Post on 17-Jan-2016
07 Oct 2011 Prof. R. Shanthini
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Cellular kinetics and associated reactor design:
Reactor Design for Cell Growth
CP504 – Lecture 7
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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.
07 Oct 2011 Prof. R. Shanthini
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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)
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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.
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
( )( )))( (
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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).
07 Oct 2011 Prof. R. Shanthini
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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:
07 Oct 2011 Prof. R. Shanthini
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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:
07 Oct 2011 Prof. R. Shanthini
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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:
07 Oct 2011 Prof. R. Shanthini
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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]);
07 Oct 2011 Prof. R. Shanthini
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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)')
07 Oct 2011 Prof. R. Shanthini
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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')
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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)
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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)
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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)
07 Oct 2011 Prof. R. Shanthini
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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)
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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
07 Oct 2011 Prof. R. Shanthini
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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):
07 Oct 2011 Prof. R. Shanthini
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Exercise 2 worked out using the calculator/spread sheet: