Section 2

MASS TRANSFER.2LIQUID-LIQUID EXTRACTION

SECTION 2

IN CASE OF MORE THAN ONE FEED

1) Separate each one in a separate apparatus(expensive solution).

2) Mix then enter as one feed (not Engineering solution).

3) Each feed is introduced at the proper stage of the same apparatus.

INTERMEDIATE FEED

1 2 ff+1

f+2

n

L0, x0 Ln, xn

Vn+1, yn+1V1, y1 Vf+1, yf+1

LF, XF

L_

, x_Lf, xf

L_

= Lf+LF

OMB:

L0+LF+Vn+1=V1+Ln

L0+LF=LT x0, xF, xT are on the same straight line

LT +Vn+1 =V1+Ln

are on the same straight line R, yn+1, xn & R, y1, xT Vn+1 - Ln =V1 -LT =R

INTERMEDIATE FEED

1 2 ff+1

f+2

n

L0, x0 Ln, xn

Vn+1, yn+1V1, y1 Vf+1, yf+1

LF, XF

Lf, xf

OMB:

L0+LF+Vn+1=V1+Ln

Vn+1 –Ln + LF =V1 - L0

R+ LF = R’ R, XF, R’ are on the same straight line

L_

, x_

INTERMEDIATE FEED

1 2 ff+1

f+2

n

L0, x0 Ln, xn

Vn+1, yn+1V1, y1 Vf+1, yf+1

LF, XF

Lf, xf

L0+Vf+1=V1+Lf

V1-L0=Vf+1-Lf=R’ y1, x0, R’ & yf+1, xf, R’ are on the same

straight line

L_

, x_

INTERMEDIATE FEED

1 2 ff+1

f+2

n

L0, x0 Ln, xn

Vn+1, yn+1V1, y1 Vf+1, yf+1

LF, XF

Lf, xf

L_

+Vn+1=Vf+1+Ln

Vn+1-Ln=Vf+1-L_

=R R, yn+1, xn &yf+1, x_

, R are on the same straight

line

L_

, x_

INTERMEDIATE FEED

1 2 ff+1

f+2

n

L0, x0 Ln, xn

Vn+1, yn+1V1, y1 Vf+1, yf+1

LF, XF

Lf, xf

Lf +LF = L_

xf ,XF &x_

are on the same straight line

L_

, x_

X0, XF, XT

R, yn+1, xn

R, y1, XT

y1, x0, R ’

R, XF, R’

yf+1, xf, R’

yf+1, x_

, R

xf , XF ,x_

AB

SR

xn

yn+1

y1

XTXF

R’

X0

yf+1

xf

x_

AB

S

x

y

R

R’

xn

yn+1

y1

xTxFx0

yf+1

ba

First Section

Second Section

N.T.S=3 First section= a/b Second section= 3-a/b

PROBLEM (3) Givens: Multistage counter current. A:Acetone. B:Water. S:MCB. Feed contains 0.2 A. Raffinate product contains 0.01 A Saturated water phase contains 0.06 A is also fed in rate of

0.1 lb/lb of solution containing 0.2 A. Extract product is to contain 0.14 A. Required: Equilibrium stage at which second feed is introduced. N.T.S required. Wt of MCB required per lb of 0.2 wt fraction acetone

solution.

PROBLEM (3)Extract layer Raffinate layer

Chloro-benzen

eWater Aceton

e

Chloro-benzen

eWater Aceton

e

yS yB yA xS xB xA

0.9982 0.00 0.0011 0.00

0.9947 0.0521 0.0018 0.05

0.8872 0.1079 0.0021 0.1

0.8317 0.162 0.0024 0.15

0.7693 0.2223 0.0031 0.2

0.6982 0.2901 0.0042 0.25

0.6080 0.3748 0.0058 0.3

PROBLEM (3)

In this problem, the given values for raffinate & extract is too much so we will take the values that we need & let the other values.

The range of drawing is too small so; we will draw the range that we need.

Also, the values of the two layers are so narrow so we will assume that the raffinate layer is AB line & extract layer is the hypotenuse.

PROBLEM (3)

xn= 0.01

x0= 0.2

yn+1 = 0.00

XF =0.06

LF=0.1 L0

By L.A.P, we can get XT

y1=0.14 By counting, the

N.T.S = 5. First section=2.5 AB

S

x

y

45

Equilibrium

Extract

Raffinate

x0xn

yn+1

XFXT

y1

R

R’

yf+1

As LF/L0=0.1

LT=Lo+LF=Lo (1+0.1)=1.1Lo

Lo/LT=1/1.1 **

From graph: Vn+1/LT=1.102 * By dividing *&** Vn+1/ L0=1.2122lb/lb

EXTRACT WITH REFLUX

2 ff+1

n

L0, x0

Ln, xn

Vn+1, yn+1V1, y1 Vf+1, yf+1

LF, xF

L_, x_Lf, xf

SRU

1

V, y

D, xD

B:zeroS:highA:low

B:lowS:lowA:high

Second product(Lean in solute)

First product(Rich in solute)

Main Feed

ANALOGY

Addition of heat. Reboiler. Removal of heat. Condenser. Mixture of liquid and vapor. Relative volatility. Change of pressure.

Operation or Condition in Extraction

Distillation Analogy

Addition of solvent Solvent mixer Removal of solvent Solvent separator Two-phase liquid mixture Selectivity Change of temperature

ASSUMPTIONS FOR EXTRACT WITH REFLUX

Stream leaving top of SRU has no B. (D+L0) stream from SRU is saturated with

solvent (lies on raffinate layer). xf lies on raffinate layer as it is saturated with

solvent.

ADVANTAGES OF EXTRACT WITH SOLVENT

y1 with extract with definite # of stages> y1 max with infinite # of stages

LIMITATIONS FOR EXTRACT WITH REFLUX

Large amount of solvent is used. Certain type of ternary diagram is used.

r= L0/D

V1=V+L0+D

V1/(L0+D)=yx0/yy1

V1/L0=((r+1)/r)yx0/yy1L0, x0

V1, y1

SRU

1

V, y

D, xD

L0+D

V1

x0 =xD

y

y1

V1-L0=V+D R’=V+D

V1=L0+R’

V1/L0=R’x0/R’y1

R’x0/R’y1=((r+1)/r)yx0/yy1

(R’ y1+y1x0)/R’y1=((r+1)/r)yx0/yy1

R ’

x0 =xD

D

y

V

y1

x0 =xD

V1

R’ L0, x0

V1, y1

SRU

1

V, y

D, xD

L0+D

AB

S From previous slide get R’ y1. R, yn+1, xn

R, R’, XF

R’, yf+1, xf

R, x_, yf+1

yyn+1

xn

R’

yf+1

xf

R

y1

x0 =xD

XF

x_

SPECIAL CASES

1-Solvent is used as recovered. y=yn+1

AB

S

y=yn+1

xn

R’

xf

y1

x0 =xD

XF

x_

R

yf+1

SPECIAL CASES

2-Total reflux. V1=V+L0+D

V1=V+L0

V1-L0=R’=V R’=y

r=L0/0= ∞

nmin

AB

S

y=R’

yn+1

xn

xf

y1

x0 =xD

XF

x_

R

yf+1

SPECIAL CASES

3-Total reflux+ Solvent as

recovered y=yn+1=R=R’

AB

S

y=yn+1=R=R’

xn

xf

y1

x0 =xD

XF

yf+1

SPECIAL CASES

4-Minimum reflux ratio (rmin)

R’minx0/R’miny1=

((rmin+1)/rmin)yx0/yy1AB

S

x

y

y1

xf

yf+1max

x0 =xD

R’min

PROBLEM (4) Givens: A:MCP B:N-Hex S:Aniline. Feed contains 0.4 mole fraction A. Solvent contains 0.05 mole fraction A…as recovered!!. Raffinate product contains 0.15 mole fraction A. Extract product contains 0.7 A. XF =0.4

yn+1 = 0.05….As Recovered!!!

xn= 0.15

x0 = xD = 0.7 Required: N.T.S if r=10. Min r = ?

PROBLEM (4)

As usual we will draw the raffinate & extract layer.

Also, draw the equilibrium relation.

1-Solvent is used as recovered. y=yn+1

(R’ y1+y1x0)/R’y1=

((r+1)/r)yx0/yy1

AB

S

y=yn+1

xn

R’

xf

y1

x0 =xD

XF

x_

R

yf+1

PROBLEM (4)

As solvent is used as recovered….y=yn+1

(R’y1+y1x0)/R’y1=((r+1)/r)yx0/yy1

As we know y1x0(10.6 cm) & yy1(1.75 cm) we can get R’y1(1.58 cm) ..so , now we have R’.

N.T.S=3

AB

Sx

y

xn

xf x0 =xDXF

y=yn+1

y1

R’

R

yf+1

PROBLEM (4)

R’minx0/R’miny1=

((rmin+1)/rmin)yx0/yy1

Measure R’minx0(10.3cm), R’miny1 (1.6cm), yx0 (10.6 cm) & yy1 (1.75 cm) then get rmin =(7.48)

AB

Sx

y

xn

xf x0 =xDXF

y=yn+1

y1

R’min

R

yf+1 min

Thank you for your attention!Any Questions?