Cellular kinetics and associated reactor design: Reactor Design for Cell Growth
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Transcript of Cellular kinetics and associated reactor design: Reactor Design for Cell Growth
Prof. R. Shanthini 30 Nov 2012
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Cellular kinetics and associated reactor design:
Reactor Design for Cell Growth
CP504 – ppt_Set 07
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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
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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
(44)
Batch Fermenter
dCX / dt = rX
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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.
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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)
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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
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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:
(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.
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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
(50)dCS / dt = -(-rS)
Batch Fermenter
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(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
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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
( )( )))( (
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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 (48).
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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() (53)
where(CX – CXi) = YX/S (CSi – CS) (54)
Plug-flow Fermenter at steady-state
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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
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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 (55)
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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)
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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
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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
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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) (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)
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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
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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)
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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
(67)
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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 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
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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:
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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.