1. Bioseparation Engineering
Crystallization

2. Crystallization crystal formation 1. Supersaturation
Supersaturated solution:
thermodynamically unstable
Metastable region:
solute will deposit on existing crystals but no new crystal nuclei formed
Intermediate zone:
both growth of the existing crystals and the formation of new nuclei occur simultaneously
Labile zone:
nuclei are formed
Figure 1. Regions of supersaturation.

3. crystal growth; genesis of new crystal
2. Purity
is important
3. Nucleation
crystal growth; genesis of new crystal
Homogeneous nucleation: result of supersaturation
Heterogeneous nucleation: insoluble material initiates crystal growth
Secondary nucleation: induced by the contact between different crystals
Nucleation Rate
[1]
where:
c – concentration in solution
c* – concentration at saturation
kn and i – empirical parameters
crystal formation equation
cluster embryo nucleus crystal

4. in nonagitated system, diffusion limited
4. Single Crystal Growth
in nonagitated system, diffusion limited
[2]
where:
A – crystal area
k – mass transfer coefficient
Agitation increase the relative velocity between
the solution and the crystal
[3]
κ – surface reaction rate (affected by cooling)
k – mass transfer coefficient (affected by viscosity and agitation)
A – crystal area
M – crystal mass
where:

5. Characteristic crystal length
(assume cubic crystal)
For cubic crystal of side, s
For spherical particle, l is the sphere diameter
Define:
φi - geometric factors characteristic of the crystal shape
crystal growth equation

6. Batch Crystallization
Figure 2. Stirred tank batch crystallizers.
solution is cooled to produce supersaturation
seeding crystal can be added
cooling rate can be controlled
(cooling curve) – key idea

7. Batch crystallization happens at metastable zone,
How to obtain the cooling curve?
Assume: a single crystal size
change in
supersaturation
change from
altered temperature
crystal growth
nucleation = +
Batch crystallization happens at metastable zone,
thus – change in supersaturation is small
– nucleation occurs by seeding
A – total crystal area
c* – saturation concentration = f(temp.)
where:

8. Integrating MS – total mass of the seed crystal
lS – initial size of the seed crystal
where:
Integrating

9. or in the form of TP – final temperature
T0 – temperature at which crystal begins to form
MP = [(T0 – TP)V dc*/dt]
Maximum mass of crystalline product minus that in the seed
η = (lp – lS) /lS
Fractional increase in product size per seed size
τ – actual time divided by the total time
where:

10. Batch Scale-up Important factor in large scale batch type,
secondary nucleation
Scaling up needs experiences and
secondary nucleation should be considered

11. Example 3.12
Briefly analyze the large-scale purification and crystallization of lipase
from Geotricbum candidum.
심세나

12. Introduction
Lipase from Geotrichum candidum
Two step isolation
Isoelectric focusing
Specificity for oleic acid
Crystallization
X- ray

13. Method: Purification of lipase
Chromatographic two-step purification
Q-Sepharose FF column (4 X 30 cm)
- Anion exchange chromatography
2) Phenyl-Sepharose CL-4B column (4 X 30 cm)
-Hydrophobic interaction chromatography

14. Results: Lab-scale isolation
Raw enzyme 12 g
- Q-Sepharose FF column (4 X 30 cm)
- Phenyl-Sepharose CL-4B column (4 X 30 cm)

15. Results: Lab-scale isolation
Step
Total protein
(mg)
Total activity
Specific activity
(U/ mg protein)
Yield (%)
Purification factor
Raw enzyme
1180
94700 80
100
1.0
Q-Sepharose
183
93000
508 98
6.4
Phenyl-Sepharose 38
39500
1052 42
13.2

16. Results: Pilot-scale isolation
Raw enzyme 300 g
- DEAE-Sepharose FF column (19 X 25.2 cm)
- Phenyl-Sepharose CL-4B column (11.3 X 17 cm)

17. Results: Isoelectric focusing

18. Specificity for oleic acid
At mole conversion 27
Specificity E is maximum
= prefer methyl oleate

19. Crystallization of GC-4 lipase
- crystallized in the presence of polyethylene glycol.
- size of crystals is dependent on the molecular weight of agent.
11 % PEG4000
at pH 4~5.5
5 % PEG20000
11 % PEG20000 at pH 4~5.5

20. X-ray
a = 53.1 Å
b = 83.5 Å
c = 57.8 Å
ß = 100°
∴ space-filling coefficient
= 2.3 Å3 per dalton
(based on MW)

21. Biological Engineering
Prof. Young Je Yoo
Bioseparation Engineering
: presentation
Example ,
Chang Hyeon Song
서울대학교
화학생물공학부
School of Chemical and
Biological Engineering

22. Example 10.1-1 Crystallization of adipic acid
90℃
35℃
solubilized
filtered
crystallized
Adipic acid 10kg
Water kg
10% water evaporated
0.05 kg adipic acid/kg water
Question ) Determine the weight of crystals recovered in this operation

23. Example 10.1-1 Crystallization of adipic acid
Question ) Determine the weight of crystals recovered in this operation
1. Set up two mass balance equation
water : water in = water in liquor + water evaporated
adipic acid : crystals in = crystals formed + remaining in mother liquor
2. Calculation of water in liquor
water in = water in liquor + water evaporated
13.1 kg = water in liquor kg×0.1
∴ water in liquor = kg
3. Calculation of crystals formed
crystals in = crystals formed + remaining in mother liquor
10 kg = crystals formed ×11.79 kg
crystals formed = kg

24. Example 10.1-2 Separation of soy sterols
sitosterol
25.4 %
sitosterol
13.5 %
Stigmasterol
74.6%
Stigmasterol
86.5 %
crystallization
sitosterol
3.4 %
2040 kg of
Stigmasterol & sitosterol
Stigmasterol
96.6 %
Question ) Determine the β value for this separation

25. Example 10.1-2 Separation of soy sterols
Question ) Determine the β value for this separation
1. Calculation of E value for stigmasterol and sitosterol
According to the equation,
2. Calculation of βvalue for this separation (purification)
β value for this separation is 9.6 and is quite large so separation is effective

26. 2011 Bioseparation engineering
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