16-Civ-A5 Hydraulic Engineering · December 2016

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

Paper format. National Exams, December 2016 — 98-Civ-A5 Hydraulic Engineering. Closed book, 3 hours, one aid sheet permitted. Six questions of equal value (20 marks each); candidates answer any five. All six are solved here as a study resource. Take water density ρ = 1000 kg/m³, kinematic viscosity ν = 1.31 × 10⁻⁶ m²/s; local losses and velocity head are negligible unless stated.

Reference texts. Mays, Water Resources Engineering (Wiley) — pipe/pump systems & distribution networks; Chow, Open-Channel Hydraulics (McGraw-Hill) — normal/critical depth, specific energy, unsteady flow; Transportation Association of Canada, Geometric Design Guide for Canadian Roads — gutter/roadway drainage. Permitted equation set (from the exam cover): Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$; Manning $Q = \tfrac{1}{n}A R^{2/3} S^{1/2}$; total dynamic head $\text{TDH} = H_s + H_f$.

Question 6: Normal Depth, Critical Depth and Specific Energy (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.

Given.

Given data — Question 6
Quantity	Symbol	Value
Discharge	$Q$	3.0 m³/s
Channel width	$b$	11 m
Wall height	—	2.5 m
Manning’s $n$	$n$	0.015
Longitudinal slope	$S_o$	0.002

Find. (a) normal depth $y_n$; (b) critical depth $y_c$; (c) sub- or super-critical classification upstream; (d) the specific-energy diagram showing the approach to critical at the weir.

Approach. Solve Manning implicitly for $y_n$; compute $y_c$ from the unit discharge; compare (and confirm with the Froude number); then sketch the $E$–$y$ curve and mark the drawdown from the upstream state to the critical point.

Normal depth (Manning). For a rectangular section $A=by$, $P=b+2y$: $$Q = \frac{1}{n}\,\frac{(by)^{5/3}}{(b+2y)^{2/3}}\,S_o^{1/2}.$$ Substituting $Q=3.0$, $b=11$, $n=0.015$, $S_o=0.002$ and solving iteratively gives $\boxed{y_n = 0.243\ \text{m}}$ (well below the 2.5 m walls).
Unit discharge and critical depth. $q = Q/b = 3.0/11 = 0.2727\ \text{m}^2/\text{s}$, and for a rectangular channel $$y_c = \left(\frac{q^2}{g}\right)^{1/3} = \left(\frac{0.2727^2}{9.81}\right)^{1/3} = \boxed{0.196\ \text{m}}.$$
Classification (part c). Since $y_n = 0.243\ \text{m} \gt y_c = 0.196\ \text{m}$, the slope is mild and the uniform flow well upstream is sub-critical. Confirm with the Froude number: $V = Q/(by_n) = 3.0/(11\times0.243) = 1.12\ \text{m/s}$, $$Fr = \frac{V}{\sqrt{g\,y_n}} = \frac{1.12}{\sqrt{9.81(0.243)}} = 0.73 \;\lt\; 1 \quad\Rightarrow\ \text{sub-critical.}$$
Specific energy and the approach to the weir (part d). With $E = y + \dfrac{q^2}{2gy^2}$, the minimum energy occurs at critical depth, $E_{\min} = 1.5\,y_c = 0.295\ \text{m}$. Upstream the flow sits on the upper (sub-critical) limb at $y_n=0.243$ m, $E = 0.307$ m. As the broad-crested weir is approached the surface draws down toward critical: $E$ decreases to $E_{\min}$ and the depth falls from 0.243 m to $y_c = 0.196$ m, sliding down the sub-critical limb to the nose of the curve.

Specific-energy curve for $q=0.273\ \text{m}^2/\text{s}$. The upstream sub-critical flow ($y_n=0.243$ m) draws down along the upper limb to the critical nose ($y_c=0.196$ m, $E_{\min}=0.295$ m) as the broad-crested weir is approached.

Final results — Question 6
Quantity	Result
Normal depth (a)	$y_n = 0.243$ m
Critical depth (b)	$y_c = 0.196$ m
Froude number at normal depth	$Fr = 0.73$
Flow classification (c)	sub-critical (mild slope, $y_n \gt y_c$)
Minimum specific energy (d)	$E_{\min} = 1.5\,y_c = 0.295$ m

Topic. Uniform & critical flow in a rectangular channel; specific energy and channel controls.

Key relations. Manning $Q=\frac1n A R^{2/3}S_o^{1/2}$; critical depth $y_c=(q^2/g)^{1/3}$; Froude $Fr=V/\sqrt{gy}$; specific energy $E=y+q^2/(2gy^2)$, $E_{\min}=1.5y_c$.

Why this works. Normal depth is where friction slope equals bed slope (uniform flow); critical depth is where specific energy is minimum and $Fr=1$. Comparing them classifies the slope: $y_n\gt y_c$ is mild (sub-critical uniform flow). A broad-crested weir is a control that forces the flow through critical depth; upstream sub-critical flow therefore accelerates and draws down toward $y_c$, which on the $E$–$y$ diagram is a slide down the upper limb to the nose as energy is minimised at the control.

Common pitfall. Reading critical depth off the total discharge instead of the unit discharge $q=Q/b$; or expecting the depth to rise at the weir — a sub-critical approach falls to critical.

Source. Chow, Open-Channel Hydraulics, ch. 3–6 (uniform flow, critical flow, specific energy); Mays, Water Resources Engineering, ch. 8.

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