16-Civ-A5 Hydraulic Engineering · December 2014

Question 2 of 6: Transmission main with in-line valve

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 2: Transmission main with in-line valve (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. One long main (valve 4,000 m from the intake, 1,000 m before the outfall) with an in-line control valve that dissipates head $\Delta h_{\text{valve}} = (Q/\tau E_s)^2$ across itself.

Given data (Question 2)
Upstream level $h_A$	105 m
Pipe $L$ / $D$ / $C$	5,000 m / 500 mm / 100
Valve constant $E_s$	0.35 m$^{5/2}$/s
Valve openings $\tau$	0.6 (part a), 0.3 (part b)

Find. (a) the downstream reservoir level $h_B$ at $\tau = 0.6$; (b) the discharge at $\tau = 0.3$ with $h_B$ held at the value from (a).

Upstream reservoir, 5,000 m main with an in-line valve, downstream reservoir. Head is dissipated by pipe friction (full 5,000 m) plus the valve.

Check (assumption per NOTE 1). As printed, part (a) supplies only the valve opening $\tau = 0.6$. The steady state of a series pipe–valve line has two unknowns — the discharge $Q$ and the downstream level $h_B$ — joined by a single energy balance, so one more datum is required to obtain a number. I therefore anchor part (a) on the standard economic transmission-main velocity, $V \approx 1.5\ \text{m/s}$, giving $Q_a = V\!\cdot\!A = 1.5(0.196) \approx 0.30\ \text{m}^3/\text{s}$, and state this assumption explicitly. The method below is exact; only the numerical anchor for $Q_a$ is assumed.

Approach. Write the energy balance from A to B as pipe friction over the full length plus the valve headloss; evaluate it at the anchored discharge to get $h_B$ (part a); then hold $h_B$ and re-solve the same balance for the new discharge at $\tau = 0.3$ (part b).

Governing energy balance. With the same $Q$ everywhere, $$h_A - h_B = h_{f,\text{pipe}}(Q) + \Delta h_{\text{valve}}, \qquad \Delta h_{\text{valve}} = \left(\dfrac{Q}{\tau E_s}\right)^2.$$
(a) Pipe friction at the anchored discharge. For $Q_a = 0.30\ \text{m}^3/\text{s}$ (Hazen–Williams, full 5,000 m): $$h_{f,\text{pipe}} = 5000\left(\dfrac{0.30}{0.278(100)(0.500)^{2.63}}\right)^{1/0.54} = 33.3\ \text{m}.$$
(a) Valve headloss and downstream level. At $\tau = 0.6$: $\Delta h_{\text{valve}} = \left(\dfrac{0.30}{0.6(0.35)}\right)^2 = 2.04\ \text{m}$. Therefore $$h_B = 105 - 33.3 - 2.04 = \boxed{69.6\ \text{m}.}$$
(b) Re-solve with the valve tightened. Hold $h_B = 69.6$ m, so the available head is $h_A - h_B = 35.4$ m. With $\tau = 0.3$ the balance becomes $$35.4 = 5000\left(\dfrac{Q}{0.278(100)(0.5)^{2.63}}\right)^{1/0.54} + \left(\dfrac{Q}{0.3(0.35)}\right)^2,$$ solved iteratively (the valve term now dominates more, but friction still leads).
(b) Discharge. Iteration converges to $$Q = \boxed{0.276\ \text{m}^3/\text{s}} \quad (h_{f,\text{pipe}} = 28.5\ \text{m},\ \Delta h_{\text{valve}} = 6.9\ \text{m}).$$ Halving $\tau$ (0.6 → 0.3) drops the discharge only from 0.30 to 0.276 m³/s — about 8 % — because this long, small-diameter main is friction-dominated: the valve loss (2–7 m) is small beside the 28–33 m of pipe friction, so throttling the valve is a weak control here.

Question 2 — results (with the anchored $Q_a$)
Quantity	Value
(a) Pipe friction / valve loss at $\tau=0.6$	33.3 m / 2.04 m
(a) Downstream level $h_B$	69.6 m
(b) Discharge at $\tau=0.3$ (with $h_B$ fixed)	0.276 m³/s

Topic. Series pipe–valve system between reservoirs; valve as a head-loss element.

Key relations. Energy balance $h_A - h_B = h_{f,\text{pipe}} + \Delta h_{\text{valve}}$; valve law $Q = \tau E_s\sqrt{\Delta h_{\text{valve}}}$, i.e. $\Delta h_{\text{valve}} = (Q/\tau E_s)^2$.

Why this works. In steady flow the discharge is uniform along the main, so the pipe and the valve act as two resistances in series and their head losses add to the reservoir-surface difference. The valve law behaves like an orifice: at fixed opening $\tau$, its loss grows with $Q^2$, and closing the valve (smaller $\tau$) raises the loss for a given $Q$, shifting the operating point to a lower discharge. Whether the valve is an effective control depends on how its loss compares with pipe friction — here friction dominates, so the valve is a weak throttle.

Common pitfall. Treating the valve law with the full reservoir difference (double-counting head against the pipe), or forgetting that at fixed $\tau$ the valve loss scales as $Q^2$ so the balance must be solved iteratively rather than in one pass.

Source. Mays, Water Resources Engineering, ch. on pipe systems and control valves; valve law as supplied on the exam.

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