liquid-liquid extraction otk

8/13/2019 Liquid-Liquid Extraction OTK

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Chapter 5 Liquid-Liquid ExtractionSubject: 1304 332 Unit Operation in Heat transfer

Instructor: Chakkrit Umpuch

Department of Chemical Engineering

Faculty of Engineering

Ubon Ratchathani University


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Here is what you will learn in this chapter.5.1 Introduction to Extraction Processes

5.2 Equilibrium Relations in Extraction

5.3 Single- Stage Equilibrium Extraction

5.4 Equipment for Liquid-Liquid Extraction

5.5 Continuous Multistage Countercurrent

Extraction

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Liquid-liquid extraction principle

When Liquid-liquid extraction is carried out in a test tube or flask the

two immiscible phases are shaken together to allow molecules topartition (dissolve) into the preferred solvent phase.

5.1 Introduction to Extraction Processes

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An example of extraction:

5.1 Introduction to Extraction processes

Acetic acid in H2O

+Ethyl acetate

Extract

Organic layer contains most of acetic acid inethyl acetate with a small amount of water. RaffinateAqueous layer contains a weak acetic acid

solution with a small amount of ethylacetate.

The amount of water in the extract and ethyl acetate in the raffinatedepends upon their solubilites in one another.

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Liquid C dissolves completely in A or in B.

Liquid A is only slightly soluble in B and B slightly soluble in A.

The two-phase region is included inside below the curved envelope.

An original mixture of composition M will separate into two phases a and b which are on

the equilibrium tie line through point M.

The two phases are identical at point P, the Plait point.

Liquid-Liquid phase diagram where components A and B are partiallymiscible.

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Point A = 100% Water

Point B = 100% Ethylene Glycol

Point C = 100% Furfural

Point M = 30% glycol, 40% water, 30% furfural

Point E = 41.8% glycol, 10% water, 48.2% furfural

Point R = 11.5% glycol, 81.5% water, 7% furfural

The miscibility limits for the furfural-water binary

system are at point D and G.

Point P (Plait point), the two liquid phases have

identical compositions.

DEPRG is saturation curve; for example, if feed

50% solution of furfural and glycol, the secondphase occurs when mixture composition is 10%

water, 45% furfural, 45% glycol or on the

saturation curve.

Liquid-Liquid equilibrium, ethylene glycol-furfural-water, 25C,101 kPa.

Ex 5.1 Define the composition of point A, B, C, M, E, R, P and DEPRG inthe ternary-mixture.

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Equilibrium data on rectangular coordinates

The system acetic acid (A) water (B) isopropyl ether

solvent (C). The solvent pair Band C are partially miscible.xB= 1.0 - xA- xCyB= 1.0 - yA- yC

Liquid-liquid phase diagram

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EX 5.2 An original mixture weighing 100 kg and containing 30 kg ofisopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is

equilibrated and the equilibrium phases separated. What are the

compositions of the two equilibrium phases?

Solution:

Composition of original mixture is xc= 0.3, xA= 0.10, and xB= 0.60.

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The solvent pairs B and C and also A and C are partially miscible.

Phase diagram where the solvent pairs B-C and A-C are partially miscible.

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Derivation of lever-arm rule for graphical addition

AAM

AMA

xx

xy

V

L

CMC

MCC

xx

xy

V

L

AAM

CMC

AMA

MCC

yx

yx

xx

xx

ML

MV

kgV

kgL

)(

)(

VL

MV

kgM

kgL

)(

)(

(5.4)

(5.5)

(5.6)

Sub 5.1 into 5.2

Sub 5.1 into 5.3

Sub 5.1 into 5.3

(5.7)

(5.8)

Lever arms rule

Eqn. 5.6 shows that points L, M, and V must lie on a straight line.

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Ex 5.3 The compositions of the two equilibrium layers in Example 5.1 are forthe extract layer (V) yA= 0.04, yB= 0.02, and yC= 0.94, and for the raffinate

layer (L) xA= 0.12, xB= 0.86, and xC= 0.02. The original mixture contained100 kg and xAM= 0.10. Determine the amounts of V and L.

Solution: Substituting into eq. 5.1

Substituting into eq. 5.2, where M = 100 kg and xAM= 0.10,

Solving the two equations simultaneously, L = 75.0 and V = 25.0. Alternatively, using

the lever-arm rule, the distance hg in Figure below is measured as 4.2 units and gi

as 5.8 units. Then by eq. 5.8,

Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the

material-balance method.

100 MLV

)10.0(100)12.0()04.0( LV

8.5

2.4

100 ig

ghL

M

L

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5.2 Single-stage liquid-liquid extraction processesSingle-state equilibrium extraction

MVLVL 1120

AMAAAA MxyVxLyVxL 11112200

MCCCCC MxyVxLyVxL 11112200

We now study the separation of A from a mixture of A and B by a solvent C in a singleequilibrium stage.

0.1 CBA xxx

An overall mass balance:

A balance on A:A balance on C:

5.9

5.105.11

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Ex 5.4 A mixture weighing 1000 kg contains 23.5 wt% acetic acid (A) and76.5 wt% water (B) and is to be extracted by 500 kg isopropyl ether (C) in a

single-stage extraction. Determine the amounts and compositions of theextract and raffinate phases.

Solution Given: kgVandkgL 5001000 20

AMx)1500()0)(500()235.0)(1000(

kgMVL 1500500100020

0765.0,235.0200

ABA

yandxx

Given:

157.0AMx

MCCC MxyVxL 2200

Given: 0765.0235.00.11 000 BAc xxx

AMAA MxyVxL 2200

MCx)1500()1)(500()0)(1000(

33.0CMx 19


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M

V2 (0,1) = (yA2, yC2)V1(0.1,0.89) = (yA1, yC1)

L1(0.2,0.03) = (xA1, xC1)

L0(0.235,0) = (xA0, xC0)

M(0.157,0.33) = (xAM, xCM)

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(1)

AMAA MxyVxL 1111

MCCC MxyVxL 1111

)157.0)(1500()1.0()2.0( 11 VL

)33.0)(1500()89.0()03.0( 11 VL

From the graph: xC1= 0.03 and yC1= 0.89;

From the graph: xA1= 0.2 and yA1= 0.1;

(2)

5.177,15.0 11 VL

500,1667.29 11 VL

Solving eq(2) and eq(3) to get L1and V1;

kgVandkgL 28.52586.914 11

89.003.0,1.0,2.0 1111 CCAA yandxyx Answer

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5.3 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 highdegree 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 equipmentMixing by mechanicalagitationMixing by fluid flowthemselves

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Mixer-Settles for Extraction

Separate mixer-settler Combined mixer-settler

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Plate and Agitated Tower Contactors for Extraction

Perforated plate tower Agitated extraction tower

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Packed and Spray Extraction Towers

Spray-type extraction tower Packed extraction tower

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5.4 Continuous multistage countercurrent extractionCountercurrent process and overall balance

1. Usually, L0 and VN+1 are known and

the desired exit composition xANis set.

2. Plot points L0, VN+1, and M as in the

figure, a straight line must connect thesethree points.

3. LN, M, and V1 must lie on one line.

Also, LN and V1 must also lie on the

phase envelope.

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Ex 5.5 Pure solvent isopropyl ether at the rate of VN+1= 600 kg/h is beingused to extract an aqueous solution of L0=200 kg/h containing 30 wt% acetic

acid (A) by countercurrent multistage extraction. The desired exit acetic acid

concentration in the aqueous phase is 4%. Calculate the compositions andamounts of the ether extract V1and the aqueous raffinate LN. Use equilibriumdata from the table.

Solution: The given values are VN+1= 600kg/h, yAN+1= 0, yCN+1= 1.0, L0= 200kg/h,

xA0= 0.30, xB0= 0.70, xC0= 0, and xAN= 0.04.

In figure below, VN+1and L0are plotted. Also, since LNis on the phase

boundary, it can be plotted at xAN= 0.04. For the mixture point M,substituting into eqs. below,

75.0600200

)0.1(600)0(200

10

1100

N

NCNC

MC VL

yVxLx

075.0600200

)0(600)30.0(200

10

1100

N

ANNAMA

VL

yVxLx

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Using these coordinates,

1) Point M is plotted in Figure below.

2) We locate V1by drawing a line from LNthrough M and extending it until

it intersects the phase boundary. This gives yA1= 0.08 and yC1= 0.90.

3) For LNa value of xCN= 0.017 is obtained. By substituting into Eqs. 5.12and 5.13 and solving, LN= 136 kg/h and V1= 664 kg/h.

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Stage-to-stage calculations for countercurrent extraction.

1120 VLVL

nnnn VLVL 11

2110 VLVL

....1110 NNnn VLVLVL

...11111100 NNNNnnnn yVxLyVxLyVxLx

Total mass balance on stage 1Total mass balance on stage n

From 5.16 obtain differencein flows

5.16

5.17

5.185.19

5.20

is constant and for all stages

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Stage-to-stage calculations for countercurrent extraction.

1

11

1

11

10

1100

NN

NNNN

nn

nnnn

VL

yVxL

VL

yVxL

VL

yVxLx

10 VL 1 nn VL 1 NN VL

5.21

5.22

x is the x coordinate of point

5.18 and 5.19 can be written as

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Stage-to-stage calculations for countercurrent extraction.1.is a point common to all streams passing each

other, such as L0and V1, Lnand Vn+1, Lnand Vn+1,LNand VN+1, and so on.

2. This coordinates to locate thisoperating point

are given for x cand x A in eqn. 5.21. Since the

end points VN+1, LNor V1, and L0are known, xcan

be calculated and point located.

3. Alternatively, thepoint is located graphically in

the figure as the intersection of lines L0V1and LNVN+1.

4. In order to step off the number of stages using

eqn. 5.22 we start at L0 and draw the line L0,

which locates V1on the phase boundary.

5. Next a tie line through V1locates L1, which is in

equilibrium with V1.

6. Then line L1 is drawn giving V2. The tie line

V2L2 is drawn. This stepwise procedure is

repeated until the desired raffinate composition LN

is reached. The number of stages N is obtained to

perform the extraction.

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Ex 5.6 Pure isopropyl ether of 450 kg/h is being used to extract an aqueoussolution of 150 kg/h with 30 wt% acetic acid (A) by countercurrent multistage

extraction. The exit acid concentration in the aqueous phase is 10 wt%.

Calculate the number of stages required.Solution: The known values are VN+1= 450, yAN+1= 0, yCN+1= 1.0, L0= 150, xA0

= 0.30, xB0= 0.70, xC0= 0, and xAN= 0.10.

1. The points VN+1, L0, and LNare plotted in Fig. below. For the mixture point M,

substituting into eqs. 5.12 and 5.13, xCM= 0.75 and xAM= 0.075.

2. The point M is plotted and V1is located at the intersection of line LNM with the

phase boundary to give yA1= 0.072 and yC1= 0.895. This construction is not shown.

3. The lines L0V1and LNVN+1are drawn and the intersection is the operating point

as shown.

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1. Alternatively, the coordinates of can

be calculated from eq. 5.21 to locate

point.

2. Starting at L0

we draw line L0

, which

locates V1. Then a tie line through V1

locates L1 in equilibrium with V1. (The

tie-line data are obtained from an

enlarged plot.)

3. Line L1is next drawn locating V2. A tie

line through V2gives L2.

4. A line L2is next drawn locating V2. Atie line through V2gives L2.

5. A line L2gives V3.

6. A final tie line gives L3, which has gone

beyond the desired LN. Hence, about

2.5 theoretical stages are needed.


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Solution: The given values are L0= 100 kg/h, x0= 0.010, VN+1= 200 kg/h, yN+1=0.0005, xN= 0.0010. The inert streams are

hrwaterkgxLxLL /0.99)010.01(100)1()1( 00

hrosenekgyVyVV NN /ker9.199)0005.01(200)1()1( 11/

Making an overall balance on A using eq. 5.23 and solving, y1= 0.00497.

These end points on the operating line are plotted in Fig. below. Since the

solutions are quite dilute, the line is straight. The equilibrium line is alsoshown. The number of stages are stepped off, giving N = 3.8 theoreticalstages.

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