Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 16-Civ-A5, Hydraulic Engineering. Closed book (one aid sheet; approved calculator); 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. Each question 20 marks; parts of equal value.
Governing relations supplied on the exam cover sheet: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI: $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^{5}}Q^{2}$; total dynamic head $TDH=H_s+H_f$. Unless stated, local losses and velocity head are negligible, diameters are nominal, and the fluid is water ($\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$).
Question 5: Momentum across a hydraulic jump (20 marks)
Given. Rectangular channel, width $b=11\ \text{m}$, discharge $Q=1.2\ \text{m}^3/\text{s}$; one conjugate depth stated as $0.3\ \text{m}$. Unit discharge $q=Q/b=0.1091\ \text{m}^2/\text{s}$.
Given data
Quantity
Value
Channel width $b$
11 m
Discharge $Q$
1.2 m³/s
Unit discharge $q=Q/b$
0.1091 m²/s
Stated depth
0.3 m
Find. (a) the momentum (specific-force) equation across the jump; (b) the fluid velocity on the other (conjugate) side.
Figure 5. Hydraulic jump between conjugate depths. The stated 0.30 m is subcritical (Fr = 0.21); its supercritical conjugate is 0.0249 m.
Approach. A hydraulic jump conserves momentum, not energy. Equate the specific force (pressure + momentum flux) on the two sides to relate the conjugate depths, then get the velocity by continuity. First check the Froude number of the stated depth — it decides which side of the jump 0.30 m actually is.
Part (a): momentum / specific-force equation. Per unit width, summing hydrostatic pressure force and momentum flux and setting the two sides equal:
$$\frac{q^{2}}{g\,y_1}+\frac{y_1^{2}}{2}=\frac{q^{2}}{g\,y_2}+\frac{y_2^{2}}{2}.$$
Equivalently, in conjugate-depth form,
$$\frac{y_2}{y_1}=\tfrac{1}{2}\left(\sqrt{1+8\,Fr_1^{2}}-1\right),\qquad Fr_1=\frac{V_1}{\sqrt{g\,y_1}}.$$
Classify the stated depth. At $y=0.30\ \text{m}$, $V=q/y=0.1091/0.30=0.364\ \text{m/s}$ and
$$Fr=\frac{0.364}{\sqrt{9.81(0.30)}}=0.21\;(\lt 1)\;\Rightarrow\;\text{subcritical}.$$
So 0.30 m is the downstream (post-jump) depth; the supercritical conjugate is the unknown.
Conjugate depth from the momentum equation. Solving the specific-force balance for the other root (with $q=0.1091$):
$$y_{sc}=\tfrac{y}{2}\!\left(\sqrt{1+8Fr^{2}}-1\right)\ \text{applied to the subcritical side gives}\ y_{sc}=0.0249\ \text{m}.$$
(Check: specific force $M=q^{2}/gy+y^{2}/2=0.0490\ \text{m}^2$ matches on both depths.)
Part (b): downstream (supercritical-side) velocity. By continuity,
$$\boxed{V_{sc}=\frac{q}{y_{sc}}=\frac{0.1091}{0.0249}\approx 4.38\ \text{m/s}\quad(Fr=8.9).}$$
The subcritical-side velocity is $V=0.364\ \text{m/s}$.
Check / interpretation. The problem labels 0.30 m as “upstream”, but its Froude number (0.21) is subcritical, so a jump cannot begin there — physically 0.30 m is the downstream depth. The momentum equation is symmetric in the two conjugate depths, so it still returns the matching supercritical depth (0.0249 m) and its velocity 4.38 m/s. Per exam Note 1 this reading is stated explicitly; the boxed velocity is the sequent (supercritical-side) value the momentum balance produces.
