2005reading7 - 2.016 Hydrodynamics Reading#7 2.016...

Info icon This preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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: G Velocity Field: V ( x , y , z , t ) = ∇ φ ( x , y , z , t ) ( x y t ) Free surface elevation: η , , Pressure field: p ( x y z t ) , , , version 3.0 updated 8/30/2005 -1- © 2005 A. Techet
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
2.016 Hydrodynamics Reading #7 Next we need to set up the equations and conditions that govern the problem: Continuity (Conservation of Mass): 2 φ = 0 for z < η (Laplace’s Equation) (7.1) Bernoulli’s Equation (given some φ ): p p a φ 1 + φ 2 + ρ + gz = 0 for z < η (7.2) t 2 No disturbance far away: φ φ , 0 and p = p ρ gz (7.3) t a 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 = p on z = η . atm ⎧∂ φ ( ) = p atm . (7.4) p = − ρ t 1 2 V 2 gz + c t ( Choosing a suitable integration constant, c t ) = p , the boundary condition on atm z = η becomes ρ { φ 1 V 2 + + g η } = 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 = η ( x p , t ) p η η z + δ z = η ( x p + δ x , t + δ t ) = η ( x p , t ) + δ x + δ t (7.6) p p p p x t On the surface, where z p = η , we can reduce the above equation to ∂η ∂η δ z = u δ t + δ t (7.7) p x t and substitute δ z = w δ t and δ x = u t to show that the normal velocity follows δ p p the particle: version 3.0 updated 8/30/2005 -2- © 2005 A. Techet
Image of page 2
2.016 Hydrodynamics Reading #7 η η w u + on z = η . (7.8) = x t ( , , , (3) On an impervious body boundary B x y z t ) = 0 . Velocity of the fluid normal to the body must be equal to the body velocity in that direction: G φ JG G G v n ˆ ( ) ˆ( ) = U on B = 0 . (7.9) = φ n ˆ = = U x t n x t 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 t t such that = o G G , ( , B ( x t ) = 0 , then B x t ) = 0 for all t , (7.10) o o so that if we were to follow P then B = 0 always. Therefore: DB B ( = + φ ⋅∇ ) B = 0 on B = 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
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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.
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern