Chapter 11: Product Recovery and Purification

Home , Ultrafiltration

David Shonnard Department of Chemical Engineering Michigan Technological University

1 David R. Shonnard Michigan Technological University

Presentation Outline:

G Overview of Bioseparations

G Separation of Insoluble Products

G Primary Isolation / Concentration of Product

G Purification / Removal of Contaminant Materials

G Product Preparation

2 David R. Shonnard Michigan Technological University

1 Introduction to Bioseparations

Characteristics of Bioseparations vs Chemical Separations

Characteristics Biochemical Chemical Environment Aqueous Media Organic Media Concentration Range v. Dilute Product Concentrated Product Temperature Sensitivity Product Vulnerable Product Not Vulnerable

Traditional chemical separations are unsuitable or must be augmented

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Biochemical Separations Technologies

Technologies

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

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2 Figure 11.2 Major Steps in Separating a Protein Product

Fermentation Filtration Centrifugation Cell Removal and Coagulation / 1. Concentration Flocculation Separation of Insoluble Ultrasonic Vibrators Products, Cell Disruption French Press Product Bead or Ball Mills Concentration (0.2 - 2.0%) Removal of Cell Ultracentrifugation Debris Ultrafiltration

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Figure 11.2 Major Steps in Separating a Protein Product

Protein 2. Precipitation or Aqueous Two- Product Other methods for Phase Extraction Isolation, product isolation: (removal of Liquid-liquid extraction water) Adsorption (1 - 10%) Ultrafiltration

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3 Figure 11.2 Major Steps in Separating a Protein Product

Chromatographic Purification Other methods for product purification: 3. Reverse Osmosis Purification / Solvent Cross-flow Removal of Precipitation Ultrafiltration Contaminant Electrophoresis Chemicals, Electrodialysis (50 - 80%) Dialysis

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Figure 11.2 Major Steps in Separating a Protein Product

4. Product Other methods: Preparation, Lyophilization Crystallization (90 - 100%)

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4 1. Removal of Insoluble Products Rotary Vacuum Filtration

• coagulation agents/ (filter aids) added

• vacuum applied ('P)

“Bioprocess Engineering: Basic Concepts” Cell Solution Shuler and Kargi, Prentice Hall, 2002

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1. Removal of Insoluble Products Rotary Vacuum Filtration

Filter Equation Filter area dV g 'P A = c Viscosity of filtrate (water) dt (r + r)P Volume m c Filter cake resistance filtered Filter medium resistance (a constant) C V r = D c A C = wt. of cells per volume filtrate (g/L) ce D = average specific resistance of filter (assume it is a constant)

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5 1. Removal of Insoluble Products Rotary Vacuum Filtration

Integrate Filter Equation: V=0 at t=0.

2 V + 2VVo = Kt Ruth Equation where

rm Vo = A D C § kg•m· 1 § 2 · ¨ 2 ¸ 2 A ' s K = ¨ ¸ P g gc = ¨ ¸ ©D C P¹ c ¨ N ¸ © ¹

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1. Removal of Insoluble Products Rotary Vacuum Filtration

Rearrange Ruth Equation

t 1 = (V+ 2V ) V K o §2 A2 · D = slope•¨ ¸'P g ©C P ¹ c

1 §2 A2 · “Bioprocess Engineering: = ¨ ¸'P g Basic Concepts” K ©C P ¹ c Shuler and Kargi, Prentice Hall, 2002 D K K C Vo = y- intercept• rm = y- intercept• 2 2 A

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6 Rotary Vacuum Filtration

Effect of pH and time on volume filtered

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

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Rotary Vacuum Filtration

Effect of filter aid and time on volume filtered

Typical Filter Conditions • pH = 3.6 • 2% - 3% filter aid • heat treatment, T=80Û& “Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002 14 David R. Shonnard Michigan Technological University

7 1. Removal of Insoluble Products Rotary Vacuum Filtration; Example 11.1

Yeast Cell suspension Filtration Rotary Vacuum Filtration….

0.12

y = 6.7E-05x + 0.0272 0.10 R2 = 0.9906 0.08

0.06

0.04

0.02

0.00 0 200 400 600 800 1000 1200 V (Liters of Filtrate)

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1. Removal of Insoluble Products Rotary Vacuum Filtration; Example 11.1

1 slope = = 6.7x10-5 (min/L2 ); solve forD if 'P is 2.3x10-4 N/m2 K 1 §2 A2 · K = = 1.5x104 L2 /min = ¨ ¸'P g 6.7x10-5 ©D C P ¹ c § 2 A2 · D ' * not 9.8 kg /(kg s2) = ¨ ¸ P gc m f ©K C P¹ as in the book

2 2 -4 2 kg m 2(.28m) (2.3x10 N/m)(1 2 /N) = s L2 10-3 m3 kg kg 1 min (1.5x104 )( )(19.2 )(2.9x10-3 )( ) min L m3 m s 60 sec = 2.59 m/kg * 1,920 as in the book

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8 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

-9 -5 • for particle size range ~ 10 to 10 m = dp • purpose o to concentrate a cell suspension o to recover dissolved solutes / proteins

Cross-Flow Filtration “Retentate” (concentrated suspension)

Filtrate or “Permeate” Feed Tank (dissolved solutes / proteins)

Membrane module

Microfiltration membrane pump 17 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products Microfiltration for Removal of Cells

• Mass balance model for separation of cells Chandrasekaran, R., (cells are retained in the feed tank) MS Thesis, Dept. of qR,CR Chemical Engineering Retentate MTU MF Membrane

Permeate qP, CP Feed Mass Balance Assumptions Tank VF(t)

MF Permeate 1. Feed tank is well mixed. Cassette Tank Feed 2. Permeate tank is well mixed. VP(t) qF, CF 3. Volume of fluid in MF cassette is negligibl e.

4. Densities of each stream are equal. 18 David R. Shonnard Michigan Technological University

9 1. Removal of Insoluble Products Microfiltration for Removal of Cells

Chandrasekaran, R., Feed Tank MS Thesis, Dept. of Chemical Engineering MTU

A total mass balance assuming constant stream densities leads to equation

[1] for the change in feed tank volume, VF (t) .

dV (t) F q q q ÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉ[1 ] dt R F P

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1. Removal of Insoluble Products Microfiltration for Removal of Cells

And similarly for entering and exit streams for the membrane cassette,

where qF , qR , and qP the volum etric flow rates of the feed, retentate, and permeate streams.

qF qR qP ÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉ ..[2]

Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU

20 David R. Shonnard Michigan Technological University

10 1. Removal of Insoluble Products Microfiltration for Removal of Cells

A cell mass balance on the feed tank results in equation [3], where

CF , CR , and CP are the concentrations of the proteincells in the feed, retentate, and permeate streams.

d (C V (t)) q C q C ÉÉÉÉÉÉÉÉÉÉÉÉÉÉ[3 ] dt F F R R F F Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU

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1. Removal of Insoluble Products Microfiltration for Removal of Cells

A cell mass balance on the cassette results in equation [4],

qF CF qR CR qPCP ÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉ .É.[4 ]

For a perfectly retained cell: CP 0 , and equation [4] becomes [5]

qRCR qF CF ÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉÉ .É..É.[5 ]

Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU 22 David R. Shonnard Michigan Technological University

11 1. Removal of Insoluble Products Microfiltration for Removal of Cells

Substituting [5] into [3] (for a perfectly retained cell)

Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering d d MTU (C V (t)) m q C q C dt F F dt F R R F F

qF CF qF CF 0

d m 0 where m is mass of proteincells in feed tank ( m =)C V (t )É.[7] dt F F F F F

23 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products Microfiltration for Removal of Cells

Chandrasekaran, R., mFo MS Thesis, Dept. of Chemical Engineering MTU mF

G " Integrating; dmdmFF = 0dtt mF Constant

At t 0, m m (m is the initial mass of cells in the feed tank) F F0 F0

? m m for all time t Òperfectly retained cellÓÉÉÉÉ[8 ] F F0

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12 1. Removal of Insoluble Products Microfiltration for Removal of Cells

Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU Cell Concentration, CF (t)

³ d(CFVF(t)) ³ 0 dt

CFVF (t) constant mFo CF m m C Fo Fo F VF(t) VFo qPt

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Microfiltration of Skim Milk to Separate Casein Protein (CP) from Whey Protein (WP)

Protein mass versus concentration factor, CF = VFo/VF Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU

26 David R. Shonnard Michigan Technological University

13 Microfiltration of Skim Milk to Separate Casein Protein (CP) from Whey Protein (WP)

Protein concentration versus concentration factor, CF = VFo/VF Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU

27 David R. Shonnard Michigan Technological University

Microfiltration of Skim Milk to Separate Casein Protein (CP) from Whey Protein (WP)

Comparison of model with experimental data Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU

28 David R. Shonnard Michigan Technological University

14 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Water (Permeate) Velocity Equation CW

' V'S J = KP ( PM - ) where J = water (permeate) velocity ' K = membrane permeability PM P Retentate ' Flow PM = pressure drop across membrane V = "reflection coefficient" 'S = osmotic pressure (RTCW )

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002 29 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Concentration Polarization - relating CW to CB

In the liquid film; dC Gel Formation J C = D When J and/or CB are high dx enough, a gel layer will form G at the membrane surface, x = C = CW causing an additional

x = 0 C= CB resistance (RG) to solute integrating: flux, J. D C C J = ln W = k ln W G CB CB where D is the diffusivity of solute in the liquid

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15 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Concentration Polarization - relating CW to CB

Example of Protein Ultrafiltration;

J 1.3x10-3 cm / sec D = 9.5x10-7 cm2 / sec (protein diffusivity) G = 180x10-4 cm D C 9.5x10-7 cm2 / sec C J = ln W 1.3x10-3 cm / sec = ln W G -4 CB 180x10 cm CB

CW = 1.3 or CW is 30% > than C B CB

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1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Microfiltration Design - time for filtration dV = -A J dt V = volume of solution remaining to be filtered A = membrane filter area | if we assume no concentration polarization, CW CB dV = -A K ('P-VRTC ) dt P B V for total reflection of solute, = 1 and n = C BV and is constant, where n is total solute mass (cells) dV § [RTn / 'P· = -A K 'P 1- P © ¹ dt V 32 David R. Shonnard Michigan Technological University

16 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Microfiltration Design - time for filtration (cont.) dV § [RTn / 'P· = -A K 'P 1- dt P © V ¹

at t = 0 V = Vo (initial volume of solution) integrating ª 1 º §R T n· §V - RTn / 'P·½ t = « »®(V -V)+ ln o ¾ ' o © ' ¹ © ' ¹ ¬A KP P¼¯ P V- RTn / P ¿ RTn often << (V -V) 'P o ª 1 º t | « »(V -V) Time to filter from Vo to V. ¬A K 'P¼ o P

33 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Microfiltration Design - Example, Cell Microfiltration

2 Vo = 1000 liters, A = 10 m

Xo = 1 g dcw / L concentration to X = 10 g dcw / L ' -4 K P P = initial water flux = 5.7x10 cm / sec X §1 g dcw / L · V = V o = 1000 L ¨ ¸ = 100 L o X ©10 g dcw / L¹ ª 1 º §103 cm3 · t « »(1000 -100)L¨ ¸ ¬(10 m2)(100 cm2 /m2 ) (5.7x10-4 cm / sec)¼ © L ¹

4 = 1.58x10 sec = 4.4 hours

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17 “Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002 35 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Modes of Operation 1. Concentration

Retentate

Permeate

2. One-Pass

Retentate

Permeate

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18 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Modes of Operation 3. Total Recycle Mode - membrane system characterization

Retentate

Permeate Post-Processing: Microfiltration of cells is often followed by conventional filtration of retentate or centrifugation. Then, cell disruption for recovery of intracellular proteins occurs.

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1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Water (Permeate) Velocity Equation CW

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002 38 David R. Shonnard Michigan Technological University

19 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Concentration Polarization - relating CW to CB

In the liquid film; dC Gel Formation J C = D When J and/or CB are high dx enough, a gel layer will form G at the membrane surface, x = C = CW causing an additional

x = 0 C= CB resistance (RG) to solute integrating: flux, J. D C C J = ln W = k ln W G CB CB where D is the diffusivity of solute in the liquid

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1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Permeate Flow Rate

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002 1 'P = P - (P -P ) M i 2 i o 'P J = M R +R G M

Gel resistance Membrane resistance

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20 1. Removal of Insoluble Products Microfiltration/Ultrafiltration

Permeate Flow Rate

' @ constant PM @ constant TFR

R dominant Flux, M Flux, J J RG increasing

RG dominant Increasing protein but decreasing concentration ' ' Tangential Flow Rate (TFR) P Across Membrane ( PM)

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1. Removal of Insoluble Products Centrifugation

Continuous Centrifuges “Biochemical Engineering” Blanch and Clark, Marcel Dekker, 1997 Cell-free liquid exit

Cell solids

Cell solids exit

Tubular Bowl or Disk 42 David R. Shonnard Michigan Technological University

21 1. Removal of Insoluble Products Tubular Centrifugation

SP 1 FD = drag force= 3 DPUoc gc r S 1 F = centrifugal force= D3 U rZ 2 A 6 P P g FB c

S 3 2 1 F F F = bouyant force= D U rZ L A D B P f 6 gc

FD = FA - FB U S OC 3SPD U = D3 (U U )rZ 2 P oc 6 P P f Z 2 2 U U r DP( P f ) y U = oc P 18 43 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products Tubular Centrifugation

Design Equations Radial Travel Time = Axial Travel Time y V = c Uoc Fc

Vc = centrifuge liquid volume; Fc = volumetric flow rate 6 Solving for Fc; Fc = 2 Uo gD2 (U U ) where U = P P f settling velocity under gravity o 18P 2SLZ2 §3 1 · 6 = r2 + r2 g ©4 2 4 1 ¹ 2 F c is proportional to L and r2 44 David R. Shonnard Michigan Technological University

22 1. Removal of Insoluble Products “Biochemical Engineering” Blanch and Clark, Cell Disruption ; 11.3 Marcel Dekker, 1997

If the desired product is intracellular, an effective method to break open the cell wall is needed in order to release the products.

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1. Removal of Insoluble Products Cell Disruption Equipment

Exposure of cells to high liquid shear rates by passing cells through a restricted orifice under high pressure rod for “Biochemical Engineering” valve Blanch and Clark, adjustment Marcel Dekker, 1997

valve impact valve handwheel seat ring

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23 1. Removal of Insoluble Products Cell Disruption Equipment

Rapid agitation of a microbial cell suspension with glass beads or similar abrasives circulating pump temperature jacket

shaft

variable v-belt

“Biochemical Engineering” Blanch and Clark, Marcel Dekker, 1997 agitator disk drive motor handwheel 47 David R. Shonnard Michigan Technological University

1. Removal of Insoluble Products“Biochemical Engineering” Blanch and Clark, Cell Disruption Equipment Marcel Dekker, 1997

Problem 6.1 Blanch and Clark textbook Protein release from yeast using disruption by an industrial homogenizer

Protein release depends upon the pressure, P, and number of recycle passes, N

Design Equation § R · log¨ m ¸ = KNPc © ¹ R m -R

R m = maximum protein conc. (mg/ L) R = protein conc. (mg / L) K = constant C = constant

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24 1. Removal of Insoluble Products Cell Disruption Equipment

Problem 6.1 Blanch and Clark textbook (cont.) Determine K and C slope = KPc or ln(slope) = ln(K) + C ln(P)

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2. Primary Isolation/Concentration of Product: 11.4 Separation Objectives • Remove water from fermentation broth • Dilute solute (product) o more concentrated solute • Often these steps concentrate chemically similar byproducts (other proteins / biomolecules)

Separation Methods A. Extraction (liquid-liquid) B. Adsorption not very selective C. Precipitation for desired product

None the less, these methods are often applied prior to purification

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25 2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction

Liquid-liquid extraction is commonly used, especially in antibiotic fermentations to recover product from broth.

Features of liquid extractant 1. nontoxic 2. inexpensive 3. highly selective toward the product 4. immiscible with the fermentation broth

Other Applications 1. removal of inhibitory fermentation products (ethanol and acetone - butanol).

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2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction

Liquid-liquid extraction is commonly used, especially in antibiotic fermentations to recover products from fermentation broth

Fermentation broth + penicillin Butyl acetate solvent

low pH

Fermentation broth - penicillin Butyl acetate + penicillin neutral pH Buffered phosphate Aqueous solution Butyl acetate

Purified aqueous drying step Solution with penicillin penicillin

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26 2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction - Equilibrium

Liquid-liquid extraction takes advantage of solute equilibrium partitioning between the fermentation broth (heavy, H) phase and a light (L) extractant phase.

Y K = , where K is a distribution coefficie D X D Y is the concentration, mass, or mole fraction solute in the light phase X is the concentration, mass, or mole fraction solute in the heavy phase

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Partitioning is a function of pH for many solutes Y K D = X Solvent is Amylacetate

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

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27 2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction

Mass balance on a single equilibrium stage Assumptions: dilute solute and immiscible phases (negligible

change in H and L) and constant KD. L H(X -X ) = LY or X = X - Y o 1 1 1 o H 1

Y1 LK D Since KD = , X1 = Xo - X 1 X 1 H X 1 1 or 1 = = X o 1+ (LK D /H) 1+E

“Bioprocess Engineering: where E= LK D /H is the extraction fac Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

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2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction

Mass balance on a multiple equilibrium stages Assumptions: dilute solute and immiscible phases (negligible

change in H and L) and constant KD.

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002 L Stage N, H(X -X ) = L(Y -Y ) or X = X + Y , N-1 N N N+1 N-1 N H N Y LK Since K = N , X = X + D X , or X = (1+ E)X D X N-1 N H N N-1 N N

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28 2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction

Stage N-1, H(X N-2 -XN-1 ) = L(Y N-1 - YN ) or

L Y N-1 Y N X N-2 =XN-1 + (YN-1 -YN ), Since KD = = , H X N-1 X N LK X = X + D (X -X ), or X = (1+ E)X N-2 N-1 H N-1 N N-1 N LK X =X + D (X -X ) = (1+ E)X -EX = N-2 N-1 H N-1 N N-1 N 2 X N-2 = (1+ E)(1+ E)X N -EXN = (1+ E) X N -EXN 2 X N-2 =(1+E+E )X N §E N+1 -1· All Stages, X = ¨ ¸X See Figure 11.9 o © ¹ N E-1

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2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction - Figure 11.9

Relates

XN/Xo to E and Number of Stages , N.

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

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29 2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction - Figure 11.9

Example 11.2 Penicillin Extraction using Isoamylacetate L = isoamylacetate flow rate = 10 L/min H = aqueous broth flow rate = 100 L/min

KD = 50, Xo = 20 g/L, XN = .1 g/L How many stages are required to achieve this separation?

Solution: XN / Xo = 0.1/20 = .005

E = LKD/H = (10)(50)/100 = 5

From Figure 11.9, we see that the required number is stages is between 3 and 4, call it 4 equilibrium stages.

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2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction - Figure 11.9

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

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30 2. Primary Isolation/Concentration of Product: Liquid-Liquid Extraction - Equipment

Podbielniak centrifugal extractor

The separation is very rapid, allowing for short residence times “Bioprocess Engineering: which benefits Basic Concepts” Shuler and Kargi, unstable products Prentice Hall, 2002

(especially pH- sensitive antibiotics.

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2. Primary Isolation/Concentration of Product: Precipitation

A very common first step after cell disruption for recovery of intracellular proteins.

Water-protein interactions are key to understanding protein precipitation / solubility in water.

Salting-Out o addition of (NH4)2SO4 or Na2SO4 up to high concentrations 1 to 3 Molar!

- + COO -- NH4 salts exclude water from the surface leading to protein protein-protein interactions and precipitation + 2- NH4 -- SO4

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31 2. Primary Isolation/Concentration of Product: Precipitation

Protein solubility is a function of ionic strength (salt concentration). §S · log¨ ¸ = -K’ I © ¹ S So S = protein solubility (g/L)

So = protein solubility at 0 ionic strength, I (g/L) ’ -1 K S = a salting out constant (mole/L) (function of pH and temperature) 1 I = ionic strength= ¦CZ 2 (mole/L) 2 i i

Ci = molar concentration of salt (mole/L)

Z i = charge on salt ions

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2. Primary Isolation/Concentration of Product: Precipitation

Organic Solvent Addition can also reduce protein-water interactions and promote protein- protein interactions leading to precipitation.

Isoelectric Precipitation at the pH of the isoelectric point, a protein is uncharged, reducing protein-water interactions which leads to precipitation. Warning: extremes in pH may denature the protein product.

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32 2. Primary Isolation/Concentration of Product: Isoelectric Precipitation

Belter, Cussler & Hu, 1988 Effects of pH on the charge of protein functional groups

+ o + NH2 + H NH3

COOH o COO- + H+

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3. Product Purification /Contaminant Removal:

Contaminants often remain with product after primary isolation.

Chromatography: is the most important separation method for biochemical products.

Basic Concepts: 1. Separation is based on differential affinities of solutes toward a solid adsorbent material.

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33 3. Product Purification /Contaminant Removal: (cont.)

2. Different kinds of affinity * o electric charge … ion exchange chromatography o van der Waals force … adsorption chromatography o solubility in liquid … liquid-liquid partitioning chromatog. o solute size/diffusion … gel filtration chromatography * o receptor - ligand … affinity chromatography o hydrophobic interactions … hydrophobic chromatography

* most common usage

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3. Product Purification /Contaminant Removal: Adsorption - 11.4.4

Definition: the removal of selected chemicals from a mobile fluid phase into an immobile solid phase.

Adsorbents: solid materials to which the chemicals (solutes, adsorbates) adhere. These are the immobile phase.

Examples: activated carbon ion exchange resins alumina silica gel other gels: dextran or agarose

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34 3. Product Purification /Contaminant Removal:

Adsorption - 11.4.4 (cont.) “Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Fixed-Bed Adsorption Prentice Hall, 2002

3 Zones Saturated zone, solute is present in both fluid and solid at maximum concentration.

Adsorption zone, concentrations of solute in fluid & solid are in trans- ition.

Virgin zone, concentrations near zero.

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3. Product Purification /Contaminant Removal: Adsorption Equilibrium

Freundlich Isotherm: an isotherm describes the partitioning of a solute between the solid and liquid phases at equilibrium.

Ion exchange resin Graver Technologies * *(1/n) www.gravertech.com CS = KF CL § mass solute · C* = equilibrium conc. of solute on adsorbent ¨ ¸ S ©mass dry adsorbent¹ § mass solute · C* = equilibrium conc. of solute in fluid L ©volume of fluid¹

K F = equilibrium constant (units depent on exponent) n = a constant

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35 3. Product Purification /Contaminant Removal: Adsorption Equilibrium- Freundlich Isotherm

Freundlich Isotherm:

Unfavorable Adsorption

* C L

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3. Product Purification /Contaminant Removal: Adsorption Equilibrium- Langmuir Isotherm

* * Langmuir Isotherm: * CS,max CL CS = * K L + CL

C* S,max Maximum Adsorption * C S Capacity

* C L

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36 3. Product Purification /Contaminant Removal: Adsorption Equilibrium - Adsorbent Capacity

Example Problem:

Calculate the capacity of ion exchange resin to adsorb protein given that:

• m = mass of dry resin in a column = 1 kg • H = porosity of the fixed-bed = 0.40 cm3 fluid/cm3 bed volume U 3 • r = resin density = 1.2 g dry resin/cm resin • n = 1 in the Freundlich Isotherm • per unit bed volume, there is 100 times more protein adsorbed as there is in the fluid at equilibrium. 3 C*L = 1 mg protein/cm fluid at equilibrium

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3. Product Purification /Contaminant Removal: Adsorption Equilibrium - Adsorbent Capacity

Problem Solution:

1. First, calculate KF in the Freundlich Isotherm.

* * CS = KF CL is the Freundlich Isotherm for= 1.n Basis of 1 cm3 bed volume perform a solute mass balance " mass adsorbed to resin in 31 bed cm volume= 100 mass of protein in flu * U H 3 * H CS r (1- )(1 cm) = 100 CL 100 C* H 100 H or C* = L = K C* K = S U H F L F U H r (1- ) r (1- ) 100 (0.4) cm3 fluid K F = = 55.6 1.2 (1-0.4) g dry resin

74 David R. Shonnard Michigan Technological University

37 3. Product Purification /Contaminant Removal: Adsorption Equilibrium - Adsorbent Capacity

Problem Solution: 2. Use the Freundlich Isotherm plus m= 1 kg resin to calculate capacity.

* * Capacity = m CS = m K F CL § cm3 fluid ·§ mg protein· = (1,000 g dry resin)¨ 55.6 ¸ 1 © g dry resin ¹© cm3 fluid ¹ = 55,555.6 mg Protein For each kg dry resin

75 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Adsorption Equilibrium - Batch Adsorption

Example Problem 2: Batch Adsorption

An aqueous solution of protein (10 mg/cm3) of volume 1000 cm3 is contacted with 10 g of the resin (from the prior example problem). What is the concentration remaining in the aqueous phase after equilibrium is achieved?

• m = mass of dry resin in a column = 10 g • n = 1 in the Freundlich Isotherm • V = volume of aqueous solution = 1,000 cm3. 3 KF = 55.6 cm solution/g dry resin Mass Balance on Protein

* * mg protein CS m + C L V = (10 ) V cm3 76 David R. Shonnard Michigan Technological University

38 3. Product Purification /Contaminant Removal: Adsorption Equilibrium - Batch Adsorption

Example Problem 2: Batch Adsorption (cont.)

Equilibrium * *(1/n) * CS = KF CL = KF CL for n = 1 Mass Balance Equation becomes: * * 3 4 K F C L (100 g resin) + CL (1,000 cm ) = 10 mg Protein * 3 4 C L ((100 g resin) KF + 1,000 cm ) = 10 mg Protein 104 mg Protein C* = L cm3 fluid ((100 g resin) (55.6 ) + 1,000 cm3 ) g resin * 3 C L = 1.52 mg Protein / cm

77 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Adsorption Equilibrium - Batch Adsorption

Example Problem 2: Batch Adsorption (cont.)

§ C* · % Recovery of Protein = ¨ 1 - L ¸100 © ¹ CLo § 1.52· = © 1 - ¹100 = 84.76% 10

78 David R. Shonnard Michigan Technological University

39 3. Product Purification /Contaminant Removal: Fixed Bed Adsorption

Theory of Solute Movement in Fixed-Bed Adsorption Columns: (Blanch and Clark, “Biochemical Engineering”, pg 514-517) H = column viod fraction (between particles)

= VL /(VL +VS ) E dz = particle void fraction (voids within particles) z §F · u = superficial velocity = S ©A¹ where F = volumetric flow rate 0 A = cross sectional area of column 79 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Fixed Bed Adsorption

Theory of Solute Movement in Fixed-Bed Adsorption Columns: (cont.) w(V C ) w(V s ) w(VC ) w2 (VC ) L L + S + u L = D L wt wt S wz L wz2 (accumulation (accumulation (convective (axial in liquid) in solid) flow) dispersion) E U s (t, z) = CLi + P C S - - - avg. concentration inside particle R S 2 ³ C Li(t, r,z)4 r dr 3 R C = 0 = ³ r2C (t, r, z)dr Li 4 3 0 Li SR2 R 3 3 R C = ³ r 2C (t, r,z)dr D = axial dispersion coefficient (cm2 /s) S R 3 0 Si L 80 David R. Shonnard Michigan Technological University

40 3. Product Purification /Contaminant Removal: Fixed Bed Adsorption

Theory of Solute Movement in Fixed-Bed Adsorption Columns: (cont.) Assumptions: !! # U CSi CLi so s P C S # Neglect Dispersion, DL 0 wC wC §1 H·wC L + u L + U S = 0 wt i wz P © H ¹ wt Another Assumption: instantaneous equilibrium,

CSi is uniform in the particles

CS = CS = f(CL ) wC wC §wC ·§wC · §wC · so S = S = ¨ S ¸ L = f ’(C ) L wt wt ©wC ¹© wt ¹ L © wt ¹ L

81 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Fixed Bed Adsorption

Theory of Solute Movement in Fixed-Bed Adsorption Columns: (cont.) Therefore

wC u wC L + i L = 0 wt ª §1-H· º wz «1+ U f ’(C )» ¬ P © H ¹ L ¼

This is the form of a kinematic wave. t2 t t 1 3 Dispersion effects C L

z 82 David R. Shonnard Michigan Technological University

41 3. Product Purification /Contaminant Removal: Fixed Bed Adsorption

Theory of Solute Movement in Fixed-Bed Adsorption Columns: (cont.) The velocity of solute propagation is w § CL · dz © wt ¹ u = = i dt §wC · ª §1- H · º L «1+ U f ’(C )» © wz ¹ ¬ P© H ¹ L ¼

the mean retention time of solute in the c L ª §1- H · º t = «1+ U f ’(C )» ¬ P © H ¹ L ¼ ui

83 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Basics of Chromatography

• a solution containing a mixture “Bioprocess Engineering: Basic Concepts” of solutes (in a small volume) Shuler and Kargi, Prentice Hall, 2002 is added to the top of the column.

• a solvent (volume 'V) is added to the top of the column.

• the solvent flow carries the solutes toward the bottom of the column.

84 David R. Shonnard Michigan Technological University

42 3. Product Purification /Contaminant Removal: Basics of Chromatography

• each solute is carried along at a different apparent velocity, depending upon the strength of interaction with the column packing.

• ideally, each solute exits the column as a discrete band of material.

“Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

time 85 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Basics of Chromatography

A technique to separate components in a mixture based upon differential affinity for solutes for the adsorbent.

The affinity is quantified by the adsorption isotherm, CS* = f(CL*), and in particular the derivative, f ‘(CL*).

The affinity could also include size selection as in gel permeation or molecular sieve chromatography.

86 David R. Shonnard Michigan Technological University

43 3. Product Purification /Contaminant Removal: Theory of Chromatography

A Theory of Solute Movement How much solvent ('V) is needed to move a solute a distance 'x?

Solute balance over a differential column height 'x

ª§wC · º §wC · §wC ’ · -« L 'x»'V = H A 'x L 'V + A 'x ¨ S ¸'V ¬© wx ¹ ¼ ©wV ¹ ©wV ¹ rate of solute rate of solute rate of solute removal by removal from removal from solvent flow void space solid phase

87 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Theory of Chromatography (cont.)

Simplifying Yields: wC § wC wC ’ · L + A ¨H L S ¸ = 0 wx © wV wV ¹ linear adsorption isotherm: ’ CS = M f (CL )

amount of adsorbed a function of CL solute per unit mass of adsorbent per unit volume of column volume of column

88 David R. Shonnard Michigan Technological University

44 3. Product Purification /Contaminant Removal: Theory of Chromatography (cont.)

Simplifying Yields: wC wC L = A H M f ’(C ) L wx L wV rearranging §wV· = A H M f ’(C ) ©wx ¹ L

Integrating from xo to x and Vo to V 'V 'x = H A M f ’(C L ) distance that elution volume of solute band solvent moves 89 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Theory of Chromatography (cont.)

• The stronger the adsorption interaction, the shorter the travel distance, 'x, for a ' given elution volume, V. adsorption isotherm

• a stronger adsorption ’ C S interactions means a M f ’(CL ) greater value of M f‘(CL).

C L

90 David R. Shonnard Michigan Technological University

45 3. Product Purification /Contaminant Removal: Theory of Chromatography (cont.)

Example: Solute A, Adsorbent B Adsorption isotherm: C = k (C )3 ª mg A º S 1 L « » ¬ mL solution¼ ªmg A adsorbedº ªmg A adsorbedº ª§ mg A ·3 º « » « »/ « » ¬ mg B ¼ ¬ ¼ © ¹ mg B ¬ mL solution ¼

k1 = 0.2

CL = 0.05 mg A/mL solution H = 0.35 M = 5 g adsorbent B/100 mL column volume 'V = volume of solvent added = 250 mL [cm3] A = column cross-sectional area = 10 cm2

91 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Theory of Chromatography (cont.)

Find 'x

3 f (CL ) = k1 CL therefore 2 f ’ (C L ) = 3k1 C L ª§mg A ads.·º «¨ ¸» «© mg B ¹» = (3)(2)(0.05) = .0015 «§ mg A · » «© ¹ » ¬« mL soln. ¼» 5 g B 50 mg B M = = 100 mL column volume mL column volume

92 David R. Shonnard Michigan Technological University

46 3. Product Purification /Contaminant Removal: Theory of Chromatography (cont.)

Find 'x

'V 'x = = H A M f ’(CL) §1 cm3 soln.· 250 mL ¨ ¸ © mL soln. ¹ ª § cm3 soln. mg B § mg A / mg B ··º «(10 cm2 )¨ 0.35 50 ¨.0015 ¸¸» ¬« © cm3 coln vol cm3 coln vol © mg A / cm3 soln.¹¹¼»

' x = 58.5 cm

93 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Examples of Chromatography

Bailey and Ollis, 1986, Fig. 11.18

94 David R. Shonnard Michigan Technological University

47 3. Product Purification /Contaminant Removal: Examples of Chromatography

Bailey and Ollis, 1986, Fig. 11.19a, cation exchange chromatography

95 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Examples of Chromatography

Bailey and Ollis, 1986, Fig. 11.19b, cation exchange chromatography

96 David R. Shonnard Michigan Technological University

48 3. Product Purification /Contaminant Removal: Examples of Chromatography

Gel Permeation Chromatography: separations based on molecular size

Equivalent Equilibrium Constant S 2 K av,i = exp(- L(rg +ri ) ) where L = concentration of gel fiber (cm/ cm3 )

rg = radius of a gel fiber (cm)

ri = radius of a spherical molecule of species, i (cm)

dz K is equivalent to f ’(C ) in calculating t or av,i L dt 97 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Examples of Chromatography

Molecule radii estimated based on protein diffusion coefficients

Bailey and Ollis, 1986,

98 David R. Shonnard Michigan Technological University

49 3. Product Purification /Contaminant Removal: Examples of Chromatography

Bailey and Ollis, 1986, Fig. 11.21, gel permeation chromatography

99 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Examples of Chromatography

Bailey and Ollis, 1986, Fig. 11.22, molecular sieve chromatography

100 David R. Shonnard Michigan Technological University

50 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids

Charged Amino Acid R Groups at Neutral pH

“Principles of Biochemistry” Lehninger, Worth, 1982 101 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Polar Amino Acid R Groups at Neutral pH

“Principles of Biochemistry” Lehninger, Worth, 1982

102 David R. Shonnard Michigan Technological University

51 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

“Principles of Biochemistry” Nonpolar Amino Acid R Groups at Neutral pH Lehninger, Worth, 1982

103 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Nonpolar and Polar R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999

104 David R. Shonnard Michigan Technological University

52 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Nonpolar and Polar R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999 105 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Nonpolar and Polar R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999 106 David R. Shonnard Michigan Technological University

53 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Negatively Charged R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999

107 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Negatively Charged R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999 108 David R. Shonnard Michigan Technological University

54 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Positively Charged R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999

109 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Positively Charged R Groups:

“Bioseparaion Process Science” Garcia et al., Blackwell Science, 1999 110 David R. Shonnard Michigan Technological University

55 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Acid dissociation reactions - Stoichiometry of COOH = HA: [H+ ][A-] HA m o A- + H+ K = a [HA] K a [A-] log K = log [H+ ] + log a [HA] + pH = - log [H ] and pK a = - log Ka [A-] [A-] pH = pK + log or = 10(pH-pKa) a [HA] [HA] - - but [HA]o = [HA] + [A ] or [HA] = [HA]o - [A ]

- - (pH-pKa) [A ] (pH-pKa) [A ] 10 - = 10 and = (pH-pKa) [HA] o - [A ] [HA]o 1 10 111 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

+ + Acid dissociation reactions - Stoichiometry of NH3 = HA : [H+ ][A] HA+ m o A + H+ K = a [HA+ ] K a [A] log K = log [H+ ] + log a [HA+ ] + pH = - log [H ] and pK a = - log Ka [A] [A] pH = pK + log or = 10(pH-pKa) a [HA + ] [HA+ ] + + + + but [HA ]o = [HA ] + [A] or [A] = [HA ]o - [HA ] [HA+ ] - [HA+ ] [HA+ ] 1 o = 10(pH-pKa) and = + + (pH-pKa) [HA ] [HA ]o 1 10

112 David R. Shonnard Michigan Technological University

56 3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Charge on amino acid groups:

D D D D - amine group -NH( 2 ) - carboxylate group- COOH( ) D-c +1 (-1)10(pH-pKa ) Charge= D-a Charge= D-c 1 10(pH-pKa ) 1 10(pH-pKa )

Charge on R Groups R-c +1 (-1)10(pH-pKa ) Charge= R- a Charge= R-c 1 10(pH-pKa ) 110(pH-pKa )

113 David R. Shonnard Michigan Technological University

3. Product Purification /Contaminant Removal: Ion Exchange of Amino Acids (cont.)

Net Charge is sum of all reactions:

-Charged R Group

D-c R-c +1 (-1)10(pH-pKa ) (-1)10(pH-pKa ) net Charge= D + D + (pH-pK -a) (pH-pK -c ) R-c 1 10 a 1 10 a 1 10(pH-pKa )

+Charged R Group D-c +1 (-1)10(pH-pKa ) +1 net Charge= (pH-pKD-a) + (pH-pKD-c ) + R- a 1 10 a 1 10 a 1 10(pH-pKa )

114 David R. Shonnard Michigan Technological University

57 3. Product Purification /Nonideal effects on Chromatographic Separations

Dispersion, Wall Effects, and Nonequilibrium:

Gaussian Peak

VÃ i tmax,i = standard deviation

Resolution of Peaks t t R max, j max,i S 1/2(t t ) w,i w, j “Bioprocess Engineering: Basic Concepts” Shuler and Kargi, Prentice Hall, 2002

115 David R. Shonnard Michigan Technological University

3. Product Purification /Nonideal Effects on Chromatographic Separations

Prediction of Peak Width:

ª (t t )2 º y y exp« - max,i » i max,i « 2(V t )2 » ¬ max,i ¼

VÃ depends on dispersion and adsorption kinetics

v V 2 v superficial velocity, kal ka surfaceadsorption reaction rate

l column length

116 David R. Shonnard Michigan Technological University

58 3. Product Purification /Nonideal Effects on Chromatographic Separations

Prediction of Peak Height:

y is inversely proportional to V t max,i max,i

VÃ may depend on other processes

vd2 V 2 v internal diffusion control, l v1/2 / d 3/2 V 2 v external film control, l vd2 V 2 v Taylor dispersion (laminar flow), Dl

117 David R. Shonnard Michigan Technological University

3. Product Purification /Scale Up of Chromatographic Separations

To Handle Increased Amount of Product:

1. Increase solute concentration using same column (may saturate column, leading to reduced purity)

2. Increase column cross sectional area, A, and particle diameter, d (maintains flow patterns, but V increases if d increases)

3. Fix d but increase v and l, but maintain ratio of v to l constant (V will be unchanged, but pressure drop will increase)

4. Increase A and volumetric flow rate, such that v is constant (V remains constant, the desired outcome!)

118 David R. Shonnard Michigan Technological University

59 3. Product Purification /Scale Up of Chromatographic Separations

Recent Advances in Chromatographic Packing:

1. Rigid beads with macropores inside particles

2. Allows higher flowrates without bead compression

3. Allows higher flowrates without excessive pressure drop

4. Good mass transfer is maintained between macropores and micropore within particles.

119 David R. Shonnard Michigan Technological University

3. Chromatographic Separation of Proteins from Cheese Whey Raw Milk

Cream Separator Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering MTU

Heat Treatment Standardize Ultrafiltration Acid Rennet/Heat (Optional) Cheese Milk UF Milk Evaporator Whey Curds Rennet/Acid Evaporator/ Spray Dryer Spray Dryer Wash Evaporator/ NFDM Curds Whey MPC Spray Dryer Evaporator/ Cheese Spray Dryer Ultrafiltration Dry Whole Evaporator/ Whey Spray Dryer Casein Evaporator/ Spray Dryer Dry Whole Whey 120 David R. Shonnard WPCMichigan Technological University

60 3. Chromatographic Separation of Proteins from Cheese Whey (cont.)

Chandrasekaran, R., MS Thesis, Dept. of Chemical Engineering Table 1.2 Composition of Whey (Weight %) (Kosikowski et al., 1997) MTU

Fluid Sweet Whey Water 93.7 Total Solid 6.35 Fat 0.5 Protein 0.8 Lactose 4.85 Ash 0.5 Lactic Acid 0.05

121 David R. Shonnard Michigan Technological University

3. Chromatographic Separation of Proteins from Cheese Whey (cont.)

Whey proteins are finding increasing application in the fields of nutrition (protein powder), as an antibiotic, and in other pharmaceutical applications. Individual whey proteins can be separated using cation exchange chromatography, using pH change during elution to recover individual proteins.

Table 1. Isoelectric Points of Major Whey Proteins [1] Whey Protein Isoelectric Point E-lactoglobulin 5.35-5.49 D-lactalbumin 4.2-4.5 Bovine Serum Albumin 5.13 Immunoglobulins 5.5-8.3 Lactoferrin 7.8-8.0 Lactoperoxidase 9.2-9.9

122 David R. Shonnard Michigan Technological University

61 3. Chromatographic Separation of Proteins from Cheese Whey (cont.)

Whey proteins have a range of molecular weights.

Table 2. Major Whey Protein Molecular Weights [1] Whey Protein Molecular Weight -lactoglobulin 18,300 -lactalbumin 14,000 Bovine Serum Albumin 69,000 Immunoglobulins 150,000 Lactoferrin 77,000 Lactoperoxidase 77,500

123 David R. Shonnard Michigan Technological University

3. Chromatographic Separation of Proteins from Cheese Whey (cont.)

pH 6.5 to 11 step (Trial 1, 4/3/03) 20 mL HiPrep 16/10 SP XL 0.9 1 Flow Rate = 3 mL/min 0.8

0.7 Temp. = 4 deg. C

0.6 300 mL of 5 g/L protein

0.5 solution loaded pH 6.5

0.4 4 Elution Absorbance 0.3 60 mL pH 6.5 buffer

0.2 to 3 0.1 1 min. gradient 2 to 0 0 2040608010012014060 mL pH 11 buffer mL collected

124 David R. Shonnard Michigan Technological University

62 3. Chromatographic Separation of Proteins from Cheese Whey (cont.)

Well Sample 1 MW Markers 2 Lactoferrin Standard 3 Peak 2a Colostrum pH 6.5 to 11 4 Peak 2b Colostrum pH 6.5 to 11 5 Peak 2c Colostrum pH 6.5 to 11 6 Peak 1 Trial 1 4/3/03 7 Peak 2 Trial 1 4/3/03 8 Peak 3 Trial 1 4/3/03 9 Peak 4 Trial 1 4/3/03

Lactoperoxidase and/or Lactoferrin appear to be in Lane 9, Peak 4.

125 David R. Shonnard Michigan Technological University

3. Effects of pH Gradient on Peak Resolution

500 ml of a solution of 5 g/L whey protein powder were loaded in the column HiPrep 16/10 SP XL and eluted using gradients from 0 to 85% pH11 (+ 15% pH 6.5 yielding pH 8.5 solution) in 2, 4, 6, 8, 10, 12 and 14 min, using program 2. Program 2 Breakpoint Conc %B Flow rate Fraction Tube A Tube B Valve (min) (ml/min) volume position (ml) 0 0 3 5 pH 6.5 pH 11 Load 20 0 3 5 pH 6.5 pH 11 Load (20+x) 85 3 5 pH 6.5 pH 11 Load (40+x) 85 3 5 pH 6.5 pH 11 Load (43+x) 100 3 5 pH 6.5 pH 11 Load (58+x) 100 3 5 pH 6.5 pH 11 Load Where x is the time for the pH gradient from 0% pH 11 to 85% pH 11. 126 David R. Shonnard Michigan Technological University

63 2 1 1 2

Figure 2. Change of 0% pH 11 to 85% pH 11 in 2 min. Figure 3. Change of 0% pH 11 to 85% pH 11 in 4 min. 127 David R. Shonnard Michigan Technological University

3 1 2

Figure 4. Change of 0% pH 11 to 85% pH 11 in 6 min. Figure 5. Change of 0% pH 11 to 85% pH 11 in 8 min 128 David R. Shonnard Michigan Technological University

64 3 3

2 2 1 1

Figure 6. Change of 0% pH 11 to 85% pH 11 in 10 min. Figure 7. Change of 0% pH 11 to 85% pH 11 in 12 min. 129 David R. Shonnard Michigan Technological University

2 1

130 David R. ShonnardFigure 8. ChangeMichigan of 0% Technological pH 11 to 85% pHUniversity 11 in 14 min.

65 4. Product Preparation / Crystallization

Crystallization is a nucleation process started from a concentrated solution:

1. Occurs when concentration exceeds saturation

2. Crystals have a well-defined morphology, large particle size

3. Homogeneous nucleation occurs when a solid interface is absent

4. Heterogeneous nucleation occurs when a foreign interface is present.

5. Secondary nucleation occurs in the presence of a crystal interface of the same solute

131 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization

Critical cluster or nucleus is the largest cluster of molecules just prior to spontaneous nucleation:

1. n* is the number of molecules in the critical nucleus.

2. Subcritical clusters refers to when, n < n*

3. Supercritical clusters refers to when n > n*

4. An embryo is a cluster having n = n*.

5. An embryo or critical nucleus can range from 10 nm to several Pm in size.

132 David R. Shonnard Michigan Technological University

66 4. Product Preparation / Crystallization

Steps in nucleation and crystal growth

B + B B2 + B B3 + B …..

Bn-1 + B Bn a critical cluster is formed

Bn + B Bn+1 which undergoes nucleation

Bn+1 + B which undergoes crystal growth

133 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization

Characteristic zones of crystallization

New nuclei form spontaneously

No crystallization Growth of existing crystals, occurs Formation of new nuclei

“Bioseparation Process Science” Garcia et al., Blackwell Science, 1999

134 David R. Shonnard Michigan Technological University

67 4. Product Preparation / Crystallization

Transport Processes During Crystallization Growth Rate of Crystals, dL G = = k 'C dt g where 'C = C* C is Molecules in the degree of saturation solution

Molecules in crystal

“Bioseparation Process Science” Garcia et al., Blackwell Science, 1999

135 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Thermodynamics of Homogeneous Nucleation

Free Energy Change for Homogeneous Nucleation ' ' ' GHomogeneous GSurface formation GClustering “Bioseparation Process Science” Garcia et al., Blackwell Science, 1999, ' S 2J Pages127-140 GSurface formation 4 r sl J where sl is the surface tension of the solid / liquid interface S 3 ' § C · 4/3 r GClustering RT ln© * ¹ C Vmolar,solid ' The critical nucleus, rc, is where there is a maximum in GHomogeneous d'G §C · 4S r 2 Homogeneous S J c = 0 = 8 rc sl RT ln© * ¹ dr C Vmolar, solid 2J V r sl molar, solid c §C · Useful calculation when seeding a RT ln Crystallization process ©C* ¹ 136 David R. Shonnard Michigan Technological University

68 4. Product Preparation / Crystallization Rate of Formation of Nuclei,dN/dt

Nucleation is analogous to reaction kinetics,

§ · ¨ ¸ ¨ 16S J 3 V 2 ¸ A exp¨- sl molar,solid ¸ § ·2 ¨ 3 3 § C · ¸ ¨ 3R T ¨ln¨ ¸¸ ¸ © © © C* ¹¹ ¹ 137 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Batch Crystallization, Solid Phase Balance

1. Cummulative Number of Crystals, N versus size, L or 2. Population Density, n Slope of N vs L curve

A balance on n tracks the number of crystals entering and leaving a specific size range due to crystal growth.

138 David R. Shonnard Michigan Technological University

69 4. Product Preparation / Crystallization Batch Crystallization, Population Balance Equation

ªNumber of º ªNumber of º ªNumber of º ªNumber of º « » « » « » « » «crystals initially » «crystals growing» «crystals at end » «crystals growing» ¬«within range, L¼» ¬«into range, L ¼» ¬«within range, L¼» ¬«out of range, L¼» ' ' ' ' V ninitial L V G1 n1 t V nfinal L V G2 n2 t V is volume, 'L is size range, G is growth rate of crystal size (dL/dt), 't is a small time step. subscript 1 is a smaller size range, subscript 2 is size range for 'L.

139 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Batch Crystallization, Population Balance Equation (cont.)

divide by V , 'L , and 't and allow 'L and 't to go to 0. dn d(Gn) 0 dt dL

Assuming G is a constant over all L dn dn G 0 dt dL

140 David R. Shonnard Michigan Technological University

70 4. Product Preparation / Crystallization Batch Crystallization, Population Balance Equation (cont.) boundary conditions (BCs) for nucleation at t 0, n 0 B0 at L 0, n G as L o f, n is finite solve population balance equation and BCs using Laplace Transforms B0 § Ls · n exp¨ ¸ in the Laplace Domain Gs © G ¹ § L · n B0 u¨t ¸ © G ¹ 141 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Batch Crystallization, Cumulative Crystal Mass

M is cumulative crystal mass per unit volume L U 3 M c kv ³ nL dL 0 U where c is density of crystal solid and kv is a shape factor

and as L o f, 1 M W U k B0 G3 t 4 4 c v

142 David R. Shonnard Michigan Technological University

71 4. Product Preparation / Crystallization Batch Crystallization, Cooling Curve

Determine the time - temperature relationship to achieve a constant degree of supersaturation during batch crystallization

rate of change of solute concentration - rate of change of W dC dW dt dt

dC U k B0 G3 t 3 dt c v

143 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Batch Crystallization, Cooling Curve (cont.) to achieve a constant degree of supersaturation, the rate of dC temperature change must be proportional to dt dC dT k U k B0 G3 t 3 dt T dt c v integrating from the temperature that crystals start to form, T0 , at t 0, we find that U k B0 G3 t 4 c v T0-T 4kT

144 David R. Shonnard Michigan Technological University

72 4. Product Preparation / Crystallization Continuous Crystallization, Solid Phase Balances

ªÃNumber of ºÃ ªÃNumber of ºÃ «Ã »Ã ªÃNumber of ºÃ «Ã »Ã ªÃNumber of ºÃ «Ãcrystals growing»Ã «Ã »Ã «Ãcrystals growing »Ã «Ã »Ã «Ãcrystals entering »Ã «Ãcrystals leaving »Ã «Ãinto range, 'L, »Ã «Ãout of range, 'L,»Ã «Ã »Ã ¬Ã«Ãrange, 'L, by flow¼Ã»Ã «Ã »Ã ¬Ã«Ãrange, 'L, by flow¼Ã»Ã «Ã ' »Ã «Ã ' »Ã ¬Ã over a time, t ¼Ã ¬Ãover a time, t ¼Ã ' ' ' ' ' ' VG1 n1 t Qnin L t VG2 n2 t Vn L t V is volume, 'L is size range, G is growth rate of crystal size (dL / dt), 't is a small time step, Q is volumetric flow rate through crystallizer. subscript 1 is a smaller size range, subscript 2 is size range for 'L, subscript in is for inlet conditions. 145 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Continuous Crystallization, Solid Phase Balances

divide by 'L , and 't and allow 'L and 't to go to 0,

and assuming that no crystals are entering, nin = 0, and that G is constant. dn VG Qn 0 dL

V Restating in terms of residence time, W = Q dn n 0 dL GW Bo Boundary Condition, L = 0, n = no = G 146 David R. Shonnard Michigan Technological University

73 4. Product Preparation / Crystallization Continuous Crystallization, Solid Phase Balances

L U 3 M c kv ³ nL dL 0 U where c is density of crystal solid and kv is a shape factor § § 1 1 · § L ·· M 6 U k no GW ¨G3W 3 G3W 3 G 2W 2 L GWL2 L3 exp - ¸ c v © © 2 6 ¹ © GW ¹¹ and as L o f, M W U k n o G 4 W 4 6 c v 147 David R. Shonnard Michigan Technological University

4. Product Preparation / Crystallization Continuous Crystallization, Advantages

Advantages:

1. Input of solute helps to maintain a constant degree of saturation, 'C

2. Desirable for determining growth rates and other kinetic parameters, but are not popular in industrial applications.

148 David R. Shonnard Michigan Technological University