= Hydraulic jump head loss
{title2=$\Delta H=\frac{(h_2-h_1)^3}{4h_1h_2}$}
In a stationary rectangular-channel <hydraulic jump>, orient the discharge $q=hu>0$ from side 1 to side 2. <Momentum conservation> gives $q^2=gh_1h_2(h_1+h_2)/2$ for unequal depths. The specific mechanical <energy> is $B=u^2/2+gh$, so
$$
B_1-B_2=\frac{g(h_2-h_1)^3}{4h_1h_2}.
$$
Dividing 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.
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