Chap 1.3 - Leaching
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Transcript of Chap 1.3 - Leaching
CHAPTER 3: SOLID – LIQUID EXTRACTION /
LEACHING
CHAPTER / CONTENT
Introduction to Leaching Process
Types of Equipments for Leaching
Equilibrium Relations in Leaching
Rates of Leaching
Calculation in Leaching
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 or also known as solid – liquid exraction involves dissolving soluble material from its mixture with an insoluble solid.
Many biological inorganic or biological organic substances occur in a different mixture of different components in a solid.
In order to separate the desired solute constituent or remove the undesirable solute component from the solid phase, the solid is contacted with a liquid phase.
Introduction to Leaching Process
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.
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.
Rates of Leaching
PRINCIPLES OF LEACHING
RATE OF LEACHING WHEN DISSOLVING A SOLID
METHODS OF OPERATING IN LEACHING
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 and solid structure with the solute and solvent in the pores in the solid, the diffusion through the porous solid can be described by an effective diffusivity.
In biological or natural substances, additional complexity occurs because of the cell present, in the leaching of thin sugar beet slices, one – fifth of the cells are ruptured in the slicing of the beets.
The leaching of sugar is then similar to the washing process, where in the remaining cells, sugar must diffuse out through the cell walls. With soybeans, whole bean cannot be leached effectively. The rolling and flaking of the soybeans ruptures cell walls so that the solvent can more easily penetrate by capillary action.
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
When a material is being dissolved from the solid to the solvent solution, however, the rate of mass transfer from the solid surface to the liquid is the controlling factor.
There is essentially no resistance in the solid phase if it is a pure materials.
The equation for this can be derived as follows from a batch system.
The rate of mass transfer of the solute A being dissolved to the solution of volume V in m3 is:
Rate of leaching when dissolving a solid
1Eq.
AASLA cck
A
N
3
A
3
AS
L
2
mol/m kg
in sec time at solution the in A of ionconcentrat c
mol/m kg in solution
the in A solute solid the of solubility saturation c
m/s in tcoefficien tansfer mass k
m in particles of area surface the
second per solution the to dissolving A of mol kg
where
t
A
N A
By material balance, the rate of accumulation of A in the solution is equal to equation (1) times the area:
Rate of leaching when dissolving a solid
AASLAA ccAkN
dt
dcV
Integrating from time t = 0 and cA = cA0 to t = t and cA = cA:
tVAk
AAS
AAS
t
t
L
c
c AAS
A
L
A
A
ecc
cc
dtV
Ak
cc
dc
0
00
The solution approaches a saturated condition exponentially.
Often the interfacial area A will increase during the extraction if the external surface becomes very irregular.
If the soluble materials forms a high proportion of the total solid, disintegration of the particles may occur.
Example 12.8 – 1. Prediction of Time for Batch Leaching
Rate of leaching when dissolving a solid
Particles having an average diameter of approximately 2.0 – mm are leached in a batch type apparatus with a large volume of solvent.
The concentration of the solute A in the solvent is kept approximately constant.
A time of 3.11 hour is needed to leach 80% of the available solute from the solid.
Assuming that diffusion in the solid is controlling and the effective diffusivity is constant, calculate the time of leaching if the particle size is reduced to 1.5 mm.
Solution 12.8 – 1. Prediction of Time for Batch Leaching
Rate of leaching when dissolving a solid
For 80% extraction, the fraction unextracted ES is 0.20.
Using Figure 5.3 – 13 for a sphere, for ES = 0.20, a value of DAeff t/a2 = 0.112 is obtained, where DAeff is the effective diffusivity in mm2/s, t is time in s, and a is radius in mm.
For the same fraction ES, the value of DAeff t/a2 is constant for a different size. Hence,
3Eq. 21
221
2 a
att
where t2 is time for leaching with a particle size a2. Substituting in equation above:
h 75.120.2
25.111.3 2
2
2 t
Methods of operating in leaching
Can be carried out in batch or unsteady state conditions, continuous and steady state conditions.
Both continuous and stage wise types of equipment are used in steady or unsteady state operations.
In steady state leaching a common method used in the mineral industries is in – place leaching, where the solvent is allowed to percolate through the actual ore body.
For example, copper is leached by sulfuric acid from sulfides ore by leach liquor is pumped over a pile of crushed ore and collected at the ground level as it drains from the heap.
Types of Equipments for Leaching
FIXED – BED LEACHING
MOVING BED LEACHING
AGITATED SOLID LEACHING
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
SINGLE – STAGE LEACHING
COUNTER – CURRENT MULTISTAGE LEACHING
Calculation in 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.
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
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 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