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2.016 Hydrodynamics Free Surface Water Waves 2.016 Hydrodynamics Reading #7 2.016 Hydrodynamics Prof. A.H. Techet Fall 2005 Free Surface Water Waves I. Problem setup 1. Free surface water wave problem. In order to determine an exact equation for the problem of free surface gravity waves we will assume potential theory (ideal flow) and ignore the effects of viscosity. Waves in the ocean are not typically uni-directional, but often approach structures from many directions. This complicates the problem of free surface wave analysis, but can be overcome through a series of assumptions. To setup the exact solution to the free surface gravity wave problem we first specify our unknowns: • Velocity Field: V (x, y, z,t ) = ∇φ(x, y, z,t ) • Free surface elevation: η(x, yt, ) • Pressure field: p(xy,,z , t ) version 3.0 updated 8/30/2005 -1- ©2005 A. Techet 2.016 Hydrodynamics Reading #7 Next we need to set up the equations and conditions that govern the problem: • Continuity (Conservation of Mass): ∇=2φ 0 for z <η (Laplace’s Equation) (7.1) • Bernoulli’s Equation (given some φ ): 2 ∂φ 1 pp− a ∂t +∇2 φ + ρ + gz = 0 for z <η (7.2) • No disturbance far away: ∂φ ∂t , ∇→φ 0 and p = pa − ρ gz (7.3) Finally we need to dictate the boundary conditions at the free surface, seafloor and on any body in the water: (1) Pressure is constant across the free surface interface: p = patm on z =η . ⎧∂φ 1 2 ⎫ p =−ρ ⎨ − V − gz⎬ + c ()t= patm . (7.4) ⎩ ∂t 2 ⎭ Choosing a suitable integration constant, ct( ) = patm , the boundary condition on z =η becomes ∂φ 1 ρ{ + V 2 + gη} = 0. (7.5) ∂t 2 (2) Once a particle is on the free surface, it remains there always. Similarly, the normal velocity of a particle on the surface follows the normal velocity of the surface itself. z p =η (xp , t) ∂η ∂η z +δ z =η (x +δ x , t +δ t) =η (x , t) + δ x + δ t (7.6) p p p p p ∂x p ∂t On the surface, where z p =η , we can reduce the above equation to ∂η ∂η δz = uδt + δt (7.7) p ∂x ∂t and substitute δ z p = w δ t and δ x p = utδ to show that the normal velocity follows the particle: version 3.0 updated 8/30/2005 -2- © 2005 A. Techet 2.016 Hydrodynamics Reading #7 ∂η ∂η wu= + onz= η . (7.8) ∂x ∂t (3) On an impervious body boundary Bx( , y,,z t ) = 0 . Velocity of the fluid normal to the body must be equal to the body velocity in that direction: ∂φ vn⋅=ˆ φ ⋅ nˆ = = U() x,,t ⋅ nˆ() x t = U onB= 0. (7.9) ∂n n Alternately a particle P on B remains on B always; ie. B is a material surface. For example: if P is on B at some time tt= o such that B(x, t o ) = 0, then B ( x, t o ) = 0 for all t , (7.10) so that if we were to follow P then B = 0 always. Therefore: DB ∂B =+ (∇φ ⋅∇)B = 0 onB= 0. (7.11) Dt ∂t Take for example a flat bottom at z = −H : ∂φ/∂z = 0 on z = −H (7.12) II. Linear Waves 2. Linearized Wave Problem. version 3.0 updated 8/30/2005 -3- ©2005 A. Techet 2.016 Hydrodynamics Reading #7 To simplify the complex problem of ocean waves we will consider only small amplitude waves (such that the slope of the free surface is small). This means that the wave amplitude is much smaller than the wavelength of the waves. As a general rule of thumb for wave height to wavelength ratios ( h / λ , where h is twice the wave amplitude) less than 17/ we can linearize the free surface boundary conditions. Non-dimensional variables can be used to assess which terms can be dropped. Non-dimensional variables: η = aη ∗ ω tt= ∗ ua= ω u∗ x = λ x∗ wa= ω w∗ φ = aωλ φ ∗ Looking back to equation (7.5) we can evaluate the relative magnitude of each term by plugging in the non-dimensional versions of each variable. For example: dφ = aωλ dφ ∗ dt =/1 ω dt∗ dx = λ dx∗ ∂φ 2 ∂φ 2 Compare ∂t and V ∼ ()∂x to determine which terms can be dropped from equation (7.5): ∗ ∗ ∂φ 2 2 2 ∂φ ∂φ () a ω ()∗ a ()∗ ∂x = ∂x = ∂x . (7.13) ∂φ aωλ2 ∂φ∗ λ ∂φ∗ ∂t ∂t ∗ ∂t ∗ Here we see that if h /<<λ 17/ , then ∂φ ∂φ ( ) 2 << , (7.14) ∂x ∂t and we can drop the smaller term resulting in linearized boundary conditions ∂η ∂φ ∂t = ∂z , ∂φ ∂t + gη = 0 , which are applicable on z =η . version 3.0 updated 8/30/2005 -4- ©2005 A. Techet 2.016 Hydrodynamics Reading #7 Throughout this discussion we have assumed that the wave height is small compared to the wavelength. Along these lines we can also see why η is also quite small. If we expand φ (,,x zt) about z = 0 using Taylor series: ∂φ φ(xz, = η, t) = φ( x, 0, t) + η + ... , (7.15) ∂z ∂φ it can be readily shown that, the second term, ∂z η and the subsequent higher order terms are very small and can be ignored in our linear equations allowing us to rewrite the boundary conditions at z =η as boundary conditions on z = 0 . 0.2in 2 ∂ φ + g ∂φ = 0 ∂t2 ∂z (7.16) 1 ∂φ η =−g ∂t (7.17) on z = 0. A Solution to the Linear Wave Problem The complete boundary value wave problem consists of the differential equation, specifically Laplace’s Equation, ∂2φ ∂ 2φ ∇2φ (,,xzt ) = + = 0 (7.18) ∂x2 ∂z2 with the following boundary conditions: 1) Bottom Boundary Condition: ∂φ = 0 on z =−H (7.19) ∂z 2) Free Surface Dynamic Boundary Condition (FSDBC): 1 ∂φ η = ( ) on z =.0 (7.20) g ∂t version 3.0 updated 8/30/2005 -5- ©2005 A. Techet 2.016 Hydrodynamics Reading #7 3) Free Surface Kinematic Boundary Condition (FSKBC): ∂2φ ∂φ + g = 0 (7.21) ∂t 2 ∂z Through separation of variables we can solve Laplace’s Equation for η and φ . η ()xt, = a cos( kx− ωψt+ ) (7.22) aω φ (xz,,t ) =− f( z) sin( kx − ωψ t+ ) (7.23) k ux(,,z t ) =a ω f( z) cos( kx − ωψ t+ ) (7.24) wx(,, zt ) =−a ω f1( z) sin( kx − ωψ t+ ) (7.25) cosh[ kz( + H)] fz()= (7.26) sinh( kH ) sinh[ kz( + H)] fz()= (7.27) 1 sinh( kH ) ω 2 = gk tanh( kH ) ⇒ dispersion relation (7.28) where a, ω , k, ψ are integration constants with physical interpretation: a is the wave amplitude, ω the wave frequency, k is the wavenumber ( k = 2π/λ ), and ψ simply adds a phase shift. III. Dispersion Relation The dispersion relationship uniquely relates the wave frequency and wave number given the depth of the water. The chosen potential function, φ , MUST satisfy the free surface boundary conditions (equation (7.20) and (7.21)) such that plugging φ in to the FSKBC (eq. (7.21)) we get: −ω 2 cosh kH + gk sinh kH = 0 (7.29) Resulting in the Dispersion relationship: ω 2 = gk tanh kH ; (7.30) version 3.0 updated 8/30/2005 -6- ©2005 A. Techet 2.016 Hydrodynamics Reading #7 For deep water where H →∞ tanh kH →1 so that the dispersion relationship in deep water is: ω 2 = gk (7.31) In general, k ↑ as ω ↑ or λ ↑ as T ↑ . Phase and Group Speed Phase speed, C p , of a wave (velocity that a wave crest is traveling at) is found in general using: g λ = C =ω = tanh kH T p k k . (7.32) This simplifies for the case of deep water such that g C = (7.33) p k Solution to the dispersion relationship in general form can be found graphically. IV. Pressure under a Wave The pressure under a wave can be found using the unsteady form of Bernoulli’s equation and the wave potential, φ(,x zt ,) : ∂φ 1 p =− ρ − ρV 2 − ρgz (7.34) ∂t 2 hydrostatic unsteady 2nd Order pressure fluctuation term DynamicPressure Since we are only considering the case for LINEAR free surface waves we can neglect all higher order terms. Dropping the second order term in the dynamic pressure from equation (7.34), the pressure under a wave is simply ∂φ px( ,,z t) =− ρ (7.35) d ∂t version 3.0 updated 8/30/2005 -7- ©2005 A. Techet 2.016 Hydrodynamics Reading #7 aω 2 = ρ fz()cos( ωt − kx− ψ ) (7.36) k ω 2 = ρ f ()z η(x,t) , (7.37) k where ω 2 cosh[kz( + H )] cosh[k( z+ H )] fz( ) = g tanh( kH) = g . (7.38) k sinh( kH) cosh( kH) Therefore the dynamic pressure for all depths becomes cosh[ kz( + H )] pxd ( ,,z t) = ρηg (x,t) cosh(kH ) (7.39) V. Motion of Fluid Particles below a Wave We can define the motion of fluid particles underneath a linear progressive wave to have a horizontal motion,ζ p and a vertical motion, η p , such that ζ p (xz,,t ) = a f( z)sin(kx − ωt) +ψ ) (7.40) η p (xz,,t ) = a f1 ( z)cos(ω t −k − x +ψ ) (7.41) The orbital pattern is given by the equation for an ellipse: 2 2 ⎡ ζ p ⎤ ⎡ ζ p ⎤ 2 ⎢ ⎥ + ⎢ ⎥ = a (7.42) ⎣ fz()⎦ ⎣ f1()z⎦ where a is the amplitude of the waves.
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