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ERT 313 BIOSEPARATION ENGINEERING EXTRACTION

ERT 313 BIOSEPARATION ENGINEERING EXTRACTION. Prepared by: Pn. Hairul Nazirah Abdul Halim. Definition of Extraction .

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ERT 313 BIOSEPARATION ENGINEERING EXTRACTION

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  1. ERT 313BIOSEPARATION ENGINEERINGEXTRACTION Prepared by: Pn. Hairul Nazirah Abdul Halim

  2. Definition of Extraction • Liquid-Liquid extraction is a mass transfer operation in which a liquid solution (the feed) is contacted with an immiscible or nearly immiscible liquid (solvent) that exhibits preferential affinity or selectivity towards one or more of the components in the feed.

  3. Purpose of Extraction • To separate closed-boiling point mixture • Mixture that cannot withstand high temperature of distillation • Example: - recovery of penicillin from fermentation broth solvent: butyl acetate - recovery of acetic acid (b.p 1180c) from dilute aqueous (b.p 1000c) solutions solvent: ethyl-acetate

  4. Solvent Extraction Large volume of an aqueous solution containing solute + Small volume of non-miscible organic solvent

  5. The solute originally present in the aqueous phase gets distributed in both phases. • If solute has preferential solubility in the organic solvent, more solute would be present in the organic phase at equilibrium • The extraction is said to be more efficient • Extract – the layer of solvent + extracted solute • Raffinate – the layer from which solute has been removed

  6. The distribution of solute between two phases is express quantitatively by distribution coefficient, KD. • Higher value of KD indicates higher extraction efficiency.

  7. Operating Modes of Extraction • Batch or continuous extractions • Batch extraction – single stage or multiple stage • Continuous extraction – co-current or countercurrent extraction

  8. Batch Extraction • The aqueous feed is mixed with the organic solvent • After equilibration, the extract phase containing the desired solute is separated out for further processing. • A schematic representation of a single batch operation: Feed Raffinate Single stage extraction Extract Solvent

  9. Batch Extraction in Multiple Stage Schematic representation of a two-stage batch extraction

  10. Continuous Extraction

  11. Principles of Extraction • Extraction of dilute solutions • Extraction of concentrated solutions; phase equilibria

  12. Extraction of Dilute Solution • Extraction factor is defined as: • Where: E = extraction factor KD = distribution coefficient V = volume of solvent L = volume of aqueous

  13. For a single-stage extraction with pure solvent; - the fraction of solute remaining is - the fraction recovered is

  14. Extraction of Concentrated Solution • Equilibrium relationship are more complicated – 3 or more components present in each phase • Equilibrium data are often presented on a triangular diagram such as Fig 23.7 and 23.8

  15. Consider Fig 23.7 • Line ACE shows extract phase • Line BDE shows raffinate phase • Point E is the plait point – the composition of extract & raffinate phases approach each other • Tie line – a straight line joining the composition of extract & raffinate phases. • Tie line in Fig 23.7 slope up to the left – extract phase is richer in acetone than the raffinate phase. • This suggest that most of the acetone could be extract from water phase using moderate amount of solvent.

  16. Consider Fig 23.8 • Line AD shows extract phase • Line BC shows raffinate phase • Tie line in Fig 23.8 slope up to the right – extraction would still be possible • But more solvent would have to use. • The final extract would not be as rich in desired component (MCH)

  17. How to obtain the phase composition using the triangular diagram? - Point M: 0.2 Acetone, 0.3 water, 0.5 MIK - Draw a new tie line - Extract phase: 0.232 acetone, 0.043 water, 0.725 MIK - Raffinate phase: 0.132 acetone, 0.845 water, 0.023 MIK - Ratio of acetone to water in the product = 0.232/0.043 = 5.4 - Ratio of acetone to water in the raffinate = 0.132/0.845 =0.156

  18. Let’s compare with Fig 23.8. Which system is more effective?

  19. Extraction Equipment • Batchwise or continuous operation • Feed liquid + solvent (put in agitated vessel) = layers (to be settled and separated) • Extract – the layer of solvent + extracted solute • Raffinate – the layer from which solute has been removed • Extract may be lighter or heavier than raffinate. • Continuous flow – more economical for more than one contact process

  20. Extraction Equipments: - Mixer settlers - Packed extraction towers - Perforated plate towers - Baffle towers - Agitated tower extractors • Auxiliary equipment: - stills, evaporators, heaters and condenser

  21. Aqueous Two Phase Extraction • Use widely in separation of proteins, enzymes, viruses, cells and cell organels. • not denature the biological entities as they might be in organic solvents. • The proteins are partitioning between two aqueous phases which contains mutually incompatible polymers or other solutes.

  22. For example; - light phase is water + 10% polyethylene glycol (PEG) and 0.5% dextran • heavy phase is water + 1% glycol and 15% dextran • Proteins are partitioned between phases with distribution coefficient (KD) that depends on the pH. • KD can vary from 0.01 to more than 100.

  23. Factors that affect protein partitioning in Aqueous Two Phase System: 1. Protein molecular weight 2. Protein charge, surface properties 3. Polymer(s) molecular weight 4. Phase composition, tie-line length 5. Salt effects 6. Affinity ligands attached to polymers

  24. ERT 313BIOSEPARATION ENGINEERINGTUTORIAL 4 - EXTRACTION Prepared by: Pn. Hairul Nazirah Abdul Halim

  25. QUESTION 1 Batch Extraction EXAMPLE 23.2. Penicillin F is recovered from a dilute aqueous fermentation broth by extraction with amyl acetate, using 6 volumes of solvent per 100 volumes of the aqueous phase. At pH 3.2 the distribution coefficient KD is 80. (a) What fraction of the penicillin would be recovered in a single ideal stage? (b) What would be the recovery with two- stage extraction using fresh solvent in both stages?

  26. QUESTION 2 Continuous Single Stage Extraction An inlet water solution of 100 kg/h containing 0.010 wt fraction nicotine in water is stripped with a kerosene stream of 200 kg/h containing 0.0005 wt fraction nicotine in a single stage extraction unit. It is desired to reduce the concentration of the exit water to 0.0010 wt fraction nicotine. Calculate the flow rate of the nicotine in both of the exit streams.

  27. SOLUTION 1. Nicotine in the feed solution = 100 (0.01) = 1 kg/h nicotine Water in feed = 100 (1 - 0.01) = 99 kg/h water • Nicotine in solvent = 200 (0.0005) = 0.1 kg/h nicotine Kerosene = 200 (1 – 0.0005) = 199.9 kg/h kerosene • Exit stream of aqueous phase, L1 Water = 99 kg/h = (1 – 0.0010) L1 L1 = 99.099 kg/h (nicotine + water) Nicotine = 99.099 – 99 = 0.099 kg/h nicotine in exit stream

  28. 4. Exit stream of solvent phase, V1 Solvent = 199.9 kg/h Nicotine in solvent = 0.1 + (1 – 0.099) = 1.001 kg/h in exit stream Solvent + Nicotine = 199.9 + 1.001 = 200.9 kg/h

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