Chapter 4 Leaching

8/12/2019 Chapter 4 Leaching

1/31

CHAPTER 4 : LEACHING


2/31

INTRODUCTION

When two phases (solid andliquid phase) are in intimatecontact and the solute or

solutes can diffuse from thesolid to the liquid phase, it

will separate thecomponents originally in the

solid.

This separationprocess is called:

liquid-solid leaching orLeaching or extraction

Definewashing


3/31

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


4/31

Leaching

processBiological and

foodprocessing

Examples:

Leaching of sugar from sugarbeets

Production of vegetable oilsfrom peanuts, soybeans,

sunflower seeds.

Pharmaceuticals product byleaching from roots, leaves

and stems.

Production of soluble instantcoffee, soluble tea

Inorganic andorganic

materials

Examples:

Metal processingindustries

Gold is leached from itsore using an aqueous

sodium cyanide solution


5/31

Fixed-bed leaching

T =344 K to 350 K


6/31


7/31

Moving-bed leaching


8/31

Exercise: Explain the process of

leaching in Agitated solid leaching


9/31

LEACHING PRINCIPLES


10/31

EQUILIBRIUM AND SINGLE STAGELEACHING1. State the components involved in leaching.

2. When is equilibrium reached in leaching?

3. State all the assumptions made for leachingprocess?

4. Explain underflow and overflow.

5. Describe equilibrium line in leaching.6. List factors affecting stage calculations


11/31

LEACHING EQUILIBRIUM

SOLVENT FORMS A LIQUID SOLUTION -CARRIER MAY BE TOTALLY IMMISCIBLE

SOME SOLUTION IS NORMALLY RETAINED

BY THE SOLID - HAS THE SAMECOMPOSITION AS THE LIQUID SOLUTION.

WHEN THE AMOUNT OF RETAINEDSOLUTION IS CONSTANT THE SYSTEM HASCONSTANT SOLUTION UNDERFLOW.

VARIABLE UNDERFLOW EXISTS WHEN THEAMOUNT RETAINED IS A FUNCTION OFCONCENTRATION


12/31

TYPICAL EQUILIBRIUM

DIAGRAM

NOTE THE SOLID PHASE ISREPRESENTED BY THE UPPER LINE

SOLUTION EQUILIBRIUM ON McCABE

IS x = y LINE FOR THIS SITUATION

Y = B / ( A + C )

xA, yA0 1

SOLID PHASE

LIQUID PHASE


13/31

SINGLE STAGE LEACHING

MODELED LIKE A SINGLE STAGE LLX

Y = B / ( A + C )

xA, yA0 1

xa,La

yb,Vb

xb,Lb

ya,Va

MULTIPLE CROSS CURRENT


14/31


LEACHING

MODEL IS BASED ON AMOUNT IN

EACH PHASE

Y = B / ( A + C )

xA, yA0 1

xa,La

yb,Vb

(xb,Lb)1= (xa, La)2

(ya,Va)1

(ya,Va)2

(xb,Lb)2

M1

M2


15/31


LEACHING EXAMPLE

https://portal.navf

ac.navy.mil/portal/

pls/portal/docs/1/3

196547.JPG


16/31

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 USINGABSORPTION FACTORS

FOR ALL STAGES

AFTER THE FIRST MIX


17/31

COUNTERCURRENT

LEACHING CONFIGURATION

http://beta.cheresources.com/articles/basics-of-leaching.html


18/31

MULTISTAGE COUNTERCURRENT LEACHING

MODEL

http://beta.cheresources.

com/articles/basics-of-

leaching.html


19/31

Single stage leachingSOLID FEED, L0 LEACHED SOLID, L1B = SOLID B = SOLID

N0, yA0 N1, yA1

SOLVENT FEED EXTRACT

V2, xA2 V1, xA1

)69.12(

)59.12(

)49.12(

1100

11112200

1120

MNLNLNB

MxVxLyVxLy

MVLVL

M

AMAAAA


20/31

V2, x2

L1, N1, y1, BLo, No, yo, B

V1, x1

slurryslurry

Figure 1 Process flow diagram for single stage extraction


21/31

Graphical solution

Equilibrium line indicatesthe solute concentration inthe solvent is the same inboth the solid underflow

and liquid overflow.

20

2200

20

VL

xVyLx

M

B

VL

BN

AAAM

M


22/31

For feed with no solvent CONCENTRATION IN SOLID IS yA0= 1


23/31

Example 1

In a single-stage leaching of soybean oil fromflaked soybeans with hexane, 100 kg of soybeans

containing 20 wt% oil is leached with 100 kg offresh hexane solvent. The value of N for the slurryunderflow is essentially constant at 1.5 kginsoluble solid/kg solution retained. Calculate theamounts and compositions of the overflow V1 andthe underflow slurry L1 leaving the stage.

V2, x2

L1, N1, y1, BLo, No, yo, B

V1, x1

slurryslurry


24/31

Solution

Given:

V2= 100 kg

xA2= 0

xC2= 1B = 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

yA0= 1


25/31

Find point M from MB

L0+ V2= 20 + 100 = 120 kg = M eq(1)

L0yA0+ V2xA2= 20 (1.0) + 100 (0) = 120 xAM eq(2)

Solving simultaneously eq 1 and 2, xAM= 0.167

B = N0L0= 4.0(20) = 80= NM(120)

So, NM= 0.667


26/31

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. L1MV1and L0MV2must lie on a straight line

2. L1and V1must lie on a vertical line.

3. Point M is the intersection of the two lines.


27/31

Countercurrent multistage leaching

Overall balance: Vn+1+ L0= V1+ Ln

Component balance on solute A: Vn+1xn+1+ L0y0= V1x1+ Lnyn


28/31

i) Variable Underflow in Countercurrent Multistage Leaching

Overall balance:

VN+1+ L0= V1+ LN= M

Component balance on A:

VN+1xAN+1+ L0yA0= V1xA1+ LNyAN= MxAM

Total solids balance on B:

B = N0L0= NNLN= NMM

Coordinate M =(xAM

, NM

)


29/31

Remember:

L0MVN+1must lie on a straight line

V1MLNmust also lie on a straight line

A balance on solute A gives:

A balance on solids gives:

Coordinate operating

point = (xA, N)


30/31

Point is locatedgraphically as theintersection of lines L0V1and LNVN+1

To determine number ofstages Locate L0 Draw line L0to locate V1A tie line through V1

locates L1 Line L1is drawn given V2A tie line gives L2 This is continued until the

desired LNis reached


31/31

THANK YOU

Chapter 4 Leaching

Documents

Transcript of Chapter 4 Leaching