Environmental Engineering Reference
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and hence the oxygen deficit as a function of travel time,
t
, is given by Equation (4.95) as
10
8
C
(
)
+
ai
[
(
)
−
(
)
]
D t D
N
( )
=
exp
−
k t
exp
−
k t
exp
−
k t
0
a
ai
a
k
−
k
6
a
ai
with
inhibition
C
[
]
oa
(
)
−
(
)
+
exp
−
k t
exp
−
k t
oa
a
k
−
k
4
a
o
a
C
in
[
(
)
−
(
)
]
+
exp
−
k t
exp
−
k t
2
without inhibition
in
a
k
−
k
a
in
(
)
0
= (2.1)
exp
−
0.8
t
0
5
10
15
20
25
distance downstream (km)
56.52
0.8 0.3
[
(
)
−
(
)
]
+
exp
−
0.3
t
exp
−
0.8
t
Figure
4.8.
variation of dissolved oxygen due
to
−
nitrification.
−
65.74
[
(
)
−
(
)
]
+
exp
−
0.4
t
exp
−
0.8
t
0.8 0.4
−
c
=
c D
−
=
9.1 5.95
−
=
3.15
mg/L
min
s
max
11.26
0.8 0.5
x Vt
c
=
=
(1728)(4.346)
=
7510
m
=
7.51
km
[
(
)
−
(
)
]
+
exp
−
0.5
t
exp
−
0.8
t
c
−
(4.100)
Hence the minimum oxygen concentration is
3.15 mg/L and occurs 7.51 km downstream of the mixing
zone. The Do profiles for both noninhibited and inhib-
ited nitrification as given by Equation (4.100) (with
appropriate parameters) are shown in Figure 4.8. From
these results, it is apparent that inhibited nitrification
yields a higher minimum Do (3.15 vs. 2.51 mg/L) at a
location farther downstream (7.51 vs. 7.11 km) com-
pared with noninhibited nitrification.
The concentration of the various nitrogen species can
be calculated using Equations (4.91-4.94). For ammonia
nitrogen (nH
3
-n) at
t
= 4.115 days (when the minimum
oxygen level occurs), Equation (4.91) gives
Equation (4.100) gives a maximum oxygen deficit of
6.59 mg/L at
t
=
t
c
= 4.115 days, and hence the minimum
oxygen concentration,
c
min
, and the critical location,
x
c
,
are given by
c
=
c D
−
=
9.1 6.59
−
=
2.51
mg/L
min
s
max
x Vt
c
=
=
(1728)(4.115)
=
7110
m
=
7.11
km
c
Hence, the minimum oxygen concentration is 2.51 mg/L
and occurs 7.11 km downstream of the mixing zone. If
nitrification inhibition occurs, then
k
ai
and
k
in
are both
adjusted by the factor
f
o
which is related to the oxygen
concentration,
c
, by Equation (4.99), hence
k c
k
[
]
oa o
0
c
=
c e
−
k t
+
e
−
k t
−
e
−
k t
ai
oa
ai
a
ao
−
k
ai
oa
f
= −
1
exp
(
−
k c
)
= −
1
exp
( 0.6 )
−
c
o
n
(0.4)(6)
0.3 0.4
[
]
−
(0.3)(4.115
)
−
(0.4)(4.115)
−
(0.3)(4.115)
=
(6)
e
+
e
−
e
k
′
=
f k
=
0.3[1
−
exp
( 0.6 )]
−
c
−
ai
of all
=
4.10
mg/L
k
′
=
f k
=
0.5[1
−
exp
( 0.6 )]
−
c
in
of in
where the primed quantities
k
ai
and
′
k
in
are used in the
′
Therefore, the nH
3
-n concentration at the location
of minimum Do is 4.10 mg/L. The concentration of all
nitrogen species with distance downstream of the mixing
zone, as given by Equations (4.91-4.94), are shown in
Figure 4.9. From these results, it is apparent that the
organic-n and ammonia-n concentrations will both
decrease with downstream distance, the nitrite-n con-
centration will peak and then go to zero, and the
nitrate-n concentration will continuously increase.
computations. However, since
k
ai
and
′
k
in
are functions
′
k
in
in Equa-
tion (4.100) produces an implicit equation for the
oxygen deficit,
D
n
(
t
), as a function of time
t
. An approxi-
mate solution can be obtained by taking very small time
steps and using the the oxygen concentration at the
previous time step to calculate
f
o
, which then yields an
explicit equation for determining the oxygen deficit at
each time step. Using this approach, Equation (4.100)
yields a maximum oxygen deficit of 5.95 mg/L at
t
=
t
c
= 4.346 d, and hence the minimum oxygen concen-
tration,
c
min
, and the critical location,
x
c
, are given by
of the oxygen concentration, using
k
ai
and
′
′
A particular advantage of the species model is that
it provides an indication of the concentrations of various
nitrogen species that are associated with different water-
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