Environmental Engineering Reference
In-Depth Information
EXAMPLE 10.28
µ
µ
1
µ
y
≈
(10.155)
2
The probability distribution of the flow rate in a river is
estimated to have a mean and standard deviation of 1.24
and 0.911 m
3
/s, respectively. The concentration of a toxic
contaminant in the river is estimated to have a mean
and standard deviation of 10.4 and 5.26 kg/m
3
, respec-
tively. Assuming that the contaminant concentration is
independent of the flow rate, estimate the mean and
standard deviation of the contaminant mass flux in the
river.
1
µ
µ
2
µ
µ
≈
+
1
1
3
σ
2
σ
2
σ
2
−
2
σ
(10.156)
y
1
2
12
2
4
µ
where
μ
i
and
σ
i
2
are the mean and variance of
x
i
,
respectively, and σ
12
is the covariance between
x
1
and
x
2
.
In cases where
x
i
's are independent (uncorrelated)
random variables, the mean and variance relationships
can be approximated by
Solution
µ
µ
1
From the given data:
μ
Q
= 1.24 m
3
/s,
σ
Q
= 0.911 m
3
/s,
μ
c
= 10.4 kg/m
3
, and
σ
c
= 5.26 kg/m
3
. The mass flux,
m
, is
given by
µ
y
≈
(10.157)
2
≈
1
+
µ
µ
2
1
σ
2
σ
2
σ
2
(10.158)
µ
2
4
2
2
m Qc
=
where
Q
and
c
are the flow rate and contaminant
concentration, respectively. Since
Q
and
c
are indepen-
dent, then the mean and variance of
m
, denoted by
μ
m
and
σ
2
, respectively, are given by Equations (10.152)
and (10.153) as
EXAMPLE 10.29
The flow rate in a river is measured with an coarse
instrument such that the flow rate is characterized by a
mean and standard deviation of 3.23 and 1.50 m
3
/s,
respectively. A separate measurement of the flow area
is also uncertain such that the flow area is characterized
by a mean of 30 m
2
and standard deviation of 6 m
2
.
Assuming that errors in the flow rate and flow area are
uncorrelated, what is the mean and standard deviation
of the estimated average velocity in the river?
µ
=
µ µ
=
( .
1 24 10 4
)(
. )
=
12 9
.
kg/s
m
Q c
(
)
(
)
−
2
2
2
2
2
2
2
σ
=
µ
+
σ
µ σ
+
µ µ
m
Q
Q
c
c
Q c
2
2
2
2
2
2
=
( .
1 24
+
0 911
.
)(
10 4
.
+
5 26
.
)
−
( .
1 24
) (
10 4
. )
(kg/s)
2
=
155 3
.
Solution
. . kg/s
. Hence, the mass flux of
contaminant in the river has a mean and standard devia-
tion of 12.9 and 12.5 kg/s, respectively. It is noteworthy
that the coefficient of variation of the mass flux is greater
than the coefficients of variation of both the flow rate
and the contaminant concentration.
where
σ
m
=
155 3
=
12 5
From the given data:
μ
Q
= 3.23 m
3
/s,
σ
Q
= 1.50 m
3
/s,
μ
A
= 30 m
2
, and
σ
A
= 6 m
2
. The average velocity,
V
, in
the river, is given by
Q
A
V
=
where
Q
and
A
are the flow rate and flow area, respec-
tively. Since
Q
and
A
are assumed to be uncorrelated,
then the mean and variance of
V
, denoted by
μ
V
and
σ
2
, respectively, are given by Equations (10.157) and
(10.158) as
10.11.3 Division
Consider the case of division of two random variables,
x
1
and
x
2
, such that
x
x
1
y
=
(10.154)
µ
µ
3 23
30
.
1
µ
V
≈
=
=
0 108
.
m/s
2
2
There is not an exact analytical expression for the
mean and variance of the quotient; however, the follow-
ing approximate expressions can be used when the vari-
ances and covariances are small,
2
2
≈
1
+
µ
µ
1
6
+
3 23
30
.
=
σ
2
σ
2
σ
2
1 5
.
2
6
2
V
1
2
µ
2
4
2
4
2
2
(m/s)
2
=
0.290
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