Environmental Engineering Reference
In-Depth Information
water longwave: Taking ε = 0.97 yields
with the unknown being the steady-state temperature,
T
, of the lake. In Equation (7.54), Term 1 is the inflow
energy, Term 2 is the outflow energy, and Term 3 is
the energy added at the lake surface. Based on the
given data, the following derived variables can be
determined,
4
4
−
9
4
6
A
εσ(
T
+
273
)
=
(
10
)( .
0 97 4 903 10
)( .
×
)(
T
+
273
) (
10
)
s
s
273
4
) J/d
=
47 6
. (
T
+
(7.59)
conduction: Taking
c
1
= 0.47 yields
A LW
s
=
=
(
100 100
)(
)
=
10
4
m
2
A c f V T T
(
)(
−
)
=
(
10
4
)( .
0 47 29 5
)(
. )(
T
−
24
)
T
17 27
237 3
.
s
1
w
s
air
air
e
=
0 6108
.
exp
sat
T
+
.
=
1 39 10
.
×
5
(
T
−
24
)
W
(7.60)
air
. ( )
24 237 3
17 27 24
=
1 20 10
.
×
10
(
T
−
24
)
J/d
=
0 6108
.
exp
+
.
=
2 98
.
kPa
=
22 4
.
mm Hg
evaporation:
RH
70
100
A f V e
(
)(
−
e
)
e
=
e
=
(
22 4
. )
=
15 7
.
mm Hg
s
w
s
air
air
sat
100
17 27
237 3
.
T
−
4
=
(
10
)(
29 5 4 58
. )
.
exp
15
.
7
W
2
2
f V
(
)
=
19 0 0 95
.
+
.
V
=
19 0 0 95 3 33
.
+
.
( .
)
=
29 5
.
(7.61)
T
+
.
w
w
17 27
237 3
.
T
−
17 27
237 3
.
T
10
=
2 55 10
.
×
4 58
.
exp
15 7
.
J/
d
e
=
0 6108
.
exp
kPa
T
+
.
s
T
+
.
17 27
237 3
.
T
mm Hg
Substituting Equations (7.55) to (7.61) into Equation
(7.54) gives
=
4 58
.
exp
T
+
.
the
following
steady-state energy
equation,
Each of the terms in Equation (7.54) can be calculated
separately as follows:
Term 1:
Assume that
ρ
= 998 kg/m
3
(which is exact at
20°C), and
c
p
= 4186 J/kg °C, hence
2 63 10
.
×
11
−
1 46
.
T
×
10
10
+
5 53 10
.
×
11
+
1 93 10
.
×
11
−
47 6
. (
T
+
273
)
4
−
1 20
.
×
10
10
(
T
T
−
24
)
17 27
237 3
.
−
2 55 10
.
×
10
4 58
.
exp
−
15 7
.
T
+
.
2 63 10
11
Q c T
ρ
p in
=
(
3500 998 4186 18
)(
)(
)(
)
=
.
×
J/d
= 0
(7.55)
Solving this equation (numerically) for
T
yields
T
= 25.7°C. Therefore, the steady-state temperature in
the lake is 25.7°C.
Term 2:
10
10
Q c T
ρ
p
=
(
3500 998 4186
)(
)(
)( )
T
=
1 46
.
T
×
J/d
(7.56)
7.5.2 One-Dimensional (Vertical) Models
Term 3:
The components of this term are given in Equa-
tion (7.50) and will be calculated separately using typical
values for the constants:
net solar:
As in the case of zero-dimensional (well-mixed) models,
the fate and transport of both mass and heat are of
interest in cases where mass concentrations and tem-
perature vary with depth below the lake surface. Mass
models are used to predict concentration distributions
of dissolved substances and are based on the advection-
diffusion equation, while heat models are used to predict
temperature distributions and are based on the law of
conservation of energy.
4
6
11
A J
=
(
10
)(
640
)
=
6 40 10
.
×
W
=
5 53 10
.
×
J/d
s
sn
(7.57)
atmospheric longwave: Taking
σ
= [= 4.903 × 10
−9
MJ
m
−2
k
−4
d
−1
and
R
L
= 0.3 yields
A T
σ(
+
273
) (
4
A
+
0 031
.
e
)(
1
−
R
)
7.5.2.1 Conservation of Mass Model.
In cases where
there is significant variability in water quality over the
depth of a lake, a one-dimensional (vertical) model is
frequently used to simulate the fate and transport of
water-quality constituents. Such vertical variations in
s
air
air
L
=
(
10
4
)( .
4 903 10
×
−
9
)(
24 273
+
) ( .
4
0 6 0 031 15 7
+
.
. )
1 0 3 10
1 93 10
(
−
. )(
6
)
=
.
×
11
J/d
(7.58)
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