16-Civ-A5 Hydraulic Engineering · December 2014

Question 6 of 6: Rectangular channel — normal & critical depth

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2014 — 98-Civ-A5 Hydraulic Engineering, 3 hours, closed book (one aid sheet). Six questions of equal value (20 marks each); candidates complete any five. All six are solved here as a study resource. ‘cms’ = m³/s; water ρ = 1000 kg/m³, ν = 1.31×10⁻⁶ m²/s; local losses and velocity head neglected unless stated.

Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, valves, parallel/loop networks); Chow, Open-Channel Hydraulics (Manning uniform flow, critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance, St-Venant equations). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI), Manning $Q = \tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach, and $\text{TDH} = H_s + H_f$. In inverted form the friction loss carried by a known discharge is $h_f = L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}$.

Question 6: Rectangular channel — normal & critical depth (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. A wide rectangular channel in uniform approach flow leading to a broad-crested weir.

Given data (Question 6)
Discharge $Q$	3.0 m³/s
Width $b$ / wall height	11 m / 2 m
Manning $n$	0.013
Longitudinal slope $S_0$	0.001

Find. (a) normal depth; (b) critical depth at the weir; (c) the flow regime well upstream; (d) the specific-energy diagram showing the approach to critical.

Approach. Solve Manning’s uniform-flow equation for normal depth; get critical depth from the rectangular-channel formula; compare the two (and the Froude number) for the regime; then sketch specific energy $E = y + q^2/(2gy^2)$.

(a) Normal depth (Manning). With unit-width flow the equation $Q = \tfrac{1}{n}A R^{2/3}S_0^{1/2}$, $A = 11y$, $R = 11y/(11+2y)$, is solved iteratively: $$3.0 = \dfrac{1}{0.013}(11y)\!\left(\dfrac{11y}{11+2y}\right)^{2/3}(0.001)^{1/2} \;\Rightarrow\; \boxed{y_n = 0.274\ \text{m}.}$$
(b) Critical depth. For a rectangular channel $y_c = (q^2/g)^{1/3}$ with unit discharge $q = Q/b = 3.0/11 = 0.273\ \text{m}^2/\text{s}$: $$y_c = \left(\dfrac{0.273^2}{9.81}\right)^{1/3} = \boxed{0.196\ \text{m}.}$$
(c) Regime well upstream. Since $y_n = 0.274\ \text{m} > y_c = 0.196\ \text{m}$, the uniform flow is deeper than critical, i.e. sub-critical. Confirming with the Froude number at normal depth, $V_n = Q/(b\,y_n) = 3.0/(11\!\cdot\!0.274) = 0.99\ \text{m/s}$, $$Fr = \dfrac{V_n}{\sqrt{g\,y_n}} = \dfrac{0.99}{\sqrt{9.81(0.274)}} = 0.61 < 1 \Rightarrow \text{sub-critical.}$$
(d) Specific-energy progression. The specific energy is $E = y + q^2/(2gy^2)$, minimum $E_{\min} = 1.5\,y_c = 0.295\ \text{m}$ at $y = y_c$. The approach flow sits on the upper (sub-critical) limb at $y_n = 0.274$ m ($E = 0.324$ m); as the water accelerates toward the broad-crested weir it draws down along that limb to the critical point ($y_c = 0.196$ m, $E_{\min}$), losing depth and specific energy until control is reached at the crest.

Specific-energy curve $E = y + q^2/(2gy^2)$. Sub-critical approach (green, $y_n=0.27$ m) draws down along the upper limb to the critical nose ($y_c=0.20$ m, $E_{\min}=0.29$ m) at the weir.

Question 6 — results
Quantity	Value
(a) Normal depth $y_n$	0.274 m
(b) Critical depth $y_c$	0.196 m
(c) Regime upstream	Sub-critical ($Fr = 0.61$)
(d) Minimum specific energy $E_{\min}$	0.295 m at $y_c$

Topic. Uniform (normal) flow, critical depth, and specific energy in a rectangular channel.

Key relations. Manning $Q = \tfrac{1}{n}A R^{2/3}S_0^{1/2}$; critical depth $y_c = (q^2/g)^{1/3}$; specific energy $E = y + q^2/(2gy^2)$, $E_{\min} = 1.5\,y_c$; Froude $Fr = V/\sqrt{gy}$.

Why this works. Normal depth is where gravity balances friction (uniform flow); critical depth is where specific energy is minimum and $Fr = 1$. Comparing them classifies the slope: $y_n > y_c$ is a mild slope carrying sub-critical flow, which is exactly what a Froude number below one confirms. A broad-crested weir acts as a control that forces critical depth at its crest, so the sub-critical approach must accelerate and draw down along the upper limb of the $E$–$y$ curve to reach that control.

Common pitfall. Using the full width for the hydraulic radius (ignoring $R = A/P$) in Manning, or confusing which limb is sub-critical — the upper limb ($y > y_c$) is sub-critical, the lower limb super-critical. Note the 2 m walls never come into play; the depth is well below them.

Source. Chow, Open-Channel Hydraulics, ch. 3–4 (uniform flow, critical flow, specific energy).

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