Geoscience Reference
In-Depth Information
suspended-load concentration and bed-load transport rate of uniform sediment with
the same size as
d
k
, taking into consideration, however, the hiding and exposure effects
in non-uniform bed material.
In analogy to Eq. (2.159), the 1-D fractional bed change equation is
∂
k
=
αω
s
B
A
b
∂
1
L
(
p
m
)
(
−
(
C
k
−
C
∗
k
)
+
Q
bk
−
Q
b
∗
k
)
1
(5.30)
t
where
k
is the rate of change in bed area due to size class
k
.
The total rate of change in bed area,
(∂
A
b
/∂
t
)
∂
A
b
/∂
t
, is determined by
∂
N
∂
A
b
∂
A
b
∂
=
(5.31)
t
t
k
k
=
1
As described in Section 2.7.2, the bed material is divided into layers. The temporal
variation of the mixing-layer bed-material gradation
p
bk
is determined by Eq. (2.161),
which is rewritten in the 1-D model as follows:
∂
k
+
p
bk
∂
∂(
A
m
p
bk
)
A
b
∂
A
m
∂
−
∂
A
b
∂
=
(5.32)
∂
t
t
t
t
where
A
m
is the cross-sectional area of the mixing layer, and
p
bk
is
p
bk
when
∂
A
b
/∂
t
−
∂
A
m
/∂
t
≥
0 and the fraction of size class
k
in the second layer of bed material when
∂
0. Accordingly, the bed material sorting equation (2.162) in the
second layer is rewritten as
A
b
/∂
t
−
∂
A
m
/∂
t
<
p
bk
∂
∂(
A
sub
p
sbk
)
∂
A
m
∂
−
∂
A
b
∂
=−
(5.33)
t
t
t
where
p
sbk
is the fraction of size class
k
in the second layer of bed material, and
A
sub
is
the cross-sectional area of the second layer. Note that Eq. (5.33) assumes no exchange
between the second and third layers.
Eqs. (5.27)-(5.33) constitute the governing equations of the total-load transport
model that discerns bed load and suspended load. This model provides the ratio of
bed load and suspended load. However, many reliable bed-material load transport
capacity formulas, such as the Ackers-White (1973), Engelund-Hansen (1967), and
Yang (1973) formulas, cannot be used directly in this approach.
1-D bed-material load transport model
When the bed-material (total) load transport is simulated without separating bed
load and suspended load, introducing Eq. (2.149) into Eq. (2.111) and consid-
ering the lateral exchange with banks and tributaries yields the bed-material load
transport equation:
Q
tk
β
tk
U
∂
∂
+
∂
Q
tk
∂
1
L
t
(
x
=
Q
t
∗
k
−
Q
tk
)
+
q
tlk
(
=
...
)
k
1, 2,
,
N
(5.34)
t
