2.II.36A

Fluid Dynamics II | Part II, 2008

Viscous fluid with dynamic viscosity μ\mu flows with velocity (ux,uy,uz)(uH,uz)\left(u_{x}, u_{y}, u_{z}\right) \equiv\left(\mathbf{u}_{H}, u_{z}\right) (in cartesian coordinates x,y,z)x, y, z) in a shallow container with a free surface at z=0z=0. The base of the container is rigid, and is at z=h(x,y)z=-h(x, y). A horizontal stress S(x,y)\mathbf{S}(x, y) is applied at the free surface. Gravity may be neglected.

Using lubrication theory (conditions for the validity of which should be clearly stated), show that the horizontal volume flux q(x,y)h0uHdz\mathbf{q}(x, y) \equiv \int_{-h}^{0} \mathbf{u}_{H} d z satisfies the equations

q=0,μq=13h3p+12h2S\nabla \cdot \mathbf{q}=0, \quad \mu \mathbf{q}=-\frac{1}{3} h^{3} \nabla p+\frac{1}{2} h^{2} \mathbf{S}

where p(x,y)p(x, y) is the pressure. Find also an expression for the surface velocity u0(x,y)\mathbf{u}_{0}(x, y) \equiv uH(x,y,0)\mathbf{u}_{H}(x, y, 0) in terms of S,q\mathbf{S}, \mathbf{q} and hh.

Now suppose that the container is cylindrical with boundary at x2+y2=a2x^{2}+y^{2}=a^{2}, where aha \gg h, and that the surface stress is uniform and in the xx-direction, so S=(S0,0)\mathbf{S}=\left(S_{0}, 0\right) with S0S_{0} constant. It can be assumed that the correct boundary condition to apply at x2+y2=a2x^{2}+y^{2}=a^{2} is qn=0\mathbf{q} \cdot \mathbf{n}=0, where n\mathbf{n} is the unit normal.

Write q=ψ(x,y)×z^\mathbf{q}=\nabla \psi(x, y) \times \hat{\mathbf{z}}, and show that ψ\psi satisfies the equation

(1h3ψ)=S02μh2hy\nabla \cdot\left(\frac{1}{h^{3}} \nabla \psi\right)=-\frac{S_{0}}{2 \mu h^{2}} \frac{\partial h}{\partial y}

Deduce that if h=h0h=h_{0} (constant) then q=0\mathbf{q}=\mathbf{0}. Find u0\mathbf{u}_{\mathbf{0}} in this case.

Now suppose that h=h0(1+ϵy/a)h=h_{0}(1+\epsilon y / a), where ϵ1\epsilon \ll 1. Verify that to leading order in ϵ,ψ=ϵC(x2+y2a2)\epsilon, \psi=\epsilon C\left(x^{2}+y^{2}-a^{2}\right) for some constant CC to be determined. Hence determine u0\mathbf{u}_{0} up to and including terms of order ϵ\epsilon.

[Hint: ×(A×z^)=z^Az^A\nabla \times(\mathbf{A} \times \hat{\mathbf{z}})=\hat{\mathbf{z}} \cdot \nabla \mathbf{A}-\hat{\mathbf{z}} \nabla \cdot \mathbf{A} for any vector field A.]\mathbf{A} .]

Typos? Please submit corrections to this page on GitHub.