Environmental Engineering Reference
In-Depth Information
Substituting these parameters and the other given data
into Equation (8.16) gives
water
13 6
/
1 2
/
0 600
2
.
+
y
(
y
+
6 184
.
)
−
0 600 6 084
100
( .
+
.
)
H
0
crest
1
1
( .
0 625 30
)(
)
ow
=
0 30
.
Q
0
which yields
y
1
= 0.678 m, and the assumption that
y
≥ 0.3 m is confirmed. Hence the stage at the inflow
section is 6.184 m + 0.678 m = 6.862 m. This inflow
stage is greater than that calculated using Equations
(8.10-8.13) (6.730 m). Therefore, it is appropriate to
make the conservative estimate that the design stage at
the inflow section is 6.862 m.
H
w
downstream
water surface
weir
Figure 8.13.
Flow over a weir.
The head differential between the water surface at the
inlet and at the outlet of a wetland provides the energy
required to overcome the frictional resistance.
Although constructing a wetland with a sloping bottom
can provide some of the head differential, the pre-
ferred approach is to construct a bottom with minimal
slope that still allows complete drainage when needed
and to provide outlet structures that allow the water
level to be adjusted to compensate for the resistance
that may increase with time. Water levels near the exit
in FWS wetlands are usually controlled by a sharp-
crested weir structures, where the water level is related
to the outflow by the (dimensionally nonhomoge-
neous) relation
where Δ (L) is height of the downstream water surface
above the crest of the weir.
EXAMPLE 8.2
A sharp-crested weir located at the exit of a wetland is
30 cm high and 30 m long. If the design flow rate through
the wetland is 1.7 m
3
/s and the downstream water
surface is below the crest of the weir, estimate the depth
of water just upstream of the weir. By how much would
this depth change if the water surface downstream of
the weir is 5 cm above the crest of the weir?
Solution
3
2
(8.18)
Q C L H H
0
=
(
−
)
w
0
w
From the given data:
H
w
= 30 cm = 0.30 m,
L
= 30 m,
and
Q
0
= 1.7 m
3
/s. In the case where the downstream
water surface is below the crest of the weir, and assum-
ing that (
H
0
−
H
w
)/
H
w
≤ 0.4 so
C
w
= 1.83, Equation
(8.18) requires that
where
Q
0
is the outflow rate (m
3
/s),
C
w
is the weir
coefficient (dimensionless),
L
is the weir length (m),
H
0
is the water stage at the wetland outlet (m),
and
H
w
is the elevation of the crest of the outlet
weir (m). A schematic diagram of a typical weir
structure is shown in Figure 8.13. As long as the
downstream water surface remains below the crest
of the weir, Equation (8.18) is applicable, and as long
as (
H
0
−
H
w
)/
H
w
≤ 0.4 (which is typically the case),
the weir coefficient can be taken as 1.83 (Chin, 2013).
In cases where the downstream water surface rises
above the crest of the weir, the flow rate over the
weir,
Q
(L
3
T
−1
), is given by
Q C L H H
0
=
(
−
)
/
3 2
w
0
w
1 7
.
=
( .
1 83 30
)(
)(
H
−
0 30
.
)
/
3 2
0
which yields
H
0
= 0.40 m. Since (
H
0
−
H
w
)/
H
w
=
0.33 ≤ 0.4, the assumption that
C
w
= 1.83 is justified.
Hence, the depth of flow just upstream of the weir is
0.40 m.
If the downstream water depth rises 5 cm above the
crest of the weir and the flow rate through the wetland
remains at 1.7 m
3
/s, then Δ = 0.05 m,
Q
= 1.7 m
3
/s, and
the combination of Equations (8.18) and (8.19) requires
that
0 385
.
3 2
/
Q
Q
∆
=
1
−
(8.19)
H H
−
0
0
w
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