16-Civ-A5 Hydraulic Engineering · May 2013

Question 2 of 6: Ten-pipe network between two reservoirs

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

Notes on this paper

Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, May 2013 · 3 hours, closed book (one aid sheet). Six questions; any five constitute a complete paper — all six are solved here as a study resource. All parts of a question are of equal value.

Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, pumps, network analysis); Chow, Open-Channel Hydraulics (Manning flow, compound sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (energy/continuity). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected, and water has $\rho=1000\ \text{kg/m}^3$.

Check (Q4 data consistency): the pipe/valve data in Question 4 are internally inconsistent — the stated “initial valve flow = 400 L/s” corresponds to a node head of only 9.5 m, but the two supply pipes driven by the 96 m and 89 m tank levels deliver far more than that at 9.5 m. Continuity at the node fixes a network-consistent initial discharge of ≈693 L/s at a node head of ≈28.4 m. The simulation below is run from that physically consistent state, with the discrepancy noted (per Note 1, candidates may state assumptions).

Question 2: Ten-pipe network between two reservoirs (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. Ten identical pipes ($D=300$ mm, $L=250$ m, $C=140$) form the series–parallel network of Figure 2 between reservoir A (90 m) and B (85 m). Node ground elevations from the dashed contours: J1–J3 at 90 m, J4–J5 at 80 m, J6 at 75 m.

Find. (a) total flow A→B; (b) maximum and minimum pressure head in the system.

Figure 2. Pipe labels P1–P10. Two 2-pipe branches (P1+P4, P2+P5) run in parallel A→J4, then P8 to J6; the lower route is P3 + (P6∥P7) + P9 to J6; P10 closes to B.

Approach. Because all pipes are identical, reduce the network by series–parallel combination in the Hazen–Williams head-loss form $h_f=k\,Q^{1.852}$ (same $k$ for every pipe). Solve the two-path parallel balance for the flow split, then trace HGL from A to each node and subtract ground elevation for pressure head.

Per-pipe resistance. For $C=140$, $D=0.300$ m, $L=250$ m, $$0.278\,C\,D^{2.63}=0.278(140)(0.300)^{2.63}=1.640,\quad h_f=\frac{250}{1.640^{\,1.852}}Q^{1.852}=100\,Q^{1.852}.$$ So each pipe: $h_f=100\,Q^{1.852}$ (Q in m$^3$/s).
Reduce the two super-paths A→J6. Upper (through J4): two identical 2-pipe branches in parallel carry $Q_U/2$ each, then P8 carries $Q_U$: $$\Delta H_U=2\big(Q_U/2\big)^{1.852}+Q_U^{1.852}\ \ (\times\,100).$$ Lower (through J5): P3 carries $Q_L$, P6∥P7 carry $Q_L/2$ each, P9 carries $Q_L$: $$\Delta H_L=2\,Q_L^{1.852}+\big(Q_L/2\big)^{1.852}\ \ (\times\,100).$$
Parallel-path balance. The two paths share the same end nodes, so $\Delta H_U=\Delta H_L$, giving the split for any total $Q=Q_U+Q_L$.
Overall energy balance. $$H_A-H_B=90-85=5=100\,\Delta H_{A\to J6}+100\,Q^{1.852},$$ where the last term is P10. Solving the coupled system: $$\boxed{Q_{\text{total}}\approx 0.158\ \text{m}^3/\text{s}=158\ \text{L/s}},$$ with $Q_U\approx 87.4$ L/s (upper) and $Q_L\approx 71.1$ L/s (lower). Hence P1=P2=P4=P5≈43.7 L/s, P8≈87.4 L/s, P3=P9≈71.1 L/s, P6=P7≈35.6 L/s, P10≈158 L/s.
Trace the HGL. Starting at $H_A=90$ and subtracting $100\,Q^{1.852}$ per pipe: J1=J2=89.70, J3=89.25, J4=89.39, J5=89.05, J6=88.30 m (both routes agree; P10 then drops to 85.0 m = B).
Pressure head = HGL − ground elevation. J1,J2: $89.70-90=-0.30$; J3: $89.25-90=-0.75$; J4: $89.39-80=+9.39$; J5: $89.05-80=+9.05$; J6: $88.30-75=+13.30$ m. Therefore $$\boxed{p_{\max}\approx 13.3\ \text{m at J6},\qquad p_{\min}\approx -0.75\ \text{m at J3}.}$$

Question 2 — results
Quantity	Value
Total system flow	≈ 158 L/s (0.158 m³/s)
Upper-path / lower-path split	87.4 L/s / 71.1 L/s
Maximum pressure head	+13.3 m at node J6 (ground 75 m)
Minimum pressure head	−0.75 m at node J3 (ground 90 m)

Check: the slightly negative pressure heads at the near-A nodes (ground 90 m, HGL just below 90 m) indicate sub-atmospheric pressure there — a real design flag (air release / cavitation risk), not an arithmetic error. It arises because the ground contour equals the reservoir level while friction has already consumed a little head.

Topic. Series–parallel pipe-network reduction with Hazen–Williams resistance.

Key relations. $h_f=k\,Q^{n}$ with $n=1/0.54=1.852$; pipes in series add $h_f$ at equal $Q$; identical pipes in parallel share $Q$ equally; parallel paths carry equal $\Delta H$.

Why this works. Identical pipes let the whole graph collapse to two nested parallel paths, so a full Hardy–Cross iteration is unnecessary — the equal-$\Delta H$ condition plus one energy equation give the split directly. HGL is then a bookkeeping walk from A, and pressure head is simply HGL minus the local ground contour.

Common pitfall. Adding parallel resistances as if flows were equal in series, or forgetting that pressure head uses ground elevation (not datum). A node with high HGL can still have low pressure head if it sits on high ground.

Source. Mays, Water Resources Engineering, 3rd ed., Ch. 5 (pipe networks; equivalent pipes; Hardy–Cross).

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