Transcript Solvent Extraction Liquid-Liquid Extraction Prepared by Dr.Nagwa

Solvent Extraction
Liquid-Liquid Extraction
Prepared by Dr.Nagwa El- mansy
Cairo University
Chemical Engineering Department
Liquid-Liquid Extraction
“The separation of the components of a liquid
mixture by treatment with a solvent in which
one or more of the desired components is
preferentially soluble is known as liquid–liquid
extraction.”
In this operation, it is essential that the liquidmixture feed and solvent are at least partially
miscible ( some time completely immiscible).
Extraction
1. Extraction is an operation in which
constituents of the liquid mixture are
separated by using an insoluble liquid solvent
Distillation
1. Constituents of the liquid mixture are
separated by using thermal energy
2. Extraction utilizes the differences in solubilties
of the components to effect separation
2. Utilizes the differences in vapor pressures
of the components to effect separation
3. Selectivity is used as a measure of degree of
separation
3. Relative volatility is used as a measure of
degree of separation
4. A new insoluble liquid phase is created by
addition of solvent to the original mixture
4. A new phase is created by addition of heat
5. Phases are hard to mix and harder to separate
5. Mixing and separation of phases is easy
and rapid
Extraction
Distillation
6. Extraction does not give pure product
and needs further processing
6. Gives almost pure products
7. Offers more flexibility in choice of
operating conditions
7. Less flexibility in choice of operating
conditions
8. Requires mechanical energy for mixing
and separation
8. Requires thermal energy
9. Does not need heating and cooling
9. Requires heating and cooling
10. Often a secondary choice for separation
of components of liquid mixture
10. Usually the primary choice for
separation of components of liquid mixture
The mechanism of extraction involves two steps which
are:First Step:Contacting Step:Bringing the feed mixture and the solvent into intimat
contact.
Second Step:Separation Step:Separation of the resulting two phases.
It is done by:- distillation, evaporation, Crystallization.
It is possible to combine first two stages into a single
piece of equipment such as a column which is then
operated continuously. Such an operation is known
as differential contacting.
Liquid–liquid extraction is also carried out in stagewise equipment, the prime example being a mixer–
settler unit in which the main features are the
mixing of the two liquid phases by agitation,
followed by settling in a separate vessel by gravity.
Applications of liquid-liquid extraction in petroleum
Field:1- Removal of aromatics from Kerosene (Edeleanu
Process) , It is one of the most important processes for
refining Kerosene which improves the smoke point of
Kerosene and Jet fuels by removing the aromatics
content by Liquid sulphur dioxide.(Smoke point is the
temperature at which it gives off smoke).
2-Removal of asphaltenes from lube oil which cause
friction by using propane as a solvent.
3-Dewaxing of lube oil by using propane as a solvent.
4-Improving kinematic viscosity index (KVI) of lube oil
By removing aromatics and naphthens using phenol or
furfural. [KVI measures the variation of viscosity with
temperature. Paraffin has high (KVI) so presence of
aromatics decreases(KVI)].
( KV = µ/ρ = --- cm2/sec = stock )
Separation Factor (Selectivity) :- β
β =
yA yB
xA xB
y A = m ole fraction of A in extract.
x A = m ole fraction of A in raffinate.
y B = m ole fraction of B in extract.
x B = m ole fraction of B in raffinate.
β = It is a measure of effectiveness of separation.
The solvent is said to be effective or selective when β
exceeds unity.
β=
yA yB
xA xB
β = 1

yA xA
> 1
(solvent is selective)
yB xB
(no separation)
The higher the selectivity the higher the separation.
Choice of solvent:1- High selectivity ( high β → high yA/xA and low yB/xB)
2-Easy to be recovered.
3-Density:- The higher the density difference between
Extract and raffinate the easier the segregation of both
Phases.
4-Surface tension:- The larger the surface tension the
higher the power needed to mix the two liquids with
the solvent but the easier the segregation.
5- Other criteria:- non-corrosive , chemically stable ,
non-toxic , low viscosity , low vapor pressure.
Equilibrium Relations:In liquid-liquid extraction we have three component
system:1- solute (A)→liquid.
2- Inert (B)→liquid.
3- Solvent (S)→liquid.
Representing the three component system on right
angle triangle as follows :-
Types of phase diagram = Solubility diagram = Ternary
diagram.
Two main diagram :A- Closed Ternary System:-
B- Open Ternary system:-
Effect of Temperature on Solubility Diagram:-
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'
Methods of Operation(Types of contact):(1) Simple Single Stage:- W here:V 0 = am ount of solvent (S )
y 0A = 0
, for pure solvent.
or has som e solute traces (A + S )
y 0A  m ole fraction of solvent =
A
A+S
L 0 = Feed am ount = M eal = (A + B )
x 0 = m ole fraction of m eal.
x0A 
A
,
A+B
x 0B 
B
A+B
B ut if feed has traces of solvent
x0A 
x 0S 
A
A+B+S
S
A+B+S
, x 0B 
B
A+B+S
M aterial B alance on S ingle S tage:O M B :-
V 0 + L 0 = V1 + L 1 = M
C M B :- V 0 y 0 + L 0 x 0 = V1 y 1 + L 1 x 1 = M x M
 y0 , x0 , x M
and y 1 , x 1 , x M

lies on a straight line.
 lies on a straigh t line.
 x M is the point of intersection betw een the tw o straight lines.
T o draw this single stage w hich lies bet w een x 1 and y 1 trial and
error procedures m ust be done to obtain the m ost convenient tie
line w hich go through x M .
T o obtain the flow rates (am ounts) apply lever arm principle
as follow s:-
y0M
L0
=
x 0M

a
=
V0
L0
b
L0
=
L 0 + V0
y0M
=
M
y0x 0
a
=
a+ b
get a & b
A lso:y1M
=
x 1M

L1
L1
d
=
V1

L 1 + V1
get L 1 & V1
L1
M
c
=
y1M
x 1M
=
d
d+ c
(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.
T o draw this single stage w hich lies betw een x 1 and y 1 trial
and error procedures m ust be done to obtain the m ost convenient
tieline w h ich go through x M 1 .T o obtain the flow rate s
(am ounts V1 , L 1 ) apply lever arm principle
M aterial B alance on second S tage:O M B :-
V0 + L 1 = V 2 + L 2 = M
C M B :-
V0 y 0 + L 1x 1 = V 2 y 2 + L 2 x 2 = M x M 2
 y0 , x1, x M2 
lies on a straight line.
& y 1 , x 1 , x M 2  lies on a straight line.
 x M 2 is the point of intersection betw een the tw o straight lines.T o draw this
single stage w hich lies betw een x 2 and y 2 trial and error procedures m ust be
done to obtain the m ost conven ient tieline w hich go through x M 2 .
T o obtain the flow rates (am ounts V 2 , L 2 ) apply lever arm principle
(3) Multi-stage counter current contact:-
M aterial B alance on B attery :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
 y n+1 , x 0 , x M 
lies on a straight line.
& y 1 , x n , x M  lies on a straight line.
 x M is the point of intersection betw een the tw o straight lines.
T o obtain the flow rates (am ounts V1 , L n ) apply lever arm principle
O perating conditions on the B attery:V n + 1 + L 0 = V1 + L n = M
 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 lin e.
& y 1 , x 0 , R  lies on a straight line.
 R is the operating polar , it is the po int of intersection
betw een the tw o operating lines .
T he num ber of stages is calculated by us ing the m aterial balance
and the oper ating equations.
T here are tw o m ethods for calculating th e num ber of theoretical
stages.
Continuous contact(packed or spray columns) in
liquid-liquid extraction:In continuous contact using packed column, one of the
phases is dispersed into the other phase and allowed to
flow counter currently past the other phase(either upwards or down-wards). Separation of the phases is not
accomplished until the outlets are reached.
Since contacting and separation do not take place at a
number of discrete stages, equilibrium conditions are
never reached with this type of equipment.
The effectiveness of a continuous contact tower, may
be expressed as:-
Height of packing =
(number of stages) x (height equivalent
to theoretical stage).
H = n x HETP
Where,
HETP :- It’s the height
of packing make the same
separation of an ideal stage
i.e two stream leaving are
in equilibrium.
No. of stages is calculated by the
same methods of stage-wise contact
Performance of a given number of stages:In many practical problems, the extraction equipment
with known number of stages is already available and
It’s required to find out the composition of the extract
and raffinate streams obtained when using a certain
(S/F) ratio or to find out the composition of extract(y1)
or raffinate (xn).
A- When(S/F) ratio is known, the products compositions
are calculated as follows:No.of stages = √ , (S/F) = √ , xn = ? , y1 = ?
1- Locate M according to the given (S/F).
2- Assume xn on raffinate locus.
3-Extend xnM to obtain y1 on the extract locus.
4- Obtain R at the intersection of x0y1 , xnyn+1 and
proceed to obtain number of theoretical stages.
5- If the no. of stages founds corresponds to the
specified no., the assumption of xn is correct and
correspond to this composition (xn).
If → n calculated > or < n given , a new xn is assumed.
6-The same procedure is repeated until the assumed xn
yield the specified no. of stages.
7- y1 is obtained by connecting xnM and extend to cut
extract locus at y1.
B-When the composition of the product raffinate
stream is known, the problem will be to calculate the
amount of solvent and the corresponding extract
composition.
Given:- xn = √ , no. of stages = n = √ , (S/F) = ? , y1 = ? ,
yn+1= √.
• The trials procedure goes along the same line of
material balance (position of point M).
• The number of stages obtained is plotted against (S/F)
• The required amount of solvent for the given no. of
stage is obtained from the curve.
• The correct M point is obtained .
• Connect xn with M then extend to cut the extract
locus at y1.
Minimum solvent requirements:As the (S/F) ratio decreases:1- Number of stages increases.
2-y1 increases to y1’ and y1” .
3- M moves from M’ and M”.
4- R moves from R’ and R’’ i.e slopes of M.B lines range
from -ve to ∞ to +ve.
5- For ∞ no. of stages , same minimum solvent
requirements and maximum extract conditions.
(S/F) min → ∞ no. of stages →y1 max
When tie line (xi , yi) coincide with operating line (xi ,yi+1)
With R {pinch occur when extension of tie line passes
Through x0 R}.
The conditions of minimum solvent requirements may
be plotted on x-y diagram, the coincidence of a tie line
and a material balance line on polar plot ,meaning that
yi = yi+1 →the operating line and equilibrium curve
having a common point.
As S/F is reduced the operating curve moves from
position(1) to (2) and(3) getting nearer to the equilibrium
curve until it cuts it or touches it (position(3)) which
corresponds to minimum solvent requirements and
maximum extract composition(y1max) as shown in the
following Figure.
Optimum (S/F) ratio:When designing a new extraction battery the number of
stages and (S/F) ratio are unknown. The optimum
conditions of no. of stages and (S/F) ratio is determined
as (S/F) increases, no. of stages decreases , but capacity
of equipment to
handle the increasing
amount of solvent
increases. As (S/F)
increases the solution
become more dilute
which increase the cost
of solvent removal unit.
Intermediate Feed:In many cases mixtures consisting of same components
but having different compositions are produced from
various parts of a plant and have to be separated by
solvent extraction to give the same terminal stream
compositions.
Three alternatives may be used:1- Each mixture is extracted in a separate apparatus.
2- All mixtures are put to gather and then extracted in
one and the same apparatus.
3-Each feed is introduced at the proper stages of one
and the same extraction equipment.
The last solution is the most economic and corresponds
to lower number of stages.
The feed stream richer in solute is introduced at
stage(1) while the other feed which contains less
amount of solute is introduced at later stages so that
the composition of each feed is closest to the
composition of the raffinate stream to which it is
added.
O M B on the B attery:1- L 0 + L F + V n+ 1 = V1 + L n = M
L0 + LF = LT 
x 0 , x F , x T lies on a str. line.
 L T + V n+ 1 = V1 + L n = M
L T x T + V n+ 1 y n+ 1 = V1 y 1 + L n x n = M x M
T o get the am ounts:-
V n+ 1
L0 + LF
=
V n+ 1
LT
=
xT xM
y n+ 1 x M
2- C onsider R is the overall operating p olar:V n + 1 - L n + L F = V1 - L 0
 R + LF =
R '  R , x F ,R '
A lso V n + 1 - L n = V1 - (L 0  L F )
R = Vn+1 - L n

R , y n+1 , x n
R = V1 - ( L 0 + L F )
 R = V1 - L T  R , y 1 , x T
R = V f+ 1 - ( L f + L F )
 R = V f+ 1 - L  R , y f+ 1 , x
3  C onsider R ' is the operating polar of the first section:R ' = V1 - L 0
 R ' , y1 , x 0
= V f+1 - L f  R ' , y f+1 , x f
 T he line connecting x F w ith S cuts the ex traction locus at x f
 T he line connecting x f w ith R ' cuts the extraction locus at y f+1
 L = Lf + LF  x , xf , xF
extending R y f+1 line to cut x f x F at point x
Extract with reflux:The maximum composition of extract product from
Countercurrent multi-stage(y1max) occurs when the
(S/F)min is used.
This maximum composition at best situation at the end
of the tie line passing through feed point, and even in
that case an ∞ no. of stages is required .
In practice it is always desirable to have an extract as
rich as possible in the solute, richer than the composition
of a layer in-equilibrium with the liquid feed.
This could be possible by using “Enriching with reflux” or
“Extract reflux” .
Which means:y 1 with reflux > y 1 max
Although extract reflux enrich the extract product it
requires using large amount of Solvent/unit quantity of
Feed.
The use of reflux is not always possible, it depends on
physical equilibrium relation governing the solubility of
The system at hand(limitation).
For systems having a phase diagram of type(I), the length of
tie lines get shorter as the composition of the extract phases
get richer in solute. Which means that the effluent
compositions are closer to one another, which makes the
separation of the two phases difficult. This limitation is not
present in type(II).
Graphical Representation of extract with reflux:Assumption concerning the behavior of the solvent removal
unit:1- Nothing of the raffinate component (B) entering unit,
Leaves with the separated solvent stream V, all (B) goes to
L0 and D. Accordingly (V) is a mixture of (S+A) and its
composition (y) on the hypotenuse.
2-(L0 + D ) stream from (SRU) is saturated with solvent
and lies on raffinate stream (x0 = xD lies on raffinate).
3- The feed stream ( xf ) lies on raffinate locus.
(note:- These simplifications are not essential and the
extract problem can be solved without them).
The ratio of reflux L0 to final extract product D is termed
the reflux ratio. The higher the ratio the higher the
enrichment of the extract with a given number of stages.
• If a feed stream LF with composition xF is to be
separated into extract with composition(xD = x0) and
raffinate composition xn , using extract reflux as shown in
the previous Figure. The rich solvent stream leaving the
top of (SRU) has composition(y).
N um ber of stages calculations:1  Locate x n , x f , x D = x 0 , y and y n+ 1 on the diagram .
2- P oint y 1 is located at the intersectio n of x 0 y extract branch
3  It is clear that:- V1  V + ( L 0 + D )
V1
x0 y
=
L 0 +D

x0 y
y y1
V1

y y1
L 0 (1+
V1

L0(
r +1
D
)

)
L 0 (1+
L0

V1

V1
L0
S ince r is know n,
L0
m ay be calculated.
)
r
= (
r
V1
1
r +1
r
)
x0y
y y1
4
R ' = V1 - L 0 = V f+ 1 - L f
 R ' lies on the extention of x 0 y 1 so that
x0 R'
=
R ' y1
V1
L0
A lso R ' lies betw een y and y 1 .
5- x f is located on the raffinate (notice that x f coinside w ith x F ).
C onnecting x f t o R ' it cuts the extract locus at y f+ 1 w hic h is also
the extract com position leaving the seco nd section.
6-
R = V n+ 1 - L n = V f+ 1 - ( L f + L F ) = R ' - L F
C onnecting x F to R ', x F R ' intersects x n y n+ 1 a t R .
7- T he num ber of stages in each section m ay be calculated by usual
polar construction using points R ' and R respectively.
8- T he operating curve m ay be projected on x-y diagram and
the num ber of stages can be obtained.
V1
=
L0
x 0R'
R 'y 1
=
r +1 x 0 y
r
y y1
R ' lies betw een y & y 1
A lso
V1  V + L 0  D
R ' = V1 - L 0 = V + D
Special cases:1- Solvent used as recovered:y = y n+1
2-Total reflux:D = 0 , reflux = r
= L0/D = ∞
→n min
V1 = V + D + L0
V1 = V + L0
V1 - L0 = R’ = V
R’ on y
3-Total reflux , used as recovered:-
4-Minimum reflux ratio:-
Optimum reflux ratio:The number of stages required increases as reflux ratio
decreases. As reflux ratio decreases R’ lies on a tie line
(xf y1max ) where reflux is minimum, infinite number
of stages, (S/F)min. As reflux ratio increased, number of
stages decreased. For total reflux, no product is
withdrawn( D = 0 ) and
reflux ratio = ∞ .
Also R’ lies on y
which means Zero
Production rate.
Solvent-free coordinate:Solvent free coordinates can be used in designing
solvent extraction systems. The construction is in
general less crowded than in the case of triangular
diagram, but the same procedure described above may
be followed. The amounts of various streams should be
expressed on solvent -free basis.
S om e notes on the diagram :P ure A = 100% , B = 0 , S = 0
x or y =
A
A
=
A + B
= 1
,
S
N =
A + 0

0
A + B
100  0
S
0
0
P ure B = 100% , A = 0 , S = 0
x or y =
A
0
=
A + B
= 0 , N =
0 + 100

A + B
P ure S = 100% , A = 0 , B = 0
x or y =
A
A + B
=
0
0
, N =
S
A + B

100
00

0  10 0
0
T o calculate the am ounts on solvent free basis diagram :L 0  A +B + S
= (A + B ) + S
L 0' + S
=
L0
=
A + B
A + B
A + B
A + B
1
=
 L 0' =
+
S
+
=
L0
L 0'
N0
L0
1 + N0
also V1 ' =
V1
1+ N 1
&
L1 ' 
L1
1+ N 1
R ' is located on x 0 y so that
& T he ratio of
thus:-
V
'
L0
= (
r+ 1
r
R ' y1
'
=
V1
'
L0
'
x0y
=
y y1
'
1
x 0R'
V1
'
L0 + D
)
'
x0y
y y1
T he operating lines of the tw o sections is projected dow n
to x-y digram and num ber of stages are determ ined by the
usual step-w ise construction.
Special cases:1- Solvent used as recovered → y = yn+1
(not pure Solvent)
2-Total reflux:- R’ = y → nmin
3- Solvent used as recovered & total reflux
(in case of pure solvent)→ n min
4-Minimum reflux ratio:- ( y = yn+1 = pure solvent →∞ )
'
x 0 R m in
y1R
'
m in
= (
rm in + 1
x0y
)
rm in
= (
y y1
rm in + 1
x 0 y1 + y y1
)
rm in
= (
y y1
rm in + 1
) (
rm in
L im t
yy 1  
(

y1R
'
m in
1 )
y y1
x 0 y1
)= 0
y y1
'
x 0 R m in
x 0 y1
(
rm in + 1
rm in
)  get rm in
Complete immiscibility:When the solvent and raffinate liquid are completely
Insoluble in each other, the extract locus will be the
hypotenuse and the raffinate locus the horizontal side
of the triangle. This simplifies the calculations because
each phase will consist of two components only. The
calculations is done on a “solute-free basis.”
L’= amount of B in raffinate (L’0 = L’1= L’2=------= Ln= L’)
V’ = amount of S in extract ( V’n+1 = V’n = V’1 = ----= V’)
X = A/B = (mass of solute/mass of inert) in raffinate.
Y = A/S = ( mass of solute/ mass of solvent) in extract.
Types of contact:1- Simple single stage:V ' Y in + L' X in = V ' Y out + L' X out
V ' ( Y in – Y out ) = L' (X out – X in )
-
L'
V'
=
(Y in - Y out )
(X in - X out )
operating line equation
2-Multi-stage cross current contact:1
st
S tage:-
V ' Y in + L' X in = V ' Y1 + L' X 1
V ' ( Y in – Y1 ) = L' (X 1 – X in )
L'
-
=
V'
2
nd
=
V'
3
-
rd
(X in - X 1 )
S tage:L'
-
(Yin - Y1 )
(Yin - Y 2 )
(X 1 - X 2 )
S tage:-
L'
V'
=
(Yin - Y3 )
(X 2 - X 3 )
3- Multi-stage counter current contact:-
V ' Y in + L' X in = V ' Y out + L' X out
V ' ( Y in – Y out ) = L' (X out – X in )
L'
V'
=
(Y in - Y out )
operating line equation
(X out - X in )
For (L’/V’)Min = Pinch point conditions = infinite number of
stages(see graph).
To calculate the number of stages use design range for Vopr.
Equipments in liquid extraction:can be classified according to the methods applied
for inter-dispersing the phases and producing the
counter-current flow pattern.
Both of these can be achieved either by force of
gravity acting on the density difference between
the phases, or by applying centrifugal force.
The "gravity method" can be further divided into 2
broad categories according to the nature of their
operation, namely stage-wise contact (e.g. in mixer
settler) and differential contact (spray or packed
column).
In commercial applications, distinction is
normally made between the light phase and the
heavy phase, and between the dispersed phase
and the continuous phase. The choice of the
dispersed phase depends on flow rates,
viscosities and wetting characteristics of both
phases and is usually based on experience. For
example, depending on circumstances, the
solvent can be the heavy phase or the
continuous phase -- but it will always be the
extract phase.
The location of the principal interface between
the extract and the raffinate depends upon
which phase is dispersed. When the light phase
is dispersed, the interface is located at the top of
the extractor. When the heavy phase is
dispersed, the interface is located at the bottom.
Normally the raffinate and/or extract needs to
be separated by other means.
Examples of Extraction Equipment some of the
following equipment will be briefly discussed :
(1) Mixer-settler:- Mixer-settlers are still widely
used because of their reliability, operating flexibility,
and high capacity. They can handle difficult-todisperse systems, such as those having high
interfacial tension and/or large phase density
difference. They can also used with highly viscous
liquids and solid-liquid slurries. The main
disadvantages are their size and the inventory of
material held up in the equipment. For multiple unit
operations, considerable capital costs may be
needed for pumping and piping.
Mixer - Settler
A mixer-settler device ordinarily consisted of
two parts: a mixer for contacting the two liquid
phases to bring about mass transfer, and a
settler for their mechanical separation. The
mixer and settler can be integral or separate.
The operation may be continuous or batch wise.
Mixer-settler can be single-stage or multi-stage
(cascade). For multi-stage system, also known as
mixer-settler battery (MSB) . System expansion
is easy by the addition of extra stages to existing
system.
Multi-Stage Counter Current Contact
The stages are normally staggered in height, so that
one phase can flow from one stage to another,
while the other phase is pumped.
Unagitated Column Extractors showed 3 types of
unagitated column extractors:
SPRAY COLUMN
Column extractors typically have the two phases
flowing in countercurrent pattern. For the unagitated
units shown the light phase being dispersed or
distributed (hence the heavy phase continuous), i.e.
the light liquid enters at the bottom of the column and
evolve as small droplets at the nozzle distributor. The
droplets of light liquid rise through the mass of heavier
liquid, which flows downward as a continuous stream.
The droplets are collected at the top and form the
stream of light liquid leaving the top of the column. The
heavy liquid on the other hand leaves the bottom of
the column. The choice may be reversed, whereby the
heavy stream is introduced into the light phase at the
top of the column and falls as dispersed droplets
through a continuous stream of light liquid.
(1) Spray Column
This set-up consists of an empty shell with
provisions at the end for introducing and
removing the liquids. Its construction is the
simplest but suffers from low efficiency due to
poor phase contacting and excessive backmixing in the continuous phase. Because of their
simple construction, spray columns are still used
in the industry for simple operations such as
washing and neutralization.
(2) Packed Column:The packing in extraction is similar to the ones
used in distillation. They include both random
and structured packing. Packing offer better
efficiency because of improved contacting and
reduced back-mixing.
It is important that the packing material be
wetted by the continuous phase to avoid
coalescence of the dispersed phase. To reduce
the effects of channeling, re-distribution of the
liquids at fixed intervals is normally required in
taller columns.
(3) Sieve-Tray (Perforated-Plate) Multi-Stage
Column:This is also an improvement over the spray column.
It is particularly suitable for corrosive systems
where absence of mechanical moving parts is an
advantage. Either the heavy liquid or the light liquid
may be dispersed. If the light phase is dispersed,
the light liquid flows through the perforations of
each plate and is dispersed into drops which rise
through the continuous phase. The continuous
phase flows horizontally across each plate and
passes to the plate below through the down comer
If the heavy phase is dispersed, the column is
reversed and up comers are used for the
continuous phase.
Mechanically Agitated Extractors:As an example of these extractors:Rotating Disk Contactor (RDC)
In this system, horizontal disks
are used as agitating elements,
which are mounted on a
centrally supported shaft.
Mounted on the column
wall and offset against the
agitator disks are the stator
rings. This device uses the
shearing action of the rapidly
rotating disks to inter-disperse
the phases.