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6.3 Exercise 15: Coastal Kelvin Waves 121η(x, y, t) = η o exp (−y/R)sin(kx − ωt) (6.4)u(x, y, t) =√ gH η o exp (−y/R)sin(kx − ωt) (6.5)where k = 2π/λ (λ is wavelength), ω = 2π/T (T is wave period), and R is theRossby radius of deformation, define by:√ gHR =| f |(6.6)It can also be shown that these waves travel with a phase speed of long surfacegravity waves of c = √ gH along the coast with the coast on their right in thenorthern hemisphere and on their left in the southern hemisphere. Their amplitudeis maximum at the coast and decreases exponentially away from the coast on alengthscale of the deformation radius.6.3 Exercise 15: Coastal Kelvin Waves6.3.1 AimThe aim of this exercise is to simulate the structure and dynamics of coastal Kelvinwaves.6.3.2 Task DescriptionConsider a rectangular model domain with a length of 400 km, a width of 100 km,and a depth of 10 m. Lateral grid spacing is set to Δx = Δy = 2 km. The time stepis set to Δt = 10 s. All boundaries are treated as coasts. An unrealistically high valueof the Coriolis parameter of f =+5×10 −4 s −1 is chosen to reduce the deformationradius to R ≈ 20 km and to keep the total simulation time within a reasonable limit.The associated inertial period is 3.5 h. The smart reader will jump on the table andclaim that the minimum inertial period on Earth is 12 hours. This is true, but with amodel we can do a little bit of science fiction can’t we?A wave paddle is located near the lower left corner of the model domain oscillatingthe sea level with an amplitude of 1 m and a period of 2 hours. This period isnot far away from the fictiona inertial period, so that an influenc by the Coriolisforce can be anticipated. The task is to simulate the resultant wave fiel over half aday (or 3.4 inertial periods) with outputs of sea-level elevation and velocity field atevery 10 mins.The f rst-order Shapiro filte with a smoothing parameter of ɛ = 0.05 should beapplied. For simplicity, wind-stress forcing, bottom friction and lateral momentum