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ERT 313
BIOSEPARATION ENGINEERING
EXTRACTION
Prepared by:
Pn. Hairul Nazirah Abdul Halim
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.
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
Solvent Extraction
Large volume of an
aqueous solution containing solute
+
Small volume of non-miscible organic solvent
• 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
• The distribution of solute between two phases is
express quantitatively by distribution coefficient,
KD.
solute concentration in extract phase
KD 
solute concentration in raffinate phase
• Higher value of KD indicates higher extraction
efficiency.
Organic solvent
KD (mol/L) at 250C
n-butanol
n-butanol
n-butanol
n-butanol
n-butanol
0.01
0.02
0.02
0.20
0.07
Antibiotics
Erythromycin
Novobiocin
Amyl acetate
Butyl acetate
Penicillin F
Amyl acetate
Penicillin K
Amyl 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
Solute
Amino acids
Glycine
Alanine
2-aminobutyric acid
Lysine
Glutamic acid
Operating Modes of Extraction
• Batch or continuous extractions
• Batch extraction – single stage or multiple stage
• Continuous extraction – co-current or
countercurrent extraction
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
Solvent
extraction
Raffinate
Extract
Batch Extraction in Multiple Stage
Schematic representation of a two-stage batch extraction
Continuous Extraction
Principles of Extraction
• Extraction of dilute solutions
• Extraction of concentrated solutions; phase
equilibria
Extraction of Dilute Solution
• Extraction factor is defined as:
• Where:
E = extraction factor
KD = distribution coefficient
V = volume of solvent
L = volume of aqueous
solute concentration in extract phase
KD 
solute concentration in raffinate phase
•For a single-stage extraction with pure solvent;
- the fraction of solute remaining is
- the fraction recovered is
E
1 E
1
1 E
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
•
•
•
•
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.
•
•
•
•
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)
• 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
• Let’s compare with Fig 23.8. Which system is more
effective?
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
• Extraction Equipments:
- Mixer settlers
- Packed extraction towers
- Perforated plate towers
- Baffle towers
- Agitated tower extractors
• Auxiliary equipment:
- stills, evaporators, heaters and condenser
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.
• 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.
• 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
ERT 313
BIOSEPARATION ENGINEERING
TUTORIAL 4 - EXTRACTION
Prepared by:
Pn. Hairul Nazirah Abdul Halim
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 twostage extraction using fresh solvent in
both stages?
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.
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
2. Nicotine in solvent = 200 (0.0005) = 0.1 kg/h
nicotine
Kerosene = 200 (1 – 0.0005) = 199.9 kg/h
kerosene
3. 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
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