Past exam of the fluids and waves course of the University of Oxford / 2014 / 1

Choose the planar streamfunction convention $u={\psi }_y$, $v=-{\psi }_x$. For an irrotational flow, a velocity potential gives $\mathbf{u}=\nabla \varphi$. ThusThe streamfunction enforces incompressibility, since $u_x+v_y={\psi }_{yx}-{\psi }_{xy}=0$. Conversely, incompressibility permits a local streamfunction, while zero curl permits a local velocity potential. These are the Cauchy-Riemann equations for the complex potential of a two-dimensional potential flow $w=\varphi +i\psi$.

The gradients of the two functions satisfyEach gradient is normal to its regular level curves. Therefore regular curves of constant velocity potential and constant streamfunction meet orthogonally. At a stagnation point, both gradients vanish and this tangent-direction argument needs a limiting interpretation. The conclusion concerns regular portions of the two families of curves.

Use instantaneous polar coordinates centered on the moving circular cylinder, but measure the fluid velocity in the laboratory frame. Impermeability equates the normal components of body and fluid velocity:Since ${\mathbf{e}}_x\cdot {\mathbf{e}}_r=\cos \theta$ and $u_r={\varphi }_r$, the no-penetration boundary condition isThe velocity potential obeys the exterior Laplace equation, by incompressibility. The boundary data have only the first cosine harmonic. A decaying harmonic potential of this form is $\varphi =C\cos \theta /r$. Matching its radial derivative gives $-C/a^2=U$, so $C=-U{a}^2$. The zero circulation condition rules out an angular potential term, and the integrated boundary condition gives zero net volume flux, ruling out a radial logarithmic source term. The remaining decaying harmonics have zero coefficients by the boundary data. HenceUsing the streamfunction in polar coordinates, $u_r=r^{-1}{\psi }_{\theta }$, $u_{\theta }=-{\psi }_r$, givesOne may also obtain both functions as the real and imaginary parts of $w(z)=-U{a}^2/z$, a two-dimensional potential doublet. The velocity decays as $r^{-2}$ and the source of this singular representation lies inside the excluded body. These are the fields used to calculate the added mass of a circular cylinder.

The product rule, the Laplace equation and $\mathbf{u}=\nabla \varphi$ giveFirst truncate the exterior fluid domain by a large circle $C_R$. Let ${\mathbf{n}}_B$ point outward from the body into the fluid. The outward normal of the exterior fluid domain at the body is $-{\mathbf{n}}_B$. The planar divergence theorem, equivalently Green's theorem, therefore givesOrient $\partial B$ counterclockwise around the body. Then ${\mathbf{n}}_{B}ds=(dy,-dx)$ andFor the finite-energy exterior rigid-body flow intended here, the outer boundary term vanishes as $R\to \infty$. In particular, zero net volume flux and zero circulation eliminate the $1/z$ term in the exterior Laurent series of the complex velocity. Thus $\mathbf{u}=O(R^{-2})$ and, after choosing the additive potential constant, $\varphi =O(R^{-1})$. The outer integral is then $O(R^{-2})$.

Multiplying by half the constant mass density gives the boundary integral for kinetic energy of planar potential flow:The minus sign comes from the fluid's inner-boundary orientation, rather than a negative physical kinetic energy.

There is a necessary qualification to the printed hypotheses. Decaying velocity and zero circulation alone do not ensure a finite kinetic energy: $\varphi =A\log r$ gives $\mathbf{u}=A{\mathbf{e}}_r/r$, zero circulation, and logarithmically divergent exterior kinetic energy. It carries a nonzero volume flux from the body and is not the flow of an impermeable rigid body. The finite-energy condition for planar exterior potential flow makes explicit the source-free assumption needed to discard the outer boundary term.

On the translating circular cylinder $r=a$, the preceding velocity potential and streamfunction giveUsing the boundary integral for kinetic energy of planar potential flow,ThereforeAs a direct check, $|\mathbf{u}{|}^2=U^2{a}^4/r^4$, and integration over $r>a$ gives the same result.

The fluid kinetic energy has the same dependence on body speed as the kinetic energy of an extra mass $M$ carried by the body. Thus $M$ is its added mass, here equal to the mass of displaced fluid per unit cylinder length. For a cylinder of physical mass per length $m$, the combined kinetic energy is $\frac{1}{2}(m+M)U^2$. Accelerating the body requires accelerating the surrounding fluid, producing the inertial reaction $-M\dot{U}$ along the direction of motion. This is the added mass of a circular cylinder.