232 Computational Modelling in Hydraulic and Coastal Engineering
where y(x,t) is the shoreline retreat (erosion) or advancement (accretion)
perpendicularly to the original shoreline, and h max is the maximum water
depth that sediment transport is affected. That depth is usually related to
the wave breaking height (h max = 2.5H b ).
Equations 8.35 and 8.36 are solved for the two unknown variables
y(x,t) and Q s (x,t) while the equations feed back to each other as the wave
breaking angles change in time from the initial distribution φ ο = φ(x, t =
0) as
ϕ
ϕ
( , )
a rctan
x t
dy
dx
o
=
−
(8.37)
The angle φ is measured counterclockwise. In practice the alongshore sediment transport process can be modified/interrupted due to various reasons
such as
• Presence of a groin interrupting the sediment flow
• Presence of a detached breakwater creating a wave ‘shadow’ on the
coast and thus changing the values of the Q s , h max and φ
• Unavailability of sediment source, etc.
The numerical solution of the system of Equations 8.35 to 8.37 is done
by using a centred finite differences scheme on a staggered discretization
one-dimensional grid. Thus, the coastline is divided into equal-length segments and the retreat or extension of the beach along the y-axis (seaward)
is calculated at the centre of the segment from the sediment fluxes defined
at the sides of the segment.
Example 8.7
This application investigates the evolution of an initially straight
coastline around a groin by using the one-line model coupled with
the alongshore sediment transport formula. The beach is subject to a
continuous attack from incoming waves approaching the beach at an
angle φ. The data used for the simulation are as follows:
Reference water depth = 2.5 m
Sediment transport coefficient (Equation 8.35) = 0.02
Empirical exponent (Equation 8.35) = 2.5
Wave heights = 1.0 m (modified near the groin as in the input file)
Angle of incident wave = 0.5 radians (28.66°) (modified near the
groin as in the input file)
Initial beach profile = Straight line
Location of the groin = 500 m (centre of the solution domain)
Length of the beach = 1000 m
where y(x,t) is the shoreline retreat (erosion) or advancement (accretion)
perpendicularly to the original shoreline, and h max is the maximum water
depth that sediment transport is affected. That depth is usually related to
the wave breaking height (h max = 2.5H b ).
Equations 8.35 and 8.36 are solved for the two unknown variables
y(x,t) and Q s (x,t) while the equations feed back to each other as the wave
breaking angles change in time from the initial distribution φ ο = φ(x, t =
0) as
ϕ
ϕ
( , )
a rctan
x t
dy
dx
o
=
−
(8.37)
The angle φ is measured counterclockwise. In practice the alongshore sediment transport process can be modified/interrupted due to various reasons
such as
• Presence of a groin interrupting the sediment flow
• Presence of a detached breakwater creating a wave ‘shadow’ on the
coast and thus changing the values of the Q s , h max and φ
• Unavailability of sediment source, etc.
The numerical solution of the system of Equations 8.35 to 8.37 is done
by using a centred finite differences scheme on a staggered discretization
one-dimensional grid. Thus, the coastline is divided into equal-length segments and the retreat or extension of the beach along the y-axis (seaward)
is calculated at the centre of the segment from the sediment fluxes defined
at the sides of the segment.
Example 8.7
This application investigates the evolution of an initially straight
coastline around a groin by using the one-line model coupled with
the alongshore sediment transport formula. The beach is subject to a
continuous attack from incoming waves approaching the beach at an
angle φ. The data used for the simulation are as follows:
Reference water depth = 2.5 m
Sediment transport coefficient (Equation 8.35) = 0.02
Empirical exponent (Equation 8.35) = 2.5
Wave heights = 1.0 m (modified near the groin as in the input file)
Angle of incident wave = 0.5 radians (28.66°) (modified near the
groin as in the input file)
Initial beach profile = Straight line
Location of the groin = 500 m (centre of the solution domain)
Length of the beach = 1000 m