Solid-Liquid Extraction

Solid liquid extraction(leaching) means the removal of aconstituent from a mixture of solids by bringing thesolid material into contact with a liquid solvent thatdissolves this particular constituent. Applications:-1- leaching of soybean oil from flaked soybeans withhexane.2- Leaching of sugar from sugar beets with hot water.

3- Production of vegetable oils with organic solventssuch as hexane , acetone and ether by extraction the oil peanuts , soybeans, sunflower seeds, cotton seeds andhalibut livers.4- Soluble tea is produced by water leaching of teaLeaves.5- Gold is leached from its ore using an aqueoussodium cyanide solution.6- Extraction of copper oxide from low grade ores withdilute H2SO4 acid.

Mechanism of leaching:-Extraction involves two steps which are:-1- Contacting step:- of solvent and the material to beTreated, so as to transfer soluble constituent to theSolvent.2- Separation step:- of the solution formed from therelatively exhausted solids.The above two steps may be conducted in separateEquipment or in one and the same equipment.Solution resulting from separation step is termedOverflow, Solids left over are termed Underflow.

Extraction Terminology:-1-Contacting Step:-Basically it’s a mass transfer step , it aims at transferringThe soluble constituent from the solid phase into theliquid phase by diffusion and dissolution.The solute is first dissolved from the surface of thesolid, then passes into the main body of the solution bydiffusion.This process may result in the formation of pores in thesolid material which exposes fresh(new) surfaces tosubsequent solvent penetration to such surfaces.

An ideal contacting (mixing) stage yields a product in Thermodynamic Equilibrium→ No mass transfer. No heat transfer. No momentum transfer.Why its difficult to reach an ideal stage behavior,thermodynamic equilibrium ?* The rate of mass transfer is slow due to the poreresistance:- the rate of diffusion of solute out of thepores(capillaries) into the bulk solution are very slow.* Some part of the contained solute is not exposed tothe solvent(inside a non-porous layer or closed pores).

* Sometimes the adsorption of the solute on thesurface of the solid is > than the solubility in the solvent,so the realization of an ideal stage needs rather longtimes if the operations is carried out batch wise or anExcessively large apparatus if the operation is continuous. How to increase the rate of mass transfer?1- Increase the agitation speed, thickness of the boundary Layer decreases. 2- As the temperature increase, the solubility increase,Hence, the diffusion increases , the viscosity decreasesleading to decrease in the film thickness , hence the poreresistance decreases.

3-Size reduction of solid to increase the exposed masstransfer area.2-Separation step:-This is a momentum transfer step that can be carriedout by (settling or filtration).Calculations:-The design problem of calculating number oftheoretical stages require simultaneous solution ofmaterial balance and equilibrium relations.We have three component system:-1- solute (A)2- Inert solid(B)3- Solvent (S)

Representing the three component system on rightangle triangle:-

Addition of two streams:-

AP AQ AR

BP BQ BR

SP SQ SR

P + Q = RP x + Q x = R x

P x + Q x = R x

P x + Q x = R x

By using lever arm principle,the length and amountsare calculated as follows:-PR Q a = = RQ P b

Subtraction of two streams:-

AN AM AK

BN BM BK

SN SM SK

N - M = KN x - M x = K xN x - M x = K xN x - M x = K xBy using lever arm principle,the length and amountsare calculated as follows:-NK M b' = = MK N a' + b'

Equilibrium Relations:-A-Locus of under flow:-1- The mass of solution retained/unit mass of inertinsoluble solids is obtained experimentally as a functionof solution composition YA. The data is usually availablein tabular form.

(Kg solution / Kg solids ) = (A+S)/B YA = ( A/ A+ S)

0.30 0.1

0.32 0.2

0.35 0.3

0.4 0.4

2- In some cases the solution retained/kg solids isapproximated constant. i.e independent of composition.This means that the locus of underflow is a line parallelto the hypotenuse.

3-Locus is a straight line:-Locus between two points ( 0, 0.8 ) & ( 0.4, 0 )

4- …… Ib solvent retained/ Ib solute-free solid:-

S

B

S

A S

S A S A S

S A

S A

S A

A

S = r = 1B

SxA+B+S r

B xA+B+S

x = r1 - ( x + x )

x = r - r (x + x ) = r - r x + r x(1+ r) x = r - r x

r r x = - x ( It's a striaght line equation)1+ r 1+ r

If x = 0 x =1If x = 0

S x = 1/2

B- Locus of overflow:-Also the overflow consists generally of solution only ,i.e(solid-free or inert-free) its locus is:-1- The hypotenuse.2- In some cases the overflow contains 10% solids and90% solution. B : A + S10 : 90xB = B / ( A + B + S ) = 0.1

3- ………..and the overflow from the first stage contains10% from the solids in the feed.

Ideal Extraction Stage:-

Ideal extraction stage indicateswhether or not we are closeto thermodynamic equilibrium.The ideal stage in leaching is represented as a straightline going through point (B) and cutting the two locusin x1 and y1.

Methods of Operation (Types of contact):-1-Simple single stage:-Considering a theoretical extraction stage where L0

with composition x0 is brought into contact with a solvent V0 having composition y0 . The amount L1 and Composition x1 of the product underflow as well as the amount V1 and comp. y1

of the overflow are to be determined. determ

0 0 1 1

0 0 0 0 1 1 1 1 M

0 0 M

1 1 M

0 M 0

0 M 0

0 0 0 M

0 0 0 0

OMB:- L + V = L + V = M CMB:- L x + V y = L x + V y = M x x , y , x lies on straight line and x , y , x lies on straight line

y x L a = = x x V b

L L y x = = L +V M y x

1 M 1

1 M 1

1 1 1 M1 1

1 1 1 M

a = get a & ba+b

Also:-y x L d = = x x V c

L L y x d = = get L &V L +V M x x d+c

M

M

1 1

1 1

x is the point of intersectionbetween the two straight lines,Connect x with (B) to cut the two loci in x and y .To obtainthe flow rates (amounts V , L ) apply lever arm principle.

2- Multi-stage cross current contact:-

1

1

1

0 0 1 1

0 0 0 0 1 1 1 1 M

0 0 M

1 1 M

Material Balance on first Stage:-

OMB:- V + L = V + L = M CMB:- V y + L x = V y + L x = M x

y , x , x lies on a straight line.

& y , x , x lies on a s

traight line.

1

1

M

M 1 1

1 1

x is the point of intersection between the two straight lines,

connect x with (B) to cut the two loci in x and y .To obtain

the flow rates (amounts V , L ) apply lever arm principle. Material Bal

2

2

2

0 0 2 2

0 0 0 0 2 2 2 2 M

0 0 M

1 1 M

ance on second Stage:-

OMB:- V + L = V + L = M CMB:- V y + L x = V y + L x = M x

y , x , x lies on a straight line.

& y , x , x lies on a straight lin

2

2

M

M 2 2

2 2

e.

x is the point of intersection between the two straight lines.

Again connect x with (B) to cut the two loci in x and y .

To obtain the flow rates (amounts V , L ) apply lever arm principle

3- Multi-stage counter current contact:-Considering a feed with rate L0 and composition x0

which is to be treated with a solvent having acomposition yn+1 at a rate of Vn+1. The weight fractionin the U.F shouldn’t exceed a given value xnA.

To determine the number of ideal stages required toachieve this extraction duty use the following steps:-The overall M.B on the n stages may be written as:-

* x0 , yn+1 , and xn are located from the specificationsof the problem.* xM , x0 and yn+1 lies on the same straight line where xM

Divides the distance between x0 and yn+1 in the ratio of Vn+1 / L0 .

n+1 0 1 n

n+1 n+1 0 0 1 1 n n M

OMB:- V + L = V + L = M CMB:- V y + L x = V y + L x = M x

M being the sum of Ln and V1, the points xn,xM,y1 lieson one straight line. Hence y1 may be located byExtrapolating xn xM to meet the locus of the overflow.

n+1 n 1 0

n+1 n+1 1 1 0 0

n+1 n

1 0

The M.B on each stage can be written as:-

V - L V - L = RAlso V y - L x = V y - L x = R

y , x , R lies on a straight line.& y , x , R lies on a straight line.

R is lo

n n

0 1 n n+1

1 1 2 2 n n

1 1 2 2

cated by extrapolating x y and x y .Tie line is drawn between y x , y x ,.........., y xby joining ( 0 , x ,y ) , (0 , x ,y ) and so on.

Condition of infinite number of stages:-To increase overflow composition the number of stagesmust be increased. the number of stagesincreases till reachingthe maximum overflowcomposition (y1max).Steps:-Connect yn+1 with (B)To cut U.F locus in x n min

connect it with xM thenextend the line to cut O.Fin y 1max.

Stage Efficiency:-A deviation between the behavior of actual andTheoretical stages is expected.In theoretical stage:-(Solvent/ Solute) O.F = (Solvent/ Solute) U.FActual stage is less efficient because:-1- Slow rates of mass transfer resulting inunattainment of T.D.E (complete dissolution) in thecontact time provided for the stage.2-The solute is adsorbed by the solid to a highercontent than the solvent.

3- The pore structure of the solid-solute mixture is suchthat only a portion of the solute is accessible to thesolvent.Calculation of actual number of stages:-A- Construction when overflow efficiency is known:-

The tie line representingan actual stage passes noLonger through the origin (B).

n+1 n actO.F *

n+1 n

y - yη = 0.8y - y

Actual change in concentrationηTheoretical change in concentration

B- Construction when underflow efficiency is known:-

The tie line representingan actual stage passes noLonger through the origin (B).

n-1 n actU.F *

n-1 n

- xη = 0.8 - x

xx

Actual change in concentrationηTheoretical change in concentration

Solids-free coordinates:-The construction is in general less crowded than inthe case of triangular diagram. No calculations areneeded before plotting the equilibrium relations ,butthe same procedures described before are followed. The amounts of various streams should be expressedon solid-free basis.Some notes on Rectangular Diagram System:- The abscissa X (weight fraction of solute on a solid-freebasis in the U.F) and Y (weight fraction of solute on asolid-free basis in the O.F) is plotted against the amount N(concentration of insoluble solids on the solid-free basis).

Equilibrium Relations:-The concentration of inert (B) in the solution mixture

is expressed in Kg units:-

kg B Kg solid Ib solidN = or KgA+Kg S Kg solution Ib solution

There will be a value of N for the overflow where (N = 0)and

A

for the underflow N will have different values, depending upon the solute concentration in the liquid.The compositions of solute in the underflow& overflow will be expressed as wt fractions:-

Kg Ay = K

A

Kg solute = ( overflow liquid)g A+Kg S Kg solution

Kg A Kg solutex = = ( underflow liquid)Kg A+Kg S Kg solution

Some notes on the diagram:-

Pure A = 100% , B = 0 , S = 0A A B 0x or y = = , N = 0

A + S A + 0 A + S 100 0

Pure S = 100% , A = 0 , B = 0A 0 B 0x or y = = = 0 , N = 0

A + S 0 + 100 A + S 0 100

P

ure B = 100% , A = 0 , S = 0A 0 0 B 100x or y = = = , N =

A + S 0 + 0 0 A + S 0 0

A typical equilibrium diagram is shown in the followingFigure. The lower curve of N vs yA , where N = 0 on theaxis, represents the locus of overflow(where all thesolids has been removed). The upper curve N vs xA , Represents the locus ofunderflow.The tie lines are vertical, and on x-y diagram, theEquilibrium line is yA = xA

0

0 n+1 n+1

Kg inert solidFor the entering solid feed to be leached, N =Kg solute A

and x = 1.0 . And for pure solvent N = 0 , y = 0 .

Types of contact:-(1) Simple Single Stage:-The application of materialbalance equations in conjunction with such diagram ,the quantities of various streams should be expressedon solid-free basis.L0 = mass of feed. = (A+B) or (A+B+S).L0’= mass of feed on solidFree-basis = (A) or (A+S).V0= mass of solvent = (S+A)V0’= mass of solvent on solidFree-basis = (S+A).

0 0 1 1

0 0 0 0 1 1 1 1 M

0 0 M

1 1 M

0 M 0

0 M 0

0

0 0

OMB:- L '+ V ' = L '+ V ' = M'CMB:- L 'x + V ' y = L 'x + V ' y = M' x

x , y , x lies on straight line and x , y , x lies on straight line

y x L ' a = = x x V ' b

L ' = L '+V '

0 0 M

0 0

1 M 1

1 M 1

1 1 1 M1 1

1 1 1 M

L ' y x a = = get a & bM y x a+b

Also:-y x L ' d = = x x V ' c

L ' L ' y x d = = get L '&V ' L '+V ' M' x x d+c

0

0

0

00

0

00

0

1 11 1

1 1

L A + S B = (A+S) + B = L ' + B

L A + S B = + A + S A + S A + S

L

= 1 + NL '

L L ' = 1 + N

V Lalso V ' = & L '1+N 1+ N

Single stage for the other type of equilibrium:-

3- Multi-stage counter current contact:-

n+1 0 1 n

n+1 0 1 n

n+1 n+1 0 0 1 1 n n M

n+1 n 1 0 2 1

n+1 n+

OMB:- V + L = V + L = M V ' + L ' = V ' + L ' = M'CMB:- V ' y + L ' x = V ' y + L ' x = M' xOperating relations:- V ' - L ' V ' L ' =V ' - L ' RV ' y

1 n n 1 1 0 0 2 2 1 1 R - L 'x =V 'y -L 'x = V 'y - L 'x = R x

Equipments in leaching:-1- Fixed-bed leaching:-

Fixed bed leaching is used in the beet sugar and is alsofor extraction of pharmaceuticals from barks and seeds,and for other processes. In the Figure, the cover is removable so sugar beetslices can be dumped into the Bed. Hot water at 71 to77◦C flows into the bed to leach out the sugar. The leached sugar solution flows out the bottom ontothe next tank in series. The top and bottom covers areremovable so that the leached beets can be removedand a fresh charge added. About 95% Of the sugar inthe beets is leached.

2-Bollman bucket-type extractor:-

Dry solids are added at the upper right side to aPerforated basket or bucket. The solid leached by adilute solution called half miscella.The liquid downward through the moving buckets andis collected at the bottom as the strong solution or fullmiscella.The buckets moving upward on the left are leachedcounter currently by fresh solvent sprayed on the topbucket. The wet flasks are dumped as shown in theFigure and removed continuously.

3-Hildebrandt screw-conveyor extractor:-

Consists of three screw conveyers arranged in a Ushape. The solids are charged at the top right conveyeddownward, across the bottom , and then up the otherLeg . The solvent flows counter currently to the solid.