= Solution
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 discharge
$$
q_j=\frac Q{b_j},\qquad q_j^2=\frac{g'H^3}{4}.
$$
Before 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 $(1+\sqrt5)/4$; the former is supercritical. Thus
$$
h_1=\frac H2,\qquad u_1=\sqrt{g'H},\qquad F_1^2=2.
$$
Mass conservation gives $h_1u_1=h_2u_2=q_j$. The <hydraulic jump> momentum balance conserves $q_j^2/h+g'h^2/2$. Factoring out the non-jump solution $h_2=h_1$ gives the conjugate-depth relation
$$
\frac{h_2}{h_1}=\frac{\sqrt{1+8F_1^2}-1}{2}.
$$
Consequently the conditions immediately after the jump are
$$
\boxed{h_2=\frac{\sqrt{17}-1}{4}H,\qquad
u_2=\frac{2\sqrt{g'H}}{\sqrt{17}-1}
=\frac{\sqrt{17}+1}{8}\sqrt{g'H}.}
$$
Its <Froude number> is below one. The specific-energy loss is
$$
\boxed{E_1-E_2=\frac{(h_2-h_1)^3}{4h_1h_2}>0.}
$$
Selecting the deeper root of the unchanged energy equation would be wrong: that would conserve the energy across a turbulent jump instead of the momentum.
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