section 2
TRANSCRIPT
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MASS TRANSFER.2LIQUID-LIQUID EXTRACTION
SECTION 2
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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.
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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
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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_
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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_
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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_
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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_
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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_
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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
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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.
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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
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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.
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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
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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
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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
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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
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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.
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ADVANTAGES OF EXTRACT WITH SOLVENT
y1 with extract with definite # of stages> y1 max with infinite # of stages
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LIMITATIONS FOR EXTRACT WITH REFLUX
Large amount of solvent is used. Certain type of ternary diagram is used.
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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
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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
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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_
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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
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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
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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
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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
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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 = ?
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PROBLEM (4)
As usual we will draw the raffinate & extract layer.
Also, draw the equilibrium relation.
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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
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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
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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
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Thank you for your attention!Any Questions?