numerical calculation was halted if z reached z0/2 (assumed project depth) or x reached
W0/2 (minimum allowable width of channel).
Example 1: (fine sand) Q = 150,000 m3/year, ad = 0.5, ab = 0.1.
The sediment at this site is fine sand, and experience with sensitivity testing of the
model and comparison to limited data gives ad = 0.5, with little bedload (ab = 0.1). Larson
and Kraus (2001) give a procedure for estimating ad consistent with the present simplified
approach. This example simulates a shallow-draft channel at an inlet located on a sandy
shore, so most of the sand is deposited into the channel or passes over the channel
( as = 1 - ad - ab = 0.4 ). If material is deposited into the channel, it can be readily
resuspended. The effective channel length is 1,000 m, so q = 150,000/1,000 =
Figures 3 and 4 compare calculations with the numerical model and the linearized
model. For short elapsed time there is agreement, with deviations occurring after about 0.6
to 0.8 year for this example. After 1.7 years, project depth (z = z0) was reached, and the
numerical model stopped.
Figure 5 shows the time evolution of the channel infilling rate qc and the bypassing rate
qy normalized by the input q. The total adds to unity at any given time. The rate of
bypassing exceeds the channel infilling rate approximately 0.6 years after dredging.
Fig. 3. Increase in elevation (decrease in depth) in channel on sand shore: comparison of non-
linear model (numerical solution) and linearized model (analytical solution)
Kraus and Larson