recovery and purification of fermentation products

8/13/2019 RECOVERY AND PURIFICATION OF FERMENTATION PRODUCTS

1/152

1Michigan Technological UniversityDavid R. Shonnard

Chapter 11: Product Recovery and Purification

David Shonnard

Department of Chemical Engineering

Michigan Technological University


2/152


Presentation Outline:

l Overview of Bioseparations

l Separation of Insoluble Products

l Primary Isolation / Concentration of Product

l Purification / Removal of Contaminant Materials

l Product Preparation


3/152


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

Introduction to Bioseparations


4/152


Biochemical Separations Technologies

Technologies

Bioprocess Engineering:

Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


5/152


Figure 11.2

Major Steps in Separating a Protein Product

Fermentation

Cell Removal and

Concentration

Cell Disruption

Removal of Cell

Debris

1.Separation

of Insoluble

Products,

Product

Concentration(0.2 - 2.0%)

Filtration

Centrifugation

Coagulation /

Flocculation

Ultrasonic Vibrators

French Press

Bead or Ball Mills

Ultracentrifugation

Ultrafiltration


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7/152


Figure 11.2


Chromatographic

Purification

SolventPrecipitation

Dialysis

3.

Purification /

Removal of

Contaminant

Chemicals,

(50 - 80%)

Other methods for

product purification:

Reverse Osmosis

Cross-flowUltrafiltration

Electrophoresis

Electrodialysis


8/152


Figure 11.2


Lyophilization

4.

Product

Preparation,

(90 - 100%)

Other methods:

Crystallization


9/152


1. Removal of Insoluble ProductsRotary Vacuum Filtration

Cell Solution

coagulation agents/

(filter aids) added

vacuum applied to

rotating drum (P)


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


10/152



)(

cm

c

rr

APg

dt

dV

+

=

Filter medium resistance (a constant)

Filter cake resistance

Viscosity of filtrate (water)

Filter area

Volumefiltered

Note: rc increases with the volume filtered, V

A

VCrc

=

C = wt. of cells per volume filtrate (g cells/L)

= average specific resistance of filter cake


11/152



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

V2+ 2VV

o= K

where

Vo=

rm

CA

K =

2 A2

C

!

"#

$

%&P gc

Ruth Equation

gc=

1kgm

s2

N

!

"

#

##

$

%

&

&&

N = Newton


12/152



Rearrange Ruth Equation


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002

CKr

KV

gPC

A

K

gPC

A

VVKV

t

mo

c

c

o

2intercept-y

2intercept-y

2

1

2slope

)2(1

2

2

='=

&&%

$

##"

!

=

&&%

$##"

!=

+=


13/152


Rotary VacuumFiltration

Effect of pHand time onvolume filtered


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


14/152


15/152


1. Removal of Insoluble ProductsRotary Vacuum Filtration; Example 11.1

Yeast Cell suspension FiltrationRotary Vacuum Filtration.Rotary Vacuum Filtration.Rotary Vacuum Filtration.Rotary Vacuum Filtration.

y = 6.7E-05x + 0.0272

R2= 0.9906

0.00

0.02

0.04

0.06

0.08

0.10

0.12

0 200 400 600 800 1000 1200

V (Liters of Filtrate)


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* not 9.8 kgm/(kgfs2)

as in the book

kg

m2.59

s60

min1

sm

kg109.2m

kg2.19L

m10

min

L105.1

Ns

mkg1

m

N103.2)m2(.28

2

2/min)(L105.1

107.6

1

)(min/L107.61

slope

3

3

2332

4

22

422

2

224

5

25

=

&%

$#"

!&%

$#"

!

&%

$#"

!&&%

$

##"

!

&&%

$

##"

!

&%

$#"

!

&%

$#"

!

=

&&%

$

##"

!

=

&&%

$##"

!===

==

xx

x

gPCK

A

gPC

Ax

xK

xK

c

c

1. Removal of Insoluble ProductsRotary Vacuum Filtration; Example 11.1

* 1,920 as in the book


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1. Removal of Insoluble ProductsMicrofiltration for Removal of Cells

Mass balance model for separation of cells

(cells are retained in the feed tank)

MF

Membrane

Permeate q P, CP

Retentate

Feed

Tank VF(t)

qR,CR

VP(t)

MF

Cassette Permeate

TankFeed

qF, CF

Mass Balance Assumptions

1. Feed tank is well mixed.

2. Permeate tank is well mixed.

3. Volume of fluid in MF cassette is negligible.

4. Densities of each stream are equal.

Chandrasekaran, R.,

MS Thesis, Dept. of

Chemical EngineeringMTU


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20/152



And similarly for entering and exit streams for the membrane

cassette, where ,, RF qq and Pq are the volumetric flow rates of

the feed, retentate, and permeate streams.

PRF qqq += ..[2]

Chandrasekaran, R.,

MS Thesis, Dept. of

Chemical Engineering

MTU


21/152



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

where ,, RF CC and PC are the concentrations of the cells in the

feed, retentate, and permeate streams.

FFRRFF CqCqtVCdt

d=))(( [3]

Chandrasekaran, R.,

MS Thesis, Dept. of


MTU


22/152



Chandrasekaran, R.,

MS Thesis, Dept. of


MTU

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

PPRRFF CqCqCq += ..[4]

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

FFRR

CqCq = ....[5]


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Chandrasekaran, R.,MS Thesis, Dept. of


MTU

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

FFRRFFF CqCqm

dt

dtVC

dt

d==))((

0== FFFF CqCq

0=Fmdtd where Fm is mass of cells in feed tank ( Fm = )(tVC FF ).[7]


24/152


Integrating; ( (= dtdmF 0 ' =Fm Constant

At0

,0 FF mmt == ( 0Fm is the initial mass of cells in the feed tank)

0FF

mm = for all time t perfectly retained cell[8]


Chandrasekaran, R.,

MS Thesis, Dept. of


MTU

t

mF

mFo


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MTU

Cell Concentration, CF(t)

d(CFVF( t)) =( 0 dt(CFVF(t) = constant = mFo

CF =

mFoVF(t) =

mFoVFo qPt

t

CF


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MTU

t

mF

+= tCqm FPF 0

0,0 FF mmt == ' 0Fm=

Constant.[11]

ConstantAt

Perfectly Permeating Cell (or Protein)

CF

Fo

o

Fo

o

Po

o

PFo

Po

o

FoPFo

F

o

FoFo

Po

FoPFoFF

CV

m

tV

qV

tV

qm

tqV

tV

mqm

tC

V

mC

tqV

tCqm

tV

tmtC

==

=

=

=

==

)1(

)1(

)(

thatnote

)(

)()(


27/152


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

Chandrasekaran, R.,

MS Thesis, Dept. of


MTU

Protein mass versus concentration factor, CF = VFo/VF


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30/152


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1. Removal of Insoluble ProductsMicrofiltration/Ultrafiltration

Water (Permeate) Velocity Equation CCWW

RetentateFlow

PM

J = KP (PM -)

where

J = water (permeate) velocity

KP = membrane permeability

PM = pressure drop across membrane

= "reflection coefficient"

= osmotic pressure (RTCW )


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


32/152


filmin thesoluteofydiffusivittheiswhere

ln

gintegratin

0

D

CCDJ

CCxCCx

dx

dCDJ

B

W

B

W

=

====

=

1. Removal of Insoluble Products

Microfiltration/Ultrafiltration

Concentration Polarization - relating CW to CB

In the liquid film;

Gel Formation

When J and/or CB are high

enough, a gel layer will form at

the membrane surface, causing

an additional resistance (RG) to

solute flux, J.


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Permeate Flow Rate

PM = Pi -1

2(Pi - Po )

J =PM

RG

+ RM

Gel resistance

Membrane resistance

Bioprocess Engineering:Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


34/152



Permeate Flow Rate

Flux,J

Tangential Flow Rate (TFR)

@ constant PM

Flux,J

P Across Membrane (PM)

@ constant TFR

RG dominant

but decreasing

RM dominant

Increasing protein

concentration

OsmoticPressureEffectIncreasing


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Concentration Polarization - relating CW to CB

Example of Protein Ultrafiltration;

J = 1.3x10-3 cm / sec

D = 9.5x10-7 cm2 / sec (protein diffusivity)

= 180x10-4

cm

J =D

ln

CWC

B

' 1.3x10-3 cm / sec =9.5x10

-7cm

2/ sec

180x10-4 cmln

CWC

B

CW

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


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Microfiltration Design - time for filtration

dV

dt = - A J

V = volume of solution remaining to be filtered

A = membrane filter area

if we assume no concentration polarization, CW CB

dV

dt = - A K P (P -RTCB )

for total reflection of solute, = 1 and n = CBV

and is constant, where n is total solute mass (cells)

dV

dt = - A K

PP 1-

[RTn / P

V

!"

$%


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Microfiltration Design - time for filtration (cont.)

dV

dt = - A K P P 1-

[RTn / P

V

!"

$%

at t = 0 V = Vo(initial volume of solution)

integrating

t =1

A KPP)

*+

,

-. (Vo - V ) +

R T n

P!"

$%

lnVo - RTn / P

V- RTn / P!"

$%

/01

234

oftenRTn

P


38/152


39/152



Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


40/152



Modes of Operation

1. Concentration

2. One-Pass

Retentate

Permeate

Retentate

Permeate


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Modes of Operation

3. Total Recycle Mode - membrane system characterization

Post-Processing:Microfiltration of cells is often followed by conventional filtration of

retentate or centrifugation. Then, cell disruption for recovery of

intracellular proteins occurs.

Retentate

Permeate


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1. Removal of Insoluble ProductsCentrifugation

Continuous Centrifuges

Bowl or DiskTubular

Cell-free liquid exit

Cell solids exit

Cell solids

Biochemical Engineering

Blanch and Clark,

Marcel Dekker, 1997


43/152


18

)(

)(

6

3

-

16

cellonforcebouyancy

1

6cellonforcelcentrifuga

1

3cellonforcedrag

22

23

23

23

fPP

OC

fPPOCP

BAD

c

fPB

c

PPA

c

OCPD

DrU

rDUD

FFF

grDF

grDF

gUDF

=

=

=

==

==

==


Tubular Centrifugation

r

FA FD

FB

UOC

y

L


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1. Removal of Insoluble ProductsTubular Centrifugation

Design Equations

Radial Travel Time = Axial Travel Time

y

Uoc =

Vc

Fc

Vc = centrifuge liquid volume; Fc = volumetric flow rate

Solving for Fc; Fc = 2 Uo

where Uo =gDP

2 (P f)

18 settling velocity under gravity

= 2L2

g

3

4r2

2+ 1

4r1

2!"

$%

Fcis proportional to L and r22


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1. Removal of Insoluble ProductsCell Disruption ; 11.3

If the desired

product is intra-

cellular, an

effective method

to break open

the cell wall is

needed in order

to release the

products.

Biochemical EngineeringBlanch and Clark,

Marcel Dekker, 1997


46/152


1. Removal of Insoluble ProductsCell Disruption Equipment

Exposure of cells to high liquid shear rates by passing cells through a

restricted orifice under high pressure

handwheel

rod forvalveadjustment

valve

valveseat

impactring


Blanch and Clark,

Marcel Dekker, 1997


47/152



Rapid agitation of a microbial cell suspension with glass beads or

similar abrasives

handwheeldrive motor

variable v-belt

shaft

agitatordisk

circulating pump

temperaturejacket


Blanch and Clark,

Marcel Dekker, 1997


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Problem 6.1 Blanch and Clark textbook

Protein release from yeast using disruption by an industrial homogenizer

Protein release depends upon thepressure, P, and number of recyclepasses, N

Design Equation

logRm

Rm - R

!

"#

$

%& = KNP

c

Rm = maximum protein conc. (mg / L)

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

C = constant


Blanch and Clark,Marcel Dekker, 1997


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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) more concentrated solute

Often these steps concentrate chemically similar byproducts

(other proteins / biomolecules)

Separation Methods

A. Extraction (liquid-liquid)

B. Adsorption

C. Precipitation

not very selective

for desired product

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


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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 Applications1. removal of inhibitory fermentation products (ethanol and

acetone - butanol).


52/152

52

Michigan Technological UniversityDavid R. Shonnard



Liquid-liquid extraction is commonly used, especially in antibiotic

fermentations to recover products from fermentation broth

Fermentation broth +

penicillin Butyl acetate solvent

Butyl acetate + penicillinFermentation broth -

penicillin

low pH

Buffered phosphate

Aqueous solution

Purified aqueous

Solution with penicillin

Butyl acetate

neutral pH

drying step

penicillin

l / f d


53/152

53



Liquid-Liquid Extraction - Equilibrium

Liquid-liquid extraction takes advantage of solute equilibrium

partitioning between the fermentation broth (heavy, H) phase and alight (L) extractant phase.

phaseheavyin thesoluteoffractionmoleormassion,concentrattheis

phaselightin thesolute

offractionmoleormassion,concentrattheis

tcoefficienondistributiaiswhere,

X

Y

KX

YK DD =


54/152

2 P i I l ti /C t ti f P d t


55/152

55


factorextractiontheiswhere

1

1

)/(1

1or

,Since

or)(

1

11

1

1

1111

o

D

Do

DoD

oo

H

LKE

EHLKX

X

XH

LKXXXYK

YH

LXXLYXXH

=

+=

+=

==

==



Mass balance on a single equilibrium stage

Assumptions: dilute solute and immiscible phases (negligible change inH and L) and constant KD.


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002



56/152

56


NNND

NN

N

ND

NNNNNNN

XEXXH

LKXX

X

YK

Y

H

LXXYYLXXH

)1(or,,Since

or)()(N,Stage

11

111

+=+==

+==

+



Mass balance on a multiple equilibrium stages

Assumptions: dilute solute and immiscible phases (negligible change in Hand L) and constant KD.

Bioprocess Engineering: Basic Concepts

Shuler and Kargi, Prentice Hall, 2002



57/152

57


11.9Figuresee1

1Stags,All

)1(

)1()1)(1(

)1()(

)1( with),(

,Since),(

or)()(1,-NStage

!

2

2

2

2

1112

1112

1

1112

112

N

N

o

NN

NNNNN

NNNND

NN

NNNND

NN

N

N

N

NDNNNN

NNNN

XE

EX

XEEX

EXXEEXXEEX

EXXEXXH

LKXX

XEXXXH

LKXX

XY

XYKYY

HLXX

YYLXXH

&&%

$##"

!

=

++=

+=++=

+=+=

+=+=

==+=

=

+




58/152



59/152

59



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.


60/152



61/152

61



Liquid-Liquid Extraction - Equipment

Podbielniak

centrifugalextractor

The separation

is very rapid,

allowing for short

residence times

which benefits

unstable products

(especially pH-

sensitive antibiotics.


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002


62/152

2 Primary Isolation/Concentration of Product:


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63



Precipitation

Protein solubility is a function of ionic strength (salt concentration).

ionsaltoncharge

(mole/L)ionsaltofionconcentratmolar

(mole/L)2

1

stsrengthionic

re)temperatuandpHoffunction(a

(moles/L)constantoutsaltinga

(g/L)sstrength,ionic0atsolubilityprotein

(g/L)solubilityprotein

SSlog

2

'

'

o

=

===

=

=

=

=&&%$

##"!

5

i

i

ii

S

o

S

Z

C

ZCI

K

S

S

IK



64/152

64



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.



65/152

65



Isoelectric Precipitation

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

NH2 + H+ NH3

+

COOH COO- + H+


66/152

66


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.



67/152

67


(cont.)

2. Different kinds of affinity

* electric charge ion exchange chromatography

van der Waals force adsorption chromatography

solubility in liquid liquid-liquid partitioning chromatog.

solute size/diffusion gel filtration chromatography

* receptor - ligand affinity chromatography hydrophobic interactions hydrophobic chromatography

* most common usage



68/152

68


Adsorption - 11.4.4

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

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

Examples: activated carbonion exchange resins

alumina

silica gel

other gels: dextran or agarose



69/152

69


Adsorption - 11.4.4 (cont.)

Fixed-Bed Adsorption

3 Zones

Saturated zone, solute

is present in both fluid

and solid at maximum

concentration.

Adsorption zone, conc-

entrations of solute in

fluid & solid are in trans-

ition.

Virgin zone, concentra-

tions near zero.


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002



70/152

70


Adsorption Equilibrium

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

CS* = KF CL

*(1/n)

CS* = equilibrium conc. of solute on adsorbent mass solute

mass dry adsorbent!"#

$%&

CL*

= equilibrium conc. of solute in fluidmass solute

volume of fluid

!"

$%

KF = equilibrium constant (units depent on exponent)

n = a constant

Ion exchange resin

Graver Technologies

www.gravertech.com



71/152

71


Adsorption Equilibrium- Freundlich Isotherm

Freundlich Isotherm:

CS*

CL*

n = 1

n > 1

0 < n < 1

Favorable

Adsorption

Unfavorable

Adsorption


72/152



73/152

73


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

= porosity of the fixed-bed = 0.40 cm3 fluid/cm3 bed volume

r= resin density = 1.2 g dry resin/cm3 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.

C*L = 1 mg protein/cm3 fluid at equilibrium



74/152

74



Problem Solution:

1. First, calculate KF in the Freundlich Isotherm.

resindryg

fluidcm55.6

)4.01)(2.1(

)4.0(100

)1(

100

)1(

100or

100)cm1)(1(volume"bedcm1in thefluidin theproteinofmassthetimes100

volumebedcm1inresintoabsorbedmass"

balancemasssoluteaperformvolumebedcm1ofBasis

1nforisothermFreundlichtheis

3

**

*

*3*

3

3

3

**

=

=

='=

=

=

=

'

==

F

r

FLF

r

LS

LrS

LFS

K

KCKC

C

CC

CKC



75/152

75



Problem Solution:

2. Use the Freundlich Isotherm plus m= 1 kg resin to calculate

capacity.

Capacity = m CS* = m KF CL

*

= (1,000 g dry resin) 55.6cm3fluid

g dry resin

!

"#$

%&1

mg protein

cm3fluid

!"

$%

= 55,555.6 mg Protein

For each kg dry resin



76/152

76


Adsorption Equilibrium - Batch Adsorption

Example Problem 2: Batch Adsorption

An aqueous solution of protein (10 mg/cm3) of volume 1000 cm3 iscontacted 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.

KF = 55.6 cm3 solution/g dry resin

Mass Balance on Protein

CS* m + CL

* V = (10mg protein

cm3) V


77/152



78/152

78


Adsorption Equilibrium - Batch Adsorption

Example Problem 2: Batch Adsorption (cont.)

% Recovery of Protein = 1 -CL

*

CLo

!

"#

$

%&100

= 1 -

1.52

10

!

"$

%100 = 84.76%



79/152

79


Fixed Bed Adsorption

Theory of Solute Movement in Fixed-Bed Adsorption Columns:

(Blanch and Clark, Biochemical Engineering, pg 514-517)

= column viod fraction

(between particles)

= VL / (VL + VS)

= particle void fraction

(voids within particles)

uS

= superficial velocity =F

A

!

"

$

%where F = volumetric flow rate

A = cross sectional area of column

dz

z

0



80/152

80




(cont.)

(VLCL )t

+ (VSs )t

+ uS (VCL )

z = DL

2

(VCL )z2

(accumulation (accumulation (convective (axial

in liquid) in solid) flow) dispersion)

s (t, z) = CLi

+ PC

S - - - avg. concentration inside particle

CLi

=C

Li(t,r,z)4r2dr

0

R

(4

3R2

=3

R3r2C

Li(t, r, z)dr

0

R

(

CS =

3

R3r2C

Si(t,r,z)dr

0

R

( DL = axial dispersion coefficient (cm2 / s)



81/152

81




(cont.)

Assumptions:

CSi >> CLi so s P CS

Neglect Dispersion, DL 0

CL

t

+ u iC

L

z

+ P1

!"

$%

CS

t

= 0

Another Assumption: instantaneous equilibrium,

CSi

is uniform in the particles

CS = CS = f(CL )

soC

S

t =

CS

t =

CS

CL

!

"#$

%&C

L

t

!"

$%

= f '(CL )C

L

t

!"

$%



82/152

82




(cont.)

Therefore

CLt

+u i

1+P

1-

!"

$%

f '(CL)

)

*+

,

-.

CLz

= 0

This is the form of a kinematic wave.

CL

z

t1 t2

t3 Dispersion effects



83/152

83




(cont.)

The velocity of solute propag

dz

dt=

CL

t

!"

$%

CL

z

!

"

$

%

=ui

1+P

1-

!

"

$

%

f '(CL)

)

*+

,

-.

the mean retention time of solute

t =

L

u 1+P

1-

!"

$%f '(CL)

)

*+,

-.



84/152

84


Basics of Chromatography a solution containing a mixture

of solutes (in a small volume)

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.


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002



85/152

85


Basics of Chromatography each solute is carried along

at a different apparent

velocity, depending uponthe strength of interaction

with the column packing.

ideally, each solute exits

the column as a discrete

band of material.


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002

time



86/152


Basics of ChromatographyA 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 gelpermeation or molecular sieve chromatography.


Th f Ch h


87/152


Theory of ChromatographyA 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

-C

Lx

!"

$%x

)

*+,

-.V = A x

CL

V

!"

$%V + A x

CS

'

V

!

"#$

%&V

rate of solute rate of solute rate of solute

removal by removal from removal from

solvent flow void space solid phase


Th f Ch h ( )


88/152


Theory of Chromatography (cont.)

Simplifying Yields:CLx

+ A CLV

+CS

'

V

!

"#

$

%& = 0

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


Th f Ch t h ( t )


89/152


Theory of Chromatography (cont.)

Simplifying Yields:

C

L

x = A + M f '(CL )( )

CL

Vrearranging

V

x

!"

$%

= A + M f '(CL

)( )

Integrating from xoto x and Voto V

x =V

A +M f ' (CL )( )

distance that elution volume ofsolute band solvent

moves


90/152


Th f Ch t h ( t )


91/152


Theory of Chromatography (cont.)Example: Solute A, Adsorbent B

Adsorption isotherm: CS = k1 (CL)3

k1

= 0.2

CL = 0.05 mg A/mL solution

= 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

mg A adsorbed

mg B

)

*+

,

-.

mg A

mL solution

)*

+,-

.

mg A adsorbed

mg B

)

*+

,

-./

mg A

mL solution

!"

$%

3)

*+

,

-.


Theory of Chromatography (cont )


92/152


Theory of Chromatography (cont.)Find x

f (CL

) = k1C

L

3 therefore

f ' (CL ) = 3k 1CL2

= (3)(2)(0.05) = .0015

mg A ads.

mg B

!

"#$

%&

mg A

mL soln.!"

$%

)

*

++

+++

,

-

.

.

..

.

M =5 g B

100 mL column volume

=50 mg B

mL column volume


Theory of Chromatography (cont )


93/152


Theory of Chromatography (cont.)Find x

x = =V

A + M f ' (CL)( )

250 mL1 cm3soln.

mL soln.

!

"#

$

%&

(10 cm2 ) 0.35 cm3soln.

cm3coln vol+ 50 mg B

cm3coln vol.0015 mg A / mg B

mg A / cm3soln.!"# $

%&!

"# $

%&)

*++

,

-..

x = 58.5 cm


Examples of Chromatography


94/152


Examples of ChromatographyBailey and Ollis, 1986, Fig. 11.18




95/152


Examples of ChromatographyBailey and Ollis, 1986, Fig. 11.19a,

cation exchange chromatography




96/152


Examples of ChromatographyBailey

and Ollis,

1986,Fig. 11.19b,

cation

exchange

chromato-

graphy




97/152


Examples of ChromatographyGel Permeation Chromatography: is based on the penetration of

solute molecules into small pores of packing particles on thebasis of molecular size and the shape of the solute molecules. It is

also known as size exclusion chromatography.

Equivalent Equilibrium Constant

Kav,i = exp(-L(rg + ri )2)

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)

Kav,i

is equivalent to f' (CL) in calculating t or

dz

dt




98/152


Examples of ChromatographyMolecule radii estimated based on protein diffusion coefficients

Bailey

and Ollis,

1986,




99/152



and Ollis,

1986,Fig. 11.21,

gel

permeation

chromato-graphy




100/152



and Ollis,

1986,Fig. 11.22,

molecular

sieve

chromato-graphy


101/152


102/152

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


103/152


Nonpolar Amino Acid R Groups at Neutral pH Principles of BiochemistryLehninger, Worth, 1982



104/152


Acid dissociation reactions - Nonpolar and Polar R Groups:

Bioseparaion Process Science

Garcia et al., Blackwell Science, 1999



105/152







106/152







107/152


Acid dissociation reactions - Negatively Charged R Groups:





108/152


Acid dissociation reactions - Negatively Charged R Groups:





109/152


Acid dissociation reactions - Positively Charged R Groups:





110/152


Acid dissociation reactions - Positively Charged R Groups:





111/152


Acid dissociation reactions - Stoichiometry of COOH = HA:

HA

K a

A-+ H+ ' Ka=

[H+][A]

[HA]

log Ka= log [H

+] + log

[A-]

[HA]

pH = -log [H+] andpK

a

= -log Ka

pH = pKa+ log

[A-]

[HA] or

[A-]

[HA]= 10

(pH-pKa)

but [HA]o= [HA]+ [A

-] or [HA]= [HA]

o- [A

-]

[A-]

[HA]o- [A

-]= 10

(pH-pKa) and[A

-]

[HA]o

=10

(pH-pKa)

1 +10(pH-pKa)



112/152


Acid dissociation reactions - Stoichiometry of NH3+ = HA+ :

HA+

Ka

A+ H+ ' Ka=

[H+][A

[HA+

]

log Ka= log [H

+] + log

[A]

[HA+]

pH = -log [H+] andpK

a

= -log Ka

pH = pKa+ log

[A]

[HA+] or

[A]

[HA+]= 10

(pH-pKa)

but [HA+]o= [HA

+] + [A] or [A]= [HA

+]o- [HA

+]

[HA+]o- [HA+]

[HA+]

= 10(pH-pKa) and

[HA+]

[HA+]o

=1

1 +10(pH-pK a)


Ch i id


113/152


Charge on amino acid groups:( )

( )

( )

( )

( )

( )

( )( )sa

sa

aa

ca

pKpH

pKpH

pKpH

pKpH

+

+=

+

=

+

=+

+=

101

11chainsideofCharge

10

11

11chainsideofCharge

10

11

11groupcarboxyl-ofCharge

101

11groupamino-ofCharge


N t Ch i f ll ti


114/152


Net Charge is sum of all reactions:

( )( ) ( )

( )

( )( )

( )

( )

( )

( )

( )( )

( )

( )

( )( )sa

aa

ca

sa

aa

ca

aa

ca

pKpH

pKpH

pKpH

pKpHpKpH

pKpH

pKpH

pKpH

+

++

+

+

+

+=+

+

+

+

+

+

+=

++

++=

101

11

10

11

11

101

11R)(withacidaminoofchargeNet

10

11

11

10

11

11

101

11(-R)withacidaminoofchargeNet

10

11

11101

11acidaminoofchargeNet



115/152


3. Product Purification /Contaminant Removal:Ion Exchange of Proteins

A computer algorithm for computing charge of proteins


116/152


A computer algorithm for computing charge of proteins

(Genetics Computer Group, Inc. 1993)

.-,+

*).

-,+

*)+.

-,+

*).

-,+

*)=.

-,+

*)

rminicarboxy te

eddeprotonatof#

terminiamino

protonatedof#

residuescharged

negativelyof#

residuescharged

positivelyof#

arg ech

Net

[ ] )()()( NKHH

tNpN += +

+

N(p) is the number of protonated residues

N(t) is the total number of residues of a specific type

[H+] is the hydrogen ion concentrationK(N) is the aminoacid dissociation constant

3. Product Purification /Contaminant Removal:Ion Exchange of Proteins


117/152


.

-

,+

*

).

-

,+

*

)+.

-

,+

*

).

-

,+

*

)=.

-

,+

*

)

rminicarboxy te

eddeprotonatof#

terminiamino

protonatedof#

residuescharged

negativelyof#

residuescharged

positivelyof#

arg ech

Net

Calculate the net charge of the following peptide NH2-Lys-Pro-Lys-COOH

68.10

12.9

19.2

chargedpositivelyLysine

nInformatio

=

=

=

sa

a

a

c

a

pK

pK

pK

41.10

9.1

aminoacidnonpolarProline

nInformatio

=

=a

a

c

a

pK

pK

2 0 1 from lys 1 from lys

,),),),),) eddeprotonatof#protonatedof#negativelyof#positivelyof#Net

68.10

12.9

19.2

chargedpositivelyLysine

nInformatio

=

=

=

sa

a

a

c

a

pK

pK

pK

41.10

9.1

aminoacidnonpolarProline

nInformatio

=

=a

a

c

a

pK

pK


118/152


.-

,+*

).

-

,+*

)+.

-

,+*

).

-

,+*

)=.

-

,+*

)

rminicarboxy te

eddeprotonatof#

terminiamino

protonatedof#

residuescharged

negativelyof#

residuescharged

positivelyof#

arg ech

Net

2 0 1 from lys 1 from lys

[ ][ ]

[ ][ ]

[ ][ ] 19.212.968.10 101010

2charge

Net+

+

+

+

+

+

+

++

+=.

-

,+*

)

H

H

H

H

H

H

0

0.5

1

1.5

2

2.5

3

3.5

0 5 10 15

pH

Netcharge

3. Product Purification /Nonideal effects onChromatographic Separations

Dispersion Wall Effects and Nonequilibrium:


119/152


Dispersion, Wall Effects, and Nonequilibrium:


Basic Concepts

Shuler and Kargi,

Prentice Hall, 2002

Gaussian Peak

i tmax,i = standard deviation

Resolution of Peaks

RS=

tmax,j tmax,i

1 / 2 (tw,i + tw,j)

3. Product Purification /Nonideal Effects onChromatographic Separations

Prediction of Peak Width:


120/152


Prediction of Peak Width:

yi =ymax,i exp -( t tmax,i)

2

2(tmax,i )2

)

*++

,

-..

depends on dispersion and adsorption kinetics

2 =v

kalv =superficial velocity,

ka =surfaceadsorption reaction rate

l=column length

3. Product Purification /Nonideal Effects onChromatographic Separations

Prediction of Peak Height:


121/152


Prediction of Peak Height:

ymax,i is inversely proportional to tmax,i

may depend on other processes

2 vd2

linternal diffusion control,

2

v1/ 2 / d3/ 2

lexternal film control,

2vd

2

DlTaylor dispersion (laminar flow),

3. Product Purification /Scale Up ofChromatographic Separations

To Handle Increased Amount of Product:


122/152


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 increases if dincreases)

3. Fix dbut increase vand l, but maintain ratio of vto lconstant

( will be unchanged, but pressure drop will increase)

4. IncreaseA and volumetric flow rate, such that vis constant

( remains constant, the desired outcome!)

3. Product Purification /Scale Up ofChromatographic Separations

Recent Advances in Chromatographic Packing:


123/152


g p g

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.

3. Chromatographic Separation of Proteins fromCheese Whey

Cream Separator

Raw Milk

Chandrasekaran, R.,


124/152


Heat Treatment

(Optional)

Evaporator

Spray Dryer

NFDM

Standardize

Cheese Milk

Rennet/Acid

Curds

Cheese

Whey

Ultrafiltration

Evaporator/

Spray Dryer

Evaporator/

Spray Dryer

WPC

Dry Whole

Whey

Ultrafiltration

UF Milk

Evaporator/Spray Dryer

MPC

Acid Rennet/Heat

Whey Curds

Evaporator/

Spray Dryer

Dry WholeWhey

Wash

Evaporator/

Spray Dryer

Casein

Chandrasekaran, R.,

MS Thesis, Dept. of


MTU

3. Chromatographic Separation of Proteins fromCheese Whey (cont.)

Chandrasekaran, R.,


125/152


Fluid Sweet Whey

Water 93.7

Total Solid 6.35

Fat 0.5

Protein 0.8Lactose 4.85

Ash 0.5

Lactic Acid 0.05

Table 1.2 Composition of Whey (Weight %) (Kosikowski et al., 1997)

, ,

MS Thesis, Dept. of


MTU

Whey proteins are finding increasing application in the fields of



126/152


y p g g pp

nutrition (protein powder), as an antibiotic, and in other

pharmaceutical applications. Individual whey proteins can be

separated using cation exchange chromatography, using pHchange during elution to recover individual proteins.

Table 1. Isoelectric Points of Major Whey Proteins [1]

Whey Protein Isoelectric Point

-lactoglobulin 5.35-5.49

-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

Whey proteins have a range of molecular weights.



127/152


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



128/152


pH 6.5 to 11 step (Trial 1, 4/3/03)

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

0 20 40 60 80 100 120 140

mL collected

Absorb

ance

1

2

3

4

20 mL HiPrep 16/10 SP XL

Flow Rate = 3 mL/min

Temp. = 4 deg. C

300 mL of 5 g/L protein

solution loaded pH 6.5

Elution

60 mL pH 6.5 buffer

to

1 min. gradient

to

60 mL pH 11 buffer



129/152


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 bein Lane 9, Peak 4.

3. Effects of pH Gradient on Peak Resolution


130/152


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 from0 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

(min)

Conc %B Flow rate

(ml/min)

Fraction

volume

(ml)

Tube A Tube B Valve

position

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.


131/152


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.

1

2

1 2


132/152


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

1

2

1

23


133/152


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.

1

2

3

1

2

3


134/152


Figure 8. Change of 0% pH 11 to 85% pH 11 in 14 min.

1

2

3


135/152

4. Product Preparation / Crystallization

Critical cluster or nucleus is the largest cluster of molecules just

prior to spontaneous nucleation:


136/152


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 m in size.


Steps in nucleation and crystal growth


137/152


B + B B2 + B B3 + B ..

Bn-1 + B Bn a critical cluster is formed

Bn + B Bn+1 6 which undergoes nucleation

Bn+1 + B ' which undergoes crystal growth


Characteristic zones of crystallization


138/152


No crystallizationoccursGrowth of existing crystals,

Formation of new nuclei

New nuclei formspontaneously

Bioseparation Process Science



Transport Processes During Crystallization

Growth Rate of Crystals


139/152


Bioseparation Process Science


Growth Rate of Crystals,

G = dLdt

= kgC

where C = C* Cis

the degree of saturation

Molecules in

crystal

Molecules insolution

4. Product Preparation / CrystallizationThermodynamics of Homogeneous Nucleation

Free Energy Change for Homogeneous Nucleation

GHomogeneous = GSurface formation + GClustering Bioseparation Process Science


140/152


GHomogeneous GSurface formation GClustering

GSurface formation = 4r2

sl

where slis the surface tension of the solid / liquid interface

GClustering = RTlnC

C*

!"

$%

4 / 3r3

Vmolar,solid

The critical nucleus, rc, is where there is a maximum in GHomogeneous

dGHomogeneous

dr = 0 = 8rcsl RTln

C

C*!"

$%

4rc2

Vmolar,solid

rc =2slVmolar,solid

RTlnC

C*

!"

$%

Useful calculation when seeding aCrystallization process

p

Garcia et al., Blackwell Science, 1999,

Pages127-140

=

4. Product Preparation / CrystallizationRate of Formation of Nuclei,dN/dt

kinetics,reactiontoanalogousisNucleation


141/152


&&&&&

%

$

#####

"

!

&%$#

"! &

%$#

"!

=

&&%

$##"

! ==

2

*

33

2

solidmolar,

3

max0

ln3

16-exp

-exp

C

CTR

VA

TR

GA

dt

dNB

sl

4. Product Preparation / CrystallizationBatch Crystallization, Solid Phase Balance

i

Crystals,ofNumbereCummulativ1.

L

N


142/152


growth.crystaltoduerangesizespecificaleavingandentering

crystalsofnumberthetracksonbalanceA

curvevsofSlope

Density,Population2.

or

size,versus

n

LN

n

L

4. Product Preparation / CrystallizationBatch Crystallization, Population Balance Equation

,),),),) ofNumberofNumberofNumberofNumber


143/152


..

.

-

,

++

+

*

)

+

..

.

-

,

++

+

*

)

=

..

.

-

,

++

+

*

)

+

..

.

-

,

++

+

*

)

LLLL range,ofout

growingcrystals

ofNumber

range,within

endatcrystals

ofNumber

range,into

growingcrystals

ofNumber

range,within

initiallycrystals

ofNumber

.forrangesizeis2subscript

range,sizesmallerais1subscript

step.timesmallais(dL/dt),sizecrystalof

rategrowthisGrange,sizeisvolume,is

22final11initial

L

t

LV

tnGVLnVtnGVLnV

+=+

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

0ttdllddbdi id tLtLV


144/152


0

alloverconstantaisGAssuming

0)(

0.togotoandallowandand,,bydivide

=+

=+

dL

dnG

dt

dn

L

dL

Gnd

dt

dn

tLtLV

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

nucleationfor(BCs)conditionsboundary


145/152


&%

$#"

!=

&%

$#"

!=

==

==

G

LtuBn

G

sL

sG

Bn

nL

G

BnL

nt

DomainLaplacein theexp

TransformsLaplaceusingBCsandequationbalancepopulationsolve

finiteis,as

0,at

00,at

0

0

0

4. Product Preparation / CrystallizationBatch Crystallization, Cumulative Crystal Mass

eunit volumpermasscrystalcumulativeis


146/152


430

c

c

3

0

c

4

1

,asand

factorshapeaisandsolidcrystalofdensityiswhere

py

tGBkWM

L

k

dLLnkM

v

v

L

v

==

= (

4. Product Preparation / CrystallizationBatch Crystallization, Cooling Curve

illib hd iifd

achievetoiprelationshetemperatur-timetheDetermine


147/152


330

c

Wofchangeofrate-ionconcentratsoluteofchangeofrate

ationcrystallizbatchduringationsupersaturofdegreeconstanta

tGBkdt

dC

dt

dW

dt

dC

v=

=

=

4. Product Preparation / CrystallizationBatch Crystallization, Cooling Curve (cont.)

ofratetheation,supersaturofdegreeconstantaachieveto


148/152


4

thatfindwe0,at

,form,start tocrystalsthatretemperatuthefromgintegratin

toalproportionbemustchangeetemperatur

430

c

0

0

330

c

T

v

vT

k

tGBk

-TT

t

T

tGBkdt

dTk

dt

dC

dt

dC

=

=

==

4. Product Preparation / CrystallizationContinuous Crystallization, Solid Phase Balances

Number of

crystals growing

)++

,..

Number of)+

,.

Number of

crystals growing

)++

,..

Number of)+

,.


149/152


crystals growing

into range, L,

over a time, t*

+

+++ -

.

..

.

+ crystals entering

range, L, by flow*

+

++ -

.

..=

crystals growing

out of range, L,

over a time, t*

+

+++ -

.

..

.

+ crystals leaving

range, L, by flow*

+

++ -

.

..

V G1 n1t+ Qnin Lt= V G2 n2 t+ V nLt

Vis volume, Lis size range, Gis growth rate

of crystal size (dL / dt), tis a small time step,

Qis 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.


divide by L, and tand allow Land tto go to 0,

and assuming that no crystals are entering, nin

= 0,


150/152


and that Gis constant.

V Gdn

dL+ Qn = 0

Restating in terms of residence time, = VQ

dn

dL+

n

G= 0

Boundary Condition,L = 0, n = no =B

o

G


Population density solution,

n = no exp L!

#$&


151/152


n n exp

G"

#%

&

M = c kv n0

L

( L3

dL

wherecis density of crystal solid and kvis a shape factor

M = 6c kv noG G

3

3 G

3

3+ G

2

2L +

1

2GL

2+

1

6L

3!"

$%exp -

L

G

!"

$%

!"#

$%&

and asL ,

M = W = 6c kv no G44

4. Product Preparation / CrystallizationContinuous Crystallization, Advantages

Advantages:

1. Input of solute helps to maintain a constant degree of


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saturation, C

2. Desirable for determining growth rates and other kinetic

parameters, but are not popular in industrial applications.

recovery and purification of fermentation products

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