 Solution

Model the specified drag by a prescribed head loss, $(E+H{)}_x=-\gamma$. A modified conserved head is consequentlywith an arbitrary reference constant absorbed into ${\mathcal{E}}_{\gamma }$. More generally the last term is ${\int }_0^x\gamma (s)ds$. The effective bed elevation for the shallow-water specific energy calculation is $H+\gamma x$. The critical momentum equation becomesThis is the head-loss correction to hydraulic control.

When $\beta =0$, $b=1$, and regularity at $F=1$ gives $-2\eta x_c+\gamma =0$. Therefore, for $\eta ,\gamma >0$,The effective bed $-\eta x^2+\gamma x$ has a strict maximum there, with elevation increment $\Delta ={\gamma }^2/(4\eta )$ relative to $x=0$.

Let $d=h(0)$ and require $h_c=2d$. At the control $Q^2=g{h}_c^3=8g{d}^3$. At $x=0$ the shallow-water specific energy is $d+Q^2/(2g{d}^2)=5d$, while at the control it is $3h_c/2=3d$. Equality of the modified head gives $5d=3d+\Delta$. HenceThe inlet has $F(0{)}^2=8$ and is supercritical flow; the branch whose depth increases into the control makes the required smooth transition. At the control the increasing branch has slope $h_x=\sqrt{2\eta h_c/3}$, as follows by expanding the modified head to second order.

The numerical discharge uses a constant loss rate on the entire segment from $0$ to $x_c$. The wording specifies that rate only near the control, which is sufficient to locate $x_c$ but does not by itself determine the inlet-to-control head loss. If the loss elsewhere is unspecified, write ${\Lambda }_c={\int }_0^{x_c}\gamma (s)ds$. The same calculation gives $h_c=H(x_c)-H(0)+{\Lambda }_c$ and $Q=\sqrt{g}{h}_c^{3/2}$; the boxed value is the constant-loss specialization.