chapter 2 - lle edited
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CHAPTER 2: LIQUID – LIQUID EXTRACTION
CHAPTER / CONTENT
Definition & Application
LLE for Partially Miscible Solvent
LLE for Immiscible Solvent
Liquid – liquid extraction equipment
LLE= 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.
LLE= Separation process of the components of a liquid mixture by treatment with 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
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
In some operations, the solutes are the desired product, hence the extract stream is the desirable stream. In other applications, the solutes my be contaminants that need to be 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 to separate 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, where emphasis is on the separation of antibiotics and protein recovery.
In the inorganic chemical industry, they are used to recover high – boiling components such as phosphoric acid, boric acid and sodium hydroxide from aqueous solution.
Definition & Application
Examples:
Extraction of nitrobenzene after reaction of HNO3 with toluene in H2SO4
Extraction of methylacrylate from organic solution with perchlorethylene
Extraction of benzylalcohol from a salt solution with toluene.
Removing of H2S from LPG with MDEA
Extraction of caprolactam from ammonium sulfate solution with benzene
Extraction of acrylic acid from wastewater with butanol
Removing residual alkalis from dichlorohydrazobenzene with water
Definition & Application
Examples:
Extraction of methanol from LPG with water
Extraction of chloroacetic acid from methylchloroacetate with water.
Definition & Application
The difference between LLE and distillation process in the separation of liquid mixtures:
LLE depends on solubilities between the liquid components and 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 heat addition.
Definition & Application
Advantages of LLE over distillation process:
Where distillation requires excessive amount of heat
Presence of azeotropes or low relative volatilities are involved (α value near unity and distillation cannot be used)
Removal of a component present in small concentrations, e.g hormones in animal oil.
Recovery of a high – boiling point component present in small quantities in waste stream, e.g acetic acid from cellulose acetate.
Recovery of heat – sensitive materials (e.g food) where low to moderate processing temperatures are needed. Thermal decomposition might occur.
Solvent recovery is easy and energy savings can be realized.
SINGLE STAGE CALCULATIONS
MULTISTAGE COUNTER CURRENT SYSTEM
LLE for Partially Miscible Solvent
Solvent and the solution are in contact with each other only once and thus the raffinate and extract are in equilibrium only once.
The solution – normally binary solution containing solute (A) dissolved in a diluent or carrier (B). The extracting solvent can be either pure solvent C or may content little A. Raffinate (R) is the exiting phase rich in carrier (B) while extract is exiting phase rich in solvent (C).
When liquid solution mixed with solvent (C), an intermediate phase M momentarily forms as the light liquid moves through the heavy liquid in the form of bubbles. These bubbles provide a large surface area for contact between the solution and the solvent that speed up mass transfer process.
The raffinate and extract are in equilibrium with each other.
Single – stage calculations
Single – stage calculations
Liquid-Liquid Extraction
y* (A)ys (A) Intermediate, M
Raffinate phase, Rx* (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. Don’t have to draw it when answering the question
Feed Solution, FxF (A)
Extracting Solvent, S
In most single – extraction, we are interested to determine the equilibrium composition and masses of raffinate and extract phases by using 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 material balance for solute A :
MxSyFx MSF Eq. (2)
Use both Eq. 1 and 2 to find xM value
On a right angle triangular diagram or equilateral triangular diagram for A-B-C system:
Single – stage calculations
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.
Draw a new tie line that pass through point M. This new tie line must take shape of the nearest given tie lines.
From the new tie line, you can locate point E and R and hence you can determine the composition of raffinate, R and extract, E that are in equilibrium.
Once you have determine composition of R and E, you can determine the masses 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
Example 1
Single – stage calculations
100 kg of a solution containing 0.4 mass fraction of ethylene glycol (EG) in water is to be extracted with equal mass of furfural 250C and 101 kPa. Using the ternary phase 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 following equilibrium tie – line to construct the ternary phase diagram
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 material balance for solute A:
MxSyFx MSF 200x100010040 M . 20xM .
Locate point F & S, draw line FS. Locate point xM on FS line. Draw new tie line that pass through point xM. From that tie line, locate point E and R hence you 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)
Single – stage calculations
Right angle method
F
S
M E
R
Single – stage calculations
Equilateral method
S
ME
R
F
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
100400
14135260100
Fx
Ey
F
% of EG extracted =
Solution for point 3
Tutorial
1. 12.5-2
2. 12.5-4
Tutorial1. 12.5-2 (Textbook-page 832)A single stage extraction is performed in which 400kg of a
solution containing 35 wt% acetic acid in water is
contacted with 400 kg of pure isopropyl ether. By using
equilateral diagram, determine
the composition of raffinate and extract phases
the mass of extract and raffinate
the percent acetic acid extracted
Tutorial2. 12.5-4 (Textbook-page 832)A mixture weighing 1000kg contains 23.5 wt% acetone
and 75.5 wt% water and is to be extracted by 500 kg
methyl isobutyl ketone in a single stage extraction. By
using equilateral diagram, determine
the composition of raffinate and extract phases
the mass of extract and raffinate
the percent acetone extracted
Solvent and solution which flow opposite (countercurrent) to each other, come into contact more than once and 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 is stage 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)
The analysis of multistage extraction can be performed using right – angle or equilateral triangular diagram to determine the number of ideal stages required 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 balance for solute A :
MxSyFx MSF Eq. (2)
Use both Eq. 1 and 2 to find xM value
On a right angle triangular diagram or equilateral triangular diagram for A-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 E1RN line).
Multi – stage counter current system
Operating Points and Lines.
Locate the Operating Point by finding the intersection of operating lines for 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.
Multi – stage counter current system
Acetone
TCE Water
Plait Point
Carrier
Solute
Feed
RN
M
E1
S
Operating PointP
Multi – stage counter current system
Operating Lines and Tie Lines: Stepping Off Stages:
Locate point R1 from the tie line intersecting E1
Draw a line from the operating point P through R1 to the extract side of the equilibrium curve. The intersection locates E2
Locate point R2 from a tie line.
Repeat Steps 2 and 3 until RN is obtained
Summary: E1 → R1 : Tie line, R1 → E2 : Operating line. Stop until E value is slightly below RN value
Multi – stage counter current system
Acetone
TCE Water
Plait Point
Solvent C
Carrier
Solute
E1
R1
Feed
RN
E2E3
E4
E5
E6
M
Operating PointP
Minimum solvent amount / minimum solvent flow rate
Minimum solvent flow rate is the lowest rate / amount at which solvent could be theoretically used for a specified extraction.
Occurs when operating line touches the equilibrium curve at which the separation requires infinite number of 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
Multi – stage counter current system
On a right angle triangular diagram or equilateral triangular diagram for A-B-C system:
Locate point F (xF) and S (yS)
Draw a best tie line that originate from F. The intersection of this line with extract half – dome is point Emin (minimum extract flow rate / amount).
Draw a straight line from Emin to point R. The intersection of this line with FS gives point Mmin. From point Mmin you can read the value of xmin.
Use the value of xmin and 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%
Example 2
5300 kg/h of a solution containing 30% by weight of ethylene glycol (EG) in water is to be reduced to 4.5% (solvent free) by a continuous extraction in a countercurrent column using recycled furfural that contains 1.5% EG as the extracting 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 ideal stages are required?
Determine the number of real stages if the overall efficiency of the column is 60%
Multi – stage counter current system
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 kg M 58.6409
Determine mass fraction of solute A in intermediate M using material balance for solute A:
From material balance:
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, no of ideal stages = 5
Number of real stages = 5 / 0.60 = 8.33 = 9 stages. Solution for point 3
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, no of ideal stages = 5
Number of real stages = 5 / 0.60 = 8.33 = 9 stages. Solution for point 3
Exercise1. 12.7-3 (Textbook, pages 833)An aqueous feed solution of 1000 kg/hr containing
23.5 wt% acetone and 76.5 wt% water is being
extracted in a countercurrent multistage extraction
system using pure methylisobutyl ketone solvent at
298-299 K. The outlet water raffinate will contain 2.5
wt% acetone. By using equilateral method,
a) Determine the minimum solvent that can be used (Answer: Xmin = 0.18, Smin = 305.56 kg/hr , Mmin = 1305.56 kg/hr)
b) Using a solvent flow rate of 1.5 times the minimum, determine the number of theoretical stages. (Answer: S= 458.34 kg/hr, Xm=0.16 , 5 stages)
LLE for Immiscible Solvent
Sometimes extraction use a solvent C that is only slightly soluble in B or the solvent C used is in range where the solubility in B is so low that for all practice purpose, it can be assumed to be completely insoluble / immiscible in B and vice versa.
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
SINGLE STAGE CALCULATIONS
MULTISTAGE COUNTER CURRENT SYSTEM
LLE for Immiscible Solvent
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
Example 3
Single – stage calculations
An aqueous solution of acetic acid is to be extracted in a single extractor with isopropyl 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 acid will 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
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
Single – stage calculations
Solution 3
x' - y' diagram for system acetic acid-water-isopropyl 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'
The principle is quite same with the partially miscible solvent, but in this case the extraction process involved the immiscible solvent.
Multi – stage counter current system
B kg/h pure diluent B in feed solution
x’2 kg A/kg pure B
B kg/h pure diluent B in raffinate
x’n kg A/kg pure B
S kg/h pure solvent C
yn’ kg A/kg pure solvent
S kg/h pure solvent in extract phase
y’2 kg A/kg pure solvent1
2
3
n
N-1
N
For immiscible solvents, analysis of the extraction process become much simpler since 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 nn
22 ''''
Form of y = mx + c
The operating line gives a relationship between mass fraction of A (x’n) in raffinate phase coming out of the nth stage, with mass fraction of A (yn) in extract phase entering the nth stage.
The operating line can be plotted by locating points (x’1,y’1) and (x’2,y’2) and draw a straight line through these points.
Using the method similar to Mc Cabe – Thiele diagram, the number of ideal stages can be determined by drawing the triangular steps connecting equilibrium line and the operating line.
Multi – stage counter current system
Example 4
8000 kg/h of an acetic acid-water solution containing 20% acid by mass is to be counter currently extracted with isopropyl ether to reduce the concentration of acid to 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/kg water)
0.000
0.025
0.050
0.100
0.150
0.200
0.220
0.240
0.260
0.300
y’ (kg acid/kg isopropyl
ether)
0.000
0.005
0.013
0.030
0.046
0.063
0.070
0.078
0.086
0.106
Multi – stage counter current system
Solution 4
Since acid is to be extracted, acid is the absorbable component A, while water is component B and isopropyl ether is component C.
BOTTOM (1)
TOP (2)
0.8(8000) kg/h of water in
x’2
20,000 kg/h isopropyl ether (C) in
y’1
0.8(8000) kg/h of water (B) in raffinate0.02 kg A/kg (solvent free)
x’1
20,000 kg/h isopropyl ether (C) in extract phase
y’2
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 (x’1,y’1) = (0.0204, 0)
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 - x’1x B = 1600 – 0.0204 x 6400= 1469.44 kg/h
Mass of solvent C = 20 000 kg/h
Multi – stage counter current system
Solution 4
For location 2,
Extract
25006400
1600x2 .' 07350
20000
441469y2 .
.'
Coordinates for (x’2,y’2) = (0.250, 0.0735)
The points for operating line are (0.0204, 0) and (0.250, 0.0735).
No. of theoretical stages = 14 stages
Multi – stage counter current system
Solution 4
TOP (0.250,0.0735)
BOTTOM(0.0204,0)
Minimum solvent amount / flow rate: immiscible solvent
Minimum solvent flow rate, Smin is defined as the flow rate of solvent at which the number of stages approaches infinity.
Smin can be determined graphically by drawing a straight line originating from the bottom (x’1,y’1) until it becomes tangential to the equilibrium curve. The slope of this line corresponds to minimum slope (mmin):
Multi – stage counter current system
minmin S
Bm
minmin m
BS
Multi – stage counter current system
The slope of the tangential line = 0.343 3430
800080
m
BS
.
.
minmin
= 18 658.89 kg/h.
Multi – stage counter current system
Example 5
Determine the minimum flow rate of isopropyl ether for the problem in Example 4.
Multi – stage counter current system
Solution 4
TOP (0.250,0.0735)
BOTTOM(0.0204,0)
Multi – stage counter current system
The slope of the tangential line = 0.343 3430
800080
m
BS
.
.
minmin
= 18 658.89 kg/h.
Tutorial12.7-6 (Textbook)A water solution of 1000kg/hr containing 1.5 wt%
nicotine in water is stripped with a kerosene stream
of 2000kg/hr containing 0.05 wt% nicotine in a
countercurrent stage tower. The exit water is to
contain only 10% of the original nicotine, that is,
90% is removed.
1.Determine the number of theoretical stages needed.
2. Determine the minimum solvent(kerosene) flow rate
Liquid – liquid extraction equipment
Two main classes of solvent – extraction equipment:
1) Vessels in which mechanical agitation is provided for mixing
2) Vessels in which the mixing is done by the flow of the fluid themselves.
The extraction equipment can be operated batch or continuous.
Liquid – liquid extraction equipment
Mixer – Settlers for ExtractionA mechanical mixer is often used to provide intimate contact between the two 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 entirely separate 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 extraction of uranium salts or copper salts from aqueous solution.
Liquid – liquid extraction equipment
Spray Extraction Towers
In Figure 12.6 – 2 the heavy liquid enters at the top of the spray tower, fills the tower as the continuous phase, and flows out through the bottom.
The light liquid enters through a nozzle distributor at the bottom, which disperses or sprays the droplets upward.
The light liquid coalesces at the top and flows out.
In some cases the heavy liquid is sprayed downward into a rising, light continuous phase.
Liquid – liquid extraction equipment
Packed Extraction TowersMore effective tower – made by packing the column with random packing such as Raschig rings, Berl saddles, Pall rings and so on
Packings cause the droplets to coalesce and redisperse at frequent intervals through the tower.
Packed tower is more efficient than spray tower.
Table 12.6 – 1 shows typical performance for several types of commercial extraction towers.