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)