The document discusses liquid-liquid extraction as a separation technique. It provides three main points: 1. Liquid-liquid extraction separates species according to differences in solubility between two immiscible liquids, rather than volatility as in distillation. 2. The key components in liquid-liquid extraction are a solute, solvent, and inert components that form two separate liquid phases - an extract and raffinate. 3. There are two main types of systems - one forming one pair of partially miscible liquids leading to a closed ternary diagram, and one forming two pairs forming an open ternary diagram. The document then discusses various aspects of setting up counter-current multi-

1. MASS TRANSFER.2
LIQUID-LIQUID EXTRACTION
SECTION 1

2. Willkommen Bienvenue
Welcome
yôkoso welkom
Benvenuto
Bienvenida tervetuloa

3. Separation
Techniques
Distillation
Crystallization
Salting Out
Solvent
Extraction

4. WHY LIQUID-LIQUID EXTRACTION
 When nearly equal 1.
 For azeotropic mixtures.
 To avoid thermal decomposition.
 Some notes:
1- Separation is done according to different in solubility
(not volatility).
2- Problem with L-L extraction that we mix then we
separate.

5. WHAT IS LIQUID-LIQUID EXTRACTION
Species involved: Vo: Mass flow rate of fresh solvent.
yo: Mass fraction of solute in solvent
 Solute + Inert
Lo: Mass flow rate of liquid feed mixture
 Solvent
xo: Mass fraction of solute in meal
Each stage contain: V1: Mass flow rate of Extract layer
 Contact y1: Mass fraction of solute in Extract
 Separation L1: Mass flow rate of Raffinate layer
x1: Mass fraction of solute in Raffinate
Solvent Extract
Vo, yo V1, y1
Contact Separation
Feed Raffinate
Lo, xo L1, x1

6.  Extract:
A
B
S
o Raffinate:
A
B
S

7. TYPES OF TERNARY SYSTEMS
 Formation of one pair of partially miscible liquids.
[closed ternary diagram]
 Formation of two pair of partially miscible liquids.
[open ternary diagram]

8. FORMATION OF ONE PAIR OF PARTIALLY
MISCIBLE LIQUIDS
 AS&BS are completely soluble.
S
 AB is partially soluble.
 M will be separated to two
compositions; N & L I
 If S is solvent for A so as S P
increases, L goes up till P(Plait
Point) at which A,B & S become
one phase. This is done due to L
the increase in mutual solubility
between A&B. N
M
II
A B

9. FORMATION OF ONE PAIR OF PARTIALLY
MISCIBLE LIQUIDS
 Equilibrium relation.
y
P
x

10. FORMATION OF TWO PAIR OF PARTIALLY
MISCIBLE LIQUIDS
S
 AS&AB are partially soluble.
 B is completely soluble in S. I
y
I II
A B
x

11. STAGE DEFINITION
 It is a mechanical device or series that allow the
solvent & solution to contact and separate. So;
stage is:

12. S y0
SINGLE STAGE
y1
V0, y0 V1, y1
M
L0, x0 1 L1, x1
x1
x0
B A
y
V0+L0=V1+L1=M
V0y0+L0x0=V1y1+L1x1=MxM
y0, x0, xM are on the same straight line
y1, x1, xM are on the same straight line
x

13. S y0
MULTI STAGE CROSS CURRENT
y2
V0, y0 V0, y0 y1
M2
M1
L0, x0 L1, x1 L2, x2 x2 x
1 2 1
x0
B A
y
V1, y1 V2, y2
are y0, x0, xM1 are on the same straight line
y1, x1, xM1 on the same straight line
are y0, x1, xM2 are on the same straight line
y2, x2, xM2 on the same straight line
x

14. S yn+1
MULTI STAGE COUNTER CURRENT
y3
y2
M y1
V1, y1 Vn, yn Vn+1, yn+1
xn x3 x x
1 n 2 1
x0
B A
L0, x0 L1, x1 Ln, xn
y
Vn+1+L0=V1+Ln=M
Vn+1yn+1+L0x0=V1y1+Lnxn=MxM
yn+1
L0-V1=Ln-Vn+1=R R
No. of stages= n+(a/a+b)
But get a & b from AB line
xn+1 b a
xn-1
xn x

15. S yn+1
PROJECTION OF OPERATING LINE
M y1
xn
x0
B A
y
R
Equilibrium
Operating
x

16. EXTRACT & RAFFINATE EFFICIENCY
y y
Equilibrium Equilibrium
H L
0.9H
0.8L
New
Equilibrium
New
Equilibrium
Operating Operating
x x
Extract efficiency 90% Raffinate efficiency 80%

17. PROBLEM (1)
 Givens:
 A counter current-multi stage system.
 We need to recover Pyridine from aqueous solution by Chloro-
benzene.
 xn=0.01
 x0=0.25
 Vn+1/L0=10
 Required:
1. Maximum pyridine content that may be obtained in the
extract phase leaving the system.
 For ratio of solvent to feed of 10 determine:
1. Composition and quantity (expressed as pounds per pound
of feed) of extract phase leaving the system.
2. The number of equilibrium stages required.

18. PROBLEM (1)
Raffinate layer Extract layer
Chloro- Chloro-
Pyridine Water Pyridine Water
benzene benzene
xA xB xS yA yB yS
0.25 0.73 0.02 0.28 0.03 0.69
0.12 0.88 0 0.18 0.02 0.8
0.03 0.97 0 0.07 0.02 0.91
0.01 0.99 0 0.03 0.01 0.96

19. S yn+1
PROBLEM (1)
Raffinate Extract
Raffinate layer Extract layer B x0
Chloro- Chloro-
xn A
Pyridine Water
benzene
Pyridine Water
benzene yA
xA xB xS yA yB yS
0.25 0.73 0.02 0.28 0.03 0.69
0.12 0.88 0 0.18 0.02 0.8
0.03 0.97 0 0.07 0.02 0.91
0.01 0.99 0 0.03 0.01 0.96
Equilibrium xA

20. S yn+1
PROBLEM (1) y1 max
 To get maximum extract
concentration, it will be at
equilibrium with feed
concentration (x0).
 y1max=0.27 B
xn x0 A
yA
Note
 Here we make an approx. as
equilibrium not complete
(otherwise we have to work like
the next problem)
xA

21. S yn+1
y1=0.05
yn+1
M
b
PROBLEM (1)
 As S/F=Vn+1/L0=10
a
 Vn+1/(Vn+1+L0)=a/(a+b)
x0
 10/11=a/(a+b)
 a+b=10.1, a=9.18
B x0
xn A
 Put Point(M). yA
xA

22. R
yn+1 y1
PROBLEM (1)
B x0 A
yA xn
xA

23. R
yn+1 y1
PROBLEM (1)
 N.T.S=1.
B x0 A
yA xn
Equilibrium
Operating
xA

24. S yn+1
y1
y1
M
PROBLEM (1) b
 As S/F=Vn+1/L0=10
a
 V1/(V1+Ln)=a/(a+b)
xn
 9/9.6=a/(a+b)
 V1/(V1+Ln)=9/9.6
B x0
xn A
 L0/(Vn+1+L0)=1/11……..* yA
 V1+Ln= Vn+1+L0
 V1 /(Vn+1+L0)=9/9.6……**
 By dividing *&**
 V1/ L0=(9*11)/(9.6*1)
 V1/ L0=10.3125 lb/lb
xA

25. PROBLEM (2)
 Givens:
 A:Diphenyl hexane.
 B:Decosane.
 S:Furfural.
 Feed contains 0.2 A & 0.8 B.
 Multistage counter current system.
 Solvent contains 0.005 A.
 Raffinate product contains 0.01 A.
 Required:
 N.T.S if mass ratio of solvent to feed is 1.65.
 Ratio of solvent to feed when N.T.S equals 3.
 Maximum concentration of A that will be obtained.

26. PROBLEM (2)
 Here the solubility saturation curve is given
between xS , xA &xB so, this means that the
solubility curve is closed.
xS xA xB
 Also the equilibrium curve is given:
xA xB xS yA yB yS
Equilibrium

27. y
yn+1
b
PROBLEM (2) R
 xn= 0.01
a
 x0= 0.2
x0
 yn+1 = 0.005
 Vn+1/L0=1.65
 Vn+1/(Vn+1+L0)=1.65/2.65=0 yn+1
x
S
.62
 Vn+1/(Vn+1+L0)=a/(a+b) y1
 0.62=a/9.5
 a= 5.89 M
 Put Point M
 y1=0.15
 N.T.S=4
xn X0=0.2
B A

28. y
PROBLEM (2)
 To get S/F that makes
N.T.S=3.
 We will assume y1. As
N.T.S decreases, y1 R
decreases too so;
yn+1 x
assume it less than y1
S
of N.T.S=4 y1 assumed
 y1 assumed =0.1
M
 What a luck!!
 Vn+1/L0=1.8
xn x0 X0=0.2
B A

29. y
PROBLEM (2) R
 To get y1max;
a) Assume y1max
b) From the equilibrium,
get x1 on the raffinate x
S yn+1
locus
y1 y 1max assumed
c) If the line connecting
y1max and x1 passes by M
xo, then the assumption
is true. If not, repeat the
steps
X1 X =0.2
xn 0
B A

30. Thank you for
your attention!
Any Questions?