Paper 4 , Section II, B

Fluid Dynamics II | Part II, 2016

A thin layer of fluid of viscosity μ\mu occupies the gap between a rigid flat plate at y=0y=0 and a flexible no-slip boundary at y=h(x,t)y=h(x, t). The flat plate moves with constant velocity UexU \mathbf{e}_{x} and the flexible boundary moves with no component of velocity in the xx-direction.

State the two-dimensional lubrication equations governing the dynamics of the thin layer of fluid. Given a pressure gradient dp/dx\mathrm{d} p / \mathrm{d} x, solve for the velocity profile u(x,y,t)u(x, y, t) in the fluid and calculate the flux q(x,t)q(x, t). Deduce that the pressure gradient satisfies

∂∂x(h312μdp dx)=∂h∂t+U2∂h∂x\frac{\partial}{\partial x}\left(\frac{h^{3}}{12 \mu} \frac{\mathrm{d} p}{\mathrm{~d} x}\right)=\frac{\partial h}{\partial t}+\frac{U}{2} \frac{\partial h}{\partial x}

The shape of the flexible boundary is a periodic travelling wave, i.e. h(x,t)=h(x, t)= h(x−ct)h(x-c t) and h(ξ+L)=h(ξ)h(\xi+L)=h(\xi), where cc and LL are constants. There is no applied average pressure gradient, so the pressure is also periodic with p(ξ+L)=p(ξ)p(\xi+L)=p(\xi). Show that

dp dx=6μ(U−2c)(1h2−⟨h−2⟩⟨h−3⟩1h3)\frac{\mathrm{d} p}{\mathrm{~d} x}=6 \mu(U-2 c)\left(\frac{1}{h^{2}}-\frac{\left\langle h^{-2}\right\rangle}{\left\langle h^{-3}\right\rangle} \frac{1}{h^{3}}\right)

where ⟨…⟩=1L∫0L…dx\langle\ldots\rangle=\frac{1}{L} \int_{0}^{L} \ldots \mathrm{d} x denotes the average over a period. Calculate the shear stress σxy\sigma_{x y} on the plate.

The speed UU is such that there is no need to apply an external tangential force to the plate in order to maintain its motion. Show that

U=6c⟨h−2⟩⟨h−2⟩−⟨h−1⟩⟨h−3⟩3⟨h−2⟩⟨h−2⟩−4⟨h−1⟩⟨h−3⟩.U=6 c \frac{\left\langle h^{-2}\right\rangle\left\langle h^{-2}\right\rangle-\left\langle h^{-1}\right\rangle\left\langle h^{-3}\right\rangle}{3\left\langle h^{-2}\right\rangle\left\langle h^{-2}\right\rangle-4\left\langle h^{-1}\right\rangle\left\langle h^{-3}\right\rangle} .

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