Paper 3, Section II, E

Fluid Dynamics II | Part II, 2009

An axisymmetric incompressible Stokes flow has the Stokes stream function Ψ(R,θ)\Psi(R, \theta) in spherical polar coordinates (R,θ,ϕ)(R, \theta, \phi). Give expressions for the components uRu_{R} and uθu_{\theta} of the flow field in terms of Ψ\Psi, and show that

×u=(0,0,D2ΨRsinθ)\nabla \times \mathbf{u}=\left(0,0,-\frac{D^{2} \Psi}{R \sin \theta}\right)

where

D2Ψ=2ΨR2+sinθR2θ(1sinθΨθ)D^{2} \Psi=\frac{\partial^{2} \Psi}{\partial R^{2}}+\frac{\sin \theta}{R^{2}} \frac{\partial}{\partial \theta}\left(\frac{1}{\sin \theta} \frac{\partial \Psi}{\partial \theta}\right)

Write down the equation satisfied by Ψ\Psi.

Verify that the Stokes stream function

Ψ(R,θ)=12Usin2θ(R232aR+12a3R)\Psi(R, \theta)=\frac{1}{2} U \sin ^{2} \theta\left(R^{2}-\frac{3}{2} a R+\frac{1}{2} \frac{a^{3}}{R}\right)

represents the Stokes flow past a stationary sphere of radius aa, when the fluid at large distance from the sphere moves at speed UU along the axis of symmetry.

A sphere of radius a moves vertically upwards in the zz direction at speed UU through fluid of density ρ\rho and dynamic viscosity μ\mu, towards a free surface at z=0z=0. Its distance dd from the surface is much greater than aa. Assuming that the surface remains flat, show that the conditions of zero vertical velocity and zero tangential stress at z=0z=0 can be approximately met for large d/ad / a by combining the Stokes flow for the sphere with that of an image sphere of the same radius located symmetrically above the free surface. Hence determine the leading-order behaviour of the horizontal flow on the free surface as a function of rr, the horizontal distance from the sphere's centre line.

What is the size of the next correction to your answer as a power of a/d?a / d ? [Detailed calculation is not required.]

[Hint: For an axisymmetric vector field u\mathbf{u},

×u=(1Rsinθθ(uϕsinθ),1RR(Ruϕ),1RR(Ruθ)1RuRθ)\nabla \times \mathbf{u}=\left(\frac{1}{R \sin \theta} \frac{\partial}{\partial \theta}\left(u_{\phi} \sin \theta\right),-\frac{1}{R} \frac{\partial}{\partial R}\left(R u_{\phi}\right), \frac{1}{R} \frac{\partial}{\partial R}\left(R u_{\theta}\right)-\frac{1}{R} \frac{\partial u_{R}}{\partial \theta}\right)

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