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 Technical Note DOER-N1
April 1998
This TN documents the layered bed processes and combined current-wave shear stress processes
that have been incorporated into the cohesive sediment transport sub-model of LTFATE. Future
TNs will address additional LTFATE model improvements, including bed armoring, combined
sand/clay sediment bed processes, cohesive sediment flocculation, and cohesive sediment settling
speeds.
LAYERED SEDIMENT BED MODEL
As previously stated, the rate and method by which cohesive sediments erode depend on several
factors, including grain-size distribution, organic content, pore water content, and mineralogy,
among others. Therefore, erosion of cohesive sediments varies significantly from location to
location as well as with depth. A commonly used method to relate erosion to bottom shear stress
has been incorporated into LTFATE. This method relates erosion to a function of shear stress to
some exponential power. The equation for the erosion rate ∈ in pounds (mass) per square foot per
second is:
m
τ - τ
∈ = A0 bm cr
τcr
(1)
where A0 and m are site-specific parameters, τbm (pounds (force) per square foot) is the near-bottom
shear stress, and τcr is the critical shear stress below which no erosion occurs. To predict erosion
at a given location accurately, several values of A0 and τcr must be used in the vertical direction.
Therefore, a vertically layered sediment bed has been incorporated into LTFATE, which includes
varying values of A0 and τcr. These parameters can vary significantly from layer to layer, and erosion
rate experiments on sediments extracted from the site are the best method for determining the
vertically varying values. If such data are not available, reasonable values from similar sediments
should be used. The figure below provides an example of the user-provided distribution of these
parameters as well as possible initial thicknesses of each layer.
Water Column
Sediment Bed
A0 = 8 x 10-6
-4
2
τcr = 5 x 10 lbf/ft
Initial thickness = 0.025 ft
A0 = 4 x 10-6
-3
2
τcr = 1 x 10 lbf/ft
Initial thickness = 0.025 ft
A0 = 4 x 10-6
-3
2
τcr = 5 x 10 lbf/ft
Initial thickness = 0.050 ft
A0 = 5 x 10-7
-2
2
τcr = 1 x 10 lbf/ft
Initial thickness = 0.050 ft
A0 = 3.75 x 10-7
-2
2
τcr = 2 x 10 lbf/ft
Initial thickness = 0.050 ft
A0 = 2.5 x 10-7
-2
2
τcr = 2 x 10 lbf/ft
Initial thickness = 0.100 ft
A0 = 1 x 10-7
-2
2
τcr = 2 x 10 lbf/ft
Initial thickness = 0.100 ft
A0 = 1 x 10-7
-2
2
τcr = 2 x 10 lbf/ft
Initial thickness = 0.100 ft
2
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