Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 90 / 1

Assume the shallow-water approximation: the horizontal scale is long compared with the depth, the hydrostatic approximation applies, and the cross-section-averaged velocity is approximately uniform. Neglect viscosity and drag for this part. Volume conservation gives the constant discharge $Q=bhu$. The steady horizontal momentum equation isso its integral is the Bernoulli equation head $\mathcal{H}=H+h+u^2/(2g)$. The shallow-water specific energy measured above the bed is thereforeThe Froude number compares the flow velocity with the long-wave speed $\sqrt{gh}$:At fixed $Q,b$, the shallow-water specific energy tends to infinity as $h$ tends to zero or infinity. Its unique minimum occurs at the critical depthThe two depth branches above that minimum are subcritical flow, $F<1$ with $h>h_c$, and supercritical flow, $F>1$ with $h<h_c$. The long-wave characteristic speeds relative to the bed are $u\pm \sqrt{gh}$. Thus a subcritical flow can receive information from downstream, whereas both characteristic curves travel downstream in a positive supercritical flow.

A hydraulic control is a critical section where one characteristic speed vanishes and a smooth flow can pass between the two branches. Regularity there supplies a condition relating $Q$, the depth and the geometry; together with reservoir or inlet data, it can determine the discharge. Merely finding $F=1$ is not enough: a finite depth gradient also requires the geometric forcing to vanish in the critical momentum equation.

Differentiate the Bernoulli equation at fixed discharge, using $u_x/u=-b_x/b-h_x/h$. This givesAt a smooth hydraulic control, $F=1$ and the right-hand side must vanish. For the specified geometry,For the intended positive parameters $\beta ,\eta ,a>0$, positive finite depth requiresThe control lies between the bed summit and the width minimum, rather than necessarily at either one. For $a<0$ the corresponding interval is $a<x_c<0$. If $a=0$, the coincident summit and throat admit $x_c=0$ without imposing a depth through this first-derivative relation. The limiting cases $\beta =0$ or $\eta =0$ must be handled before division: their control is at the bed summit or throat, respectively.

At $x_c=a/2$, the critical-depth formula simplifies toThis establishes the requested depth without assuming that the bed summit and throat coincide.

There is a further regularity qualification if a critical candidate is to be a simple, smooth transcritical hydraulic control. At fixed $Q$, define the critical-head envelopeReal depths require $\mathcal{H}\ge T(x)$. At a control $\mathcal{H}=T(x_c)$, so $T$ must have a local maximum, not a local minimum. The first condition $T_x=0$ is precisely the compatibility equation above, andFor the positive-parameter case, a nondegenerate control therefore additionally satisfiesIndeed $T_{xx}=-2\eta [3a+\beta (a-x_c{)}^2(3a-10x_c)]/[3(a-x_c)b(x_c)]$. The interval $0<x_c<a$ gives all positive-depth candidates, while this inequality excludes candidates with no smooth critical crossing. At $a/2$ it becomes $\beta a^2<6$ for a simple control; the requested depth relation remains necessary whenever such a control exists. Equality is a degenerate critical point requiring higher-order analysis. For example, $a=1$, $\beta =16$ makes the midpoint a local minimum of $T$, so positivity of the depth alone would incorrectly identify it as a regular control.

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.