1. Paul Ashall 2007 Separation processes - general Mechanical separations e.g. filtration of a solid from a suspension in a liquid, centrifugation, screening etc Mass transfer operations e.g. distillation, extraction etc

2. Paul Ashall 2007 Mass transfer operations – nature of interface between phases Gas-liquid contact e.g. absorption, evaporation, distillation etc Liquid-liquid contact e.g. extraction Liquid-solid contact e.g. crystallisation, adsorption Gas-solid contact e.g. adsorption, drying etc

3. Paul Ashall 2007 Mass transfer operations – controlling transport phenomenon Mass transfer controlling e.g.distillation, absorption, extraction, adsorption etc Mass transfer and heat transfer controlling e.g. drying, crystallisation Heat transfer controlling e.g. evaporation

4. Paul Ashall 2007 Methods of operation Non steady state – concentration changes with time e.g. batch processes Steady state Stage Differential contact

5. Paul Ashall 2007 When both phases are flowing: Co-current contact Cross flow Counter-current flow Stage 1Stage 2 12 12 etc

6. Paul Ashall 2007 Choice of separation process Factors to be considered: Feasibility Product value Cost Product quality selectivity

7. Paul Ashall 2007 Liquid-liquid extraction principles Feed phase contains a component, i, which is to be removed. Addition of a second phase (solvent phase) which is immiscible with feed phase but component i is soluble in both phases. Some of component i (solute) is transferred from the feed phase to the solvent phase. After extraction the feed and solvent phases are called the raffinate (R) and extract (E) phases respectively.

8. Paul Ashall 2007 continued Normally one of the two phases is an organic phase while the other is an aqueous phase. Under equilibrium conditions the distribution of solute i over the two phases is determined by the distribution law. After the extraction the two phases can be separated because of their immiscibility.Component i is then separated from the extract phase by a technique such as distillation and the solvent is regenerated.Further extractions may be carried out to remove more component i.Liquid liquid extraction can also be used to remove a component from an organic phase by adding an aqueous phase.

9. Paul Ashall 2007 Example - Penicillin G 6-aminopenicillanic acid (6-APA) is manufactured by GSK in Irvine. It is used to manufacture amoxicillin and ‘Augmentin’. Fermentation products (penicillin G broth) are filtered (microfiltration) and extracted at low pH with amyl acetate or methyl isobutyl ketone. The penicillin G is then extracted further at a higher pH into an aqueous phosphate buffer.

10. Paul Ashall 2007 Extractants The efficiency of a liquid liquid extraction can be enhanced by adding one or more extractants to the solvent phase. The extractant interacts with component i increasing the capacity of the solvent for i.To recover the solute from the extract phase the extractant-solute complex has to be degraded.

11. Paul Ashall 2007 Distribution coefficient K = mass fraction solute in E phase mass fraction solute in R phase

12. Paul Ashall 2007 Immiscible liquids e.g. water – chloroform Consider a feed of water/acetone(solute). K = mass fraction acetone in chloroform phase mass fraction acetone in water phase K = kg acetone/kg chloroform = y/x kg acetone/kg water K = 1.72 i.e. acetone is preferentially soluble in the chloroform phase

13. Paul Ashall 2007 Partially miscible liquids E.g. water – MIBK Consider a solute acetone. Need to use a triangular phase diagram to show equilibrium compositions of MIBK- acetone-water mixtures. Characteristics are single phase and two phase regions, tie lines connecting equilibrium phase compositions in two phase region.

14. Paul Ashall 2007 Triangular phase diagrams Each apex of triangle represents 100% pure component B A S P %A %S %B

15. Paul Ashall 2007 continued A mixture of overall composition M will split into two phases – E & R. R phase is in equilibrium with E phase R/E = line ME/line MR B A S R E M

16. Paul Ashall 2007 Distribution curve Plot of y (kgsoluteB/kgsolventS) in E phase v x (kgsoluteB/kgA) in R phase

17. Paul Ashall 2007 Example

18. Paul Ashall 2007 Choice of solvent Factors to be considered: Selectivity Distribution coefficient Insolubility of solvent Recoverability of solute from solvent Density difference between liquid phases Interfacial tension Chemical reactivity Cost Viscosity, vapour pressure Flammability, toxicity

19. Paul Ashall 2007 Selectivity β = (mass fraction B in E)/(mass fraction A in E) (mass fraction B in R)/(mass fraction A in R) β > 1

20. Paul Ashall 2007 Distribution coefficient K = y/x Large values are desirable since less solvent is required for a given degree of extraction

21. Paul Ashall 2007 Insolubility of solvent B A S B A S Preferred solvent – A and S have limited solubility S very soluble in A and A very soluble in S

22. Paul Ashall 2007 Recoverability of solvent and solute No azeotrope formed between solvent and solute Mixtures should have a high relative volatility Solvent should have a small latent heat of vapourisation

23. Paul Ashall 2007 Density A density difference is required between the two phases.

24. Paul Ashall 2007 Interfacial tension The larger the interfacial tension between the two phases the more readily coalescence of emulsions will occur to give two distinct liquid phases but the more difficult will be the dispersion of one liquid in the other to give efficient solute extraction.

25. Paul Ashall 2007 Chemical reactivity Solvent should be stable and inert.

26. Paul Ashall 2007 Physical properties For material handling: Low viscosity Low vapour pressure Non-flammable (high flash point) Non-toxic

27. Paul Ashall 2007 Mass balances For counter-current contact with immiscible solvents a simple mass balance for solute B at steady state gives the operating line: y n+1 = a/s(x n – x F ) + y 1,where y n+1 = kgB/kgS in solvent feed a = mass component A s = mass solvent x n = kgB/kgA after n stages x F = kgB/kgA in feed y 1 = kgB/kgS in extract after first stage

28. Paul Ashall 2007 continued A graphical procedure may be used to analyse these systems. The number of theoretical stages (n) required to pass from x F to xn is found by drawing in ‘steps’ between the operating line and the equilibrium curve (yn, xn). In practice equilibrium conditions may not be attained and extraction efficiency will be less than 100% thus requiring more stages in practice than the above analysis would suggest. Also partial miscibility of the solvents has to be considered in the separation process.

29. Paul Ashall 2007 continued y x y1,xF yn+1,xn

30. Paul Ashall 2007 Operation Batch Continuous Single/multi stage contact

31. Paul Ashall 2007 Equipment Mixer-settler units Columns Centrifugal contactors

32. Paul Ashall 2007