CP504 – ppt_Set 07
Cellular kinetics and associated reactor design:
Reactor Design for Cell Growth
Prof. R. Shanthini
30 Nov 2012
1
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
30 Nov 2012
2
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
30 Nov 2012
3
Batch Fermenter
Combining (43) and (44), we get
dCX
dt
= μ CX
(45)
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
4
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
(47)
μ=
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
30 Nov 2012
5
Batch Fermenter
Substituting μ in (45) by the Monod Model given by (47), we get
dCX
dt
=
μm CS
KS + CS
CX
(48)
Equation (48) could be integrated only if we know how CS
changes with either CX or t.
How to do that?
Prof. R. Shanthini
30 Nov 2012
6
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
30 Nov 2012
7
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
30 Nov 2012
8
Batch Fermenter
YX/S =
rX
- rS
(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)
Prof. R. Shanthini
30 Nov 2012
(51)
9
Batch Fermenter
Substituting CS from (51) in (49) and integrating, we get
(
(
μm (t - t0) =
KS YX/S
CX0 + CS0YX/S
KS YX/S
+
)( )
)( )
+1
CX0 + CS0YX/S
ln
CS0
CS
ln
CX
CX0
(52)
where
(CX – CX0) = YX/S (CS0 – CS)
Prof. R. Shanthini
30 Nov 2012
(51)
10
Batch Fermenter
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.
Prof. R. Shanthini
30 Nov 2012
11
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 = 1 + 0.6 (10 – CS)
t is calculated using (50) as follows:
0.935 t =
(
0.71 x 0.6
1 + 10 x 0.6
(
)( )
)( )
+1
0.71 x 0.6
+
1 + 10 x 0.6
ln
CX
1
ln
10
CS
CX is calculated using (48).
Prof. R. Shanthini
30 Nov 2012
12
Exercise 1 worked out using the calculator/spread sheet:
specify
CS
Calculate t
using (50)
10
Calculate
CX using
(49)
1
Calculate
dCX/dt
using (46)
9.95
1.03
0.0317
0.9335
9.8
1.06
0.0624
0.9332
9.85
1.09
0.0923
0.9329
0
Continue
until CS
becomes 0
Prof. R. Shanthini
30 Nov 2012
13
Exercise 1 worked out using the calculator/spread sheet:
12
10
8
CS
CX
6
4
2
0
0
Prof. R. Shanthini
30 Nov 2012
1
Time (in hr)
3
2
14
Exercise 1 worked out using the calculator/spread sheet:
12
dCx/dt
10
8
CS
CX
6
4
2
0
0
Prof. R. Shanthini
30 Nov 2012
1
Time (in hr)
3
2
15
Exercise 1 worked out using an ODE solver:
Programme written in MATLAB
[t,y] = ode45(@CP504Lecture_07,[0:0.01:3],[1; 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
30 Nov 2012
16
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
17
Exercise 1 worked out using an ODE solver:
mumax = 0.935;
plot(t,y(:,1),'b',t,y(:,2),'r')
Ks
= 0.71;
legend('Cell','Substrate')
mu= mumax*y(:,2)./(Ks+y(:,2));
ylabel('Concentration (g/L)')
rX = mu.*y(:,1);
xlabel('Time (h)')
plot(t,rX,'g')
Prof. R. Shanthini
30 Nov 2012
18
Plug-flow Fermenter at steady-state
F
CXi, CSi
F
CX, CS
θ = V/F
(
)()
( )( )
KS YX/S
μm θ =
CXi + CSiYX/S
KS YX/S
+
CXi + CSiYX/S
ln
+1
CSi
CS
ln
CX
CXi
(53)
where
(CX – CXi) = YX/S (CSi – CS)
Prof. R. Shanthini
30 Nov 2012
(54)
19
Continuous Stirred Tank Fermenter (CSTF) at steady-state
F
CXi, CSi
- also known as chemostat
V
CX, CS
F
CX, CS
- 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
30 Nov 2012
20
Continuous Stirred Tank Fermenter (CSTF) at steady-state
F
CXi, CSi
V
CX, CS
F
CX, CS
Mass balance for cells over V:
FCXi + rX V = FCX
Prof. R. Shanthini
30 Nov 2012
(55)
21
Continuous Stirred Tank Fermenter (CSTF) at steady-state
Equation (55) gives
CX - CXi
V
F
=
rX
(56)
Introducing Dilution Rate D as
1
F
D =
V
=
θ
(57)
in (56), we get
CX - CXi
1
D
Prof. R. Shanthini
30 Nov 2012
=
rX
(58)
22
Continuous Stirred Tank Fermenter (CSTF) at steady-state
Since rX = μ CX, (58) becomes
CX - CXi
1
D
=
μ CX
(59)
If the feed is sterile (i.e., CXi = 0), (59) gives
(60)
CX (D – μ) = 0
which means either CX = 0
Prof. R. Shanthini
30 Nov 2012
or
D=μ
23
Continuous Stirred Tank Fermenter (CSTF) at steady-state
If D = μ, then
D
=
μ
=
μm CS
(61)
KS + CS
(61) can be rearranged to give CS as
CS =
KS D
μm - D
(62)
To determine CX, we need to write the mass balance for
substrate over the CSTF
Prof. R. Shanthini
30 Nov 2012
24
Continuous Stirred Tank Fermenter (CSTF) at steady-state
F
CXi, CSi
V
CX, CS
F
CX, CS
Mass balance for substrate over V:
FCSi = FCS + (-rS) V
Prof. R. Shanthini
30 Nov 2012
25
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)
Prof. R. Shanthini
30 Nov 2012
(64)
26
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
CX
= YX/S
Prof. R. Shanthini
30 Nov 2012
(
CSi -
KS D
μm - D
)
(66)
27
Continuous Stirred Tank Fermenter (CSTF) at steady-state
CS =
KS D
(62)
μm - D
which is valid only when D < μm
CX = YX/S
(
CSi -
KS D
μm - D
)
(66)
which is valid only when
CSi > KS D / (μm - D)
D < CSi μm / (KS + CSi)
Prof. R. Shanthini
30 Nov 2012
28
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)
Prof. R. Shanthini
30 Nov 2012
(67)
29
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
30 Nov 2012
30
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
30 Nov 2012
31
Continuous Stirred Tank Fermenter (CSTF) at steady-state
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.
Prof. R. Shanthini
30 Nov 2012
32
Exercise 2 worked out using the calculator/spread sheet:
Plot the following using excel / MATLAB
From (60): CS =
0.71 D
0.935 - D
(
From (64): CX = 0.6 10 -
g/L
0.71 D
)
g/L
0.935 - D
From (65): DC = CSi μm / (KS + CSi)
= 10 x 0.935 / (0.71+10) = 0.873 per h
Prof. R. Shanthini
30 Nov 2012
33
Exercise 2 worked out using the calculator/spread sheet:
Prof. R. Shanthini
30 Nov 2012
DC = 0.873
34
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.
Prof. R. Shanthini
30 Nov 2012
DC = 0.873
35