Past exam of the fluids course of the University of Oxford / 2022 / 2

At a fixed point beside a river, the Eulerian coordinate description measures $\mathbf{u}(\mathbf{x},t)$. Its local acceleration ${\partial }_t\mathbf{u}$ measures change at that fixed point. Following a floating parcel gives a Lagrangian trajectory $\mathbf{X}(t)$ satisfying $\dot{\mathbf{X}}=\mathbf{u}(\mathbf{X},t)$. The chain rule gives its physical material acceleration:The extra term is the convective acceleration. For a river that narrows steadily, an observer at one fixed position measures a constant velocity, yet a parcel speeds up as it reaches the narrowing. A parcel following a steady bend also experiences a centripetal acceleration even when its speed is constant. An Eulerian calculation of the physical acceleration must include the same convective acceleration; the two descriptions do not predict different forces.

Choose the stream function convention $u=-{\psi }_y$, $v={\psi }_x$, so that this is an incoming stagnation point flow rather than the reversed flow. The velocity field isAt the plate $v(x,0)=0$, satisfying the no-penetration boundary condition. The nonzero tangential velocity $u(x,0)=Ax$ is permitted for an inviscid flow; the no-slip boundary condition is not imposed on the Euler equations for an inviscid fluid.

The steady convective acceleration is $(A^{2}x,A^{2}y)$. Thus the Euler equations for an inviscid fluid requireIntegrating givesThe printed formula omits $\rho$ from its quadratic term. The source uses physical pressure, not pressure divided by mass density, as is clear from its $\rho gy$ term and its Navier-Stokes equation. The omitted factor is also required by dimensional analysis. For a concrete substitution check, the printed expression gives $-p_x/\rho =A^{2}x/\rho$ rather than the required $A^{2}x$ when $\rho \ne 1$ in the chosen numerical units.

A stationary viscous wall imposes the no-slip boundary condition, $u=v=0$, whereas the outer inviscid flow has $u=Ax$. At large Reynolds number, the transition is confined to a thin viscous boundary layer, where normal velocity gradients amplify the otherwise small viscous term.

Take a horizontal scale $L$, an outer speed $U=AL$, and a boundary layer thickness $\delta \ll L$. The continuity equation gives $v=O(U\delta /L)$. Write $ϵ=\delta /L$ and ${\mathrm{Re}}_L=UL/\nu$. The tangential Navier-Stokes equation has scalesBalancing convective acceleration with normal viscous diffusion gives $ϵ=O({\mathrm{Re}}_L^{-1/2})$. The leading tangential equation and incompressible flow constraint areHere and below $\nu$ is the kinematic viscosity, while $u$ is the horizontal velocity.

For the normal equation, first remove the hydrostatic pressure by writing $P=p+\rho gy$. Normal inertia and normal viscous terms are $O(\rho U^2\delta /L^2)$, so the change of $P$ across the layer is only $O(\rho U^2{ϵ}^2)$. Matching to the outer pressure therefore makes the leading $O(\rho U^2)$ excess pressure independent of $y$:The outer expression similarly has $(p_{\infty }{)}_y=-\rho g-\rho A^{2}y$. Thus the printed conditionis a leading-order boundary layer matching statement, with the smaller normal dynamic terms discarded on both sides. It is not an exact identity at the next order. In particular, retaining the outer $-\rho A^{2}y$ while dropping normal inertia and normal viscous terms inconsistently mixes orders. The tangential driving term is unambiguous: $-(p_{\infty }{)}_x/\rho =A^{2}x$.

Use the same stream function convention and denote differentiation with respect to the dimensionless variable $\eta =y/\delta$ by a prime. The proposed similarity solution givesIts continuity equation holds identically. Substitution in the tangential Prandtl boundary-layer equation givesChoose the constant scale $\delta =\sqrt{\nu /A}$. The Hiemenz flow equation and its boundary conditions areThe first two conditions impose the no-penetration boundary condition and no-slip boundary condition. The last matches the tangential outer velocity. One also has $f^{\prime \prime }\to 0$ and $f(\eta )/\eta \to 1$. It is too strong to impose $f-\eta \to 0$: the displacement thickness generally permits a constant offset, $f\sim \eta -B$.

The normal Navier-Stokes equation makes the approximation in part (c) explicit. With this similarity solution, it givesAt the wall this differs from $(p_{\infty }{)}_y=-\rho g$ by $-\rho A^2\delta f^{\prime \prime }(0)$. This is a concrete next-order counterexample to interpreting the printed normal pressure gradient equality exactly; the tangential Hiemenz flow equation remains valid at its stated order.

The Hiemenz flow profile changes on $\eta =O(1)$, so its boundary layer thickness isA specified edge convention, such as $u/(Ax)=0.99$, introduces an order-one numerical multiplier requiring the profile to be solved. Scaling alone does not determine that multiplier. The result has no separate dependence on $x$, $y$, $\rho$ or $p_0$ at fixed kinematic viscosity. At fixed dynamic viscosity, it is $\sqrt{\eta /(\rho A)}$.

The viscous diffusion time across a thickness $\delta$ is ${\delta }^2/\nu$; the imposed strain has time scale $A^{-1}$. Their balance sets the thickness. Increasing kinematic viscosity spreads the layer, while increasing the imposed velocity gradient presses the layer toward the wall. The outward tangential velocity increases as $Ax$, so the decreasing travel time per unit distance offsets the usual downstream broadening.

For a conventional estimate of transition to turbulence, form the local horizontal Reynolds number:If transition is assumed to occur at an empirical value ${\mathrm{Re}}_{\mathrm{crit}}$, thenA rough borrowed threshold of order $10^5$–$10^6$ would give $l$ of order $10^2$–$10^3\sqrt{\nu /A}$. This is a heuristic estimate, not a universal stability prediction: the accelerating outer flow and favorable pressure gradient can delay transition, and disturbance amplitudes and wall roughness matter. Simply setting ${\mathrm{Re}}_x\sim 1$ locates failure of the thin-layer approximation near the stagnation point, not onset of turbulence.