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Recovery and Purification of Bio-Products (Chapter 11, M. Shular, textbook) - Strategies to recovery and purify bioproducts - Solid-liquid separation - Cell disruption - Separation of soluble products - Finishing steps for purification Recovery and Purification of Bio-Products - Strategies to recovery and purify bio-products Unique characteristics of bioseparation products: - the products are in concentration in an aqueous medium. e.g therapeutic protein 0.01mg/l. - The products are usually - There is a - The products can be inclusion bodies. sensitive. of products to be separated. , often as insoluble - The physical and chemical properties of products are similar to contaminants. - Extremely purity and homogeneity may be needed for human care. Recovery and Purification of Bio-Products - Strategies to recovery and purify bio-products Fementer Solid-liquid separation Cell products Cells Supernatant Cell rupture Recovery Cell debris Purification Crystallization and drying Recovery and Purification of Bio-Products - Strategies to recovery and purify bio-products Fementer Solid-liquid separation Cell products Cells Supernatant Cell rupture Recovery Cell debris Purification Crystallization and drying Recovery and Purification of Bio-Products - Liquid and solid separation - solid particles: mainly cellular mass, specific gravity 1.05-1.1. size (diameter): bacterial cells: 0.5-3 µm yeast cells: 5 -10 µm mold: 5-15 µm in diameter and 50-500 µm in length animal cells: 10 µm plant cells: 20 µm Methods: . Liquid-Solid Separation Filtration Physical separation of solid particles from liquid or gas. a porous medium: allow fluid to pass through solid particles to be retained. Filter cake Filter medium Slurry flow Filtrate Liquid-Solid Separation Filtration • . – Particle size: greater than 10 µm, yeast, mold, animal or plant cells. i.e. mycelium separation for antibiotics production or waste water treatment • . – Particle size: 0.1 - 10 µm, bacterial and yeast cells. • . – Size: 10-200 Å, Cell debris, macromolecules Rotary Vacuum Filter A rotary vacuum filter is a continuous filter partially submerged in the slurry. - A drum is covered with a filter medium. -Vacuum is applied to within the drum - As the drum rotates, the solid constituent is separated by retained on the porous medium The liquid is drawn through the cake into the inner filtrate pipes. Each revolution consists of cake formation, cake washing (if required), drying and cake discharge. http://www.solidliquid-separation.com/ VacuumFilters/vacuum.htm http://www.komline.com/Products_Services/ Filtration/RotaryVac.html Rotary Vacuum Filter The rate of filtration (the flow of the filtrate) for (vaccum) filtration operation can be determined by (Bennet &Myers, Momentum, Heat and Mass Transfer, 1974, p221, the equation is from the mass balance of the cake.) gc pA dV dt (rm rc ) where V is the volumeof filtrate(m3 ), A is thesurface area of thefilter(m2 ), p is thepressure drop through he t cake and thefiler medium (N/m2 ), is the viscosityof thefiltrate(kg/m - s), rm is theresistanceof thefilter medium (m-1 ), rc is theresistanceof thecake (m-1 ). g c is 9.8 kg m kg f s 2 Rotary Vacuum Filter T hecake resistancerc is given by W CV rc , A A where W CV is theaveragespecificresisitance of thecake (m/kg). W is the tot alweight of thecake on filt er(kg). C is the weight of thecake deposit per unit volume of filt rate(kg/m3 ). T hen t heequation of thefilt rationrate wit h constantA becomes g c pA dV CV dt (r ) m A Rotary Vacuum Filter Assuming incompressible cake: constant α & constant pressure. Integrating the following equation (V at t, V=0 at t=0) yields Agc p dV CV dt (r ) m A V 2 2VV0 Kt (Rut h equat ion) rm V0 A, C 2 A2 pgc K C V0 and K can be determinedby experimental data. Rotary Vacuum Filter T heexpermenta l data pointsV ~ t are obtained. Ruth equation can be rearrangedas t 1 (V 2V0 ) V K t Plot versus V, theslopeis 1/K, V and theinterceptis 2V0 / K . t/V V0 , K determineparameters: rm , , or p V Rotary Vacuum Filter To design a scaled-up rotary vaccum filter If given a total volume of fermentation broth Vb and required time tb to complete the filtration task at the large scale, determine the filter surface area. Based on the results from the smaller filter (incompressible cake: same α, medium & pressure drop): Vb 2 2Vb V0 Kt b (Ruth equation) 2A 2 rm V0 Ab , K b pg c C C Substitute V0 and K in the Ruth equation by the above two expression s, A b , the surface area of the bigger filter,can be determined. Rotary Vacuum Filter For incompressible cake: constant α. If filtration rate is constant, dV q0 (constant ) V q0t , V 0 at t 0 dt Agc p dV q0 dt r CV m A p K1q0 t K2q0 2 rm C K1 , K2 2 gc A gc A Liquid-Solid Separation Filtration • Rotary vacuum filtration – Particle size: greater than 10 µm, yeast, mold, animal or plant cells. i.e. mycelium separation for antibiotics production or waste water treatment • Microfiltration – Particle size: 0.1 - 10 µm, bacterial and yeast cells. • Ultrafiltration – Size: 10-200 Å, Cell debris, macromolecules (antibiotics, proteins, polysaccharides) Liquid-Solid Separation Filtration • Microfiltration & Ultrafiltration Use membrane as porous medium for filtration. Challenge: gel formation on the surface of membrane. Solution: cross-flow (tangential flow filtration) Pressure P1 Feed in Pressure P2 Feed out Liquid-Solid Separation Filtration • Centrifugation - Particle size: 100-0.1 µm - more expensive than filtration - limited for scale-up - drive force: centrifugal force Example The following data were obtained in a constant-pressure unit for filtration of a yeast suspension: t(min) 4 20 48 76 120 V(L) 115 365 680 850 1130 Characteristics of the filter are as follows: A=0.28m2, C=1920kg/m3, μ=2.9X10-3 kg/m-s, α=4m/kg Determine: a) The pressure drop across the filter. b) The filter medium resistance. c) The size of the filter for the same pressure drop to process 4000L of cell suspension in 20 min.