Environmental Engineering Reference
In-Depth Information
this is where Do conditions are the worst. If the value
of
x
c
calculated using Equation (4.72) is less than or
equal to zero, the minimum oxygen deficit occurs at the
discharge location. This is a typical occurrence when
well-treated domestic sewage is discharged into small
streams (Lung and Sobeck, 1999).
It is important to keep in mind that the oxygen deficit,
D
, calculated using the Streeter-Phelps equation, Equa-
tion (4.71), can (mathematically) exceed the saturation
concentration of Do, even though this is physically
impossible. If the oxygen deficit calculated from Equa-
tion (4.71) is greater than the saturation concentration
of Do, this means that all the oxygen was depleted at
some earlier time. Equation (4.71) is only valid up to
the travel time at which the oxygen concentration is
greater than or equal to zero; beyond this point, the
following approach is recommended (Gundelach and
Castillo, 1970). Let
t
* be the time at which the oxygen
concentration is equal to zero, and hence Equation
(4.71) gives
EXAMPLE 4.11
A diffuser discharges wastewater into a slow-moving
river that has a mean velocity of 3 cm/s. After initial
mixing, the Do concentration in the river is 9.5 mg/L
and the temperature is 15°C. If the ultimate BoD of the
mixed river water is 30 mg/L, the rate constant for BoD
at 20°C is 0.6 d
−1
, the reaeration rate constant at 20°C is
0.8 d
−1
, and the removal of BoD by sedimentation can
be neglected, estimate the minimum Do and the critical
location in the river.
k L
k
[
(
)
−
(
)
]
+
(
)
d
0
*
*
*
c
s
=
exp
−
k t
exp
−
k t
D
exp
−
k t
r
a
0
a
−
k
a
r
(4.74)
Solution
where
c
s
is the saturation concentration of oxygen, and
Equation (4.74) is an implicit equation for
t
* that
requires a numerical solution to obtain
t
*. Beyond a
travel time
t
*, oxygen depletion can no longer proceed
at the rate of
k
1
L
, but will be limited by the rate or
reaeration such that
At
T
= 15°C, the saturation concentration of oxygen is
10.1 mg/L (Table 2.2), and hence the initial oxygen
deficit,
D
0
, is 10.1 − 9.5 = 0.6 mg/L. The BoD rate con-
stant at 15°C,
k
d
15
, is given by
k
=
k
20
1.04
T
−
20
=
(0.6)1.04
15 20
−
=
0.48
d
−
1
d
d
15
dL
dt
(4.75)
The reaeration rate constant at 15°C,
k
a
15
, is given by
= −
k c
a
s
k
=
k
20
1.024
T
−
20
=
(0.8)1.024
15 20
−
=
0.72
d
−
1
a
a
15
Using the initial condition that
L
=
L
* at
t
=
t
*, the
BoD beyond a travel time
t
* is given by
Since
L
0
= 30 mg/L,
V
= 3 cm/s = 2592 m/d, and
k
r
=
k
d
,
Equation (4.72) gives the location,
x
c
, of the critical
oxygen deficit as
L L k c t
=
*
−
(
−
t
*
)
(4.76)
a
s
V
k
k
D k
(
−
k
)
a
0
a
r
x
c
=
ln
1
−
As long as the BoD is just being met by the oxygen
provided by reaeration, the concentration of oxygen in
the water remains equal to zero. At the point at which
the oxygen concentration rises above zero, the rate at
which oxygen is supplied by reaeration is equal to the
unrestricted rate of BoD,
k
d
L
, in which case
k
−
k
k L
a
r
r
d
0
2592
0.72 0.48
0.72
0.48
0.6(0.72 0.48)
(0.48)(30)
−
=
ln
1
−
−
=
4270
m
and Equation (4.73) gives the critical oxygen deficit,
D
c
,
as
k c
s
=
k L
(4.77)
a
d
k
k
L
k x
V
Substituting Equation (4.77) into (4.76) gives the travel
time,
t
**, at which the oxygen concentration will rise
above zero as
d
r
c
D
c
=
exp
−
0
a
0.48
0.72
(30)
0.48 4270
2592
×
=
exp
−
*
1
k L k c
k c
−
= 9.0 mg/L
d
a
s
t
**
=
t
*
+
(4.78)
k
d
a
s
Hence the minimum Do level in the stream is
10.1 − 9.0 = 1.1 mg/L. This level of Do will be devastat-
ing to the aquatic ecosystem.
and the corresponding BoD remaining at this time,
L
**, is given by Equation (4.76) as
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