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Liquid-Liquid-Extraction.pdf

This document provides an overview of liquid-liquid extraction (LLE) processes. It defines key terms used in LLE such as solute, feed, solvent, extract, and raffinate. It describes common applications of LLE in industries such as bioseparations, mining, and oil sands processing. The document discusses different types of equipment used for LLE including mixer-settlers, column-type units, and rotating devices. It also covers concepts such as distribution coefficients, phase diagrams, and the lever rule for determining phase compositions. Cross-current and counter-current flow arrangements are compared. The overall aim of LLE processes is to efficiently separate and concentrate a solute into the extract stream.

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Separation Processes: Liquid-liquid extraction ChE 4M3 © Kevin Dunn, 2013 kevin.dunn@mcmaster.ca http://learnche.mcmaster.ca/4M3 Overall revision number: 256 (November 2013) 1
Copyright, sharing, and attribution notice This work is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License. To view a copy of this license, please visit http://creativecommons.org/licenses/by-sa/3.0/ This license allows you: I to share - to copy, distribute and transmit the work I to adapt - but you must distribute the new result under the same or similar license to this one I commercialize - you are allowed to use this work for commercial purposes I attribution - but you must attribute the work as follows: I “Portions of this work are the copyright of Kevin Dunn”, or I “This work is the copyright of Kevin Dunn” (when used without modification) 2
We appreciate: I if you let us know about any errors in the slides I any suggestions to improve the notes All of the above can be done by writing to kevin.dunn@mcmaster.ca or anonymous messages can be sent to Kevin Dunn at http://learnche.mcmaster.ca/feedback-questions If reporting errors/updates, please quote the current revision number: 256 3
Liquid-liquid extraction (LLE) [Flickr# 3453475667] 4
References I Schweitzer, “Handbook of Separation Techniques for Chemical Engineers”, chapter 1.9 I Seader, Henley and Roper, “Separation Process Principles”, 3rd edition, chapter 8 I Richardson and Harker, “Chemical Engineering, Volume 2”, 5th edition, chapter 13 I Geankoplis, “Transport Processes and Separation Process Principles”, 4th edition, chapter 12.5 and 12.6 I Ghosh, “Principles of Bioseparation Engineering”, chapter 7 I Uhlmann’s Encyclopedia, “Liquid-Liquid Extraction”, DOI:10.1002/14356007.b03 06.pub2 5
Definitions I solute: species we aim to recover (A) from the feed I feed or “feed solvent”: one of the liquids in the system (“carrier”) I solvent: MSA (by convention: the “added” liquid) I extract: solvent (not solute) mostly present in this layer. yE,A = concentration of A, the solute, in extract. We aim for the solute (A) to be mostly in the extract stream. I raffinate: residual solute in this layer = xR,A I distribution: how the solute partitions itself = DA = yE,A xR,A I measure of affinity of solute I DA = µ0 R − µ0 E RT = chemical potential difference (R)(temperature) 6
Where/why LLE is used Where? I Bioseparations I Nuclear (uranium recovery) I Mining: nickel/cobalt; copper/iron I Perfumes, fragrances and essential oils I Fine and specialty chemicals I Oil sands: extract bitumen away from water Why? I Temperature sensitive products I High purity requirements I High-boiling point species in low quantity I Need to separate by species type (rather than relative volatility) I Close-boiling points, but high solubility difference I Azeotrope-forming mixtures 7
Extractor types: 3 major steps required 1. Mixing/contacting: I turbulent contact between liquid phases I small droplet dispersion in a continuous phase I which phase is dispersed? I mass-transfer between phases I limited by solute loading in solvent 2. Phase separation: I reverse of above mixing step I drops come together and coalesce I relies on density difference 3. Collection of phases leaving the unit I split the raffinate from the extract 8
What are we aiming for? Main aims I High recovery of solute overall (low xR and high yE ) I Concentrated solute in extract (high yE ) How to achieve this? I Cocurrent mixer-settlers I Cross-current mixer settlers I Counter-current mixer-settlers in series I High interfacial area during mixing I Reduce mass-transfer resistance I Promote mass transfer: aim is to move solute in feed to the solvent in extract stream I molecular diffusion ← microscopic movement I eddy diffusion ← orders of magnitude greater 9
Co-, cross-, counter-current operations [Hanson article, Chemical Engineering, posted to course website] 10
Some photos of equipment [http://www.flickr.com/photos/51009184@N06/5062827346/] 11
Some photos of equipment [http://www.rousselet-robatel.com/images/products/Mixer-settler-pic-2lg.jpg] 12
Equipment for LLE 1. Mixer-settlers (integrated unit) I mix: impellers I mix: nozzles I mix: feeds meet directly in the pump I mix: geared-teeth devices I main aim: good contact; avoid droplets smaller than 2 µm I settle: baffles, membranes I settle: ultrasound I settle: chemical treatment I settle: centrifuges 2. Column-type units contain: I (a) nothing or I (b) trays and/or I (c) packing and/or I (d) pulsating and/or I (e) agitation 3. Rotating devices Important point: LLE is an equilibrium-limited separation (as opposed to rate-limited separations seen up to now). 13
Mixer-settlers [Richardson and Harker, p 745] Common in mining industry: requirements ∼40000 L/min flows 14
Mixer-settlers KnitMesh coalescer: consistency of “steel wool” [Richardson and Harker, p 747] 15
Horizontal gravity settling vessel [Seader, 3ed, p302] 16
Spray columns: separation principle is gravity [Richardson and Harker, p 751] 17
Tray columns [Richardson and Harker, p 749] I coalescence on each tray I tray holes: ∼ 3mm I breaks gradient formation (axial dispersion) 18
Tray columns with mechanical agitation I shearing to create dispersion I can have alternating layers of packing (coalescence) I some column designs pulsate ⇑⇓ [Seader, 3ed, p302] 19
Rotating devices [Seader, 3ed, p 306] I “white” = lighter liquid I “grey” = heavier liquid Used when foams and emulsions would easily form: i.e. gentle mass transfer. Look up in your own time: centrifuge mixer-settlers 20
Linking up units (more on this later) [Richardson and Harker, p 723] 21
Integration with downstream units [Schweitzer, p 1-257] 22
Selecting a solvent Schweitzer: “The choice of solvent for a LLE process can often have a more significant impact on the process economics than any other design decision that has to be made”. Which properties of a solvent influence our aims with LLE? I High distribution coefficient (selectivity) for solute I Low distribution coefficient for carrier I Reasonable volatility difference with solute and carrier I Reasonable surface tension: easy to disperse and coalesce I High density difference: separates rapidly by gravity I Stability to maximize its reuse I Inert to materials of construction I Low viscosity: maximizes mass transfer I Safe: non-toxic, non-flammable I Cheap, and easily available I Compatible with carrier and solute: avoid contamination I Doesn’t foam, form emulsions, scum layers at interface 23
Calculating the distribution coefficient (in the lab only) Mass balance: FxF + SyS = EyE + RxR D = yE xR If F = S = E = R and ys = 0, then only measure xR: D = xF xR − 1 ← only in this special case I Capital letters refer to mass amounts I y2 ← refers to mass fractions in solvent and extract layers I x2 ← refers to mass fractions in carrier and raffinate layers Once D is determined, we can obtain phase diagrams to understand how the process will operate. Also: see Perry’s for many values of D 24
Triangular phase diagrams: from laboratory studies [Flickr# 3453475667] 25
Using a triangular phase diagrams [http://www.youtube.com/watch?v=gGYHXhcKM5s] 26
Lever rule Mix P and Q I mixture = K I PK KQ = mass Q mass P I The converse applies also: when a settled mixture separates into P and Q I Applies anywhere: even in the miscible region 27
Q1: Using the lever rule Which is a more flexible system: (a) or (b)? I S = pure solvent used I F = feed concentration point (more correctly it is xF ) Answer: range of feed concentrations (xF ) is wider, i.e. more desirable, for (a). Difference between (a) and (b): I due to solvent choice I due to different temperatures I due to pH modification, etc 28
Q2: Using the lever rule Mix a feed stream, F, containing C and A (i.e. xF ) with a pure solvent stream S (i.e. yS = 0). Composition of the mixture? 29
Q2 solution: Using the lever rule Composition of the mixture? we need more information (e.g. amount of F and S must be given to find point M) 30
Q3: Going to equilibrium Let that mixture M achieve equilibrium. What is the composition of the raffinate and extract? 31
Q3 solution: Going to equilibrium What is the composition of the raffinate and extract? Use the tie lines [solid lines]; or interpolate between existing ones. 32
Q4: Altering flows Recovery = fraction of solute recovered For 1 stage 1 − (xR)(R) (xF )(F) Specifically 1− (xR,A)(R) (xF,A)(F) Same system, but now lower solvent flow rate (to try save money!). What happens to (a) extract concentration and (b) solute recovery? 33
Q4 solution: Altering flows Recovery = fraction of solute recovered For 1 stage 1 − (xR)(R) (xF )(F) Specifically 1− (xR,A)(R) (xF,A)(F) (a) extract concentration increases: (A at E*) > (A at E): yE∗ > yE (b) solute recovery drops: (A at R*) > (A at R): xR∗ > xR 34
Q5: Composition of the mixture, M? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 Answer: M = xM,A = xM,C = xM,S = 35
Q5 solution: Composition of the mixture, M? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 M = F + S Answer: M = 350kg; xM,A = 0.17; xM,C = 0.54; xM,S = 0.29 36
Q6: Composition of the 2 phases leaving in equilibrium? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 R1 = xR1,A = xR1,C = xR1,S = E1 = yE1,A = yE1,C = yE1,S = 37
Q6 solution: Composition of the 2 phases in equilibrium? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 M = E1 + R1 R1 = 222kg; xR1,A = 0.10; xR1,C = 0.82; xR1,S = 0.08 E1 = 128kg; yE1,A = 0.33; yE1,C = 0.06; yE1,S = 0.61 38
Link units in series [Richardson and Harker, p 723] Introduce fresh solvent each time 39
Q7: send raffinate from Q6 to second mixer-settler Question: how much solvent should we use in the second stage? 40
Q7 solution: send raffinate from Q6 to second mixer-settler Compare extract: Compare volumes: Answer: equilibrium from point B (most solvent), C, D (least solvent) will each be different. Trade-off: higher extraction vs lower recovery 41
Phase diagram: furfural, water, ethylene glycol Feed Solvent F = 100 kg S = 200 kg xF,A = 0.45 xS,A = 0.0 xF,C = 0.55 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 I A = ethylene glycol (solute) I C = water (carrier) I S = furfural (solvent) AIM: to remove ethylene glycol (solute) from water (carrier) into solvent (furfural) 1. Calculate the mixture composition, M 2. Calculate the equilibrium compositions in E1 and R1 Note: extract is defined as “the solvent-rich stream leaving the system” 42
43
Solution: Phase diagram: furfural, water, ethylene glycol Feed Solvent F = 100 kg S = 200 kg xF,A = 0.45 xS,A = 0.0 xF,C = 0.55 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 I A = ethylene glycol solute I C = water (carrier) I S = furfural solvent M = 300kg; xM,A = 0.15; xM,C = 0.18; xM,S = 0.67 R1 = 82kg; xR1,A = 0.33; xR1,C = 0.57; xR1,S = 0.10 E1 = 218kg; yE1,A = 0.09; yE1,C = 0.04; yE1,S = 0.87 44
Recap: Cross-flow arrangements N = 3 in this illustration [Schweitzer, p 1-263] I Recovery = fraction of solute recovered 1 − (xRN )(RN) (xF )(F) I Concentration of overall extract = solute leaving in each extract stream, divided by total extract flow rate N X n (yEn )(En) N X n En 45
Review from last time 46
Review from last time 47
Review from last time 48
Cross-current vs counter-current Cross-current (N = 2 stages) Counter-current (N = 2 stages) I We combine multiple extract streams I (Only 2 in illustration) I In general: yE1 > yE2 > . . . I Fresh solvent added at each stage I “Re-use” the solvent, so I Far lower solvent flows I Recovery = 1 − (xRN )(RN) (xF )(F) I Concentration = yE1 I How many stages? What solvent flow? You will have an assignment question to compare and contrast these two configurations 49
What we are aiming for General approach: 1. Use ternary diagrams to determine operating lines 2. Estimate number of “theoretical plates” or “theoretical stages” 3. Convert “theoretical stages” to actual equipment size. E.g. assume we calculate that we need N ≈ 6 theoretical stages. I does not mean we require 6 mixer-settlers (though we could do that, but costly) I it means we need a column which has equivalent operation of 6 counter-current mixer-settlers that fully reach equilibrium I at this point we resort to correlations and vendor assistance I vendors: provide HETS = height equivalent to a theoretical stage I use that to size the column I unit height (or size) = HETS × number of theoretical stages stage efficiency 50
For example [WINTRAY (Japanese company; newly patented design)] 51
Two counter-current units Reference for this section: Seader textbook, 3rd ed, p 312 to 324. Consider N = 2 stages for now. Steady state mass balance: F + E2 = R1 + E1 R1 + S = R2 + E2 Rearrange: F − E1 = R1 − E2 R1 − E2 = R2 − S (F − E1) = (R1 − E2) = (R2 − S) = P Note: each difference is equal to P (look on the flow sheet above where those differences are). 52
Counter-current graphical solution: 2 units Rearranging again: F = E1 + P R1 = E2 + P R2 = S + P Interpretation: P is a fictitious operating point on the ternary diagram (from lever rule) I F is on the line that connects E1 and P I R1 is on the line that connects E2 and P I R2 is on the line that connects S and P 53
Counter-current graphical solution: 2 units Step 1 Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 Overall balance gives: M = S + F = E1 + R2 For example, let’s require xR2,A = 0.05 (solute concentration in raffinate). Given an S flow rate, what is yE1,A? (concentration of solute in extract) 54
Counter-current graphical solution: 2 units Step 2 Note: the line connecting E1 to R2 is not a tie line. We use the lever rule and an overall mass balance (F + S = E1 + R2) to solve for all flows and compositions of F, S, E1, and R2. yE1,A ≈ 0.38 is found from an overall mass balance, through M. Simply connect R2 and M and project out to E1. 55
Counter-current graphical solution: 2 units Step 3 Recall: F = E1 + P F is on the line that connects E1 and P R2 = S + P R2 is on the line that connects S and P Extrapolate through these lines until intersection at point P. 56
Counter-current graphical solution: 2 units Step 4 Once we have E1, we can start: note that in stage 1 the R1 and E1 streams leave in equilibrium and can be connected with a tie line. 57
Counter-current graphical solution: 2 units Step 5 Again recall: R1 = E2 + P R1 is on the line that connects E2 and P Since we have point P and R1 we can bring the operating line back and locate point E2 58
Counter-current graphical solution: 2 units Step 6 The last unit in a cascade is a special case: we already know RN=2, but we could have also calculated it from the tie line with E2. We aim for some overshoot of RN. (Good agreement in this example.) 59
In general: Counter-current units F + E2 = E1 + R1 E2 + R2 = E3 + R1 En+Rn = En+1+Rn−1 Rearrange: F − E1 = R1 − E2 R1 − E2 = R2 − E3 Rn−1−En = Rn−En+1 (F − E1) = (R1 − E2) = . . . = (Rn−1 − En) = (Rn − En+1) = . . . = (RN − S) = P Notes: 1. each difference is equal to P (the difference between flows) 2. En and Rn are in equilibrium, leaving each stage [via tie line] 60
Counter-current graphical solution 1. We know F and S; connect with a line and locate “mixture” M 2. Either specify E1 or RN (we will always know one of them) 3. Connect a straight line through M passing through the one specified 4. Solve for unspecified one [via tie line] 5. Connect S through RN and extrapolate 6. Connect E1 through F and extrapolate; cross lines at P 7. Locate P by intersection of 2 lines 8. In general: connect En and Rn via equilibrium tie lines 61
Tutorial-style question Consider a system for which you have been given the ternary diagram (see next slides). A = solute, S = solvent (100% pure), C = carrier. The feed, F enters at 112 kg/hr with composition of 25 wt% solute and 75 wt% carrier. 1. Calculate the flow and composition of the extract and raffinate from: I 1st cross-current stage, using a pure solvent flow of 50 kg/hr. I 2nd cross-current stage, with an additional solvent flow of 50 kg/hr. 2. For the overall 2-stage cross-current system, find the: I overall recovery [answer: ∼93%] I overall concentration of combined extract streams [answer: ∼21%] 3. The objective now is to have a counter-current system so the raffinate leaving in the Nth stage, RN has yRN = 0.025 I Show the construction on the ternary diagram for the number of equili- brium stages to achieve xRN = 0.025, given a solvent flow of 28 kg/hr. I Calculate the overall recovery and concentration of the extract stream. I Plot on the same axes the concentrations in the extract and raffinate streams. 62
Tutorial solution: step 1 63
Tutorial solution: step 2 64
Tutorial solution: step 3 65
Tutorial solution: step 4 66
Tutorial solution: step 5 67
Tutorial solution: step 6 68
Tutorial solution: concentration profile 69
For practice (A) 70
For practice (B) 71
Counter-current graphical solution: 2 units Step 3(b) Recall: F + P = E1 R2 + P = S Thought experiment: What is the minimal achievable E1 concentration? mentally move point M towards S. What happens to P as solvent flow S is increased? Alternative explanation next. 72
Counter-current graphical solution: maximum solvent flow Step 3(b) Recall: F + P = E1 R2 + P = S Subtle point: minimal achievable Emin 1 concentration: I occurs at a certain maximum solvent flow rate indicated by ◦ I note that R2 is fixed (specified) in this example 73
Safety concerns Liquid-liquid (solvent) extraction units safety concerns can be reduced by: I using a high-flash point solvent I avoid static electricity generation (sparking) I slow liquid flow rates in pipes < 1 m/s to avoid static build-up I use conductive piping, not plastics or rubbers I feed organic phase from the bottom of tank, not the side, to avoid splashing I avoid any areas for air pockets to form I avoid generating mists of solvent I electrical circuits must be rated explosion proof I strict rules related to mobile electronics, welding, etc in the neighbourhood I conduct full HAZOP and fire assessment I do not allow trenches/dips for the solvent phase to accumulate, when spills occur [Courtesy of “ALTA Solvent Extraction Short Course”, Alan Taylor] 74

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Liquid-Liquid-Extraction.pdf

  • 1.
    Separation Processes: Liquid-liquid extraction ChE4M3 © Kevin Dunn, 2013 kevin.dunn@mcmaster.ca http://learnche.mcmaster.ca/4M3 Overall revision number: 256 (November 2013) 1
  • 2.
    Copyright, sharing, andattribution notice This work is licensed under the Creative Commons Attribution-ShareAlike 3.0 Unported License. To view a copy of this license, please visit http://creativecommons.org/licenses/by-sa/3.0/ This license allows you: I to share - to copy, distribute and transmit the work I to adapt - but you must distribute the new result under the same or similar license to this one I commercialize - you are allowed to use this work for commercial purposes I attribution - but you must attribute the work as follows: I “Portions of this work are the copyright of Kevin Dunn”, or I “This work is the copyright of Kevin Dunn” (when used without modification) 2
  • 3.
    We appreciate: I ifyou let us know about any errors in the slides I any suggestions to improve the notes All of the above can be done by writing to kevin.dunn@mcmaster.ca or anonymous messages can be sent to Kevin Dunn at http://learnche.mcmaster.ca/feedback-questions If reporting errors/updates, please quote the current revision number: 256 3
  • 4.
  • 5.
    References I Schweitzer, “Handbookof Separation Techniques for Chemical Engineers”, chapter 1.9 I Seader, Henley and Roper, “Separation Process Principles”, 3rd edition, chapter 8 I Richardson and Harker, “Chemical Engineering, Volume 2”, 5th edition, chapter 13 I Geankoplis, “Transport Processes and Separation Process Principles”, 4th edition, chapter 12.5 and 12.6 I Ghosh, “Principles of Bioseparation Engineering”, chapter 7 I Uhlmann’s Encyclopedia, “Liquid-Liquid Extraction”, DOI:10.1002/14356007.b03 06.pub2 5
  • 6.
    Definitions I solute: specieswe aim to recover (A) from the feed I feed or “feed solvent”: one of the liquids in the system (“carrier”) I solvent: MSA (by convention: the “added” liquid) I extract: solvent (not solute) mostly present in this layer. yE,A = concentration of A, the solute, in extract. We aim for the solute (A) to be mostly in the extract stream. I raffinate: residual solute in this layer = xR,A I distribution: how the solute partitions itself = DA = yE,A xR,A I measure of affinity of solute I DA = µ0 R − µ0 E RT = chemical potential difference (R)(temperature) 6
  • 7.
    Where/why LLE isused Where? I Bioseparations I Nuclear (uranium recovery) I Mining: nickel/cobalt; copper/iron I Perfumes, fragrances and essential oils I Fine and specialty chemicals I Oil sands: extract bitumen away from water Why? I Temperature sensitive products I High purity requirements I High-boiling point species in low quantity I Need to separate by species type (rather than relative volatility) I Close-boiling points, but high solubility difference I Azeotrope-forming mixtures 7
  • 8.
    Extractor types: 3major steps required 1. Mixing/contacting: I turbulent contact between liquid phases I small droplet dispersion in a continuous phase I which phase is dispersed? I mass-transfer between phases I limited by solute loading in solvent 2. Phase separation: I reverse of above mixing step I drops come together and coalesce I relies on density difference 3. Collection of phases leaving the unit I split the raffinate from the extract 8
  • 9.
    What are weaiming for? Main aims I High recovery of solute overall (low xR and high yE ) I Concentrated solute in extract (high yE ) How to achieve this? I Cocurrent mixer-settlers I Cross-current mixer settlers I Counter-current mixer-settlers in series I High interfacial area during mixing I Reduce mass-transfer resistance I Promote mass transfer: aim is to move solute in feed to the solvent in extract stream I molecular diffusion ← microscopic movement I eddy diffusion ← orders of magnitude greater 9
  • 10.
    Co-, cross-, counter-currentoperations [Hanson article, Chemical Engineering, posted to course website] 10
  • 11.
    Some photos ofequipment [http://www.flickr.com/photos/51009184@N06/5062827346/] 11
  • 12.
    Some photos ofequipment [http://www.rousselet-robatel.com/images/products/Mixer-settler-pic-2lg.jpg] 12
  • 13.
    Equipment for LLE 1.Mixer-settlers (integrated unit) I mix: impellers I mix: nozzles I mix: feeds meet directly in the pump I mix: geared-teeth devices I main aim: good contact; avoid droplets smaller than 2 µm I settle: baffles, membranes I settle: ultrasound I settle: chemical treatment I settle: centrifuges 2. Column-type units contain: I (a) nothing or I (b) trays and/or I (c) packing and/or I (d) pulsating and/or I (e) agitation 3. Rotating devices Important point: LLE is an equilibrium-limited separation (as opposed to rate-limited separations seen up to now). 13
  • 14.
    Mixer-settlers [Richardson and Harker,p 745] Common in mining industry: requirements ∼40000 L/min flows 14
  • 15.
    Mixer-settlers KnitMesh coalescer: consistencyof “steel wool” [Richardson and Harker, p 747] 15
  • 16.
    Horizontal gravity settlingvessel [Seader, 3ed, p302] 16
  • 17.
    Spray columns: separationprinciple is gravity [Richardson and Harker, p 751] 17
  • 18.
    Tray columns [Richardson andHarker, p 749] I coalescence on each tray I tray holes: ∼ 3mm I breaks gradient formation (axial dispersion) 18
  • 19.
    Tray columns withmechanical agitation I shearing to create dispersion I can have alternating layers of packing (coalescence) I some column designs pulsate ⇑⇓ [Seader, 3ed, p302] 19
  • 20.
    Rotating devices [Seader, 3ed,p 306] I “white” = lighter liquid I “grey” = heavier liquid Used when foams and emulsions would easily form: i.e. gentle mass transfer. Look up in your own time: centrifuge mixer-settlers 20
  • 21.
    Linking up units(more on this later) [Richardson and Harker, p 723] 21
  • 22.
    Integration with downstreamunits [Schweitzer, p 1-257] 22
  • 23.
    Selecting a solvent Schweitzer:“The choice of solvent for a LLE process can often have a more significant impact on the process economics than any other design decision that has to be made”. Which properties of a solvent influence our aims with LLE? I High distribution coefficient (selectivity) for solute I Low distribution coefficient for carrier I Reasonable volatility difference with solute and carrier I Reasonable surface tension: easy to disperse and coalesce I High density difference: separates rapidly by gravity I Stability to maximize its reuse I Inert to materials of construction I Low viscosity: maximizes mass transfer I Safe: non-toxic, non-flammable I Cheap, and easily available I Compatible with carrier and solute: avoid contamination I Doesn’t foam, form emulsions, scum layers at interface 23
  • 24.
    Calculating the distributioncoefficient (in the lab only) Mass balance: FxF + SyS = EyE + RxR D = yE xR If F = S = E = R and ys = 0, then only measure xR: D = xF xR − 1 ← only in this special case I Capital letters refer to mass amounts I y2 ← refers to mass fractions in solvent and extract layers I x2 ← refers to mass fractions in carrier and raffinate layers Once D is determined, we can obtain phase diagrams to understand how the process will operate. Also: see Perry’s for many values of D 24
  • 25.
    Triangular phase diagrams:from laboratory studies [Flickr# 3453475667] 25
  • 26.
    Using a triangularphase diagrams [http://www.youtube.com/watch?v=gGYHXhcKM5s] 26
  • 27.
    Lever rule Mix Pand Q I mixture = K I PK KQ = mass Q mass P I The converse applies also: when a settled mixture separates into P and Q I Applies anywhere: even in the miscible region 27
  • 28.
    Q1: Using thelever rule Which is a more flexible system: (a) or (b)? I S = pure solvent used I F = feed concentration point (more correctly it is xF ) Answer: range of feed concentrations (xF ) is wider, i.e. more desirable, for (a). Difference between (a) and (b): I due to solvent choice I due to different temperatures I due to pH modification, etc 28
  • 29.
    Q2: Using thelever rule Mix a feed stream, F, containing C and A (i.e. xF ) with a pure solvent stream S (i.e. yS = 0). Composition of the mixture? 29
  • 30.
    Q2 solution: Usingthe lever rule Composition of the mixture? we need more information (e.g. amount of F and S must be given to find point M) 30
  • 31.
    Q3: Going toequilibrium Let that mixture M achieve equilibrium. What is the composition of the raffinate and extract? 31
  • 32.
    Q3 solution: Goingto equilibrium What is the composition of the raffinate and extract? Use the tie lines [solid lines]; or interpolate between existing ones. 32
  • 33.
    Q4: Altering flows Recovery= fraction of solute recovered For 1 stage 1 − (xR)(R) (xF )(F) Specifically 1− (xR,A)(R) (xF,A)(F) Same system, but now lower solvent flow rate (to try save money!). What happens to (a) extract concentration and (b) solute recovery? 33
  • 34.
    Q4 solution: Alteringflows Recovery = fraction of solute recovered For 1 stage 1 − (xR)(R) (xF )(F) Specifically 1− (xR,A)(R) (xF,A)(F) (a) extract concentration increases: (A at E*) > (A at E): yE∗ > yE (b) solute recovery drops: (A at R*) > (A at R): xR∗ > xR 34
  • 35.
    Q5: Composition ofthe mixture, M? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 Answer: M = xM,A = xM,C = xM,S = 35
  • 36.
    Q5 solution: Compositionof the mixture, M? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 M = F + S Answer: M = 350kg; xM,A = 0.17; xM,C = 0.54; xM,S = 0.29 36
  • 37.
    Q6: Composition ofthe 2 phases leaving in equilibrium? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 R1 = xR1,A = xR1,C = xR1,S = E1 = yE1,A = yE1,C = yE1,S = 37
  • 38.
    Q6 solution: Compositionof the 2 phases in equilibrium? Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 M = E1 + R1 R1 = 222kg; xR1,A = 0.10; xR1,C = 0.82; xR1,S = 0.08 E1 = 128kg; yE1,A = 0.33; yE1,C = 0.06; yE1,S = 0.61 38
  • 39.
    Link units inseries [Richardson and Harker, p 723] Introduce fresh solvent each time 39
  • 40.
    Q7: send raffinatefrom Q6 to second mixer-settler Question: how much solvent should we use in the second stage? 40
  • 41.
    Q7 solution: sendraffinate from Q6 to second mixer-settler Compare extract: Compare volumes: Answer: equilibrium from point B (most solvent), C, D (least solvent) will each be different. Trade-off: higher extraction vs lower recovery 41
  • 42.
    Phase diagram: furfural,water, ethylene glycol Feed Solvent F = 100 kg S = 200 kg xF,A = 0.45 xS,A = 0.0 xF,C = 0.55 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 I A = ethylene glycol (solute) I C = water (carrier) I S = furfural (solvent) AIM: to remove ethylene glycol (solute) from water (carrier) into solvent (furfural) 1. Calculate the mixture composition, M 2. Calculate the equilibrium compositions in E1 and R1 Note: extract is defined as “the solvent-rich stream leaving the system” 42
  • 43.
  • 44.
    Solution: Phase diagram:furfural, water, ethylene glycol Feed Solvent F = 100 kg S = 200 kg xF,A = 0.45 xS,A = 0.0 xF,C = 0.55 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 I A = ethylene glycol solute I C = water (carrier) I S = furfural solvent M = 300kg; xM,A = 0.15; xM,C = 0.18; xM,S = 0.67 R1 = 82kg; xR1,A = 0.33; xR1,C = 0.57; xR1,S = 0.10 E1 = 218kg; yE1,A = 0.09; yE1,C = 0.04; yE1,S = 0.87 44
  • 45.
    Recap: Cross-flow arrangements N= 3 in this illustration [Schweitzer, p 1-263] I Recovery = fraction of solute recovered 1 − (xRN )(RN) (xF )(F) I Concentration of overall extract = solute leaving in each extract stream, divided by total extract flow rate N X n (yEn )(En) N X n En 45
  • 46.
  • 47.
  • 48.
  • 49.
    Cross-current vs counter-current Cross-current(N = 2 stages) Counter-current (N = 2 stages) I We combine multiple extract streams I (Only 2 in illustration) I In general: yE1 > yE2 > . . . I Fresh solvent added at each stage I “Re-use” the solvent, so I Far lower solvent flows I Recovery = 1 − (xRN )(RN) (xF )(F) I Concentration = yE1 I How many stages? What solvent flow? You will have an assignment question to compare and contrast these two configurations 49
  • 50.
    What we areaiming for General approach: 1. Use ternary diagrams to determine operating lines 2. Estimate number of “theoretical plates” or “theoretical stages” 3. Convert “theoretical stages” to actual equipment size. E.g. assume we calculate that we need N ≈ 6 theoretical stages. I does not mean we require 6 mixer-settlers (though we could do that, but costly) I it means we need a column which has equivalent operation of 6 counter-current mixer-settlers that fully reach equilibrium I at this point we resort to correlations and vendor assistance I vendors: provide HETS = height equivalent to a theoretical stage I use that to size the column I unit height (or size) = HETS × number of theoretical stages stage efficiency 50
  • 51.
    For example [WINTRAY (Japanesecompany; newly patented design)] 51
  • 52.
    Two counter-current units Referencefor this section: Seader textbook, 3rd ed, p 312 to 324. Consider N = 2 stages for now. Steady state mass balance: F + E2 = R1 + E1 R1 + S = R2 + E2 Rearrange: F − E1 = R1 − E2 R1 − E2 = R2 − S (F − E1) = (R1 − E2) = (R2 − S) = P Note: each difference is equal to P (look on the flow sheet above where those differences are). 52
  • 53.
    Counter-current graphical solution:2 units Rearranging again: F = E1 + P R1 = E2 + P R2 = S + P Interpretation: P is a fictitious operating point on the ternary diagram (from lever rule) I F is on the line that connects E1 and P I R1 is on the line that connects E2 and P I R2 is on the line that connects S and P 53
  • 54.
    Counter-current graphical solution:2 units Step 1 Feed Solvent F = 250 kg S = 100 kg xF,A = 0.24 xS,A = 0.0 xF,C = 0.76 xS,C = 0.0 xF,S = 0.00 xS,S = 1.0 Overall balance gives: M = S + F = E1 + R2 For example, let’s require xR2,A = 0.05 (solute concentration in raffinate). Given an S flow rate, what is yE1,A? (concentration of solute in extract) 54
  • 55.
    Counter-current graphical solution:2 units Step 2 Note: the line connecting E1 to R2 is not a tie line. We use the lever rule and an overall mass balance (F + S = E1 + R2) to solve for all flows and compositions of F, S, E1, and R2. yE1,A ≈ 0.38 is found from an overall mass balance, through M. Simply connect R2 and M and project out to E1. 55
  • 56.
    Counter-current graphical solution:2 units Step 3 Recall: F = E1 + P F is on the line that connects E1 and P R2 = S + P R2 is on the line that connects S and P Extrapolate through these lines until intersection at point P. 56
  • 57.
    Counter-current graphical solution:2 units Step 4 Once we have E1, we can start: note that in stage 1 the R1 and E1 streams leave in equilibrium and can be connected with a tie line. 57
  • 58.
    Counter-current graphical solution:2 units Step 5 Again recall: R1 = E2 + P R1 is on the line that connects E2 and P Since we have point P and R1 we can bring the operating line back and locate point E2 58
  • 59.
    Counter-current graphical solution:2 units Step 6 The last unit in a cascade is a special case: we already know RN=2, but we could have also calculated it from the tie line with E2. We aim for some overshoot of RN. (Good agreement in this example.) 59
  • 60.
    In general: Counter-currentunits F + E2 = E1 + R1 E2 + R2 = E3 + R1 En+Rn = En+1+Rn−1 Rearrange: F − E1 = R1 − E2 R1 − E2 = R2 − E3 Rn−1−En = Rn−En+1 (F − E1) = (R1 − E2) = . . . = (Rn−1 − En) = (Rn − En+1) = . . . = (RN − S) = P Notes: 1. each difference is equal to P (the difference between flows) 2. En and Rn are in equilibrium, leaving each stage [via tie line] 60
  • 61.
    Counter-current graphical solution 1.We know F and S; connect with a line and locate “mixture” M 2. Either specify E1 or RN (we will always know one of them) 3. Connect a straight line through M passing through the one specified 4. Solve for unspecified one [via tie line] 5. Connect S through RN and extrapolate 6. Connect E1 through F and extrapolate; cross lines at P 7. Locate P by intersection of 2 lines 8. In general: connect En and Rn via equilibrium tie lines 61
  • 62.
    Tutorial-style question Consider asystem for which you have been given the ternary diagram (see next slides). A = solute, S = solvent (100% pure), C = carrier. The feed, F enters at 112 kg/hr with composition of 25 wt% solute and 75 wt% carrier. 1. Calculate the flow and composition of the extract and raffinate from: I 1st cross-current stage, using a pure solvent flow of 50 kg/hr. I 2nd cross-current stage, with an additional solvent flow of 50 kg/hr. 2. For the overall 2-stage cross-current system, find the: I overall recovery [answer: ∼93%] I overall concentration of combined extract streams [answer: ∼21%] 3. The objective now is to have a counter-current system so the raffinate leaving in the Nth stage, RN has yRN = 0.025 I Show the construction on the ternary diagram for the number of equili- brium stages to achieve xRN = 0.025, given a solvent flow of 28 kg/hr. I Calculate the overall recovery and concentration of the extract stream. I Plot on the same axes the concentrations in the extract and raffinate streams. 62
  • 63.
  • 64.
  • 65.
  • 66.
  • 67.
  • 68.
  • 69.
  • 70.
  • 71.
  • 72.
    Counter-current graphical solution:2 units Step 3(b) Recall: F + P = E1 R2 + P = S Thought experiment: What is the minimal achievable E1 concentration? mentally move point M towards S. What happens to P as solvent flow S is increased? Alternative explanation next. 72
  • 73.
    Counter-current graphical solution:maximum solvent flow Step 3(b) Recall: F + P = E1 R2 + P = S Subtle point: minimal achievable Emin 1 concentration: I occurs at a certain maximum solvent flow rate indicated by ◦ I note that R2 is fixed (specified) in this example 73
  • 74.
    Safety concerns Liquid-liquid (solvent)extraction units safety concerns can be reduced by: I using a high-flash point solvent I avoid static electricity generation (sparking) I slow liquid flow rates in pipes < 1 m/s to avoid static build-up I use conductive piping, not plastics or rubbers I feed organic phase from the bottom of tank, not the side, to avoid splashing I avoid any areas for air pockets to form I avoid generating mists of solvent I electrical circuits must be rated explosion proof I strict rules related to mobile electronics, welding, etc in the neighbourhood I conduct full HAZOP and fire assessment I do not allow trenches/dips for the solvent phase to accumulate, when spills occur [Courtesy of “ALTA Solvent Extraction Short Course”, Alan Taylor] 74