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

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.