Past exam of the waves and compressible flow course of the University of Oxford / 2023 / 3

Take a unit-width control volume spanning the stationary hydraulic jump. Mass conservation makes the discharge $q=hu$ constant. The horizontal momentum flux is $\rho h{u}^2$, and the integrated hydrostatic pressure force on a vertical cross-section is ${\int }_0^h\rho g(h-z)dz=\rho g{h}^2/2$. Momentum conservation therefore givesThese are the Rankine-Hugoniot conditions for the shallow water equations in the stationary shock frame. With depths $h_1\ne h_2$ on the two sides, eliminating the velocity gives

Orient the flow so $q>0$, and let subscripts 1 and 2 denote upstream and downstream. The mechanical energy flux is $\rho qB$, where $B=u^2/2+gh$. Inserting the relation for $q^2$ in i gives the hydraulic jump head loss,An admissible jump dissipates mechanical energy, so $\rho q(B_1-B_2)>0$. The denominator is positive, henceThe zero depth-change case is continuous flow. A depth-decreasing discontinuity would require an energy input at the jump; it is not the dissipative hydraulic jump selected by this model.

Put $R=h_2/h_1>1$. The squared Froude numbers follow directly from the discharge relation:Thus the jump connects supercritical flow to subcritical flow. Long surface gravity waves have speed $\sqrt{gh}$ relative to the fluid, so the characteristic speeds in the stationary frame are $u\pm \sqrt{gh}$. Upstream both point with the flow; downstream one points against it. The jump changes whether long-wave information can propagate upstream through the local flow.

Let $H$ be the depth behind the bore. The wall's kinematic boundary condition makes the velocity there $U$; ahead of the bore the depth and velocity are $h_0$ and zero. Across a rightward hydraulic bore moving at $V$, the Rankine-Hugoniot conditions areThe first gives $R=H/h_0=V/(V-U)=v/(v-1)>1$. Substituting into the second gives $F^{2}v=(R^2-1)/2$, where $v=V/U$ and $F=U/\sqrt{g{h}_0}$. Hence the piston-driven shallow-water bore obeysThe positive depth increase is consistent with the energy-loss condition in ii, now viewed in the bore's own shock frame.

For fixed $F>0$, the left side $L(v)=F^2{v}^2$ is a strictly increasing function on $v>1$, with limit $F^2$ at $1$ and unbounded growth. The right side $R(v)=1+3/[2(v-1)]+1/[2(v-1{)}^2]$ is a strictly decreasing function, sinceIt tends to infinity as $v\downarrow 1$ and to one as $v\to \infty$. Their continuity and the intermediate value theorem give a crossing, and strict opposite monotonicity makes it unique. There is exactly one admissible bore speed $V>U$ for every $F>0$.

The illustrated case is $F=1$; changing $F>0$ moves the increasing curve without changing the existence or uniqueness argument. The vertical asymptote at $v=1$ expresses the growing depth ratio $H/h_0=v/(v-1)$.