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16-Civ-A5 Hydraulic Engineering · December 2015

Question 2 of 6: Valve-controlled transmission main

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

Notes on this paper

National Exams — 98-Civ-A5 Hydraulic Engineering, December 2015. Full worked solutions. Closed-book format, 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. All questions are of equal value (20 marks); where a question has parts, each part is of equal value.

Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe systems, equivalent pipes, valves, quasi-steady network simulation); Chow, Open-Channel Hydraulics (Manning uniform flow, gutter/triangular sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (unsteady open-channel flow, St. Venant equations, kinematic vs. dynamic waves). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI units), Manning $Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, valve law $Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$, and Total Dynamic Head $\mathrm{TDH}=H_S+H_f$. Unless stated, local losses and velocity head are neglected, $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.

Convention. Inverting Hazen–Williams gives a single-pipe resistance $h_f = k\,Q^{1.852}$ with $k=L\big/(0.278\,C\,D^{2.63})^{1.852}$ (SI, $Q$ in m³/s). Identical pipes in parallel between two nodes share the flow equally; pipes in series add their losses at a common flow. This collapses these "N identical pipe" networks by equivalent-pipe reduction, so no Hardy–Cross iteration is needed.

Question 2: Valve-controlled transmission main (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. $L=5000\text{ m}$, $C=120$, $D=450\text{ mm}=0.45\text{ m}$, $h_A=105\text{ m}$, $E_s=0.45\ \text{m}^{5/2}/\text{s}$. Part (a): $h_B=95\text{ m}$, $\tau=0.7$. Part (b): $h_B=85\text{ m}$, hold $Q$ at the part-(a) value.

Find. (a) the discharge $Q$; (b) the new valve setting $\tau$ and whether it opens or closes.

h_A=105 h_B valve τ 4,000 m 1,000 m L = 5,000 m, D = 450 mm, C = 120
Figure 2. Transmission main with an in-line valve; the valve loss is in series with the full-length pipe friction.

Approach. The pipe friction (over the whole 5,000 m) and the valve loss are in series and share the same $Q$, so the total available head equals their sum: $h_A-h_B = k\,Q^{1.852}+\Delta H_{\text{valve}}$ with $\Delta H_{\text{valve}}=\left(Q/(\tau E_s)\right)^2$.

  1. Pipe resistance. $$k=\frac{5000}{(0.278\cdot120\cdot0.45^{2.63})^{1.852}}=369.1\ \text{(SI)},\qquad h_{f}=369.1\,Q^{1.852}.$$
  2. (a) Energy balance. With $h_A-h_B = 105-95 = 10\text{ m}$ and $\Delta H_{\text{valve}}=(Q/(0.7\cdot0.45))^2$, $$369.1\,Q^{1.852} + \left(\frac{Q}{0.315}\right)^2 = 10.$$ Solving iteratively, $$\boxed{Q = 0.141\ \text{m}^3/\text{s}\approx 141\ \text{L/s}.}$$ Here friction dominates: $h_f=9.80\text{ m}$ against only $\Delta H_{\text{valve}}=0.20\text{ m}$ (this 5,000 m / 450 mm line is friction-controlled).
  3. (b) Hold $Q$, re-balance. $Q$ unchanged means $h_f$ unchanged at 9.80 m, but the available head is now $105-85=20\text{ m}$, so the valve must dissipate the rest: $$\Delta H_{\text{valve,new}} = 20 - 9.80 = 10.2\text{ m}.$$
  4. New valve setting. From $Q=\tau E_s\sqrt{\Delta H_{\text{valve}}}$, $$\tau=\frac{Q}{E_s\sqrt{\Delta H_{\text{valve,new}}}}=\frac{0.141}{0.45\sqrt{10.2}}=\boxed{0.098\approx 0.10.}$$ The downstream drop increased the driving head, which would otherwise raise $Q$; to keep $Q$ fixed the valve must absorb far more head, so it is closed further (from $\tau=0.7$ to $\approx0.10$).
Final results — Question 2
QuantityValue
(a) Discharge at $\tau=0.7$0.141 m³/s (141 L/s)
(a) Pipe friction / valve loss9.80 m / 0.20 m
(b) New valve loss to hold $Q$10.2 m
(b) New valve setting $\tau$0.098 (valve closes further)