Hydraulic jump head loss ($\Delta H=\frac{(h_2-h_1{)}^3}{4h_1{h}_2}$)

In a stationary rectangular-channel hydraulic jump, orient the discharge $q=hu>0$ from side 1 to side 2. Momentum conservation gives $q^2=g{h}_1{h}_2(h_1+h_2)/2$ for unequal depths. The specific mechanical energy is $B=u^2/2+gh$, soDividing by $g$ gives the displayed loss of head, measured as a length. The mechanical energy flux lost is $\rho q(B_1-B_2)$, so a dissipative jump must have $h_2>h_1$. With $R=h_2/h_1>1$, the squared Froude numbers are $R(1+R)/2>1$ upstream and $(1+R)/(2R^2)<1$ downstream. Thus the shallow water equations connect supercritical flow to subcritical flow through the admissible jump.