Past exam of the mathematics course of the University of Cambridge / 2016 / ib / Paper 1 / 17C / c

Write the conservative body force per unit mass as $-\nabla \Phi$. For a fluid of constant density $\rho$, the Euler equations for an inviscid fluid and the incompressible flow condition areUsing part (a), these becomeFor irrotational flow, locally choose a velocity potential with $\mathbf{u}=\nabla \varphi$. Then $\nabla ({\partial }_t\varphi +H)=0$, so the unsteady Bernoulli integral isThe constant is spatially uniform on a connected region where the velocity potential is defined; its time dependence can be absorbed into the additive time-dependent gauge of $\varphi$. A global single-valued velocity potential requires an appropriate topological condition, such as a simply connected domain; without it, the integral is locally valid. In a connected steady irrotational region, $\nabla H=0$ directly gives one spatial Bernoulli constant even without choosing a global potential.

In steady flow, the Euler equations for an inviscid fluid and part (a) giveTaking the dot product with $\mathbf{u}$ yields $\mathbf{u}\cdot \nabla H=0$, because a cross product is perpendicular to its first factor. Part (b) therefore gives Bernoulli's quantity along streamlines:For rotational flow these constants can differ between streamlines, since $\mathbf{u}\times \mathbf{\omega }$ need not vanish and therefore $\nabla H$ need not be zero. Special rotational flows can still have one global constant when that cross product vanishes; the general conclusion is only constancy along individual streamlines.

Let $\mathbf{v}$ denote the velocity relative to the uniformly rotating frame, and write $r_{\perp }^2=x^2+y^2$. For rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$, the Euler equations for an inviscid fluid in that frame areThe negative quadratic term is the centrifugal potential, whose negative gradient is the outward centrifugal acceleration per unit mass. In steady flow, dotting the equation with $\mathbf{v}$ eliminates the Coriolis force, since $\mathbf{v}\cdot (\mathbf{\Omega }\times \mathbf{v})=0$. The remaining acceleration term is $\mathbf{v}\cdot \nabla (|\mathbf{v}{|}^2/2)$. ThusBy part (b), the steady rotating-frame Bernoulli integral isSteadiness here is measured in the rotating frame, including the body-force potential used in that frame.