ERDC/CHL CHETN-IV-35
June 2001
It can be seen from Equations 14 and 15 that this solution is valid for relatively short times after
t = 0 because the term proportional to exp(r1t) diverges for long elapsed time (r1 > 0).
Substituting (14) into (9) and integrating gives
C1
abR
C2
x = 2 qR ( exp(r1t ) - 1) +
( exp(r2t ) - 1)
(17)
r1
r2
z0
For small t, (14) can be expanded to give (retaining leading order in t, adR, and abR),
ε a
ε a
z = rd dR qR t - rd dR ( adR + abR ) qR t 2
2
(18)
2W 2 z0
W0
0
indicating that the channel starts filling linearly with time. If abR = 0 (no bed-load transport),
then x = 0 for all time, and Equation 18 reduces to
-εrd adR
z = z0 1 - exp
(19)
qR t
W0 z0
which indicates exponential filling of the channel.
Similarly, for small t, Equation 17 yields
εrd abR adR 2 2
abR
x=
q t+
qR t
(20)
z0 R
2
2W0 z
0
showing that if adR = 0 (no suspended sediment), the channel fills in linear manner with time by
growth and intrusion of the updrift side into the channel.
Channel Infilling Rate: The rate of channel infilling, how rate at which the bottom is
shoaling, is Rz = dz/dt and can be from Equation 14. For a time shortly after dredging,
Equation 18 gives
a ε
a ε
Rz = dR rd qR - dR rd ( adR + abR ) qR t
2
(21)
2
W0
W0 z0
The leading-order term is independent of z0, so the rate of channel infilling depends more
strongly on W0 than on z0. The solution thus indicates that the rate of channel infilling can be
reduced more by increasing channel width than by increasing channel depth.
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