Downstream Processing (Textbook Chapter 11 & Supplemental

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Downstream Processing (Textbook Chapter 11 & Supplemental Chapter 11) Topics Downstream processing needed to separate cell mass & other byproducts from the water solvent Major categories . Separation of solids & liquids • Filtration • Centrifugation . Drying of solids • Dryness limited by the dryness of air & adhesion of water to solid . Separation of soluble components • Adsorption – chromatography as special case

Updated: November 27, 2017 2 John Jechura ([email protected]) Typical process to manufacture enzymes

Updated: November 27, 2017 3 John Jechura ([email protected]) Need for downstream processing

Fermentation products are formed in dilute solutions . Remove water without harming the products formed . Byproducts? Biological products are labile (i.e., sensitive to temperature, solvents, etc.) Fermentation broths are susceptible to contamination

Updated: November 27, 2017 4 John Jechura ([email protected]) General processing schemes

1. Cell removal 2. Cell disruption & cell debris . If the cells are the product then removal little additional processing may . Required if product is within the be required cells . Removal of the cells may make • Cell debris must then be the separation of liquid species removed easier . Option . Options.. • High pressure homogenization • Filtration • Microfiltration • Centrifugation

Updated: November 27, 2017 John Jechura ([email protected]) General processing schemes (cont.)

3. Primary isolation 4. Product enrichment . Remove components that are very . Separation of product for high different from the product concentrations • May not be selective but . Option… overall help the efficiency of • Chromatography other separation . Options… 5. Final isolation • Solvent extraction . Depends on product, phase, & final purity • Two-phase liquid extraction . Options… • Adsorption • Ultrafiltration (liquid) • Precipitation • Crystallization followed by • Ultrafiltration centrifugation, filtration, and drying (solids)

Updated: November 27, 2017 John Jechura ([email protected]) Generalized downstream processing schemes

Updated: November 27, 2017 7 John Jechura ([email protected]) Water Content & Drying

Updated: November 27, 2017 John Jechura ([email protected]) Drying

Remove relatively small amounts of residual water and/or solvent . May be necessary to minimize chemical or physical degradation Relative terms – “dry” may range form 0 to 20 wt% water Energy intensive operation

Updated: November 27, 2017 9 John Jechura ([email protected]) Water Content of Air

Absolute humidity defined as the ratio of the mass of water in dry air

mmWW Humidity wW mmmair W air Relative humidity is the actual amount of water compared to the maximum amount at the same pressure & temperature

pW Relative humiditysat 100% where pPWW y pW

May also discuss properties of “wet bulb” temperature, …

Updated: November 27, 2017 10 John Jechura ([email protected]) Water Content of Air

Example: What is the wet bulb temperature for air @ 80oF & 60% relative humidity? Answer: 70oF

Ref: GPSA Data Book, 13th ed. Updated: November 27, 2017 11 John Jechura ([email protected]) Water Content of Solids

Updated: November 27, 2017 12 John Jechura ([email protected]) Mechanisms of Drying

Moisture transport in solids . Molecular diffusion of liquid water . Capillary flow of liquid water within porous solids . Molecular diffusion of vapor evaporated within the solid . Convective transport of vapor evaporated within the solid When the drying rate is controlled by the vaporization of water from the surface then the heat transfer & mass transfer effects are balanced

hAsh T a T NkaXX cGa Hvap Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 13 John Jechura ([email protected]) Drying Time

Can define the rate of drying to the water content of the solids

dX mw Nm s where X  dt ms

ms dX nA  Adt

Over the period of constant rate of drying

X ms Nm cs   t  XX01  tN c

Updated: November 27, 2017 14 John Jechura ([email protected]) Drying Example – Humidity of Air

Dry sodium benzyl penicillin to a moisture content of 20 wt% (wet basis). Use air at 25oC & 1 atm pressure in a fluidized bed drier. Assume equilibrium (open circles) Target dry basis water content: 20 kg water X 0.25  25 0 80 100 kg dry mass Required relative humidity – 65% (from chart) Mole fraction water – water’s vapor pressure @ 25oC is 3.17 kPa p 0.65 Psat 0.65 3.17 y w  0.020 w PP 101.28

Updated: November 27, 2017 15 John Jechura ([email protected]) Drying Time Example

Want to dry 10 kg washed filter cake (wet basis) from 15% to 8% water (both wet bases) under constant drying conditions . Drying area 1.2 m² . Air temperature 35oC & surface temperature of solids 28oC . Heat transfer coefficient 25 J/m²·sec·oC . Drying time? From initial conditions:

mmww,0 ,0 0.15 mmws,0  1.5 and  8.5 mmws,0  100 1.5 X 0.176 0 8.5

Updated: November 27, 2017 16 John Jechura ([email protected]) Drying Time Example (cont.)

Final conditions

mw ,1 0.08 8.5 0.08mw ,1  0.74 mw,1 8.5 1 0.08 0.74 X 0.087 1 8.5 Rate of drying

hAsh T a T 25 1.2 35 28 5 kg N c 3 8.62 10 Hvap 2435.4 10 sec Drying time 8.5 t 0.176  0.087  8770 sec  2.4 hr 8.62 105

Updated: November 27, 2017 17 John Jechura ([email protected]) Freeze Drying

Wet solids are frozen then exposed to vacuum – water directly sublimes from solid to vapor state Steps . Freezing . Primary drying (sublimation) . Secondary drying (desorption)

Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 18 John Jechura ([email protected]) Cell Removal

Updated: November 27, 2017 John Jechura ([email protected]) Filtration

Solid particles removed from a fluid suspension by forcing the fluid through a filter media which holds back the solid particles Ease of filtration depends on the properties of the solid & fluid . Crystalline incompressible solids fairly easy to filter, … . Fermentation broths not so much • Small particle size & gelatinous nature May use filter aids to improve efficiency of filtration

Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 20 John Jechura ([email protected]) Filtration Equipment

Plate & frame filter press . Batch operation . Accumulates solids. Must be opened & cleaned . https://www.youtube.com/watch?v=M4wBd1_CvNw . https://www.youtube.com/watch?v=aOQPr7Eomek Rotary drum filters . Can be operated continuously . Can be operated at pressure with controlled atmosphere . https://www.youtube.com/watch?v=zSflxzE0nIo Horizontal belt filters . Continuous operation . Typically pull a vacuum below the filter . https://www.youtube.com/watch?v=6voXE1HxYsY

Updated: November 27, 2017 21 John Jechura ([email protected]) Filtration Theory

Rate of filtration depends on the pressure drop across the cake and the filter media 1 dV PPP f  Adt mcVcV solids  f  f fm R fm R f  A  A  A

. Rm is the resistance due to the filter media & is typically negligible compared to the resistance due to the cake . Various correlations to relate  (the specific cake resistance) to the cake properties • Compressible cake Rigid particles 2 s Kav 1   P  3 p

Updated: November 27, 2017 22 John Jechura ([email protected]) Filtration Theory

Can integrate the reciprocal form…

1 dVfffmPdt V R Acf  AdtcVf dVf  AP P fm R A VV t ffVR A dt cffm dV  dV fff  00AP 0 P 2 ffcV  fm R At Vf 2 AP P Linearized form

t ffmcR  2 Vf VAPAPf 2 

Updated: November 27, 2017 23 John Jechura ([email protected]) Filtration Example

Can filter 30 mL broth from penicillin fermentation on 3 cm² filter in 4.5 min using 5 psi pressure drop. Neglect media resistance & use s=0.5: s 0.5  PP   

Lab conditions to “tune” filter equation

0.5 0.5 2 0.5 t  Pc2234.55 AtP2       psi0.5 min f Vc 0.201 22ff 2 2 VAPff2cm V 30

Filter 500 L broth in 1 hour using 10 psi pressure drop

0.5 t  Pc  c ffVAV 11500 cm22 1.15 m 2 ff0.5 VAPf 2  2tP

Updated: November 27, 2017 24 John Jechura ([email protected]) Filtration Example (cont.)

What if the cake is much less compressible? Let’s resize with s=0.1:

s 0.1  PP    Lab conditions to “tune” filter equation

0.5 0.1 2 0.9 t  Pc2234.55 AtP2       psi0.9 min f Vc 0.383 22ff 2 2 VAPff2cm V 30 Filter 500 L broth in 1 hour using 10 psi pressure drop

0.1 t  Pc  c ffVAV 10000 cm22 1.0 m 2 ff0.9 VAPf 2  2tP Compressible cake will require more surface area

Updated: November 27, 2017 25 John Jechura ([email protected]) Centrifugation

Increase the “gravitational” field experienced by particles to promote faster settling times Stoke’s law:

pL222  pL udgudrgpCp 18LL 18 Bioprocess Engineering Principles, 2nd ed Z is the ratio of the acceleration due Pauline Doran, Elsevier Science & Technology to the centrifuge relative to the acceleration due to gravity (the g-force), 2r/g . Industrial centrifuges can achieve 16,000, small lab centrifuges up to 500,000

Updated: November 27, 2017 26 John Jechura ([email protected]) Equipment – Tubular Bowl Centrifuge

http://slideplayer.es/slide/7225569/24/images/11/EQUIPOS+DE+CENTRIFUGACI%C3%93N.jpg

Updated: November 27, 2017 27 John Jechura ([email protected]) Equipment – Disc Stack Bowl Centrifuge

Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 28 John Jechura ([email protected]) Equipment – Disc Stack Bowl Centrifuge

Can be run in batch or continuous modes . Batch mode – equipment stopped, opening up, & solids removed . Continuous – either side discharge or movement of solids out the top • Disadvantage – the solids must remain wet to allow them to move Video: Alfa Laval PX Disc Stack Centrifuge . https://www.youtube.com/watch?v=

ZmTMz4VkAwI http://www.alfalaval.com/globalassets/images/products/se paration/centrifugal-separators/disc-stack-separators/sx- bowl_440x360.png

Updated: November 27, 2017 29 John Jechura ([email protected]) Equipment – Decanter Separator

Andritz 2-phase decanter with CIP (clean in place) . Video: https://www.youtube.com/watch?v=OqEODWcJwnY

http://www.gn-decanter-centrifuge.com/mud/wp-content/uploads/2012/09/decanter-1.png

Updated: November 27, 2017 30 John Jechura ([email protected]) Centrifuge Performance

Relate the performance based Disc stack bowl centrifuge on the centrifuge’s cross- sectional area to that of a 212 N   rr33  gravity separator 3tang  21 Q  Tubular bowl centrifuge 2ug

Two centrifuges that operate 222bbr2  3rr22   with equal effectiveness 2gg21

QQ 12 12

Updated: November 27, 2017 31 John Jechura ([email protected]) Centrifuge Example

Use continuous disc stack centrifuge @ 5000 rpm to recover 50% baker’s yeast cells at 60 L/min. . Found that at constant rpm the solids recovery factor is inversely proportional to flow rate At 5000 rpm what flow rate will give 90% solids recovery?

10.5Y1 YQQ21  60  33.3 L/min QY 2 0.9 What rpm necessary to get 90% recovery @ 60 L/min? . At equal effectiveness by adjusting flowrate:

QQ Q 33.3 12 2 2 0.556 12  1Q 160 . By adjusting rotational speed: 2 22  1 0.5562 2 6710 rpm 110.556 Updated: November 27, 2017 32 John Jechura ([email protected]) Separation of Soluble Components

Updated: November 27, 2017 John Jechura ([email protected]) Separation of Soluble Components

Liquid-liquid extraction Precipitation . Followed by solid/liquid separation & possibly drying Adsorption . Chromatography a special case Membrane separations . Dialysis (separation by selective membrane) . Reverse Osmosis – transport of water from low to higher concentration . Ultrafiltration & microfiltration . Electrodialysis Crystallization

Updated: November 27, 2017 34 John Jechura ([email protected]) Absorption vs Adsorption

Absorption Adsorption

From Prof. Art Kidnay

Updated: November 27, 2017 35 John Jechura ([email protected]) Adsorption

Chemical species cling to the surface of the adsorbent allowing species that don’t adsorb to pass on by Equilibrium relationship between components in liquid & to surface

* 1/ n ** *CKCLSm L L CKCSFL (Freundlich) C S * (Langmuir) 1 KCLL Differential material balance in a packed bed column CC C u LL 1  S zt  t Including overall mass transfer coefficient dC CC S Ka C C** uLL1 Ka C C dtLLL z t LLL

Updated: November 27, 2017 36 John Jechura ([email protected]) Adsorption (cont.)

Approximate steady flow through fixed-bed adsorption column as a steady-state moving-bed operation dC uKaCCL 1   * dz LLL Can integrate to determine the height of a column for a particular degree of separation

H u CL0 dC Hdz L 1 Ka C C* 0 LLLCL Need the equilibrium relationship to fully integrate

Updated: November 27, 2017 37 John Jechura ([email protected]) Adsorption (cont.)

Mass balance considerations on the adsorbent provides an operating line & relates the concentrations in the liquid & on the solid

* FCLS0000 BC 

B CL CL0 * FCSS C0 F is the liquid volumetric flow (uA) & B is the resin’s volumetric flow

Updated: November 27, 2017 38 John Jechura ([email protected]) Adsorption (cont.)

Fixed beds can be thought to have three zones: . Saturated zone . Adsorption zone – also called the mass transfer zone . Virgin zone We would stop flow before the adsorption zone gets to the exit (i.e., before the virgin zone goes completely to zero) If we let the zones travel through the column then we would get a typical breakthrough curve

Updated: November 27, 2017 39 John Jechura ([email protected]) Adsorption (cont.)

Updated: November 27, 2017 40 John Jechura ([email protected]) Adsorption Example

Separate cephalosporin on to an ion-exchange resin . Bed is 4 cm diameter . Resin density is 1.3 g/cm³ & packed to 0.8 cm³ resin per cm³ bed

. Feed solution CL0 = 5 g/L . Broth superficial velocity u = 1.5 m/h & broth to resin feed ratio F/B = 10 * 1/2 -1 . Equilibrium relationship CS = 25 (C L) & mass transfer KLa = 15 h

. What is the height necessary to get CL = 0.2 g/L?

B CL CL0 1 * *  CCSL 10 and C S00  10 C L FCSS C0 10

1/ 2 112 CC**25 C * C *2 10 C 0.16 C 2 SL L625 S 625 L L

Updated: November 27, 2017 41 John Jechura ([email protected]) Adsorption Example (cont.)

Substitute into integral equation for the height

u CL0 dC H  L 1Ka C  C* LLLCL 1.5 5 dC  L  2 0.8 150.2 CCLL 0.16 5   1.5 CL  ln 0.8 15 1 0.16C  L 0.2 0.6 m Amount of resin needed? 2 D2 4 mH 1.3 60  980 g resin resin 44

Updated: November 27, 2017 42 John Jechura ([email protected]) Chromatography

Form of adsorption where chemical species adsorb & desorb at different rates . Those species that interact most strongly with the stationary phase will elute later than those species that weakly interact . Separation of a “pulse” of mixture • Followed up by more solvent • Recovered material separated from the other species of the original mixture but diluted with more solvent • Can obtain multiple products from each pulse We’ll focus on chromatography as a batch process . Continuous & semi-continuous options • Moving Bed Chromatography • Simulated Moving Bed Chromatography

Updated: November 27, 2017 43 John Jechura ([email protected]) Chromatography

Concentrations in the separation of mixture with 3 solutes Species that adsorb onto the stationary phase the least will elute the earliest

Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 44 John Jechura ([email protected]) Chromatography Methods

Adsorption chromatography (ADC) Affinity chromatography (AFC) . Controlled by weak van der Waals . Based on specific chemical forces & steric interactions interactions between solute molecules & ligands Liquid-liquid partition . Possible to get very high resolution chromatography (LLC) (degree of separation) but expensive Ion-exchange chromatography (IEC) sorbent . Adsorption of ions on resin particles Hydrophobic chromatography (HC) by electrostatic forces . Widely used, especially for protein High-pressure liquid chromatography recovery (HPLC) Gel Permeation Chromatography – Size Exclusion Chromatography . Smaller molecules wander into the pores of the solid

Updated: November 27, 2017 45 John Jechura ([email protected]) Chromatography Separation of Peaks

The species will move through the chromatography column slower than the rest of the solvent – differential migration Given enough column length the species will elute from column with little to no overlap

Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 46 John Jechura ([email protected]) Chromatography Zone Spreading

Elution bands spread out so each solute takes time to pass across the end of the column Factors that lead to zone spreading . Axial diffusion . Eddy diffusion . Local nonequilibrium effects Minimize zone spreading by improving mass transfer more closely approach equilibrium between the phases . Good: increasing the particle surface area per unit volume . Bad: increasing liquid flow rate

Updated: November 27, 2017 47 John Jechura ([email protected]) Chromatography Separation of Peaks

Descriptors

. Ve – volume of eluting solvent needed to carry the solute through column

. Vo – void volume in the column (·Vempty) • In gel chromatography (size exclusion) there is also a volume term for the interstitial volume only accessible by the smaller molecules

. ki – capacity factor for species “i”

VVei,  o ki  Vo .  – selectivity between components “1” & “2” k  1 k2

Updated: November 27, 2017 48 John Jechura ([email protected]) Chromatography Separation of Peaks

Descriptors based on solute peaks as normal distributions

. ymax,i – Maximum concentration of eluting peak

. tmax,i – Time at which peak elutes .  – Standard deviation of peak shape

2 tt  yyexp max,i iimax, 2  2 tmax,i

Updated: November 27, 2017 49 John Jechura ([email protected]) Chromatography Separation of Peaks

Descriptors based on solute peaks as normal distributions

. Yi –Yield of solute (ratio amount recovered to total in original mixture & recoverable) t Vy dt   i 0 Yti,   Vy dt   i 0

. Cumulative yield for normal distribution   1tt max,i  Y 1erf  ti, 2 2 t  max,i  . Incremental yield the difference between the cumulative yields @ two times

Updated: November 27, 2017 50 John Jechura ([email protected]) Chromatography Separation of Peaks

Descriptors based on solute peaks as normal distributions . Excel has functions for normal Gaussian distributions – scaled so that the cumulative amount is 1

Cumulative: NORM.DIST( x, mean, standard_dev, TRUE )

Differential: NORM.DIST( x, mean, standard_dev, FALSE )

Updated: November 27, 2017 51 John Jechura ([email protected]) Chromatography Separation of Peaks

Descriptors based on solute peaks as normal distributions . 2 is essentially a Peclet number – ratio of flowrate to reaction rate uVA / 2  kLaa kL

• Value depends on the rate controlling step ud2 2 Internal diffusion control L ud1/ 2/ 3 / 2 2 External film control L ud2 2 Taylor diffusion DL

Updated: November 27, 2017 52 John Jechura ([email protected]) Chromatography Separation of Peaks

Descriptors based on HETP concept . N – Number of theoretical stages (theoretical plates)

2 Ve N  16 w . HETP – Height of a theoretical plate (L = length of column) L H  N

Bioprocess Engineering Principles, 2nd ed Pauline Doran, Elsevier Science & Technology

Updated: November 27, 2017 53 John Jechura ([email protected]) Chromatography Separation of Peaks

Resolution – ratio distance of solute to solvent . Resolution between two adjacent peaks

tt R  max,ji max, S 1 tt 2 wj,, wi tt 2VV max,ji max, ee,2 ,1 1 ww 444tt21 2 jjmax, i max, i . Assume the spreading is about the same (w2 = w1) & use concept of number of theoretical stages…

VVee,2 ,1111 VV ee ,2 ,1  k2 RNNS   wV44e,2  k 2  1

Updated: November 27, 2017 54 John Jechura ([email protected]) Chromatography Material Balance

Assume downward flow through a fixed-bed chromatography column – adsorbed species come to very quick equilibrium with the adsorbent Following a small pulse (moving reference system)

CCLLCS A  0 xVV Integrated form… V x  AMC f L Uses equilibrium relationship…

CMCSLf and f C L ddC f/ L

Updated: November 27, 2017 55 John Jechura ([email protected]) Chromatography Material Balance Example #1

Chromatography column 10 cm long with cross sectional. Packed with adsorbent, =0.35 & M=50 mg adsorbent per mL column volume. Adsorption isotherm given by:

3 f0.2CCLL Determine position of the solute band & solvent front when ΔV=250 mL &

CL=0.05 mg/mL . Solvent front: V 250 cm3 x  71.4 cm A 10 cm2  0.35 . Solute:

2 2 fCCLL 0.6 f C L 0.6 0.05  0.0015 V 250 x   58.8 cm  AMC fL 10 0.35  50 0.0015

Updated: November 27, 2017 56 John Jechura ([email protected]) Chromatography Material Balance Example #2

Ion-exchange chromatography to purify 20 g of Protein A.

Operate column @ 20 cm/hr. Peak exits at 80 min (tmax) with 10 min standard deviation ( tmax). How long must the column be run to get 95% yield?

 1tt max,i Y erf 1 ti, 2 2 t max,i 1t  80 erf 1 0.95 2 210

. By iterative solution t = 96.4 min

Updated: November 27, 2017 57 John Jechura ([email protected]) Chromatography Material Balance Example #3

Let’s double the flowrate to 40 cm/hr. Assume the peak spreading depends on Taylor diffusion. How long must the column be run to get 95% yield?

. Adjust the tmax &  parameters

tmax,2 u1 20 cm/hr  0.5 tmax,2  40 min tumax,1 2 40 cm/hr 2 22u 10 2 2220.177 21   11u 80 . By iterative solution t = 51.6 min 1t  40 Yti, erf 1 0.95 2 20.17740

Updated: November 27, 2017 58 John Jechura ([email protected]) Material Balance Examples #2 & #3

Example #3 requires relatively more time because of the change in peak shape at the two velocities

Updated: November 27, 2017 59 John Jechura ([email protected]) Summary

Updated: November 27, 2017 John Jechura ([email protected]) Summary

Downstream processing needed to separate cell mass & other byproducts from the water solvent Major categories . Separation of solids & liquids • Bulk separation – chemical species remain in their respective phases • Types o Filtration o Centrifugation . Drying of solids • Dryness limited by the dryness of air & adhesion of water to solid . Separation of soluble components • Adsorption – chromatography as special case

Updated: November 27, 2017 61 John Jechura ([email protected])