Geoscience Reference
In-Depth Information
be expected with highly suspended material. The
-values approximate unity, which
indicates a nearly uniform sediment concentration profile. Together with the zero-
values of
a
, the nodal-point relation (
2
) approximates the proportional relation (
1
).
In the exponential nodal-point relation (
3
), the
m
-values are slightly higher than
unity for channel 2 and 3, indicating a weak nonlinearity. For channel 4, the value
m
b
1 shows a fully linear dependence. The negative
n
-values suggest a depen-
dence on the widths of the branches, which is quite expectable for bed-load
problems. However, conclusions cannot be drawn because the influence of the
channel widths has not been investigated. The theoretical relation
n
¼
m
(see (
3
)) is not satisfied. It should be noticed that Wang et al. (
1993
) recommend
an empirical determination of
n
.
For the suspended load problems, the values
m
¼
1
0 show a nearly full
proportionality of the nodal-point relation (
3
) approximating (
S
1
/
S
2
¼
1 and
n
Q
1
/
Q
2
), and
an independence of the width ratio
B
1
/
B
2
. This result was to be expected and
satisfies the theoretical relation
n
m
.
Resuming the foregoing, it can be concluded that for each of the bed-load
problems, relatively more sediment is attracted to the diverting channel, whereas
the sediment distribution over the branches for suspended load is practically
proportional to the discharge distribution.
¼
1
4.1
Influence of Diverting Angle
The relation between the sediment transport ratio
S
1
/
S
2
and the angle of diversion is
presented in Fig.
7
. It shows that for bed-load problems, a critical angle exists,
causing a maximum sediment load in the diverting branch. With an increasing
angle, the sediment load decreases again behind this point for both discharge ratios.
A possible explanation could be that the particles cannot follow the sharply bending
Fig. 7 Sediment distribution of bed and suspended loads as a function of diverting angle for
discharges
Q
1
/
Q
2
¼
0.2 and
Q
1
/
Q
2
¼
0.5
