Environmental Engineering Reference
In-Depth Information
ammonia concentration at the location where the Do
concentration is a minimum?
k c
k
k c
k
oa o
0
ai a
0
C
=
r k c
−
r k
+
r k
ai
oa
ai a
0
oa
ai
oi
in
−
k
−
k
ai
oa
in
ai
(4.96)
k k c
ai oa o
0
Solution
−
r k
oi
in
(
k
−
k
)(
k
−
k
)
ai
oa
in
ai
From the given data:
c
0
= 7 mg/L,
c
o
0
= 6 mg/L,
c
a
0
=
6 mg/L,
T
= 20°C,
V
= 2 cm/s = 1728 m/d,
k
a
= 0.8 d
−1
,
k
oa
= 0.4 d
−1
,
k
ai
= 0.3 d
−1
, and
k
in
= 0.5 d
−1
. At 20°C, the
saturation concentration of Do,
c
s
, is 9.1 mg/L, and so
the initial oxygen deficit,
D
0
, is 9.1 mg/L − 7 mg/L =
2.1 mg/L. The oxygen consumption rate in converting
ammonia to nitrite can be assumed as
r
oa
= 3.43 go/gn,
and the consumption rate for converting nitrite to
nitrate can be assumed as
r
oi
= 1.14 go/gn. Substituting
these parameters into Equations (4.96-4.98) yields
k c
k
k k c
oa o
0
ai oa o
0
C
=
r k
+
r k
oa
oa
ai
oi
in
−
k
(
k
−
k
)(
k
−
k
)
ai
oa
ai
oa
in
oa
(4.97)
k c
k
2
k k c
ai a
0
ai oa o
0
C
=
r
−
r k
in
oa
oi
in
−
k
(
k
−
k
)(
k
−
k
)
ai
in
ai
oa
in
oa
(4.98)
k k c
ai oa o
0
+
r
k
o
i
in
(
k
−
k
)(
k
−
k
)
ai
oa
in
ai
Equation (4.95) gives the oxygen deficit in terms
the travel time from the source,
t
, which can be con-
verted to the distance from the source,
x
, using the rela-
tionship
x
=
Vt
, where
V
is the average velocity in the
stream.
The nitrification process can be inhibited by oxygen,
since as the available oxygen decreases so does the rate
of nitrification. one way to account for the role of
oxygen on nitrification is to multiply the nitrification
rates
k
ai
and
k
in
each by a factor,
f
o
, that is related to the
amount of oxygen. A possible expression for
f
o
is given
by (Brown and Barnwell, 1987)
k c
k
k c
k
oa o
0
ai a
0
C
=
r k c
−
r k
+
r k
ai
oa
ai a
0
oa
ai
oi
in
−
k
−
k
ai
oa
in
ai
k k c
ai oa o
0
−
r k
oi
in
(
k
−
k
)(
k
−
k
)
ai
oa
in
ai
(3.43)(0.3)
(
0.4)(6)
0.3 0.4
=
(3.43)(0.3)(6)
−
+
(1.14)(0.5)
−
(0.3)(6)
0.5 0.3
)(0.4)(6)
(0.3 0.4)(0.5 0.3)
(0.3
−
(1.14)(0.5)
−
−
−
f
= −
1
exp
(
−
k c
)
(4.99)
=
56.52
mg/L
o
n
where
k
n
is the nitrification inhibition coefficient (L
3
M
−1
),
and
c
is the concentration of Do (ML
−3
). A typical value
of
k
n
is 0.6 L/mg, in which case inhibition becomes sig-
nificant when the oxygen concentration,
c
, is less than
3 mg/L, and under these circumstances it is appropriate
to use the modified nitrification constants
f
o
k
ai
and
f
o
k
in
.
If inhibition is not accounted for when it exists, the
predicted oxygen concentrations will be lower than they
actually are.
k c
k
k k c
oa o
0
ai oa o
0
C
=
r k
+
r k
oa
oa
ai
oi
in
−
k
(
k
−
k
)(
k
−
k
)
ai
oa
ai
oa
in
oa
(3.43)(0.3)
(0.4)(6)
0.3 0.4
=
−
(0.3)(0.4)(6)
(0.3 0
+
(1.14)(0.5)
− .4)(0.5 0.4)
−
= −
65.74
mg/L
EXAMPLE 4.14
2
k c
k
k k c
Wastewater is discharged into a river, and after initial
mixing, the concentration of Do is 7 mg/L, and the
concentrations of organic nitrogen and ammonia nitro-
gen are both equal to 6 mg/L. The river has a tempera-
ture of 20°C, an average velocity of 2 cm/s, and an
aeration rate constant of 0.8 d
−1
. Laboratory experi-
ments show that the rate constants in the nitrification
process are
k
oa
= 0.4 d
−1
,
k
ai
= 0.3 d
−1
, and
k
in
= 0.5 d
−1
.
Determine the minimum Do in the stream and how the
minimum oxygen concentration would be affected if
nitrification is inhibited with an inhibition coefficient
of 0.6 L/mg. In the absence of inhibition, what is the
ai a
0
ai oa o
0
C
=
r
−
r k
in
oa
oi
in
−
k
(
k
−
k
)(
k
−
k
)
ai
in
ai
oa
in
oa
k k c
ai oa o
0
+
r
k
o
i
in
(
k
−
k
)(
k
−
k
)
ai
oa
in
ai
2
(3.43)
(0.3) (6)
0.3 0.5
(0.3)(0.4)(6)
(0.3 0.4)(0.5 0.4)
=
− .14)(0.5)
(1
−
−
−
(0.4)(6)
(0.3 0.4)(0.5 0.3)
(0.3)
+
(1.14)(0.5)
−
−
=
11.26
mg/L
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