Question 4 of 6: Hydraulic jump in a rectangular channel
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 4: Hydraulic jump in a rectangular channel (20 marks)
Given. A supercritical spillway discharge enters a rectangular channel and passes through a
hydraulic jump to a subcritical downstream depth.
Given data
Quantity
Symbol
Value
Channel width
B
6.5 m
Discharge
Q
20 m³/s
Upstream (pre-jump) depth
$z_1$
0.20 m
Find. (a) the momentum (specific-force) equation of the jump; (b) the sequent depth $z_2$;
(c) why momentum, not energy, governs the jump.
Hydraulic jump (Figure 3): shallow, fast supercritical inflow rises abruptly to a deep, slow subcritical depth with strong turbulent energy loss.
Approach. Compute the unit discharge and upstream Froude number, apply conservation of
specific force across the jump to obtain the sequent-depth relation, and evaluate the energy dissipated.
Unit discharge, velocity and Froude number.
$$q=\frac{Q}{B}=\frac{20}{6.5}=3.077\ \text{m}^2/\text{s},\quad V_1=\frac{q}{z_1}=15.38\ \text{m/s},\quad
Fr_1=\frac{V_1}{\sqrt{g z_1}}=10.98.$$ Since $Fr_1\gt 1$ the inflow is strongly supercritical — a jump forms.
(a) Momentum equation. Neglecting bed friction over the short jump, the specific force
(pressure + momentum flux) per unit width is conserved, $M_1 = M_2$:
$$\frac{z_1^{2}}{2}+\frac{q^{2}}{g\,z_1}=\frac{z_2^{2}}{2}+\frac{q^{2}}{g\,z_2}.$$
(b) Sequent depth. Solving the specific-force balance gives the Belanger conjugate-depth relation:
$$z_2=\frac{z_1}{2}\left(\sqrt{1+8\,Fr_1^{2}}-1\right)=\frac{0.20}{2}\left(\sqrt{1+8(10.98)^2}-1\right)=\boxed{3.01\ \text{m}}.$$
Downstream state and energy loss. $V_2=q/z_2=1.02\ \text{m/s}$, $Fr_2=0.19\lt 1$ (subcritical), and
the jump dissipates
$$\Delta E=\frac{(z_2-z_1)^{3}}{4\,z_1 z_2}=\frac{(2.81)^3}{4(0.20)(3.01)}=9.2\ \text{m}.$$
(c) Momentum vs energy. The jump is a region of intense, self-generated turbulence that destroys
a large but a-priori-unknown amount of mechanical energy (here 9.2 m). The energy equation cannot be applied
without that loss as an input. Momentum, by contrast, involves only the external pressure and boundary forces —
which are negligible over the short jump — so specific force is conserved and yields $z_2$ directly.
