CHAPTER 4 : LEACHING
INTRODUCTION
When two phases (solid and liquid phase) are in intimate contact and the solute or solutes can diffuse from the solid to the liquid phase, it will separate the components originally in the solid.
This separation process is called: liquid-solid leaching or Leaching or extraction
Define washing Define washing
INTRODUCTION
When two phases (solid and liquid phase) are in intimate contact and the solute or solutes can diffuse from the solid to the liquid phase, it will separate the components originally in the solid.
This separation process is called: liquid-solid leaching or Leaching or extraction
Define washing Define washing
LEACHING PROCESS DESCRIPTION •
PROCESS TO CONTACT A SOLID WITH A LIQUID PHASE.
•
SIMILAR TO EXTRACTION BECAUSE OF TWO IMMISCIBLE PHASES
•
SIMILAR TO ABSORPTION BECAUSE TWO PHASES ARE NORMALLY PRESENT
•
DISSIMILAR TO OTHER SEPARATION PROCESS BECAUSE ONLY EQUILIBRIUM IN LIQUID PHASES IS CONSIDERED
Leaching process Biological and food processing
Inorganic and organic materials
Examples: Leaching of sugar from sugar beets
Examples:
Production of vegetable oils from peanuts, soybeans, sunflower seeds.
Metal processing industries
Pharmaceuticals product by leaching from roots, leaves and stems.
Gold is leached from its ore using an aqueous sodium cyanide solution
Production of soluble instant coffee, soluble tea
Fixed-bed leaching T =344 K to 350 K
LEACHING PRINCIPLES
EQUILIBRIUM AND SINGLE STAGE LEACHING 1. State the components involved in leaching. 2. When is equilibrium reached in leaching? 3. State all the assumptions made for leaching process? 4. Explain underflow and overflow. 5. Describe equilibrium line in leaching. 6. List factors affecting stage calculations
LEACHING EQUILIBRIUM • •
•
•
SOLVENT FORMS A LIQUID SOLUTION CARRIER MAY BE TOTALLY IMMISCIBLE SOME SOLUTION IS NORMALLY RETAINED BY THE SOLID - HAS THE SAME COMPOSITION AS THE LIQUID SOLUTION. WHEN THE AMOUNT OF RETAINED SOLUTION IS CONSTANT THE SYSTEM HAS CONSTANT SOLUTION UNDERFLOW . VARIABLE UNDERFLOW EXISTS WHEN THE AMOUNT RETAINED IS A FUNCTION OF CONCENTRATION
TYPICAL EQUILIBRIUM DIAGRAM •
NOTE THE SOLID PHASE IS REPRESENTED BY THE UPPER LINE
•
SOLUTION EQUILIBRIUM ON McCABE IS x = y LINE FOR THIS SITUATION SOLID PHASE
Y=B/(A+C)
LIQUID PHASE
0
xA, yA
1
SINGLE STAGE LEACHING •
MODELED LIKE A SINGLE STAGE LLX xa,La
xb,Lb
Y=B/(A+C)
yb,Vb
0
ya,Va
xA, yA
1
MULTIPLE CROSS CURRENT LEACHING •
MODEL IS BASED ON AMOUNT IN EACH PHASE (xb,Lb)2
xa,La
(xb,Lb)1 = (xa, La)2
Y=B/(A+C) M2 M1 yb,Vb
0
(ya,Va)1 xA, yA (ya,Va)2
1
MULTIPLE CROSS CURRENT LEACHING EXAMPLE https://portal.navf ac.navy.mil/portal/ pls/portal/docs/1/3 196547.JPG
MULTISTAGE COUNTERCURRENT LEACHING •
RESULTS FOR DESIGN ARE SIMILAR TO SHOWN IN FIG. 12.10-2
•
FOR SYSTEM WITH CONSTANT L/V RATIO, THE APPROACH IS TO MODEL USING ABSORPTION FACTORS FOR ALL STAGES AFTER THE FIRST MIX
COUNTERCURRENT LEACHING CONFIGURATION
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Multi – stage counter current Leaching Process flow Overflow solution
Solvent Feed
V 1, x 1
V N+1, x N+1 Underflow solution
Feed Slurry
LN , N N , y N , B
L0 , N 0 , y 0 , B V L B
Mass of overflow solution Mass of liquid in slurry solution Mass of dry, solute – free solid.
x A y A
Composition of A at overflow solution Composition of A at slurry solution
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 L0 V N 1 L N V 1 M
Total solution balance
L0 y A0 V N 1 x AN 1 L N y AN V 1 x A1 Mx AM
B N 0 L0
N N L N N M M
Comp. A balance
Solid balance
Multi – stage counter current Leaching 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 Overflow solution
Solvent Feed
V 1, x 1
V N+1, x N+1 Underflow solution
Feed Slurry
LN , N N , y N , B
L0 , N 0 , y 0 , B
Information given: Feed slurry (L 0): A = 800 kg/h
B = 2000 kg/h
C = 50 kg/h
Entering solvent (V N+1 ) A = 20 kg/h
B = 0 kg/h
C = 1310 kg/h
Multi – stage counter current Leaching Solution 2 Overflow solution
Solvent Feed
V 1, x 1
V N+1, x N+1 Underflow solution
Feed Slurry
LN , N N , y N , B
L0 , N 0 , y 0 , B
Information given: 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. Mass of A = 800 kg/h Mass of B = 2000 kg/h Mass of C = 50 kg/h y A0 N 0
A
L0
B L0
A
800
A C
B A C
800
800
50
2000
2000
800
50
0.94
850
2.35
850
Coordinate for L0 (yA0 , N 0) = (0.94 , 2.35)
Multi – stage counter current Leaching Solution 2
Find coordinate at V N+1. Mass of A = 20 kg/h Mass of B = 0 kg/h Mass of C = 1310 kg/h x N 1
A
V N 1
N N 1
A
B V N 1
20
A C
B A C
20
20 1310
0
0.015
1330
0
20 1310
Coordinate for V N+1(xN+1 , N N+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 Slope of graph,
N N y N
B N N y N
L N A
B A
2000
16.67
120
N
L N
If x = 0.1, N = 16.67 x 0.1 = 1.67 Plot New Coordinate(x , N ) = (0.1 , 1.67)
16.67 y
Multi – stage counter current Leaching Solution 2 3 2
L N L0
1 0 -0.4 N
-0.2
-1 0
V N 1
0.2
-2 -3 -4 -5 -6 -7 x A, y A
0.4
0.6
0.8
1
Multi – stage counter current Leaching Solution 2
From material balance calculations: Total solution balance: L0 V N 1 L N V 1 M
L0 V N 1 M
850 1330
M
M 2180 kg
Component A balance:
L0 y A0 V N 1 x N 1 Mx AM
850 0.94 1330 0.015 2180 x AM x AM 0.376
Multi – stage counter current Leaching Solution 2
From material balance calculations: Solid balance: B N 0 L0 N 0 L0
N N L N N M M
N M M
2.35850 N M 2180
N M
0.916
Coordinate for M (xM , N M) = (0.376 , 0.916) Plot coordinate M in the graph. Construct line from point L N to point M until it cross at x – axis. Point at x – axis = V 1
Multi – stage counter current Leaching 3 L N
2
L0 M
1
V 1
0 -0.4
-0.2
N
-1
V N 1
0
0.2
0.4
-2 -3 -4 -5 -6 -7
From figure,
x A, y A
Coordinate for V 1 (x1 , N 1) = (0.592 , 0)
0.6
0.8
1
Multi – stage counter current Leaching Solution 2
From material balance calculations: Total solution balance: V 1 2180 L N
Eq.1
Component A balance: L N y N V 1 x1 Mx AM L N 0.12 V 1 0.592 2180 0.376
Insert Eq. 1 into equation above L N 0.12 2180 L N 0.592 2180 0.376 0.12 L N 1290 .56 0.592 L N 0.472 L N
470 .88
819 .68
L N 997.62 kg
Multi – stage counter current Leaching Solution 2
From material balance calculations: Total solution balance: V 1 2180 L N
Eq.1
V 1 2180 997.62
V 1 1182 .38 kg
Construct operating point: Connect L0 with V 1 & LN with V N+1. The cross line – operating point. Total stages: 4 stages
Point Δ is located graphically as the intersection of lines L 0V1 and LNVN+1 To determine number of stages –Locate
L0 –Draw line L0 Δ to locate V1 – A tie line through V1 locates L1 –Line L1 Δ is drawn given V2 – A tie line gives L2 –This is continued until the desired LN is reached
Multi – stage counter current Leaching Solution 2
Construct the stages: 3 L N
2
L3
L2
L1
L0
M
1
V 1
0
V N 1
-0.4
-0.2
N
-1
0
0.2
-2 -3 -4 -5 P
-6 -7 x A, y A
0.4
0.6
0.8
1
MULTISTAGE COUNTERCURRENT LEACHING MODEL
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LEACHING LIQUID-SOLID LEACHING
Moving-bed leaching
Exercise: Explain the process of leaching in Agitated solid leaching
Single stage leaching SOLID FEED, L0 B = SOLID N0, y A0
LEACHED SOLID, L1 B = SOLID N1, y A1
SOLVENT FEED V 2, x A2
EXTRACT V 1, x A1
L0
V 2
y A0L0
L1 V 1
x A 2V 2
B N 0L0
M
(12 .9 4)
y A1L1 x A1V 1 x AM M (12.9 5)
N 1L1 N M M (12.9 6)
V1, x1
V2, x2
slurry
slurry
Lo, No, yo, B
L1, N1, y1, B
Figure 1 Process flow diagram for single stage extraction
Graphical solution Equilibrium line indicates the solute concentration in the solvent is the same in both the solid underflow and liquid overflow.
N M
B
x AM
L0
V 2
B M
L0 y A0
V 2 x A 2
L0
V 2
For feed with no solvent •
CONCENTRATION IN SOLID IS y A0 = 1
Example 1
V1, x1
V2, x2
slurry
slurry
Lo, No, yo, B
L1, N1, y1, B
In a single-stage leaching of soybean oil from flaked soybeans with hexane, 100 kg of soybeans 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.
Solution Given: V2 = 100 kg x A2 = 0 xC2 = 1 B = 100 (1.0 – 0.2) = 80 kg insoluble solid L0 = 100 (1.0 – 0.8) = 20 kg A N0 = 80/20 = 4 kg solid/kg solution y A0 = 1
Find point M from MB L0 + V2 = 20 + 100 = 120 kg = M L0y A0 + V2x A2 = 20 (1.0) + 100 (0) = 120 x AM
Solving simultaneously eq 1 and 2, x AM = 0.167 B = N0L0 = 4.0(20) = 80 = NM(120) So, NM = 0.667
eq(1) eq(2)
The coordinates for the points: Lo = (y0,N0) L1 = (y1,N1) V1 = (x1,0) V2 = (x2,0) M = (xM,NM)
L0 = (1,4) L1 = (read from graph,1.5) V1 = (read from graph,0) V2 = (0,0) M = (0.167, 0.667)
Please take note: 1. L1MV1 and L0MV2 must lie on a straight line 2. L1 and V1 must lie on a vertical line. 3. Point M is the intersection of the two lines.
NOW CHECK UNDERSTANDING….. 1.
State the components involved in leaching.
2.
Describe about inert materials in leaching.
3.
Solute-free solid is insoluble in the solvent Sufficient solvent is present No adsorption of the solute by the solid in the leaching
Explain underflow and overflow.
6.
When the solute is dissolved in the liquid.
State all the assumptions made for leaching process?
5.
The inert or insoluble solute B in the slurry is denote with N. Is kg B over total kg A plus C
State condition when the equilibrium is reached in leaching?
4.
Solute (A), Inert/Leached Solid (B) and Solvent (C)
Underflow-A slurry stream/solid stream Overflow – Liquid steam/solvent stream
List factors affecting stage calculations
Concentrations of the solute in the solution The retention time Temperature of process
Countercurrent multistage leaching
Overall balance: V n+1 + L0 = V 1 + Ln Component balance on solute A: V n+1 xn+1 + L0 y 0 = V 1x1 + Ln y n
i) Variable Underflow in Countercurrent Multistage Leaching
Overall balance: V N+1 + L0 = V 1 + LN = M Component balance on A: V N+1 x AN+1 + L0 y A0 = V 1x A1 + LN y AN = Mx AM Total solids balance on B: B = N0L0 = NNLN = NMM Coordinate M =(x AM , NM)
Remember: •
L0MV N+1 must lie on a straight line
• V 1MLN
must also lie on a straight line
A balance on solute A gives:
A balance on solids gives: Coordinate operating point Δ = (x A Δ , NΔ)
