Report on
Simulation of Fermentation Processes
By,
Vijaya Krishna Bodla
(s041492)
Analysis of the production of penicillin at steady state in a chemostat
Chemostat is a continuous bioreactor operating with a continuous inflow and outflow. The
substrate concentration inside the reactor is maintained constant all through the process. The biomass growth is
assumed to follow the monod kinetics. The rate of penicillin production is given by the following kinetics:
rp = kp*Cs/(Cs+Kp+Cs2/Ki). The typical of a chemostat that biomass growth rate is equal to the dilution rate and
can be proved through the mass balance over the biomass production. So, from D all the others have been
calculated and tabulated as:
D or mu
Cs
rs
rp
0,010000
22,222222
0,025000 0,000211
0,020000
50,000000
0,045000 0,000435
0,030000
85,714286
0,065000 0,000650
0,040000
133,333333
0,085000 0,000822
0,050000
200,000000
0,105000 0,000909
0,060000
300,000000
0,125000 0,000882
0,070000
466,666667
0,145000 0,000742
0,080000
800,000000
0,165000 0,000519
0,090000 1800,000000
0,185000 0,000259
0,098000 9800,000000
0,201000 0,000050
Y_sp
Y_sx
0,008431
0,400000
0,009662
0,444444
0,010002
0,461538
0,009670
0,470588
0,008658
0,476190
0,007059
0,480000
0,005118
0,482759
0,003148
0,484848
0,001402
0,486486
0,000251
0,487562
X
3991,111111
4422,222222
4575,824176
4643,137255
4666,666667
4656,000000
4602,298851
4460,606061
3989,189189
97,512438
Qp
0,841218
1,922705
2,974994
3,816277
4,242424
4,108235
3,415133
2,317198
1,034660
0,004922
D vs Y_sp and Qp
4,500000
0,012000
Qp
4,000000
Y_sp
0,010000
3,500000
Y_sp;Qp
3,000000
0,008000
2,500000
0,006000
2,000000
1,500000
0,004000
1,000000
0,002000
0,500000
0,000000
0,000000
0,020000
0,040000
0,060000
0,080000
0,100000
0,000000
0,120000
D
Chart1: Plot of Y and Q as function of dilution rate
sp
p
D vs rp
0,001000
0,000900
0,000800
0,000700
rp
0,000600
0,000500
0,000400
0,000300
0,000200
0,000100
0,000000
0,000000
0,020000
0,040000
0,060000
0,080000
0,100000
0,120000
D
Chart2: Plot of r as function of dilution rate
p
2
The graphs show a nonlinear relation between the dilution rate or the specific growth rate and the product yield
and volumetric productivity. The rate of product production depends on the production kinetics. In this case of
penicillin production with the given kinetics it can observed from the graphs that the maximum yield and
maximum volumetric productivity cannot be attained at the same dilution rate. If the yield is of main concern,
which normally is in an industry, then low dilution rates would lead to an optimal production process.
Productivity on the other hand becomes significant for a large scale production of the plant or if the selling price
of the product is relatively high. However its better to choose the specific dilution rate that leads to an optimal
production process.
Maximum Yield (Y_sp) is obtained at µ = 0.03.
Maximum Volumetric productivity (Qp) is attained at µ = 0.052.
Production of penicillin in a Fed-Batch Reactor
Conditions:
Fin  F 
dv
F 1 dv
; D   * ; Fout  0
dt
V V dt
Cs=constant.
General Mass balances:
Accumulation =In –out +formation –consumption.
Glucose:
 d(Cs*V)/dt = F*Cs - Fout*Cs+0-rs*X*V
 Cs*dV/dt+V*dCs/dt = F*cfs-0*Cs+0-rs*.X*V
 Cs*F+V*dCs/dt = F*Cfs - rs*X*V
 V*dCs/dt = F*(Cfs - Cs) - rs*X*V
 dCs/dt = F/V*(Cfs - Cs) - rs*X
 dCs/dt = D*( Cfs - Cs) - rs*X
(1)
Biomass:
 d(X*V)/dt = 0-Fout*X +µ*X*V-0
 d(X*V )/dt = µ*( X*V)
 V*dX/dt + X*dV/dt = µ*X*V
 V*dX/dt= µ*X*V – X*D*V (since D=1/VdV/dt)
 dX/dt=( µ-D).X
(2)
3
Penicillin (Product):
 d (cp.v)/dt = F. Cpf - Fout..Cp+rp.x.v-0
 V.dcp/dt=F.Cpf-0. Cp +rp.x.v
 V.dcp/dt=F.Cpf_F. Cp + rp.x.v
 F.(Cpf-Cp)+ rp.x.v
 dcp/dt =D.( Cpf_Cp)+ rp.x
(3)
In a Fed Batch process Cs is constant.
 dcs/dt=0
 0 = D.(cfs-cs)- rs.x
 D= rs.x/(cfs-cs)
(4)
Defining yield co-efficient
 Yxs=rs/µ0
 rs= Yxs. µ0
From (4),
D = Yxs . µ0 .X / (cfs-cs)
(5)
Now, d(xv)/(xv) = µ*dt
d(lnxv)= µ dt
Integrating…
Ln(xv)-ln(x0v0) = µ0 . t
xv = x0v0 e(µ0.t)
(6)
From (5),
F(t)/V = Yxs . µ0 .X / (cfs-cs)
(7)
F(t) = Yxs . µ0 .(X.V) / (cfs-cs)
Substiting for X.V from (6)
F(t) = Yxs . µ0 .(X0.V0).e e(µ0.t)/ (cfs-cs)
(8)
Different strategies of feed addition to a fed-batch reactor
The Fed batch is the most common kind of operation in the industries as the environmental conditions can be
maintained. Also enables maintaining of very high titers of metabolites which plays an important role in the down
stream processing.
Different feeding strategies lead to different design aspects of the process. A possible strategy is to maintain a
constant substrate concentration inside the reactor. This can obtained by an exponential feeding profile as the
4
biomass grows exponentially inside the reactor which can also be derived from the mass balances. This strategy
helps in maintaining potentially low concentrations of glucose inside the reactor and can be used in cases where
there is glucose or substrate repression for product formation. But in this case there might be problems with the
oxygen transfer inside the reactor as the biomass grows exponentially.
The other design parameter which is a typical of most of the present processes in the industry maintaining a
constant µ*X (product of specific growth rate and biomass concentration). This strategy can be obtained either by
a constant feed rate or a linear feed rate. Both the heat production and the oxygen transfer inside the reactor are
like directly proportional to this product and they can thus be maintained constant. The strategy involves the
proportional decrease in the specific growth rate with the increase in the biomass concentration. The volumetric
uptake rates of the oxygen and the substrate can be maintained constant.
Probably a better feeding strategy would be starting with an exponential feed for some time and then making the
feed rate constant or linear.
Reactor:
The oxygen consumption pr. g biomass formed is given by Y = 60 mmol oxygen (g DW)
-1
xo
Initial Condition:
V0 = 30 m3
S0 = 0
Limitations:
Vmax = 80 m3
Sf,max = 500 g/l
X0,max =5 g/l
Kla = 600 h-1
Exponential Feeding:
As discussed before, the exponential feeding profile with constant growth rate (µ) gives a nearly constant
substrate concentration inside the reactor.
The feeding profile is thus given as:
F(t) = Yxs . µ0 . X0 . V0. eµ0. t/(Csf – Cs);
The volume and Biomass are given by:
V(t) = V0 (1 – ax0 + ax0 eµ0. t);
X(t) = X0 . eµ0. t / (1 – aX0 +aX0 eµ0. t);
Where a = Yxs / (Csf – Cs)
5
Rate of product formation is assumed to be the same as in chemostat as the product production is less compared to
the substrate consumption. The concentration of the product Cp is thus given by:
Cp(t) = (rp/ µ).(X(t)-X(0));
Kla is given by:
Kla = (µ.X.Yxo)/(C*0 - C0)
From the results obtained from the chemostat,
µ = 0.03 for the maximum Yeild Y_sp;
µ = 0.052 for the maximum volumetric productivity Qp;
Calculations for the fedbatch have been made using these µ values.
Using Excel, the optimal solution has been obtained by manupulating the varibles X0, Csf.
Csf is found to have a prominent effect on the overall yield and the productivity over X0. So, X0 has been
optimised and following table has been made:
mu = 0.03
Csf
t (hr)
100
93
200
83
300
97
400
105
500
95
X0=5
X(t)
30,6559
37,7115
56,4209
72,701
63,8915
V(t)
79663,314
47974,574
48803,198
48148,052
40587,058
F(t)
1587,4
587,987
596,596
568,817
337,112
Kla
282,9778
348,1058
520,8082
671,0858
589,7675
Cp
0,66437
0,81728
1,22274
1,57556
1,38465
Qp
0,0071438
0,0098467
0,0126056
0,0150054
0,0145752
Cp*V
Cs consumed
52925,92 4966331,436
39208,48 3594914,739
59673,8 5640959,534
75860,27 7259220,989
56198,67 5293529,098
Overall Yeild
0,01065694
0,01090665
0,01057866
0,01045019
0,01061648
It can be seen that for Csf = 400, gives the optimal overall yield, productivity and product formed.
Biomass Vs time
time Vs volume
70
50000
45000
60
40000
50
biomass
30000
25000
20000
15000
40
30
20
X
10000
10
5000
0
0
0
10
20
30
40
50
60
70
80
90
100
0
10
20
30
40
time
50
60
70
80
90
100
time
time Vs Cp
1,4
1,2
1
0,8
Cp
volume
35000
0,6
0,4
0,2
0
0
10
20
30
40
50
60
70
80
90
100
time
Chart3: Evolution of Biomass, Volume and Product as a function of time for µ=0.03
6
mu = 0.052
Csf
t (hr)
100
54
200
47
300
43
400
41
500
41
X0=5
X(t)
31,484
37,1285
36,2091
35,2873
36,4763
V(t)
78976,839
46537,121
38757,556
35841,755
34673,404
F(t)
2710,3
941,68
509,893
344,646
275,717
Kla
503,7446
594,0561
579,3451
564,596
583,6206
Cp
0,553
0,65214
0,63599
0,6198
0,64068
Qp
0,0102407
0,0138753
0,0147905
0,0151171
0,0156264
Cp*V
Cs consumed Overall Yeild
43674,06 4897683,872 0,00891729
30348,72 3307424,148 0,00917594
24649,44 2627266,725 0,00938216
22214,7
2336702,19 0,00950686
22214,7
2336702,19 0,00950686
It can be seen that for Csf = 200, gives the optimal overall yield, productivity and product formed.
time Vs volume
Biomass Vs time
50000
40
45000
35
X
40000
30
25
30000
biomass
volume
35000
25000
20000
15000
20
15
10
10000
5
5000
0
0
0
10
20
30
40
0
50
10
20
30
40
50
60
time
time
time Vs Cp
0,7
0,6
0,5
Cp
0,4
0,3
0,2
0,1
0
0
10
20
30
40
50
time
Chart4: Evolution of Biomass, Volume and Product as a function of time for µ=0.052
From the results for the exponential feeding profile it can be seen that with the limitations and exponentially
increasing feed and biomass, the process cannot be run for longer time. The required yield and the productivity
can still be obtained. This kind of feeding profile normally poses problems in the effective mass transfer inside the
reactor requiring a higher value of Kla. Higher value of the specific growth rate results in more of the product
formed in lesser time.
Simulations:
Simulations have been made using Matlab files provided in campus net and are done for both the feeding
strategies, Constant and Linear. The parameters Kfeed, X0 and Sf, have been manipulated to obtain the optimal
solution. It is observed that Sf has a prominent effect on the overall yield and productivity. So, the other
parameters are optimised and then Sf manipulated and the results tabulated as:
7
1) Constant Feed (F = Kfeed)
Sf
Time
160
160
15,7
15,5
15,1
100
200
300
400
500
Kfeed = 300
X
Cs
32,686
63,455
19,595
19,559
19,168
X0 = 5
Cp
0,0125
0,0129
10,084
23,099
36,112
V
0,62989
1,1913
0,00966
0,00615
0,00457
Kla
78000
78000
34699
34638
34543
Cp*V
Cs Consumed Overall Yeild Ysp Qp
59,177
49131,42
118,35
92921,4
591,2 335,14029
596,64 213,06873
586,54 157,91332
4800000
9600000
1409700
1855200
2271500
0,010235713
0,009679313
0,000237739
0,000114849
6,95194E-05
0,00393681
0,00744563
0,0006166
0,00039791
0,00030185
Optimal Conditions:
Kfeed X0
Sf
300
5
250
Optimal Solution:
Time X
Cs
Cp
V
160 78,84 0,01299
1,4584
78000
Kla
Cp*V
Cs Consumed Overall Yeild Ysp
Qp
147,94 113755,2
12000000
0,0094796
0,009115
4
80
Glucose, g/L
Biomass gDW/L
70
8
x 10
Volume, L
7.5
7
60
6.5
50
6
40
5.5
5
30
4.5
20
4
10
0
3.5
0
50
100
150
3
200
0
20
40
60
80
100
120
140
160
600
1.5
Kla
Penicillin, g/L
500
400
1
300
200
0.5
100
0
0
50
100
150
200
0
0
20
40
60
80
100
120
140
160
Chart5: Profiles of all the results at the optimal condition
8
2) Linear (F = Kfeed * time)
Sf
100
200
300
400
500
Kfeed = 5.5
X0 = 5
Time
X
Cs
Cp
V
Kla
Cp*V
Cs Consumed Overall Yeild Ysp Qp
134,9 33,123 4,56E-03
0,67637
80000
22,741 54109,6
5000000
0,01082192 0,0050139
134,84 64,358 3,37E-02
1,3453
80000
285,23 107624
10000000
0,0107624 0,009977
14,071 19,425
24,399 0,0030047
80000
592,85 240,376
15000000
1,60251E-05 0,0002135
134,86 126,87 0,0032942
2,6528
80000
63,258 212224
20000000
0,0106112 0,0196708
15,262 18,975
25,301 0,0027406
80003
579,27 219,2562
25001500
8,76972E-06 0,0001796
Optimal Conditions:
Kfeed Sf
X0
5,5
400
5
Optimal Solution:
Time X
Cs
Cp
135 126,87 0,003294
V
2,6528
80000
Kla
Cp*V
Cs Consumed Overall Yeild Ysp Qp
63,258
212224
20000000
0,0106112
0,01967077
4
140
Glucose, g/L
Biomass gDW/L
120
9
x 10
Volume, L
8
100
7
80
60
6
40
5
20
4
0
-20
0
50
100
150
3
200
3
Penicillin, g/L
2.5
0
20
40
60
80
100
120
140
160
600
Kla
500
400
2
300
1.5
200
1
100
0.5
0
0
0
50
100
150
200
-100
0
20
40
60
80
100
120
140
160
Chart6: Profiles of all the results at the optimal condition
9
Conclusion from the simulation results:
It can be observed that the optimal solution for both the feeding strategies requires some initial biomass
concentration.
From the linear feeding strategy, it can be observed that starting at a relatively low value for the feed
makes fermentor run for longer time producing more product and also good overall yield and volumetric
productivity. The product concentration is also seen to be relatively high. The optimal feed rate is found
to be at 5.5. The limitations on Kla and Volume also play an important role and can change the optimal
solution by varying them.
For both the constant and linear feeding strategy, it can be observed that the glucose concentration in the
feed has a significant influence on the results.
10