Section 1

MASS TRANSFER.2LIQUID-LIQUID EXTRACTION

SECTION 1

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Separation Technique

s

Distillation

Crystallization

Salting Out

Solvent Extraction

WHY LIQUID-LIQUID EXTRACTION

When 𝜶 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.

WHAT IS LIQUID-LIQUID EXTRACTION

Species involved: Solute + Inert Solvent

Each stage contain: Contact Separation

Solvent

Feed Raffinate

Extract

Separation

Vo, yo

Lo, xo

V1, y1

L1, x1

Contact

Vo: Mass flow rate of fresh solvent.

yo: Mass fraction of solute in solvent

Lo: Mass flow rate of liquid feed mixture

xo: Mass fraction of solute in meal

V1: Mass flow rate of Extract layer

y1: Mass fraction of solute in Extract

L1: Mass flow rate of Raffinate layer

x1: Mass fraction of solute in Raffinate

Extract: A B So Raffinate: A B S

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]

FORMATION OF ONE PAIR OF PARTIALLY MISCIBLE LIQUIDS AS&BS are completely soluble.

AB is partially soluble.

M will be separated to two compositions; N & L

If S is solvent for A so as S increases, L goes up till P(Plait Point) at which A,B & S become one phase. This is done due to the increase in mutual solubility between A&B.

S

BA

P

II

I

MN

L

FORMATION OF ONE PAIR OF PARTIALLY MISCIBLE LIQUIDS

Equilibrium relation.

x

y

P

FORMATION OF TWO PAIR OF PARTIALLY MISCIBLE LIQUIDS

AS&AB are partially soluble.

B is completely soluble in S.

S

BAI

I

II

x

y

STAGE DEFINITION

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

Contact

Separation

Stage

1

V0, y0

L1, x1L0, x0

V1, y1

V0+L0=V1+L1=M

V0y0+L0x0=V1y1+L1x1=MxM

are on the same straight line y0, x0,

xM

are on the same straight line y1, x1, xM

SINGLE STAGE

AB

S

x

y

M

x0

x1

y1

y0

MULTI STAGE CROSS CURRENT

1 2L0, x0 L1, x1 L2, x2

V0, y0 V0, y0

V2, y2V1, y1

y0, x0, xM1 are on the same straight

line are on the same straight line y1,

x1, xM1

AB

S

x0

y0

x

y

y0, x1, xM2 are on the same straight

line are on the same straight line y2,

x2, xM2

M1

x1

y1

M2

y2

x2

MULTI STAGE COUNTER CURRENT

AB

S

x0

x

y

1 n

L0, x0 L1, x1 Ln, xn

V1, y1 Vn, yn Vn+1, yn+1

Vn+1+L0=V1+Ln=M

Vn+1yn+1+L0x0=V1y1+Lnxn=MxM

L0-V1=Ln-Vn+1=R

yn+1

M

xn

y1

R

x1

y2

x2

y3

x3

xn+1 xn-1

xn

b a

No. of stages= n+(a/a+b)But get a & b from AB line

yn+1

AB

S

x0

x

y

M

xn

y1

R

PROJECTION OF OPERATING LINE

yn+1

EquilibriumOperating

EXTRACT & RAFFINATE EFFICIENCY

x

y

x

y

Equilibrium

Operating

Equilibrium

Operating

90%=Extract efficiency 80%=Raffinate efficiency

0.9HH

0.8L

L

New Equilibrium

New Equilibrium

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.

PROBLEM (1)

Extract layer Raffinate layer

Chloro-benzene Water Pyridine Chloro-

benzene Water Pyridine

yS yB yA xS xB xA

0.69 0.03 0.28 0.02 0.73 0.25

0.8 0.02 0.18 0 0.88 0.12

0.91 0.02 0.07 0 0.97 0.03

0.96 0.01 0.03 0 0.99 0.01

PROBLEM (1)

xA

Extract layer Raffinate layerChloro-

benzene Water PyridineChloro-

benzene Water Pyridine

yS yB yA xS xB xA

0.69 0.03 0.28 0.02 0.73 0.25

0.8 0.02 0.18 0 0.88 0.12

0.91 0.02 0.07 0 0.97 0.03

0.96 0.01 0.03 0 0.99 0.01

Equilibrium

yA

AB

S

ExtractRaffinate

x0

yn+1

xn

PROBLEM (1)

To get maximum extract concentration, it will be at equilibrium with feed concentration (x0).

y1max=0.27

Note Here we make an approx.

as equilibrium not complete (otherwise we have to work like the next problem)

xA

yA

AB

S

x0

yn+1

xn

y1 max

PROBLEM (1)

As S/F=Vn+1/L0=10

Vn+1/(Vn+1+L0)=a/(a+b) 10/11=a/(a+b) a+b=10.1, a=9.18 Put Point(M).

xA

yA

AB

S

x0

yn+1

xn

x0

yn+1

b

a

y1=0.05

M

PROBLEM (1)

xA

yA

AB x0

yn+1

xn

y1

R

PROBLEM (1)

N.T.S=1.

xA

yA

AB x0

yn+1

xn

y1

R

Equilibrium

Operating

PROBLEM (1)

As S/F=Vn+1/L0=10

V1/(V1+Ln)=a/(a+b) 9/9.6=a/(a+b) V1/(V1+Ln)=9/9.6

L0/(Vn+1+L0)=1/11……..*

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

yA

AB

S

x0

yn+1

xn

xn

y1

b

a

y1

M

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.

PROBLEM (2)

Here the solubility saturation curve is given between xS , xA &xB so, this means that the solubility curve is closed.

Also the equilibrium curve is given:

xB xA xS

yS yB yA xS xB xA

Equilibrium

PROBLEM (2)

xn= 0.01

x0= 0.2

yn+1 = 0.005

Vn+1/L0=1.65

Vn+1/(Vn+1+L0)=1.65/2.65=0.62

Vn+1/(Vn+1+L0)=a/(a+b) 0.62=a/9.5 a= 5.89 Put Point M y1=0.15 N.T.S=4

AB

Sx

y

yn+1

x0

b

a

yn+1

X0=0.2 xn

M

y1

R

PROBLEM (2)

To get S/F that makes N.T.S=3.

We will assume y1. As N.T.S decreases, y1 decreases too so; assume it less than y1 of N.T.S=4

y1 assumed =0.1

What a luck!! Vn+1/L0=1.8

AB

S

x

y

yn+1

X0=0.2 xn

M

x0

y1 assumed

R

PROBLEM (2)

To get y1max;

a) Assume y1max

b) From the equilibrium, get x1 on the raffinate locus

c) If the line connecting y1max and x1 passes by xo, then the assumption is true. If not, repeat the steps

AB

Sx

y

yn+1

X0=0.2 xn

M

y1

R

y 1max assumed

X1

Thank you for your attention!Any Questions?