Past exam of the flows, fluctuations and complexity and symmetry and relativity paper of Part B of the physics course of the University of Oxford / 2011 / 1

Set ${\partial }_t\mathbf{u}=0$ and $\nu =0$ in the Navier-Stokes equation. Dotting the resulting Euler equation with a displacement $d\mathbf{r}$ tangent to a streamline givesHere mass density $\rho$ is constant, and $B$ is constant along each streamline. The terms are kinetic energy per unit mass, the pressure work per unit mass available from pushing a parcel through the flow, and gravitational potential energy per unit mass. Their sum is conserved mechanical energy per unit mass; $p/\rho$ is not an additional independently conserved parcel energy. Different streamlines can have different $B$ unless, for example, the flow is also irrotational in a connected region.

For planar incompressible flow, $u_x+w_z=0$. Locally this permits a streamfunction with $u={\psi }_z$, $w=-{\psi }_x$, because equality of the mixed derivatives enforces incompressibility. The remaining component of the curl is $u_z-w_x={\psi }_{zz}+{\psi }_{xx}$, so irrotational flow requires Laplace's equation for $\psi$.

Put $K=\pi /b$. The sinusoidal contribution to the supplied streamfunction has ${\psi }_{xx}=U_{0}a{K}^2\cos (Kx)e^{-Kz}$ and ${\psi }_{zz}=-U_{0}a{K}^2\cos (Kx)e^{-Kz}$; the linear term is also harmonic. HenceAs $z\to \infty$, these approach the specified uniform flow, $(u,w)=(U_0,0)$.

The impermeability kinematic boundary condition is $w=u{z}_s^{\prime }(x)$ on $z=z_s(x)$, equivalently $\psi$ is constant on the wall. For $aK\ll 1$, evaluating the perturbed velocities at $z=0$ gives $w=-U_{0}aK\sin (Kx)=U_0{z}_s^{\prime }$ to first order. The residual is of order $U_0(aK{)}^2$. Equivalently $\psi (x,z_s)=U_{0}a\cos (Kx)[1-e^{-Ka\cos (Kx)}]=O(U_{0}K{a}^2)$. The surface velocity isThe first expression is the supplied approximate field evaluated exactly at the wall; its accuracy as a wall-flow solution is still only first order in $a/b$.

Use the rest frame of the hydrofoil, with undisturbed velocity $U_0{\mathbf{e}}_x$. Let $p_{\ast }$ be the undisturbed pressure at the lower surface level $z=0$, so the undisturbed hydrostatic pressure is $p_{\ast }-\rho gz$. The Bernoulli equation gives, with $h=a\cos (\pi x/b)$ and $ϵ=\pi a/b$,These formulas implement the prescribed neglect of the other velocity components. To the controlled first order in $a/b$, $p_{\mathrm{lower}}-p_{\mathrm{upper}}=\rho gh+\rho U_0^2ϵ\cos (\pi x/b)$. The vertical component of pressure acting on an inclined surface is integrated over its horizontal projection, soThe total upward fluid force per unit span is $L_{\mathrm{dynamic}}^{\prime }+B^{\prime }$. The lift due to motion is $L_{\mathrm{dynamic}}^{\prime }$; $B^{\prime }$ is the separate buoyancy. If the given approximate speed is squared without truncation, the formal additional dynamic term is $\rho U_0^2{\pi }^2{a}^2/(4b)$, but a controlled second-order prediction would require the omitted second-order flow and transverse-velocity contributions too.

Take positive circulation to mean counterclockwise in the $(x,z)$ diagram, with $x$ right and $z$ up. The upper surface is then traversed right to left. To first order in $a/b$,Thus the bound circulation is clockwise. Any exterior contour enclosing the hydrofoil has the same circulation in an irrotational flow; a globally completed two-dimensional flow must contain a circulation component decaying as $|\Gamma |/(2\pi r)$ at large distance. The simple local upper/lower velocity prescription cannot describe the entire exterior. For a start from rest, Kelvin's circulation theorem requires compensating shed vorticity, represented by a starting vortex in a physical wake.

Numerically,If the question's fluid force includes buoyancy, $B^{\prime }\simeq 1.25\times 10^3\mathrm{N}{\mathrm{m}}^{-1}$ and the total upward fluid force is about $2.05\times 10^3\mathrm{N}{\mathrm{m}}^{-1}$, using $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$. Weight of the hydrofoil is not specified and is not part of this fluid force. The motion-induced lift also agrees with the Kutta–Joukowski theorem, $L^{\prime }=-\rho U_0\Gamma$ in this orientation.

The chord-based Reynolds number isAt this large Reynolds number, smooth laminar flow over the whole body is unlikely in ordinary conditions; wall layers may undergo transition and separate. A thin-layer inviscid flow approximation outside them can remain useful, but does not justify globally laminar flow.