Past exam of the mathematics course of the University of Cambridge / 2008 / ib / Paper 4 / 18B (Fluid Dynamics)

For constant density $\rho$, the inviscid Euler equations with conservative body force areThe vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times (\nabla \times \mathbf{u})$ simplifies for potential flow $\mathbf{u}=\nabla \varphi$, since its vorticity vanishes. Also ${\partial }_t\mathbf{u}=\nabla {\varphi }_t$. ThereforeOn each connected fluid region the expression is spatially constant but may depend on time, giving the Unsteady Bernoulli equationThe arbitrariness of $f$ corresponds to the freedom to add a function of time to the velocity potential without changing $\mathbf{u}$. Replacing $\varphi$ by $\varphi -{\int }^{t}f(s)ds$ sets the right-hand side to zero.

Use the ideal one-dimensional pipe model: constant density, steady inviscid flow, approximately uniform endpoint speeds $u_1,u_2$, and equal body-force potential at the two ends. Conservation of mass gives the constant volumetric flow rate $q=u_1{S}_1=u_2{S}_2$, and the mass flow rate is $m=\rho q$. The steady Bernoulli equation givesFor flow from a wider to a narrower section, $S_1>S_2$ and $\Delta p>0$. Solving for the positive mass throughput gives the Venturi effect formulaThe formula requires these hydraulic assumptions; steady incompressibility alone does not impose uniform endpoint profiles or equal potential. If $V_1\ne V_2$, the same derivation replaces $\Delta p$ by $\Delta p+\rho (V_1-V_2)$. Viscous losses would also modify the pressure balance. The equal-area case is singular in this pressure-based determination: in the ideal equal-potential model its pressure difference must vanish and does not determine the throughput.