Environmental Engineering Reference
In-Depth Information
homogeneous classes and calculates the estimators sep-
arately for each class. The choices of the number and
size of the classes are crucial and depend on the number
of data and flow characteristics of the stream. It is
common to use two classes: low flows and high flows
(Donlan et al., 1981; Preston et al., 1989); however, three
classes are sometimes used (e.g., base flow, high flow,
and flood flow) when much data is available (Michaud
et al., 2004). Intervals for stratification schemes can be
statistically determined by assuming that the flow data
follows a lognormal distribution (Rousseau et al., 1987).
In estimating contaminant loads from regular flow
and irregular concentration measurements, a recom-
mended first step is to verify if any accurate correlation
can be determined between streamflow and concentra-
tion in order to generate regular concentration values,
since this method gives generally the most valuable
results. It should be recognized that correlation between
contaminant concentrations and streamflows is gener-
ally better for high flows than for low flows, and a strati-
fication approach might be useful. In cases where there
is not significant correlation, a flow stratification scheme
combined with the ratio estimator has been found to
yield adequate estimates of the contaminant load
(Quilbé et al., 2006). In cases where there is a large
amount of synoptic flow and concentration measure-
ments, more sophisticated concentration-flow relation-
ships can be developed, such as relationships that
account for the “first flush” effect and relationships
that account for seasonal differences (e.g., Wang et al.,
2011).
Estimate the mass loading of solids (in kg) into the
reservoir over the 10-day period using all available
methods, and hence determine the range of 10-day load-
ings that are indicated by the data.
Solution
From the given data:
N
= 10, Δ
t
= 1 day, and
10 40 18 20 8
5
+
+
+
+
c
=
=
19.2 mg/L
1 20 10 2.5 1
5
+
+
+
+
Q
=
=
6.9 m /s
3
1 1.5 15 100 20 10 5 2.5 1.5 1
10
+
+
+
+
+
+ +
+
+
µ
Q
=
=
15.75 m /s
3
(10)(1)
+
(40)(20)
+
(18)(10)
+
(20)(2.5)
+
(8)(1)
cQ
=
5
3
=
209.6 (mg/L)(m /s)
1
5 1
(10)(1)
[
S
cQ
=
+
(40)(20)
+
(18)(10)
+
(20)(2.5)
−
]
+
(8)(1) 5(6.9)(209.6)
−
= −
1546 (
mg/L m /s
)(
3
)
1
5 1
(1 20 10 2.5 1) 5(6.9)
[
]
S
Q
2
=
+
+
+
+
−
2
−
3
2
= −
50.89 (
m /s
)
Using these data to calculate the loads according to the
averaging and ratio estimators yields (where 86.4 is a
unit conversion factor):
EXAMPLE 4.27
A river has a significant effect on a reservoir through
the discharge of suspended solids from the river to the
reservoir. It is required to estimate the mass loading of
solids to the reservoir to determine the extent of long-
term sedimentation of solids in the reservoir, which
reduces its available volume. A brief survey over a 10-
day period yielded the following data:
L cQN t
s
=
∆
=
(19.2)(6.9)(10)(1)(86.4)
=
114, 500
kg
L
=
c N t
Q
µ ∆
=
(19.2)(15.75)(10)(1)(86.4)
=
261, 300
kg
w
L
=
cQN t
∆
=
(209.6)(10)(1)(86.4)
=
181,100
kg
a
µ
(209.6)
15.75
6.9
Q
L
=
cQ
Q
N t
∆
=
(10)(1)(86.4)
=
413, 400
kg
c
1
S
cQQ
Day
Flow (m
3
/s)
Suspended Solids (mg/L)
cQ
1
+
µ
N
1
1.0
10
Q
d
L
re
=
cQ
Q
N t
∆
S
Q
1
2
1.5
-
2
2
Q
1
+
3
15.0
-
N
d
4
100.0
-
1
5
1546
(209.6)(6.9)
−
5
20.0
40
1
+
6
10.0
18
(209.6)
15.75
=
(10)(1) 86 4
(
. )
7
5.0
-
1
5
−
50.89
6.9
2
6.9
1
+
8
2.5
20
9
1.5
-
10
1.0
8
= 413, 400 kg
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