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solnsheet10 - Fluid Dynamics 3 Solutions to Sheet 10(ii...

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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, 0 , w ) = φ satisfying 2 φ = 0. Conditions are that fluid velocities must tend to zero at large depths so that u = φ 0 , 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 + = 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 0 , z → −∞ remains with ∂ζ ∂t = ∂φ ∂z and ∂φ ∂t + = 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 = (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 0 + X ( t ) , z 0 + 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 0 + Z ) sin( k ( x 0 + X ) ωt ) 1
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and d Z d t = ∂φ ∂z = kg ω H e k ( z 0 + Z ) cos( k ( x 0 + X ) ωt ) But | X/x 0 | ≪ 1, | Z/z 0 | ≪ 1 and gk = ω 2 so write d X d t ωH e kz 0 sin( kx 0 ωt ) d Z d t ≈ − ωH e kz 0 cos( kx 0 ωt ) Now we can integrate these equations up to get X ( t ) = ωH e kz 0 cos( kx 0 ωt ) , Z ( t ) = ωH e kz 0 sin( kx 0 ωt ) , and now clearly, X 2 ( t ) + Z 2 ( t ) = H 2 e 2 kz 0 This represents circular motion with a radius which decays exponentially with depth. On z 0 = 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 and then move back, as well as the obvious upward and downward movement in the wave). The actual path of a particle, found by not making any “small circle” assumptions is a slowly drifting circle.
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