Chapter 3 - Leachingedited

CHAPTER 3: SOLID – LIQUID EXTRACTION /

LEACHING

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


Leaching process for biological substances


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


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.


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.


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.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.


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)


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)


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)


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:


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


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


Solution 1



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:


Solution 1


From Eq. (1)

kg

1Eq.

64.6636.53120

120

11

11

VV

LV

5 min break…

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.


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


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.


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


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


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)


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)


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


If x = 0.1, N = 16.67 x 0.1 = 1.67

Plot New Coordinate(x , N) = (0.1 , 1.67)

Solution 2


-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



376.0

2180015.0133094.08501100

AM

AM

AMNNA

x

x

MxxVyL

kg 2180133085010

110

MM

MVL

MVLVL

N

NN



Solution 2



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


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



1Eq. NLV 21801


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


Solution 2



kg

1Eq.

38.1182

62.9972180

2180

1

1

1

V

V

LV N


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


1NV

0LNL

M

1V

Construct the stages:

P

1L2L3L

5 min break….

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

Tutorial

12.10-3 Textbook, page 835

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:


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.


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 245 minutes

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

Chapter 3 - Leachingedited

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