16-Civ-A5 Hydraulic Engineering · December 2014

Question 1 of 6: Penstock — single and parallel discharge

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 1: Penstock — single and parallel discharge (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 single penstock links two reservoirs whose surfaces differ by a fixed elevation head.

Given data (Question 1)
Pipe diameter $D$	800 mm = 0.800 m
Hazen–Williams $C$	120
Pipe length $L$	2,000 m
Upstream / downstream level	1010 m / 998 m

Find. (a) The discharge in one penstock; (b) the total discharge when a second, identical penstock is added in parallel, with an explanation of the change.

Two reservoirs connected by the penstock; the 12 m surface-elevation difference is the head that drives the flow.

Approach. Two reservoirs fix the total head at the elevation difference; apply Hazen–Williams with $S = \Delta z/L$, then use the fact that identical parallel pipes between the same two reservoirs each carry the full head.

Available head and slope. The reservoir surfaces set the total head, so $\Delta z = 1010 - 998 = 12\ \text{m}$ and the friction slope is $S = \dfrac{\Delta z}{L} = \dfrac{12}{2000} = 0.006$.
Single-penstock discharge (Hazen–Williams). Substituting into $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$: $$Q_1 = 0.278(120)(0.800)^{2.63}(0.006)^{0.54} = \boxed{1.17\ \text{m}^3/\text{s}.}$$ The corresponding velocity $V = Q_1/A = 1.17/0.503 = 2.33\ \text{m/s}$ is reasonable for a penstock.
Parallel penstock. A second identical pipe connects the same two reservoir surfaces, so it too sees the full 12 m head and independently carries $Q_1$. Because each branch conveys the same head-driven discharge, the combined flow is $$Q_{\text{tot}} = 2\,Q_1 = 2(1.17) = \boxed{2.34\ \text{m}^3/\text{s}.}$$
Higher or lower — why. The combined flow is higher — exactly double. The two reservoir surfaces fix the head difference at 12 m regardless of how many pipes join them, so adding a parallel path does not steal head from the first pipe; it simply provides a second, equal conduit. (Flow does not double for pipes added in series, where the head is shared.)

Question 1 — results
Quantity	Value
Single-penstock discharge $Q_1$	1.17 m³/s
Two penstocks in parallel $Q_{\text{tot}}$	2.34 m³/s (double, higher)

Topic. Steady pipe flow between fixed-head reservoirs; series vs. parallel.

Key relations. Hazen–Williams $Q = 0.278\,C\,D^{2.63}(\Delta h/L)^{0.54}$; parallel branches between common nodes each carry the full head loss.

Why this works. When both ends of every pipe terminate at the same two reservoir surfaces, the head loss across each pipe is identical and equal to the surface-elevation difference. The Hazen–Williams law then returns the same discharge for each identical pipe, and continuity sums them. Head is a boundary condition set by the reservoirs, not a resource divided among parallel pipes — that is the essence of why parallel capacity adds directly.

Common pitfall. Assuming the two parallel pipes “split” the original flow (so total stays 1.17) — that is the series result. Another error is halving the slope for the second pipe; both pipes see the full 12 m.

Source. Mays, Water Resources Engineering, ch. on pipe systems (equivalent pipes, parallel/series rules).

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