Geoscience Reference
In-Depth Information
respectively. This is due to that the Reynolds number decreases as the vegetation
concentration increases.
Fig. 10.12 shows the simulated flow vectors for vegetation concentrations of 0.6%,
2.5%, and 10%. It can be seen that the vegetation forced the thread of highest veloc-
ity to meander. The meandering flow pattern became more obvious, as the vegetation
concentration increased. When the vegetation concentration was 10%, a recircula-
tion flow occurred downstream of each vegetation zone. Fig. 10.13 compares the
measured and calculated flow velocities along several cross-sections for the case with
vegetation concentration of 10%. The simulation results and measurement data match
qualitatively well.
10.3 SIMULATION OF SEDIMENT TRANSPORT
IN VEGETATED CHANNELS
10.3.1 Sediment transport models in vegetated
channels
The total load is separated as bed load and suspended load, as shown in Fig. 2.6. The
3-D transport equation of suspended load in vegetated channels is
+
∂
[
(
1
−
c
v
0
)
u
j
c
k
]
−
∂
[
(
1
−
c
v
0
)ω
sk
δ
3
j
c
k
]
∂
[
(
1
−
c
v
0
)
c
k
]
=
∂
∂
s
∂
c
k
(
1
−
c
v
0
)ε
∂
t
∂
x
j
∂
x
j
x
j
∂
x
j
(10.61)
where
c
k
is the local concentration of the
k
th size class of suspended load.
To solve Eq. (10.61), the boundary condition at the water surface is given as
Eq. (7.44), and the deposition and entrainment rates at the lower boundary of the
suspended sediment layer are
D
bk
=
ω
sk
c
bk
and
E
bk
=
ω
sk
c
b
∗
k
.
Integrating Eq. (10.61) over the flow depth yields the depth-averaged 2-D transport
equation of suspended load in vegetated channels:
∂
[
(
1
−
c
v
)
hC
k
/β
sk
]
+
∂
[
(
1
−
c
v
)
hU
x
C
k
]
+
∂
[
(
1
−
c
v
)
hU
y
C
k
]
∂
t
∂
x
∂
y
h
D
sxk
h
ε
D
syk
=
∂
∂
s
∂
C
k
∂
+
∂
∂
s
∂
C
k
∂
(
1
−
c
v
)
ε
+
(
1
−
c
v
)
+
x
x
y
y
+
αω
sk
(
1
−
c
v
)(
C
∗
k
−
C
k
)
(10.62)
where
C
k
is the depth-averaged concentration of the
k
th size class of suspended load.
Integrating Eq. (10.62) over the channel width yields the 1-D transport equation of
suspended load in vegetated channels:
∂
[
(
1
−
c
v
)
AC
k
/β
]
+
∂
[
(
1
−
c
v
)
AUC
k
]
sk
=
α(
1
−
c
v
)ω
sk
B
(
C
−
C
k
)
+
q
lsk
(10.63)
∗
k
∂
∂
t
x
