCarrier Rich Liquid out

E, A, B, C

Solvent Feed

S, C

Define the raffinate as the exiting phase rich in carrier.Define the extract as the exiting phase rich in solvent.



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:

The raffinate is the exiting phase rich in carrier.The extract is the exiting phase rich in solvent.

Liquid Feed

Extract out


F, XA(F), XB, T, PR, XA (R), XB (R), XC (R), T, P

Raffinate out

E, XA (E), XB (E), XC (E), T, P

Solvent FeedS, XC(S),T, P



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:

Liquid Feed

Extract out


F, XA(F), XB, T, PR, XA (R), XB (R), T, P

Raffinate out

E, XB (E), XC (E), T, P


The raffinate is the exiting phase rich in carrier.The extract is the exiting phase rich in solvent.

All of carrier exitsin the raffinate

All of solvent exitsin the extract



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.

Liquid Feed

Extract out

F, XA(F), XB, T, P R, XA (R), XB (R), T, PRaffinate out

E, XB (E), XC (E), T, P


Mass Ratio Example:

FB XBF FA

EB XBE EC XB

E S

RB XBR RA XB

R FASB XB

S 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.


Material Balances

Liquid Feed

Extract out

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

E, XB (E), XC (E)

Solvent FeedS, XC(S)

Solute Material Balance:

FB XBF FA

EB XBE EC XB

E S

RB XBR RA XB

R FASB XB

S 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 FAB

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

' SFA

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

1KDB

' SFA

1

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


Tie Line

• Tie line adalah garis yang menghubungkan satu titik pada rafinat dan pada extraktan

• Campuran yang komposisinya terletak dalam tie line akan memiliki kesetimbangan sama yaitu pada kedua titik diujung tie line

• Tie line dibuat dengan cara membiarkan suatu campuran yang diketahui komposisinya mencapai kesetimbangan 2 fasa.


Ekstraksi Kesetimbangan Satu tahap

• Counter current satu tahap seperti dilukiskan pada gambar sebelah kiri bawah ini tidaklah ada dalam aplikasinya (bisa anda jelaskan mengapa?). Namun berguna dalam memahami neraca masa perhitungan selanjutnya

CampuranRafinat R

Ekstrak E

L0 L1

V1 V2

L + V = ML.xA + V.yA = M.xAM

L.xC + V.yC = M.xCM

L0 + V2 = L1 + V1 = M L.xA + V.yA = M.xAML.xA + V.yA = M.xAM

L.xA + V.yA = M.xAML.xC + V.yC = M.xCM


Ekstraksi jamak

M

L0

1. Buatlah garis yang menghubungkan VN+1, M dan L0

2. Buat garis yg menghubungkan LN dan VN+1

3. Buat pertemuan kedua garis tsb (1 & 2) >

4. Buat tie line yg mengenai M,

5. Pada sisi amplop fasa > L1 & V1

6. Hubungkan L1 dengan , perpotongan pada satu sisi amplop menjadi V2

7. Buat tie line kedua dari V2,

8. Dst sampai melewati LN,

9. Jumlah tahap = jumlah garis yg menemui

LN

VN+1

L1

V1

V2





Partially Soluble Ternary SystemsExample: 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 SF S

0.3200kg 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


Differential extractors

Raffinate

Feed

Extract

Feed

Extracting solvent

Extract

Raffinate

Feed

Extracting solvent

Extract

Raffinate

Feed

Extract

Extracting solvent

Raffinate


Ekstraksi Diferensial

• Continuous process• Feed and extracting solvent flow past

one another• One phase is dispersed in the other• Contacting and phase separation takes

place within one unit• Phases are not in equilibrium except

locally i.e at the interface

Raffinate

Extract

Extracting solvent

Feed


Transport zat terlarut pada ekstraksi

Feed

Extracting solvent

Raffinate

Raffinate side film

Extract side film

Raffinate Extract

Extract side film

Raffinate side film

Interface

Concentration profile

x

xi

yi

y


Pemilihan Pelarut (Solvent)

• SelektivitasAdalah kemampuan pelarut untuk melarutkan suatu zat yang dikehendaki. Selektivitas

ditentukan dengan menghitung , yaitu• Koefisien distribusi• Ketidaklarutan solven• Rekoverabilitas• Densitas• Tegangan permukaan• Reaktivitas kimia• Viskositas, tekanan uap dan titik beku• Toksisitas, nonflammable dan murah



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


ALTERNATE VERSION

• USING SOLVENT-FREE DIAGRAM


MASS BALANCE EQUATIONS

• OVERALL & COMPONENT

• CAN BE REARRANGED TO SOLVE FOR Va AND Lb ON THE BASIS OF La AND Vb FED TO PROCESS

MxVyLxVyLxMVLVL

maabbbbaa

abba

RULELEVERFROMxxxy

VL

am

mb

b

a


MULTIPLE CROSS-CURRENT EXTRACTION

A

BS

xA,La

[yb,Vb]1,2

[ya,Va]1

[xb,Lb]1 = [xa,La]2

[xm,M]1

[ya,Va]2[xb,Lb]2

[xm,M]2


COUNTER-CURRENT MULTISTAGE EXTRACTION

• MINIMUM SOLVENT DETERMINATION BASED ON EQUILIBRIUM LINE THROUGH xa

A

BS

xa

ya,min

xbyb

P,min

xj,min


DESIGN SOLVENT RATES

• GRAPHICAL CALCULATION OF NUMBER OF STAGES


TRANSFER TO McCABE-THIELE

• OPERATING LINE SLOPE BASED ON L/V

• PERRY’S P. 15-18.

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01 Liqliq Extraction