Cellular kinetics and associated reactor design: Reactor Design for Cell Growth

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07 Oct 2011 Prof. R. Shanthini 1 Cellular kinetics and associated reactor design: Reactor Design for Cell Growth CP504 – Lecture 7

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CP504 – Lecture 7. Cellular kinetics and associated reactor design: Reactor Design for Cell Growth. r X. (41). = μ C X. Cell Growth Kinetics. Using the population growth model, we could write the cell growth rate (r X ) as. where μ : specific growth rate (per time) - PowerPoint PPT Presentation

Transcript of Cellular kinetics and associated reactor design: Reactor Design for Cell Growth

Page 1: Cellular kinetics and associated reactor design: Reactor Design for Cell Growth

07 Oct 2011 Prof. R. Shanthini

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Cellular kinetics and associated reactor design:

Reactor Design for Cell Growth

CP504 – Lecture 7

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Cell Growth Kinetics

rX = μ CX (41)

where

μ : specific growth rate (per time)

CX : cell concentration (dry cell weight per unit volume)

Using the population growth model, we could write the cell growth rate (rX) as

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V for volume of the reacting mixture at time t

CX for concentration of the cells in V at time t

(rX) for cell growth rate in V at time t

Mass balance for the cell:

0 + (rX) V = 0 + d(VCX) / dt

which for a batch reactor with constant volume reacting mixture gives

(42)

Batch Fermenter

dCX / dt = rX

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Combining (41) and (42), we get

dCX= μ CX

dt(43)

Batch Fermenter

If μ is a constant then integrating (43) gives,

CX = CX0 exp[μ(t-t0)] (44)

where CX = CX0 when t = t0.

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Cell Growth Kinetics

where μmax and KS are known as the Monod kinetic parameters.

Mostly, however, μ is not a constant with time. It depends on CS, the substrate concentration.

The most commonly used model for μ is given by the Monod model:

Monod Model is an over simplification of the complicated mechanism of cell growth.

However, it adequately describes the kinetics when the concentrations of inhibitors to cell growth are low.

μ = KS + CS

μm CS (45)

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μm CS =

KS + CS

(46)CX

dCX

dt

Substituting μ in (43) by the Monod Model given by (45), we get

Equation (46) could be integrated only if we know how CS changes with either CX or t.

How to do that?

Batch Fermenter

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Stoichiometry could have helped. But we don’t have such a relationship in the case of cellular kinetics.

Therefore, we introduce a yield factor (YX/S) as the ratio between cell growth rate (rX) and substrate consumption rate (-rS) as follows:

(47)YX/S = rX / (-rS)

It is done as follows:Batch Fermenter

We know (rX) from (41) and/or (42). But we don’t know (-rS). Therefore obtain an expression for (-rS) as shown in the next slide.

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V for volume of the reacting mixture at time t

CS for concentration of the Cells in V at time t

(rS) for substrate utilization rate in V at time t

Mass balance for substrate:

0 = 0 + (-rS) V + d(VCS) / dt

which for a batch reactor with constant volume reacting mixture gives

(48)dCS / dt = -(-rS)

Batch Fermenter

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(48)dCS / dt = -(-rS)

YX/S = - rS

rX(47)

(42)dCX / dt = rX

Combining the above equations, we get

dCX / dCS = -YX/S

which upon integration gives

(CX – CX0) = YX/S (CS0 – CS) (49)

Batch Fermenter

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Substituting CS from (49) in (47) and integrating, we get

μm (t - t0) = KS YX/S

CX0 + CS0YX/S

+ 1 lnCX0

CX

+KS YX/S

CX0 + CS0YX/S

lnCS

CS0 (50)

where

(CX – CX0) = YX/S (CS0 – CS) (49)

Batch Fermenter

( )( )))( (

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Exercise 1:

The growth rate of E. coli be expressed by Monod kinetics with μm = 0.935 hr-1 and KS = 0.71 g/L.

Assume that YX/S is 0.6 g dry cells per g substrate.

CX0 is 1 g/L and CS0 = 10 g/L when the cells start to grow exponentially (i.e., at t = 0). show how CX, CS, and dCX/dt change with respect to time.

Batch Fermenter

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CS is varied from 10 g/L to 0.

( )( ) 0.935 t = 0.71 x 0.6

1 + 10 x 0.6+ 1 ln

1

CX

+0.71 x 0.6

1 + 10 x 0.6( )lnCS

10( )

CX is calculated using (49) as

Exercise 1 worked out using the calculator/spread sheet:

CX = 1 + 0.6 (10 – CS)

t is calculated using (50) as follows:

CX is calculated using (46).

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specify

CS

Calculate CX using

(49)

Calculate t using (50)

Calculate dCX/dt

using (46)

10 1 0

9.95 1.03 0.0317 0.9335

9.8 1.06 0.0624 0.9332

9.85 1.09 0.0923 0.9329

Continue until CS

becomes 0

Exercise 1 worked out using the calculator/spread sheet:

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0

2

4

6

8

10

12

0 1 2 3Time (in hr)

CS

CX

Exercise 1 worked out using the calculator/spread sheet:

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0

2

4

6

8

10

12

0 1 2 3Time (in hr)

dCx/dt

CS

CX

Exercise 1 worked out using the calculator/spread sheet:

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Programme written in MATLAB

Exercise 1 worked out using an ODE solver:

function dydt =CP504Lecture_07(t,y)%data givenmumax = 0.935; % per hrKs = 0.71; % g/LYXS = 0.6; %Monod modelmu = mumax*y(2)/(Ks+y(2));%rate equationsrX = mu*y(1);rS = -rX/YXS;dydt=[rX; rS]

[t,y] = ode45(@CP504Lecture_07,[0:0.01:3],[1; 10]);

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Exercise 1 worked out using an ODE solver:

plot(t,y(:,1),'b',t,y(:,2),'r')legend('Cell','Substrate')ylabel('Concentration (g/L)')xlabel('Time (h)')

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Exercise 1 worked out using an ODE solver:

plot(t,y(:,1),'b',t,y(:,2),'r')legend('Cell','Substrate')ylabel('Concentration (g/L)')xlabel('Time (h)')

mumax = 0.935;Ks = 0.71;mu= mumax*y(:,2)./(Ks+y(:,2));rX = mu.*y(:,1);plot(t,rX,'g')

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F FCXi, CSi CX, CS

θ = V/F

μm θ = KS YX/S

CXi + CSiYX/S( + 1)lnCXi

CX()+

KS YX/S

CXi + CSiYX/S( )lnCS

CSi() (51)

where(CX – CXi) = YX/S (CSi – CS) (52)

Plug-flow Fermenter at steady-state

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FCXi, CSi

FCX, CS

VCX, CS

Continuous Stirred Tank Fermenter (CSTF) at steady-state

Mass balance for cells over V:

FCXi + rX V = FCX (53)

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Equation (53) gives

V

F =

CX - CXi

rX

(54)

Continuous Stirred Tank Fermenter (CSTF) at steady-state

Introducing Dilution Rate D as

= (55)F

VD =

1

θ

in (54), we get

1

D =

CX - CXi

rX

(56)

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Since rX = μ CX, (56) becomes

1

D =

CX - CXi

μ CX

(57)

Continuous Stirred Tank Fermenter (CSTF) at steady-state

If the feed is sterile (i.e., CXi = 0), (57) gives

CX (D – μ) = 0 (58)

which means either CX = 0 or D = μ

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CS = (60)μm - D

KS D

(59) can be rearranged to give CS as

D = μ (59)μm CS

KS + CS

=

If D = μ, then

To determine CX, we need to write the mass balance for substrate over the CSTF

Continuous Stirred Tank Fermenter (CSTF) at steady-state

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FCXi, CSi

FCX, CS

VCX, CS

Mass balance for substrate over V:

FCSi = FCS + (-rS) V

Continuous Stirred Tank Fermenter (CSTF) at steady-state

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which is rearranged to give

(-rS) = D (CSi - CS) (61)

Continuous Stirred Tank Fermenter (CSTF) at steady-state

rX = D (CX - CXi )

(56) gives

Using the above equations in the definition of yield factor, we get

(CX – CXi) = YX/S (CSi – CS) (62)

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Since the feed is sterile, (62) gives

CX = YX/S (CSi – CS) (63)

(60) is

Therefore, we have

CX = (64)YX/S (CSi - )

Continuous Stirred Tank Fermenter (CSTF) at steady-state

CS = (60)μm - D

KS D

μm - D

KS D

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which is valid only when D < μm

which is valid only when

D < CSi μm / (KS + CSi)

CS = (60)

CX = (64)YX/S (CSi - )

Continuous Stirred Tank Fermenter (CSTF) at steady-state

μm - D

KS D

μm - D

KS D

CSi > KS D / (μm - D)

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Since D < CSi μm / (KS + CSi) < μm

DC = CSi μm / (KS + CSi)

critical value of the Dilution Rate is as follows:

Continuous Stirred Tank Fermenter (CSTF) at steady-state

(65)

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If μm equals or less than DC, then CX is negative.

That is impossible.

We need to take the solution CX = 0 of (58), not D = μ

So, when μm equals or less than DC,

Substituting CX = 0 in CX = YX/S (CSi – CS) gives

CS = CSi

Continuous Stirred Tank Fermenter (CSTF) at steady-state

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CX = 0 means no cell in the reactor.

Since the CSTF has a sterile feed (CXi = 0), no reaction takes place unless we inoculate with the cells once again.

So, CSTF gets into a WASHED OUT situation.

To avoid CSTF getting into WASHED OUT situation, we need to maintain D = F / V < DC

CS = CSi means substrate is not utilised.

Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Exercise 2

The growth rate of E. coli be expressed by Monod kinetics with μm = 0.935 hr-1 and KS = 0.71 g/L.

Assume that YX/S is 0.6 g dry cells per g substrate.

The feed is sterile (CXi = 0) and CSi = 10 g/L. show CX and CS changes with dilution rate.

Continuous Stirred Tank Fermenter (CSTF) at steady-state

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Exercise 2 worked out using the calculator/spread sheet:

DC = CSi μm / (KS + CSi)

= 10 x 0.935 / (0.71+10) = 0.873 per h

CS = From (60):

CX = 0.6(10 - )0.935 - D

0.71 D

0.935 - D

0.71 D

Plot the following using excel / MATLAB

From (64):

g/L

g/L

From (65):

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Exercise 2 worked out using the calculator/spread sheet: