Question 6 of 6: Rectangular channel — normal depth, critical depth, regime
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 98-Civ-A5 Hydraulic Engineering — May 2016.
Closed book; one aid sheet and a non-communicating calculator permitted. Six questions of equal value (20 marks each);
candidates answer any five. All six are solved here as a study resource. Exam-supplied relations: Hazen–Williams
$Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI, $Q$ in m$^3$/s, $D,L$ in m), Manning
$Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, Darcy–Weisbach $\Delta h = 0.0826\,f\,L\,Q^2/D^5$, valve law
$Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$, and $\mathrm{TDH}=H_s+H_f$. Water: $\rho=1000$ kg/m$^3$,
$\nu=1.31\times10^{-6}$ m$^2$/s.
Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe
networks, branched systems, valves, transmission mains); Chow, Open-Channel Hydraulics (Manning uniform flow,
critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance,
laminar film flow).
Given. Rectangular channel, discharge $Q=3.0$ m$^3$/s, width $b=11$ m, wall height 2 m,
Manning $n=0.013$, bed slope $S_0=0.001$.
Find. (a) normal depth $y_n$; (b) critical depth $y_c$; (c) the flow regime well upstream of the weir;
(d) the specific-energy diagram.
Approach. Solve Manning’s uniform-flow equation for the normal depth, compute the
critical depth from the unit discharge, compare the two (with the Froude number) to classify the regime, and sketch the
specific-energy curve $E=y+q^2/(2gy^2)$ with the operating points.
Normal depth (a). Manning $Q=\dfrac1n A R^{2/3}S_0^{1/2}$ with $A=by$, $P=b+2y$, $R=A/P$:
$$3.0=\frac{1}{0.013}(11y)\!\left(\frac{11y}{11+2y}\right)^{2/3}\!(0.001)^{1/2}.$$
Solving iteratively gives $\boxed{y_n=0.274\ \text{m}}$ (velocity $V=Q/(by_n)=0.99$ m/s).
Critical depth (b). Unit discharge $q=Q/b=3.0/11=0.273$ m$^2$/s. For a rectangular section
$$y_c=\left(\frac{q^2}{g}\right)^{1/3}=\left(\frac{0.273^2}{9.81}\right)^{1/3}=\boxed{0.196\ \text{m}}.$$
Regime (c). Since $y_n=0.274\text{ m}\;\gt\;y_c=0.196\text{ m}$, the uniform flow is
sub-critical. Confirm with the Froude number $Fr=V/\sqrt{g\,y_n}=0.99/\sqrt{9.81(0.274)}=0.61\lt1$: the slope is
hydraulically mild, so well upstream the flow is sub-critical.
Specific-energy picture (d). $E=y+\dfrac{q^2}{2gy^2}$ has a minimum
$E_{min}=\tfrac{3}{2}y_c=0.295$ m at $y=y_c$. Upstream the flow rides the upper (sub-critical) limb at $y_n=0.274$ m; as it
approaches the broad-crested weir the specific energy decreases toward $E_{min}$ and the depth is drawn down to $y_c$ at the
crest (control section).
Specific-energy curve E = y + q²/(2gy²): sub-critical normal depth on the upper limb draws down to critical depth (nose) at the weir.
