Biomedical Engineering Reference
In-Depth Information
At steady state,
Q
Q
2
D
2
S
1
þ
1
Q
Q
2
D
2
S
02
D
2
S
2
m
G
2
X
2
YF
X
=
S
¼ 0
(12.44)
where
m
G
2
¼
m
max
S
2
K
S
þ S
2
¼ m
net
2
þ k
d
(12.45)
Substitute
Eqn (12.42)
into
Eqn (12.44)
and rearrange, one can obtain
Q
Q
2
D
2
S
1
þ
1
Q
Q
2
D
2
S
02
D
2
S
2
m
G
2
¼
ðD
2
þ k
d
Þ¼0
(12.46)
Q
Q
2
YF
X
=
S
þ
Q
D
2
X
1
1
Q
Q
2
Q
2
D
2
S
1
þ
D
2
S
02
D
2
S
2
Combining
Eqns (12.46) and (12.45)
and solving for
S
2
,
s
b
2
4D
2
ðm
max
D
2
k
d
ÞðD
2
þ k
d
Þ
Q
Q
2
D
2
S
1
þ
1
Q
Q
2
b
D
2
S
02
S
2
¼
(12.47)
2D
2
ðm
max
D
2
k
d
Þ
where
b ¼ðD
2
þ k
d
ÞD
2
K
S
þ m
max
Q
D
2
S
1
YF
X
=
S
þðm
max
D
2
k
d
Þ
1
Q
Q
2
D
2
S
02
þ
Q
Q
2
D
2
S
1
Q
2
(12.48)
Substituting
Eqn (12.46)
into
(12.42)
, one obtain
Q
Q
2
YF
X
=
S
þ
Q
D
2
X
1
1
Q
Q
2
Q
2
D
2
S
1
þ
D
2
S
02
D
2
S
2
X
2
¼ YF
X
=
S
(12.49)
D
2
þ k
d
Thus, a two-stage chemostat is fully described with the reactor size, reaction kinetics, and
feed conditions.
We can generalize these equations for a system with no additional streams added or with
more subsequent units. For the
m
th chemostat unit, the mass balances at steady state yield:
Q
m1
Q
m
D
m
X
m1
D
m
X
m
þ m
net
m
X
m
¼ 0
(12.50)
Q
m1
Q
m
1
Q
m1
Q
m
D
m
S
0m
D
m
S
m
m
G
m
X
m
D
m
S
m1
þ
YF
X
=
S
¼ 0
(12.51)
where parameters with a subscript
m
denote at the exit of effluent of stage
m
,
S
0
m
is the
substrate concentration in the fresh (or extra) feed stream to stage
m
. The difference of the
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