MASS TRANSFER.2LIQUID-LIQUID EXTRACTION

SECTION 3

SOLVENT-FREE COORDINATES

100%A, 0%B, 0%S

x, y= A/(A+B)=100/(100+0)=1

N=S/(A+B)=0/(100+0)=0 0%A, 100%B, 0%S

x, y= A/(A+B)=0/(100+0)=0

N=S/(A+B)=0/(0+100)=0 0%A, 0%B, 100%S

x, y= A/(A+B)=0/(0+0)=0/0

N=S/(A+B)=100/(0+0)=Inf

Extract (y,N)

Raffinate (x,N)

N=S/(A+B)

(1,0)((0,0

100%A0%B

100%B0%A A/(A+B)

x, y

AB

S On Extract with reflux: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_

y

yn+1

xn

R

xf

R’yf+1 y1

x0(R’y1+y1x0/)R’y1=((r+1)/

r).yx0/yy1

r= L0/DGet R’

1-Not pure solvent

2-Pure Solvent yn+1= ∞ N=S/(A+B) A&B=Zero so; N= ∞ R’x0/R’y1=((r+1)/r).yx0/yy1

y

yn+1

xn

R

xf

R’

yf+1 y1

x0

∞

SPECIAL CASES

2-Pure Solvent & Total Reflux

yn+1= ∞ R’=y

yn+1

xn

R

y1

y=R’∞

∞xf

yf+1

SPECIAL CASES 3-Pure Solvent & Total Reflux &

Solvent used as recovered :

V1=V+L0+D

V1=V+L0

V1-L0=R’=V

r=L0/D=L0/0=∞

n=nmin

y=yn+1=R=R’= ∞

yn+1

xn

Rxf

R’

y1

y

x0

∞

∞

∞ ∞

x1

y2

x2

y3

Operating

ANOTHER WAY

yn+1

xn

Rxf

R’

y1

y

x0

∞

∞

∞ ∞

Operating

projection of operating curve that will

be line 45 here.

SPECIAL CASES

4-Pure Solvent & Min Reflux (n=∞) R’x0/R’y1=((r+1)/

r).yx0/yy1

=((r+1)/r).(x0y1+ yy1)/yy1

=((r+1)/r).(0+1) [by taking the limit when yy1 approaches infinity].

=R’minx0/R’miny1

=((rmin+1)/rmin)

Get rmin

xf

y1

x0

yf+1 max

R’min

y∞yn+1

xn

∞

PROBLEM (5)

Givens: Ethyl Benzene(Inert liquid) Styrene(Solute)…..XF=0.5 DEG(Solvent) Extract with reflux as r is mentioned. x0=0.9 as product contains this %. xn=10% Required: Min N.T.S. Min reflux ratio (rmin).

N.T.S when r=1.5 rmin

PROBLEM (5)

N Y N X

lb glycol/lb Hydrocarbon

lb styrene/lb Hydrocarbon

lb glycol/lb Hydrocarbon

lb styrene/lb Hydrocarbon

Raffinate

Equilibrium

Extract

PROBLEM (5)

As nothing is mentioned about y& yn+1, take them equal to zero.

To get R’ min:

=R’min.x0/R’min.y1

=((rmin+1)/rmin)

Get rmin= 4.7

xf=0.5

y1

x0=0.9

yf+1 max R’min

y∞

xn=0.1

PROBLEM (5)

To get Minimum N.T.S, 45 line will be the operating line so count down the stages.

N.T.S=10.5

xf=0.5

y1

x0=0.7

R’min

y∞

xn=0.1

yn+1

R

R’∞ ∞

∞

PROBLEM (5)

When r=1.5 rmin ,r=7. So now we will we will

place the new R’ point by: (1+y1x0/R’y1 )=((r+1)/r)

Total N.T.S= 21.2

xf=0.5

y1

x0=0.7

R’

y∞

xn=0.1

yn+1

R

∞

COMPLETE IMMISCIBILITY

Say that we have a solvent that dissolves A only & doesn’t dissolve B so; at this case the extract will be the A-S line while the raffinate will be B-A line.

S

ABRaffinate

Extract

COMPLETE IMMISCIBILITY

We will use mass ratios (not mass fractions) as all compositions will be located at two component lines.

We will work on solute-free basis.

X=A/B Y=A/S

S

ABRaffinate

Extract

COMPLETE IMMISCIBILITY

As we assumed solute-free basis, amount of solids in feed will be the same as amount in product so;

L0=L1=Ln=L’=B It is the same for the

solvent Vn+1=Vn=V1=V’=S

Men el a7’er ABSORPTION

1 n

L0, x0 Ln, xn

V1, y1 Vn+1, yn+1

SINGLE STAGE

1

V‘, Y0

L‘, X1L‘, X0

V‘, Y1

V‘ Y0+ L‘ X0= V‘ Y1+ L‘ X1

V‘(Y0 - Y1)= L‘(X1 -X0 )

-(L‘/V(= )‘Y0 - Y1( / )X0 -X1 )

Operating line equation

between two points(X0,Y0)

& (X1,Y1) with slope of (-

L‘/V‘)

Y=A/S

X=A/B

Equilibrium

Operating

X0,Y0

X1,Y1

-L‘/V‘

MULTI STAGE CROSS CURRENT

As we said before, the inlet points will be at the same straight lines and the same for the outlet points.

1 2L‘, X0 L‘, X1 L‘, X2

V‘, Y0 V‘, Y0

V‘, Y2V‘, Y1

So; we will have operating lines with the same number of given stages..at this case we will have two operating lines.

COMPLETE IMMISCIBILITY

First line: (X0,Y0), (X1,Y1) & (- L‘/V‘)

Second line: (X1,Y0), (X2,Y2) & (- L‘/V‘)

1 2L‘, X0 L‘, X1 L‘, X2

V‘, Y0 V‘, Y0

V‘, Y2V‘, Y1

Y=A/S

X=A/B

Equilibrium

Operating 1

X0,Y0

X1,Y1

-L‘/V‘ Operating 2 -L‘/V‘

X1,Y0

X2,Y2

MULTI STAGE COUNTER CURRENT

V‘ Yn+1+ L‘ X0= V‘ Y1+ L‘ Xn

V‘(Yn+1 - Y1)= L‘(Xn -X0 )

(L‘/V(= )‘Yn+1 - Y1( / )Xn –X0 )

Operating line equation between

two points(X0,Y1) & (Xn,Yn+1) with

slope of (L‘/V‘)

1 n

L‘, X0

V’, Yn+1V’, Y1

L‘, Xn

Y=A/S

X=A/B

Equilibrium

Operating

L‘/V‘

Xn,Yn+1 X0

Y1

MAY 2008,Q.2 Nicotine (A) in Water (B) solution containing 1% is to be

extracted with Kerosene (S). Water and Kerosene are essentially insoluble.

Equilibrium data:

i)Determine the % extraction of nicotine if 100 lb of feed solution is extracted once with 150 lb solvent.

ii) If 100 lb/hr of nicotine-water solution containing 1% nicotine is to be counter-currently extracted with kerosene to reduce nicotine content to 0.1%...Determine:

a- Minimum kerosene rate. b- N.T.S if 150 lb/hr kerosene is used.

0.02 0.00998 0.00751 0.00502 0.00246 0.001011 0 X

0.0087 0.00913 0.00686 0.00465 0.001961 0.000807 0 Y

MAY 2008,Q.2

x0= 0.01

X0=x0/(1-x0)

X0=0.0101 L=100 lb L‘=L(1-x0) L’=99 lb V=150 lb V=V’=150 (- L‘/V‘) =(Y0 - Y1) / (X0 -X1 )

-99/150=(0-Y1) / (0.0101-X1) Assume X=0.005, Y=0.0033

1

V‘, Y0

L‘, X1L‘, X0

V‘, Y1

MAY 2008,Q.2

0 0.005 0.01 0.015 0.02 0.0250

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

xo ,yo

y1

x1

MAY 2008,Q.2 i) To get X1 &Y1, extend the

operating till it cut the equilibrium in X1 & Y1.

X1=0.0046 %Extraction=(0.0101- -

0.0046)/(0.0101) %Extraction= 54% 0 0.005 0.01 0.015 0.02 0.025

00.0010.0020.0030.0040.0050.0060.0070.0080.0090.01

EquilibriumOperating

MAY 2008,Q.2

ii) X0=0.01, Xn=0.001. From 0.001 draw tangent to the

equilibrium to get the pinch point, at which the solvent has its minimum value.(0.0036,0.003)

Slope (@pinch)=1.15 L’/V’min=1.15

V’min=86 lb

0 0.005 0.01 0.015 0.02 0.0250

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

EquilibriumPinch

MAY 2008,Q.2 If V’=150 Slope=L’/V’ Slope=99/150=0.66 0.66 =(0 - Y1) / (0.001 –0.0101 )

Get Y1 from previous equation

Y1=0.0059..then draw operating curve N.T.S=4

0 0.005 0.01 0.015 0.02 0.0250

0.001

0.002

0.003

0.004

0.005

0.006

0.007

0.008

0.009

0.01

EquilibriumPinchOperating 2

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