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ERT 320 Bio-Separation Engineering Semester 2 2012/2013 Huzairy Hassan School of Bioprocess Engineering UniMAP.

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Presentation on theme: "ERT 320 Bio-Separation Engineering Semester 2 2012/2013 Huzairy Hassan School of Bioprocess Engineering UniMAP."— Presentation transcript:

1 ERT 320 Bio-Separation Engineering Semester 2 2012/2013 Huzairy Hassan School of Bioprocess Engineering UniMAP

2 Product Formulation and Finishing Operations

3 “ Ability to evaluate process and polishing important parameters involved in purification and polishing steps of bio-products for selected bio-separation units ” Product Formulation and Finishing Operations

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5 Introduction Product Formulation In pharmacy, a formulation is a mixture or a structure such as a capsule, a pill, tablet, or an emulsion, prepared according to a specific procedure (called a “formula”). Formulations are a very important aspect of creating medicines (or foods, cosmetics, fuels, fertilizer, etc), since they are essential to ensuring that the active part of the drug is delivered to the correct part of the body, in the right concentration, and at the right rate (not too fast and not too slowly).

6 Introduction Product Finishing Surface finishing is a broad range of industrial processes that alter the surface of a manufactured item to achieve a certain property. Finishing processes may be employed to: improve appearance, adhesion or wettability, solderability, corrosion resistance, tarnish resistance, chemical resistance, wear resistance, hardness, modify electrical conductivity, remove burrs and other surface flaws, and control the surface friction. In bio-separation principle, product finishing is the polishing process (normally the last stage in bio-separation scheme) to obtain a great surface structure of final product, for example in crystalline or powder form.

7 CRYSTALLIZATION

8 A process of producing crystals from homogeneous phase, i.e. homogeneous phase is always a solution Crystallization

9 Solid particles are formed from a solution, highly structured in 3-D arrays (space lattices) Occur at low level of supersaturation and nucleation rates Wide range, medium to high solubility Precipitation Solid particles are formed from a solution, has poor defined morphology (amorphous) Occur at high level of supersaturation and nucleation rates Low solubility

10 CRYSTALLIZATION

11 -Crystallization is capable of producing bioproducts at very high purity (say, 99.9%), and is considered to be both a polishing step and purification step. -Polishing: a process needed to put the bioproduct in its final form for use. -Ex: Antibiotics; normally the final form must be in crystalline. -May replace other more expensive purification steps such as chromatography.

12 Crystallization Principle 1) Crystals -When allowed to form freely, crystals appear as polyhedrons, or solid formed by plane faces,. -Although the relative sizes of the faces differ, the angles made by the corresponding faces of the same material do not vary  determine the characteristics of the substance. -Difference on crystal relatives sizes, results in a variety of crystal shapes called modification of habit.

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14 Crystallization Principle 2) Nucleation -The generation of ultramicroscopic particles in the process of nucleation is the sum of contributions by primary nucleation and secondary nucleation. -Primary nucleation: occurs in the absence of crystals. -Secondary nucleation: is attributed to the influence of existing crystals.

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19 Primary nucleation:  Can be either homogeneous or heterogeneous  The rate of primary nucleation modeled by power law expression: Crystallization Principle No foreign particles present Foreign particles present B = no. of nuclei formed per unit volume per unit time N = no. of nuclei per unit volume k n = rate constant c = instantaneous solute conc. c* = solute conc. at saturation (c – c*) = supersaturation n = exponent typically 3 to 4.

20 Secondary nucleation (usually predominates) :  The rate of secondary nucleation: Crystallization Principle Shear nucleation Contact nucleation K 1 = rate constant M T = suspension density b = exponent typically 2. j= 1 (most probable value) Occurs as a result of fluid shear on growing crystal faces Occurs because of crystals colliding with each other and with impeller and other vessel internal surfaces

21 -As in precipitation, the solution must be supersaturated in order for particles to form crystal  (c – c*) in both Eqs. -Supersaturation must be above a certain value before nucleation will begin. Crystallization Principle

22 1 In Metastable region, the supersaturation is so low that nucleation will not start. 2 The supersaturation is raised to Labile region, nucleation can begin. At this point, crystals begin to grow, and the supersaturation decreases. Point is possible way of carrying out a crystallization. 3 If the supersaturation becomes too high, the nucleation rate will be too great, and an amorphous precipitate will result. Crystallization Principle

23 3) Crystal Growth -Crystal growth is the post-nucleation process in which the molecules in solution are added to the surface of existing crystals. -For designing a crystallizer, the useful relationship describing the rate of mass deposition, R during crystal growth is: W = mass of crystals per volume of solvent A = surface area of crystals per volume of solvent k G = overall mass transfer coefficient (depend on T, crystal size, hydrodynamic conditions, and presence of impurities.) g = the order usually between 0 – 2.5 (near unity is most common)

24 Crystallization Principle 3)Crystal Growth -Also, to express overall linear growth rate as ( delta L law ): L = characteristic single dimension of the crystal, such as length  It is shown that geometrically similar crystals of the same material grow at the rate described by Eq 9.2.4.

25 Crystallization Principle 3)Crystal Growth -Crystal growth is actually a process that consists of two steps in series. (1) Solute molecules must reach the crystal surface by means of diffusion. (2) At the surface, the solute must be integrated into the crystal lattice.

26 Crystallization Principle 3)Crystal Growth Where c i = concentration at the interface between liquid and solid phase. k d & k r = mass transfer coefficients. -When the exponents are unity, combining Eqs (9.2.3), (9.2.5) & (9.2.6) gives: Thus, if surface integration is very fast compared with bulk diffusion, then, and

27 Crystallization Principle 4) Crystallization Kinetics from Batch Experiment

28 Process Crystallization of Protein Why use crystallization for purification of protein? 1)One or more expensive chromatography steps possibly can be eliminated. 2)Relatively inexpensive to carry out, since costly adsorbents are not required (as in chromatography) 3)Proteins crystals often can be stored for long periods at low T without being degraded or denatured after the addition of stabilizing agents such as ammonium sulfate, glycerol, or sucrose.

29 Process Crystallization of Protein How to crystallize protein? -Adjustment of pH to isoelectric point ( isoelectric precipitation ) -Addition of organic solvents -Addition of salts (salting-out) -Addition of non-ionic polymers. Another way to crystallize protein? -Reduction of ionic strength by dialysis or diafiltration, which relies on the limited solubility of many proteins at low ionic strength (reverse of “salting-in” effect). -Key strategy: move the system very slowly to a state of min. solubility of the desired protein until a limited degree of supersaturation is reached.

30 Process Crystallization of Protein Study case: 1)Alcohol oxidase enzyme - from Pichia pastoris yeast - grown in a 100-L pilot plant fermentor - crystallized by lowering ionic strength by diafiltration with deionized water - crystallization was preceded by lysis of yeast cells, diafiltration with microfilter to obtain alcohol oxidase in the permeate, and concentration and then diafiltration with an ultrafilter.

31 Process Crystallization of Protein Study case: 2)Ovalbumin - crystallized in the presence of conalbumin and lysozyme by addition of 2.5 µm seed crystals to a solution (600 mL) that had been made supersaturated by slowly adding ammonium sulfate solution. - the supersaturation was kept in the metastable region to avoid nucleation. - the crytals growth rate is a second-order dependence on ovalbumin supersaturation, and the presence of the other 2 proteins did not affect the growth rate constant.

32 Crystallizer Scale-up & Design Impurity is a problem in crystallization, can be in the following locations in crystal sample: 1)Deposited on crystal surfaces due to incomplete removal of impure mother liquor. 2)Trapped within voids between separate crystals in materials that agglomerate. 3)Contained in inclusions of mother liquor within individual crystals. 4)Distributed throughout the crystals by molecular substitution at the lattice sites.

33 Crystallizer Scale-up & Design  It is recommended that geometric similarity and constant power per volume be used in scale-up of crystallizers.  For turbulent flow in vessels (Reynolds number > 10,000), constant power per volume:  For an agitated tank, the Reynolds number is: N i = impeller rotation rate d i = impeller diameter N i = rev. per min ρ = fluid density µ = fluid viscosity

34 Crystallizer Scale-up & Design  2 additional strategies also used in Scale-up: 1)Maintaining constant impeller tip speed; 2)Scale-up at the minimum speed required for particle suspension;

35 Example 9.2: Scale-up of Crystallization based on Constant Power per Volume It is desired to scale up a batch crystallization of an antibiotic based on experiments with a 1-L crystallizer. The use of a 3 cm diameter impeller at a speed of 800 rpm led to good crystallization results. For maintaining power per volume constant upon scale-up to 300 L, what should be the diameter and speed of the large-scale impeller? The solvent has the same density and viscosity as water.

36 Practice 1: Solubility in Crystallization Given solubility data for Phthalic acid as follows: 18 g / 100 mL water at 100 ºC 0.54 g / 100 mL water at 15 ºC A student obtained 3.2 g of crude phthalic acid. After recrystallization and drying, 2.5 g of pure acid was obtained. Calculate: a)The percent recovery b)Amount of water needed for recrystallization c)Amount of product lost to the filtrate.

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