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16-Civ-A5 Hydraulic Engineering · December 2015

Question 3 of 6: Three-pipe distribution network

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 3: Three-pipe distribution network (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. Reservoir (water level 90 m) feeds node N1 through P1 and node N2 through P2; P3 links N1–N2. All pipes $D=250\text{ mm}$, $C=110$, $L=500\text{ m}$. Demands $q_{N1}=q_{N2}=5\text{ L/s}$.

Find. the flow in each of P1, P2, P3 and the pressure head at N1 and N2.

Reservoir 90 m N1 (5 L/s) N2 (5 L/s) P1P2P3
Figure 3. Single-loop distribution network. By symmetry the tie P3 carries no flow.

Approach. The two feeders are identical and the two demands are equal, so the network is symmetric about the P1↔P2 mirror; the two nodes reach the same head, which forces P3 (the tie between them) to a zero flow — the pipe-network analogue of a zero-force truss member. Each feeder then simply carries its own node's demand.

  1. Symmetry ⇒ zero-flow tie. Identical P1, P2 and equal demands give $\text{HGL}_{N1}=\text{HGL}_{N2}$. With no head difference across P3, $$\boxed{Q_{P3}=0.}$$
  2. Feeder flows by continuity. Each node draws 5 L/s and is fed only by its own feeder, so $$\boxed{Q_{P1}=Q_{P2}=5\ \text{L/s}=0.005\ \text{m}^3/\text{s}.}$$ The reservoir delivers $10$ L/s total.
  3. Friction loss in a feeder. $k=500/(0.278\cdot110\cdot0.25^{2.63})^{1.852}=759.3$, so $$h_{f,P1}=759.3\,(0.005)^{1.852}=0.042\text{ m}.$$ The loss is only $\approx 4$ cm — these lightly loaded mains are almost frictionless.
  4. Pressure heads. $\text{HGL}_{N1}=\text{HGL}_{N2}=90-0.042=89.96\text{ m}$. Node ground elevations are not given on Figure 3, so taking them at a common datum ($z=0$), $$\boxed{p/\gamma\big|_{N1}=p/\gamma\big|_{N2}\approx 89.96\text{ m (above datum).}}$$ If a node elevation $z$ is specified, subtract it: $p/\gamma = 89.96 - z$.
Check: Figure 3 shows no node elevations, so the pressure heads are reported relative to a common datum. If the exam figure carries node elevations, deduct each from the 89.96 m HGL; the flows are unaffected.
Final results — Question 3
QuantityValue
Flow in P1, P25 L/s each (0.005 m³/s)
Flow in P3 (tie)0 (symmetry)
HGL at N1, N289.96 m
Pressure head at N1, N289.96 m above datum