Environmental Engineering Reference
In-Depth Information
which yields
t
* = 0.380 d, which corresponds to
x
=
x
* =
Vt
* = 1641 m. The BoD remaining at
t
=
t
*,
denoted by
L
*, is given by Equation (4.70) as
L
**
=
L k c t
*
−
(
**
−
t
*
)
(4.79)
a
s
The condition that
L
=
L
** at
t
=
t
** is then used
as the initial condition in the the Streeter-Phelps equa-
tion, Equation (4.71), to calculate the Do further
downstream.
[
]
=
[
]
=
*
*
L L
=
exp
−
k t
r
(25)
exp
−
(1.0)(0.380)
17.1
mg/L
0
The Do level remains at 0 mg/L for
t
* ≤
t
≤
t**
, where
t
** is given by Equation (4.78) as
EXAMPLE 4.12
1
k L k c
k c
*
−
d
a
s
t
**
=
t
*
+
k
A wastewater is discharged into a river with a flow
velocity of 5 cm/s, an aeration coefficient of 0.4 d
−1
, and
a temperature of 20°C. After initial mixing with the
wastewater, the river has a Do concentration of 7 mg/L,
an ultimate BoD of 25 mg/L, and a BoD decay con-
stant of 1.0 d
−1
. The effect of sedimentation on BoD
removal is negligible. Determine how far downstream
from the discharge location before the Do level returns
to its initial value of 7 mg/L.
d
a
s
1
1.0
(1.0)(17.1)
−
(0.4)(9.1)
=
0.380
+
=
4.077
days
(0
.4)(9.1)
The
location
corresponding
to
t
=
**
is
, m km
. The BoD remain-
ing at
t
=
t**
, denoted by
L**
, is given by Equation
(4.79) as
x
=
x
**
=
Vt
**
=
17 613
=
17 6
**
*
**
*
L
=
L k c t
−
(
−
t
)
=
(17.1)
−
(0.4)(9.1)(4.077 0.380)
−
Solution
s
mg/L
a
=
3.64
From the given data:
V
= 5 cm/s = 4320 m/d,
k
a
= 0.4 d
−1
,
T
= 20°C,
c
0
= 7 mg/L,
L
0
= 25 mg/L, and
k
r
=
k
d
= 1.0 d
−1
.
At 20°C, the saturation concentration of oxygen,
c
s
, is
equal to 9.1 mg/L, so
D
0
=
c
s
−
c
0
= 9.1 mg/L − 7 mg/L
= 2.1 mg/L. The location of the critical point,
x
c
, and the
corresponding oxygen deficit,
D
c
, are given by Equa-
tions (4.72) and (4.73) as
The Do deficit at
t
=
t**
, denoted by
D
**
, is 9.1 mg/L,
and so the concentration of Do in the river for
t
>
t**
(i.e.,
x
> 17.6 km) is given by Equation (4.71) as
k L
k
**
k x
(
−
x
**
)
k x
(
−
x
**
)
exp
−
d
r
a
D x
( )
=
−
exp
−
−
k
V
V
a
r
k x
(
−
x
**
)
V
k
k
D k
(
−
k
)
a
**
+
D
exp
−
a
0
a
r
x
c
=
ln
1
−
V
k
−
k
k L
a
r
r
d
0
4320
0.4 1.0
0.
4
1.0
2.1(0.4 1.0)
(1.0)(25)
−
=
=
ln
1
−
6243
m
and if
x***
is the location where the Do concentration
is equal to 7.0 mg/L, then
−
k
k
L
k x
V
1.0
0.4
(25)
(1.0)(6243)
4320
=
d
r
c
D
=
exp
−
exp
−
***
c
0
(1.0)(3.64)
0.4 1.0
(1.0)(
x
17613)
4320
−
9 1 7.0
.
−
=
exp
−
a
−
=
1
4.7 mg/L
(0.4)(
x
***
17 613)
4320
−
,
−
exp
−
Since the critical oxygen deficit (14.7 mg/L) is greater
than the saturation concentration of Do (9.1 mg/L),
then the Do level must become zero at some point. If
the travel time for zero Do is
t
*, then
t
* must satisfy
Equation (4.74) such that
(0.4)(
x
**
*
17 613)
4320
−
,
+
(9.1)
exp
−
which yields
x***
= 38,733 m = 38.73 km. This is quite a
long distance for the Do level to be depressed below
its initial value. The variation in Do with distance from
the initial discharge is shown in Figure 4.7.
Every effort should be made to prevent the occur-
rence of the anoxic region in the river, possibly by
restricting the volume and/or strength of the wastewater
discharge.
k L
k
[
(
)
−
(
)
]
+
(
)
d
0
*
*
*
c
s
=
exp
−
k t
exp
−
k t
D
exp
−
k t
r
a
0
a
−
k
a
r
( . )(
1 0 25
0 4 1 0
)
9 1
.
=
[exp( ( . ) ) exp( ( . ) )]
−
1 0
t
*
−
−
0 4
t
*
.
−
.
*
+
( . )exp
2 1
( ( . ) )
− 0 4
t
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