Transcript Fermentation Technology - Just for sharing ideas

Chapter V
Black box growth
1
Growth without non-catabolic product
Single substrate limited growth
2
Hyperbolic kinetic equation for specific
substrate uptake rate (qS)
q max
S
-qS
0.5 qSmax
0
KS
(qS )  qSmax
CS
CS
K S  CS
KS: substrate affinity [kg substrate/m3]
qSmax: maximal uptake rate [kg S consumed/ kg X present-h]
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Substrate uptake for maintenance
Some part of substrate is used for producing energy needed for
cells maintenance. This part is catabolized with a rate mS.
Hence, mS = kg substrate catabolized for maintenance / kg
biomass present–hour.
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Herbert-Pirt relation for substrate
(qS ) 
μ
max
YSX
 (mS )
It means the substrate taken into cells are used for growth and
maintenance only. Both mS and YSXmax are model parameters.
-qS
1
max
YSX
(-mS)
0
Herbert-Pirt linear plot for
substrate, non-catabolic
product is absent

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Combination of the qS equation with HerbertPirt relation


 
 - qSmax CS
μ
  mS
 K S  CS
 max
YSX

• CS = 0 ,  = mSYSXmax = -kd
• CS >>>,  = [-qSmax-(mS)]YSXmax = max
•  = 0, (-qS) = (-mS), which occurs at CS = CSmin.
The specific growth rate equation based on single substrate is
then:
C  C min
μ  μ max
S
S
K S  CS
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The plot of  - CS
max

-kd
CSmin
For mS = 0, μ  μ
CS
max
CS
K S  CS
(Monod equation)
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Using Herbert-Pirt relation to calculate other
rates
Example:
The case of aerobic growth on glucose, N-source is ammonia
Suppose –qS = 0.3125  + 0.0015
Growth reaction:
-0.3125 C6H12O6 + aNH4+ + bO2 + cH+ + dH2O +
C1H1.8O0.5N0.2 + eCO2
Use the yield and elemental balance to obtain a, b, c, d, e
The catabolic reaction is: -C6H12O6 - 6O2 + 6CO2 + 6H2O
What are (-qNH4+), (-qO2), qCO2, qH+, qH2O?
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How to measure the kinetic parameters?
Obtaining max and YSXmax.
Using batch fermentation at constant volume where CS >>
Hence  = max (constant, during the exponential phase)
qS = qSmax (constant),
YSX = YSX max = (CX-CXo)/(CSo-CS)
Obtaining KS and mS.
Using chemostats,  is the variable to get -qS. Apply HerbertPirt relation and plot –qS vs.  to get YSXmax and mS.
Plot –qS vs. CS to obtain KS and qSmax.
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Further reading
• The Herbert-Pirt relation including non catabolic product
formation
• Extended Herbert-Pirt relation
• Different qP -  function
• Effect of temperature and pH.
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Chapter VI
Growth and product formation in
bioreactors
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Fermentation processes
substrate
fermenter
Downstream
processing
products
wastes
The upstream processing costs about 20 – 50 %, whereas the
downstream processing costs about 50-80 %
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The type of reactors and processes
The reactors systems:
• Aerated stirred tank reactor 0.1 – 1000 m3 with cooling
equipment
• Bubble column 0.1 – 10,000 m3 with airlift and cooling
equipment
• Bioreactors for immobilized cells/biocatalysts (packed bed,
fluidized bed, trickle bed)
The types of processes used are batch, continuous, and fed
batch systems. Mostly, batch and fed batch reactors are used in
industry whereas continuous reactors (chemostats) are used in
laboratories.
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Fermenter utilities
Sterilization:
1. Autoclaving
2. Live steam batch
3. Membrane
4. Heat shock for continuous system (less odor, energy,
degradation, dilution, time)
Mixing by stirrer
mix 20 – 200 seconds, power 1-3 kW/m3.
Oxygen input
50 – 200 mol/m3-h.
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Fermenter utilities (cont’d)
Air filtration, using membrane / bed filter with dry air about 60
m3 air/m3 broth-h.
Compressor, 2 bar, 150oC.
Heat removal
•
Cooling jacket
•
Coil
With cool water temperature at 10oC and capacity is about
50,000 kJ/m3 reactor-h.
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Laboratory chemostat and industrial fermenter
Applikon laboratory chemostat at
Kluyverlab, Delft University of
Technology
Labatt Breweries - London, Ontario
New Vertical Fermenter
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The world largest fermenter
There may be biological waste treatment in larger vessels, but the world's largest
fermenter is shown in these photos taken from Chemical and Engineering News. The
fermenter is 200' high and 25 ft diam. The first photo (Chem. Eng. News, 10-Apr-78)
shows the fermenter being transported on vehicles with tank treads.
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Packed bed fermenter
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What do we want?
High yield of biomass
High max
High temperature tolerant
Grow on low cost substrate
High qP at low 
Extra cellular products
Low viscous biomass
Generally regarded as safe (GRAS)
Mineral medium
Thus, it is important to
analyze the kinetics of
growth and product
formation, designing the
optimal feed profile, role
of transport processes,
maybe use bigger
reactors.
Remember!
qS or  completely
determines the microbial
behavior. It must be
controlled at an optimal
value
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Batch, continuous, and fed batch processes
In batch reactor:  and qS are at the maximum value and not
controlled.
In chemostat:  is controlled at  optimum
In fed batch reactor: qS is controlled.
batch
chemostat
fed
batch
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Constant volume batch reactor
The steps:
1. Add sterile growth medium solution (substrate consist of
C, N, P source, K+, Mg+, salts, vitamin, trace elements)
and choose the right T and pH.
2. Add electron acceptors (O2, NO3-, etc)
3. Inoculate microorganism at initial time with concentration
CXo.
4. Run the fermentation and harvest the broth, proceed to the
downstream processing.
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Constant volume batch reactor
What is the plot CS and CX with time?
Derive from the mass balance, one can obtain that the growth
is exponential, the slope of the curve CX vs. time is rX.
Parameters
State
variables
reactor
operator
micro
organism
V
CS0
CX0
qSmax
max
YSXmax
mS
KS
CS
CX
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Substrate concentration profile
At constant volume:
dCS
 rS  qS C X
dt
For batch reactor:
dCS
max
max
μ max t
 qS C X  qS C Xoe
dt
t = 0, CS = Cso ; CSo  CS 
qSmax
μ


max
C
exp(
μ
t)  1
max Xo
Hence, by putting CS = 0, t end of batch is obtained
The slope in CS vs. t plot is rS.
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Try to sketch the curve CS, CX, , -qS, YSX, rX, -rS versus
time for batch reactor!
Conclusion for the batch reactor
• , YSX, and qS are constant at the maximum values
• Microbial model parameters are determined by combining the
exponential equation with CS and CX data versus time.
• , the most important rate, can not be controlled
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Chemostat
Steady state: constant volume, constant in and out flow rates.
Constant CS and CX.
Parameters
State
variables
reactor
operator
micro
organism
V
CS,in
L,in
qSmax
YSXmax
mS
KS
L,out
CS
CX
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Steps in operating continuous culture
•
Sterilized reactor is filled with sterile growth medium
solution containing substrate at CS,in
•
Air sparging starts to provide electron acceptor
•
Add a small amount of microorganism (inoculum)
•
Start the medium in flow (often L,in and L,out are nearly
equal, not always)
•
Wait until steady state is achieved
The main property of chemostat:
1. CS and CX are independently manipulated by the user by
manipulating transport (to and from the reactor) of substrate
and biomass.
2. Excellent experimental tool to study microbial kinetics,
stoichiometry, under controlled conditions.
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Biomass mass balance in chemostat
In – out + gen = acc
0  C X  L,out  rX V  0
C X  L,out  μC X V
μ
 L,out
V
D
The µ is chosen by manipulating exit flow rate
Hence, all other qi-values are set (stoichiometric coupling)
Does CS depend on CS,in? Use µ(CS) to prove it.
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Critical dilution rate
The maximal value for D (=µ) attainable in the chemostat, is
achieved when the maximal CS in achieved. At critical dilution
rate, CS CS,in.
Dcrit  μ
max CS ,in
 CS ,min
KS  CS ,in
 μ max
Note: CS,in normally in order 10000 mg/l, CS,min order 1 mg/l, KS
order 10 mg/l.
Wash out
If D > Dcrit, the cell will be washed out
Then, sketch the graph CS vs. D.
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Substrate mass balance
 L,inCS ,in   L,out CS ,out  rSV  0
 rS  αCS,in  CS D
α
 L,in
 L,out
This equation together with Herbert-Pirt relation provide rS and
(-qS) from measured concentration, flows, and volume.
What about CX?
CX


D αCS ,in  CS
 rS


 YSX αCS ,in  CS
D
 qS
 (mS )
max
YSX


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The application of chemostat
1. Kinetic studies, at different µ = D one can varies qS, CS to
obtain YSXmax, mS, qSmax, KS.
2. Physiological and genomic micro array study studies of
microorganisms under defined steady state, I.e. substrate,
electron acceptor, N-source, type of limitation for growth.
3. Waste water treatment
4. Industrial fermentation, which is not widely applied
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Chemostat optimization
Normally, the maximal rX and –rS are needed as the economic
parameter
Try to find the optimum D where drS/dD and drX/dD = 0.
Note that rS has a maximum at higher D.
Chemostat wash out dynamic
What happen when D is close to Dcrit?
What happen to CX when D = D’>Dcrit?
dC X
 μC X  D' C X
dt


C X  C X , ss exp μ max  D' t
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Pro and cons of chemostat
(+)
Excellent experimental tool because µ is defined
(-)
Low biomass and product concentration
Loss of biomass in outflow
Relatively prone to be contaminated compare to batch or fed
batch reactors
Microbial selection for non-producing mutants
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Fed batch fermentation
In batch reactor, CS and CX are high. No transport of S or X and
no control on µ.
In chemostat, CS and CX are low. Transport of S or X and
control on µ.
In fed batch reactor. Substrate transport in, not out. No biomass
transport.
Why fed batch?
1. Low CS  no toxicity / osmotic problem
2. High CX  high CP  easier downstream processing
3. Control of µ?
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Fed batch fermentation
Start feeding
CSO
CS
Batch phase
Feeding phase under substrate
limited conditions
CS = 1 – 50 mg/l.
CSO  5000 –
20000 mg/l
time
In substrate limited feeding phase, CS is very low. Thus, one
can use the pseudo steady state condition for substrate mass
balance
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Substrate mass balance in fed batch reactor
d VCS 
  S ,inCS ,in  0  rSV  0
dt
 S ,inCS ,in
 rS 
V
Hence, in a fed batch reactor the substrate conversion (-rS) is
controlled by the operator.
Through controlled rS, µ is controlled.
It is obvious that the reactor volume changes with time.
However, since the change is very small, for simplicity we can
assume constant volume, constant density.
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Rates in fed batch reactor
We start with an assumption that volume is a variable
d VC X 
Biomass mass balance
 rX V  μ VC X 
dt
d VC X  1
μ
dt VC X
Pseudo steady state substrate mass balance
Product mass balance
rS s,inCs,in
 qS 

CX
VCX
d VCP 
 qP VC X 
dt
d VCP  1
qP 
dt VC X
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Possible substrate feeding strategies in a fed
batch fermentation
1. Substrate input (S,inCS,in) is constant
2. µ is maintained at µ optimum that gives maximal qP or
maximal YSP.
3. Substrate input is determined by other known reactor
limitations (oxygen, heat, etc)
Apply Herbert-Pirt equation and biomass mass balance to
calculate rX and CX.
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Biomass mass balance in fed batch reactor
Assume constant volume (DV <<)
dC X
 rX
dt
Combine with Herbert-Pirt relation and eliminate –rS using
its fed batch relation:
dC X
max  S ,inCS ,in
max
 mS C X
 YSX
 YSX
dt
V
This is homogeneous differential equation in CX.
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Biomass mass balance in fed batch reactor
dC X
max  S ,inCS ,in
max
 mS C X
 YSX
 YSX
dt
V
The first explains growth, the later is about maintenance.
Initially CX is low in a fed batch reactor. By neglecting
maintenance:
 C
dC X
max
 YSX
dt
S ,in S ,in
V
 constant
CX increases linearly with time
Later, CX increases, more and more substrate needed for
maintenance  dCX/dt decreases.
End, all substrate needed for maintenance, dCX/dt = 0
C Xmax

 S ,inCS ,in / V

 mS 
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Fed batch reactor with constant feeding rate
The analytical solution for CX(t) in feeding phase.

max

C
1

exp

Y
S
,
in
S
,
in
max
SX  mS t
C X t   YSX
max
YSX
V
 mS 
 C
Xf

max
 mS t
exp  YSX

t = time after start feeding
CXf = biomass concentration at start feeding
Note:
For t >>> 0, CX 
S ,inCS ,in
V (mS )
For t <<< (YSXmax(-mS))-1,
linear growth

max
 mS t
1  exp  YSX

max
 mS 
YSX
t

max
 mS t  1
exp  YSX
max
C X  C Xf  YSX
 S ,inCS ,in
V
t
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Fed batch reactor with constant feeding rate
dC X
max  S ,inCS ,in
max
 mS CX (t )
 YSX
 YSX
dt
V
 rX decreases
rX t  
μt  
CX t  
rS
- qS t  
CX t  
 µ sharply decreases
 qS decreases
 CS decreases slowly
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Fed batch reactor with constant optimum µ
• µ = µopt = constant  qi, Yij are all constant.
• qS constant at qSopt  CS constant.
• Biomass concentration increases exponentially
dC X
 rX  μ opt C X
dt
• All rates increases exponentially
ri t   qiopt C X t 
• Substrate feeding rate increases exponentially
S ,inCS ,in t    rS V  S ,inCS ,in t f expμopt t 
tf = time when feeding starts
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