Biomedical Engineering Reference
In-Depth Information
12.29. Cell growth with uncompetitive substrate inhibition is taking place in a chemostat.
The cell growth rate law for this system is
m
max
S
K
S
þ Sð1 þ S=K
I
Þ
m
G
¼
1.5 h
1
,
K
S
¼
with
m
max
¼
1 g/L,
K
I
¼
50 g/L,
S
0
¼
30 g/L, YF
X/S
¼
0.08 g-cell/g-S,
0.5 g/L.
(a) Make a plot of the steady-state cell concentration
X
as a function of D.
(b) Make a plot of the substrate concentration
S
as a function of D on the same graph
as that used for part (a).
12.30. The data in
Table P12.30
were obtained for
Pyrodictium occultum
at 98
C. Run 1 was
carried out in the absence of yeast extract and run 2 with yeast extract. Both runs
initially contained Na
2
S. The vol % of the growth product H
2
S collected above the
broth was reported as a function of time as shown in
Table P12.30
.
X
0
¼
TABLE P12.30
Run 1
Time, h
0
10
15
20
30
40
50
60
70
Cell Density,
cells/L
0.27
0.28
1.5
7.0
40.0
60.0
71.5
60.0
52.5
%H
2
S
0.5
0.8
1.0
1.2
6.8
4.7
7.5
8.0
8.2
Run 2
Time, h
0
5
10
15
20
30
40
50
60
Cell Density,
cells/L
0.27
0.70
1.1
8.0
25.0
35.0
35.0
25.0
%H
2
S
0.1
0.7
0.7
0.8
1.2
4.3
7.5
11.0
12.3
(a) What is the lag time with and without the yeast extract?
(b) What is the difference in the maximum specific growth rates,
m
max
, of the bacteria
with and without the yeast extract?
(c) How long is the stationary phase?
(d) During which phase does the majority production of H
2
S occur?
(e) The liquid reactor volume in which these batch experiments were carried out was
0.2 L. If this reactor were converted to a chemostat, what would be the
corresponding washout rate?
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