152 Computational Modelling in Hydraulic and Coastal Engineering
PROBLEM 6.3
Conduct the following simulations by making the suggested changes while
keeping all of the other data constant:
1. Change the longitudinal length to 4000 m, then compare and explain the
simulation results after running the model for 600 and 3600 time steps.
2. Using the same seismic seafloor spatial segment, change the oscillation mode to a sudden rise of the floor to a height of 0.1 m at time
t = 0.1 s followed by an additional rise to 0.2 m at t = 0.2 s and then
falling back to zero elevation at t = 0.3 s. Repeat the problem using a
sudden fall to a depth of –0.1 m at t = 0.1 s and an additional fall to
–0.2 m at t = 0.2 s and then returning the seafloor elevation back to
zero level at t = 0.3 s. Run the two simulations, and then compare and
comment on the results.
3. For the right half of the solution domain, change the bed friction to
f bo = 0.1 s –1 . Conduct the simulation and comment on the results.
4. Assume that the water over the 20 m seismic section is 5 m deep,
while the water the left of that section the is 6 m deep and to the right
is 4 m deep. Run the computer program, and explain and comment
on the simulation results.
5. If the anticipated seismic characteristic parameters are zbmax = 4 m
and nb = 2, how much artificial bed friction is needed so that the tsunami wave does not exceed 1 m over the equilibrium mean sea level?
6.3 TWO-DIMENSIONAL LINEAR
GRAVITY LONG WAVES
For a horizontal, two-dimensional domain, the linear non-dispersive long
wave equation involving bed friction and wave-breaking energy losses
takes the form
∂
∂
=
∂
∂
∂
∂

 

  +
∂
∂
∂
∂

 

  −
∂
∂
+
2
2
2
2
ζ
ζ
ζ
λ
ζ
t
x
c x
y
c y
t
o
o
N N
t x
t y
b
∂
∂
∂
∂





 +
∂
∂
∂
∂














2
2
2
2
ζ
ζ
(6.24)
If wave dispersion effects are to be accounted for, then the equation becomes
∂
∂
=
∂
∂
∂
∂

 

  +
∂
∂
∂
∂

 

  −
−
2
2
2
ζ
ζ
ζ
σ
t
x
c c x
y
c c y
o g
o g
(
k k c c
t
N
t x
t y
o g
b
2
2
2
2
2
)ζ λ
ζ
ζ
ζ
−
∂
∂
+
∂
∂
∂
∂

 

  +
∂
∂
∂
∂




 









(6.25)
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