Past exam of the mathematics course of the University of Cambridge / 2013 / ib / Paper 1 / 17A / Solution

For inviscid flow with a conservative body-force potential $\Phi$, the Euler momentum equation and the identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ apply. In an irrotational region write $\mathbf{u}=\nabla \varphi$. For constant density they giveThis is the Unsteady Bernoulli equation; its constant is spatially uniform, with time-dependent gauge absorbed into $\varphi$ if desired.

For the intended linearly elastic balloon discharge model, take an initially stationary, inviscid plug of water in the constant-area tube. Put $B=\pi a^2$ and $p_{\mathrm{in}}-p_{\mathrm{atm}}=KV$ with $K>0$. At the two tube ends the speeds are equal, so their kinetic terms cancel in the Bernoulli difference. Since $\varphi (L)-\varphi (0)=Lu$, this givesHence $\ddot{V}+{\Omega }^{2}V=0$, where ${\Omega }^2=KB/(\rho L)$. With $V(0)=V_0$ and $u(0)=0$,ThusAcceleration stays positive until emptying, so the maximum speed is reached at that instant. The time is independent of initial volume, while the maximum speed is not.

The initial-rest condition is implicit in the claimed volume independence, rather than specified in the text. For an independently prescribed positive initial speed $u_0$, the first zero instead occurs at ${\Omega }^{-1}\arctan [\Omega V_0/(B{u}_0)]$. The calculation also takes the given balloon pressure as the pressure directly at the tube entrance; a reservoir-to-exit kinetic-head or entrance-loss model would add other terms and is not this equal-speed tube model.