Geoscience Reference
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time intervals. At each time interval, the flow and sediment transport are assumed to
be steady, but the bed change and bed material sorting are still calculated, and thus
the temporal evolution of channel morphology is simulated.
The time intervals are generally used as time steps in the calculation of bed change
and bed material sorting, and should thus be restricted by the stability criteria of
the sediment transport model. In addition, the time intervals should be defined in
such a way that the temporal variations of flow discharge, water stage, and sediment
discharge are well represented. This means that shorter intervals should be used for
high flow periods and longer intervals may be used for low flow periods. Normally,
the time intervals can be hours or days.
5.3.2.2 Discretization of quasi-steady sediment transport
equations
As discussed in Section 5.1.2.1, two approaches may be used in the simulation of total-
load transport. The approach that computes bed load and suspended load separately
is adopted here.
Under steady flow conditions, the suspended-load and bed-load transport equations
(5.27) and (5.28) without side sediment discharges are written as
d
(
QC k )
dx
= αω sk B
(
C
C k )
(5.128)
k
dQ bk
dx
1
L (
=
Q b k
Q bk )
(5.129)
Eqs. (5.128) and (5.129) are first-order ordinary differential equations. They can
be discretized using many numerical schemes, such as the Euler scheme, central dif-
ference scheme, and the Runge-Kutta method. Han (1980) established the following
exponential difference scheme for Eq. (5.128), based on its analytical solution:
exp
αω sk B i + 1 / 2
x i + 1 / 2
C k , i + 1 =
C
+ (
C k , i
C
)
k , i
+
1
k , i
Q i + 1 / 2
Q i + 1 / 2
+ (
C
C
)
k , i
k , i
+
1
αω
sk B i + 1 / 2
x i + 1 / 2
1
exp
αω
sk B i + 1 / 2
x i + 1 / 2
×
(5.130)
Q i + 1 / 2
The Han scheme (5.130) is very stable, but it is not strictly conservative. However,
many tests have shown that it has good accuracy.
Similarly, Eq. (5.129) can be discretized using the following exponential difference
scheme:
exp
x i + 1 / 2
L
Q bk , i + 1 =
Q b k , i + 1 + (
Q bk , i
Q b k , i )
1
exp
L
x i + 1 / 2
L
+ (
Q b k , i
Q b k , i + 1
)
(5.131)
x i + 1 / 2
 
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