Environmental Engineering Reference
In-Depth Information
The critical DO deficit,
D
C
, can then be computed using the following equation:
L
L
0
dC
Kt
0
De
f
(9.9)
C
^
`
1/(
f
1)
>
@
ff
1(
f
( / )
DL
0
0
If Eq. 9.9 is reworked, the allowable upstream CBOD concentration,
L
0
*
, needed to keep the critical deficit
below a specified value, i.e. the difference between
C
s
and the DO concentration standard, can be determined
as follows:
^
`
1/(
f
1)
LDf
*
f
ª
1(
f
( / )
DL
*
º
(9.10)
¬
¼
0
C
0
0
Knowing the allowable upstream CBOD concentration, the allowable effluent discharge CBOD concentration
can be determined by mass balance of flows upstream and downstream of the effluent discharge point.
That is,
L
0
is computed as follows:
CQ
CQ
uu
ee
L
(9.11)
0
QQ
u
e
where
C
u
= the CBOD concentration in flow upstream of the effluent discharge point,
x
= 0, in milligrams
per liter,
C
e
= the CBOD concentration in the wastewater treatment plant effluent in milligrams per liter,
Q
u
= flow discharge upstream of the effluent discharge point in cubic meters per second, and
Q
e
=
wastewater treatment plant effluent discharge in cubic meters per second.
Reworking Eq. (9.11) the allowable CBOD concentration,
C
e
*
, in the effluent discharge may be computed
as follows:
LQQ CQ
0
*(
)
C
*
u
e
u
u
(9.12)
e
Q
e
Streeter-phelps example
—A stream has a reaeration-rate coefficient of 2.76 d
-1
(representative of the
25
th
percentile of the U.S. Geological Survey (USGS) national reaeration-rate coefficient database (Melching
and Flores, 1999)) and a deoxygenation rate of 0.5 d
-1
(representative of raw sewage, see Table 9.1). The
water temperature is 20
ć
. The initial oxygen deficit is 3 mg/L. Find (1) the DO deficit 48 hours later and
the value of the critical time and critical deficit for an initial CBOD concentration of 40 mg/L, and (2) the
allowable loading if a minimum DO concentration of 5 mg/L is required. If the mean stream velocity is
0.6 m/s, plot the DO sag curve versus distance along the stream.
Solution:
In this example, the self-purification ratio form of the oxygen-sag curve (Eq. (9.7)) is used.
The self-purification ratio,
1
1
f
KK
/
2.76
d
/ 0.5
d
5.52.
Entering the appropriate values to
a
d
Eq. (9.7) yields:
ª
º
40 mg / L
§
3 mg / L
·
0.5
t
(5.52 1)0.5
t
D
e
1e
1( . 2 )
«
¨
¸
»
5.52
1
40 mg / L
©
¹
¬
¼
which yields:
0.5
t
2.26
t
D
(8.85 mg / L) e
(1
0.661 e
)
For
t
= 2 days,
x x
D
. For water with a temperature of
20
ć
, Eq. 9.4 yields the DO saturation concentration as 9.022 mg/L. Assuming a stream flow velocity of
0.6 m/s and using the above equation and Eq. (9.6), the oxygen-sag curve for the example problem was
computed and is shown in Fig. 9.2.
The critical time is computed by entering the appropriate values into Eq. (9.8) as follows:
(8.85 mg / L) e
0.5
2
(1
0.661e
2.26
2
)
3.23 mg / L
ª
º
1
§
3 mg / L
·
t
ln 5.52 1
(5.52
1)
0.573 days
«
¨
¸
»
(5.52
1) 0.5
˄˅
d
1
40 mg / L
©
¹
¬
¼
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