1. ERT 313 BIOSEPARATION ENGINEERING EXTRACTION
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. aqueous solution containing solute
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. Solute
Organic solvent
KD (mol/L) at 250C
Amino acids
Glycine
Alanine
2-aminobutyric acid
Lysine
Glutamic acid
n-butanol
0.01
0.02
0.20
0.07
Antibiotics
Erythromycin
Novobiocin
Penicillin F
Penicillin K
Amyl acetate
Butyl acetate
120
100 at pH 7.0
0.01 at pH 10.5
32 at pH 4.0
0.06 at pH 6.0
12 at pH 4.0
0.1 at pH 6.0

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

9. 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
Single stage
extraction
Raffinate
Extract
Solvent

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

11. Continuous Extraction

13. Principles of Extraction
Extraction of dilute solutions
Extraction of concentrated solutions; phase equilibria

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

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

16. 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

18. 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.

20. 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)

21. 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: acetone, water, MIK
- Raffinate phase: acetone, water, 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

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

23. 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

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

27. 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.

28. 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.

29. 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

30. ERT 313 BIOSEPARATION ENGINEERING TUTORIAL 4 - EXTRACTION
Prepared by:
Pn. Hairul Nazirah Abdul Halim

31. 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?

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

35. 1. Nicotine in the feed solution = 100 (0.01) = 1 kg/h nicotine
Water in feed = 100 ( ) = 99 kg/h water
Nicotine in solvent = 200 (0.0005) = 0.1 kg/h nicotine
Kerosene = 200 (1 – ) = kg/h kerosene
Exit stream of aqueous phase, L1
Water = 99 kg/h = (1 – ) L1
L1 = kg/h (nicotine + water)
Nicotine = – 99 = kg/h nicotine in exit stream

36. 4. Exit stream of solvent phase, V1
Solvent = kg/h
Nicotine in solvent = (1 – 0.099) = kg/h in exit stream
Solvent + Nicotine = = kg/h