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
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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
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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 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.
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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
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,
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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.
M = 100 kg and xAM= 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
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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)
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 V1+L1=V2+L2=M V1yA1+L1xA1=V2yA2+L2xA2=MxAM V1yB1+L1xB1=V2yB2+L2xB2=MxBM V1yC1+L1xC1=V2yC2+L2xC2=MxCM L1 V1 yA2 xA2 V1+L1=V2+L2=M V1yA1+L1xA1=V2yA2+L2xA2=MxAM
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
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L1 V1 yA2 xA2Material 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.
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Pure solvent isopropyl ether (C) VN+1= 600 kg/h yAN+1=0 yCN+1=1.0
B: aqueous solution of L0 = 200 kg/h containing 30 wt % acetic acid (A) exit
acetic acid concentration in the aqueous phase is 4%.
xA0= 0.30, xB0=0.70, xC0=0, xAN=0.04
L0xA0+VN+1yAN+1=LNxAN+V1yA1=MxAM
VN+1= 600 kg/h yAN+1=0 yCN+1=1.0
VN+1= 600 kg/h yAN+1=0 yCN+1=1.0
L0 = 200 kg/h xA0= 0.30 xB0=0.70, xC0=0, xAN=0.04 xCM=0.75, xAM= 0.075
L0xA0+VN+1yAN+1=LNxAN+V1yA1=MxAM
VN+1 and L0 are plotted
LN is on phase boundary
It can be plotted at xAN=0.04 For the mixture point M xCM=0.075, xAM=0.75
Draw a line from the point of LN and M at equilibrium of extract phase gives V1, yA1=0.08 and yC1=0.90 , yCN+1, yAN+1 yC1, yA1 xCN, xAN xC0, xA0 VN+1= 600 kg/h yAN+1=0 yCN+1=1.0 L0 = 200 kg/h xA0= 0.30 xB0=0.70, xC0=0, xAN=0.04 xCM=0.75, xAM= 0.075 yA1=0.08 and yC1=0.90 LN= 136 kg/h V1= 664 kg/h
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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 LN
total balance on stage 1
total balance on stage n
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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. VN+1=450, yAN+1=0, yCN+1=1.0 L0=150, xA0=0.30, xB0=0.70, xC0=0, xAN=0.10 (150 0) (450 1.0) 0.75 150 450 x x (150 0.30) (450 0) 0.075 150 450 VN+1=450, yAN+1=0, yCN+1=1.0 L0=150, xA0=0.30, xB0=0.70, xC0=0, xAN=0.10
Countercurrent-Stage Extraction with Immiscible Liquids
If the solvent stream VN+l contains components A(solute) and C(solvent) and
the feed stream L0 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 L0 to VN+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
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EXAMPLE Extraction of Nicotine with Immiscible Liquids. An inlet water solution of100 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.
L0=100 kg/h, x0=0.010 VN+1= 200 kg/h, yN+1= 0.0005, xN=0,0010 L’=L(1-x)=L0(1-x0)=100(1-0.010)=99.0 kg water/h
V’=V(1-y)=VN+1(1-yN+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
y1=0.00497 N=3.8 theoretical plate L0=100 kg/h, x0=0.010 VN+1= 200 kg/h, yN+1= 0.0005, xN=0,0010 y1=0.0049726.02.2012
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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 L0 represents the condition for minimum solvent flow rate.
This gives V1 min . Then draw lines LN V1 min and L0 VN+1 to give the mixture point Mmin and the coordinate xAMmin, find VN+ 1 min the minimum value of the solvent flow rate VN+ 1]
b) Using a solvent flow rate of 1.5 times the minimum, calculate the number of theoretical stages.
L0=1000 kg/h xA0=0.235 xC0=0 VN+1=? kg/h yAN+1=0 yCN+1=1.0 xAN=0.025 A:Acetone C:MIK
Tie line on L0(xA0), V1(y1)min
xA yA xA0 y1min L0, xA0 VN+1 A C LN, xAN=0.025 V1 min y1 min Dmin Mmin Tie line xAmin=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 NL x
V
y
V
x
L
V
V
V
kg h
VN+1=1.5(VN+1 min)=1.5(340)=510kg/k L0=1000 kg/h xA0=0.235 xC0=0 yAN+1=0 yCN+1=1.0 0 0 1 1 0 11000(0.235) 510(0)
1000 510
0.156
A N AN AM N AML x
V
y
x
L
V
x
0 0 1 1 0 11000(0) 510(1.0)
1000 510
0.337
C N CN CM N CML x
V
y
x
L
V
x
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M(xAM and xCM) M(0.156, 0.337) xAN=0.025 given LN-M line gives V1, y1 yA1=0.290, yC1=0.645 xCN=0.020 from graph Lo+VN+1=LN+V1=M 1000+510=LN+V1 V1=1510-LN
1 1 1(0.025)
(1510
)(0.290)
0.156
1510
N AN A AM N N NL x
V y
x
L
V
L
L
LN=763, V1=747 To calculate D point 0 0 1 1 0 1 0 0 1 1 0 11000(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 CL x
V y
x
L
V
x
L x
V y
x
L
V
x
D D D D
L0, xA0 A LN, xAN V1 , y1 M VN+1 C xA yA x1 y1
Draw tie line y1 (V1), x1(L1) Draw L1D,
Draw tie line y2 (V2), x2(L2)
Draw L2D
Draw tie line y3 (V3), x3(L3)
Repeat steps to LN N= 5 stage V2 , y2 y2 x2 L1 L2 V3, y3
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.)
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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.