1. Reactor Design for Cell Growth
CP504 – ppt_Set 07
Cellular kinetics and associated reactor design:
Reactor Design for Cell Growth
Prof. R. Shanthini Nov 2012

2. Cell Growth Kinetics
Using the population growth model, we could write the cell growth rate (rX) as rX
= μ CX
(43)
where
μ : specific growth rate (per time)
CX : cell concentration (dry cell weight per unit volume)
Prof. R. Shanthini Nov 2012

3. Batch Fermenter Mass balance for the cell:
0 + (rX) V = 0 + d(VCX) / dt
which for a batch reactor with constant volume reacting mixture gives
dCX / dt = rX
(44)
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
Prof. R. Shanthini Nov 2012

4. Batch Fermenter Combining (43) and (44), we get dCX = μ CX (45) dt
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 Nov 2012

5. Cell Growth Kinetics
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:
μm CS
μ =
(47)
KS + CS
where μm and KS are known as the Monod kinetic parameters.
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.
Prof. R. Shanthini Nov 2012

6. Batch Fermenter
Substituting μ in (45) by the Monod Model given by (47), we get
μm CS =
KS + CS
(48) CX
dCX dt
Equation (48) could be integrated only if we know how CS changes with either CX or t.
How to do that?
Prof. R. Shanthini Nov 2012

7. Batch Fermenter It is done as follows:
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:
YX/S = rX / (-rS)
(49)
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 Nov 2012

8. Batch Fermenter Mass balance for substrate:
0 = 0 + (-rS) V + d(VCS) / dt
which for a batch reactor with constant volume reacting mixture gives
dCS / dt = -(-rS)
(50)
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
Prof. R. Shanthini Nov 2012

9. Batch Fermenter YX/S = - rS rX (49) dCX / dt = rX (44)
dCS / dt = -(-rS)
(50)
Combining the above equations, we get
dCX / dCS = -YX/S
which upon integration gives
(CX – CX0) = YX/S (CS0 – CS)
(51)
Prof. R. Shanthini Nov 2012

10. ( ) ( ) ( ) ( ) Batch Fermenter
Substituting CS from (51) in (49) and integrating, we get ( ) ( )
KS YX/S CX
μm (t - t0) =
+ 1 ln
CX0 + CS0YX/S
CX0 ( ) ( )
KS YX/S
CS0 ln
(52) +
CX0 + CS0YX/S CS
where
(CX – CX0) = YX/S (CS0 – CS)
(51)
Prof. R. Shanthini Nov 2012

11. Batch Fermenter Exercise 1:
The growth rate of E. coli be expressed by Monod kinetics with μm = 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.
Prof. R. Shanthini Nov 2012

12. Exercise 1 worked out using the calculator/spread sheet:
CS is varied from 10 g/L to 0.
CX is calculated using (49) as
CX = (10 – CS)
t is calculated using (50) as follows: ( ) ( )
0.71 x 0.6 CX
0.935 t =
+ 1 ln
x 0.6 1 ( ) ( )
0.71 x 0.6 10 + ln CS
x 0.6
CX is calculated using (48).
Prof. R. Shanthini Nov 2012

13. Calculate dCX/dt using (46)
Exercise 1 worked out using the calculator/spread sheet:
specify CS
Calculate CX using (49)
Calculate t using (50)
Calculate dCX/dt using (46) 10 1
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
Prof. R. Shanthini Nov 2012

14. CS CX Exercise 1 worked out using the calculator/spread sheet:
Prof. R. Shanthini Nov 2012

15. CS CX Exercise 1 worked out using the calculator/spread sheet:
Prof. R. Shanthini Nov 2012

16. Exercise 1 worked out using an ODE solver:
Programme written in MATLAB
[t,y] = 10]);
function dydt =CP504Lecture_07(t,y)
%data given
mumax = 0.935; % per hr
Ks = 0.71; % g/L
YXS = 0.6;
%Monod model
mu = mumax*y(2)/(Ks+y(2));
%rate equations
rX = mu*y(1);
rS = -rX/YXS;
dydt=[rX; rS]
Prof. R. Shanthini 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 Nov 2012

18. Exercise 1 worked out using an ODE solver:
mumax = 0.935;
Ks = 0.71;
mu= mumax*y(:,2)./(Ks+y(:,2));
rX = mu.*y(:,1);
plot(t,rX,'g')
plot(t,y(:,1),'b',t,y(:,2),'r')
legend('Cell','Substrate')
ylabel('Concentration (g/L)')
xlabel('Time (h)')
Prof. R. Shanthini Nov 2012

19. ( ) ( ) ( ) ( ) Plug-flow Fermenter at steady-state F F θ = V/F
CXi, CSi
CX, CS
KS YX/S
CXi + CSiYX/S (
+ 1 ) ln
CXi CX ( )
μm θ = +
KS YX/S
CXi + CSiYX/S ( ) ln CS
CSi ( )
(53)
where
(CX – CXi) = YX/S (CSi – CS)
(54)
Prof. R. Shanthini Nov 2012

20. Continuous Stirred Tank Fermenter (CSTF) at steady-state
CXi, CSi
CX, CS V
- also known as chemostat
- 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
Prof. R. Shanthini Nov 2012

21. Continuous Stirred Tank Fermenter (CSTF) at steady-state
CXi, CSi
CX, CS V
Mass balance for cells over V:
FCXi + rX V = FCX
(55)
Prof. R. Shanthini Nov 2012

22. Continuous Stirred Tank Fermenter (CSTF) at steady-state
Equation (55) gives V
CX - CXi
(56) = rX F
Introducing Dilution Rate D as =
(57) F V
D = 1 θ
in (56), we get 1 D =
CX - CXi rX
(58)
Prof. R. Shanthini Nov 2012

23. Continuous Stirred Tank Fermenter (CSTF) at steady-state
Since rX = μ CX, (58) becomes 1 D =
CX - CXi
μ CX
(59)
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 Nov 2012

24. Continuous Stirred Tank Fermenter (CSTF) at steady-state
If D = μ, then
μm CS
KS + CS D = μ =
(61)
(61) can be rearranged to give CS as
KS D CS =
(62)
μm - D
To determine CX, we need to write the mass balance for substrate over the CSTF
Prof. R. Shanthini Nov 2012

25. Continuous Stirred Tank Fermenter (CSTF) at steady-state
CXi, CSi
CX, CS V
Mass balance for substrate over V:
FCSi = FCS + (-rS) V
Prof. R. Shanthini Nov 2012

26. Continuous Stirred Tank Fermenter (CSTF) at steady-state
which is rearranged to give
(-rS) = D (CSi - CS)
(63)
(58) gives
rX = D (CX - CXi )
Using the above equations in the definition of yield factor, we get
(CX – CXi) = YX/S (CSi – CS)
(64)
Prof. R. Shanthini Nov 2012

27. ) ( Continuous Stirred Tank Fermenter (CSTF) at steady-state
Since the feed is sterile, (6 4) gives
CX = YX/S (CSi – CS)
(65)
(62) is
KS D CS =
(62)
μm - D
Therefore, we have ( )
KS D CX =
YX/S
CSi -
(66)
μm - D
Prof. R. Shanthini Nov 2012

28. ( ) Continuous Stirred Tank Fermenter (CSTF) at steady-state KS D CS =
(62)
μm - D
which is valid only when D < μm ( )
KS D CX =
YX/S
CSi -
(66)
μm - D
which is valid only when
D < CSi μm / (KS + CSi)
CSi > KS D / (μm - D)
Prof. R. Shanthini Nov 2012

29. Since D < CSi μm / (KS + CSi) < μm
Continuous Stirred Tank Fermenter (CSTF) at steady-state
Since D < CSi μm / (KS + CSi) < μm
critical value of the Dilution Rate is as follows:
DC = CSi μm / (KS + CSi)
(67)
Prof. R. Shanthini Nov 2012

30. Continuous Stirred Tank Fermenter (CSTF) at steady-state
If μm equals or less than DC, then CX is negative.
That is impossible.
So, when μm equals or less than DC,
We need to take the solution
CX = 0 of (58), not D = μ
Substituting CX = 0 in CX = YX/S (CSi – CS) gives
CS = CSi
Prof. R. Shanthini Nov 2012

31. Continuous Stirred Tank Fermenter (CSTF) at steady-state
CX = 0 means no cell in the reactor.
CS = CSi means substrate is not utilised.
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
Prof. R. Shanthini Nov 2012

32. Continuous Stirred Tank Fermenter (CSTF) at steady-state
Exercise 2
The growth rate of E. coli be expressed by Monod kinetics with μm = 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.
Prof. R. Shanthini Nov 2012

33. ( ) Exercise 2 worked out using the calculator/spread sheet:
Plot the following using excel / MATLAB
0.71 D
From (60):
g/L CS = D ( )
0.71 D
From (64): CX =
0.6 10 -
g/L D
From (65):
DC = CSi μm / (KS + CSi)
= 10 x / ( ) = per h
Prof. R. Shanthini Nov 2012

34. Exercise 2 worked out using the calculator/spread sheet:
DC = 0.873
Prof. R. Shanthini Nov 2012

35. 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.
DC = 0.873
Prof. R. Shanthini Nov 2012