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leaching
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CHAPTER / CONTENT
Introduction to Leaching Process
Principles of Leaching
Single Stage Calculation
Multi Stage Countercurrent system
Leaching equipment
Widely used in the metallurgical, natural product and food industries under batch, semi – continuous or continuous condition.
The major difference between Leaching and LLE centers about the difficulty to transport the solid or the solid slurry from stage to stage.
Leaching is also known as solid – liquid exraction
In leaching, to separate the desired solute constituent or remove the undesirable solute component from the solid phase, the solid is contacted with a liquid phase.
The two phases are in intimate contact and the solute or solutes can diffuse from the solid to the liquid phase, which causes a separation of the components originally in the solid.
This process is called liquid – solid leaching or simply leaching
Introduction to Leaching Process
Leaching process for biological substances
Introduction to Leaching Process
An important such process is to leach sugar from sugar beets with hot water.
In production of vegetable oils, organic solvents such as hexane, acetone and ether are used to extract the oil from peanuts, soybeans, flax seeds, castor beans, sunflower seeds, etc.
In pharmaceutical industry, many different pharmaceutical products are obtained by leaching plan roots, leaves and stems.
For ‘instant’ coffee, ground roasted coffee is leached with water and soluble tea is produced by water leaching of tea leaves.
Tannin is removed from tea barks by leaching with water.
Leaching process for inorganic and organic materials
Introduction to Leaching Process
Used in metal – processing industries
In metal ores, the desired metal components usually occur with a large amount of undesirable constituents and leaching is used to obtain these metal components in the form of metal salts.
E.g.: Copper salts are leached by dissolving raw copper ores by using sulfuric acid or ammoniacal solutions.
E.g.: Nickel salts are leached using sulfuric acid – ammonia – oxygen mixures.
Gold is leached using an aqueous sodium cyanide solution.
The solvent must be transferred from the bulk solvent solution to the surface of the solids.
Next, the solvent must penetrate or diffuse into the solids.
The solute then diffuses through the solid solvent mixture to the surface of the particle.
Finally, the solute is transferred to the bulk solution.
The rate of the solvent transfer from the bulk solution to the solid surface is quite rapid.
However, the rate of transfer of the solvent into the solid can be rather slow or rapid.
This solvent transfer usually occurs initially when the particle are first contacted with the solvent.
Principles of Leaching
The rate of diffusion of the solute through the solid and solvent to the surface of the solid is often the controlling resistance in the overall leaching process and can depend on a number of different factors.
If the solid is made of porous the diffusion through the porous solid can be described by an effective diffusivity.
The resistance to mass transfer to the solute from the solid surface to the bulk solvent is generally quite small compared to the resistance to the diffusion within the solid itself.
Principles of Leaching
Single – stage Leaching
Process flow
Feed Slurry
L0, N0, y0, B
Overflow solution
V1, x1
Underflow solution
L1, N1, y1, B
V2, x2
Solvent Feed
V Mass of overflow solution xA Composition of A at overflow solutionL Mass of liquid in slurry solution yA Composition of A at slurry solutionB Mass of dry, solute – free solid.N Mass of dry,solute (B)/Mass of solution retained (L) Material balance is divided into 3 parts:
balanceSolid
balanceA Comp.
balancesolution Total
MNLNLNB
MxxVyLxVyL
MVLVL
M
AMAAAA
00 1100
11112200
1120
Example 1
Single – stage calculations
In a single – stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybean containing 20 wt% oil is leached with 100 kg of fresh hexane solvent.
The value of N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained.
Calculate the amounts and compositions of the overflow V1 and the underflow slurry L1 leaving the stage.
Single – stage calculations
Solution 1
Information given:
Entering solvent, V2 = 100 kg
Feed Slurry
L0, N0, y0, B
Overflow solution
V1, x1
Underflow solution
L1, N1, y1, B
V2, x2
Solvent Feed
Feed slurry = 100 kg containing 20 wt% oil
N = 1.5 kg B/kg (A+C)
Single – stage calculations
Solution 1
Find coordinate at L0.
Coordinate for L0
Mass of A = 0.20 x 100 A = 20 kg
Mass of B = 0.80 x 100 B = 80 kg
Mass of C = 0 kg C = 0 kg
0.402
80
0.102
20
00
00
0
0
CA
B
L
BN
CA
A
L
AyA
(yA0 , N0) = (1.0 , 4.0)
Single – stage calculations
Solution 1
Find coordinate at V2.
Coordinate for V2
Mass of A = 0 A = 0 kg
Mass of B = 0 B = 0 kg
Mass of C = 100 kg C = 100 kg
01
0
000
0
22
22
00 0
10
CA
B
V
BN
CA
A
V
Ax
(x2 , N2) = (0 , 0)
Single – stage calculations
Solution 1
From material balance calculations:
Total solution balance:
167.0
12001000.1202200
AM
AM
AMAA
x
x
MxxVyL
kg 120100201120
MM
MVLVL
Component A balance:
Single – stage calculations
Solution 1
667.012020400
1100
MM
M
M
NN
MNLN
MNLNLNB
Solid balance:
Coordinate for M (xM , NM) = (0.167 , 0.667)
Plot coordinate M in the graph.
Construct straight vertical line through point M in order to find value V1 and L1
Single – stage calculations
Solution 1
From figure,
0
0.5
1
1.5
2
2.5
3
3.5
4
0 0.2 0.4 0.6 0.8 1
xA, yA
N
1L
1V AxN versus
AyN versus
Coordinate for V1 (x1 , N1) = (0.167 , 0)
Coordinate for L1 (y1 , N1) = (0.167 , 1.5)
0L
2V
M
Single – stage calculations
Solution 1
From material balance calculations:
Total solution balance:
1Eq.
11
11
11
120
120
LV
VL
MVL
kg 53.36
11
11
1100
120667.05.1 LL
MNLN
MNLNLNB
M
M
Solid balance:
Single – stage calculations
Solution 1
From material balance calculations:
From Eq. (1)
kg
1Eq.
64.6636.53120
120
11
11
VV
LV
Tutorial:
12.9-2-Textbook, page 835
A slurry of flakes soybeans weighing a total of
100kg contains 75kg of inert solids and 25kg
of solution with 10wt% oil and 90wt% solvent
hexane. This slurry contacted with 100kg of
pure hexane in a single stage so that the
value of N for the outlet underflow is 1.5kg
insoluble solid/kg solution retained. Determine
the amount and compositions of the overflow
V1 and the underflow L1 leaving the stage.
Answer:
Coordinate Lo (0.1, 3.0)
Coordinate V2 (0,0)
Coordinate M (0.02, 0.6)
Coordinate V1 (0.02,0)
Coordinate L1 (0.02,1.5)
L1= 50 kg
V1=75 kg
Solution balance;
L1 + V1 = M
V1= 125-L1
Solid balance;
N1.L1=B
1.5 (L1)=75
L1=50 kg
Therefore
V1=125-50
= 75kg
Example 12.9-1In a single – stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybean containing 22 wt% oil is leached with hexane. The solvent feed is 80 kg of solvent containing 3 wt% of soybean oil. The value of N for the slurry underflow is essentially constant at 1.5 kg insoluble solid/kg solution retained.Calculate the amounts and compositions of the overflow V1 and the underflow slurry L1 leaving the stage.
Multi – stage counter current Leaching
Process flow
Feed Slurry
L0, N0, y0, B
Overflow solution
V1, x1
Underflow solution
LN, NN, yN, B
VN+1, xN+1
Solvent Feed
V Mass of overflow solution xA Composition of A at overflow solutionL Mass of liquid in slurry solution yA Composition of A at slurry solutionB Mass of dry, solute – free solid.
Multi – stage counter current Leaching
The ideal stages are numbered in the direction of the solids or underflow stream.
The solvent (C) – solute (A) phase or V phase is the liquid phase that overflows continuously from stage to stage countercurrently to the solid phase, and it dissolves solute as it moves along.
The slurry phase L composed of inert solid (B) and liquid phase of A and C is the continuous underflow from each stage.
Composition of V – denoted by x
Composition of L – denoted by y
Assumption: The solid B is insoluble and is not lost in the liquid V phase.
The flow rate of solid is constant throughout the process
Multi – stage counter current Leaching
balanceSolid
balanceA Comp.
balancesolution Total
MNLNLNB
MxxVyLxVyL
MVLVL
MNN
AMAANNANNA
NN
00
111100
110
Example 2
A continuous countercurrent multistage system is to be used to leach oil from meal by benzene solvent (B3).
The process is to treat 2000 kg/h of inert solid meal (B) containing 800 kg oil (A) and also 50 kg benzene (C).
The inlet flow per hour of fresh solvent mixture contains 1310 kg benzene and 20 kg oil. The leached solids are to contain 120 kg oil.
Data (B3) are tabulated below as N kg inert solid B/kg solution and yA
kg oil A/kg solution
Calculate the amounts and concentrations of the stream leaving the process and the number of stages required.
Multi – stage counter current Leaching
Solution 2
Information given:
Entering solvent (VN+1 )
A = 20 kg/h B = 0 kg/h C = 1310 kg/h
Feed Slurry
L0, N0, y0, B
Overflow solution
V1, x1
Underflow solution
LN, NN, yN, B
VN+1, xN+1
Solvent Feed
Feed slurry (L0): A = 800 kg/h B = 2000 kg/h C = 50
kg/h
Multi – stage counter current Leaching
Solution 2
Information given:
Feed Slurry
L0, N0, y0, B
Overflow solution
V1, x1
Underflow solution
LN, NN, yN, B
VN+1, xN+1
Solvent Feed
Underflow solution (LN): A =120 kg/h B = 2000 kg/h C = ??
kg/h
Multi – stage counter current Leaching
Solution 2
Find coordinate at L0.
Coordinate for L0
Mass of A = 800 kg/h
Mass of B = 2000 kg/h
Mass of C = 50 kg/h
35.2850
2000
08
2000
94.0850
800
08
800
00
00
5 00
5 00
CA
B
L
BN
CA
A
L
AyA
(yA0 , N0) = (0.94 , 2.35)
Multi – stage counter current Leaching
Solution 2
Find coordinate at VN+1.
Coordinate for VN+1
Mass of A = 20 kg/h
Mass of B = 0 kg/h
Mass of C = 1310 kg/h
0131020
0
015.01330
20
131020
20
11
11
CA
B
V
BN
CA
A
V
Ax
NN
NN
(xN+1 , NN+1) = (0.015 , 0)
Multi – stage counter current Leaching
Solution 2
Find coordinate at LN.
Mass of A = 120 kg/h
Mass of B = 2000 kg/h
Mass of C = ?? kg/h
y. N A
B
LA
LB
y
N
y
N
N
N
N
N
N
N
671667.16120
2000
graph, of Slope
Multi – stage counter current Leaching
If x = 0.1, N = 16.67 x 0.1 = 1.67
Plot New Coordinate(x , N) = (0.1 , 1.67)
Solution 2
Multi – stage counter current Leaching
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1
x A, y A
N
1NV
0LNL
Solution 2
From material balance calculations:
Total solution balance:
376.0
2180015.0133094.08501100
AM
AM
AMNNA
x
x
MxxVyL
kg 2180133085010
110
MM
MVL
MVLVL
N
NN
Component A balance:
Multi – stage counter current Leaching
Solution 2
From material balance calculations:
Multi – stage counter current Leaching
916.0218085035.200
00
MM
M
MNN
NN
MNLN
MNLNLNB
Solid balance:
Coordinate for M (xM , NM) = (0.376 , 0.916)
Plot coordinate M in the graph.
Construct line from point LN to point M until it cross at x – axis. Point at x – axis = V1
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1
x A, y A
N
Multi – stage counter current Leaching
1NV
0LNL
M
1V
From figure,
Coordinate for V1 (x1 , N1) = (0.592 , 0)
Coordinate for LN (y1 , N1) = (0.12 , 2.0)
Solution 2
From material balance calculations:
Total solution balance:
1Eq. NLV 21801
Multi – stage counter current Leaching
kg 997.62
above equation into 1 Eq. Insert
NN
NN
NN
N
AMNN
LL
LL
LL
VL
MxxVyL
88.470472.0
68.819592.056.129012.0
376.02180592.0218012.0
376.02180592.012.0 1
11
Component A balance:
Solution 2
From material balance calculations:
Total solution balance:
kg
1Eq.
38.1182
62.9972180
2180
1
1
1
V
V
LV N
Multi – stage counter current Leaching
Connect L0 with V1 & LN with VN+1. The cross line – operating point.
Total stages: 4 stages
Construct operating point:
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
-0.4 -0.2 0 0.2 0.4 0.6 0.8 1
x A, y A
N Solution 2
Multi – stage counter current Leaching
1NV
0LNL
M
1V
Construct the stages:
P
1L2L3L
Tutorial
12.10-4 Textbook, page 835Fresh halibut livers containing 25.7 wt% oil
are to be extracted with pure ethyl ether to
remove 95 % of the oil in a countercurrent
multistage leaching process. The feed rate is
1000 kg of fresh livers per hour. The final exit
overflow solution is to contain 70 wt% oil. The
retention of solution by inert solids (oil free
liver) of the liver varies as follow, where N is
kg inert solid/kg solution retained and yA is kg
oil/kg solution
Fixed – Bed Leaching
Used in beet sugar industry and is also used for extraction of tanning extracts from the tanbark, extraction of pharmaceuticals from barks and seeds and other processes.
Figure 12.8-1 shows a typical sugar beet diffuser or extractor. The cover is removable so
that sugar beet slices called cossettes can be dumped into the bed.
Heated water at 344 K to 350 K flows into the bed to leach out the sugar.
The leached sugar solution flows out the bottom onto the next tank in series.
About 95% of the sugar in beets is leached to yield an outlet solution from the system of about 12 wt%.
Moving – Bed Leaching
There are number of devices for stagewise countercurrent leaching where the bed or stages moves.
Used widely in extracting oil from vegetable seeds such as cottonseeds, peanuts and soybeans.
The seeds are usually dehulled first, sometimes precooked, often partially dried and rolled or flaked.
The solvents used are particularly hydrocarbons such as hexane and the final solvent – vegetable solution called miscella may contain some finely divided solids.
Agitated Solid Leaching
When the solid can be ground fine abou 200 mesh (0.074 mm), it can kept in suspension by small amounts of agitation.
Continuous countercurrent leaching can be accomplished by placing the number of agitator in series, with setttling tanks or thickeners between each agitator.
Sometimes thickeners are used as combination contactor – agitators and settlers – shown in Figure 12.8-3.
To analyze single – stage and countercurrent – stage leaching, an operating line equation, or material balance relation and the equilibrium relations between the two streams are needed as in LLE.
Assumptions made by achieving the equilibrium relations:
Sufficient solvent is present so that all the solute in the entering solid dissolved in the solvent.
The solute in the entering solid dissolved completely in the first stage.
No adsorption of the solute by the solid. * This means the solution in the liquid phase leaving a stage is the same as the solution that remains with the solid matrix in the settled slurry leaving the stage.
Equilibrium Relations in Leaching
The settled solid leaving a stage always contains some liquid in which dissolved solids is present.
The solid – liquid stream is called underflow or slurry stream.
Consequently, the concentration of oil or solute in the liquid or overflow stream is equal to the concentration of solute in the liquid solution accompanying the slurry or underflow stream.
The amount of solution retained with the solids in the settling portion of each stage may depend the density and viscosity of liquid in which the solid is suspended.
Equilibrium diagrams for leaching:
Equilibrium Relations in Leaching
The concentration of inert or insoluble solid B in the solution mixture or the slurry mixture can be expressed in kg (lbm) units:
solution lbsolid lb
solution kgsolid kg
kg kg kg
CA
BN
For overflow, N = 0
For underflow, N value depending on the solute concentration in the liquid.
Equilibrium Relations in Leaching
The composition of solute A in liquid will be expressed as wt fractions:
liquid underflow
or slurry in liquid
solution kgsolute kg
kg kg kg
liquid overflow solution kgsolute kg
kg kg kg
CA
Ay
CA
Ax
A
A
QUIZ 2 (45 minutes)
A continuous countercurrent multistage system is to be used to leach oil from meal by benzene solvent. The process is to treat 2000 kg/h of inert solid meal containing 800 kg oil and also 50kg benzene. The inlet flow per hour of fresh solvent mixture contains 1310 kg benzene and 20kg oil. The leached solids are to contain 120 kg oil. If the value of N for the outlet underflow is constant at 1.85kg solid/kg solution, determine
a)The exit flows compositions (xA1, yAN, overflow V1 and the underflow LN leaving the stage.
b)Number of stages required