Trapezoidal Channel Uniform Flow, Specific Energy Over a Bump, and the Shields Criterion: 18-Env-A2 Hydrology and Municipal Hydraulics Engineering, May 2016, Q7

Problem 7: Trapezoidal Channel Uniform Flow, Specific Energy Over a Bump, and the Shields Criterion (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(i) Trapezoidal Channel: Discharge and Reynolds Number

Given.

Given data
Quantity	Symbol	Value
Normal depth	$y$	4 m
Base width	$b$	15 m
Side slope (V:H)	$z$	1:4 $\Rightarrow z=4$ (horizontal per unit vertical)
Bed slope	$S_0$	0.05 (5%)
Lining	—	excavated earth

Trapezoidal channel cross-section: base $b=15$ m, side slope $z=4$, normal depth $y=4$ m.

Find. (a) $Q$ in m³/s; (b) $Re$ and flow type.

Approach. Compute the trapezoidal geometry ($A$, wetted perimeter $P$, hydraulic radius $R=A/P$), select a Manning's n appropriate to an excavated earth channel, apply Manning's equation for $Q$, then classify the flow with $Re=VR/\nu$.

Check: Manning's n is not printed on this page of the source; $n=0.022$ is used, a standard textbook design value for a clean, weathered excavated earth channel (Chow's Open-Channel Hydraulics table 5-6). The 5% bed slope combined with a 15 m base width produces a very high computed velocity/Froude number (supercritical, $Fr\approx3.8$) well outside typical municipal channel practice; used here exactly as printed on the exam.

Geometry. $$A=(b+zy)y=(15+4\times4)\times4=\boxed{124.0\ \text{m}^2},$$ $$P=b+2y\sqrt{1+z^2}=15+2(4)\sqrt{1+16}=\boxed{47.98\ \text{m}},\qquad R=\frac{A}{P}=\boxed{2.585\ \text{m}}.$$
Manning's equation. $$Q=\frac{1}{n}AR^{2/3}S_0^{1/2}=\frac{1}{0.022}(124.0)(2.585)^{2/3}(0.05)^{1/2}=\boxed{2373\ \text{m}^3/\text{s}}.$$
Velocity, Froude number and Reynolds number. $$V=\frac{Q}{A}=\frac{2373}{124.0}=19.14\ \text{m/s},\qquad Fr=\frac{V}{\sqrt{gA/T}}=3.76\ (\textbf{supercritical}),$$ $$Re=\frac{VR}{\nu}=\frac{19.14\times2.585}{1.0\times10^{-6}}=\boxed{4.95\times10^{7}}\ \Rightarrow\ \textbf{turbulent}.$$

Quantity	Value
Flow area, $A$	124.0 m²
Hydraulic radius, $R$	2.585 m
Discharge, $Q$	2373 m³/s
Mean velocity, $V$	19.14 m/s (supercritical, $Fr=3.76$)
Reynolds number, $Re$	4.95×10&sup7; — turbulent

(ii) Depth of Flow Over the Bump (Specific Energy)

Given. Wide channel (unit-width analysis); approach velocity $V_1=2.0$ m/s; approach depth $y_1=1.2$ m; bump height $\Delta Z=0.12$ m; frictionless transition (energy conserved except for the bed rise).

Longitudinal profile: subcritical approach flow passing over a positive bed rise (bump) of height ΔZ.

Find. (a) $y_2$ over the bump; (b) the bump height that would force critical flow at the crest.

Approach. Compute the approach Froude number to confirm the flow regime, then apply specific-energy conservation $E_2=E_1-\Delta Z$ (unit discharge $q=V_1y_1$ constant) and solve $E_2=y_2+q^2/2gy_2^2$ for the physically relevant root; separately, the critical bump height is the one that drops $E$ exactly to the minimum specific energy $E_c=1.5\,y_c$.

Approach conditions. $$q=V_1y_1=2.0\times1.2=2.40\ \text{m}^2/\text{s},\qquad E_1=y_1+\frac{V_1^2}{2g}=1.2+\frac{2.0^2}{19.62}=\boxed{1.404\ \text{m}},$$ $$Fr_1=\frac{V_1}{\sqrt{gy_1}}=\frac{2.0}{\sqrt{9.81\times1.2}}=\boxed{0.583}\ (\textbf{subcritical}).$$
Critical depth and critical bump height (part b). For unit discharge $q$, the minimum specific energy occurs at $$y_c=\left(\frac{q^2}{g}\right)^{1/3}=\boxed{0.837\ \text{m}},\qquad E_c=1.5\,y_c=\boxed{1.256\ \text{m}}.$$ The bump height that would just drop the crest energy to this critical minimum is $$\Delta Z_{\text{crit}}=E_1-E_c=1.404-1.256=\boxed{0.148\ \text{m}}.$$
Depth over the actual 0.12 m bump (part a). Since $\Delta Z=0.12\ \text{m}<\Delta Z_{\text{crit}}=0.148\ \text{m}$, the bump does not choke the flow, so a subcritical crest depth $y_2>y_c$ exists: $$E_2=E_1-\Delta Z=1.404-0.12=1.284\ \text{m}=y_2+\frac{q^2}{2gy_2^2}.$$ Solving numerically for the subcritical root gives $y_2=\boxed{0.975\ \text{m}}$, with crest velocity $V_2=q/y_2=2.40/0.975=\boxed{2.46\ \text{m/s}}$.

Quantity	Value
Unit discharge, $q$	2.40 m²/s
Approach specific energy, $E_1$	1.404 m (subcritical, $Fr_1=0.583$)
Critical depth, $y_c$	0.837 m
(b) Critical bump height, $\Delta Z_{\text{crit}}$	0.148 m
(a) Depth over the 0.12 m bump, $y_2$	0.975 m (crest velocity 2.46 m/s)

(iii) The Critical Shields Stress

Given. $\tau^*=\dfrac{\tau_o}{g(\rho_p-\rho_f)D_p}$, the dimensionless criterion marking the onset of sediment particle motion.

Check: the printed equation shows only "$/g(\rho_p-\rho_f)D_p$", with no numerator; the numerator is taken as $\tau_o$, the bed shear stress, which is the only term making the equation dimensionally consistent and matches the standard Shields-parameter definition.

Find. The meaning and consistent dimensions of each term.

$\tau^*$ is the dimensionless critical (Shields) shear stress at which a bed particle of a given size just begins to move; $\tau_o$ is the bed shear stress exerted by the flow [Pa = N/m² = kg/(m·s²)]; $g$ is gravitational acceleration [m/s²]; $\rho_p$ is the sediment particle density [kg/m³]; $\rho_f$ is the fluid (water) density [kg/m³]; and $D_p$ is the representative particle diameter [m]. The denominator's units, $[\text{m/s}^2]\times[\text{kg/m}^3]\times[\text{m}]=\text{kg}/(\text{m}^2\cdot\text{s}^2)=\text{Pa}$, match the numerator's units exactly, confirming $\tau^*$ is dimensionless — the Shields curve then plots this dimensionless stress against a particle Reynolds number to give a single empirical incipient-motion criterion applicable across particle sizes.

Concept & Theory

Topic. Manning uniform flow in a trapezoidal channel, specific-energy analysis over a bed rise, and the Shields incipient-motion criterion.

Key relations. $A=(b+zy)y$, $P=b+2y\sqrt{1+z^2}$; $Q=\frac1nAR^{2/3}S_0^{1/2}$; $E=y+V^2/2g$; $y_c=(q^2/g)^{1/3}$, $E_c=1.5y_c$; $\tau^*=\tau_o/[g(\rho_p-\rho_f)D_p]$.

Why this works. Specific energy (energy measured relative to the local channel bed) is conserved across a frictionless bed-rise transition precisely because the datum shift cancels out of the total-energy equation, turning a 2-D problem into a 1-D root-find for $y_2$; a bump can only force critical flow at its crest if it removes enough specific energy to reach the minimum $E_c$ — any smaller rise leaves the flow subcritical throughout. The Shields parameter works because it non-dimensionalizes the ratio of destabilizing fluid drag to stabilizing submerged particle weight, so a single empirical curve applies across sediment sizes.

Common pitfall. Assuming a bump always lowers the water depth (true only for subcritical approach flow — a supercritical approach flow instead RISES over a positive bump); solving the specific-energy cubic and keeping the wrong (supercritical) root when the approach flow is subcritical.

Source. Chow, Open-Channel Hydraulics, uniform-flow, specific-energy and sediment-transport chapters.

Question 7 of 7: Trapezoidal Channel Uniform Flow, Specific Energy Over a Bump, and the Shields Criterion

Problem 7: Trapezoidal Channel Uniform Flow, Specific Energy Over a Bump, and the Shields Criterion (20 marks)

(i) Trapezoidal Channel: Discharge and Reynolds Number

(ii) Depth of Flow Over the Bump (Specific Energy)

(iii) The Critical Shields Stress