Question 6 of 6: Normal depth, critical depth and specific energy
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2017 · 16-Civ-A5 Hydraulic Engineering.
Three hours; closed book (one aid sheet). Six questions of equal value; candidates answer any five.
All six are solved here as a study resource. Take water ρ = 1000 kg/m³,
ν = 1.31×10−6 m²/s, g = 9.81 m/s²; local losses and velocity head are neglected (Note 6).
Reference texts (subject):
L.W. Mays, Water Resources Engineering (Wiley) — pipe flow, Hazen-Williams, distribution networks, pumps.
V.T. Chow, Open-Channel Hydraulics (McGraw-Hill) — normal & critical depth, hydraulic jump, specific energy, unsteady flow.
Munson, Young & Okiishi, Fundamentals of Fluid Mechanics — momentum/energy principles.
Question 6: Normal depth, critical depth and specific energy (20 marks)
Given. A wide rectangular channel conveys a modest discharge on a mild slope up to a
broad-crested weir control.
Given data
Quantity
Symbol
Value
Discharge
Q
3.0 m³/s
Width
b
11 m
Manning roughness
n
0.015
Longitudinal slope
$S_0$
0.002
Find. (a) normal depth $y_n$; (b) critical depth $y_c$; (c) the flow regime upstream of the weir;
(d) the specific-energy diagram.
(a) Normal depth (Manning). With $A=by$, $P=b+2y$, $R=A/P$, solve
$Q=\tfrac{1}{n}A R^{2/3}S_0^{1/2}$:
$$3.0=\frac{1}{0.015}(11y)\left(\frac{11y}{11+2y}\right)^{2/3}(0.002)^{1/2}\ \Rightarrow\ y_n=\boxed{0.243\ \text{m}}.$$
(b) Critical depth. For a rectangular section with unit discharge $q=Q/b=3.0/11=0.273\ \text{m}^2/\text{s}$,
$$y_c=\left(\frac{q^{2}}{g}\right)^{1/3}=\left(\frac{0.273^{2}}{9.81}\right)^{1/3}=\boxed{0.196\ \text{m}}.$$
(c) Regime upstream. Since $y_n = 0.243\ \text{m}\gt y_c = 0.196\ \text{m}$ (equivalently
$Fr=0.73\lt 1$), the uniform flow well upstream of the weir is sub-critical on a mild slope.
(d) Specific-energy progression. With $E=y+\dfrac{q^{2}}{2gy^{2}}$, the upstream state sits on the
upper (sub-critical) limb at $y_n=0.243\ \text{m}$, $E_n=0.307\ \text{m}$. Approaching the broad-crested weir the flow
accelerates and the depth is drawn down along the upper limb to the critical point, where energy is a minimum
$E_{min}=1.5\,y_c=0.295\ \text{m}$ at $y_c=0.196\ \text{m}$ (see figure).
Specific-energy curve $E=y+q^2/2gy^2$. The sub-critical upstream state ($y_n=0.243$ m) is drawn down along the upper limb to the critical point ($y_c=0.196$ m, $E_{min}=0.295$ m) over the weir.
