113.42 Lecture #2Linear WavesFebruary 5, 2004Alexandra H. TechetMIT Department of Ocean Engineering©A. H. Techet 20041. Recap of Linear Wave Problem2. Phase Speed3. Particle motion beneath waves2Ocean Waves3Linear Wave ProblemLinear free-surface gravity waves can be characterized by their amplitude, a, wavelength, λ, and frequency, ω.4Conditions for Linear Waves• Linear wave theory assumes that the ratio of the wave height to wavelength is less than 1/7. Above this value waves begin to exhibit non-linear behavior, eventually breaking./1/7hλ<5Conservation of Mass222220xzφφφ∂∂∇=+=∂∂Vφ=∇JKJKThe velocity potential, φ, must satisfy the Laplace Equationin order for mass to be conserved:()0V∇×=JKJKAssuming ideal flow: incompressible, inviscid & irrotationalDefine a velocity potential, φ, such that6Boundary Conditions• Body boundary condition: there is no fluid flux through (normal to) a solid surface such as a body or the ocean floor.• On the ocean floor, @ , assuming a horizontal orientation: 0Bnφ∂=∂0wzφ∂==∂zH=−ˆBn7Free Surface Conditions• Dynamic Boundary Condition: Pressure is constant across the interface:21()2atmpgzctptφρφ∂=− + ∇ + + =∂21(,) 02gxttφρφη∂+∇+ =∂211(,)2xtgtφηφ∂=− + ∇∂(,)zxtη=Let the arbitrary Bernoulli constant equal atmospheric pressure: c(t) = patm. Substitute η(x,t) for z:on8Free surface Conditions• Kinematic Boundary Condition: Once a particle on the surface, it remains there always. ()()(),,pp pp p pzz xxtt xt x txtηηδηδ δη δ δ∂∂+= + += + +∂∂(,)ppzxtη=ppzxtxtηηδδδ∂∂=+∂∂,ppxzuwttδδδδ==We can write the velocity of a particle under the wave as:wuxtηη∂∂=+∂∂(,)zxtη=, onTherefore we have the kinematic boundary condition (KBC)9Free surface Conditions• Kinematic Boundary Condition:wuxtηη∂∂=+∂∂(,)zxtη=, onSubstitute the velocity potential into the KBC:,uwxdzφφ∂∂==∂ztxxφηφη∂∂∂∂=+∂∂∂∂(,)zxtη=, on10Non-dimensional variables*aηη=Wave Elevation:*uauω=x-velocity:*wawω=z-velocity:*ttω=Time:*xxλ=Length:*aφωλ φ=Velocity Potential:*dadφωλ φ=*1dt tω=*dx dxλ=11Linear Waves/1/7hλRecall for linear waves:We need to linearize the boundary conditions. For starters lets look at the terms from the dynamic boundary condition:222****22**2**xxaaaxatttφφφωφφφωλλλ ∂∂∂ ∂∂∂ ==≈∂∂∂∂∂∂Therefore:2xtφφ∂∂∂∂12Linearized DBC & FBC:211(,)2xtgtφηφ∂=− + ∇∂ztxxφηφη∂∂∂∂=+∂∂∂∂On :(,)zxtη=DBC KBCNeglecting higher order terms, and expanding φ(x,z,t) about z=0, these conditions become:220gtzφφ∂∂+=∂∂; on 0z =1and ( , )xtgtφη∂=−∂13Where is the dispersion relationshipA Solution to the Laplace Equation222220xzφφφ∂∂∇= + =∂∂()1(,,) sinaxzt f kx tkωφωψ=− − +()()1coshsinhkz HfkH+=()()2sinh sinhkz HfkH+=()1(,,) cosduxzt a f kx tdxφωωψ==− −+()2(,,) sindwxzt a f kx tdzφωωψ== −+()(,) cosxt a kx tηωψ=−+2tanhgkkHω=sin cos , cos sinsinh cosh , cosh sinhddxxxxdx dxddxxxxdx dx==−==14Pressure Under a Wavend2hydrostatic2 order12pgztφρρφρ∂=− + ∇ +∂ Dynamic pressure due to linear surface gravity waves:() ( )21cosdapfzkxwttkφωρρψ∂=−= −+∂() ( )21,dpfzxtkωρη=2dispersion ret la atinh onshipgkHkω= ⇔15Motion of a Fluid Particle16Particle Orbits17Particle Orbits in Deep WaterCircles with exponentiallydecreasing radiusParticle motion extinct at z ≅ -λ/2()1HkH→∞ 12() ()kzfz fz e≅≅()222 kzppaeξη+=2dispersion relationshipgkω⇔=18Particle acceleration and velocity19at the free surface…The intersection between the circle on which ξpand ηplie and the elevation profile η(x,t) define the location of the particle.This applies at all depths, z: η1(x,z,t) = a ekzcos(kx-ωt+ψ) = η(x,t) ekz20Phase SpeedVelocity at which a wave crest is moving21Group Speedη(t, x = 0)tWaves packets (or envelopes) move at the Group Speedη(x,t) = a
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