Past exam of the fluids course of the University of Oxford / 2021 / 4

For the simple shear flow, the vertical vorticity isFor $\beta \ne 0$, this is not an irrotational flow. A complex potential of a two-dimensional potential flow $w=\varphi +i\psi$, analytic in the flow region, has complex velocity $w^{\prime }=v_x-i{v}_y$. The Cauchy-Riemann equations imply both incompressibility and zero vorticity, so this shear has no such complex potential. It does have a Cartesian streamfunction, $\psi =\beta y^2/2$. In the degenerate case $\beta =0$, the resting fluid has a constant complex potential.

Using $w=\varphi +i\psi$ and the complex velocity convention $w^{\prime }=v_x-i{v}_y$,The streamlines are $xy=\psi /(2\beta )$, a family of rectangular hyperbolas whose asymptotes are the coordinate axes. The origin is a stagnation point. Fluid moves inward along the $y$ axis and outward along the $x$ axis, with the same directions approached by the neighbouring streamlines. This is a planar extensional incompressible flow, illustrated in the left panel below.

Taking the imaginary part of the complex potential gives the streamfunctionThe corresponding velocity field isPut $\zeta =z{e}^{i\alpha /2}=\xi +i\eta$. Then $w=\beta {\zeta }^2$ and $\psi =2\beta \xi \eta$. Thus the streamlines are the previous rectangular hyperbolas rotated clockwise by $\alpha /2$. Their separatrices have angles $\theta =-\alpha /2+n\pi /2$; the extension axis is $\theta =-\alpha /2$ and the compression axis is perpendicular to it. The middle panel of the figure shows this rotation.

For pure advection, a marked fluid element obeysConsequently the dye circle becomes an ellipse with semiaxes $R{e}^{2\beta t}$ and $R{e}^{-2\beta t}$. Its area isStretching in one principal direction exactly compensates compression in the other. Equivalently the flow-map Jacobian determinant is one, as required by incompressibility. Negligible diffusion is needed to regard the sharp dye boundary as a material curve.

The coefficient of the growing term is fixed by the far-field complex potential: $A=\beta e^{i\alpha }$. On $z=a{e}^{i\theta }$, the no-penetration boundary condition makes the cylinder a streamline, so the imaginary part of $A{a}^2{e}^{2i\theta }+B{a}^{-2}{e}^{-2i\theta }$ must be constant. Choosing the two terms to be complex conjugates makes that constant zero, givingThis is the zero-circulation solution, also obtained by the Milne-Thomson circle theorem. Impermeability and the leading far field alone permit an additional circulation term $-i\Gamma \log z/(2\pi )$; the requested algebraic form implicitly selects $\Gamma =0$. No no-slip boundary condition is imposed on an inviscid flow.

The streamfunction in polar coordinates isIt gives $v_r(a,\theta )=0$ andThe liquid still exerts normal pressure on each surface element even though tangential viscous traction is absent. For this steady flow with zero vorticity, the Bernoulli equation has one constant $C$ throughout the connected fluid region:There is no finite uniform far-field pressure to substitute for $C$, because this straining velocity field grows with radius. Taking $\mathbf{n}=(\cos \theta ,\sin \theta )$ outward from the cylinder, the fluid force per unit axial width isIndeed $p(\theta +\pi )=p(\theta )$ while $\mathbf{n}(\theta +\pi )=-\mathbf{n}(\theta )$: diametrically opposite elements cancel exactly. Zero resultant force does not mean zero local pressure or zero deformation tendency.

Use the Joukowski map with $|c|<a$, in particular its intended range $0\le c<a/2$. On the original circle,Define the exterior inversewhere the holomorphic square root is chosen so that $\sqrt{Z^2-4c^2}\sim Z$ at infinity. Then $f(Z)\sim Z$ is the branch mapping the exterior of the ellipse to $|z|>a$. Composition of the zero-circulation complex potential from part 4(c) with this inverse givesThe ellipse is a streamline because its inverse image is the circle, and $W(Z)\sim \beta e^{i\alpha }{Z}^2$ at infinity. An additive constant in the complex potential has no effect on the velocity.

The transformed complex velocity obeys the chain rule:Since $|c|<a$, the denominator is nonzero on and outside $|z|=a$. The four stagnation points therefore come fromTheir physical locations areThey all lie on the ellipse, and their positions are independent of $\beta >0$, which only sets the overall speed scale.

The printed restriction that $c$ is real and $c<a/2$ is insufficient literally: $c=-2a$ satisfies it but makes $dZ/dz=0$ at $z=2a$ in the exterior, so there is no exterior conformal map of the claimed type; the stated minor semiaxis would also be negative. The intended positive-parameter range, or more generally $|c|<a$, is needed for the above solution. The reference to a sphere in the source is a wording slip: the geometry is two-dimensional flow around the mapped cylinder's elliptical cross-section.