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 wavesOcean Waves2Linear Wave ProblemLinear free-surface gravity waves can be characterized by their amplitude, a, wavelength, λ, and frequency, ω. Conditions 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<3Conservation 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 thatBoundary 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=−ˆBn4Free Surface Conditions• Dynamic Boundary Condition: Pressure is constant across the interface:21()2atmpgz c t ptφρφ∂=− + ∇ + + =∂21(,) 02gxttφρφη∂+∇+ =∂211(,)2xtgtφηφ∂=− + ∇∂(,)zxtη=Let the arbitrary Bernoulli constant equal atmospheric pressure: c(t) = patm. Substitute η(x,t) for z:onFree 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)5Free surface Conditions• Kinematic Boundary Condition:wuxtηη∂∂=+∂∂(,)zxtη=, onSubstitute the velocity potential into the KBC:,uwxdzφφ∂∂==∂ztxxφηφη∂∂∂∂=+∂∂∂∂(,)zxtη=, onNon-dimensional variables*aηη=Wave Elevation:*uauω=x-velocity:*wawω=z-velocity:*ttω=Time:*xxλ=Length:*aφωλ φ=Velocity Potential:*dadφωλφ=*1dt tω=*dx dxλ=6Linear 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∂∂∂∂Linearized 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∂=−∂7Where 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==−==Pressure Under a Wavend2hydrostatic2 order12pgztφρρφρ∂=− + ∇ +∂ Dynamic pressure due to linear surface gravity waves:() ( )21cosdapfzkxwttkφωρρψ∂=− = − +∂() ( )21,dpfz xtkωρη=2dispersion ret la atinh onshipgkHkω= ⇔8Motion of a Fluid ParticleParticle Orbits9Particle Orbits in Deep WaterCircles with exponentiallydecreasing radiusParticle motion extinct at z ≅ -λ/2()1HkH→∞ 12() ()kzfzfze≅≅()222 kzppaeξη+=2dispersion relationshipgkω⇔=Particle acceleration and velocity10at 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) ekzPhase SpeedVelocity at which a wave crest is moving11Group Speedη(t, x = 0)tWaves packets (or envelopes) move at the Group Speedη(x,t) = a
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