# MASS TRANSFER.2 LIQUID-LIQUID EXTRACTION SECTION 3.

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MASS TRANSFER.2 LIQUID-LIQUID EXTRACTION SECTION 3

S OLVENT - 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%A 0%B 100%B 0%A A/(A+B) x, y

A B S On Extract with reflux: get R’ y 1. R, y n+1, x n R, R’, X F R’, y f+1, x f R, x _, y f+1 y y n+1 xnxn R’ y f+1 xfxf R y1y1 x 0 =x D XFXF x_x_

y y n+1 xnxn R xfxf R’ y f+1 y1y1 x0x0 (R’y 1+ y1x 0 )/R’y 1 =((r+1)/r).yx 0 /y y 1 r= L 0 /D Get R’ 1-Not pure solvent

2-Pure Solvent y n+1 = ∞ N=S/(A+B) A&B=Zero so; N= ∞ R’x 0 /R’y 1 =((r+1)/r).yx 0 /yy 1 y y n+1 xnxn R xfxf R’ y f+1 y1y1 x0x0 ∞

S PECIAL CASES 2-Pure Solvent & Total Reflux y n+1 = ∞ R’=y y n+1 xnxn R y1y1 y=R’ ∞ ∞ xfxf y f+1

S PECIAL CASES 3-Pure Solvent & Total Reflux & Solvent used as recovered : V 1 =V+L 0 +D V 1 =V+L 0 V 1 -L 0 =R’=V r=L 0 /D=L 0 /0= ∞ n=n min y=y n+1 =R=R’= ∞ y n+1 xnxn R xfxf R’ y1y1 y x0x0 ∞ ∞ ∞ ∞ x1x1 y2y2 x2x2 y3y3 Operating

A NOTHER WAY y n+1 xnxn R xfxf R’ y1y1 y x0x0 ∞ ∞ ∞ ∞ Operating projection of operating curve that will be line 45 here.

S PECIAL CASES 4-Pure Solvent & Min Reflux (n=∞) R’x 0 /R’y 1 =((r+1)/r).yx 0 /yy 1 =((r+1)/r).(x 0 y 1 + yy 1 )/yy 1 =((r+1)/r).(0+1) [by taking the limit when yy 1 approaches infinity]. =R’ min x 0 /R’ min y 1 =((r min +1)/r min ) Get r min xfxf y1y1 x0x0 y f+1 max R’ min y∞y∞ y n+1 xnxn ∞

P ROBLEM (5) Givens:  Ethyl Benzene(Inert liquid)  Styrene(Solute)…..X F =0.5  DEG(Solvent)  Extract with reflux as r is mentioned.  x 0 =0.9 as product contains this %. x n =10% Required:  Min N.T.S.  Min reflux ratio (r min ).  N.T.S when r=1.5 r min

P ROBLEM (5) NYNX lb glycol /lb Hydrocarbon lb styrene /lb Hydrocarbon lb glycol /lb Hydrocarbon lb styrene /lb Hydrocarbon Raffinate Equilibrium Extract

P ROBLEM (5) As nothing is mentioned about y& y n+1, take them equal to zero. To get R’ min : =R’ min. x 0 /R’ min. y 1 =((r min +1)/r min ) Get r min = 4.7 x f=0.5 y1y1 x 0=0.9 y f+1 max R’ min y∞y∞ x n=0.1

P ROBLEM (5) To get Minimum N.T.S, 45 line will be the operating line so count down the stages. N.T.S=10.5 x f=0.5 y1y1 x 0=0.7 R’ min y∞y∞ x n=0.1 y n+1 R R’ ∞ ∞ ∞

P ROBLEM (5) When r=1.5 r min,r=7. So now we will we will place the new R’ point by: (1+y 1 x 0 /R’y 1 )=((r+1)/r) Total N.T.S= 21.2 x f=0.5 y1y1 x 0=0.7 R’ y∞y∞ x n=0.1 y n+1 R ∞

C OMPLETE 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 AB Raffinate Extract

C OMPLETE 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 AB Raffinate Extract

C OMPLETE IMMISCIBILITY As we assumed solute- free basis, amount of solids in feed will be the same as amount in product so; L 0 =L 1 =L n =L’=B It is the same for the solvent V n+1 =V n =V 1 =V’=S Men el a7’er ABSORPTION 1 n L 0, x 0 L n, x n V 1, y 1 V n+1, y n+1

S INGLE STAGE 1 V‘, Y 0 L‘, X 1 L‘, X 0 V‘, Y 1 V‘ Y 0 + L‘ X 0 = V‘ Y 1 + L‘ X 1 V‘(Y 0 - Y 1 )= L‘(X 1 -X 0 ) (- L‘/V‘) =(Y 0 - Y 1 ) / (X 0 -X 1 ) Operating line equation between two points(X 0,Y 0 ) & (X 1,Y 1 ) with slope of (- L‘/V‘) Y=A/S X=A/B Equilibrium Operating X 0,Y 0 X 1,Y 1 - 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 2 L‘, X 0 L‘, X 1 L‘, X 2 V‘, Y 0 V‘, Y 2 V‘, Y 1 So; we will have operating lines with the same number of given stages..at this case we will have two operating lines.

C OMPLETE IMMISCIBILITY First line: (X 0,Y 0 ), (X 1,Y 1 ) & (- L‘/V‘) Second line: (X 1,Y 0 ), (X 2,Y 2 ) & (- L‘/V‘) 1 2 L‘, X 0 L‘, X 1 L‘, X 2 V‘, Y 0 V‘, Y 2 V‘, Y 1 Y=A/S X=A/B Equilibrium Operating 1 X 0,Y 0 X 1,Y 1 - L‘/V‘ Operating 2 - L‘/V‘ X 1,Y 0 X 2,Y 2

MULTI STAGE COUNTER CURRENT V‘ Y n+1 + L‘ X 0 = V‘ Y 1 + L‘ X n V‘(Y n+1 - Y 1 )= L‘(X n -X 0 ) (L‘/V‘) =(Y n+1 - Y 1 ) / (X n –X 0 ) Operating line equation between two points(X 0,Y 1 ) & (X n,Y n+1 ) with slope of (L‘/V‘) 1 n L‘, X 0 V’, Y n+1 V’, Y 1 L‘, X n Y=A/S X=A/B Equilibrium Operating L‘/V‘ X n,Y n+1 X0X0 Y1Y1

M AY 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. 0.020.009980.007510.005020.002460.0010110X 0.00870.009130.006860.004650.0019610.0008070Y

M AY 2008,Q.2 x 0 = 0.01 X 0 =x 0 /(1-x 0 ) X 0 =0.0101 L=100 lb L‘=L(1-x 0 ) L’=99 lb V=150 lb V=V’=150 (- L‘/V‘) =(Y 0 - Y 1 ) / (X 0 -X 1 ) -99/150=(0-Y 1 ) / (0.0101-X 1 ) Assume X=0.005, Y=0.0033 1 V‘, Y 0 L‘, X 1 L‘, X 0 V‘, Y 1

M AY 2008,Q.2 x1x1

i) To get X 1 &Y 1, extend the operating till it cut the equilibrium in X 1 & Y 1. X1=0.0046 %Extraction=(0.0101- - 0.0046)/(0.0101) %Extraction= 54%

M AY 2008,Q.2 ii) X 0 =0.01, X n =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

M AY 2008,Q.2 If V’=150 Slope=L’/V’ Slope=99/150=0.66 0.66 =(0 - Y 1 ) / (0.001 –0.0101 ) Get Y 1 from previous equation Y 1 =0.0059..then draw operating curve N.T.S=4