{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

week10 - 5 Waves and free surface ows Free surface ows are...

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
5 Waves and free surface flows Free surface flows are in some ways very different from what we have done before. In all problems considered so far, the domain D in which to solve the problem is given (for example some box or the exterior of an aeroplane wing). A free surface, on the other hand, moves, so the domain varies in time. The key feature, however, is that the domain does not vary according to some predetermined program. Instead, it moves in response to the flow itself, that is in response to the flow solution (which of course itself depends on the domain). This leads to solutions which are quite different in character to the fixed-boundary solu- tions we were considering so far (and generally speaking much more interesting solutions)! Another nice feature of free surface flows is that they are easy to observe experimentally, one just needs to track the motion of the free surface. 5.1 Kinematic boundary condition If a fluid particle is adjacent to a boundary then we must impose a condition which links the velocity of the boundary to that of the particle. This is known as the kinematic boundary condition . Let S ( x , t ) = 0 describe the equation of a surface in (or on the boundary of) the fluid. As the flow evolves, particles remain on the surface S if DS Dt = ∂S ∂t + u · ∇ S = 0 . (50) The proof is analogous to the arguments of (1.7). Namely, according to Taylor’s theorem: 0 = S ( x , t ) S ( x + u δt, t + δt ) = δt parenleftbigg u · ∇ S + ∂S ∂t parenrightbigg . Water Oil S=0 Example : Consider two fluids bounded by an interface S ( x , t ) = 0 (e.g. water/oil). Then DS Dt = ∂S ∂t + u ( oil ) · ∇ S = 0 , on S = 0 from above DS Dt = ∂S ∂t + u ( water ) · ∇ S = 0 , on S = 0 from below Now S is normal to the surface S = const , so n = S/ |∇ S | is the unit normal to the surface S = 0. By subtracting one of the two equations from the other, it follows that u ( oil ) · n = u ( water ) · n , on S = 0 . (51) This means the normal component of the velocity on either side of the interface must be equal, which is very intuitive.
Image of page 1

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

View Full Document Right Arrow Icon
5.2 Nonlinear free-surface motion We begin with the general equations for free-surface flow, assuming that the flow inside the fluid is potential, and the fluid is incompressible. We assume for now that the flow is 2D. z=-h z x ζ( x,t) = z z=0 Density = ρ Pressure = p atm Hence the fluid flow is described by u = φ, 2 φ = 0 , where u = ( u, 0 , w ) and φ = φ ( x, z, t ). We choose z = 0 to coincide with the undisturbed free surface, and the bottom of the fluid is at z = h . Let the surface of the water in motion be given by z = ζ ( x, t ). The interesting part of the problem are its boundary conditions: (i) At the lower boundary z = h , the vertical velocity must vanish (kineamtic b.c.): 0 = n · ∇ φ = ∂φ ∂z = w. In other words, ∂φ ∂z = 0 on z = ζ. (52) (ii) On z = ζ ( x, t ), the kinematic boundary condition on a moving surface is DS Dt = 0, where the free surface is the zero set of S ( x , t ). If the surface can be represented by a height function ζ ( x, t ) (i.e. we are not allowed overhangs) we can define a function S ( x, z, t ) = z ζ ( x, t ) which is indeed zero at the free surface. Thus
Image of page 2
Image of page 3
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