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.
3. INTERMEDIATE FEED
V1, y1 Vf+1, yf+1 Vn+1, yn+1
1 2 f f+1 f+2 n
_ _
L0, x0 Lf, xf L ,x Ln, xn
_
L = Lf+LF
LF, XF
OMB:
L0+LF+Vn+1=V1+Ln
L0+LF=LT x0, xF, xT are on the same straight line
LT +Vn+1 =V1+Ln
Vn+1 - Ln =V1 -LT =R R, yn+1, xn & R, y1, xT are on the same straight line
4. INTERMEDIATE FEED
V1, y1 Vf+1, yf+1 Vn+1, yn+1
1 2 f f+1 f+2 n
_ _
L0, x0 Lf, xf L ,x Ln, xn
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
5. INTERMEDIATE FEED
V1, y1 Vf+1, yf+1 Vn+1, yn+1
1 2 f f+1 f+2 n
_ _
L0, x0 Lf, xf L ,x Ln, xn
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
6. INTERMEDIATE FEED
V1, y1 Vf+1, yf+1 Vn+1, yn+1
1 2 f f+1 f+2 n
_ _
L0, x0 Lf, xf L ,x Ln, xn
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
7. INTERMEDIATE FEED
V1, y1 Vf+1, yf+1 Vn+1, yn+1
1 2 f f+1 f+2 n
_ _
L0, x0 Lf, xf L ,x Ln, xn
LF, XF
_
Lf +LF = L
_
xf ,XF &x are on the same straight line
8. R
S
X0, XF, XT
yn+1
R, yn+1, xn
R, y1, XT
R’ y1, x0, R’
R, XF, R’
yf+1
yf+1, xf, R’
_
y1 yf+1, x , R
_
xf , XF ,x
xn
xf _
x
B A
XF XT X0
9. S R
yn+1
R’
N.T.S=3
First section= a/b yf+1 y
1
Second section= 3-a/b
xn
xF xT x0
B A
y
First Section
a
b
Second Section
x
10. 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.
11. PROBLEM (3)
Raffinate layer Extract layer
Chloro- Chloro-
Acetone Water Acetone Water
benzene benzene
xA xB xS yA yB yS
0.00 0.0011 0.00 0.9982
0.05 0.0018 0.0521 0.9947
0.1 0.0021 0.1079 0.8872
0.15 0.0024 0.162 0.8317
0.2 0.0031 0.2223 0.7693
0.25 0.0042 0.2901 0.6982
0.3 0.0058 0.3748 0.6080
12. 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.
13. R
y
Equilibrium
PROBLEM (3) 45
xn= 0.01
x0= 0.2 R’
yn+1 = 0.00
x
XF =0.06 yn+1
S yf+1 y1
LF=0.1 L0
By L.A.P, we can get Extract
XT
y1=0.14
By counting, the N.T.S
= 5.
Raffinate
First section=2.5
B A
xn XF XT x0
14. 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
15. EXTRACT WITH REFLUX
V, y B:zero
S:high
A:low
V1, y1 Vf+1, yf+1 Vn+1, yn+1
S
1 2 f f+1 n
R
U
Lf, xf L_, x_ Ln, xn
L0, x0 Second product
B:low (Lean in solute)
S:low LF, xF
A:high
Main Feed
D, xD
First product
(Rich in solute)
16. ANALOGY
Operation or Condition in
Distillation Analogy
Extraction
Addition of heat. Addition of solvent
Reboiler. Solvent mixer
Removal of heat. Removal of solvent
Condenser. Solvent separator
Mixture of liquid and vapor. Two-phase liquid mixture
Relative volatility. Selectivity
Change of pressure. Change of temperature
17. 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.
18. ADVANTAGES OF EXTRACT WITH SOLVENT
y1 with extract with definite # of stages> y1 max with infinite # of stages
19. LIMITATIONS FOR EXTRACT WITH REFLUX
Large amount of solvent is used.
Certain type of ternary diagram is used.
20. V, y
r= L0/D
V1, y1
V1=V+L0+D
S
V1/(L0+D)=yx0/yy1 R
1
V1/L0=((r+1)/r)yx0/yy1 U
y L0, x0
L0+D
y1
D, xD
V1
x0 =xD
21. y
D
V, y
V1-L0=V+D
R’=V+D
R’
V1, y1
S
1
V R
U
x0 =xD
R’ L0, x0
L0+D
V1=L0+R’
V1/L0=R’x0/R’y1
R’x0/R’y1=((r+1)/r)yx0/yy1 y1
D, xD
(R’ y1+y1x0)/R’y1=((r+1)/r)yx0/yy1
V1
x0 =xD
22. R
S From previous slide get R’ y1.
R, yn+1, xn
yn+1 y R, R’, XF
R’, yf+1, xf
R, x_, yf+1
R’
y1
yf+1
xf x0 =xD
xn x_
B A
XF
23. R
S
SPECIAL CASES y=yn+1
1-Solvent is used as R’
recovered. y1
yf+1
y=yn+1
x0 =xD
xf
xn x_
B XF A
24. R
S
SPECIAL CASES yn+1
y=R’
2-Total reflux.
V1=V+L0+D y1
yf+1
V1=V+L0
V1-L0=R’=V
R’=y
r=L0/0= ∞
nmin
xf x0 =xD
xn x_
B XF A
25. S
y=yn+1=R=R’
SPECIAL CASES
3-Total reflux+
Solvent as y1
yf+1
recovered
y=yn+1=R=R’
xf x0 =xD
xn
B XF A
26. S
R’min
SPECIAL CASES yf+1max y1
4-Minimum reflux ratio (rmin)
R’minx0/R’miny1=
((rmin+1)/rmin)yx0/yy1 x0 =xD
xf
B A
y
x
27. 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 = ?
28. PROBLEM (4)
As usual we will draw the raffinate & extract layer.
Also, draw the equilibrium relation.
29. R
S
y=yn+1
1-Solvent is used as R’
recovered. y1
yf+1
y=yn+1
(R’ y1+y1x0)/R’y1=
((r+1)/r)yx0/yy1
x0 =xD
xf
xn x_
B XF A
30. y
PROBLEM (4)
As solvent is used as
recovered….y=yn+1
(R’y1+y1x0)/R’y1=((r+1)/r)y
R
x0/yy1 x
S
As we know y1x0(10.6 y=yn+1
cm) & yy1(1.75 cm) we R’
can get R’y1(1.58 cm) ..so yf+1 y1
, now we have R’.
N.T.S=3
xf
xn x0 =xD
B XF
31. y
PROBLEM (4)
R’minx0/R’miny1=
((rmin+1)/rmin)yx0/yy1
Measure R
R’minx0(10.3cm), S
x
y=yn+1
R’miny1 (1.6cm), yx0
R’min
(10.6 cm) & yy1
(1.75 cm) then get yf+1 min y1
rmin =(7.48)
xf
xn x0 =xD
B XF
