Environmental Engineering Reference
In-Depth Information
The maximum value of
c
(1 km,
t
) for
t
> 60 minutes
can be determined numerically and is equal to
1.011 mg/L occurring at
t
= 84 minutes. The varia-
tion of
c
(1 km,
t
) with
t
, as given by Equations
(3.100) and (3.103), for
τ
= 60 minutes is shown in
Figure 3.16. Based on these results, it is apparent
that the maximum concentration at the water supply
intake is not significantly reduced by limiting the
waste discharge to 1 hour per day.
{
+
M
AV
Vx
D
(
)
c x t
( , )
=
exp
1
+
Γ
2
Γ
2
x
x Vt
D t
Γ
−
x V
+
t
D t
(
−
−
τ
τ
)
Γ
−
erf
erf
2
2
(
)
x
x
(3.108)
Vx
D
x Vt
D
(
)
exp
1
−
Γ
2
x
−
Γ
x V t
D t
x
−
(
−
−
τ
τ
)
Γ
−
erf
erf
2
t
2
(
)
x
Fixed Source Mass Flux.
The case of a fixed source mass
flux is fundamentally different from the case of a fixed
concentration at the source location. In the case of a
source releasing mass at a rate of
M
(MT
−3
) over an
area
A
(L
2
) for a finite duration
τ
(T), the resulting
concentration field is expressed separately for cases
where
t
≤
τ
and
t
>
τ
. For cases where
t
≤
τ
,
Application of the equations for a constant mass flux
from a plane source are illustrated in the following
example.
EXAMPLE 3.8
Wastewater is discharged uniformly over the cross
section of a stream such that contaminant is introduced
into the stream at a rate of 2 kg/s. The stream is approxi-
mately 5 m wide and 2 m deep, with an average flow
velocity of 20 cm/s and an estimated longitudinal diffu-
sion coefficient of 10 m
2
/s. The decay rate of the con-
taminant in the river is estimated as 0.1 min
−1
. (a) What
is the concentration 100 m downstream of the source 10
minutes after the start of the contaminant release.
(b) What is the steady-state concentration 100 m down-
stream of the source? (c) What is the concentration at
the source 10 minutes after the start of the contaminant
release? (d) What is the steady-state concentration at
the source?
M
AV
Vx
D
x Vt
D t
+
Γ
(
)
c x t
( , )
=
exp
1
+
Γ
erf
∓
1
2
Γ
2
2
x
x
Vx
D
x Vt
D t
−
Γ
(
)
−
1
−
Γ
erf
∓
1
exp
2
2
x
x
(3.104)
where Γ is defined by Equation (3.97), and the − sign is
used for
x
> 0 and the + sign for
x
< 0. Equation (3.104)
also represents the case where the source mass
flux begins at
t
= 0 and continues indefinitely. In this
case, the concentration at the source (
x
= 0) can be
expressed as
M
AV
Vt
D t
x
Γ
Solution
c
( , )
0
t
=
erf
(3.105)
Γ
2
From the given data:
M
= 2 kg/s
,
A
= 5 m × 2 m =
10 m
2
,
V
= 20 cm/s = 0.20 m/s,
D
x
= 10 m
2
/s,
and
If the source continues to release mass at a constant rate
indefinitely, then the steady-state concentration derived
from Equation (3.104) is
k
= 0.1 min
−1
= 0.00167 s
−1
.
(a) The location and time of interest are given as
x
= 100 m and
t
= 10 minutes = 600 seconds. Based
on the given data, Equation (3.97) gives
M
AV
Vx
D
x
c x
( ,
∞
) =
exp
(1
∓
Γ
)
,
(3.106)
Γ
2
kD
V
( .
0 00167 10
0 20
)(
)
=
=
x
Γ =
1 4
+
1 4
+
1 63
.
2
2
( .
)
and if longitudinal diffusion is neglected (i.e., Pe >> 1),
Equation (3.106) simplifies to
and the required concentration is given by Equa-
tion (3.104) as
M
AV
kx
V
(3.107)
c x
( ,
∞ =
)
exp
−
M
AV
Vx
D
x Vt
D t
+
Γ
(
)
c x t
( , )
=
exp
1
+
Γ
erf
∓
1
2
Γ
2
2
x
x
In cases where the mass release only lasts for a finite
time
τ
, the concentration distribution for
t
>
τ
is given
by
Vx
D
x Vt
D t
−
Γ
(
)
∓
−
exp
1
−
Γ
erf
1
2
2
x
x
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