week10 - 5 Waves and free surface flows Free surface flows...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 Taylors theorem: 0 = S ( x ,t ) S ( x + u t,t + t ) = t 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. 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 ( x,t) = z z=0 Density = Pressure = p atm Hence the fluid flow is described by u = , 2 = 0 , where u = ( u, ,w ) and = ( 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 = ( 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 = z = w....
View Full Document

This note was uploaded on 01/20/2012 for the course MATH 33200 taught by Professor Eggers during the Fall '11 term at University of Bristol.

Page1 / 10

week10 - 5 Waves and free surface flows Free surface flows...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online