cdb2013 liquid liquid extraction lecture 1
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Separation Process ICDB2013
Dr Chew Thiam Leng
Chapter –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. Discuss the principles of liquid-liquid processes andbasic design considerations.
2. Able to estimate the exit stream amounts andcompositions of single stage extractor.
3. Able to conduct total and component material balanceon multistage liquid-liquid extraction equipment.
4. Able to estimate the number of stages required toachieve the desired separation in liquid-liquidextraction processes.
Recap of pervious lesson
Determination of multiple stages required for a desired
separation using
Material balance and graphical method
Analytical Equations (Kremser Equation)
• Countercurrent Multiple-Contact Stages
• General design considerations
• Working principles of absorption and desorption (stripping)
• Gas-liquid equilibrium
Introduction
• Liquid-liquid extraction (solvent extraction): is the separation of theconstituents of a liquid solution by contact with another insolubleliquid.
Important terms
• The solution which is to be extracted is called the feed,
• the liquid with which the feed is contacted is called the solvent.
• The solvent rich product of the operation is called the extract,the residual liquid from which the solute has been removed isthe raffinate.
• It is a mass transfer operation in which a liquid solution (feed) iscontacted with an immiscible or nearly immiscible liquid (solvent)that exhibits preferential affinity or selectivity towards one or moreof the components in the feed.
Working Principles of Liquid-liquid extraction
When Liquid-liquid extraction is carried out in a test tube or flaskthe two immiscible phases are shaken together to allow moleculesto partition (dissolve) into the preferred solvent phase.
7
CSOLVENT
A+BFEED
EXTRACT
B+A+C
C+A+B
RAFFINATE
• In liquid-liquid extraction, a soluble component (the
solute) moves from one liquid phase to another.
• The two liquid phases must be either immiscible, or
partially miscible.
usually isothermal and isobaric
can be done at low temperature
(good for thermally fragile solutes, such as large organic molecules or
biomolecules)
extracting solvent is usually recycled, often by
distillation.
can be single stage (mixer-settler) or multistage
(cascade)
• An example of extraction:
Acetic acid (A)
in Water (B) +Ethyl acetate
(C)
Extract (C+A+B)Organic layer contains most of
acetic acid in ethyl acetate with a small amount of water.
Raffinate (B+C+A)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 solubility's in one another.
9
10
Laboratory separation funnel
Typical liquid-liquid extraction equipment
Mixer-settler for Extraction
Typical liquid-liquid extraction equipment
12
b) Combined mixer-settlera) Separate mixer-settler
Mixer-settler for Extraction
Typical liquid-liquid extraction equipment
13
Mixer-settler for Extraction
Typical liquid-liquid extraction equipment
14
Advantages Disadvantages
Reliable scale up techniques. Large amounts of floor space required.
Can handle high flow rates. Only systems with a few stages are economical.
Ideal for processes with slow mass transfer.
Large amounts of valuable material, such as the solvent, are tied up during the process.
Intense mixing, resulting in small droplets for good mass transfer.
Poor solvent conservation.
Spray Extraction Tower
Typical liquid-liquid extraction equipment
Spray Extraction Tower
Vessel in which the mixingis done by the flow of thefluid themselves,
mixing and settlingproceed continuously andsimultaneously.
Spray Extraction Tower
Typical liquid-liquid extraction equipment
Advantages Disadvantages
It has no internal parts except for the spray nozzles and hence eliminate the scale buildup and plugging problem.
Inefficient.
Low cost
The column consists of
• a packed bed,
• distributors for
the two liquid
Typical liquid-liquid extraction equipment
Packed Extraction Tower
Typical liquid-liquid extraction equipment
Packed Extraction Tower
Advantages Disadvantages
Highly efficient when only a few stages are needed.
Suffer from solids plugging
The random packings in the tower cause the droplets to coalesce and redisperse at frequent intervals throughout the tower
The initial dispersion is a concern for packed columns, to avoid channeling
Reduced axial mixing Hard to scale up accurately
rotating-disk contactor
a. agitator; b. stator disk
Typical liquid-liquid extraction equipment
Mechanically agitated extraction tower
VesselWalls
Shaft
Stators RotorsLightPhase In
HeavyPhase In
LightPhase Out
HeavyPhase Out
Drive Motor Gearbox
InterfaceControl
Interface
Mixer-settlers operate with a purely stage-wise contact.After every mixer there is a settler. Mixer-settlers can beoperated in a multistage, co- or countercurrent fashion.
Typical liquid-liquid extraction equipment
Mechanically agitated extraction tower
Advantages Disadvantages
Mechanical parts result in good dispersion.
Repair of internal mechanical parts can cause process delays.
Little axial mixing compared to non-agitated columns
General Design Considerations
Some factors that influence extraction processes are:
• Type of stage configurations
• Minimum solvent flow rate and actual flow as multiple ofminimum flow rate (or reflux for more than one stage system)
• Operating conditions
• Choice of liquid solvents
• Number of equilibrium stages
• Emulsification and scum formation tendencies
• Phase-density difference
• Interfacial tension
• Type of extractor
General Design Considerations
Some characteristics of an ideal solvent:
High selectivity for targeted solute to be extracted
High capacity for dissolving solute
Mininum or no corrosion problem.
Large density difference with carrier
High thermal and chemical stability
No or low toxicity
Low price
Equilibrium relations in extraction
• Extraction involves the use of systems of at least
three substances. The most important notation
scheme for extraction are:
1. Liquid B and C are pure, substantially insoluble
liquids, and A is the distributed solute.
2. Mixtures to be separated by extraction composed of A
and B where as C is the extracting solvent.
3. Equilateral triangular coordinates are often used to
represent the equilibrium data for a three component
systems as shown in the figure;
Fig.: Triangular diagram
Ternary relationship
xA + xB + xC = 0.40 + 0.20 + 0.40 = 1.0
3. Any point on the side of the triangle
represents a binary mixture
Point D for example contains a
fraction of 0.8 A and 0.2 B.
4. The perpendicular distance from any point such as M to the base ABrepresents the mass fraction of C in the mixture at M, the distance tothe base AC of B, and that to the base CB the mass fraction of A. Thus,at point M;
1. Each altitude of the trianglerepresents 100% pure componentand the distance to the three sidesrepresents the percentages or thefractions of the three components.
2. Point M represents the mixture of
the three components
Ternary relationship when A and B are partially miscible phase
• This is the most common type of systems in extraction
Example:
i. water (A)-chloroform(B)-acetone (C) and
ii. methyl isobutyl ketone (A)-water (B)- acetone (C)
iii. benzene (A)-water (B)-acetic acid (C)
26
• Liquid C dissolves completely in A and B, but Liquid A is only slightlysoluble in B and B slightly soluble in A.
• The more insoluble the liquid A and B , the nearer the apexes of the
triangle will points L and K be located.
• The two phase region is included inside below the curved envelope.
• Point M represents a mixture of the two phases.
Fig.: Liquid-liquid phase diagram where components A & B are partially miscible
• Any mixture outside the curve is ahomogenous mixture of one phase liquid.
Any ternary mixture underneath the curve, such as M, will form two
insoluble saturated liquid phase of equilibrium compositions indicated
by a (A-rich) and b (B-rich).
• 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.
• Curve LPK is a bimodal solubility
curve indicating the change in solubility
of the A-and B-rich phases upon addition
of C.
Reading ternary phase diagram
28
• Example: Define the composition of point A, B, C, M, E, R, P and DEPRGin the ternary-mixture.
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% Ethylene glycol, 40% water, 30% furfural
Point E = 43.8% glycol, 10% water, 46.2% furfural
Point R = 11% glycol, 81% water, 8% 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 or bimodal curve
29
Example 5.1Any point on the side of the triangle represents a binary mixture
30
From the diagram below determine the concentrations (mole fractions) at
Point M, E and R?
Example 5.2
31
•
•
•
Consider the point M:
Furfural content (xC):
0.19
0.20
0.61
check: xA + xB + xC = 1
Solution
water content (xA) :
Ethylene glycol content (xB):
Do the same for point E, R and P.
Equilibrium data on rectangular coordinates
Since triangular coordinates have some disadvantage due to the special
coordinates, in order to be able to expand one concentration scale relative
to the other, a more useful method of plotting the three-component data is
to use rectangular coordinates.
The solvent pair B and C are partially miscible.
The two phase region is inside the envelop.
yB = 1.0 - yA - yC
The concentration of B is obtained by differences from equation below
xB = 1.0 - xA - xC
The tie line gi is connecting the water-rich layer i
called the raffinate, x layer and the ether-rich layer, g called the extract layer, y
To construct the tie line gi using the equilibrium
yA-xA plot below the phase diagram , vertical line to g and I are drawn
Raffinate compositions are represented by
coordinates: (xA, xC)
Extract compositions are represented by
coordinates: (yA, yC)
The system acetic acid (A) – water (B) – isopropyl ether solvent (C).
Liquid-liquid Equilibrium phase diagram
34
1. Find mixing point, h
2. Find tie-line through h (trial
and error);
3. Find g and i at either end
(co-linear with h).
4. Find flow rates of g and i.
Overview of solution using
Rectangular diagram:
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. 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.
35
Example 5.3
36
1. Plot the composition of xC =
0.30, xA = 0.10 and xB on the
equilibrium diagram as at
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 –
0.12 – 0.02 = 0.86.
Solution
Equilibrium contact stages
Single stage equilibrium extraction
• As shown in Figure below the equation can be derived using rectangular coordinates.
• Two streams containing L kg and V kg containing component A, B, and C are mixed to give a resulting mixture M total mass.
• Writing an overall mass balance and balance on A and on C
VM
mixer
L
L + V = M
VyA + LxA = MxA,M = (V + L)xAM
VyC + LxC = MxCM = (V + L)xCM 5.4
5.3
5.2
AAM
AMA
xx
xy
V
L
• Similarly combining Eqs (5.2) and (5.4)
CMC
MCC
xx
xy
V
L
(5.5)
AAM
CMC
AMA
MCC
yx
yx
xx
xx
(5.6)
•Equating 5.5 and 5.6 and rearranging
• Combining Eqns (5.2) and (5.3)
V
M
mixerL
(5.7)
ML
MV
kgV
kgL
)(
)(
VL
MV
kgM
kgL
)(
)(
(5.8)
Lever arm’s rule
Eqn. 5.7 shows that points L, M, and V must lie on a straight line.
By using the properties of similar right triangles,
(5.9)
V•
M•
L•
similar triangles
The compositions of the two equilibrium layers in Example 5.1 are for the extract layer (V2) yA = 0.04, yB = 0.02, and yC = 0.94, and for the raffinate layer (L1) xA = 0.12, xB = 0.86, and xC = 0.02. The original mixture contained 100 kg and xAM = 0.10. Determine the amounts of V1 and L0.
Solution:
10012 MLV
)10.0(100)12.0()04.0( 12 LV
40
Substituting into eq. 5.3, where M = 100 kg and xAM = 0.10,
Substituting into eq. 5.2
V1V2
mixer
L0 L1
Example 5.4
Given:
extract layer (V2): yA = 0.04, yB = 0.02,
raffinate layer (L1) xA = 0.12, xB = 0.86,
M= 100 kg, xAM = 0.10
41
Solution:
)(10012 aMLV
)10.0(100)12.0()04.0( 12 LV
Substituting into eq. 5.3, where M = 100 kg and xAM = 0.10,
Substituting into eq. 5.2
V1V2
mixer
L0 L1
Given:
Extract layer (V2): yA = 0.04, yB = 0.02,
Raffinate layer (L1) xA = 0.12, xB = 0.86,
M= 100 kg, xAM = 0.10
Required: V1 and L0
)(10121.004.0 12 bLV
8.5
2.4
100
ig
ghL
M
L
Solving, L = 72.5 kg and V = 27.5 kg, which is a reasonably close check on the material-balance method.
•Solving eqns (a) and (b) simultaneously,
L1 = 75.0 and V2 = 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,
g•
h•
i•
similar triangles
Single-stage liquid-liquid extraction processes
Single-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 single equilibrium
stage.
0.1 CBA xxx
An overall mass balance:
A balance on A:
A balance on C:
5.10
5.11
5.12
43
V1V2
L0 L1
To solve the three equations, the equilibrium-phase-diagram is used.
1. L0 and V2 are known.
2. We calculate M, xAM, and xCM by using
equation 5.10-5.12.
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
5.10-5.12 or by using lever-arm rule.
44
V1V2
L0 L1
Example 5.5
45
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.
Given:
10,0500
0765.0,235.0,1000
2222
0
CBA
CoBoAo
yandyykgV
xandxxkgL
AMx)1500()0)(500()235.0)(1000(
kgMVL 1500500100020
157.0AMx
MCCC MxyVxL 2200
Similarly 0765.0235.00.11 000 BAc xxx
AMAA MxyVxL 2200
MCx)1500()1)(500()0)(1000(
33.0CMx
V1V2
L0 L1
Solution
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)
47
1. L0 and V2 are known.
2. We calculate M, xAM, and xCM
by using equation 5.10-5.12.
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 5.10-
5.12 or
• Alternatively, by using lever-
arm rule.
(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(1) and eq(2) to get L1 and V1;
kgVandkgL 28.52586.914 11
89.003.0,1.0,2.0 1111 CCAA yandxyx Answer
48
49
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? Use
equilibrium data given below.
Exercise
50
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
Draw the equilibrium diagram and Locate point,
L0, V1 and M
51
(i)
AMAA MxyVxL 1111
)175.0)(800()12.0()22.0( 11 VL
• From the graph: xA1 = 0.22 and yA1 = 0.12;
36.63654.0 11 VL
V2 (0,1) = (yA2, yC2)
V1 (0.12,0.87) = (yA1, yC1)
L1(0.22,0.03) = (xA1, xC1)
L0(0.35,0) = (xA0, x
M(0.175,0.5) = (xAM, xCM)
M
• Using trial and error, the tieline is drawn through pointM that determines thecomposition of LN and V1
MCCC MxyVxL 1111
)5.0)(800()87.0()03.0( 11 VL
From the graph:
xC1 = 0.03 and
yC1 = 0.87;
(ii) 33.333,129 11 VL
Solving eq(i) and eq(ii) to get L1 and V1;
kgVandkgL 49.2412.623 11
88.003.0,12.0,23.0 1111 CCAA yandxyx
52
53
ML
MV
kgV
kgL
1
1
1
1
)(
)(
11
11
)(
)(
VL
MV
kgM
kgL
Lever arm’s 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
)(
)(
1
1 kgV
kgL
44.086.0
38.0
)(
)(1 kgM
kgL kg352kg80044.0)kg(1 L
kg57.43481.0/kg352)(1 kgV
The amount and composition of extract are
*57.4341 kgV
74.003.0,23.0 111 BCA xandxx
54
*3251 kgL
0
88.0
,12.0
1
1
1
B
C
A
y
andy
y
The amount and composition of raffinate are
*The correction of amount of extract and raffinate depends on the
identification of points on the graph (setting the pint). The value obtained from Lever arm’s rule is more reliable.
55
The percent of acetic acid removed is about
%𝑅𝑒𝑚𝑜𝑣𝑒𝑑
=𝑎𝑚𝑜𝑢𝑛𝑡 𝑜𝑓 𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑 𝑖𝑛 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑 − 𝑎𝑚𝑜𝑢𝑛𝑡𝑜𝑓𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑𝑖𝑛 𝑡ℎ𝑒 𝑟𝑎𝑓𝑖𝑛𝑎𝑡
𝑎𝑚𝑜𝑢𝑛𝑡𝑜𝑓𝑎𝑐𝑒𝑡𝑖𝑐 𝑎𝑐𝑖𝑑𝑖𝑛 𝑡ℎ𝑒 𝑓𝑒𝑒𝑑
%𝑅𝑒𝑚𝑜𝑣𝑒𝑑 =(400x0.35−325x0.23)x100%
400𝑥0.35= 𝟒𝟔. 𝟔%
56
3. A feed mixture weighing 200 kg of unknown composition
containing water, acetic acid, and isopropyl ether is contacted
in a single stage with 280 kg of a mixture containing 40 wt %
acetic acid, 10 wt % water, and 50 wt. % isopropyl ether. The
resulting raffinate layer weighs 230 kg and contains 29.5 wt. %
acetic acid,66.5 wt. % water, and 4 wt. % isopropyl ether.
Determine the original composition of the feed mixture and the
composition of the resulting extract layer. (Use Equilibrium
data Given below
57
TABLE : Acetic acid-water-isopropanol ether liquid-liquid
equilibrium data
Water layer (wt%) Isopropyl layer (wt%)
Acetic acid Water
Isopropyl ether Acetic acid Water
Isopropyl ether
0 98.8 1.2 0 0.6 99.4
0.69 98.1 1.2 0.18 0.5 99.3
1.41 97.1 1.5 0.37 0.7 98.9
2.89 95.5 1.6 0.79 0.8 98.4
6.42 91.7 1.9 1.93 1.0 97.1
13.30 84.4 2.3 4.82 1.9 93.3
25.50 71.1 3.4 11.4 3.9 84.7
36.70 58.9 4.4 21.6 6.9 71.5
44.30 45.1 10.6 31.10 10.8 58.1
46.40 37.1 16.5 36.20 15.1 48.7