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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
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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
• • • •
•
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.
Liquid-liquid phase diagram where components A and B are partially miscible
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Phase diagram where the solvent pairs b—c and a-c are partially miscible.
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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 raffinate composition is designated by x and the extract by y
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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.
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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 auxiliary line that will allow graphical construction of tie lines.
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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
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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.
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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 the exit streams L1 and V1 leave in equilibrium with each other.
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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)
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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
V1
A: acetone
L1
V2
yA2
L1
L2 xA2 V1
B: water V1+L1=V2+L2=M V1yA1+L1xA1=V2yA2+L2xA2=MxAM V1yB1+L1xB1=V2yB2+L2xB2=MxBM V1yC1+L1xC1=V2yC2+L2xC2=MxCM
V1+L1=V2+L2=M V1yA1+L1xA1=V2yA2+L2xA2=MxAM
C: chloroform
V= C, Solvent, chloroform, ya1=0, yC1=1.0 L= Liquid mixture, A: acetone, B=water xA1=0.60, xB1=0.40
L1
yA2
xA2 V1
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
yA2
xA2
V1
Continuous Multistage Countercurrent Extraction
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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.
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L0xA0+VN+1yAN+1=LNxAN+V1yA1=MxAM 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
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
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, yCN+1, yAN+1 yC1, yA1
xC0, xA0
xCN, xAN
L0xA0+VN+1yAN+1=LNxAN+V1yA1=MxAM 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 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
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
Assumption: D (kg/h) is constant for all stages
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balance on component A, B, or C.
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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 x0) (450 x1.0) 0.75 150 450
(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
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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 L 0 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 Slope of the operating line is L’ / V’ (if y and x are dilute)
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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.
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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
y1 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 1 y1=0.00497
L0=100 kg/h, x0=0.010 VN+1= 200 kg/h, yN+1= 0.0005, xN=0,0010
y1=0.00497
N=3.8 theoretical plate
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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.
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VN+1=? kg/h yAN+1=0 yCN+1=1.0 L0=1000 kg/h xA0=0.235 xC0=0
xAN=0.025 Tie line on L0(xA0), V1(y1)min C
A:Acetone C:MIK
VN+1
V1 min y1 min
Tie line yA
Mmin
xAmin=0.175
LN, xAN=0.025 L0, xA0
A y1min
Dmin
x AM min
L0 x A0 VN 1min y AN 1 L0 VN 1min
xA0
0.175
xA
1000(0.235) VN 1min (0) 1000 VN 1min
VN 1min 340kg / h
VN+1=1.5(VN+1 min)=1.5(340)=510kg/k
xAM
L0=1000 kg/h xA0=0.235 xC0=0
yAN+1=0 yCN+1=1.0
L0 x A0 VN 1 y AN 1 1000(0.235) 510(0) L0 VN 1 1000 510
xAM 0.156 xCM
L0 xC 0 VN 1 yCN 1 1000(0) 510(1.0) L0 VN 1 1000 510
xCM 0.337
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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
x AM
Lo+VN+1=LN+V1=M 1000+510=LN+V1 V1=1510-LN
LN x AN V1 y A1 LN V 1
0.156
LN (0.025) (1510 LN )(0.290) 1510
LN=763, V1=747
To calculate D point
xDC
L0 xC 0 V1 yC1 1000(0) 747(0.645) L0 V1 1000 747
xDC 1.91 xDA
L0 x A0 V1 y A1 1000(0.235) 747(0.290) L0 V1 1000 747
xDC 0.073
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VN+1
yA C
V3, y3 V2 , y2
y2
V1 , y1
y1 x1
x2
xA
M L1
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)
LN, xAN L0, xA0
A L2
Repeat steps to LN N= 5 stage
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.
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