solnsheet10 - Fluid Dynamics 3 - Solutions to Sheet 10 1....

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Unformatted text preview: Fluid Dynamics 3 - Solutions to Sheet 10 1. This is just like lectures, but with infinite depth. All this means is a different choice of separable solution in the vertical dependence – replace cosh k ( z + h ) for a fluid with a bottom at z = − h with e kz which decays exponentially to zero as z → −∞ . (i) We have a potential φ ( x,z,t ) with u = ( u, ,w ) = ∇ φ satisfying ∇ 2 φ = 0. Conditions are that fluid velocities must tend to zero at large depths so that u = ∇ φ → , z → −∞ a point on the free surface stays there, D Dt ( z − ζ ( x,t )) = 0 , z = ζ and pressure on the surface of the fluid is at- mospheric p atm ρ + 1 2 |∇ φ | 2 + ∂φ ∂t + gζ = C ( t ) ρ , z = ζ Linearise this: Taylor expand all terms depen- dent on z about z = 0 in powers of ζ and throw away any terms which are quadratic or higher powered in either ζ or φ (because they are assumed to be small). Then ∇ φ → , z → −∞ remains with ∂ζ ∂t = ∂φ ∂z and ∂φ ∂t + gζ = C ( t ) − p atm ρ Terms on RHS only depend on t , so absorb these terms into the definition of φ because they don’t affect the velocity field. So this last eqn is ∂φ ∂t = − gζ (ii) Let ζ ( x,t ) = H sin( kx − ωt ) and adopt a similar separable form for φ ( x,z,t ) = Z ( z )cos( kx − ωt ) which fits with the boundary conditions on z = 0. Stuff this into Laplace’s and separate to give Z ′′ ( z ) − k 2 Z ( z ) = 0 and solve for Z ( z ). Could have Z ( z ) = A e kz + B e − kz but the second term is out because it is un- bounded as z → −∞ . So set B = 0. So φ = A e kz cos( kx − ωt ) Put it into the conditions on z = 0 (dynamic): gH sin( kx − ωt ) = − ωA sin( kx − ωt ) and so A = − gH/ω . Then from the Kin. B.C: − ωH cos( kx − ωt ) = kA cos( kx − ωt ) and then it almost immediately follows that ω 2 = gk (iii) Now we have ζ ( x,t ) = H sin( kx − ωt ) φ ( x,z,t ) = − gH ω cos( kx − ωt ) Now we consider a point in the fluid with po- sition vector x = ( x + X ( t ) ,z + Z ( t )), where X (0) = 0, Z (0) = 0. Now u ( x ,t ) = d x d t = ( ∇ φ ) | x Taking each component in turn, d X d t = ∂φ ∂x = kg ω H e k ( z + Z ) sin( k ( x + X ) − ωt ) 1 and d Z d t = ∂φ ∂z = − kg ω H e k ( z + Z ) cos( k ( x + X ) − ωt ) But | X/x | ≪ 1, | Z/z | ≪ 1 and gk = ω 2 so write d X d t ≈ ωH e kz sin( kx − ωt ) d Z d t ≈ − ωH e kz cos( kx − ωt ) Now we can integrate these equations up to get X ( t ) = ωH e kz cos( kx − ωt ) , Z ( t ) = ωH e kz sin( kx − ωt ) , and now clearly, X 2 ( t ) + Z 2 ( t ) = H 2 e 2 kz This represents circular motion with a radius which decays exponentially with depth. On z = 0, the radius is H , which is the height of the wave above the mean level. So if you put a marker on the surface of a wave, and monitored its position in time, you’d approximately sketch out a circle (you may have experienced this if swim in the sea – that when a wave passes you first move forward...
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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.

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solnsheet10 - Fluid Dynamics 3 - Solutions to Sheet 10 1....

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