CENG 3210
Separation Process
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Lecture 8: Extraction Learning Outcomes By the end of this lecture, you should be able to:
Understand the use of McCabe-Thiele Method for dilute and immiscable solvent Understand the Method for cross flow multistage extraction Understand the multistage extraction for partially miscable LLE.
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Liquid-Liquid Extraction
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Immiscible solvents
• McCabe-Thiel analysis is similar to the analysis for absorption and stripping • For linear system, Kremser’s equation is also applicable • Concentrations are expressed as weight fraction or weight ratio • Flow rates are expressed as mass flow rates CENG 3210_4
McCabe-Thiele Method Mass Balance Envelope for Simple Countercurrent Cascade
𝐹𝐷 = diluent flow rate, kg D/hr 𝐹𝑆 = solvent flow rate, kg S/hr
𝑋 = weight ratio solute in diluent 𝑌 =weight ratio solute in solvent
𝑅= 𝐸=
𝑥 = weight fraction solute in raffinate 𝑦 = weight fraction solute in extract
raffinate flow rate, kg/hr extract flow rate, kg/hr
}
both y and x are liquid Assumptions: 1. Isothermal (liquid-liquid absorption) 2. Isobaric 𝑦 𝑌= 3. Negligible heat of mixing 1−𝑦 4. Diluent and solvent are totally immiscible CENG 3210_5
solute free basis
𝑥 𝑋= 1−𝑥
McCabe-Thiele Method Method 1: use mass fraction (valid for dilute systems).
𝐹 = 𝑅0 = 𝑅1 … = 𝑅𝑁 = 𝑅 𝑆 = 𝐸𝑁+1 = 𝐸𝑁 … = 𝐸1 = 𝐸 Mass balance around the envelope: 𝐸𝑦𝑗+1 + 𝑅𝑥0 = 𝐸𝑦1 + 𝑅𝑥𝑗
Solving for yj+1 we obtain the operating line 𝑅 𝑅 𝑦𝑗+1 = 𝑥𝑗 + [𝑦1 − 𝑥0 ] 𝐸 𝐸
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McCabe-Thiele Method Method 2: use weight ratio Operating line
𝑌𝑗+1
𝐹𝐷 𝐹𝐷 = 𝑋𝑗 + [𝑌1 − 𝑋0 ] 𝐹𝑆 𝐹𝑆 Note: Method 1 & 2 will essentially give the same results for very dilute systems. Use method 2 if the system is concentrated (say, x & y > 0.05), and carrier & solvent are still immiscible. Kremser equation for stripping (i.e. in terms of liq. phase composition) can be used directly for extraction. Mole fraction becomes mass fraction and L/V replaced with R/E Mole fraction becomes mass ratio and L/V replaced with FD/FS
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Kremser Method for Dilute Systems (optional) Assumptions: 1) dilute system and 2) linear equilibrium relationship
𝑦𝑖 = 𝑚𝑥𝑖 + 𝑦𝑖
We can use the Kremser equations directly, but we have used different different symbols for flow rates, replace L/V with R/E.
𝑦𝑁+1 − 𝑦1 𝑁= 𝑅 𝑦1 − 𝑥0 − 𝑏 𝐸
𝑅 𝑭𝑜𝑟 =1 𝑚𝐸
𝑅 𝐹𝑜𝑟 ≠1 𝑚𝐸 𝑦1 − 𝑦 ∗1 = ∗ 𝑦𝑁+1 − 𝑦 1
𝑁= CENG 3210_8
𝑅 1 − 𝑚𝐸 𝑅 𝑁+1 1− 𝑚𝐸
𝑚𝐸 ln [ 1 − 𝑅
𝑤ℎ𝑒𝑟𝑒 𝑦 ∗1 = 𝑚𝑥0 + 𝑏
𝑦𝑁+1 − 𝑦 ∗1 𝑚𝐸 + ] 𝑦1 − 𝑦 ∗1 𝑅 𝑅 ln ( ) 𝑚𝐸
Kremser Type Plot (optional) YBE
E1
1.0 0.8 0.6
F1
E = 0.3
0.4 0.3
XBF
S1
YBS XBR
R1
E = Extraction Factor E = m (S1/F1)
XBR/XBF = Fraction Unextracted
0.2
0.1 0.08 0.06 0.04 0.03 0.02
0.01 0.008 0.006 0.004
0.003 0.002
0.001 0.0008 0.0006 0.0005
1
2
3
4 5
Number of Ideal Stages
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6
7 8 10 15 20
Example 1 A feed of 100.0 kg/minute of a 1.2 wt % mixture of acetic acid in water is to be extracted with 1-butanol at 1 atm pressure and 26.7℃. We desire an outlet concentration of 0.1 wt % acetic acid in the exiting water. We have available solvent stream 1 that is 44.0 kg/minute of pure 1-butanol and solvent stream 2 that is 30.0 kg/minute of 1-butanol that contains 0.4 wt of acetic acid. Devise a scheme
to do this separation, find the outlet flow rate and concentration of the exiting 1butanol phase, and find the number of equilibrium contacts needed. Equilibrium data: 𝑦 = 1.613𝑥.
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Cross Flow Extraction
E1
E2 B+C
A+B F
R1
B+C
R2 C
F + S = M1
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E3
E4 B+C
R3 C
R1 + S = M 2
B+C
R4 C
R 2 + S = M3
A C
R3 + S = M 4
Single-stage and cross-flow extraction One type occasionally used is the cross-flow cascade,
For dilute systems the resulting steady-state mass balance for stage j is For single-solute systems the same five assumptions made for counter-current cascades are required for the McCabe Thiele analysis.
𝑅𝑥𝑗−1 + 𝐸𝑗 𝑦𝑗,𝑖𝑛 = 𝑅𝑥𝑗 + 𝐸𝑗 𝑦𝑗 Solving for yj to obtain the operating equation,
𝑅 𝑅 𝑦𝑗 = − 𝑥𝑗 + ( 𝑥𝑗−1 + 𝑦𝑗,𝑖𝑛 ) 𝐸𝑗 𝐸𝑗
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Multiple stages with cross flow of solvent If the process liquid stream from the first stages fed into a second
extractor and mixed with more fresh solvent, as shown in the Figure, we have what is called cross-flow extraction. The process can be described the same as for a single stage. it is simply repeated again for each stage, using the raffinate phase from the upstream stage as the feed to each stage.
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Single-stage and cross-flow extraction Each stage will have a different operating equation. On a McCabe-Thiele diagram plotted as y vs. x, this is a straight line of slope –R/Ej and y intercept (yj, in , xj–1) Designer can specify values of Ej and yj, in as well as x0, R, and either xN or N.
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Example 2 We wish to extract a dilute solution of the protein alcohol dehydrogenase from a aqueous solution of 5 wt % poly (ethylene glycol) (PEG) with an aqueous solution that is 10 wt % dextran. Aqueous two-phase extraction system is a very gentle method of recovering proteins that is unlikely to denature the protein since both phases are aqueous (Albertsson et al., 1990; Harrison et al., 2003). The two phases can be considered to be essentially immiscible. The dextran phases is denser and will be the cross-flow solvent. The entering dextran phases contain no protein. The entering PEG phase flow rate is 20 kg/hr.
a. If 10 kg/hr of dextran phase is added to a single-stage extractor, find the total recovery fraction of alcohol dehydrogenase in the dextran solvent phase.
b. If 10 kg/hr of dextran phase is added to each stage of a cross-flow cascade with two stages, find the total recovery fraction of alcohol dehydrogenase in the dextran solvent phase. The protein distribution coefficient is (Harrison et al., 2003)
𝐾 = (wt frac protein in PEG, x)/(wt frac protein in dextran, y) = 0.12
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Extraction of Concentrated Solution Ternary System
Most practical situations involving liquid-liquid equilibrium involve three or more components.
Our attention is with three component systems. In this process, a solute is removed from a feed stream by contacting it with a solvent.
The solute is quite soluble in the solvent, while the other component in the feed is less soluble.
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Extraction of Concentrated Solution Extraction of Concentrated Solution
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Example 3： Single-stage extraction In an extraction process we have two entering streams (L kg and V kg). The solvent, as stream V2, enters from one side and the stream L0 enters from the other side. The two entering streams are mixed and equilibrated and then exit as streams L1 and V1,
which are in equilibrium with each other. Find the final product compositions in the two phases
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Example 4 Example 4: 30,000 kg/hr of a ternary mixture of 19 weight percent
isopropyl alcohol (IPA, component 1), 41 weight percent toluene, and 40 weight percent water (component 3) are fed into a decanter operating at 25°C. The figure gives the LLE data for the system. Determine the compositions and flow rates of the
two liquid streams leaving the decanter.
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Multistge countercurrent extraction Multistage
countercurrent
extraction
is
the
most
commonly
encountered liquid-liquid extraction process. The raffinate and solvent streams travel countercurrent to each other through N stages. The flow rate of the raffinate leaving the last stage (tray N) is
R.
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Multi-stage extraction E1
B+C A+B F
R1 B+C
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B+C
R2
C
E3
R3 E2
R4 B+C
A
E4
Multi-stage extraction The overall balance on all N stages is
𝐿0 + 𝑉𝑁+1 = 𝐿𝑁 + 𝑉1 = 𝑀 where M is the total mass (kg/h) and is a constant, L0 the inlet feed flow rate (kg/h),
VN+1 the inlet solvent flow rate (kg/h), V1 the exit extract stream, and LN the exit raffinate stream.
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Multi-stage extraction Material balance on C gives
𝐿0 𝑥𝐶,0 + 𝑉𝑁+1 𝑦𝐶,𝑁+1 = 𝐿𝑁 𝑥𝐶,𝑁 + 𝑉1 𝑦𝐶,1 = 𝑀𝑥𝐶,𝑀 xCM is obtained 𝑥𝐶,𝑀
𝐿0 𝑥𝐶,0 + 𝑉𝑁+1 𝑦𝐶,𝑁+1 𝐿𝑁 𝑥𝐶,𝑁 + 𝑉1 𝑦𝐶,1 = = 𝐿0 + 𝑉𝑁+1 𝐿𝑁 + 𝑉1
𝑥𝐴,𝑀
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𝐿0 𝑥𝐴,0 + 𝑉𝑁+1 𝑦𝐴,𝑁+1 𝐿𝑁 𝑥𝐴,𝑁 + 𝑉1 𝑦𝐴,1 = = 𝐿0 + 𝑉𝑁+1 𝐿𝑁 + 𝑉1
Multi-stage extraction So the point M, which ties together the two entering streams (usually known) and the two exit streams, can be located. The desired exit composition xAN is often set, which is on the equilibrium curve (phase boundary). Then the line LNM is extended to intersect the phase boundary of the extract phase to give V1 composition.
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Example 5 Example 5: Pure solvent isopropyl ether (C) 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) and 70 wt % water (B) 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 . The equilibrium data at 20oC, 1 atm, are given and plotted below. isopropyl ether phase (mass fraction)
Water phase (mass fraction) acetic acid (xA)
Water (xB)
isopropyl ether (xC)
acetic acid (yA)
Water (yB)
isopropyl ether (yC)
6.9e-3
0.9810
0.0120
1.8e-3
5.0e-3
0.9930
0.0141
0.9710
0.0150
3.7e-3
7.0e-3
0.9890
0.0289
0.9550
0.0160
7.9e-3
8.0e-3
0.9840
0.0642
0.9170
0.0190
0.0193
0.0100
0.9710
0.1330
0.8440
0.0230
0.0482
0.0190
0.9330
0.2550
0.7110
0.0340
0.1140
0.0390
0.8470
0.3670
0.5890
0.0440
0.2160
0.0690
0.7150
0.4430
0.4510
0.1060
0.3110
0.1080
0.5810
0.4640
0.3710
0.1650
0.3620
0.1510
0.4870
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Single stage Extraction 1. Calculate M. 2. Calculate xAM and xCM. 3. Locate the point M using xAM and xCM . 4. Find the LLE tie-line that passes through the point M. 5. The points at the two ends of the LLE tie-line give the compositions of the two phases leaving the system. 6. 6. Calculate the flow rates。
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Multi-stage extraction The next step is to go stage by stage to determine the concentrations at each stage and the total number of stages N needed to reach LN in the process.
Making a total balance on stage 1 𝐿0 + 𝑉2 = 𝐿1 + 𝑉1 on stage n
𝐿𝑛−1 + 𝑉𝑛+1 = 𝐿𝑛 + 𝑉𝑛
The above equations can be rearranged as 𝐿0 − 𝑉1 = 𝐿1 − 𝑉2 = ⋯ = 𝐿𝑛 − 𝑉𝑛+1 = 𝐿𝑁 − 𝑉𝑁+1 = Δ CENG 3210_47
Multi-stage extraction The mass balances for the first stage, the definition of Δ can be combined to give
𝐿0 − 𝑉1 = 𝐿1 − 𝑉2 = ⋯ = 𝐿𝑛 − 𝑉𝑛+1 = 𝐿𝑁 − 𝑉𝑁+1 = Δ
The two streams L1 and V2 that pass each other between the first and second stages are related to each other by the Δ point. Hence, if
we know L1 we can determine V2 by using the straight line that connects L1 and Δ.
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Multi-stage extraction The coordinates of the operating point
can be obtained by
material balances on A, B or C: 𝐿0 𝑥𝐴0 − 𝑉1 𝑦𝐴1 = ⋯ = 𝐿𝑁 𝑥𝐴𝑁 − 𝑉𝑁+1 𝑦𝐴,𝑁+1 = Δ ∗ xAΔ 𝐿0 𝑥𝐶0 − 𝑉1 𝑦𝐶1 = ⋯ = 𝐿𝑁 𝑥𝐶𝑁 − 𝑉𝑁+1 𝑦𝐶,𝑁+1 = Δ ∗ xCΔ
𝑥𝐴,Δ
(𝐿0 𝑥𝐴0 − 𝑉1 𝑦𝐴1 ) 𝐿𝑁 𝑥𝐴,𝑁 − 𝑉𝑁+1 𝑦𝐴,𝑁+1 = = 𝐿0 − 𝑉1 𝐿𝑁 − 𝑉𝑁+1
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Multi-stage extraction The pseudo composition 𝑥𝐴,Δ is entirely fictitious and, therefore, can be less than zero or greater than unity. We see that
𝐿0 − 𝑉1 = 𝐿1 − 𝑉2 = ⋯ = 𝐿𝑛 − 𝑉𝑛+1 = 𝐿𝑁 − 𝑉𝑁+1 = Δ Therefore, the Δ point must lie on two straight lines, one through the points LN and VN+1 and the other through the points L0 and V1.
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Multi-stage extraction Δ can lie either to the left or to the right of the phase diagram.
V1 LN V0
V1
LN V0
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Multi-stage extraction This point
is located
either by its coordinates as calculated or graphically as the intersection of lines
L0V1 and LNVN+1. All the operating lines (L0V1, L1V2, LnVn+1, ... , LNVN+1)
must pass through the common point .
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Example 6 Example 6: 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) and 70 wt % water (B) by countercurrent multistage
extraction. The exit acid concentration in the aqueous phase is 10 wt %. Calculate the number of stages required.
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Multi-stage extraction To graphically determine the number of stages, follow the procedures below. (1) locate L0, VN+1 and LN by their compositions. (2) draw a line L0VN+1, and locate the mixture point M (3) draw a line from LN through M & extend it until it intersects the phase boundary, where is V1. (3) extend lines L0V1, and LNVN+1, which will intersect at the common operating point . (4) start at L0 and draw a line L0 which intersects the phase boundary at V1. (5) draw an equilibrium tie line through V1 to locate L1. (6) draw a line L1 to give V2 at the phase boundary. (7) a tie line from V2 gives L2. This is continued until the desired LN is reached. CENG 3210_59
Take home message Understand the calculation of single stage extraction. Understand the graphical method of multistage cross flow extraction. Understand the graphical method of multistage countercurrent extraction.
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