1SEPARATION PROCESS ICDB 2013

January Semester 2015

Dr Sintayehu Mekuria Hailegiorgis

Liquid-liquid Extraction

Lesson outline Introduction

General design considerations

Working principles of liquid-liquid processes

Liquid-liquid equilibrium/equilibrium relation in extraction processes

Single-stage equilibrium extraction

Continuous multiple stage countercurrent extraction

Lesson outcome

At the end of the session, the students are able to:

1. Able to conduct total and component materialbalance on multistage liquid-liquid extractionequipment.

2. Able to estimate the number of stages required toachieve the desired separation in liquid-liquidextraction processes.

Recap of pervious lesson

Single stage stages liquid-liquid extraction

General design considerations and liquid-liquid extraction equipment types

Working principles of liquid-liquid extraction

Liquid-liquid equilibrium for designing extraction unit

Continuous multistage countercurrent extraction

In solvent extraction processes, in order to use less solventsand obtain more concentrated exit extracts, counter currentmultistage contacting is mostly employed.

The fundamental principles of multistage gas-liquid absorptionis similar to countercurrent multistage liquid-liquid extraction.

In multistage countercurrent extraction, the feed streamcontaining the solute A to be extracted enters at one end ofthe process and the solvent enters at the other end of theprocess.

The extract and raffinate flow counter currently from stage tostage, and the final products are the extract stream V1 leavingstage one and the raffinate stream LN leaving stage N.

Countercurrent process and overall balance

MVLVL NN 110

MCCNCNNCNC MxyVxLyVxL 111100

1

11

10

1100

VL

yVxL

VL

yVxLx

N

CNCN

N

NCNC

MC

1

11

10

1100

VL

yVxL

VL

yVxLx

N

AANN

N

ANNA

MA

An overall mass balance:

A balance on C:

Combining 5.132 and 5.14

Balance on component A gives

5.13

5.14

5.15

5.16

Countercurrent process and overall balance

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

desired exit composition xAN is set.

2.Eqs (5.15) and (5.16) can be used to calculate the coordinates of point M and phase diagram, which ties together the two entering streams Lo and VN+1 and the two exit streams V1 and LN.

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

figure, a straight line called mixing line

must connect these three points (points

are collinear).

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

called tie line. Also, LN and V1 must also

lie on the phase envelope.

8

Flow rates of LN and V1 are related by mass balance. Compositions of LN and

V1 are also related by equilibrium.

mixing line

tie-line

Pure solvent isopropyl ether at the rate of VN+1 = 600 kg/h is being used 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 and amounts of the ether extract V1 and the aqueous raffinate LN. Use equilibrium data from the table.

9

Example 5.6

Given: 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.

Solution:

1. In figure below, Plot VN+1 and L0.

2. Also, since LN is on the phase boundary, it can be plottedat xAN = 0.04.

3. For the mixture point M, substituting into eqs (5.15 and5.16), calculate the compositions of the mixing point,

Required: V1, LN, yA1, yB1, yC1 , xBN and xCN

Solution

75.0600200

)0.1(600)0(200

10

1100

N

NCNC

MCVL

yVxLx

075.0600200

)0(600)30.0(200

10

1100

N

ANNAMA

VL

yVxLx

11

Using these coordinates,

1) Point M is plotted in Figure below.

2) Draw the mixing line that connects L0,M and VN+1

3) Also, since LN is on the phase boundary, it can

be plotted at xAN = 0.04

4) We locate V1 by drawing a line from LN through Mand extending it until it intersects the phaseboundary. This gives

yA1 = 0.08 and yC1 = 0.90.

3) For LN a value of xCN = 0.017 isobtained.

4) by substituting into Eqs. 5.13 and5.14 and solving,

LN = 136 kg/h and

V1 = 664 kg/h.

Stage-to-stage calculations for countercurrent

extraction.

1120 VLVL

nnnn VLVL 11

2110 VLVL

Total mass balance on stage 1

Total mass balance on stage n

From 5.17 obtain difference in flows

5.17

5.18

5.19

12

The next step after an overall material balance has been made is to

go stage by stage to determine the concentration at each stage and

the total number of stages N needed to reach LN

111

1

11

10

1100

NN

NNNN

nn

nnnn

VL

yVxL

VL

yVxL

VL

yVxLx

10 VL 1 nn VL 1 NN VL

5.22

5.23

x is the x coordinate of point

Eqns (5.19) and 5.20) can be written as

13

....1110 NNnn VLVLVL

...11111100 NNNNnnnn yVxLyVxLyVxLx

5.20

5.21

in kg/h is constant and for all stages

Stage-to-stage calculations for countercurrent extraction.

14

1. is a point common to all streams

passing each other, such as L0 and V1,

L1 and V2, Ln and Vn+1, LN and VN+1,

and so on.

2. This coordinates to locate this

operating point are given for xc and

xA in eqn. 5.21. Since the end points

VN+1, LN or V1, and L0 are known, xcan be calculated and point located.

3. Alternatively, the point is located

graphically in the figure as the

intersection of lines L0V1 and

LN VN+1.

15

V2

V3

L2L1

6. Then line L1 is drawn giving V2. Thetie line V2L2 is drawn. This stepwiseprocedure is repeated until thedesired raffinate composition LN isreached. The number of stages N isobtained to perform the extraction.

4. In order to step off the number of stages

using eqn. 5.23 we start at L0 and draw

the line L0, which locates V1 on the

phase boundary.

5. Next a tie line through V1 locates L1,

which is in equilibrium with V1.

Example 5.7

16

Pure isopropyl ether of 450 kg/h is being used to extract an

aqueous solution 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:

Given: VN+1 = 450, yAN+1 = 0, yCN+1 = 1.0, L0 = 150, xA0 = 0.30, xB0 = 0.70, xC0 = 0, and xAN = 0.10.

Required: Calculate the number of stages required.

(yA1, yB1, yC1) (yA,N+1, yB,N+1, yC,N+1)

(xAo, xBo, xC0)(xAN, xBN, xCN)

18

4. The lines L0V1 and LNVN+1 are drawnand the intersection is the operatingpoint as shown.

1. The points VN+1, L0, and LN are plotted in Fig. below.

2. For the mixture point M, substituting into eqs. 5.12 and 5.13,

xCM = 0.75 and xAM = 0.075.

3. The point M is plotted and V1 islocated at the intersection of line LNMwith the phase boundary to give yA1 =0.072 and yC1 = 0.895. Thisconstruction is not shown.

Solution

LN the coordinates of can be calculated from

eq. 5.21 to locate point .

Alternatively,

M

19

L1L2L3 LN

Solution

5. Starting at L0 we draw line L0 ,

which locates V1. Then a tie line

through V1 locates L1 in equilibrium

with V1.

6. Line L1 is next drawn locating V2. A

tie line through V2 gives L2.

7. A line L2 is next drawn locating V2.

A tie line through V2 gives L2.

8. A line L2 gives V3.

9. A final tie line gives L3, which has

gone beyond the desired LN. Hence,

about 2.5 theoretical stages are

needed.

V2

V3

Continuous multistage countercurrent extraction

Countercurrent-Stage Extraction with Immiscible Liquids

1

1

1

1

0

0

1111 y

yV

x

xL

y

yV

x

xL

N

N

N

N

1

1

1

1

0

0

1111 y

yV

x

xL

y

yV

x

xL

n

n

n

n

If the solvent stream VN+1 contains components A and C and the feed

stream L0 contains A and B and components B and C are relatively

immiscible in each other, the stage calculations are made more easily. The

solute A is relatively dilute and is being transferred from L0 to VN+1.

Where L = kg inert B/h, V = kg inert C/h, y = mass fraction A in V stream, and

x = mass fraction A in L stream.

5.23

5.24

20

21

The number of stages arestepped off as shownpreviously for cases ofdistillation and absorption asshown in the Figure.

Eq.(5.24) is an operating-line equation whose slope L/V.

If y and x are quite dilute, the line will be straight when

plotted on an xy diagram.

An inlet water solution of 100 kg/h containing 0.010 wt fraction nicotine

(A) in water is stripped with a kerosene stream of 200 kg/h containing

0.0005 wt fraction nicotine in a countercurrent stage tower. The water and

kerosene are essentially immiscible in each other. It is desired to reduce

the concentration of the exit water to 0.0010 wt fraction nicotine.

Determine the theoretical number of stages needed. The equilibrium data

are as follows (C5), with x the weight fraction of nicotine in the water

solution and y in the kerosene.

X y x y

0.00101 0 0.000806 0.00746 0.00682

0.00246 0.001959 0.00988 0.00904

0.00500 0.00454 0.0202 0.0185

22

Ex ample 5.7

The given values are

L0 = 100 kg/h, xA0 = 0.010,

VN+1 = 200 kg/h, yAN+1 = 0.0005,

xAN = 0.0010.

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

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

Making an overall balance on A using eq. 5.23 and solving, yA1 = 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 also shown. The number of stages are stepped off,

giving N = 3.8 theoretical stages.23

Solution:

The inert streams are

24

25

End of Chapter Five