Solid * Liquid Extraction Prepared by Dr.Nagwa El

Report
Solid – Liquid Extraction
Prepared by Dr.Nagwa El-Mansy
Chemical Engineering Department
Cairo University
Solid-Liquid Extraction
Solid liquid extraction(leaching) means the removal of a
constituent from a mixture of solids by bringing the
solid material into contact with a liquid solvent that
dissolves this particular constituent.
Applications:1- leaching of soybean oil from flaked soybeans with
hexane.
2- Leaching of sugar from sugar beets with hot water.
3- Production of vegetable oils with organic solvents
such as hexane , acetone and ether by extraction the oil
peanuts , soybeans, sunflower seeds, cotton seeds and
halibut livers.
4- Soluble tea is produced by water leaching of tea
Leaves.
5- Gold is leached from its ore using an aqueous
sodium cyanide solution.
6- Extraction of copper oxide from low grade ores with
dilute H2SO4 acid.
Mechanism of leaching:Extraction involves two steps which are:1- Contacting step:- of solvent and the material to be
Treated, so as to transfer soluble constituent to the
Solvent.
2- Separation step:- of the solution formed from the
relatively exhausted solids.
The above two steps may be conducted in separate
Equipment or in one and the same equipment.
Solution resulting from separation step is termed
Overflow, Solids left over are termed Underflow.
Extraction Terminology:1-Contacting Step:Basically it’s a mass transfer step , it aims at transferring
The soluble constituent from the solid phase into the
liquid phase by diffusion and dissolution.
The solute is first dissolved from the surface of the
solid, then passes into the main body of the solution by
diffusion.
This process may result in the formation of pores in the
solid material which exposes fresh(new) surfaces to
subsequent 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 pore
resistance:- the rate of diffusion of solute out of the
pores(capillaries) into the bulk solution are very slow.
* Some part of the contained solute is not exposed to
the solvent(inside a non-porous layer or closed pores).
* Sometimes the adsorption of the solute on the
surface of the solid is > than the solubility in the solvent,
so the realization of an ideal stage needs rather long
times if the operations is carried out batch wise or an
Excessively 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 decreases
leading to decrease in the film thickness , hence the pore
resistance decreases.
3-Size reduction of solid to increase the exposed mass
transfer area.
2-Separation step:This is a momentum transfer step that can be carried
out by (settling or filtration).
Calculations:The design problem of calculating number of
theoretical stages require simultaneous solution of
material 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 right
angle triangle:-
Addition of two streams:P + Q = R
P x AP + Q x AQ = R x AR
P x BP + Q x BQ = R x BR
P x SP + Q x SQ = R x SR
B y using lever arm principle,
the length and am ounts
are calculated as follow s:PR
RQ
=
Q
P
=
a
b
Subtraction of two streams:N -M = K
N x AN - M x AM = K x AK
N x BN - M x BM = K x BK
N x SN - M x SM = K x SK
B y using lever arm principle,
the length and am ounts
are calculated as follow s:NK
MK
=
M
N
=
b'
a' + b'
Equilibrium Relations:A-Locus of under flow:1- The mass of solution retained/unit mass of inert
insoluble solids is obtained experimentally as a function
of solution composition YA. The data is usually available
in 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 is
approximated constant.
i.e independent of
composition.
This means that the
locus of underflow
is a line parallel
to 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
= r= 1
B


S
A+B+S
B
A+B+S


 r
xB

xS

xS
= r
1 - ( xA + xS)
 x S = r - r (x A + x S ) = r - r x A + r x S
(1+ r) x S = r - r x A
 xS =
r
1+ r
r
-
1+ r
x A ( It's a striaght line equation )
If x S = 0
 x A =1
If x A = 0

x S = 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 and
90% solution.
B : A+S
10 : 90
xB = B / ( A + B + S ) = 0.1
3- ………..and the overflow from the first stage contains
10% from the solids in the feed.
Ideal Extraction Stage:-
Ideal extraction stage indicates
whether or not we are close
to thermodynamic equilibrium.
The ideal stage in leaching is represented as a straight
line going through point (B) and cutting the two locus
in 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
O M B :- L 0 + V 0 = L 1 + V1 = M
C M B :- L 0 x 0 + V 0 y 0 = L 1 x 1 + V1 y 1 = M x M
 x 0 , y 0 , x M  lies on straight line
and x 1 , y 1 , x M  lies on straight line
y0x M

L0
=
a
=
x0x M
V0
b
L0
L0
y0x M
=
L 0 + V0
=
M
a
=
y0x 0

get a & b
a+ b
A lso:y1x M
L1
=
x 1x M

L1
L 1 + V1
=
V1

L1
M
d
c
=
y1x M
x 1x M
=
d
d+ c
 get L 1 & V1
x M is the point of intersection
betw een the tw o straight lines,
C onnect x M w ith (B ) to cut the
tw o loci in x 1 and y 1 .T o obtain
the flow rates (am ounts V1 , L 1 )
apply lever arm principle.
2- Multi-stage cross current contact:-
M aterial B alan ce o n first S tag e:O M B :C M B :-
V 0 + L 0 = V1 + L 1 = M
V 0 y 0 + L 0 x 0 = V1 y 1 + L 1 x 1 = M x M 1
 y 0 , x 0 , x M1 
lies o n a straig h t lin e.
& y 1 , x 1 , x M 1  lies o n a s traig h t lin e.
x M 1 is the point of intersection betw een the tw o straight lines,
connect x M 1 w ith (B ) to cut the tw o loci in x 1 and y 1 .T o obtain
the flow rates (am ounts V1 , L 1 ) apply lev er arm principle.
M aterial B al ance on second S tage:O M B :-
V0 + L 0 = V2 + L 2 = M
C M B :-
V0 y 0 + L 0 x 0 = V2 y 2 + L 2 x 2 = M x M 2
 y0, x 0, x M2 
lies on a straight line.
& y 1 , x 1 , x M 2  lies on a straight lin e.
 x M 2 is the point of intersection betw een the tw o straight lines.
A gain connect x M 2 w ith (B ) to cut the tw o loci in x 2 and y 2 .
T o obtain the flow rates (am ounts V 2 , L 2 ) 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 a
composition yn+1 at a rate of Vn+1. The weight fraction
in the U.F shouldn’t exceed a given value xnA.
To determine the number of ideal stages required to
achieve this extraction duty use the following steps:The overall M.B on the n stages may be written as:O M B:- V n+1 + L 0 = V1 + L n = M
C M B:- V n+1 y n+1 + L 0 x 0 = V1 y 1 + L n x n = M x M
* x0 , yn+1 , and xn are located from the specifications
of 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 .
M being the sum of Ln and V1, the points xn,xM,y1 lies
on one straight line. Hence y1 may be located by
Extrapolating xn xM to meet the locus of the overflow.
T he M .B on each stage can be w ritten as: V n+ 1 - L n  V1 - L 0 = R
A lso V n+ 1 y n+ 1 - L n x n = V1 y 1 - L 0 x 0 = R
 y n+ 1 , x n , R 
lies on a straight line.
& y 1 , x 0 , R  lies on a straight line.
 R is lo cated by extrapolating x 0 y 1 and x n y n+ 1 .
T ie line is draw n betw een y 1 x 1 , y 2 x 2 ,.........., y n x n
by joining ( 0 , x 1 ,y 1 ) , (0 , x 2 ,y 2 ) and so on.
Condition of infinite number of stages:To increase overflow composition the number of stages
must be increased.
the number of stages
increases till reaching
the maximum overflow
composition (y1max).
Steps:Connect yn+1 with (B)
To cut U.F locus in x n min
connect it with xM then
extend the line to cut O.F
in y 1max.
Stage Efficiency:A deviation between the behavior of actual and
Theoretical stages is expected.
In theoretical stage:(Solvent/ Solute) O.F = (Solvent/ Solute) U.F
Actual stage is less efficient because:1- Slow rates of mass transfer resulting in
unattainment of T.D.E (complete dissolution) in the
contact time provided for the stage.
2-The solute is adsorbed by the solid to a higher
content than the solvent.
3- The pore structure of the solid-solute mixture is such
that only a portion of the solute is accessible to the
solvent.
Calculation of actual number of stages:A- Construction when overflow efficiency is known:Actual change in concentration
η
T heoretical change in concentration
η O .F =
y n+1 - y n act
y n+1 - y
*
n
 0.8
The tie line representing
an actual stage passes no
Longer through the origin (B).
B- Construction when underflow efficiency is known:η
Actual change in concentration
T heoretical change in concentration
η U .F =
x n-1 - x n act
x n-1 - x
*
n
 0.8
The tie line representing
an actual stage passes no
Longer through the origin (B).
Solids-free coordinates:The construction is in general less crowded than in
the case of triangular diagram. No calculations are
needed before plotting the equilibrium relations ,but
the same procedures described before are followed.
The amounts of various streams should be expressed
on solid-free basis.
Some notes on Rectangular Diagram System:The abscissa X (weight fraction of solute on a solid-free
basis in the U.F) and Y (weight fraction of solute on a
solid-free basis in the O.F) is plotted against the amount N
(concentration of insoluble solids on the solid-free basis).
Equilibrium Relations:T he concentration of inert (B ) in the so lution m ixture
is expressed in K g units:N =
kg B
K g solid

K gA + K g S
or
K g solution
Ib solid
Ib solution
T here w ill be a value of N for the overflow w here (N = 0)
and for the underflow N w ill have different values, depending
upon the solute concentration in the liq uid.
T he com positions of solute in the underflow & overflow w ill be
expressed as w t fractions:yA =
xA =
Kg A
=
K g solute
Kg A+Kg S
K g solution
Kg A
K g solute
Kg A+Kg S
=
K g solution
( overflow liquid)
( underflow liquid)
S om e notes on the diagram :P ure A = 100% , B = 0 , S = 0
x or y =
A
=
A + S
A
,
B
N =
A + 0

A + S
0
100  0
0
P ure S = 100% , A = 0 , B = 0
x or y =
A
0
=
A + S
= 0 , N =
0 + 100
B

0
A + S
0  100
B
100
0
P ure B = 100% , A = 0 , S = 0
x or y =
A
A + S
=
0
0+ 0
=
0
0
, N =
A + S

00

For the entering solid feed to be leached, N 0 =
K g inert solid
K g solute A
and x 0 = 1.0 . A nd for pure solvent N n+1 = 0 , y n+1 = 0 .
A typical equilibrium diagram is shown in the following
Figure.
The lower curve of N vs yA , where N = 0 on the
axis, represents the locus of overflow(where all the
solids has been removed).
The upper curve N vs xA , Represents the locus of
underflow.
The tie lines are vertical, and on x-y diagram, the
Equilibrium line is yA = xA
Types of contact:(1) Simple Single Stage:-The application of material
balance equations in conjunction with such diagram ,
the quantities of various streams should be expressed
on solid-free basis.
L0 = mass of feed.
= (A+B) or (A+B+S).
L0’= mass of feed on solid
Free-basis = (A) or (A+S).
V0= mass of solvent = (S+A)
V0’= mass of solvent on solid
Free-basis = (S+A).
O M B :- L 0 '+ V 0 ' = L 1'+ V1' = M '
C M B :- L 0 'x 0 + V 0 ' y 0 = L 1'x 1 + V1' y 1 = M ' x M
 x 0 , y 0 , x M  lies on straight line
and x 1 , y 1 , x M  lies on straight line
y0x M
L0 '
=
x 0x M

a
=
V0 '
L0 '
=
b
L0 '
L 0 '+ V 0 '
y0x M
=
M
a
=
y0x 0

get a & b
a+ b
A lso:y1x M
=
x 1x M

L1 '
L 1 '+ V1 '
L1 '
d
=
V1 '

L1 '
M'
c
=
y1x M
x 1x M
=
d
d+ c
 get L 1'& V1 '
L0  A + S  B
= (A + S ) + B
= L 0' + B
L0
=
A + S
L0
A + S
+
B
A + S
= 1 + N0
L 0'
 L 0' =
A + S
L0
1 + N0
also V1 ' =
V1
1+ N 1
&
L1 ' 
L1
1+ N 1
Single stage for the other type of equilibrium:-
3- Multi-stage counter current contact:-
O M B :- V n+1 + L 0 = V1 + L n = M
V n+1 ' + L 0 ' = V1 ' + L n ' = M '
C M B :- V n+1 ' y n+1 + L 0 ' x 0 = V1 ' y 1 + L n ' x n = M ' x M
O perating relations:V n+1 ' - L n '  V1 '  L 0 ' = V 2 ' - L 1 '     R
V n+1 ' y n+ 1 - L n 'x n = V1 'y 1 -L 0 'x 0 = V 2 'y 2 - L 1 'x 1 = R x R
Equipments in leaching:1- Fixed-bed leaching:-
Fixed bed leaching is used in the beet sugar and is also
for extraction of pharmaceuticals from barks and seeds,
and for other processes.
In the Figure, the cover is removable so sugar beet
slices can be dumped into the Bed. Hot water at 71 to
77◦C flows into the bed to leach out the sugar.
The leached sugar solution flows out the bottom onto
the next tank in series. The top and bottom covers are
removable so that the leached beets can be removed
and a fresh charge added. About 95% Of the sugar in
the beets is leached.
2-Bollman bucket-type extractor:-
Dry solids are added at the upper right side to a
Perforated basket or bucket. The solid leached by a
dilute solution called half miscella.
The liquid downward through the moving buckets and
is collected at the bottom as the strong solution or full
miscella.
The buckets moving upward on the left are leached
counter currently by fresh solvent sprayed on the top
bucket. The wet flasks are dumped as shown in the
Figure and removed continuously.
3-Hildebrandt screw-conveyor extractor:-
Consists of three screw conveyers arranged in a U
shape.
The solids are charged at the top right conveyed
downward, across the bottom , and then up the other
Leg . The solvent flows counter currently to the solid.

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