1 / 32

Shallow Water Waves 2: Tsunamis and Tides

Shallow Water Waves 2: Tsunamis and Tides. MAST-602 Lecture Oct.-16, 2008 (Andreas Muenchow). Knauss (1997): p. 218-222 (tsunamis and seiches) p. 234-244 tides p. 223-228 Kelvin waves. Descriptions: Tsunamis, tides, bores Tide Generating Force Equilibrium tide Co-oscillating basins

cillian
Download Presentation

Shallow Water Waves 2: Tsunamis and Tides

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript


  1. Shallow Water Waves 2: Tsunamis and Tides MAST-602 Lecture Oct.-16, 2008 (Andreas Muenchow) Knauss (1997): p. 218-222 (tsunamis and seiches) p. 234-244 tides p. 223-228 Kelvin waves Descriptions: Tsunamis, tides, bores Tide Generating Force Equilibrium tide Co-oscillating basins Kelvin and Poincare waved

  2. 0 = pressure gradient + horizontal tide generating force 0 = g ∂ht/∂s + hTGF Equilibrium Tide ht gives ht (“bulge”) of water-covered earth, no accelerations

  3. Tidal Breaking: • Friction between the ocean’s bulge and solid earth drags the bulge in the direction of the earth’s rotation. • This frictional effect removes rotational kinetic energy from the earth, thus increasing the length of the day by about 0.0023 seconds in 100 years. • It also implies a net forward acceleration of the moon that moves it about 3.8 cm/year away from earth (lunar recession). © Richard Pogge

  4. Tidal Locking of the Moon • The early moon rotated much faster: • As earth does now, it rotated under its tidal bulge; • internal friction resulted, which slowed the moon's rotation. • the Moon's rotation slowed until it matched its orbital period around the earth (29 days), and the friction stopped. • The end result is that the moon became Tidally Locked in synchronous Rotation.Therefore the moon keeps the same face towards the earth. Its rotational and orbital periods are the same: • ---> the moon is tidally locked to the earth.

  5. Resonance: response of an oscillatory system forced close to its natural frequency Breaking wine glass ---> 1-min movie Tacoma Bridge 1940 ---> 4-minutes movie Forced string ---> Java applet

  6. Resonance: response of an oscillatory system forced close to its natural frequency d damping parameter (friction) Response Forcing Frequency/Natural Frequency

  7. 0 = pressure gradient + horizontal tide generating force 0 = g ∂ht/∂s + hTGF Equilibrium Tide ht gives ht (“bulge”) of water-covered earth, no accelerations Unrealistic as water must “instantaneously” adjust to changing forcing but known ht provides useful in the dynamics of tides

  8. Dynamics of Tides ht is a known forcing function of (x,y,t): acceleration + coriolis = pressure gradient u/t - fv = - g (h-ht)/x x-momentum v/t + fv = - g (h-ht)/y y-momentum H(u/x + v/y) + h/t = 0 continuity used p(x,y,z,t) = gr[h(x,y,t)-z] from z-momentum p/z = - rg to convert 1/r p/x ---> g h/x

  9. Ocean Basin responding to tidal forcing ht under the influence of the earth’s rotation (Coriolis); Apparent standing wave rotating around the basins (Atlantic or Pacific Oceans); These are Kelvin and Poincare Waves.

  10. L H u/t = - g h/x H u/x + h/t = 0 Tidal Co-oscillation (without Coriolis): Standing wave due to perfect reflection at wall c=l/T=(gH)1/2 and L=l/4 (quarter wavelength resonator) ---> T=4L(gH)-1/2 deep ocean tide forced by hTGF due to moon/sun shelf tide forced by small h(t) at seaward boundary

  11. Currents Sealevel Time

  12. Example of quarter-wavelength resonator: Cook Inlet, Alaska h0~5m tidal bore forms: L L ~ 290 km H ~ 50 m T=12.42 hours c=(gh)1/2~22 m/s l=c*T~12.42 hrs*22 m/s=990km

  13. Dynamics of Tides ht is a known forcing function of (x,y,t): acceleration + coriolis = pressure gradient u/t - fv = - g (h-ht)/x x-momentum v/t + fv = - g (h-ht)/y y-momentum (uh)/x + (vh)/y + h/t = 0 continuity used p(x,y,z,t) = gr[h(x,y,t)-z] from z-momentum p/z = - rg to convert 1/r p/x ---> g h/x

  14. Kelvin Wave: peculiar balance of acceleration, Coriolis, and p-grad in the presence of a coast ∂u/∂t - fv = -g∂h∂x along-shore force balance ∂v/∂t + fu = -g∂h∂y across-shore force balance COAST ∂(uh)/∂x + ∂(vh)/∂y = -∂h∂t mass balance Assume v=0 everywhere y,v x,u

  15. COAST y,v x,u Kelvin Wave: peculiar balance of acceleration, Coriolis, and p-grad in the presence of a coast ∂u/∂t - fv = -g∂h∂x along-shore force balance ∂v/∂t + fu = -g∂h∂y across-shore force balance ∂(uh)/∂x + ∂(vh)/∂y = -∂h∂t mass balance Assume v=0 everywhere

  16. High High convergence convergence Low Low u u u u divergence divergence High High Kelvin wave: geostrophic across the shore COAST COAST Time t Time t+dt convergence y,v y,v convergence x,u x,u

  17. Wave Equation ∂2u/∂t2 = c2∂2u/∂x2 EQ-1 ∂(x-mom)/∂t: ∂2u/∂t2 = -g ∂(∂h/∂x)/∂t = -g ∂(∂h/∂t)/∂x EQ-2 ∂(continuity)/∂x: H∂2u/∂x2 = -∂(∂h/∂t)/∂x Insert EQ-2 into EQ-1: ∂2u/∂t2 = gH ∂2u/∂t2 Subject to the dispersion relation or c2 = gH w2 = k2 gH

  18. Wave Equation ∂2u/∂t2 = c2∂2u/∂x2 x y Try solutions u = Y(y)*cos(kx-ct) c/f is the lateral decay scale (Rossby radius) to find that Y(y) = A e-fy/c

  19. Tidal co-oscillation with Coriolis (Taylor, 1922) h (u,v) head head

  20. © 1996 M. Tomczak

  21. Internal Kelvin Wave in a closed basin layer-1 Layer-3 h layer-2 (u,v) layer-2 h layer-3 (u,v) layer-3 from Dr. Antenucci

  22. Inertia Gravity (Poincare)Wave: balance of acceleration, Coriolis, and p-grad ∂u/∂t - fv = -g∂h∂x along-shore force balance ∂v/∂t + fu = -g∂h∂y across-shore force balance COAST ∂(uh)/∂x + ∂(vh)/∂y = -∂h∂t mass balance No assumption on v y,v x,u

  23. Wave Equation ∂2u/∂t2 = c2∂2u/∂x2 subject to the dispersion relation w2 = k2 gh + f2 w>f or c2 = w2/k2 = gh + f2/k2

  24. Progressive Poincare Wave in a Channel Sea level Horizontal velocity from Dr. Antenucci

  25. Progressive Poincare Wave in a Channel Mode-2 Sea level Horizontal velocity from Dr. Antenucci

  26. Standing Poincare Wave in a Channel Mode-1 Sea level Horizontal velocity from Dr. Antenucci

  27. Standing Poincare Wave in a Channel Mode-2 Sea level Horizontal velocity from Dr. Antenucci

  28. Internal Poincare Wave in a closed basin (vertical mode-1) layer-1 (u,v) layer-1 layer-3 h layer-2 (u,v) layer-2 h layer-2 (u,v) layer-3 from Dr. Antenucci

  29. Internal Poincare Wave in a closed basin (vertical mode-1, horizontal mode-2) layer-1 (u,v) layer-1 layer-3 h layer-2 (u,v) layer-2 h layer-2 (u,v) layer-3 from Dr. Antenucci

  30. Tidal Dynamics: Scaling Depth-integrated (averaged) continuity (mass) balance: (uh)/x + (vh)/y + h/t = 0 UH/L UH/L h0/T ---> U ~ (h0/H) (L/T) or L ~ UHT/h0 Velocity scale U Vertical length scales H (depth) and h0(sealevel amplitude) Hirozontal length scale L Time scale T

  31. Depth-integrated (averaged) force (momentum) balance: Acceleration + nonlinear advection + Coriolis = pressure gradient u/t U/T U/T 1 uu/x+vu/y U2/L U2h0/(UHT) e= h0/H << 1 fv fU fU fT ~ 1 g h/x gh0/L gH(h0 /H)2/UT (ec/U)2 ~ 1 L ~ UHT/h0 h0 ~1m, H~100m --> e~0.01<<1 --> (gH)1/2~30 m/s --> U~0.3 m/s 2p/f ~ 12-24 hours, hence Coriolis acceleration contributes as fT~1

More Related