Geoscience Reference
In-Depth Information
where
f
k
(
z
)
is the Rouse distribution of suspended-load concentration, and
δ
is the
thickness of the bed-load zone at cross-section 2.
Then,
C
k
,2
and
C
k
,3
can be determined using Eqs. (5.163) and (5.165).
The determination of bed-load transport rates
Q
bk
,2
and
Q
bk
,3
has not been well
investigated, due to the complexity. Their ratio may be assumed to be approxi-
mately equal to that of bed-load transport capacities
Q
b
∗
k
,2
and
Q
b
∗
k
,3
at cross-
sections 2 and 3:
Q
bk
,2
Q
bk
,3
≈
Q
b
∗
k
,2
Q
b
∗
k
,3
(5.166)
and
Q
bk
,2
and
Q
bk
,3
can then be determined using Eqs. (5.164) and (5.166).
Note that Eqs. (5.161)-(5.166) can be applied in both quasi-steady and unsteady
sediment transport models.
5.3.5 Lateral allocation of bed change in
1-D model
A 1-D model provides only the lumped change in bed area,
A
b
, at a cross-section. In
order to acquire a reasonable prediction for long-term river morphological evolution,
A
b
must be allocated appropriately to the local change in bed elevation,
z
b
, along
the cross-section at each time step. The obtained
z
b
is used to update the cross-section
geometry, as expressed in Eq. (5.134).
The simplest method is the uniform distribution of bed change along the cross-
section, except for water edges, where the bed change may be zero. Wu (1991)
suggested a slight modification of this method, assuming uniform deposition and ero-
sion for wide channels and horizontal deposition and uniform erosion for narrow
channels, as shown in Fig. 5.16. This method is more adequate for suspended load
(fine sediments) than for bed load (coarse sediments), because the suspended-load con-
centration tends to be relatively uniform along the cross-section while the bed load
usually moves in strips.
Figure 5.16
Allocation of bed change in cross-section: (a) wide channel and (b) narrow channel.
A more general method, used by Chang (1988), allocates deposition and erosion
along the cross-section by a power function of excess shear stress:
m
B
(τ
b
−
τ
c
)
(τ
−
τ
)
c
b
z
b
=
y
A
b
(5.167)
m
