Environmental Engineering Reference
In-Depth Information
The simultaneous solution of Equations (4.144-
4.146) yields
0.85
0.75
5.3
η φ
ηφ
−
f
=
(4.142)
k
a
=
7.5
where
η
14
k
a
=
0.114
h
−
1
=
2.7
d
−
1
θ
=
1.13
rad
γ
=
1.24
where
ϕ
and
f
are in hours and
k
a
is in d
−1
. The discrep-
ancy between the exact (Eq. 4.134) and approximate
(Eq. 4.142) relations are most pronounced for
f
≥ 17
hours and
k
a
< 1 d
−1
. The average production rate,
P
av
,
can be estimated using the following approximate rela-
tion in lieu of Equation (4.136) (McBride and Chapra,
2005):
The times
t
min
and
t
max
when the oxygen deficit is a
minimum and maximum, respectively, are given by
Equations (4.138) and (4.139) as the solution to
π
t
f
*
−
*
−
k t
a
π
cos
−
θ
(
k f
)
γ
e
=
0
a
∆
P
16
(33
=
(4.143)
π
t
*
−
1.5
η
+
k
a
)
0.114
*
−
t
π
cos
−
1.13
(0.114
)(13)(1.24)
e
=
0
av
13
Utilization of Equations (4.142) and (4.143) to estimate
k
a
and
P
av
from Do observations is called the
approxi-
mate delta method
, and development of this method is
attributed to McBride and Chapra (2005). This method
is mostly suited to water bodies with moderate reaera-
tion (
k
a
< 10 d
−1
) and moderate photoperiods (
f
= 10-14
hours).
(4.147)
where
t
* represents the multiple solutions to Equa-
tion (4.147), and
t
min
=
t
* when
t
* >
f
/2 and
t
max
=
t
*
when
t
* <
f
/2. Solutions to Equation (4.147) are
t
* = 10.4 hours and 0.575 hours; hence,
t
max
= 0.575
hours and
t
min
= 10.4 hours. Substituting into Equa-
tion (4.140) yields
EXAMPLE 4.20
π
t
+
Measurements reported by Chapra and Di Toro (1991)
taken on a slow-moving section of the Grand River
(Michigan) characterized by profuse growths of
aquatic plants showed that the diurnal Do variations
w
ere characterized by:
ϕ
= 3.93 hours, Δ = 8.4 mg/L,
D
= −0 27. mg/L
(i.e., average supersaturation), and
f
= 13 hours (0700-2000 hours). (a) Determine the
values of
k
a
,
P
av
, and
R
corresponding to these data. (b)
Compare these values with those obtained using the
approximate relations.
max
δ
= −
sin
−
θ
γ
e
−
k t
a
max
f
π
t
f
+
min
+
sin
−
θ
γ
e
−
k t
a
min
π 0.575
13
×
+
= −
sin
−
1.13
1.24
e
−
0.114 0.575
×
π 10.4
13
×
+
+
sin
−
1.13
1.24
e
−
0.114 10.4
×
=
1.04
Solution
(a) From the data given,
f
= 13 hours,
T
= 24 hours, and
Equation (4.133) gives
Substituting Δ = 8.4 mg/L,
δ
= 1.04,
k
a
= 0.114 h
−1
,
and
f
= 13 hours into Equation (4.136) gives
π
π
=
−
1
−
1
(4.144)
∆
P
πδ
θ
=
tan
tan
=
k f
13
k
a
a
2 (
k f
a
)
2
+
π
2
av
−
k T f
(
−
)
−
11
k
1
+
e
=
1
1
+
−
e
e
a
a
γ
=
sin
( )
θ
sin
( )
θ
8.4
(1.04)
2 (0.114 13)
2
π
=
1
−
e
−
k T
−
24
k
a
a
P
av
×
+
π
2
(4.145)
Also, from the given data,
ϕ
= 3.93 hours and Equa-
tion (4.134) give
which yields
P
av
=
17.9
mg/(L d)
⋅
1
2
φ
−
−
π
cos
π
+
θ
(
k f
)
γ
e
−
k
(
φ
+
f
/
2)
=
0
a
a
f
Subs
tit
uting
P
av
= 17.9 mg/(L·d),
k
a
= 0.114 h
−1
= 2.7 d
−1
,
and
D
= −0 272
1
2
3.93
13
.
mg/L
into Equation (4.141) gives
π
cos
π
+
−
θ
−
(13 )
k
γ
e
−
k
a
(3.93 13 2)
+
/
=
0
a
(4.146)
R P
=
+
k D
a
=
17.9 2.7( 0.272)
+
−
=
17.2
mg/(L d
⋅
)
av
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