Past exam of the mathematics course of the University of Cambridge / 2009 / ib / Paper 1 / 17D (Fluid Dynamics)

Use the steady inviscid, hydrostatic, slowly varying channel approximation, without a hydraulic jump. Conservation of volume flux per unit width gives $Q=u(x)h(x)=u_1{h}_1$. The free-surface Bernoulli equation isDividing by $h_1$ yieldsThe parameter $F$ here is the squared upstream Froude number. For nonzero discharge, regarding $D$ as a function of $h$, its derivative is $dD/dh=-1+Q^2/(g{h}^3)$. It has its unique maximum at $h_c=(Q^2/g{)}^{1/3}=h_1{F}^{1/3}$, where the local Froude number is one. Substitution gives $D^{\ast }=h_1(1-3F^{1/3}/2+F/2)$. This maximum is the hydraulic control over a smooth hump.

The upstream flow is subcritical flow, with $h_1>h_c$. On this branch $dD/dh<0$, so the depth decreases as the bed rises; the velocity $Q/h$ increases. Since the crest height is below $D^{\ast }$, the depth remains above $h_c$ and never becomes critical. Over the falling side of the hump the depth increases again to $h_1$ downstream. The free surface falls over the hump: $d(h+D)/dD=-{\mathrm{Fr}}^2/(1-{\mathrm{Fr}}^2)<0$ on the subcritical branch.

The upstream flow is supercritical flow, with $h_1<h_c$. Now $dD/dh>0$, so the depth increases over the rising hump and the velocity decreases, while the free surface rises. Since $D_{\max }<D^{\ast }$, the depth stays below $h_c$. On the descending bed it decreases back to $h_1$ downstream, remaining supercritical throughout.

At the crest the two depth branches meet at $h=h_c=h_1{F}^{1/3}$. For a smooth nondegenerate maximum of the bed, $D^{\ast }-D$ is proportional to $(x-x_c{)}^2$, while expansion of the depth relation gives $D^{\ast }-D\sim 3(h-h_c{)}^2/(2h_c)$. A differentiable flow through the crest therefore switches branches rather than following the cusp obtained by staying on the same branch. Subcritical upstream flow becomes supercritical downstream; supercritical upstream flow becomes subcritical downstream.

When $F<1$ and the bed returns to zero, put $y=h_{\mathrm{down}}/h_1$. The head equation becomes $2y^3-(2+F)y^2+F=0$, orThe root $y=1$ is the original subcritical branch. The other positive root is smaller and lies on the supercritical branch, givingThis is the smooth transcritical, shock-free solution. A downstream hydraulic jump would require further downstream conditions and energy loss. A degenerate flat crest can also require a branch-selection condition; the standard critical-control interpretation uses a nondegenerate smooth crest. For $F=1$, $D^{\ast }=0$, so a positive hump is incompatible with the prescribed upstream critical head. If $F=0$, the head equation is the still-water relation $h+D=h_1$; a crest of height $h_1$ dries out, so the wet transcritical-flow interpretation requires $F>0$.