Geoscience Reference
In-Depth Information
Fig. 12.15 Same as previous
figure, for the case of
a time-area function
with the shape of a
right-angled triangle
(dashed line).
2
K/t c =0.1
u
×
t c
0.2
1
0.5
1
2
0
0
1
2
3
t/t c
which yields
1
2 e t c / (2 K )
4 Ke t / K
t c
4
t c
=−
+
u ( t )
( t
K
t c )
(12.34)
It can be readily checked that Equations (12.32) and (12.34) yield the same value for u at
t
=
t c /
2, as they should. For t
>
t c , (12.33) must be integrated again, but with the upper
limit at
τ =
t c in the second and third integral, because A r is zero beyond that point; this
produces
4 Ke t / K
t c
1
e t c / K
2 e t c / (2 K )
u ( t )
=
+
(12.35)
Again, it can be seen that both (12.34) and (12.35) produce the same result at t
t c ,as
they should. The resulting instantaneous unit hydrograph, obtained by patching (12.32),
(12.34) and (12.35) together over their respective time ranges, is shown in Figure 12.14.
In principle, it should be possible to use this three-component unit response function
with any input function x ( t ) in the convolution integral (12.2), to calculate the actual
outflow y ( t ) analytically. However, because this u ( t ) consists of three parts, this is rather
involved, so that in practical applications it may be more convenient to convert u ( t ) into
tabular form and carry out the calculations numerically. An idea of the effect of the shape
of the time-area function on the resulting unit response can be gained by comparing
this result with the response function obtained with a right-angled triangle shown in
Figure 12.15.
=
Combinations of linear storage elements
In yet another class of models, the basin outflow is derived by routing the rainfall input
solely through a number of storage elements, without any formal or explicit translation in
the formulation. Equation (12.28) is used as the response function of each of the storage
elements. In what follows, a few examples are reviewed of this type of representation.
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