Prof. R. Shanthini 30 Nov 2012 1 Cellular kinetics and associated reactor design: Reactor Design for...

35
Prof. R. Shanthini 30 Nov 201 1 Cellular kinetics and associated reactor design: Reactor Design for Cell Growth CP504 – ppt_Set 07

Transcript of Prof. R. Shanthini 30 Nov 2012 1 Cellular kinetics and associated reactor design: Reactor Design for...

Prof. R. Shanthini 30 Nov 2012

1

Cellular kinetics and associated reactor design:

Reactor Design for Cell Growth

CP504 – ppt_Set 07

Prof. R. Shanthini 30 Nov 2012

2

Cell Growth Kinetics

rX = μ CX (43)

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

Prof. R. Shanthini 30 Nov 2012

3

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

(44)

Batch Fermenter

dCX / dt = rX

Prof. R. Shanthini 30 Nov 2012

4

Combining (43) and (44), we get

dCX= μ CX

dt(45)

Batch Fermenter

If μ is a constant then integrating (45) gives,

CX = CX0 exp[μ(t-t0)] (46)

where CX = CX0 when t = t0.

Prof. R. Shanthini 30 Nov 2012

5

Cell Growth Kinetics

where μm 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 (47)

Prof. R. Shanthini 30 Nov 2012

6

μm CS =

KS + CS

(48)CX

dCX

dt

Substituting μ in (45) by the Monod Model given by (47), we get

Equation (48) could be integrated only if we know how CS changes with either CX or t.

How to do that?

Batch Fermenter

Prof. R. Shanthini 30 Nov 2012

7

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:

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

It is done as follows:Batch Fermenter

We know (rX) from (43) and/or (44). But we don’t know (-rS). Therefore obtain an expression for (-rS) as shown in the next slide.

Prof. R. Shanthini 30 Nov 2012

8

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

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

Batch Fermenter

Prof. R. Shanthini 30 Nov 2012

9

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

YX/S = - rS

rX(49)

(44)dCX / dt = rX

Combining the above equations, we get

dCX / dCS = -YX/S

which upon integration gives

(CX – CX0) = YX/S (CS0 – CS) (51)

Batch Fermenter

Prof. R. Shanthini 30 Nov 2012

10

Substituting CS from (51) in (49) and integrating, we get

μm (t - t0) = KS YX/S

CX0 + CS0YX/S

+ 1 lnCX0

CX

+KS YX/S

CX0 + CS0YX/S

lnCS

CS0 (52)

where

(CX – CX0) = YX/S (CS0 – CS) (51)

Batch Fermenter

( )( )))( (

Prof. R. Shanthini 30 Nov 2012

11

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

Prof. R. Shanthini 30 Nov 2012

12

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 (48).

Prof. R. Shanthini 30 Nov 2012

13

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:

Prof. R. Shanthini 30 Nov 2012

14

0

2

4

6

8

10

12

0 1 2 3Time (in hr)

CS

CX

Exercise 1 worked out using the calculator/spread sheet:

Prof. R. Shanthini 30 Nov 2012

15

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:

Prof. R. Shanthini 30 Nov 2012

16

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]);

Prof. R. Shanthini 30 Nov 2012

17

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)')

Prof. R. Shanthini 30 Nov 2012

18

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')

Prof. R. Shanthini 30 Nov 2012

19

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() (53)

where(CX – CXi) = YX/S (CSi – CS) (54)

Plug-flow Fermenter at steady-state

Prof. R. Shanthini 30 Nov 2012

20

FCXi, CSi

FCX, CS

VCX, CS

Continuous Stirred Tank Fermenter (CSTF) at steady-state

- Mixing supplied by impellers and/or rising gas bubbles- Complete mixing is assumed (composition of any phases do not vary with position)- Liquid effluent has the same composition as the reactor contents

- also known as chemostat

Prof. R. Shanthini 30 Nov 2012

21

FCXi, CSi

FCX, CS

VCX, CS

Continuous Stirred Tank Fermenter (CSTF) at steady-state

Mass balance for cells over V:

FCXi + rX V = FCX (55)

Prof. R. Shanthini 30 Nov 2012

22

Equation (55) gives

V

F =

CX - CXi

rX

(56)

Continuous Stirred Tank Fermenter (CSTF) at steady-state

Introducing Dilution Rate D as

= (57)F

VD =

1

θ

in (56), we get

1

D =

CX - CXi

rX

(58)

Prof. R. Shanthini 30 Nov 2012

23

Since rX = μ CX, (58) becomes

1

D =

CX - CXi

μ CX

(59)

Continuous Stirred Tank Fermenter (CSTF) at steady-state

If the feed is sterile (i.e., CXi = 0), (59) gives

CX (D – μ) = 0 (60)

which means either CX = 0 or D = μ

Prof. R. Shanthini 30 Nov 2012

24

CS = (62)μm - D

KS D

(61) can be rearranged to give CS as

D = μ (61)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

Prof. R. Shanthini 30 Nov 2012

25

FCXi, CSi

FCX, CS

VCX, CS

Mass balance for substrate over V:

FCSi = FCS + (-rS) V

Continuous Stirred Tank Fermenter (CSTF) at steady-state

Prof. R. Shanthini 30 Nov 2012

26

which is rearranged to give

(-rS) = D (CSi - CS) (63)

Continuous Stirred Tank Fermenter (CSTF) at steady-state

rX = D (CX - CXi )

(58) gives

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

(CX – CXi) = YX/S (CSi – CS) (64)

Prof. R. Shanthini 30 Nov 2012

27

Since the feed is sterile, (6 4) gives

CX = YX/S (CSi – CS) (65)

(62) is

Therefore, we have

CX = (66)YX/S (CSi - )

Continuous Stirred Tank Fermenter (CSTF) at steady-state

CS = (62)μm - D

KS D

μm - D

KS D

Prof. R. Shanthini 30 Nov 2012

28

which is valid only when D < μm

which is valid only when

D < CSi μm / (KS + CSi)

CS = (62)

CX = (66)YX/S (CSi - )

Continuous Stirred Tank Fermenter (CSTF) at steady-state

μm - D

KS D

μm - D

KS D

CSi > KS D / (μm - D)

Prof. R. Shanthini 30 Nov 2012

29

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

(67)

Prof. R. Shanthini 30 Nov 2012

30

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

Prof. R. Shanthini 30 Nov 2012

31

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 WASHOUT situation.

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

CS = CSi means substrate is not utilised.

Continuous Stirred Tank Fermenter (CSTF) at steady-state

Prof. R. Shanthini 30 Nov 2012

32

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

Prof. R. Shanthini 30 Nov 2012

33

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):

Prof. R. Shanthini 30 Nov 2012

34DC = 0.873

Exercise 2 worked out using the calculator/spread sheet:

Prof. R. Shanthini 30 Nov 2012

35DC = 0.873

Exercise 2 worked out using the calculator/spread sheet:

Near washout the reactor is very sensitive to variations in D. Small change in D large shifts in X and/or S.