10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13: Biological Reactors- Chemostats
This lecture covers: theory of the chemostat, fed batch or semi-continuous fermentor
operations
Biological Reactors (Chemostat)
Concentration/Combustion constant
Biological CSTR
[S]0
F
[S], x
V
F
Figure 1. Diagram of a chemostat.
F = Volumetric flow rate
x=
biomass
volume
[S]0 = Concentration of growth limiting substrate. (for growing cells)
At steady-state, biomass balance
In – Out + Prod = Acc
Sterile feed: In=0
Steady state: Acc=0
− Fx + rx V = 0 at steady-state
Cell growth kinetics rx = μ x
− Fx + μ x V = 0
F
Solve μ =
V
F 1
=
D=Dilution rate ≡
V τ
μ=D
Biological
Mechanical
Cite as: K. Dane Wittrup, course materials for 10.37 Chemical and Biological Reaction Engineering, Spring
2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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When at steady-state, can control cell mass.
Allows precisely reproducible cell states.
Not easy to run at steady-state.
Material balance on [S] (sugar concentration)
In – Out + Prod = Acc
0
F [ S ]0 − F [ S ] −
at steady-state
1
μx V = 0
Yx
s
mass biomass created
Yield coefficient
mass substrate consumed
Divide by V
μx
D ([ S ]0 − [S]) =
Yx
change in sugar
concentration
s
At steady-state
μ=D
x = Yx ([ S ]0 − [S])
s
What is the value of
μ = f ([S])
[S] ? What more information do we need?
Å must choose a growth model to connect
μ
and
[S]
Monod growth model:
μ=
μmax [ S ]
K s + [S ]
[S ] =
Æ at steady-state Æ
Ks D
μmax − D
D=
μmax [ S ]
K s + [S ]
substitute in x equation
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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⎛
Ks D ⎞
x = Yx ⎜ [ S ]0 −
⎟
μmax − D ⎠
s ⎝
Specifying
μmax , K s , Yx , D , [ S ]0 , can predict x , [S] .
s
x < 0 is non-physical but formally in solution
μmax − D
can go to 0. If you turn knobs incorrectly: if D is too high, the cells cannot
grow fast enough to reach steady-state. Washout will occur.
so use x = 0 to find Dmax
Dmax =
μmax [ S ]0
K s + [ S ]0
For D > Dmax “washout”, no steady-state.
S
x, S
x
mass
volume
washout
D
Dmax
Figure 2. Biomass/volume versus dilution rate. Beyond the maximum dilution rate,
washout occurs.
For real systems K s << [ S ]0 . Most cell growth systems reach maximum at fairly low
concentrations; hence x is flat, then drops off sharply.
If biomass is the product, is there a best operating condition?
What should we consider?
dx
optimize x with respect to D? D=0 (no, because this would be batch reactor)
dD
Define productivity as
biomass
(reactor
volume )( time )
= xD
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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d ( xD)
= 0 for optimum. (is a maximum)
dD
x
xD
D
D
Doptimum
Figure 3. Left: Biomass/volume versus dilution rate. Right: Productivity versus
dilution rate.
⎛
Ks
Doptimum = μmax ⎜⎜1 −
K s + [ S ]0
⎝
⎞
⎟⎟
⎠
K s << [ S ]0
Doptimum ≈ μ max
≈ Dmax
Close to washout conditions.
Operability would be difficult. We would not want to run too close to washout
conditions.
Fed-batch fermentor (microbes or mammalian cells)
-used to achieve very high cell densities (e.g. hundreds of grams cell dry weight
(c.d.w)/liter)
If you want x final =
100 g
L
If Yx ≈ 0.5 , [S]0 =
wt
200 g
≈ 20%
Å Toxic, sugar content cells will die
L
volume
s
Why do we not feed all at once? Cells will die.
Calculate medium feed rate in order to hold
μ
constant.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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[S]0
F
x(t)
S(t), very small
Figure 4. Diagram of a fed-batch fermentor.
If
μ
is constant, biomass = biomass t =0 e
μt
There is a dilution term, because as we feed in fresh medium, volume will change.
Volume often doubles.
x V = x0 V0 e μt
μ x0 V0 e μt
F[S ] =
0
Feed
sugar feed
Yx
s sugar consumed
Assume all converted into biomass.
F=
x0 V0
μ e μt
[ S ]0 Yx
s
Exponential flow rate. Typically
μ
specified as “small.”
Dilution:
dV
=F
dt
⎛
⎞
x0
⎜
⎟
V(t ) = V0 ⎜1 +
e μt − 1) ⎟
(
⎜ [ S ]0 Yx
⎟
s
⎝
⎠
x=
biomass
V
x0 e μt
=
1+
x0
e μt − 1)
(
Yx [ S ]0
s
=
x0 V0 e
V
μt
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
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“Logistic equation”
x
Figure 5. Graph of logistic growth.
t
If product is something cells are making:
Product synthesis kinetics
1)
1 dP
= αμ
x dt
P≡
2)
growth associated (e.g. ethanol)
product
volume
1 dP
=β
x dt
t
P = β ∫ xdt
0
not growth associated (e.g. antibiotics, proteins, antibodies)
integrate for amount of product.
10.37 Chemical and Biological Reaction Engineering, Spring 2007
Prof. K. Dane Wittrup
Lecture 13
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2007. MIT OpenCourseWare (http://ocw.mit.edu), Massachusetts Institute of Technology. Downloaded on
[DD Month YYYY].