Liquid-Liquid Extraction
Lecture 23
26 Nov 2012
1
Overview
•Liquid-Liquid Extraction
•(solvent extraction)
•Pioneered during 1940’s (uranium purification)
•Alternative to distillation, absorption/stripping
•Energy savings
•Sometimes easier separation
•Lower temperatures
•Usually two distinct phases formed
•Usual purpose, to either purify the
•Raffinate, or
•Solute
2
Liquid-Liquid Extraction
Extract
Feed
[a+b]
[s + a] (+b)
Extractor
Solvent
[s]
a = solute
b = diluent
s = solvent
Raffinate
[b] (+ a & s)
let: x  mass fraction solute in raffinate
y  mass fraction
• Separation accomplished by
chemical differences
• Usually in two phase
• - light phase
• - heavy phase
• Usually coupled with another
separation technique
Separator could be:
column w/ stages or packing
column with moving internals
single stage mixer/settler
equilibrium stage(s)
phase
solute in extract phase
y *  in equilibriu m with associated
x
3
Example Industrial Processes
Seader & Henley (2006)
4
Typical LL
Extraction
Process
Seader & Henley (2006)
5
Equipment
Examples
Treybal (1980)
Seader & Henley (2006)
6
7
8
9
Spray Columns:
Seader & Henley (2006) 10
Packed-bed Column
Seader & Henley (2006)
Light liquid - dispersed phase
Treybal (1980)
11
Sieve-tray Extraction Column:
light phase dispersed
Treybal (1980)
12
Oldshue-Rushton (Mixco
Lightnin CMContactor) column
Scheibel column
Seader & Henley (2006)
13
Seader & Henley (2006)
14
Podbielniak Extractor
Treybal (1980)
15
Equipment
Seader & Henley (2006)
16
Equipment Examples
Seader & Henley (2006)
17
Equilateral Triangular Diagrams
[a]
R [ kg ] raffinate
E [ kg ] extract
mixture
mixture
M [ kg ] combined
mixture
Overall material balance:
RE  M
[b]
[s]
Rearrange:
[kg ]
Component material balance (on a):
xR R  yE E  xM M
R
E

yE  xM
xM  xR
Lever principle:
R
E
[s]

e
m

EM
MR
[b]
18
Equilateral Triangular Diagrams
[a]
[a]
[b]
[s]
[b]
[s]
Type I
Type II
Examples:
Example:
• water (b), ethylene glycol (a), furfural (s)
• water (b), acetone (a), chloroform (s)
• n-heptane (b), methylcyclohexane (a), aniline (s)
19
Distribution Curves
[a]
yE
[b]
[s]
xR
Type I
[a]
[b]
[s]
Type II
20
Distribution Curves
[a]
yE
[b]
[s]
xR
Type I
[a]
yE
[b]
[s]
Type II
xR
21
Distribution Curves
[a]
yE
[b]
[s]
xR
Type I
[a]
yE
[b]
[s]
Type II
xR
22
Effect of Temperature (and Pressure)
Treybal (1980)
23
Effect of Temperature (and Pressure)
Treybal (1980)
24
Choice of Solvent
•Selectivity separation factor
 ya


y b  extract

 
  1; better   1
 xa


x b  raffinate

•Distribution Coefficient better if K  1
•Insolubility of Solvent better if less soluble in R phase
•Solvent Recoverability should be easy to separate solvent from E and R
•Density large density differences between the two phases is desired
•Interfacial Tension would like large for easier coalescence of dispersed phase
•Others:
• solvent stable, inert, nontoxic, nonflammable, low cost
• low viscosity
• low vapor pressure
• low freezing point
25
Mixer – Settler (single stage extraction)
Raffinate
Feed
New
Solvent
Solvent
settler
mixer
Extract
solvent
recovery
solvent
recovery
Purified
Raffinate
Purified
Extract
Recycled Solvent
F
Black Box:
S
Raffinate
Feed
xF
Solvent
yS
1
stage
xR
Extract
yE
R
[a]
E
Material balance:
F S  RE  M
[s]
[b]
26
Mixer – Settler (single stage extraction)
given:
xF , F , yS , S
find:
xM , M , xR , yE , E , R
Component material balance (on a in feeds): x F F  y S S  x M M
xM 
xF F  yS S
F S
Component material balance (on a in products): x R R  y E E  x M M
E 
M xM  xR 
[a]
yE  xR
[s]
[b]
27
Mixer – Settler (single stage extraction)
Minimum Solvent (rate):
S min

F
FD

DS
xF  xD
Maximum Solvent (rate):
S max
F

FK
KS

[a]
xD  yS
[s]
[b]
xF  xK
xK  yS
28
Cross-Current (multi-stage extraction)
E1
F
Stage
1
Feed
xF
E2
y1
R1
Stage
2
x1
E3
y2
R2
x2
Final Extract
y3
Stage
3
Solvent S
Solvent S
Solvent S
yS
yS
yS
R3
x3
Final
Raffinate
[a]
Final Extract:
y FE 
E1  E 2  E 3
y1 E 1  y 2 E 2  y 3 E 3
E1  E 2  E 3
[s]
[b]
29
Continuous Multistage Countercurrent Extraction
Feed
F xF
Extract
E1 y1
Total MB:
1
R1
x1
R2
2
x2
E3
y3
E2
y2
RN 2
xN 2
E N 1
y N 1
N-1
R N 1
x N 1
N
EN
yN
RN
xN
Raffinate
S
yS
Solvent
F  S  E1  R N  M
Total MB on a: x M 
xF F  yS S
[a]
F S
If y1 & x N known (specified), then
flowrates E 1 & R N can be
found.
[s]
[b]
30
Continuous Multistage Countercurrent Extraction
Feed
F xF
Extract
E1 y1
1
R1
x1
R2
2
x2
E3
y3
E2
y2
Total MB: R N  S   F  E1 
Operating Point:  R  F  E 1
RN 2
xN 2
E N 1
y N 1
N-1
R N 1
x N 1
N
EN
yN
RN
xN
Raffinate
S
yS
Solvent
MB from feed to N-1 stage: R N 1  E N   F  E1 
[a]
R
[s]
[b]
31
Continuous Multistage Countercurrent Extraction
Feed
F xF
Extract
E1 y1
1
R1
x1
R2
2
x2
E3
y3
E2
y2
RN 2
xN 2
E N 1
y N 1
N-1
R N 1
x N 1
N
EN
yN
RN
xN
Raffinate
S
yS
Solvent
Now step off to find number of equilibrium
stages:
[a]
R
[s]
[b]
32
Questions?
33