A4.19

Transport Processes | Part II, 2003

A shallow layer of fluid of viscosity μ\mu, density ρ\rho and depth h(x,t)h(x, t) lies on a rigid horizontal plane y=0y=0 and is bounded by impermeable barriers at x=Lx=-L and x=Lx=L (Lh)(L \gg h). Gravity acts vertically and a wind above the layer causes a shear stress τ(x)\tau(x) to be exerted on the upper surface in the +x+x direction. Surface tension is negligible compared to gravity.

(a) Assuming that the steady flow in the layer can be analysed using lubrication theory, show that the horizontal pressure gradient pxp_{x} is given by px=ρghxp_{x}=\rho g h_{x} and hence that

hhx=32τρg.h h_{x}=\frac{3}{2} \frac{\tau}{\rho g} .

Show also that the fluid velocity at the surface y=hy=h is equal to τh/4μ\tau h / 4 \mu, and sketch the velocity profile for 0yh0 \leqslant y \leqslant h.

(b) In the case in which τ\tau is a constant, τ0\tau_{0}, and assuming that the difference between hh and its average value h0h_{0} remains small compared with h0h_{0}, show that

hh0(1+3τ0x2ρgh02)h \approx h_{0}\left(1+\frac{3 \tau_{0} x}{2 \rho g h_{0}^{2}}\right)

provided that

τ0Lρgh021\frac{\tau_{0} L}{\rho g h_{0}^{2}} \ll 1

(c) Surfactant at surface concentration Γ(x)\Gamma(x) is added to the surface, so that now

τ=τ0AΓx,\tau=\tau_{0}-A \Gamma_{x},

where AA is a positive constant. The surfactant is advected by the surface fluid velocity and also experiences a surface diffusion with diffusivity DD. Write down the equation for conservation of surfactant, and hence show that

(τ0AΓx)hΓ=4μDΓx\left(\tau_{0}-A \Gamma_{x}\right) h \Gamma=4 \mu D \Gamma_{x}

From equations (1), (2) and (3) deduce that

ΓΓ0=exp[ρg18μD(h3h03)]\frac{\Gamma}{\Gamma_{0}}=\exp \left[\frac{\rho g}{18 \mu D}\left(h^{3}-h_{0}^{3}\right)\right]

where Γ0\Gamma_{0} is a constant. Assuming once more that h1hh0h0h_{1} \equiv h-h_{0} \ll h_{0}, and that h=h0h=h_{0} at x=0x=0, show further that

h13τ0x2ρgh0[1+AΓ0h04μD]1h_{1} \approx \frac{3 \tau_{0} x}{2 \rho g h_{0}}\left[1+\frac{A \Gamma_{0} h_{0}}{4 \mu D}\right]^{-1}

provided that

τ0h0LμD1 as well as τ0Lρgh021\frac{\tau_{0} h_{0} L}{\mu D} \ll 1 \quad \text { as well as } \quad \frac{\tau_{0} L}{\rho g h_{0}^{2}} \ll 1

Typos? Please submit corrections to this page on GitHub.