Extraction

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26.02.2012
1
SEPARATION PROCESSES


LIQUID-LIQUID EXTRACTION PROCESSES

Equilibrium Relations in LL Extraction


Transport Processes and Separation Process Principles CHRISTIE J.
GEANKOPLIS
Unit operations in chemical engineering by Mccabe, Smith and Harriot
PRINCIPLES AND MODERN APPLICATIONS OF MASS TRANSFER
OPERATIONS by Jaime Benitez
Separation process principles by Seader and Henley
Separation processes by King
Chemical Engineering by Coulson and Richardson
LIQUID-LIQUID EXTRACTION PROCESSES
removing one constituent from a liquid by means of an other liquid
solvent
When separation by distillation is ineffective or very difficult (close-boiling
mixtures or substances / or heat sensitive mixtures), liquid extraction is one
of the main alternatives to consider

recovery of penicillin from the fermentation broth by extraction with a
solvent such as butyl acetate

recovery acetic acid from dilute aqueous solutions
26.02.2012
2
Equilibrium Relations in LL Extraction
Equilateral triangular coordinates are often used to represent the equilibrium
data of a three-component system, since there are three axes
A, B, or C : a pure component
point M a mixture of A, B, and C
• the perpendicular distance from the
point M to
• the base AB : the mass fraction xc of
C in the mixture at M,
• to base CB the mass fraction xA of A,
• to base AC the mass fraction xB of B
xA + xB + xc = 0.40 + 0.20 + 0.40 = 1.0
Liquid-liquid phase diagram where components A and B are partially
miscible
• C dissolves completely in A or in B.
• A is only slightly soluble in B and B slightly soluble in A.
• The two-phase region is included inside below the curved envelope.
• An original mixture of composition M will separate into two phases a and b
which are on the equilibrium tie line through point M.
• Other tie lines are also shown. The two phases are identical at point P, the Plait
point.
26.02.2012
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Phase diagram where the solvent pairs b—c and a-c are partially miscible.
26.02.2012
4
A: acetic acid, B: water, C: isopropyl ether solvent
The solvent pair B and C are partially miscible.
The concentration of the component C is plotted on the vertical axis and that
of A on the horizontal axis.
The concentration of component B is obtained by difference from
inside the envelope: two phase
Outside: one-phase region
A tie line g-i connecting the raffinate and extract layers
raffinate layer : water-rich layer
extract layer : ether-rich solvent layer g,
The raffinate composition is designated by x and the extract by y
26.02.2012
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the mass fraction of C in the extract layer : yc
the mass fraction of C in the raffinate layer and as xc
To construct the tie line gi using the equilibrium yA — xA plot below the phase
diagram, vertical lines to g and i are drawn.
26.02.2012
6
Equilibrium: System Water-Chloroform-Acetone
Equilibrium extraction data for the system
water-chloroform-acetone at 298 K and 1 atm are given in Table. Plot these
data on a right-triangular diagram, including in the diagram a conjugate or
auxiliary line that will allow graphical construction of tie lines.
26.02.2012
7
Chloroform as solvent (B), water as diluent (A), and
acetone as solute (C).
This system exhibits a type I equilibrium behavior. The
graphical construction to generate the conjugate line
from the tie-line data given (e.g., line RE) is shown, as
well as the use of the conjugate line to generate other
tie lines not included in the original data. Line LRP is
the saturated raffinate line, while line PEK is the
saturated extract line.
Single-Stage Equilibrium Extraction
Derivation of lever-arm rule for graphical addition.
two streams, L kg and V kg, containing components A, B, and C, are mixed
(added) to give a resulting mixture stream M kg total mass.
overall mass balance and a balance on A,
Graphical addition and lever-arm rule : (a) process flow, (b) graphical addition
26.02.2012
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An original mixture weighing 100 kg and containing isopropyl ether (C), acetic
acid (A), and water (B) is equilibrated and the equilibrium phases separated The
compositions of the two equilibrium layers are for the extract layer (V) yA = 0.04,
yB = 0.02, and yc = 0.94, and for the raffinate layer (L) xA =0.12, xB = 0.86, and
xc = 0.02. The original mixture contained 100 kg and xAM = 0.10. Determine the
amounts of V and L.
26.02.2012
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M = 100 kg and x
AM
= 0.10
L = 75.0 and V = 25.0
L = 72.5 kg and V = 27.5 kg
Single-stage equilibrium extraction
separation of A from a mixture of A and B by a solvent C in a single equilibrium
stage.
the solvent, as stream V2 and the stream L0 enter
The streams are mixed and equilibrated and
the exit streams L1 and V1 leave in equilibrium with each other.
26.02.2012
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Single-Stage Extraction
Pure chloroform is used to extract acetone from a feed containing 60 wt%
acetone and 40 wt% water in a co-current extractor. The feed rate is 50 kglh,
and the solvent rate is also 50 kg/h. Operation is at 298 K and 1 atm. Find the
extract and raffinate flow rates and compositions when one equilibrium stage
is used for the separation. (assume water and chloroform do not mix)
26.02.2012
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V= C, Solvent, chloroform, ya1=0, yC1=1.0
L= Liquid mixture, A: acetone, B=water
xA1=0.60, xB1=0.40

A: acetone
C: chloroform
B: water
V1
L1
V2
L2
V
1
+L
1
=V
2
+L
2
=M
V
1
y
A1
+L
1
x
A1
=V
2
y
A2
+L
2
x
A2
=Mx
AM
V
1
y
B1
+L
1
x
B1
=V
2
y
B2
+L
2
x
B2
=Mx
BM

V
1
y
C1
+L
1
x
C1
=V
2
y
C2
+L
2
x
C2
=Mx
CM




L1
V1
yA2
xA2
V
1
+L
1
=V
2
+L
2
=M
V
1
y
A1
+L
1
x
A1
=V
2
y
A2
+L
2
x
A2
=Mx
AM
V= C, Solvent, chloroform, ya1=0, yC1=1.0
L= Liquid mixture, A: acetone, B=water
xA1=0.60, xB1=0.40

L1
V1
yA2
xA2
Tie line is constructed through by trial and error, using conjugate line extract
(V2, yA2) and rafinate (L2, xA2) locations are obtained. The concentrations
yA2=0.334, xA2=0.189
26.02.2012
12
L1
V1
yA2
xA2
Continuous Multistage Countercurrent Extraction
26.02.2012
13
Material Balance for Countercurrent Stage Process Pure solvent isopropyl
ether at the rate of VN+1= 600 kg/h is being used to extract an aqueous
solution of L0 = 200 kg/h containing 30 wt % acetic acid (A) by countercurrent
multistage extraction. The desired exit acetic acid concentration in the
aqueous phase is 4%. Calculate the compositions and amounts of the ether
extract V1 and the aqueous raffinate LN.
26.02.2012
14
Pure solvent isopropyl ether (C)
V
N+1
= 600 kg/h y
AN+1
=0 y
CN+1
=1.0
B: aqueous solution of L
0
= 200 kg/h containing 30 wt % acetic acid (A) exit
acetic acid concentration in the aqueous phase is 4%.
x
A0
= 0.30, x
B0
=0.70, x
C0
=0, x
AN
=0.04
L
0
x
A0
+V
N+1
y
AN+1
=L
N
x
AN
+V
1
y
A1
=Mx
AM
V
N+1
= 600 kg/h y
AN+1
=0 y
CN+1
=1.0
L
0
= 200 kg/h x
A0
= 0.30, x
B0
=0.70, x
C0
=0, x
AN
=0.04
26.02.2012
15
V
N+1
= 600 kg/h y
AN+1
=0 y
CN+1
=1.0
L
0
= 200 kg/h x
A0
= 0.30
x
B0
=0.70, x
C0
=0, x
AN
=0.04
x
CM
=0.75, x
AM
= 0.075
L
0
x
A0
+V
N+1
y
AN+1
=L
N
x
AN
+V
1
y
A1
=Mx
AM
V
N+1
and L
0
are plotted
L
N
is on phase boundary
It can be plotted at x
AN
=0.04
For the mixture point M
x
CM
=0.075, x
AM
=0.75
Draw a line from the point of
L
N
and M at equilibrium of
extract phase gives V
1
,
y
A1
=0.08 and
y
C1
=0.90
, y
CN+1
, y
AN+1
y
C1
, y
A1
x
CN
, x
AN
x
C0
, x
A0
V
N+1
= 600 kg/h y
AN+1
=0 y
CN+1
=1.0
L
0
= 200 kg/h x
A0
= 0.30
x
B0
=0.70, x
C0
=0, x
AN
=0.04
x
CM
=0.75, x
AM
= 0.075
y
A1
=0.08 and
y
C1
=0.90

L
N
= 136 kg/h
V
1
= 664 kg/h
26.02.2012
16
Stage-to-stage calculations for countercurrent extraction
stage by stage to determine the concentrations at each stage and the total
number of stages N needed to reach L
N
total balance on stage 1
total balance on stage n
Assumption: D (kg/h) is constant for all stages
26.02.2012
17
balance on component A, B, or C.
26.02.2012
18
Number of Stages in Countercurrent Extraction
Pure isopropyl ether (C) of 450 kg/h is being used to extract an aqueous solution of
150 kg/h with 30 wt % acetic acid (A) by countercurrent multistage extraction. The
exit acid concentration in the aqueous phase (B) is 10 wt %. Calculate the number
of stages required.
V
N+1
=450, y
AN+1
=0, y
CN+1
=1.0
L
0
=150, x
A0
=0.30, x
B0
=0.70, x
C0
=0, x
AN
=0.10
(150 0) (450 1.0)
0.75
150 450
x x +
= =
+
(150 0.30) (450 0)
0.075
150 450
× + ×
= =
+
V
N+1
=450, y
AN+1
=0, y
CN+1
=1.0
L
0
=150, x
A0
=0.30, x
B0
=0.70, x
C0
=0, x
AN
=0.10
26.02.2012
19
Countercurrent-Stage Extraction with Immiscible Liquids
If the solvent stream V
N+l
contains components A(solute) and C(solvent) and
the feed stream L
0
contains A (solute) and B(feed solvent) and components
B and C are relatively immiscible in each other, the stage calculations are
made more easily.
The solute A is relatively dilute and is being transferred from L
0
to V
N+1

where L’ = kg inert B/h, V’ = kg inert C/h, y = mass fraction A in V stream, and x = mass fraction A in L
stream
Slope of the operating line is L’ / V’ (if y and x are dilute)
26.02.2012
20
EXAMPLE Extraction of Nicotine with Immiscible Liquids. An inlet water solution of
100 kg/h containing 0.010 wt fraction nicotine (A) in water is stripped with a
kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a
countercurrent stage tower. The water and kerosene are essentially immiscible in
each other. It is desired to reduce the concentration of the exit water to 0.0010 wt
fraction nicotine. Determine the theoretical number of stages needed. The
equilibrium data are as follows (C5), with x the weight fraction of nicotine in the
water solution and y in the kerosene.
26.02.2012
21
L
0
=100 kg/h, x
0
=0.010
V
N+1
= 200 kg/h, y
N+1
= 0.0005, x
N
=0,0010
L’=L(1-x)=L
0
(1-x
0
)=100(1-0.010)=99.0 kg water/h
V’=V(1-y)=V
N+1
(1-y
N+1
)=200(1-0.0005)=199.9 kg kerosen/h
1
1
0.010 0.0005 0.0010
99.0 199.9 99.0 199.9
1 0.010 1 0.0005 1 0.0010 1
y
y
| |
| | | | | |
+ = +
| | | |
÷ ÷ ÷ ÷
\ . \ . \ .
\ .
y
1
=0.00497
N=3.8 theoretical plate
L
0
=100 kg/h, x
0
=0.010
V
N+1
= 200 kg/h, y
N+1
= 0.0005, x
N
=0,0010
y
1
=0.00497
26.02.2012
22
Minimum Solvent and Countercurrent Extraction of Acetone. An aqueous
feed solution of 1000 kg/h containing 23.5 wt % acetone and 76.5 wt % water is
being extracted in a countercurrent multistage extraction system using pure
methyl isobutyl ketone solvent at 298-299 K. The outlet water raffinate will
contain 2.5 wt % acetone. Use equilibrium data from Appendix A.3.
a) Calculate the minimum solvent that can be used. [Hint: In this case the tie
line through the feed L
0
represents the condition for minimum solvent flow rate.
This gives V
1 min
. Then draw lines L
N
V
1 min
and L
0
V
N+1
to give the mixture point
M
min
and the coordinate x
AMmin
, find V
N+ 1 min
the minimum value of the solvent
flow rate V
N+ 1
]
b) Using a solvent flow rate of 1.5 times the minimum, calculate the number of
theoretical stages.
26.02.2012
23
L
0
=1000 kg/h
x
A0
=0.235
x
C0
=0
V
N+1
=? kg/h
y
AN+
1=0
y
CN+1
=1.0
x
AN
=0.025
A:Acetone
C:MIK
Tie line on L0(xA0), V1(y1)min
xA
yA
x
A0
y
1min
L
0,
x
A0
V
N+1
A

C

L
N
,
x
AN
=0.025
V
1 min
y
1 min
A
min
M
min
Tie line

x
Amin
=0.175
0 0 1min 1 1min
min
0 1min 1min
1min
1000(0.235) (0)
0.175
1000
340 /
A N AN N
AM
N N
N
L x V y V
x
L V V
V kg h
+ + +
+ +
+
+ +
= =
+ +
=
V
N+1
=1.5(V
N+1 min
)=1.5(340)=510kg/k
L
0
=1000 kg/h
x
A0
=0.235
x
C0
=0
y
AN+1
=0
y
CN+1
=1.0
0 0 1 1
0 1
1000(0.235) 510(0)
1000 510
0.156
A N AN
AM
N
AM
L x V y
x
L V
x
+ +
+
+ +
= =
+ +
=
0 0 1 1
0 1
1000(0) 510(1.0)
1000 510
0.337
C N CN
CM
N
CM
L x V y
x
L V
x
+ +
+
+ +
= =
+ +
=
26.02.2012
24
M(x
AM
and x
CM
)
M(0.156, 0.337)
x
AN
=0.025 given
L
N
-M line gives V
1
, y
1
y
A1
=0.290, y
C1
=0.645
x
CN
=0.020 from graph
Lo+V
N+1
=L
N
+V
1
=M
1000+510=L
N
+V
1
V
1
=1510-L
N
| |
1 1
1
(0.025) (1510 )(0.290)
0.156
1510
N AN A
AM
N
N N
L x V y
x
L V
L L
+
=
+
+ ÷
=
L
N
=763, V
1
=747
To calculate A point
0 0 1 1
0 1
0 0 1 1
0 1
1000(0) 747(0.645)
1000 747
1.91
1000(0.235) 747(0.290)
1000 747
0.073
C C
C
C
A A
A
C
L x V y
x
L V
x
L x V y
x
L V
x
A
A
A
A
÷ ÷
= =
÷ ÷
= ÷
÷ ÷
= =
÷ ÷
=
26.02.2012
25
L
0,
x
A0
A

L
N
, x
AN

V
1 ,
y
1
M

V
N+1
C

xA
yA
x1

y
1
Draw tie line y
1
(V
1
), x
1
(L
1
)
Draw L
1
A,
Draw tie line y
2
(V
2
), x
2
(L
2
)
Draw L
2
A
Draw tie line y
3
(V
3
), x
3
(L
3
)


Repeat steps to L
N

N= 5 stage
V
2 ,
y
2
y
2
x2

L1
L2
V
3,
y
3
Countercurrent Extraction of Acetic Acid and Minimum Solvent. An
aqueous feed solution of 1000 kg/h of acetic acid-water solution contains 30.0 wt
% acetic acid and is to be extracted in a countercurrent multistage process with
pure isopropyl ether to reduce the acid concentration to 2.0 wt % acid in the final
raffinate. (a) Calculate the minimum solvent flow rate that can be used.
(b) If 2500 kg/h of ether solvent is used, determine the number of theoretical
stages required. (Note: It may be necessary to replot on an expanded scale the
concentrations at the dilute end.)
26.02.2012
26
Extraction with Immiscible Solvents. A water solution of 1000 kg/h containing
1.5 wt % nicotine in water is stripped with a kerosene stream of 2000 kg/h
containing 0.05 wt % nicotine in a countercurrent stage tower. The exit water is to
contain only 10% of the original nicotine, i.e., 90% is removed. Calculate the
number of theoretical stages needed.
26.02.2012
27

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