Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 76 / 3 / c

Assume the jump is stationary, short compared with the scale of width variation, hydrostatic on either side, and has negligible bed stress integrated over its short length. Treat its width $b_j=\sqrt{32/27}$ as constant. It must occur on the downstream supercritical branch, at $x_j=L\sqrt{b_j-1}>0$. The discharge from the control gives the local unit-width dischargeBefore the jump the energy is still $H$. Writing $y=h/H$, its equation is $y+1/(8y^2)=1$. Its positive roots are $1/2$ and $(1+\sqrt{5})/4$; the former is supercritical. ThusMass conservation gives $h_1{u}_1=h_2{u}_2=q_j$. The hydraulic jump momentum balance conserves $q_j^2/h+g^{\prime }{h}^2/2$. Factoring out the non-jump solution $h_2=h_1$ gives the conjugate-depth relationConsequently the conditions immediately after the jump areIts Froude number is below one. The specific-energy loss isSelecting the deeper root of the unchanged energy equation would be wrong: that would conserve the energy across a turbulent jump instead of the momentum.