Transcript Continuous Bioreactors

Continuous Bioreactors –
Chemostat with Recycle
ChEE 481a
Batch Reactors
• Cell Growth
rX 
 S
dX
 X  m X
dt
KS  S
• Substrate Utilization
m S X
dS
X
rS 


dt
YX / S
K S  S YX / S
• Product (cometabolic contaminants use negative sign)
1
qp 
X
1
qp 
X
dP
 YP / X  g   g  
dt
m S
dP


dt
KS  S
Continuous Reactors
• Chemostat - CSTR - continuous stirred tank reactor for
the cultivation of cells.
– mixing supplied by impellers and rising gas bubbles
– assume complete mixing - composition of any phases do not
vary with position
– liquid effluent has the same composition as the reactor
contents
Mass Balance on Chemostat
Acc = in - out + gen - cons
dci
VR
 Fcif  Fci  VR rfi
dt
– VR - reactor volume
– F - volumetric flow rate of feed and effluent streams (they are
equal)
– ci - concentration of component i in the reactor
– cif - concentration of component i in the influent or feed stream
• If we have a steady state reactor - no changes in
composition with time
dci
then
and
dt
0
F
rfi  (ci  cif )
VR
• Define as the dilution rate -
F 1
D

VR 
– reciprocal of the mean holding or residence time
– detention time
• For cell mass, if we assume a sterile feed:
ci = X and Xf = 0 and
rx = X then X = DX
D= at SS
Chemostat with Monod Kinetics
D
m S
KS  S
S
KS
 m m 1
• The above equations only holds if mmax >1
• If mmax < 1 or D>
– washout of the cells occurs
– Cells leave the reactor faster than they are dividing.
Dmax 
m S f
KS  S f
Near washout the reactor is very sensitive to
variations in D
• Small change in D large shifts in X and/or S
•If max = 0.5 hr-1 then D< 0.4 hr-1
Intracellular Product Formation Chemostat
dci
VR
 Fcif  Fci  VR rfi
dt
• Steady State and ci = P
DPf  P YP / X X  0
• If Pf = 0
YP / X X
P
D
Substrate Balance on Chemostat Intracellular Product
• If ci = S
dci
VR
 Fcif  Fci  VR rfi
dt
FS0  FS  VR X
• At Steady State
DS0  S  
• With Monod
1
YX / S
X
YX / S
dS
 VR
dt
X  YX / S S0  S 

Ks D 

X  YX / S  S0 
m  D 

Chemostat with Extracellular
Product
• Cell Mass Balance
dX
VR
 FX 0  FX  VR X
dt
D
• Substrate Balance
FS0  FS  VR X
1
YX / S
 VR qP X
• Solve Substrate at SS for X
1
YX / P
dS
 VR
dt
Class Exercise
• Problem 6.17
• E. coli is cultivated in continuous culture under
aerobic conditions with glucose limitation. When
the system is operated at D= 0.2 hr-1, determine the
effluent glucose and biomass concentrations
assuming Monod kinetics (S0 = 5 g/l, m= 0.25 hr-1 ,
KS = 100 mg/L, Y x/s = 0.4 g/g)
Chemostat with Recycle
FR, XR
F, X2
F, X0
V, X1
F+FR, X1
•
•
•
•
•
•
F - nutrient flow rate
V - reactor volume
X1 - x concentration in reactor
X2 - X concentration in effluent
XR - X concentration in recycle
FR - recycle flow rate
Chemostat with Recycle Cell
mass equation
Acc = in - out + gen
dX 1
F X0 + FR XR - (F+ FR) X1 + VX1 =
V
dt
FR, XR
F, X2
F, X0
V, X1
F+FR, X1
Chemostat with Recycle cont.
Define
 = FR/F
recycle ratio
C = XR /X1
concentration factor
Substitutions
• F + FR = (1 + )F
• FRXR term
FR = F
XR = CX1
FRXR = CFX1
dX 1
F X0 + FR XR - (F+ FR) X1 + VX1 =
V
dt
dX 1
F X0 + CFX1- (1 + )F X1 + VX1 =
V
dt
Recycle cont
• Assume
– steady state
dX 1
dt
=0
– sterile feed X0 = 0
Then
(C - 1 -)F + V = 0
Chemostat can be
operated at higher
dilution rates than the
specific growth rate
when cell recycle is
used.
If D = F/V for recycle
 = D(1+ (1 -C))
if C > 1 (concentration of cells) then (1 - C) < 0
then  < D
Substrate balance - Recycle
FS0  FS  V
X 1
YX / S
dS
 (1   ) FS  V
dt
• At Steady state and substituting for 
YX / S ( S0  S )
X 1  YX / S ( S0  S ) 

(1    C )
D
Recycle Substrate cont.
• Assuming Monod
K S D(1    C )
S
 max  D(1    C )

YX / S
K S D(1    C ) 
X1 
S0 

(1    C ) 
 max  D(1    C ) 
In Class Exercise • Consider a 1000 L CSTR in which biomass is
being produced with glucose as the substrate. The
microbial system follows a Monod relations with
m = 0.4 hr –1, KS = 1.5 g/L, and yield factor = 0.5
g/g. If S0 = 10g/L glucose and F = 100 L/h:
– What is the specific biomass production rate (g/l-h) at
SS?
– If recycle is used with a recycle stream of 10 L/h and a
recycle biomass concentration five times as large as that
in the reactor exit, what would be the new specific
biomass production rate?
Chemostat in Series
F, S0
V1, X1,
S1
F, S1, X1
F’, S’0
V2, X2,
S2
F2, S2, X2
Chemostat in Series
(no additional feed)
• First stage (assuming Monod)
K S1D1
S1 
m1  D1
X 1  YX / S S0  S1 
• Second Stage
dX 2
FX 1  FX 2  V2  2 X 2  V2
dt
Chemostat in Series cont.
• At Steady State

X1 

 2  D2 1 
 X2 
X1
 1,  2  D2
X2
• Substrate Balance
FS1  FS2  V2  2 X 2
1
YX / S
dS
 V2
dt
Chemostat in series
• At Steady State
S 2  S1 
2 X 2
D2 YX / S
• D2 = F/V2 and could have Monod growth for 2
• Solve S and  equations simultaneously for
X2 and S2 once the value of 2 is known
Chemostat in Series
(Additional Feed in Second Stage)
• Cell balance around second stage
dX 2
F1 X 1  F ' X '( F ' F1 ) X 2  V2  2 X 2  V2
dt
• At Steady State with X’ = 0
F1 X 1
 2  D '2 
V2 X 2
D '2 
F1  F '
V2
Growth rate does not
typically follow Monod
in Second Stage if
additional feed.
Chemostat in Series cont.
• Substrate Balance if Additional Feed
F1S1  F ' S  ( F1  F ' )S2  V2 2 X 2
'
0
1
YX / S
dS
 V2
dt
• At steady state the two equations can be
solved simultaneously for S2 and V2
• Major advantage is to separate production
from growth
In Class Example – 9.2
• In a two stage chemostat system, the volumes of the
first and second reactors are 500 L and 300 L
respectively. The first reactor is used for biomass
production and the second is for a secondary
metabolite formation. The feed flow rate to the first
reactor is F = 100 L/h, and the glucose concentration is
5.0 g/L. Use the following constants for the cells.
m = 0.3 h-1, Ks = 0.1 g/L Y X/S= 0.4 g/g
• Determine the cell and glucose concentrations after the
first stage.
• Assume that growth is negligible in the second stage
and the specific rate of product formation is qP = 0.02
gP/g cell hr, and Y P/S = 0.6 gP/gS. Determine the
product and substrate concentrations in the effluent of
the second reactor.
Fed Batch Reactor
• Reactor Design Equation
V
FA0  FA  
dN A
rA dV 
dt
• No outflow FA = 0
• Good Mixing rA dV
term out of the integral
dN A d C A  V 
FA0  rA  V 

dt
dt
Fed Batch Continued
• Convert the mass (NA) to concentration. Applying
integration by parts yields
dC A
dV
FA0  rAV  V
 CA
dt
dt
• Since
dV
 FA0
dt
• Then
dC A
FA0  rAV  V
 C A FA0
dt
• Rearranging
C A FA0
dC A FA0

 rA 
dt
V
V
Fed Batch Continued
• Or
dC A FA0
1  C A   rA

dt
V
• Used when there is substrate inhibition and
for bioreactors with cells.
Fed-batch Reactors
d (Vc i )
 Vr fi  F (t )c fi
dt
dV
 F (t )
dt
Differentiation the above equation using chain rule, and substitute
for dV/dt
dci
F (t )
 rfi 
[cif  ci ]
dt
V
Fed-batch cont.
• Cell balance – sterile feed
rfi  X
dX
 (   D) X
dt
KS D
S
m  D
• This can be a steady state reactor if substrate is
consumed as fast as it enters (quasi-steady-state).
Then dX/dt = 0 and  = D, like in a chemostat.
Recall, D = F / V
Fed batch cont
• Substrate balance – no outflow (Fcout = 0), sterile feed
• St = SV and Xt = XV (mass of substrate or cells in reactor at a
given time)
• S0 = substrate in feed stream
substrate
in
dS
X
 FS0 
dt
YX / S
t
Substrate
balance
Cell
balance
substrate
consumed
t
t
dX
t
 X
dt
no substrate out
(Flow out = 0)
Fed batch cont.
• Quasi steady state for St – change in substrate
is very small in reactor and is consumed as
rapidly as fed
FS0 
X
t
YX / S
t
dX
t
 X  FYX / S S 0
dt
t
t
X  X 0  FYX / S S 0t
Integrating from t=0 to t
Fed batch cont.
• Quasi steady state for St – change in substrate
is very small in reactor and is consumed as
rapidly as fed (then, dSt/dt = 0)
X t
FS0 
YX / S
YX / S FS0
X 

dX t
 X t  FYX / SS0
dt
X t  X0t  FYX / SS0 t
t
Integrating from t=0 to t
Note, "t" was missing.
Fed batch cont.
X  X  FYX / SS0t
t
t
0
• What this means
– the total amount of cells in the reactor increases
with time
– dilution rate and  decrease with time in fed
batch culture
– Since  = D, the growth rate is controlled by
the dilution rate.
Product profiles in fed batch
• Product profiles can be obtained by using the definitions
of Yp/s or qp.
• When Yp/s is constant (at quasi-steady-state with S <<S0):
P ≈ Yp/s S0, and
FP ≈ FYp/s S0,
• When qp is constant,
t
dP
 q pX t
dt
• where Pt is the total amount of product in culture
• Substituting
X  X  FYX / SS0t
t
• yields
t
0
t
dP
 q p X m V0  Ft 
dt
Ft 

t
t
P  P0  q p X m  V0   t
2

• in terms of product concentration
V0
 V0 Dt 
P  P0
 q pX m   t
V
2 
V
Class Exercise – 9.4
• Penicillin is produced in a fed-batch culture with the
intermittent addition of glucose solution to the culture
medium. The initial culture volume at quasi-steady state
is V0= 500 L, and the glucose containing nutrient
solution is added with a flow rate of F = 50 L/h. X0 = 20
g/L, S0 = 300 g/L, m = 0.2 h-1, Ks = 0.5 g/L and Y x/s=
0.3 g/g
• Determine culture volume at t = 10 h
• Determine concentration of glucose at t = 10 h
• Determine the concentration and total amount of cells at
t = 10 h
• If qp = 0.05 g product.g cells h and P0 = 0.1 g/L,
determine the product concentration at t = 10 h