Environmental Engineering Reference
In-Depth Information
Solution
Great care should be taken in estimating
k
a
since this
is typically the dominant parameter affecting the reli-
ability of simulated Do concentrations in streams
(Brown and Barnwell, 1987; Melching and Yoon, 1997).
The most accurate estimates of
k
a
are obtained from
field measurements using the gas tracer method, and
such measurements are strongly recommended in cases
where allowable discharge estimates have significant
economic and environmental consequences.
(a) From the data given,
d
= 5 m
,
W
= 20 m,
S
0
=
S
=
0.00003,
Q
= 47 m
3
is and the average velocity,
V
, is
given by
Q
Wd
47
(20)(5)
V
=
=
=
0.47 m/s
The o'Connor and Dobbins (1958) formula and the
Melching and Flores (1999) empirical equation are
the only applicable relations in Table 4.6. According
to the o'Connor and Dobbins (1958) model,
4.4.3 Streeter-Phelps Model
0.5
0.5
V
d
3.93
(0.47)
(5)
The total flux of oxygen into river water,
S
m
(ML
−3
T
−1
),
can be estimated by adding the (de)oxygenation rate
due to biodegradation,
S
1
, to the oxygen flux due to
reaeration,
S
2
, to yield
k
a
=
3.93
=
=
0.24
d
−
1
1.5
1.5
According to the Melching and Flores (1999)
empirical equation,
S
= −
k L k c
+
(
−
c
),
(4.62)
0.528
−
0.136
m
d
a
s
k
a
=
=
596( )
596(0.47 0.00003)
VS
Q
0.528
−
0.136
×
(47)
where it is assumed that biodegradation is a first-order
process. The concentration distribution of oxygen in
rivers can be described by combining the ADE (Eq.
4.27) with the source flux given by Equation (4.62) to
yield
d
−
1
=
0.
97
According to Table 4.5, the reaeration rate calcu-
lated using the o'Connor and Dobbins (1958)
formula is typical of sluggish streams and large lakes,
whereas the reaeration rate calculated using the
Melching and Flores (1999) empirical equation is
typical of swift streams. Given that the stream veloc-
ity is fairly swift (47 cm/s), the value of
k
a
= 0.97 d
−1
given by the Melching and Flores, (1999) empirical
equation is more characteristic of the actual condi-
tions. The calculated values of
k
a
(0.24 d
−1
, 0.97 d
−1
)
exceed the range of minimum values of
k
a
given by
Equation (4.61) as 0.6/5-1.0/5 or 0.12-0.2 d
−1
.
(b) The reaeration rate,
S
2
, is given by Equation (4.59)
as
∂
∂
c
t
∂
∂
c
x
∂
∂
∂
∂
c
x
−
)
(4.63)
+
V
=
x
K
k L k c
+
(
−
c
L
d
a
s
Assuming steady-state conditions (∂
c
/∂
t
= 0) and that
the dispersive flux of oxygen is much less than both the
advective flux and the oxygen fluxes due to reaeration
and deoxygenation, Equation (4.63) becomes
∂
∂
c
x
(4.64)
V
= −
k L k c
+
(
−
c
)
d
a
s
Instead of dealing with the oxygen concentration,
c
, it is
convenient to deal with the
oxygen deficit
,
D
, defined by
S
=
k c
(
−
c
)
2
a
s
where
c
= 5 mg/L, and the saturation concentration,
c
s
, at 20°C is given in Table 2.2 as 9.1 mg/L. There-
fore, the reaeration rate,
S
2
, is given by
(4.65)
D c
=
−
c
s
Combining Equations (4.64) and (4.65) and replacing
the partial derivative by the total derivative (since
c
is
only a function of
x
) yields the following differential
equation that describes the oxygen deficit in the river:
S
2
=
(0.97)(9.1 5)
−
=
3.98
mg/L d
⋅ =
3980
mg/m d
3
⋅
(c) The volume of river water per meter is given by
dD
dx
k
V
k
V
=
−
volume of river water
=
Wd
(1)
=
(20)(5)(1)
=
100
m
3
d
a
(4.66)
L
D
Hence, the mass of oxygen added per day per
meter along the river is 3980(100) = 398,000 mg/d·m
= 398 g/d·m.
The BoD remaining at any time,
t
, since release will
follow the first-order reaction
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