Environmental Engineering Reference
In-Depth Information
second. Combining Eq. (9.29) with Eqs. (9.5), (9.20), (9.22), (9.24), and (9.27) the total DO balance
equation for a reach of river can be obtained. The following example illustrates the total DO balance for
a stream reach.
Total DO balance example
—This example will build on the previous Streeter-Phelps example. The
stream reaeration-rate coefficient remains 2.76 d
-1
(representative of the 25
th
percentile of the USGS
national reaeration-rate coefficient database (Melching and Flores, 1999)), the water temperature remains
20
ć
, and the initial oxygen deficit remains 3 mg/L. However, the previous example showed that an initial
CBOD load,
L
0
, of 40 mg/L is too high. Thus, it is assumed that as a result of the earlier computations, a
20 mg/L effluent standard for CBOD
5
was applied at a newly built wastewater treatment plant at the
upstream end of the reach of interest. The wastewater flow is 1 m
3
/s and the design river flow is 5.14 m
3
/s
and the river carries an upstream concentration of CBOD
U
of 5 mg/L. Further, since the waste flow now
receives secondary treatment the deoxygenation rate,
K
d
, becomes 0.23 d
-1
, which is representative of well
treated sewage (Table 9.1). Using Eq. (9.3) the CBOD
U
in the wastewater treatment plant effluent may be
computed as follows:
20 mg / L
5
K
CBOD
CBOD / (1
e
)
29.27 mg / L
d
U
5
1
e
u
50.23
From the mass balance, then the upstream CBOD concentration,
L
0
, is computed as 8.96 mg/L, which is
far below the upstream CBOD loading of 27.98 mg/L previously computed. But can the required DO
concentration of 5 mg/L be met if the effects of NBOD, SOD, and photosynthesis and respiration are
considered?
Thomann (1972, p. 16) reports that the approximate average NBOD concentration in untreated municipal
wastewater is 220 mg/L, but about 10% of the ammonia and organic nitrogen will be taken up in the
sludge at a wastewater treatment plant applying the activated sludge process to achieve secondary treatment.
Thus, NBOD is assumed to be 200 mg/L. The NBOD concentration of the upstream flow is assumed to
be 5 mg/L. The NBOD removal rate due only to oxidation,
K
n
, is assumed to be 0.3 d
-1
. Further, since the
stream has been subjected to years of untreated wastewater discharges, the average SOD rate for the
Chicago Waterway System in 1976 (Fig. 9.9) of 5.24 g/m
2
/day representing the result of the discharge of
minimal treatment on the SOD in a reach is applied for the reach of interest. The gross photosynthetic
production of DO for the reach will be taken as 1 g/m
2
/day as per the middle of the range for the Neuse
River System (Table 9.4) with
P
DS
and
R
DS
taken as 3 and 2 g/m
2
/day, respectively, where the subscript
DS
means photosynthesis and respiration are being treated as a distributed source and distributed sink,
respectively. Finally, the flow depth in the reach will be taken as 1 m.
Solution:
Using the mass balance the upstream concentration of NBOD is computed as:
1 m / s (200 mg / L)
3
5.14 m / s (5 mg / L)
3
L
N
36.76 mg / L
0
1 m / s
3
5.14 m / s
3
Using the previously defined rate constant and loading values the DO deficit component resulting from
the initial deficit is computed as
Kt t
DD
The DO deficit component resulting from CBOD is computed as
2.76
0
e
3e
init
LK
8.96 mg / L (0.23 d
1
)
Kt
Kt
0.23
t
2.76
t
D
0
d
e
e
e
e
d
a
CBOD
KK
2.76 d
1
0.23d
1
a
d
0.23
t
2.76
t
D
0.815 e
e
CBOD
The DO deficit component resulting from NBOD is computed as
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