single stage.pdf

8/14/2019 single stage.pdf

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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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5.1 Introduction to Extraction ProcessesWhen separation by distillation is ineffective or very difficult e.g. close-boiling mixture, liquid extraction is one of the main alternative toconsider.

What is Liquid-liquid extraction(or solvent extraction)?

Liquid-Liquid extraction is a mass transfer operation in which a liquid

solution (feed) is contacted with an immiscible or nearly immiscible liquid

(solvent) that exhibits preferential affinity or selectivity towards one or

more of the components in the feed. Two streams result from this contact:

a)Extract is the solvent rich solution containing the

desired extracted solute.

b) Raffinate is the residual feed solution containing little solute.

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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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Equilateral triangular diagram

(A and B are partially miscible.)

Triangular coordinates and equilibrium dataEach of the three corners

represents a pure component A,

B, or C.

Point M represents a mixture of

A, B, and C.

The perpendicular distance fromthe point M to the base AB

represents the mass fraction xC.

The distance to the base CB

represents xA, and the distanceto base AC represents xB.xA+ xB+ xC= 0.4 + 0.2 + 0.4 = 1xB= 1.0 - xA- xCyB= 1.0 - yA- yC

5.2 Single-stage liquid-liquid extraction processes

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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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Liquid-liquid phase diagram

1. Composition of xC = 0.30, xA

= 0.10 is plotted as point h.2. The tie line gi is drawn

through point h by trial and

error.

3. The composition of the

extract (ether) layer at g is yA

= 0.04, yC = 0.94, and yB =1.00 - 0.04 - 0.94 = 0.02

mass fraction.

4. The raffinate (water) layer

composition at i is xA = 0.12,

xC = 0.02, and xB = 1.00

0.120.02 = 0.86.

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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 addition5.3 Single-Stage Equilibrium Extraction

MLV An overall mass balance:

MCCC MxLxVy

AMAA MxLxVy A balance on A:

A balance on C:

5.1

5.2

5.3Where xAMis the mass fraction of A in the M stream.

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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

Equating 5.4 and5.5 and rearranging

(5.7)

(5.8)

Lever arms rule

Eqn. 5.6 shows that points L, M, and V must lie on a straight line. Byusing the properties of similar right triangles,

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Ex 5.3 The compositions of the two equilibrium layersin 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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To solve the three equations, the equilibrium-phase-diagram is used.

1. L0and V2are known.

2. We calculate M, xAM, and xCMby

using equation 5.9-5.11.

3. Plot L0, V2, M in the Figure.

4. Using trial and error a tie line isdrawn through the point M, which

locates the compositions of L1and V1.

5. The amounts of L1and V1can be

determined by substitution in

Equation 5.9-5.11 or by using lever-arm rule.

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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

0.1765.0,235.0200

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

MVLVL NN 110

MCCNCNNCNC MxyVxLyVxL 111100

1

11

10

1100

VLyVxL

VLyVxLx

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 5.12 and 5.13

Balance on component A gives

5.125.13

5.14

5.15

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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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5.4 Continuous multistage countercurrent extractionCountercurrent-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, thestage calculations are made more easily. The solute A is relatively dilute and is beingtransferred from L0to 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. (5.24) 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.

5.23

5.24

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Ex 5.7 An inlet water solution of 100 kg/h containing 0.010 wt fractionnicotine (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 toreduce 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 andy in the kerosene.

X y x y

0.001010 0.000806 0.00746 0.00682

0.00246 0.001959 0.00988 0.00904

0.00500 0.00454 0.0202 0.0185

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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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39

Homework No.9

1. A single-stage extraction is performed in which 400 kg of a solution

containing 35 wt% acetic acid in water is contacted with 400 kg of pure

isopropyl ether. Calculate the amounts and compositions of the extract

and raffinate layers. Solve for the amounts both algebraically and by

the lever-arm rule. What percent of the acetic acid is removed?


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Homework No.10

1. Pure isopropyl ether is to be used to extract acetic acid from 400 kg of

a feed solution containing 25 wt% acetic acid in water.

(a) If 400 kg of isopropyl is used, calculate the percent recovery in theisopropyl solution in a one-stage process.

(b) If a multiple four-stage system is used and 100 kg fresh isopropyl is

used in each stage, calculate the overall percent recovery of the acid in

the total outlet isopropyl ether. (Hint: First, calculate the outlet extract

and raffinate streams for the first stage using 400 kg of feed solution

and 100 kg of isopropyl ether. For the second stage, 100 kg of

isopropyl ether contacts the outlet aqueous phase from the first stage.

For the third stage, 100 kg of isopropyl ether contacts the outlet

aqueous phase from the first stage. For the third stage, 100 kg of

isopropyl ether contacts the outlet aqueous phase from the secondstage, and so on.)


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Solution

AMx)800()0)(400()35.0)(400(

kgMVL 80040040020

0.165.0,35.0 200 ABA yandxx

Given:

175.0AMx

MCCC MxyVxL 2200

Given: 065.035.00.11 000 BAc xxx

AMAA MxyVxL 2200

MCx)800()1)(400()0)(400(

5.0CMx

Homework No.9 (Solution)

1. A single-stage extraction is performed in which 400 kg of a solution

containing 35 wt% acetic acid in water is contacted with 400 kg of pure

isopropyl ether. Calculate the amounts and compositions of the extractand raffinate layers. Solve for the amounts both algebraically and by

the lever-arm rule. What percent of the acetic acid is removed?


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42

M

V2 (0,1) = (yA2, yC2)V1(0.12,0.87) = (yA1, yC1)

L1(0.22,0.03) = (xA1, xC1)

L0(0.35,0) = (xA0, xC0)

M(0.175,0.5) = (xAM, xCM)


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(1)

AMAA MxyVxL 1111

MCCC MxyVxL 1111

)175.0)(800()12.0()22.0( 11 VL

)5.0)(800()87.0()03.0( 11 VL



(2)

36.63654.0 11 VL

33.333,129 11 VL


kgVandkgL 49.2412.623 11

88.003.0,12.0,23.0 1111 CCAA yandxyx Answer43


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44ML

MV

kgV

kgL

)(

)(

VL

MV

kgM

kgL )(

)(1

Lever arms rule

M

V1(0.12,0.88) = (yA1, yC1)

L1(0.23,0.03) = (xA1, xC1)

M(0.175,0.5) = (xAM, xCM)

81.047.0

38.0

)(

)(

kgV

kgL

44.086.038.0

)()(1

kgM

kgL kgxkgL 35280044.0)(1

kgkgV 57.43481.0/352)(


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The amount and composition of extract are

*57.4341 kgV

74.003.0,23.0 111 BCA xandxx Answer

45

*3251 kgL

088.0,12.0 111 BCA yandyy Answer

The amount and composition of raffinate are

The percent of acetic acid removed is about

=(400x0.35-325x0.23)x100% = 46.6%

400x0.35

Answer

*The correction of amount of extract and raffinate depends on theidentification of points on the graph. The value obtained from Levers arm

rule is more reliable personally.


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46

Homework No.10

1. Pure isopropyl ether is to be used to extract acetic acid from 400 kg of

a feed solution containing 25 wt% acetic acid in water.

(a) If 400 kg of isopropyl is used, calculate the percent recovery in theisopropyl solution in a one-stage process.

(b) If a multiple four-stage system is used and 100 kg fresh isopropyl is

used in each stage, calculate the overall percent recovery of the acid in

the total outlet isopropyl ether. (Hint: First, calculate the outlet extract

and raffinate streams for the first stage using 400 kg of feed solution

and 100 kg of isopropyl ether. For the second stage, 100 kg of

isopropyl ether contacts the outlet aqueous phase from the first stage.

For the third stage, 100 kg of isopropyl ether contacts the outlet

aqueous phase from the first stage. For the third stage, 100 kg of

isopropyl ether contacts the outlet aqueous phase from the secondstage, and so on.)


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47

Solution a):

AMx)800()0)(400()25.0)(400(

kgMVL 80040040020

0.175.0,25.0 200 ABA yandxx

Given:

125.0AMx

MCCC MxyVxL 2200

Given:075.035.00.11 000

BAc xxx

AMAA MxyVxL 2200

MCx)800()1)(400()0)(400(

5.0CMx


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48

M

V2 (0,1) = (yA2, yC2)V1(0.09,0.9) = (yA1, yC1)

L1(0.17,0.03) = (xA1, xC1)

L0(0.25,0) = (xA0, xC0)

M(0.125,0.5) = (xAM, xCM)


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(1)

AMAA MxyVxL 1111

MCCC MxyVxL 1111

)125.0)(800()12.0()17.0( 11 VL

)5.0)(800()9.0()03.0( 11 VL



(2)

24.58871.0 11 VL

33.333,130 11 VL


9.003.0,09.0,17.0 1111 CCAA yandxyx Answer49

kgVandkgL 43.2518.570 11


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50

Solution b):

AMx)500()0)(100()25.0)(400(

kgMVL 50010040020

0.175.0,25.0 200 ABA yandxx

Given:

2.0AMx

MCCC MxyVxL 2200

Given:075.035.00.11 000

BAc xxx

AMAA MxyVxL 2200

MCx)500()1)(100()0)(400(

2.0CMx

First stage


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51

M

V2 (0,1) = (yA2, yC2)V1(0.12,0.86) = (yA1, yC1)

L1(0.17,0.03) = (xA1, xC1)

L0(0.25,0) = (xA0, xC0)

M(0.2,0.2) = (xAM, xCM)


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(1)

AMAA MxyVxL 1111

MCCC MxyVxL 1111

)125.0)(800()12.0()17.0( 11 VL

)5.0)(800()9.0()03.0( 11 VL



(2)

24.58871.0 11 VL

33.333,130 11 VL


9.003.0,09.0,17.0 1111 CCAA yandxyxAnswer

52

kgVandkgL 43.2518.570 11

% recovery in isopropyl solution is 96 93%

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