Chapter 5 Liquid-Liquid Extraction
Subject: 1304 332 Unit Operation in Heat transfer transf er Instructor: Chakkrit Umpuch Department of Chemical Engineering Faculty of Engineering Ubon Ratchathani University
Here is what you will learn in this chapter. 5.1 Introduction to Extraction Processes 5.2 Equilibrium Relations in Extraction 5.3 Single Single-- Stage Stage Equilibri Equilibrium um Extraction Extraction 5.4 Equipment for Liquid-Liquid Extraction 5.5 Continuous Multistage Countercurrent Extraction
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5.1 Introduction to Extraction Processes ³W hen
separation by distillation is ineffective or very difficult e.g. closeboiling mixture, liquid extraction is one of the main alternative to consider.´
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 solution containing containing little solute. solute.
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5.1 Introduction to Extraction Processes Liquid-liquid extraction principle
W hen
Liquid-liquid extraction is carried out in a test tube or flask the two immiscible phases are shaken together to allow molecules to partition (diss dissol olve ve)) into into the the pref prefer erre red d solv solven entt phas phase. e. 4
5.1 Introduction to Extraction processes An
example of extraction: Extract
Acetic
acid in H 2O
Organic layer contains most of acetic acid in ethyl acetate with a small amount of water.
+ Raffinate Ethyl acetate
Aqueous layer contains a weak acetic acid solution with a small amount of ethyl acetate.
The amount of water in the extract and ethyl acetate in the raffinate depends upon their solubilites in one another.
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5.2 Single-stage liquid-liquid extraction processes Triangular coordinates and equilibrium data Each 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 from the point M to the base AB represents the mass fraction xC. The distance to the base CB represents x A, and the distance to base AC represents xB. x A + xB + xC = 0.4 + 0.2 + 0.4 = 1
Equilateral triangular diagram
xB = 1.0 - x A - xC
( A and B are partially miscible.) yB = 1.0 - y A - yC 6
Liquid-Liquid phase diagram where components A and B are partially miscible.
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.
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Ex 5.1 Define the composition of point A, B, C, M, E, R, P and DEPRG in the ternary-mixture. 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 second phase occurs when mixture composition is 10% water, 45% furfural, 45% glycol or on the saturation curve.
Liquid-Liquid equilibrium, ethylene glycol-furfural-water, 25ºC,101 kPa. 9
Equilibrium data on rectangular coordinates The system acetic acid ( A) ± water (B) ± isopropyl ether solvent (C). The solvent pair B and C are partially miscible. xB = 1.0 - x A - xC yB = 1.0 - y A - yC
Liquid-liquid p hase diagram
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EX 5.2
An original mixture weighing 100 kg and containing 30 kg of isopropyl ether (C), 10 kg of acetic acid (A), and 60 kg water (B) is equilibrated and the equilibrium phases separated. W hat are the compositions of the two equilibrium phases? Solution: Composition of original mixture is xc= 0.3, x A = 0.10, and xB = 0.60.
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Liquid-liquid p hase diagram
1. Composition of xC = 0.30, x A = 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 y A = 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 x A = 0.12, xC = 0.02, and xB = 1.00 ± 0.12 ± 0.02 = 0.86.
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Phase diagram where the solvent pairs B-C and A-C are partially miscible. ³The
solvent pairs B and C and also A and C are partially miscible.´
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5.3 Single-Stage Equilibrium Extraction Derivation
of lever-arm rule for graphical addition Extract layer =V Raffinate Layer= L
An
overall mass balance:
A balance
on A:
W here x AM A balance
V L ! M Vy A Lx A
!
Mx A M
5.1
5.2
is the mass fraction of A in the M stream.
on C:
VyC LxC ! MxC M
5.3
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Derivation
of lever-arm rule for graphical addition L
Sub 5.1 into 5.2
Sub 5.1 into 5.3
V
L V
Sub 5.1 into 5.3
!
!
y A x A M x A M x A
y C xC M xC M x A
xC xC M x A x A M
!
xC M yC x A M y A
(5.4)
(5.5)
(5.6)
Eqn. 5.6 shows that points L, M, and V must lie on a straight line.
Lever arm¶s rule
L( kg ) V ( kg )
L (kg ) M (kg )
!
!
V M
(5.7)
L M
V M L V
(5.8) 15
Ex 5.3 The compositions of the two equilibrium layers in Example 5.1 are for the extract layer (V) y A = 0.04, yB = 0.02, and yC = 0.94, and for the raffinate layer (L) x A = 0.12, xB = 0.86, and xC = 0.02. The original mixture contained 100 kg and x AM = 0.10. Determine the amounts of V and L. Solution:
Substituting into eq. 5.1
V L ! M ! 100 Substituting into eq. 5.2, where M = 100 kg and x AM = 0.10,
V (0.04) L(0.12) ! 100(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, L L h g 4.2
M
!
100
!
g i
!
5.8
Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the material-balance method.
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5.2 Single-stage liquid-liquid extraction processes Single-state equilibrium extraction We now study the separation of A from a mixture of A and B by a solvent C in a single equilibrium stage.
An
L0
overall mass balance:
A balance
on A:
L0 x A0
A balance
on C:
L0 xC 0
V 2
!
L1
V 2 y A2
!
L1 x A1
V 2 y C 2
!
L1 xC 1
x A
x B
xC
V 1
!
!
5.9
M
V 1 y A1
!
Mx A M
5.10
V 1 y C 1
!
MxC M
5.11
1.0 17
To
solve the three equations, the equilibrium-phase-diagram is used.
1. L0 and V2 are known. 2. We calculate M, x AM, and xCM by using equation 5.9-5.11. 3. Plot L0, V2, M in the Figure. 4. Using trial and error a tie line is drawn through the point M, which locates the compositions of L1 and V1. 5. The amounts of L1 and V1 can be determined by substitution in Equation 5.9-5.11 or by using leverarm rule.
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Ex 5.4 A mixture weighing 1000 kg contains 23.5 wt% acetic acid ( A) and 76.5 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 the extract and raffinate phases. Solution Given:
L0
!
1000 kg and V 2
500kg
!
Given: L0
x A0
V 2
!
!
1000 500 ! M ! 1500 kg
0.235, x B 0
!
0.765 and y A2
L0 x A0 V 2 y A2
!
1.0
Mx A M
!
(1000)(0.235) (500)(0) ! (1500) x A M x A M Given: xc 0
!
!
1 x A0 x B 0
0.157 !
1.0 0.235 0.765 ! 0
L0 xC 0 V 2 yC 2
!
MxC M
(1000)(0) (500)(1) ! (1500) xC M
xCM
!
0.33
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V2 (0,1) = (yA2, yC2) V1 (0.1,0.89) = (yA1, yC1) M(0.157,0.33)
= (xAM, x CM)
L1(0.2,0.03) = (xA1, x C1) M
L0(0.235,0) = (xA0, xC0)
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From
the graph: xA1 = 0.2 and yA1 = 0.1;
L1 x A1 V 1 y A1 ! Mx A M L1 (0.2) V 1 (0.1) ! (1500)(0.157) L1 0.5V 1 From
!
1,177.5
(1)
the graph: xC1 = 0.03 and yC1 = 0.89;
L1 xC 1 V 1 yC 1
!
MxC M
L1 (0.03) V 1 (0.89) ! (1500 )(0.33) L1 29.67V 1
!
(2)
16,500
Solving eq(2) and eq(3) to get L 1 and V1;
L1
!
914.86kg
x A1 ! 0.2, y A1
!
and V 1 0.1, xC 1
!
!
525.28kg
0.03 and yC 1
!
0.89
Answer
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5.3 Equipment for Liquid-Liquid Extraction Introduction and Equipment Types As in the separation processes of distillation, the two phases in liquidliquid extraction must be brought into intimate contact with a high degree 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 equipment
Mixing by mechanical agitation Mixing by fluid flow themselves
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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 24
Packed and Spray Extraction Towers
Spray-type extraction tower
Packed extraction tower 25
5.4 Continuous multistage countercurrent extraction Countercurrent process and overall balance
An
L0
overall mass balance:
A balance
on C:
L0 xC 0
Combining 5.12 and 5.13
Balance on component A gives
V N
1 !
L N
V 1
!
M
V N 1 yC N
1 !
L N xC N
V 1 yC 1
xC M x A M
!
!
L0 xC 0
L0 L0 x A0
L0
V N 1 y C N 1
V N
!
1
V N 1 y A N
5.12
1
V N
1
!
!
MxC M
L N xC N
V 1 yC 1
L N
V 1
L N x A N
V 1 y A1
L N
V 1
5.13
5.14
5.15
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5.4 Continuous multistage countercurrent extraction Countercurrent process and overall balance
1. Usually, L0 and VN+1 are known and the desired exit composition x AN is set. 2. Plot points L0, VN+1, and M as in the figure, a straight line must connect these three 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 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. Solution: The given values are VN+1 = 600kg/h, y AN+1 = 0, yCN+1 = 1.0, L0 = 200kg/h, x A0 = 0.30, xB0 = 0.70, xC0 = 0, and x AN = 0.04. In figure below, VN+1 and L0 are plotted. Also, since LN is on the phase boundary, it can be plotted at x AN = 0.04. For the mixture point M, substituting into eqs. below,
xC M
!
L0 xC 0 V N 1 yC N
1
L0 V N
!
1
x A M
!
L0 x A0 V N 1 y A N
1
L0 V N
1
!
200(0) 600(1.0) 200 600
!
200(0.30) 600(0) 200 600
0.75
!
0.075
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Using these coordinates, 1) Point M is plotted in Figure below. 2) We locate V1 by drawing a line from LN through M and extending it until it intersects the phase boundary. This gives y A1 = 0.08 and yC1 = 0.90. 3) For LN a value of xCN = 0.017 is obtained. By substituting into Eqs. 5.12 and 5.13 and solving, LN = 136 kg/h and V1 = 664 kg/h.
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Stage-to-stage calculations for countercurrent extraction.
L0
Total mass balance on stage 1
Ln 1
Total mass balance on stage n
is constant and for all stages
( x (
!
L0
From 5.16 obtain difference in flows
( ! L0 V 1
L0 x0 V 1 y1
!
!
V 2
V n
Ln x n V n 1 y n
L1
1 !
V 1
Ln V n
!
!
V 1
5.16
V n
5.17
Ln
L1
V 2
!
(
5.18
1 !
L N V N
1 !
L N x N V N 1 y N
1 !
....
5.19
1 !
...
5.20
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Stage-to-stage calculations for countercurrent extraction.
x is the x coordinate of point
x (
!
L0 x 0 V 1 y1 L0 V 1
!
Ln x n V n 1 y n
Ln V n
1
1
!
L N x N V N 1 y N
1
5.21
( V N 1
5.22
L N V N
1
5.18 and 5.19 can be written as
L0
!
( V 1
Ln
!
(
V n
1
L N
!
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S tage-to-stage calculations for countercurrent extraction.
1. is a point common to all streams passing each other, such as L0 and V1, L n and Vn+1, L n and Vn+1, LN and VN+1, and so on. 2. This coordinates to locate this operating point are given for x c and x A in eqn. 5.21. Since the end points VN+1, LN or V1, and L0 are known, x can be calculated and point located. 3. Alternatively, the point is located graphically in the figure as the intersection of lines L 0 V1 and LN VN+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 V 1 on the phase boundary. 5. Next a tie line through V 1 locates L1, which is in equilibrium with V 1. 6. Then line L1 is drawn giving V2. The tie line V2L2 is drawn. This stepwise procedure is repeated until the desired raffinate composition L N is reached. The number of stages N is obtained to perform the extraction. 32
Ex 5.6 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: The known values are VN+1 = 450, y AN+1 = 0, yCN+1 = 1.0, L0 = 150, x A0 = 0.30, xB0 = 0.70, xC0 = 0, and x AN = 0.10. 1. The points V N+1, L0, and LN are plotted in Fig. below. For the mixture point M, substituting into eqs. 5.12 and 5.13, xCM = 0.75 and x AM = 0.075. 2. The point M is plotted and V 1 is located at the intersection of line LNM with the phase boundary to give y A1 = 0.072 and yC1 = 0.895. This construction is not shown. 3. The lines L0V1 and LNVN+1 are 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 L1 is next drawn locating V2. A tie line through V2 gives L2. 4. A line L2 is next drawn locating V2. A tie line through V2 gives L2. 5. A line L2 gives 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 extraction Countercurrent-Stage Extraction with Immiscible Liquids If the solvent stream V N+1 contains components A and C and the feed stream L 0 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 L 0 to VN+1.
¨ x 0 ¸ ¨ y N 1 ¸ ¨ x N ¸ ¨ y1 ¸ ¹¹ V d ©© ¹¹ ! L d ©© ¹¹ V d ©© ¹¹ 1 1 1 1 x y x y ª 0 º ª N 1 º ª N º ª 1 º
©© L d
5.23
¨ x0 ¸ ¨ y n 1 ¹¹ V d ©© ª 1 x 0 º ª 1 y n
©© L d
¸ ¨ x n ¸ ¨ y1 ¸ ¹¹ ! L d ©© ¹¹ V d ©© ¹¹ 1 º ª 1 x n º ª 1 y1 º
5.24
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. W here
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Ex 5.7 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
0.001010
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 Ld ! L(1 x ) ! L0 (1 x0 ) ! 100(1 0.010) ! 99.0 kg water / hr
V /
!
V (1 y) ! V N 1 (1 y N 1 ) ! 200(1 0.0005) ! 199.9kg ker osene / hr
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 also shown. The number of stages are stepped off, giving N = 3.8 theoretical stages.
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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. W hat percent of the acetic acid is removed?
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