Environmental Engineering Reference
In-Depth Information
Application of these equations are illustrated in the fol-
lowing example.
(b) The steady-state concentration can be determined
using Equations (3.122) and (3.123), where
[
]
[
]
D x D y V D
2
+
2
2
+
4
D D k
y
x
y
x
y
β
2
=
EXAMPLE 3.10
4
D D
x
y
2
2
) (
[(
10 100
)(
)
+
(
10 0
)( ) ][( .10
0
10
)
+
A vertical diffuser discharges industrial wastewater at a
rate of 5 m
3
/s uniformly over a 5-m-deep reservoir in
which the mean velocity is 10 cm/s and the diffusion
coefficient is isotropic and equal to 10 m
2
/s. The waste-
water contains 50 mg/L of a toxic contaminant, which
has a first-order decay constant of 0.05 min
−1
. (a) Deter-
mine an expression for the concentration of the con-
taminant as a function of time at a distance 100 m
downstream of the diffuser. (b) What is the steady-state
concentration at this location? (c) Compare the exact
steady-state concentration with the approximate steady-
state concentration obtained by neglecting longitudinal
diffusion.
4 10 10 0 000833
4 10 10
(
)(
)( .
)]
=
(
)(
)
=
0 520
.
Substituting into Equation (3.122) gives the steady-
state concentration as
M
L D D
Vx
D
[
]
c x y
( ,
)
=
exp
K
2
β
0
2
2
2
π
x
x
y
.
( ) (
0 25
c
(
100 0
, )
=
2
π
5
10 10
0 10 100
2 10
)(
)
Solution
( .
)(
)
[
]
exp
K
2 0 5
( .
20
)
0
(
)
From the given data:
Q
= 5 m
3
/s,
c
= 50 mg/L,
k
= 0.05 min
−1
= 0.000833 s
−1
,
L
= 5 m,
V
= 10 cm/s =
0.10 m/s,
D
x
=
D
y
= 10 m
2
/s,
x
= 100 m, and
y
= 0 m. The
contaminant mass flux,
3
=
0 000521
.
kg/m
=
521
µ
g/L
M
, is given by
Hence the steady-state concentration 100 m down-
stream from the vertical line source is 521
μ
g/L.
(c) If longitudinal diffusion is neglected, the steady-
state concentration is given by Equation (3.124) as
M Qc
=
=
( )(
5 50 10
×
−
3
)
=
0 25
.
kg/s
(a) The concentration distribution as a function of time
is given by Equation (3.121) as
2
M
Vy
D x
kx
V
c x y
( ,
)
=
exp
−
−
4
L
4
π
xVD
τ τ
( )
m d
t
y
y
∫
(
)
=
c x y t
,
,
(
)
4
π
t
−
τ
L D D
0
x
y
0 25
.
c
(
100 0
, )
=
2
(
)
x Vt
D t
−
y
D t
2
( )
5
4
π
100 0 10 10
0 10 0
4 10 100
(
)( .
)(
)
(
)
exp
−
)
−
)
−
k t
−
τ
(
(
4
−
τ
4
−
τ
x
y
2
( .
)( )
( .
0 000833 100
0 10
)(
)
exp
−
−
0 25
.
t
∫
(
)
=
(
)(
)
.
c
100 0
,
,
t
(
)( )
(
)(
)
4
π
t
−
τ
5
10 10
0
3
=
0 000613
.
kg/m
=
613
µ
g/L
(
)
2
00 0 1
4 10
1
−
.
t
exp
−
(
)
−
(
)
τ
t
Therefore, the approximate steady-state solution
is 613
μ
g/L. This differs by 18% from the exact
steady-state solution of 521
μ
g/L. This discrepancy
can be attributed to the fact that Pe =
Vx
/
D
x
= (0.10)
(100)/(10) = 1, and hence the Pe >> 1 condition
required to apply Equation (3.124) is not met in this
case.
2
0
4 10
)
(
−
)
−
0 000833
.
t
−
τ
d
τ
(
)
−
(
t
τ
1
t
∫
(
)
=
c
1
00 0
,
,
t
0 000398
.
exp
(
)
t
t
−
τ
0
(
)
2
100 0 1
40
−
.
(
)
τkg/m
3
−
−
0 000833
.
t
−
τ
d
(
)
t
−
τ
3.3.2.3 Continuous Plane Sources.
Continuous plane
sources that are well mixed or of infinite extent in one
coordinate direction will undergo two-dimensional
diffusion. Two configurations that are encountered in
This equation will require numerical integration
to determine the concentration as a function if
time.
Search WWH ::
Custom Search