Geoscience Reference
In-Depth Information
5.1.2.4 Boundary and initial conditions of sediment
The fractional sediment discharges for all size classes must be imposed at the inflow
boundary in each time step, but no sediment boundary condition is required at the
outflow boundary in the 1-D model. The initial sediment discharge, channel topog-
raphy, and bed-material gradation should be provided for the simulation of unsteady
sediment transport and channel morphological evolution.
5.2 1-D CALCULATION OF OPEN-CHANNEL FLOW
5.2.1 1-D steady flow calculation
5.2.1.1 Discretization of steady flow equations
For steady open-channel flow without side inflow or outflow, Eq. (5.1) reduces to
∂
0 and leads to a constant flow discharge along the study reach, while
Eq. (5.2) can be rewritten as the energy equation:
Q
/∂
x
=
β
Q
2
2
A
2
∂
∂
g
∂
z
s
g
Q
|
Q
|
+
x
+
=
0
(5.40)
x
∂
K
2
β
is the correction factor for kinetic energy due to the non-uniformity of stream-
wise velocity over the cross-section. For the compound cross-section shown in Fig. 5.4,
β
can be determined using the discharge-weighted average kinetic energy:
where
1
QU
2
(
β
=
Q
LF
U
LF
+
Q
MC
U
2
MC
+
Q
RF
U
RF
)
K
LF
A
2
K
3
K
3
MC
K
RF
A
RF
=
A
LF
+
A
2
MC
+
(5.41)
where all parameters are the same as those in Eq. (5.26).
Suppose that the computational domain of a single channel is divided into
I
1
reaches by
I
cross-sections (computational points), as shown in Fig. 5.7. The cross-
sections are numbered 1 through
I
in the downstream direction. Each cross-section
is represented by an adequate number of points (stations), as shown in Fig. 5.4, with
each point characterized by a pair of values of the distance to the left bank and the
bed elevation. In the longitudinal direction, each reach is characterized by its length.
For a simple channel, the reach length measures the path of the main flow or channel
thalweg. For a compound channel, the flow paths in the main channel and floodplains
may be significantly different, and an average, such as the discharge-weighted average,
of their lengths should be used as the reach length.
Applying the standard step method to discretize Eq. (5.40) yields
−
Q
i
+
1
|
(5.42)
z
s
,
i
=
β
i
+
1
Q
i
+
1
2
gA
i
+
1
β
i
Q
i
z
s
,
i
+
1
+
x
i
+
1
/
2
2
Q
i
+
1
|
K
i
+
1
Q
i
|
Q
i
|
K
i
2
gA
i
+
+
+
