Bioreactor Analysis and Operation
Chapter 9&10 (textbook)
- Overview of bioreactors
- Modified batch and continuous reactors
- Scale up/down
- Immobilized cell systems
- Operation consideration
- Sterilization
- Bioreactor Instrumentation and control
Bioreactor Analysis and Operation
- Fermentation processes
- solid state: water content: 40~ 80%, mostly mold
fermentation on agriculture products and food:
rice, wheat, barley, corn and soybean.
e.g.rotary drum fermentator
- submerged systems: water content > 95%
e.g. bacteria, yeast.
Bioreactor Analysis and Operation
- Overview of bioreactors for submerged system
- Classification:
operation modes:
- batch: stirred tank
- continuous: chemostat, fluidized-bed
- modified types of the above modes:
fed-batch, chemostat with recycle,
multi-stage continuous reactors
Oxygen supply:
- aerobic: airlift
- anaerobic
Form of biocatalyst:
- free cell (enzyme)
- immobilized cell (enzyme)
packed-bed, membrane reactor
Industrial Bioreactor
Glacial Lakes Energy in Watertown, South Dakota
47+ million gallon per year ethanol production .
World's Largest Industrial Fermenter (Chem. Eng. News,10-Apr-78)
The fermenter is 200' high and 25 ft diam.
Requirements for Cultivation Methods
• Biomass concentration which must remain high
• Sterile conditions being maintained
• Effective agitation so that the distribution of
substances in the reaction is uniform
• Heat removal
• Creation of the correct shear conditions - high
may damage cells, low may lead to flocculation
or growth on wall and stirrer
Chemostat with Cell Recycle
- To keep the cell concentration higher than the normal
steady-state level in a chemostat.
- To increase the cell and growth-associated product yield.
- For low-product-value processes: e.g. waste treatment.
fuel ethanol
,
X1, S
v
Chemostat with Cell Recycle
Cell mass balance (qp=0, kd ≈0, X0=0, Monod equation is applied):
where µ=µnet=µg-kd
A chemostat can be operated at dilution rates higher than
the specific growth rate when cell recycle is used.
Chemostat with Cell Recycle
When kd=0
Chemostat with Cell Recycle
Mass balance on growth-limiting substrate (qp=0, kd ≈0, X0=0,
Monod equation is applied):
FS0  FS  V g X 1
1
dS
 (1   ) FS  V
dt
M
YX / S
At steady state, dS/dt  0,
D M
X1 
Y X / S (S0  S )
g
Since  g  [1   (1  C )]D,
Y X / S (S0  S )
M
M
K s D(1    C )
S
 m  D(1    C )
K s D(1    C )
X1 
, X1 
[S0 
]
1    C
[1   (1  C )]
 m  D(1    C )
YX / S
Chemostat with Cell Recycle
No recycle
µm=1.00h-1, S0=2.0g/l, Ks=0.01 g/l, Yx/s=0.5 g/g,
concentration factor C=2.0 and recycle ratio α=0.5
Chemostat with Cell Recycle
Cell mass balance around the cell separator.
X1, S
v
(1   ) FX1  FX 2  FCX1  X 2
The average residence time in cell separator
Vcell separator

(1   ) F
Example-Chemostat with Cell Recycle
Organisms are cultured in a chemostat with cell recycle. The
system is operated under glucose limitation.
F  100 ml/h, V  1000ml, S 0  10 g glucose/L
M
YX/S
 0.5gdw cells/g substrate; μ m  0.2h 1 ,
Ks  1g/L, C  1.5, α  0.7, X 0  0, K d  0
Determine specific growth rate μnet, S in the reactor
effluent, cell concentration in the recycle stream (CX1)
and in the concentrator effluent (X2)
If the concentrator has a volume of 300 mL, what is the
residence time in it?
Fed-Batch
Nutrients are continuously or semi-continuously fed,
while effluent is removed discontinuously.
- overcome substrate inhibition or catabolite repression by
intermittent feeding of substrate by maintaining low
substrate concentration.
for production of secondary metabolites e.g. antibiotics,
lactic acid, E. Coli making proteins from recombinant DNA
technology.
Fed-Batch
Analysis of fed-batch with substrate continuously fed
and no output: t=0, V0=0, F is constant.
Volume:
dV
 F  V  Vo  Ft
dt
At quasi steady state, S added=S consumed,
X, S, P concentrations are constant.
Cell mass balance: FX  V X  d ( XV )
O
net
dt
since
d ( XV )
dX
dV
V
X
, then
dt
dt
dt
FX O  V net X  V
V net X  X
dX
dV
X
dt
dt
dV
1 dV F
  net 
 D
dt
V dt V
Do
F
F
 net  

V V0  Ft 1  D0 t
 net  D 
m S
Ks  S
if K d  0
Then S 
(Monod growth model applied)
Ks D
m  D
Fed-Batch
M
where S≈0
assuming X≈Xm
M
where Xt=Xt0 at t=0
Fed-Batch
(g)
qp (i.e. g product/g cells-min)
at Pt=Pt0, t=0
Fed-Batch
Example: Fed- Batch
In a fed-batch culture operating with intermittent
addition of glucose, the value of V is given at time
t=2hr, when the system is at quasi-steady state.
dV
F
 200 ml/h, V  1000ml, S0  100 g glucose/L
dt
M
YX/S
 0.5gdw cells/g glucose; μ m  0.3h 1 ,
Ks  0.1g/L, X t 0  30 g cells
Determine V0.
At t=2 hr, find S, X and Xt and P at quasi-steady state if
qp=0.2 g product/g cells-h, P0=0.