Bioseparation Engineering Crystallization

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.

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

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:

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

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

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:

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

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:

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

Example 3.12 Briefly analyze the large-scale purification and crystallization of lipase from Geotricbum candidum. 2011.11.14 심세나

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

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

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)

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

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)

Results: Isoelectric focusing

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

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

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 61600 MW)

Biological Engineering 2011-2 Prof. Young Je Yoo Bioseparation Engineering : presentation Example 10.1-1 , 10.1-2 Chang Hyeon Song 2011-21042 서울대학교 화학생물공학부 School of Chemical and Biological Engineering

Example 10.1-1 Crystallization of adipic acid 90℃ 35℃ solubilized filtered crystallized Adipic acid 10kg Water 13.1 kg 10% water evaporated 0.05 kg adipic acid/kg water Question ) Determine the weight of crystals recovered in this operation

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 + 13.1 kg×0.1 ∴ water in liquor = 11.79 kg 3. Calculation of crystals formed crystals in = crystals formed + remaining in mother liquor 10 kg = crystals formed + 0.05×11.79 kg crystals formed = 9.41 kg

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

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

2011 Bioseparation engineering The end - Thank you -