ChE 221 / ChE 516 Mass Transfer Operations
Chapter 8 - Extraction and Leaching Main references: McCabe, 7th Ed, Ch 23; Seader, 3rd Ed, Ch 8 & 16
Part I: Liquid-Liquid Extraction Based on equilateral triangular diagram
Liquid-liquid extraction The method to remove one constituent from a solid or liquid by means of a liquid solvent
Liquid extraction – extraction – to recover recover a valuable product from from a multicomponent solution by contact with an immiscible solvent that has a high affinity for the product
When separation by distillation is ineffective or very difficult, liquid-liquid extraction (or solvent extraction) is one of the t he main alternatives to consider.
Close-boiling mixtures or substances that cannot withstand the temperature of distillation, even under a vacuum, may often be separated from impurities by extraction, which utilises chemical differences differences instead of vapour pressure differences. 3
An industrial example The feed of 30260 1b/h of 22wt% acetic acid in water is in contact with 71100 1b/h of ethyl-acetate ethyl-acetate solvent in a singlesection extraction column. The low-density, solvent rich extract exits from top of the extractor ex tractor with 99.8% acetic acid in the feed. The high density, carrierrich raffinate exiting from bottom contains only 0.05wt% acetic acid.
Acetic acid: b.p.=118.1 b.p.=118.1oC Water: Water: b.p.=100oC Ethyl-acetate: b.p.=77.1oC 4
4 major steps in an extraction process
Bringing the feed and the solvent into intimate contact by dispersing one phase into the other o ther as droplets
Separation of the extract and the raffinate phases that have different densities
The extract is the layer of solvent plus extracted solute
The raffinate is the layer from which solute has been removed.
Removal and recovery recovery of the solute from the extract phase in a relatively pure form (by distillation, evaporation, crystallization, etc)
Removal and recovery of the solvent from each phase, usually by distillation 5
Extraction preferred preferred over distillation for
Dissolved or complexed complexed inorganic substances in organic or aqueous solutions
Removal of a contaminant present in small concentration
A high-boiling component present in relatively small quantities in an aqueous waste steam
Recovery of heat-sensitive materials, where extraction may be less expensive than vacuum distillation
Separation of mixtures according to chemical type rather than relative relative volatility
Separation of close-melting or close-boiling liquids, where solubility differences can be exploited
Separation of mixtures that form azeotropes azeotropes 6
Industrial liquid-liquid leaching process Solute
Carrier
Solvent
Acetic acid
Water
Ethyl acetate
Ammonia
Butenes
Water
Aromatics
Paraffins
Furfural
Aromatics
Kerosene
Sulfur dioxide
Benzoic acid
Water
Benzene
Fatty acids
Oil
Propane
Acetic acid
Water
Methyl isobutyl ketone
Formic acid
Water
Tetrahydrofuran etrahydrofu ran
Penicillin
Broth
Butyl acetate
Vanilla
Oxidized Oxidiz ed liquors liquor s
Toluene oluen e
Vitamin A
Fish-liver oil
Propane
7
Principles of Extraction
Since most continuous extraction methods use countercurrent contacts between two phases, many of the fundamentals of countercur counte rcurrent rent gas absorption and of distillation carry over into the study of liquid extraction.
Thus questions about ideal stages, stage efficiency, minimum ratio between the two streams, and size of equipment have have the same importance impor tance in extraction as in distillation.
8
Design Considerations Some factors influencing extraction: Types of stage configuration Min solvent flow rate and actual flow as multiple of min flow rate (or reflux for more than one stage system) Operating conditions Emulsification and scum formation tendencies difference Phase density difference Interfacial tension extractor or Types of extract
9
Desirable properties of a solvent Some characteristics of an ideal solvent:
High selectivity for solute / low for carrier
High capacity for dissolving solute
Volatility difference with solute - recoverability
Large density difference with carrier
Lack tendency to form scum layer layer at interface i nterface
Low viscosity – promote phase separation
Stable – minimise make-up solvent
Availability Availability and low cost 10
Types of extractor extractor – Mixer-settlers Mixer-settlers
For batchwise extraction the mixer and settler may be the same unit.
For continuous flow the mixer and settler are usually separate pieces of equipment.
11
The mixer may be a small agitated tank provided with inlets and drawoff line and baffles to prevent short-circuiting, or it may be other flow mixer;
The settler is often a simple continuous gravity decanter
The extract may be lighter or heavier than the raffinate, and so the extract may may come from the top or the bottom of the equipment
Perforated-plate Perforated-pla te towers
Perforations in horizontal plates
12
Cascade weir tray with mixing and settling zones
The perforations are typically 1½ to 4½ mm in diameter and plate spacings range from 150 to 600 mm;
Usually the light liquid is the dispersed phase, and downcomers carry the heavy continuous phase from one plate to the next;
Depending on gravity flow both for mixing and for separation.
Agitated tower extractors
Rotating-disk unit 13
York-Scheibel extractor
Mechanical energy is provided provided by internal turbines or other agitators, mounted on a central rotating shaft.
Types of configuration configuration
(a) Single-section cascade; (b) two-section two-section cascade; (c) dual dual solvent with twosection cascade. 14
15
The equilateral triangular triangular diagram
Ternary phase diagram 16
The equilateral triangular triangular diagram
The basis of the technique of presentation of equilibrium data on an equilateral triangular graph paper is that the sum of the distances of a point within an equilateral triangle from from the three sides is equal to the height of the triangle;
On the equilateral triangle ABC triangle ABC , each apex is a pure component of the mixture, A mixture, A,, B and C , respectively;
Each edge is a mixture of the two pure components at the terminal apexes of the side, A-B side, A-B,, B-C and and C-A, C-A, respectively.
Consider the point M point M,, for example: MN1+ MN2+ MN3 = CT ; M represents M represents mixture of 16% A, A, 24%B 24%B and 60%C 60%C [100-(16+24)=60%]. [100-(16+24)=60%].
17
Exercise: Exercise: 73.27%A , 6.54%B, 20.19%C (Raffinate phase, x, rich in A )
C 100%
(Print this slide in full size.)
5.2%A, 91.8%B , 3.0%C (Extract phase, y, rich in B ) A: Carrier B: Solvent C: Solute
0% 10%
73.27% A
20.19% C 10% 100% 18
Construct ternary phase diagram Equilibrium miscibility data
Mutual equilibrium (tie-line) data
Point
Furfurall Furfura
Glycol
Water
A
95.0
0.0
5.0
B
90.3
5.2
4.5
C
86.1
10.0
3.9
D
75.1
20.0
E
66.7
F
Point
Glycol in water layer, wt%
Glycol in furfural layer, wt%
P
41.5
41.5
4.9
1
52.5
27.5
27.5
5.8
2
51.5
20.0
49.0
41.5
9.5
3
47.5
15.0
G
34.3
50.5
15.2
4
40.0
10.0
H
27.5
52.5
20.0
5
30.0
7.5
I
13.9
47.5
39.6
6
20.0
6.2
J
11.0
40.0
49.0
7
7.3
2.5
K
9.7
30.0
60.3
L
8.4
15.0
76.6
M
7.7
0.0
92.3
19
Ethylene Glycol (B) Print this slide in full size for Class Problem - 1 0.9
0.1
0.8
0.2
0.7
0.3
0.6
0.4
0.5
0.5
0.4
0.6
0.3
0.7
0.2
0.8
0.1
0.9
Furfural (C) 20
Water (A) 0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
Water-EG-Furfural phase diagram
Water-EG and furfuralEG are completely miscible pairs
Furfural-water is a partially miscible pair
At plait point, point, P , the two liquid phases have have identical compositions
Miscibility limit for the furfural-water binary system are D and G
Thus, furfural can be used as solvent to remove EG from water
21
Furfural-rich phase is the extract and waterrich phase is raffinate
A single-stage extraction solute
What is composition of mixture represented by point M? point M?
What is the composition of extract phase and raffinate phase?
H2O = 0.19
Mixture, M: 20% EG; 19% Water; Water; 61% Furfural.
•
Extract, E: 10% EG; 4% Water Water;; 86% Furfural.
•
EG = 0.2
Raffinate, Raffina te, R: 40% EG; 49% Water; Water; 11% Furfural.
•
solvent
22
carrier
Liquid-liquid equilibria (LLE)
Ternary ernar y systems syste ms comprising compri sing a carrier A carrier A,, a solvent B and a solute C . Three binary binar y mixtures can be formed out of these components, A-B, A-B, B-C , and C-A. C-A. The mutual miscibility behaviour behaviour of the components in each of these binaries determines the nature of the equilibrium diagram for the ternary system.
Here, the solute C is is miscible with A and A and B in all proportions, but the carrier and the solvent are only partially miscible. The curve RPS is the equilibrium diagram in the equilateral triangular coordinate system.
Solute C
Plait point,
Equilibrium or binodal curve
P
•
Carrier A
23
Solvent B
Separation range
The larger the two-phase region on line R-S, the greater the immiscibility of carrier and solvent.
The closer the top of the two-phase region is to apex A, A, the greater the range of feed composition, along the line R-C , that can be separated with solvent S.
Solute C Feed composition range to be separated
F1
F2 D •
M
Carrier A
24
Solvent B
Limiting cases.
LLE…continue
The point R on the line AB line AB represents represents the solubility of the solvent B in the carrier A carrier A;; the point S represents that of the carrier A carrier A in in the solvent B.
The pair of points G 1 and H1 on the curve cur ve represents represents two liquid phases in equilibrium: G 1 is rich in the carrier (raffinate phase) and H1 is rich in the solvent (extract phase). The line G 1H1 is called tie line. Usually they are not parallel. paralle l. The point P , that demarcates the raffinate and the extract sides of the equilibrium curve is called the plait point.. The type of equilibrium point equili brium RPS is RPS is called binodal because because it has two arms RP and and PS PS,, representing the raffinate and the extract sides. 25
Phase splitting of ternary mixtures Extract E Component B, C Solvent S Component B
Feed F Component A, C
Solvent S Component B
Raffinate R Component A, C
(a) Components A and B mutually insoluble.
26
Extract E Component A, B, C
Feed F Component A, C
Raffinate R Component A, B, C
(b) Components A and B partially soluble.
Reverse-level-arm rule
C
E
M •
A
P
• •
S L
R•
•
•
•
•
•
F
N
K
B
The mixing rule in a triangular diagram
Level rule or mixing rule for ternary liquid systems asserts that the composition that results when two liquid solutions are mixed lies on a straight line connecting their compositio compo sitions. ns. Thus, Thus , if R kg of a mixture represented by point R is combined with E kg of a solution located at point E, the resulting composition M will M will lie on a straight line connecting points R and E. Furthermore, the location of M of M will will be such that line segments it defines stand in the ratio of the weights of the parent solutions:
R E
27
=
EM RM
(8.1)
Validation of reverse-leve reverse-level-arm l-arm rule
First we have:
M is mass of mixture, and for the component mass balance
R
RxCR
Re-arrange eq.(8.4)
E
=
MxCM
(8.3)
( R + E ) xCM
(8.4)
xCR − xCM xCM − xCE
=
xCM − xCR xCE − xCM
(8.5)
Because xCM = MN ; xCE = EK ; xCR = RF E
+ ExCE =
+ ExCE =
R
(8.2)
M
Combine eq. (8.2) and (8.3) yields RxCR
+ E =
=
MN − RF
=
SL
EK − MN ES (8.6) Because triangle MPR is similar to triangle ESM, consequently
Thus
R
E R 28
=
SL ES
=
MR EM
(8.1)
Type I - Acetone-water-MIK Acetone-water-MIK system solute
solvent
29
carrier
The solute and solvent are miscible in all compositions.
Partial miscibility of the solvent (methyl isobutyl ketone, MIK) and the carrier (water): one immiscible pair
The tie lines slope up to the left and the extract is richer in acetone than the raffinate phase.
The most common type.
Type II - Aniline-n Aniline-n -heptane-MCH -heptane-MCH system solute
solvent
30
carrier
The solvent (aniline) is only partially miscible with both the solute (methylcyclohexane, MCH) and carrier n-heptane: n-heptane: two immiscible pairs. pairs.
The tie lines slope up to the right → more solvent would be required since the final extract would not be as rich in the desired component.
In Type II system, syste m, there is no plait point. point.
Single stage extraction Feed F
Locate the feed and solvent compositions at points F and S, respectively.
Define mixing point M as M as M = = F + + S = E + R.
Apply the inverse-level-arm inverse-level-arm rule. Let w i (E) be the species spec ies I in the extract, w i (R) be the species I in the raffinate, raffinate , and w i (M) be the species I in the feed+solvent phase. phase . For solvent C
Extract E
Solvent S
Raffinate R
Solute B
A
F S ( F + S ) w M = + Fw Sw c c c
B
F•
•
E
F S
SM =
MF
•
R
•
M •
Carrier A
31
C
Obtain R and E phase compositions.
The inverse-level-arm inverse-level-arm rule applies to obtain weight fractions of R and E phases.
S C Solvent
Multistage cross-flow extraction Extract E1
Extract E2
Raffinate R1
Feed goes through stages in series and to each stage, fresh fresh solvent is added. This process is an extension of single-stage extraction.
For multistage with cross-flow, cross-flow, raffinate from each stage is contacted with following stage with pure solvent. Extracts are withdrawn from each stage.
The concentration of solute (B) in raffinate and extract decreases from stage to stage.
Apply material balance and inverse-levelarm rules, similar to single-stage sing le-stage extraction calculation procedures.
R2
Feed F Solvent S1
Solvent S2 Solute B
A
F
B •
M1 •
R1 •
32
E1 •
E2
•
R2 • Carrier A
•
M2 C
•
S C Solvent
Right-triangle Right-triang le diagrams
33
Ternary, countercurrent extraction calculations can also be made on a right-triangle right-triangle diagram.
A right-triangle diagram can be developed from an equilateraltriangle diagram. Rectangular coordinate are in mass or mole fractions, fractio ns, xA and xC, where whe re A A is is the solute and C the the carrier. carrier.
The advantage of the right-triangle diagram is i s that th at ordinary ordinar y, rectangular-coordinate graph paper can be used and either one of the coordinates can be expanded.
Right-triangle Right-triang le diagrams
L
0.15
L
0.84
Point L : Furfural-8.4%; EG-15% (water=100-8.4-15=76.6) 34
Distribution diagram
Plots of solute weight fraction in two phases against each other → two phase behaviour.
Plait point is located on the 45o diagonal → equality of phase concentrations
Compositions that lie above the diagonal are richer in solute content than those of the companion phase, while points below it denote a depletion in solute.
Distribution coefficient is defined as the solute mole-fraction ratio in two phases,: m = x = x CB/ x x CA 35
Part II: Solid-Liquid Extraction
Solid-liquid extraction or leaching
Leaching involves involves the removal of a soluble fraction (the solute or leachant) leachant) of a solid material by a liquid solvent
Effluents from a leaching stage are essentially solid-free so lid-free liquid, called the overflow , and , and wet solids, the underflow . To reduce the concen concentration tration of solute in the liquid portion of the underflow, leaching is often accompanied by countercurrent-flow countercurrent-flow washing stages.
The combined process produces a final overflow, called extract, extract, and and a final underf underflow low,, the extracted or or leaching solids. solids.
Ideally, the soluble solids are perfectly separated from the insoluble solids, but solvent is distributed to both products
37
Additional processing processing of extract and leaching solids is necessary necessar y to recover recov er solvent for recycle.
Industrial applications of leaching
Inorganic and organic materials
Metal processing industries e.g. removal as soluble salts
Removal of copper from ore using sulfuric acid
Recovery of gold from ore using sodium-cyanide solution
Biological substances
38
Biological and food industries
Extraction of sugar from sugar beets using hot water as solvent
Removal of caffeine from f rom green coffee beans using supercritical supercritical CO2
Vegetable oils from nuts and seeds using organic solvents (hexane, acetone, ether)
Pharmaceuticall products from plant roots, leav Pharmaceutica leaves, es, stems
An industrial example Soybean flakes (10.67%moisture, 0.2675g oil/dry flake): 6.375 1b/h
Extract of 7.313 1b/h contains 15.35% oil after 11 h operation
Solvent flow: 10.844 1b/h
Kennedy extractor
Leached solid contains 0.0151 g oil/dry oil, free flake
Soybeans contain highest percentage of edible oil
For soybeans, whose oil content is typically less than 0.3 1b per 1b of dry and oilfree solids, leaching is more mo re desirable than expression.
The ideal solvent for commercial leaching of soybeans should s hould have a high oil solubility; a high volatility; nonflammability; low cost; chemical stability; low toxicity.
Nonflammable trichloroethylene is an ideal solvent, but is classified as a hazardous, toxic chemical.
The favoured solvent is the commercial n-hexane, which presents fire hazard but has a low toxicity. toxicity. 39
Espresso machine
40
A batch-leaching machine
In the machine, machin e, 7-9 grams of coffee c offee beans are ground to a powder of particle size of 250-750 µm
Water is pumped p umped to a pressure of 915 atm and heated to 88-92oC. The high pressure is required required for pressure infusion of hot water through the bed of coffee powder, so extraction can proceed rapidly
During a period of 20-30 20 -30 s, hot water is percolated through the coffee powder to produce a 45 mL shot.
Maximise extraction of the flavourflavourand-aroma chemicals and minimise extraction of the chemicals associated with bitterness such as quinine and caffeine.
Moving-bed leaching: Bollman extractor
Buckets are loaded with flaky seeds and beans such as soybeans and are sprayed sprayed with intermediate solvent, half miscella. miscella.
As solids and solvent flow concurrently concurrently down the right-hand side of the machine, the solvent extracts more oil from the beans.
As the partially extracted beans rise through the left side of the machine, a stream of pure solvent so lvent percolates countercurrently countercurrently through them.
Fully extracted beans are dumped at the top of the elevator.
41
Each basket contains ~350kg of solids and ~200,000 kg of solids can be extracted per day
Moving-bed leaching: Rotocel extractor
Rotocel extractors are typically 3.4-11.3 m in diameter, 6.4-7.3m in height, and with bed depths of 1.8-3.0m.
42
A horizontal basket is divided into walled compartments with floor that is permeable to the liquid.
Solids are admitted to each compartment at the th e feed point; the compartments then successively pass a number of solvent sprays, a drainage drain age section, se ction, and a discharge point at which the floor of the compartment opens to discharge the extracted solids.
To give countercurrent counte rcurrent extraction, ext raction, fresh solvent is fed only to the last compartment before discharge point.
Continuous, perforated-belt extractor
43
Used to process sugar cane, sugar beets, oil seeds and apples (for apply juice)
The feed solids are fed from a hopper to a slow-moving, continuous and nonpartitioned perforated belt. Below the belt are compartments for collecting solvent
Fresh solvent is sprayed over solids and above the compartments in a countercurrent fashion, starting from the discharge end of the belt.
Units from 7 to 37m long with belts be lts from 0.5 to 9.5m wide and 0.8 to 2.6m deep can process as much as 7,000,000 kg/day kg/day of sugar cane or sugar beets.
Factors influencing leaching processes
Stages involved in leaching: (1) dissolution of solute in the solvent; (2) diffusion of the solute through the solvent; (3) transfer of the solute from the solution in contact with the particles to the bulk of the solution.
Particle size, nature of the solvent used, temperature and agitation are the important factors that influence extraction rates.
44
Countercurrent leaching cascade y N+1
x N
underflow
y L
x L
S = mass flow rate of inert solid and is constant constant from from stage to stage
= mass flow rate of entering solvent or overflow liquid (solvent + solute), V = which varies varies from from stage to stage.
L = mass flow rate of underflow liquid (solvent + solute) which varies varies from from stage to stage.
y = = mass fraction of solute in the overflow liquid
x = = mass fraction of solute in the underflow liquid
45
overflow
Alternatively, V V and and L can refer to mass flow rate of solvent on solute-free basis and symbols Y and X and X can can be used as mass ratios of solute to solvent in overflow and underflow liquids, respectively.
Equilibrium-stage Equilibrium-stag e model
Overflow contains no solid; solvent is not vaporised, adsorbed, or crystallised
Any entering solid solute is completely dissolved into liquid in the stage; composition of liquid in the stage is uniform
Solute is not adsorbed on the surface of inert iner t solid; inert solid leaving in underflow from each stage are wet with liquid, such that mass ratio of solvent in that liquid to inert solid is constant from stage to stage
Concentration of solute in the overflow is equal to that in the liquid li quid equiva lent to an portion of solute in the underflow . This is equivalent equilibrium assumption
46
Equilibrium-stage Equilibrium-stag e model
For the continuous, countercurrent system of ideal leaching stages, solute and total-liquid material balances can be used to solve problems:
Determination of ideal stages required to achieve a specified degree of washing
Determination of the effect of washing for a specified degree of washing with a certain cer tain number of ideal stages
For most problems, it is best to consider the leaching stage separately from the washing stage
Depending on the problem, either an algebraic or a graphic method (e.g., (e .g., McCabe-Thiele method) can be used. 47
Ideal leaching stage
Solid feed, F
Overflow
Insoluble A Soluble B
Liquid (B,C)
Mixer-settler Liquid solvent, S
underflow Solid (A) Liquid (B,C)
48
In an ideal leaching stage, stage, all of the solute is dissolved by the solvent, whereas the solid is not dissolved. Also the composition of the retained liquid phase in the underflow slurry is identical to the composition of the liquid overflow, and that overflow is free of solids.
Leaching and washing
Strong solution
49
When leaching is very rapid, it is common to countercurr cou ntercurrently ently wash the solids to reduce the solute concentration in the liquid adhering to the solids. This can be accomplished in a series of gravity of gravity thickeners thickeners or centrifugal thickeners, called hydroclones. hydroclones.
The arrangement ensures that a clear overflow is obtained.
Operating lines for washing stages
Material balances for the portion of the cascade consisting of the N units are as follows: Total solut solution ion:
LL + V N+1 = LN + V 1
(8.7)
Solute:
L x L L + V N+1 y N+1 = LN x N + V 1 y 1
(8.8)
Solving for yN+1 gives the operating line equation:
y N +1 =
L N V N +1
x N +
V 1 y1 − L L x L V N +1
(8.9)
As usual, the operating line passes through the two end points ( X X L, Y 1) and ( X X N, Y N+1), and if the flow rates are constant, the slope is L/V .
50
McCabe-Smith algebraic method
McCabe and McCabe and Smith derived Smith derived an algebraic method based on Kremser equation equation for the calculation of number of ideal stages, N x N − y N +1 y L − y L
log N =
log( L / V )
The term log(L/V) can be written in terms of end points L V
(8.10)
=
y1 − y N +1 x L − x N
=
y1 − y N +1 y L − x N
(8.11)
Combining eq(8.10) and (8.11) gives x N − y N +1 y − y L L N = y − y N +1 log 1 y − x L N log
(8.12)
When V and and L are liquid flows of solute-free solvent, x solvent, x and y and y in in (8.10) and (8.11) are replaced by X and and Y solute solute mass ratios. 51
Tutorial 9
Study questions
When a binary feed is contacted with a solvent to form two equilibrium liquid phases, which is the extract and which is the raffinate? What are the characteristics of triangular diagrams? On such a diagram, what are the miscibility boundary, plait point, and tie lines? What are the conditions for an ideal, equilibrium leaching stage? What are the similarities and differences differences between extraction and distillation? What is inverse-lev inverse-level-arm el-arm rule and its application in extraction? 52
Class problem - 1
Construction of equilateral triangular diagram for Furfural-Ethylene Glycol-Water system at 25oC and 101 kPa.
53
Class problem - 2
A mixture containing 40wt% acetone and 60wt% water is contacted with an equal amount of MIK
(a) What fraction of the acetone can be extracted in a singlestage process?
(b) What fraction of the acetone could be extracted if the fresh solvent were divided into two parts and two successive extractions used?
Answers: (a) fraction of acetone extracted = 78.9%; (b) Fraction of acetone extracted = 86.3%. 54
Class problem - 3
4000 1b per day of waxed paper containing 25wt% soluble wax and 75wt% insoluble pulp are to be dewaxed dewaxed by leaching with kerosene kerosene in the continuous, continuo us, countercurrent count ercurrent system. The wax wa x is complet c ompletely ely dissolved dissol ved by kerosene in the leaching stage, L. Subsequent washing stages reduce the wax content in the liquid adhering to the pulp leaving the last stage, s tage, N, to 0.2 1b wax/100 1b pulp. Kerosene entering the system is recycled from a solvent-recovery solvent-recovery system and contains 0.05 1b wax/100 1b kerosene. kerosene. The final extract is to contain 5 1b wax/100 1b kerosene. Experiments show that underflow from each stage contains 2 1b kerosene/1b kerosene/1b insoluble pulp. Determine the required washing stage using McCabe-Thiele graphic method. Y 1, X L
Y N+1, X N
55
