B1.25

Fluid Dynamics II
Part II, 2002

State the minimum dissipation theorem for Stokes flow in a bounded domain.

Fluid of density ρ\rho and viscosity μ\mu fills an infinite cylindrical annulus arba \leq r \leq b between a fixed cylinder r=ar=a and a cylinder r=br=b which rotates about its axis with constant angular velocity Ω\Omega. In cylindrical polar coordinates (r,θ,z)(r, \theta, z), the fluid velocity is u=(0,v(r),0)\mathbf{u}=(0, v(r), 0). The Reynolds number ρΩb2/μ\rho \Omega b^{2} / \mu is not necessarily small. Show that v(r)=Ar+B/rv(r)=A r+B / r, where AA and BB are constants to be determined.

[You may assume that 2u=(0,2vv/r2,0)\nabla^{2} \mathbf{u}=\left(0, \nabla^{2} v-v / r^{2}, 0\right) and (u)u=(v2/r,0,0).(\mathbf{u} \cdot \nabla) \mathbf{u}=\left(-v^{2} / r, 0,0\right) . ]

Show that the outer cylinder exerts a couple G0G_{0} per unit length on the fluid, where

G0=4πμΩa2b2b2a2.G_{0}=\frac{4 \pi \mu \Omega a^{2} b^{2}}{b^{2}-a^{2}} .

[You may assume that, in standard notation, erθ=r2ddr(vr)e_{r \theta}=\frac{r}{2} \frac{d}{d r}\left(\frac{v}{r}\right).]

Suppose now that b2ab \geq \sqrt{2} a and that the cylinder r=ar=a is replaced by a fixed cylinder whose cross-section is a square of side 2a2 a centred on r=0r=0, all other conditions being unchanged. The flow may still be assumed steady. Explaining your argument carefully, show that the couple GG now required to maintain the motion of the outer cylinder is greater than G0G_{0}.