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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