16-Civ-A5 Hydraulic Engineering · May 2017

Question 1 of 6: Penstock — single and parallel flow

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

Notes on this paper

National Exams — May 2017 · 16-Civ-A5 Hydraulic Engineering. Three hours; closed book (one aid sheet). Six questions of equal value; candidates answer any five. All six are solved here as a study resource. Take water ρ = 1000 kg/m³, ν = 1.31×10⁻⁶ m²/s, g = 9.81 m/s²; local losses and velocity head are neglected (Note 6).

Reference texts (subject):

L.W. Mays, Water Resources Engineering (Wiley) — pipe flow, Hazen-Williams, distribution networks, pumps.
V.T. Chow, Open-Channel Hydraulics (McGraw-Hill) — normal & critical depth, hydraulic jump, specific energy, unsteady flow.
Munson, Young & Okiishi, Fundamentals of Fluid Mechanics — momentum/energy principles.

Question 1: Penstock — single and parallel flow (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 gravity penstock links two constant-level reservoirs, and the whole available elevation difference is dissipated as pipe friction.

Given data
Quantity	Symbol	Value
Diameter	D	0.700 m
Hazen-Williams coefficient	C	125
Length	L	4,000 m
Upstream level	z_u	1010 m
Downstream level	z_d	998 m

Find. (a) the discharge in one penstock; (b) the combined discharge with a second identical penstock in parallel, and whether it rises or falls.

Penstock between two fixed-level reservoirs. The full 12 m elevation drop is available as friction head.

Approach. With both reservoir surfaces fixed, the friction slope is set by the elevation drop; apply Hazen-Williams, then note that a parallel pipe sees the same fixed head.

Available head and friction slope. All of the elevation difference drives friction: $$\Delta H = z_u - z_d = 1010 - 998 = 12\ \text{m}, \qquad S = \frac{\Delta H}{L} = \frac{12}{4000} = 0.003.$$
Single-penstock discharge (Hazen-Williams). With $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$, $$Q_1 = 0.278(125)(0.700)^{2.63}(0.003)^{0.54} = \boxed{0.591\ \text{m}^3/\text{s}}.$$
Second penstock in parallel. Because both reservoir surfaces are fixed, each pipe independently experiences the full $\Delta H = 12\ \text{m}$ (identical $S$), so each carries the same $0.591\ \text{m}^3/\text{s}$: $$Q_{\text{tot}} = 2\,Q_1 = \boxed{1.182\ \text{m}^3/\text{s}}.$$
Higher or lower — why. The combined flow is higher: it doubles. Adding a pipe in parallel between the same two fixed-level reservoirs adds conveyance area without lowering the driving head, so the system simply carries twice the discharge (contrast a series addition, which would add friction and reduce flow).

Final results — Question 1
Result	Value
Single penstock, $Q_1$	0.591 m³/s
Two parallel penstocks, $Q_{\text{tot}}$	1.182 m³/s (double)
Change	Higher — flow doubles

Topic: Steady pipe flow & the Hazen-Williams equation.

Key relations: $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = \Delta h/L$; identical pipes in parallel between fixed heads carry equal, additive discharges.

Why this works: Between two reservoirs whose surfaces do not move, the total head loss is pinned at the elevation difference regardless of discharge. A second parallel pipe does not change that fixed head; it only offers a second, equally-resisting path, so the flows add. This is the pipe-network dual of resistors in parallel.

Common pitfall: Assuming the parallel pipe “splits” the flow so each carries half. The split-flow rule applies when a fixed total discharge is shared; here the head is fixed, so the discharge increases instead.

Source: Mays, Water Resources Engineering, Ch. 3–4 (pipe flow, Hazen-Williams).

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