Chapter3 Leaching 131018031158 Phpapp02

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Leaching

Transcript of Chapter3 Leaching 131018031158 Phpapp02

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CHAPTER 3: SOLID – LIQUIDEXTRACTION / LEACHING

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CHAPTER / CONTENT

Introduction to Leaching Process

Principles of Leaching

Single Stage Calculation

Multi Stage Countercurrent system

Leaching equipment

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  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 thedifficulty to transport the solid or the solid slurry from stage to stage. Leaching is also known assolid – liquid exraction

  In leaching, to separate the desired solute constituent or remove theundesirable solute component from the solid phase, the solid is contacted witha liquid phase.

 The two phases are in intimate contact and the solute or solutes can diffusefrom the solid to the liquid phase, which causes a separation of thecomponents originally in the solid.

 This process is called liquid – solid leaching or simply leaching

Introduction to Leaching Process

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 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 andether are used to extract the oil from peanuts, soybeans, flax seeds, castor beans,

sunflower seeds, etc.

 In pharmaceutical industry, many different pharmaceutical products areobtained 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.

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 The solvent must be transferred from the bulk solvent solution to thesurface 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 surfaceis quite rapid.

 However, the rate of transfer of the solvent into the solid can be ratherslow or rapid.

 This solvent transfer usually occurs initially when the particle are firstcontacted with the solvent.

Principles of Leaching

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SINGLE – STAGE LEACHING

COUNTER – CURRENT MULTISTAGE LEACHING

Calculation in Leaching

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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  x A Composition of A at overflow solution

L Mass of liquid in slurry solution y A Composition of A at slurry solution

B 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 

 balanceAComp. 

 balancesolutionTotal 

 M  N  L N  L N  B

 Mx xV  y L xV  y L

 M V  LV  L

 M 

 AM  A A A A

=+=+=

=+=+=+=+

001100

11112200

1120

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 Example 1

Single – stage calculations

In a single – stage leaching of soybean oil from flaked soybeans with hexane, 100kg of soybean containing 20 wt% oil is leached with 100 kg of fresh hexane solvent.

The value ofN for the slurry underflow is essentially constant at 1.5 kg insolublesolid/kg solution retained.

Calculate the amounts and compositions of the overflow V1 and the underflow

slurryL1 leaving the stage.

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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)

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Single – stage calculations

 Solution 1

From material balance calculations:

Total solution balance:

( ) ( ) ( ) ( ) ( )

167.0

12001000.120

2200

==+

=+

 AM 

 AM 

 AM  A A

 x

 x

 Mx xV  y L  

kg  12010020

1120

==+

=+=+

 M  M 

 M V  LV  L

Component A balance:

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Single – stage calculations

 Solution 1

( ) ( ) ( )   667.0120204

00

1100

===

===

 M  M 

 M 

 M 

 N  N 

 M  N  L N 

 M  N  L N  L N  B

 

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 valueV1 and

L1 

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Single – stage calculations

 Solution 1

From material balance calculations:

From Eq. (1)

( )

kg 

1Eq. 

64.6636.53120

120

11

11

=−=

→−=

V V 

 LV 

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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% solent

he!ane" #his slurry contacted with 100kg of

$ure he!ane in a single stage so that the

alue of for the outlet underflow is 1"5kg

insoluble solid&kg solution retained" 'etermine

the amount and com$ositions of the oerflow

(1 and the underflow )1 leaing the stage"

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 Answer:

*oordinate )o +0"1, -"0.

*oordinate (2 +0,0.

*oordinate / +0"02, 0".

*oordinate (1 +0"02,0.*oordinate )1 +0"02,1"5.

)1 50 kg

(175 kg

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olution balance3

)1 4 (1 /

(1 125)1

olid balance3

1")16

1"5 +)1.75

)150 kg

#herefore(112550

  75kg

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Example 12.9-1

n a single 8 stage leaching of soybean oil from flaked

soybeans with he!ane, 100 kg of soybean containing 22 wt

% oil is leached with he!ane" #he solent feed is 0 kg of

solent containing - wt% of soybean oil" #he alue of N  for

the slurry underflow is essentially constant at 1"5 kg

insoluble solid&kg solution retained"*alculate the amounts and com$ositions of the oerflow V 1 

and the underflow slurry L1 leaing the stage"

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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  x A Composition of A at overflow solution

L Mass of liquid in slurry solution y A Composition of A at slurry solution

B Mass of dry, solute – free solid.

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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 orVphase is the liquid phase that overflowscontinuously from stage to stage countercurrently to the solid phase, and itdissolves solute as it moves along.

The slurry phase L composed of inert solid (B) and liquid phase of A and C is thecontinuous 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

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Multi – stage counter current Leaching

( )

( )

( ) balanceSolid 

 balanceAComp. 

 balancesolutionTotal 

 M  N  L N  L N  B

 Mx xV  y L xV  y L

 M V  LV  L

 M  N  N 

 AM  A AN  N  AN  N  A

 N  N 

===

=+=+

=+=+

++

+

00

111100

110

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 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 20kg oil. The leached solids are to contain 120 kg oil.

Data (B3) are tabulated below asN kg inert solid B/kg solution and yAkg oil A/kg

solution

Calculate the amounts and concentrations of the stream leaving the process andthe number of stages required.

Multi – stage counter current Leaching

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 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

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 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

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 Solution 2

Find coordinate atVN+1.

Coordinate forVN+1 

Mass of A = 20 kg/h

Mass ofB = 0 kg/h

Mass ofC = 1310 kg/h

0131020

0

015.01330

20

131020

20

11

1

1

=+=+==

==+

=+

==

++

++

C  A

 B

 B

 N 

C  A

 A

 A x

 N  N 

 N 

 N 

 

(xN+1 ,NN+1) = (0.015 , 0)

Multi – stage counter current Leaching

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 Solution 2

Find coordinate atLN.

Mass of A = 120 kg/h

Mass ofB = 2000 kg/h

Mass ofC = ?? kg/h

 y. N A

 B

 L A

 L

 B

 y

 N 

 y

 N 

 N 

 N 

 N 

 N 

 N 

 N 

671667.16

120

2000=→====  

graph,ofSlope

Multi – stage counter current Leaching

Ifx= 0.1,N = 16.67 x 0.1 = 1.67

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

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 Solution 2

Multi – stage counter current Leaching

1+ N V 

0 L

 N  L

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 Solution 2

From material balance calculations:

Total solution balance:

( ) ( ) ( ) ( ) ( )376.0

2180015.0133094.0850

1100

==+

=+  ++

 AM 

 AM 

 AM  N  N  A

 x

 x

 Mx xV  y L  

kg  21801330850

10

110

==+

=+

=+=+

+

+

 M  M  M V  L

 M V  LV  L

 N 

 N  N 

Component A balance:

Multi – stage counter current Leaching

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 Solution 2

From material balance calculations:

Multi – stage counter current Leaching

( ) ( ) ( )   916.0218085035.2

00

00

==

=

===

 M  M 

 M 

 M  N  N 

 N  N  M  N  L N 

 M  N  L N  L N  B

 

Solid balance:

Coordinate for M (xM,NM) = (0.376 , 0.916)

Plot coordinate M in the graph.

Construct line from pointLN to point M until it cross at x – axis. Point at x –

axis =V1

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Multi – stage counter current Leaching

1+ N V 

0 L

 N  L

 M 

1V 

From figure,

Coordinate forV1 (x1,N1) = (0.592 , 0)

Coordinate forLN (y1,N1) = (0.12 , 2.0)

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 Solution 2

From material balance calculations:

Total solution balance:

( )

kg

1Eq. 

38.1182

62.9972180

2180

1

1

1

=

−=

→−=

 LV   N 

Multi – stage counter current Leaching

ConnectL0 withV1 &LN withVN+1. The cross line – operating point.

Total stages: 4 stages

Construct operating point:

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 Solution 2

Multi – stage counter current Leaching

1+ N V 

0 L

 N  L

 M 

1V 

Construct the stages:

P∆

1 L2

 L3 L

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5 min break…"

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#utorial

12.10-4 Textbook, page 835 

Fresh halibt li!ers "o#tai#i#g 25.$ %t& oilare to be extra"te' %ith pre eth(l ether to

re)o!e 95 & o* the oil i# a "o#ter"rre#t

)ltistage lea"hi#g pro"ess. The *ee' rate is1000 kg o* *resh li!ers per hor. The *i#al exit

o!er*lo% soltio# is to "o#tai# $0 %t& oil. The

rete#tio# o* soltio# b( i#ert soli's +oil *reeli!er o* the li!er !aries as *ollo%, %here N is

kg i#ert soli'kg soltio# retai#e' a#' (  . is kg

oilkg soltio#

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#utorial

12.10-3 Textbook, page 835 

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Types of Equipments for Leaching

FIXED – BED LEACHING

MOVING BED LEACHING

AGITATED SOLID LEACHING

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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 andseeds and other processes.

 Figure 12.8-1 shows a typical sugar beet diffuser or extractor.

 The cover is removable so that

sugar beet slices calledcossettes can be dumped into the bed.

 Heated water at 344 K to 350 Kflows into the bed to leach out thesugar.

 The leached sugar solution flowsout the bottom onto the next tank inseries.

 About 95% of the sugar in beets isleached to yield an outlet solutionfrom the system of about 12 wt%.

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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 areparticularly hydrocarbons suchas hexane and the final solvent –vegetable solution called

miscella may contain somefinely divided solids.

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Agitated Solid Leaching

When the solid can be ground fine abou 200 mesh (0.074 mm), it can keptin suspension by small amounts of agitation.

 Continuous countercurrent leaching can be accomplished by placing thenumber of agitator in series, with setttling tanks or thickeners between eachagitator.

 Sometimes thickeners are used as combination contactor – agitators andsettlers – shown in Figure 12.8-3.

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 To analyze single – stage and countercurrent – stage leaching, an operating line

equation, or material balance relation and the equilibrium relations between thetwo 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 dissolvedsolids is present.

 The solid – liquid stream is calledunderflow orslurry stream.

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 Consequently, the concentration of oil or solute in the liquid oroverflow stream is

equal to the concentration of solute in the liquid solution accompanying the slurryor underflow stream.

 The amount of solution retained with the solids in the settling portion of eachstage may depend the density and viscosity of liquid in which the solid issuspended.

 Equilibrium diagrams for leaching:

Equilibrium Relations in Leaching

 The concentration of inert or insoluble solid B in the solution mixture or theslurry mixture can be expressed in kg (lbm) units:

solutionlb

solidlb

solutionkg

solidkg

 kgkg

 kg==

+

=

C  A

 B N 

 For overflow,N = 0

 For underflow,N value depending on the solute concentration in the liquid.

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Equilibrium Relations in Leaching

 The composition of solute A in liquid will be expressed as wt fractions:

( )

   

  

 =

+=

=+

=

liquidunderflow

orslurryinliquid 

solutionkg

solutekg

 kgkg

 kg

liquidoverflow solutionkg

solutekg

 kgkg

 kg

C  A

 A y

C  A

 A x

 A

 A

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/ 2 +45 )i#tes

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 A continuous countercurrent multistage system is to be used

to leach oil from meal by ben>ene solent" #he $rocess is to

treat 2000 kg&h of inert solid meal containing 00 kg oil andalso 50kg ben>ene" #he inlet flow $er hour of fresh solent

mi!ture contains 1-10 kg ben>ene and 20kg oil" #he

leached solids are to contain 120 kg oil" f the alue of for

the outlet underflow is constant at 1"5kg solid&kg solution,determine

a.#he e!it flows com$ositions +! A1, y A, oerflow (1 and the

underflow ) leaing the stage"

b.umber of stages re?uired