Past exam of the mathematics course of the University of Cambridge / 2020 / ii / Paper 3 / 38B (Fluid Dynamics II)

Choose $x$ tangential to the solid boundary and locally along the outer inviscid streamlines, and let $y$ be the inward normal coordinate. Write $u$ and $v$ for the corresponding tangential and normal velocity field components, $U(x)$ for the outer tangential velocity, and $\nu$ for the kinematic viscosity.

For a layer of streamwise scale $L$ and thickness $\delta \ll L$, incompressible flow givesso $u=O(U)$ and $v=O(U\delta /L)$. The normal Navier-Stokes equation then gives $p_y=0$ to leading order, while the outer Euler equations giveIn the tangential Navier-Stokes equation, streamwise viscous diffusion is smaller than normal diffusion by $O((\delta /L{)}^2)$. Retaining the leading inertial and normal-diffusion terms produces the Prandtl boundary-layer equationIt is supplemented by $u=v=0$ at a stationary wall and $u\to U(x)$ on matching to the outer flow.

Conservation of volume flux through a circular arc of radius $r$ givesso the outer flow is the sink flow in a sectorThis is a potential flow and already satisfies the no-penetration boundary condition on the radial walls. It has nonzero tangential speed $Q/r$ there, however, and therefore violates the no-slip boundary condition. Thin vorticity-bearing boundary layers are needed to reduce that speed to zero at each wall.

Along $y=0$, put $x=r$ and $U(x)=-Q/x$. If $\delta (x)$ is the layer thickness, thenThe boundary-layer thickness from an advection-diffusion balance therefore givesand henceup to a constant factor absorbed into the similarity coordinate.

LetUsing the Cartesian streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$ givesAt fixed $y$,The terms proportional to $\eta f^{\prime }{f}^{\prime \prime }$ cancel, leavingSince $U=-Q/x$, $U{U}^{\prime }=-Q^2/x^3$, while $\nu u_{yy}=Q^2{f}^{\prime \prime \prime }/x^3$. The boundary-layer equation therefore becomesChoosing the wall value of the streamfunction as zero, imposing no slip, and matching to $U=-Q/x$ give

SetWriting $C=\cosh z$, direct differentiation givesOn the other hand,Thus $F^{\prime \prime }=1-F^2$, which is precisely $f^{\prime \prime \prime }=1-(f^{\prime }{)}^2$. The far-field condition is automatic because $F\to -1$. No slip requiressoFor $c=+\mathrm{arcosh}5$, $F$ decreases directly from $0$ to $-1$, matching flow everywhere toward the sink. For the negative root, $F$ initially becomes positive and creates a reverse-flow region next to the wall. The physically likely choice is therefore