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7/23/2019 Chapter 2 Lle
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CHAPTER 2: LIQUID LIQUID EXTRACTION
By : SYAFIZA ABDHASHIB
Credit to : Pn Siti
Wahidah
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CHAPTER / CONTENT
Definition & Process Operation
LLE for Partially Miscible Solvent
LLE for Immiscible Solvent
Liquid liquid extraction equipment
Solvent selectivity & Phase Diagram
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The separation of constituents (solutes) of a liquid solution by contact with
another insoluble liquid.
Solutes are separated based on their different solubilities in different liquid.
The separation process of the components of a liquid mixture by treatmentwith a solvent in which one or more desired components is soluble.
There are two requirements for liquid liquid extraction to be feasible:
component (s) to be removed from the feed must preferentially
distribute in the solvent.the feed and solvent phases must be substantially immiscible
Definition & Application
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The simplest LLE involves only a ternary (i.e 3 component system)
Important terms you need to know:
Feed - The solution which is to be extracted(denoted by component A)
Solvent - The liquid with which the feed is contacted(denoted by component C)
Diluent - Carrier liquid(denoted by component B)
Extract - The solvent rich product of the operation
Raffinate - The residual liquid from which solutes has been removed.
Definition & Application
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In some operations, the solutes are the desired product, hence the extract stream isthe desirable stream. In other applications, the solutes my be contaminants that need tobe removed, and in this instance the raffinate is the desirable product stream.
Extraction processes are well suited to the petroleum industry because of the need toseparate heat sensitive liquid feeds according to chemical type (e.g aromatic,naphthenic) rather than by molecular weight or vapor pressure.
Application:
Major applications exist in the biochemical or pharmaceutical industry, whereemphasis is on the separation of antibiotics and protein recovery.
In the inorganic chemical industry, they are used to recover high boilingcomponents such as phosphoric acid, boric acid and sodium hydroxide fromaqueous solution.
Definition & Application
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Examples:
1. Extraction of nitrobenzene after reaction of HNO3with toluene inH2SO4
2. Extraction of methylacrylate from organic solution withperchlorethylene
3. Extraction of benzylalcohol from a salt solution with toluene.
4. Removing of H2S from LPG with MDEA
Definition & Application
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Examples:
5. Extraction of caprolactam from ammonium sulfate solutionwith benzene
6. Extraction of acrylic acid from wastewater with butanol
7. Removing residual alkalis from dichlorohydrazobenzene withwater
8. Extraction of methanol from LPG with water
9. Extraction of chloroacetic acid from methylchloroacetate withwater.
Definition & Application
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Definition & Application
LLE VS distillation process :
LLE depends on solubilities between the liquid componentsand produces new solution which in turn has to be separated
again, whereas;
Distillation depends on the differences in relative volatilities /vapor pressures of substances. Furthermore, it requires heataddition.
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Definition & Application
Advantages of LLE over distillation process:
1. Where distillation requires excessive amount of heat
2. Presence of azeotropes or low relative volatilities are involved (valuenear unity and distillation cannot be used)
3. Removal of a component present in small concentrations, e.g
hormones in animal oil.
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Definition & Application
Advantages of LLE over distillation process:
4. Recovery of a high boiling point component present in small
quantities in waste stream, e.g acetic acid from cellulose acetate.
5. Recovery of heat sensitive materials (e.g food) where low tomoderate processing temperatures are needed. Thermaldecomposition might occur.
6. Solvent recovery is easy and energy savings can be realized.
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Solvent selectivity
Selectivity (Separation factor).
If there are more than one solutes (sat two solutes A and B), then
consideration should be given to the selectivity of the solvent forsolutes A against B.
The selectivity between two solutes A and B is defined as the ratioof the distribution coefficient of A to the distribution coefficient of B.For all useful extraction operation the selectivity must exceed unity. Ifthe selectivity is unity, no separation is possible.
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Solvent selectivity
Insolubility of Solvent.
The solvent should have low solubility in the feed solution,otherwise the separation is not clean. For example, if there issignificant solubility of solvent in the raffinate stream, anadditional separation step is required to recover the solvent.
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Solvent selectivity
Recoverability.
It is always necessary to recover the solvent for re-use, and thismust ordinarily be done by other means, e.g distillation. If distillation isto be used, the solvent should form no azeotrope with the extracted
solute and mixtures should show high relative volatility. The solventshould also be thermally stable under the distillation temperature.
Density.
A large difference in density between extract and raffinate phases
permits high capacities in equipment. This is especially important forextraction devices utilizing gravity for phase separation.
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Solvent selectivity
Chemical Reactivity.
The solvent should be stable chemically and inert toward theother components of the system and toward the commonmaterials of construction.
Other Criteria.
Toxicity and flammability of the solvent are importantoccupational health and safety consideration.
Stability of the solvent (i.e resistance to breakdown), particularlyin the recovery steps, is significant, especially if the breakdownproducts might contaminate the products of the main separation.
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Design ofLLE tower
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1. SINGLE STAGE CALCULATIONS
2. MULTISTAGE COUNTER CURRENT SYSTEM
LLE for Partially Miscible Solvent
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Solvent and the solution are in contact with each other only once andthus the raffinate and extract are in equilibrium only once.
The solution normally binary solution containing solute (A) dissolvedin a diluent or carrier (B). The extracting solvent can be either puresolvent C or may content little A. Raffinate (R) is the exiting phase rich incarrier (B) while extract is exiting phase rich in solvent (C).
Single stage calculations
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When liquid solution mixed with solvent (C), an intermediate phaseM momentarily forms as the light liquid moves through the heavyliquid in the form of bubbles. These bubbles provide a large surface
area for contact between the solution and the solvent that speed upmass transfer process.
The raffinate and extract are in equilibrium with each other.
Single stage calculations
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Single stage calculations
Liquid-Liquid Extraction
y* (A)ys(A) Intermediate, M
Raffinate phase, R
x* (A)
Extract phase, E
xM (A)
F Mass of feed solution S Mass of extracting solventE Mass of extract phase R Mass of raffinate phaseM Mass of intermediate xF Mass fraction of A in FyS Mass fraction of A in S xM Mass fraction of A in Mx* Equilibrium mass fraction y* Equilibrium mass fraction of A in E
of A in R
Note: Intermediate shown just for purpose of demonstration. Dont have to draw it whenanswering the question
Feed Solution, F
xF(A)
Extracting Solvent, S
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In most single extraction, we are interested to determine theequilibrium composition and masses of raffinate and extract phases byusing ternary phase diagram and simple material balances.
Using material balance,
Single stage calculations
Calculate the mass of intermediate M using total material balance:MSF Eq. (1)
Determine mass fraction of solute A in intermediate M using materialbalance for solute A :
MxSyFx MSF Eq. (2)
Use both Eq. 1 and 2 to find xM value
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Single stage graphical steps
1. Locate point F (xF) and S (yS)
2. Draw a straight line from F to S
3. Using the calculated value of xM, locate point M (xM) on the FS
line. Note that point M must be on FS line.
4. Draw a new tie line that pass through point M. This new tie linemust take shape of the nearest given tie lines.
5. From the new tie line, you can locate point E and R and hence
you can determine the composition of raffinate, R and extract, Ethat are in equilibrium.
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Once you have determine composition of R and E, you can determine themasses of E and R using the material balance as follows:
Single stage calculations
Using the total material balance:
Eq. (3)ERSF
Using the material balance for solute A:
EyRxSyFx SF ** Eq. (4)
Solve those Eq 3 and 4 to determine the masses of E and R
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Example 1
Single stage calculations
100 kg of a solution containing 0.4 mass fraction of ethylene glycol (EG) in water isto be extracted with equal mass of furfural 250C and 101 kPa. Using the ternaryphase equilibrium diagram method, determine the followings:
the composition of raffinate and extract phases
the mass of extract and raffinate
the percent glycol extracted
Furfural rich layer Water rich layer
% EG % water % furfural % EG % water % furfural
0.0 5.0 95.0 0.0 92.0 8.0
8.5 4.5 87.0 2.0 89.6 8.4
14.5 4.5 81.0 5.5 86.0 8.5
21.0 6.0 73.0 7.0 84.4 8.6
29.0 7.0 64.0 8.0 83.3 8.7
42.0 8.5 49.5 14.0 77.2 8.8
50.0 14.0 36.0 31.0 60.0 9.0
51.0 33.0 16.0 51.0 33.0 16.0
Use the followingequilibrium tie lineto construct theternary phasediagram
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Solution 1
Single stage calculations
Calculate the mass of intermediate M using total material balance
F = 100 kg S = 100 kg xF=0.4 yS=0
MSF M100100 200M kg
Determine mass fraction of solute A in intermediate M using materialbalance for solute A:
MxSyFx MSF 200x100010040 M . 20xM .
Locate point F & S, draw line FS. Locate point xM on FS line. Draw new tieline that pass through point xM. From that tie line, locate point E and R henceyou can determine the composition of R (x*) and E (y*) which is in equilibrium.From the graph, y* = 0.26, x* = 0.075(Solution for point 1)
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Single stage calculations
Right angle method
F
S
M E
R
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Single stage calculations
Equilateral method
S
ME
R
F
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Solution 1 (cont)
Single stage calculations
Using the total material balance
Using the material balance for solute A:
ERSF ER100100 E200R Eq. (i)
EyRxSyFx SF ** E260R0750100010040 ...
Insert eq (i) into eq above
kg8664R
14135200E200R
kg14135E
25E1850
40E260E075015
40E260E2000750
.
.
.
.
..
..
Solution for point 2
% of EG extracted = (Mass of EG in extract /Mass of EG in feed) x 100%
%.%.
..%
*887100x
10040014135260100
FxEy
F
% of EG extracted =
Solution for point 3
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Solvent and solution which flow opposite (countercurrent) to each other,come into contact more than onceand mix on stages inside the reactor.
Normally numbering of the stages begin at the top down to the bottom.Thus the top most stage is named as stage 1, stage directly below stage 1 isstage 2 and so on.
Multi stage counter current system
XR (A)
Final raffinate, RExtracting solvent, S
yS (A)
Final extract, E
xF (A)
Feed solution, F
1
2
3
n
N-1
N
yE (A)
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The analysis of multistage extraction can be performed using right angleor equilateral triangular diagram to determine the number of ideal stagesrequired for a specified separation.
Using material balance,
Multi stage counter current system
Calculate the mass of intermediate M using total material balance:
MSF Eq. (1)
Determine mass fraction of solute A in intermediate M using material balancefor solute A :
MxSyFx MSF Eq. (2)
Use both Eq. 1 and 2 to find xMvalue
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On a right angle triangulardiagram or equilateral triangulardiagram forA-B-C system:
Locate point F (xF) and S (yS)
Draw a straight line from F to S
Using the calculated value of xM, locate point M (xM) on the FS line. Note that
point M must be on FS line.
Locate point E1(Point M must be on E1RNline).
Multi stage counter current system
Operating Points and Lines.
Locate the Operating Point by finding the intersection of operating linesfor the left most and right most stage.
Draw a line through E1 and F.Draw a line through S and RN.Locate the intersection P. This point is the operating point P.
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Multi stage counter current system
Acetone
TCE Water
Plait Point
Carrier
Solute
Feed
RN
M
E1
S
Operating Point
P
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Multi stage counter current system
Operating Lines and Tie Lines: Stepping Off Stages:
Locate point R1from the tie line intersecting E1
Draw a line from the operating point P through R1 to the extract side ofthe equilibrium curve. The intersection locates E2
Locate point R2from a tie line.
Repeat Steps 2 and 3 until RNis obtained
Summary: E1 R1: Tie line, R1 E2: Operating line. Stop until E valueis slightly below RNvalue
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Multi stage counter current system
Acetone
TCE Water
Plait Point
Solvent C
Carrier
Solute
E1
R1
Feed
RN
E2
E3
E4
E5
E6
M
Operating Point
P
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Minimum solvent amount / minimum solvent flow rate
Minimum solvent flow rate is the lowest rate / amount at which solvent could betheoretically used for a specified extraction.
Occurs when operating line touches the equilibrium curve at which the
separation requires infinite numberof ideal stages.
Point M is dependent upon the solvent flow rate / amount. The larger the rate /amount, the closer is point M to point S on the FS line.
Multi stage counter current system
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Multi stage counter current system
On a right angle triangulardiagram or equilateral triangulardiagram for A-B-C system:
Locate point F (xF) and S (yS)
Draw a best tie linethat originate from F. The intersection of this line withextract half dome is point Emin (minimum extract flow rate / amount).
Draw a straight line from Emin to point R. The intersection of this line withFS gives point Mmin. From point Mminyou can read the value of xmin.
Use the value of xminand material balance to calculate the Smin.
% Overall efficiency of multi stage extraction column:
% Overall efficiency = (number of ideal stage / number of real stage) x 100%
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Example 2
5300 kg/h of a solution containing 30% by weight of ethylene glycol (EG) in wateris to be reduced to 4.5% (solvent free) by a continuous extraction in acountercurrent column using recycled furfural that contains 1.5% EG as theextracting solvent:
Determine the minimum solvent flow rate for the extraction above
If the solvent enters at 1.25 times the minimum solvent rate, how many idealstages are required?
Determine the number of real stages if the overall efficiency of the column is60%
Multi stage counter current system
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Solution 2
F = 5300 kg/hr xF= 0.30 RSF= 0.045 yS= 0.015
Multi stage counter current system
F
S
Emin
RSF
Mmin
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Multi stage counter current system
From the graph above, XMmin= 0.25
From material balance:
minmin MSF
minmin MS5300
minmin minMxSyFx MsF
minmin ... M250S0150530030 Smin=1127.66 kg/h
Solution for point 1
S = 1.25 x Smin=1.25 x 1127.66 kg/h S = 1409.58 kg/h
Calculate the mass of intermediate M using total material balance :
MSF M1409.585300 kgM 58.6409
Determine mass fraction of solute A in intermediate M using material
balance for solute A:
From material balance:
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Multi stage counter current system
Solution for point 2
MxSyFx MSF 6709.58x1409.580.01553000.3 M 0.24xM
F
S
E1
RSF
M
P
E2
E3
E4
E5
From figure above, noof ideal stages = 5
Number of real stages = 5 / 0.60 = 8.33 = 9 stages.Solution for point 3
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LLE for Immiscible Solvent
Sometimes extraction use a solvent C that is only slightly soluble in B or thesolvent C used is in range where the solubility in B is so low that for all practicepurpose, it can be assumed to be completely insoluble / immiscible in B and viceversa.
Bancroft weight fractions or mass ratio, x and y are defined as follows:
x (in raffinate phase) = mass of solute A / mass of diluent B
y (in extract phase) = mass of solute A / mass of solvent C
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SINGLE STAGE CALCULATIONS
MULTISTAGE COUNTER CURRENT SYSTEM
LLE for Immiscible Solvent
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Solvent C is used in such a range that it is considered insoluble in B.
Material balance of the solute (A) are:
Feed solutionM kg A in feed
S kg solvent C in Extracty kg A/kg solvent C
SolventN kg A in feed
B kg diluent B in Raffinatex kg A/kg diluent B
BxSyNM ''
S
NMx
S
By
''
Eq. (3)
Single stage calculations
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Example 3
Single stage calculations
An aqueous solution of acetic acid is to be extracted in a single extractor withisopropyl ether. The solution contains 24.6 kg of acetic acid and 80 kg of water.
If 100 kg of isopropyl ether is added to the solution, what weight of acetic acidwill be extracted by isopropyl ether if equilibrium is attained?
Water and isopropyl ether may be considered as completely immiscible under
the condition of extraction. The equilibrium data as follows:
x (kg acid/kg isopropyl ether) 0.030 0.046 0.063 0.070 0.078 0.086 0.106
y (kg acid/kg water) 0.10 0.15 0.20 0.22 0.24 0.26 0.30
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A = Acetic Acid B = Water C = Isopropyl ether
Single stage calculations
From mass balance,
Feed solution24.6 kg A
100 kg solvent C in Extracty kg A/kg solvent C
Solvent0 kg A in feed
80 kg diluent B in Raffinatex kg A/kg diluent B
BxSyNM '' '80100'06.24 xy 2460x800y .'.'
The equilibrium data and equation above is plotted in figure next page.
From the intersection of lines, x = 0.062, y = 0.195
Amount of acetic acid extracted = 19.5 kg
Solution 3
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Single stage calculations
Solution 3
x' - y' diagram for s ystem acetic acid-water-isopro pyl ether
0
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0 0.02 0.04 0.06 0.08 0.1 0.12 0.14
x'
y'
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The principle is quite same with the partially miscible solvent, but in this casethe extraction process involved the immiscible solvent.
Multi stage counter current system
B kg/h pure diluent B in feed solution
x2kg A/kg pure B
B kg/h pure diluent B in raffinate
xnkg A/kg pure B
S kg/h pure solvent C
yn+1 kg A/kg pure solvent
S kg/h pure solvent in extract phase
y2 kg A/kg pure solvent1
2
3
n
N-1
N
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For immiscible solvents, analysis of the extraction process become much simplersince the flow rate of pure solvents (extracting solvent and feed solvent (diluent) )are constants).
The operating line equation for multi stage liquid liquid extraction:
Multi stage counter current system
S
BxSyx
S
By 22
''''
Form of y = mx + c
The operating line gives a relationship between mass fraction of A (xn) inraffinate phase coming out of the nth stage, with mass fraction of A (yn+1) in extractphase entering the nth stage.
The operating line can be plotted by locating points (x1,y1) and (x2,y2) anddraw a straight line through these points.
Using the method similar to Mc Cabe Thiele diagram, the number of idealstages can be determined by drawing the triangular steps connecting equilibriumline and the operating line.
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Multi stage counter current system
Example 4
8000 kg/h of an acetic acid-water solution containing 20% acid by mass is to becounter currently extracted with isopropyl ether to reduce the concentration of acidto 2% in the solvent-free raffinate product.
Determine the number of theoretical stages if 20,000 kg/h solvent is used.
Use the following equilibrium data:
x (kg acid/kgwater)
0.000 0.025 0.050 0.100 0.150 0.200 0.220 0.240 0.260 0.300
y (kg acid/kgisopropyl ether)
0.000 0.005 0.013 0.030 0.046 0.063 0.070 0.078 0.086 0.106
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Multi stage counter current system
Solution 4
Since acid is to be extracted, acid is the absorbable component A, while water iscomponent B and isopropyl ether is component C.
BOTTOM (1)
TOP (2)
0.8(8000) kg/h of water in
x2
20,000 kg/h isopropyl ether (C) in
y1
0.8(8000) kg/h of water (B) in raffinate0.02 kg A/kg (solvent free)
x1
20,000 kg/h isopropyl ether (C) in extract phase
y2
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Multi stage counter current system
From the definition:
x (in raffinate phase) = mass of solute A / mass of diluent B
y (in extract phase) = mass of solute A / mass of solvent C
Solution 4
F = 8000 kg/h xF= 0.20 xBF= 0.80 S = 20,000 kg/h
Solvent feed
Mass of solute A = 0 kg/h
Mass of solvent C = 20,000 kg/h.
For location 1,
Raffinate phase
Solvent-free composition of A = 0.02, B = 0.98
02040980
020
B
BA
BA
A
BA
B
BA
A
B
Ax1 .
.
.'
020000
0y1 'Coordinates for (x1,y1) = (0.0204, 0)
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Multi stage counter current system
Solution 4
Feed
Mass of solute A = 0.20 x 8000 = 1600 kg/h
Mass of diluent B = 0.80 x 8000 = 6400 kg/h.
For location 2,
Extract
Mass of solute A = Mass of acid in Mass of acid in raffinate
Mass of solute A = 1600 - x1x B = 1600 0.0204 x 6400= 1469.44 kg/h
Mass of solvent C = 20 000 kg/h
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Multi stage counter current system
Solution 4
For location 2,
Extract
25006400
1600x2 .'
0735020000
441469y2 .
.'
Coordinates for (x2,y2) = (0.250, 0.0735)
The points for operating line are (0.0204, 0) and (0.250, 0.0735).
No. of theoretical stages = 14 stages
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Multi stage counter current system
Solution 4
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Minimum solvent amount / flow rate: immiscible solvent
Minimum solvent flow rate, Sminis defined as the flow rate of solvent at whichthe number of stages approaches infinity.
Smincan be determined graphically by drawing a straight line originating fromthe bottom (x1,y1) until it becomes tangential to the equilibrium curve. The slope
of this line corresponds to minimum slope (mmin):
Multi stage counter current system
min
minS
Bm
min
minm
BS
Example 5
Determine the minimum flow rate of isopropyl ether for the problem in
Example 4.
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Multi stage counter current system
Solution 5
The line tangential to the equilibrium curve and originates from (x1,y1)which is (0.0204, 0) is drawn in figure below:
The slope of the tangential line = 0.3433430
800080
m
BS
.
.
min
min
= 18 658.89 kg/h.
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Liquid liquid extraction equipment
Two main classes of solvent extraction equipment:
1) Vessels in which mechanical agitation is provided for mixing
2) Vesselsin which the mixing is done by the flow of the fluid themselves.
The extraction equipment can be operated batch or continuous.
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Liquid liquid extraction equipment
Mixer Settlers for Extraction
A mechanical mixer is often used to provide intimate contact between thetwo liquid phases to provide efficient mass transfer.
One phase is usually dispersed into the other in the form of small droplet.
In figure 12.6 1(a) for typical mixer settler, mixer or agitator is entirelyseparate from the settler. The feed of aqueous phase and organic phase are
mixed in the mixer, and then the mixed phases are separated in the settler.
In figure 12.6 1(b) for combined mixer settler, sometimes used in extractionof uranium salts or copper salts from aqueous solution.
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Liquid liquid extraction equipment
Spray Extraction Towers
In Figure 12.6 2 the heavy liquid enters atthe top of the spray tower, fills the tower asthe continuous phase, and flows out throughthe bottom.
The light liquid enters through a nozzledistributor at the bottom, which disperses orsprays the droplets upward.
The light liquid coalesces at the top andflows out.
In some cases the heavy liquid is sprayed
downward into a rising, light continuousphase.
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Liquid liquid extraction equipment
Packed Extraction Towers
More effective tower madeby packing the column withrandom packing such asRaschig rings, Berl saddles,Pall rings and so on
Packings cause the droplets tocoalesce and redisperse atfrequent intervals through thetower.
Packed tower is more efficientthan spray tower.
Table 12.6 1 shows typical
performance for several typesof commercial extractiontowers.