Biomedical Engineering Reference
In-Depth Information
Solution. Assume the waste is not being concentrated during the separation phase of the
biomass after the sludge reactor. Refer to
Fig. 12.12
with
S
R
¼
S
e
¼
S
, mass balance over the
sludge reactor for the biomass at steady state yields
d
ðVXÞ
d
Rc
R
QX ð1 þ RÞQX þ Vm
net
X ¼
¼ 0
(E12-5.1)
t
Since
X
s
0,
Eqn (E12-5.1)
gives
Rc
R
Qð1 þ RÞQ þ Vm
net
¼ 0
(E12-5.2)
That is,
m
net
¼
Q
V
½1 þ Rð1 c
R
Þ
(E12-5.3)
Thus
K
S
þ S
¼ m
G
¼ m
net
þ k
d
¼
Q
m
max
S
V
½1 þ Rð1 c
R
Þ þ k
d
(E12-5.4)
or
Q
V
½1 þ Rð1 c
R
Þ þ k
d
m
max
Q
K
S
ðm
net
þ k
d
Þ
S ¼
m
max
m
net
k
d
¼ K
S
(E12-5.4)
V
½1 þ Rð1 c
R
Þ k
d
Substituting the kinetic and operating parameters into
Eqn (E12-5.4)
, we obtain
Q
V
½1 þ Rð1 c
R
Þ þ k
d
m
max
Q
S ¼ K
S
V
½1 þ Rð1 c
R
Þ k
d
400
3200
½1 þ 0:4 ð1 2:5Þ þ 0:005
0:2
400
S ¼ 50
5
=
mg-BOD
L
3200
½1 þ 0:4 ð1 2:5Þ 0:005
¼ 18:97
mg-BOD
5
=
L
Therefore, the operation meet the specification as
S
20 mg/L.
We next compute the biomass concentration. Mass balance of the substrate (BOD
5
) over
the sludge reactor at steady state leads to
¼
18.97 mg/L
<
QS
0
þ RQS ð1 þ RÞQS V
m
G
X
d
ðSVÞ
d
YF
X
=
S
¼
¼ 0
(E12-5.5)
t
which gives
X ¼
Q
V
S
0
S
S
0
S
1 þ Rð1 c
R
Þþk
d
V
Q
m
G
YF
X
=
S
¼
YF
X
=
S
(E12-5.6)
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