16-Civ-A5 Hydraulic Engineering · December 2016

Question 3 of 6: Five-Pipe Distribution Network with a Control Valve

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

Question 3: Five-Pipe Distribution Network with a Control 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. All five pipes are identical ($L=450$ m, $C=110$, $D=0.305$ m). From Figure 1 the topology is: source tank $R_1$ (level 67 m) feeds node $N_3$ through the parallel pair $P_1\,\|\,P_2$ and node $N_2$ through the parallel pair $P_3\,\|\,P_4$; pipe $P_5$ links $N_3$ to $N_2$. The valve is on $P_3$ (one of the two feeders to $N_2$).

Find. (a) HGL at $N_2$ and $N_3$ with $P_3$ closed; (b) whether the pressure head at $N_3$ rises or falls when $P_3$ is opened, and why.

Approach. Express every head loss with the single-pipe Hazen–Williams law $h_f = k\,Q^{1.852}$ (here $k = 259.3$ for $Q$ in m³/s). Two identical pipes in parallel between the same two nodes split the flow equally, so each carries half. Enforce continuity at $N_2,N_3$ and equal HGL at $N_2$ by the two routes from the tank; one unknown ($Q_5$ in $P_5$) closes the loop.

1. Pipe resistance. For one pipe, $h_f = L\big(Q/(0.278\,C\,D^{2.63})\big)^{1/0.54} = 259.3\,Q^{1.852}$ m ($Q$ in m³/s).
2. Continuity with $P_3$ closed. Let $Q_5$ flow $N_3\!\to\!N_2$. Then the pair to $N_3$ carries $Q_A = 0.050 + Q_5$ (split as $Q_A/2$ each), and the single live feeder $P_4$ to $N_2$ carries $Q_B = 0.200 - Q_5$.
3. Loop (equal HGL at $N_2$). The direct route ($P_4$) and the indirect route ($P_1\|P_2$ then $P_5$) must give the same $\text{HGL}_{N_2}$: $$k\,Q_B^{1.852} = k\left(\tfrac{Q_A}{2}\right)^{1.852} + k\,Q_5^{1.852}.$$ Solving for $Q_5$: $\boxed{Q_5 = 86.7\ \text{L/s}}$, hence $Q_A = 136.7$ L/s (68.3 per pipe) and $Q_B = 113.3$ L/s.
4. HGL at the demand nodes. $$\text{HGL}_{N_3} = 67 - 259.3\,(0.0683)^{1.852} = 67 - 1.80 = \boxed{65.2\ \text{m}},$$ $$\text{HGL}_{N_2} = 67 - 259.3\,(0.1133)^{1.852} = 67 - 4.60 = \boxed{62.4\ \text{m}}.$$ Cross-check via the indirect route: $65.2 - 259.3(0.0867)^{1.852} = 62.4$ m ✓.
5. Pressure heads (elevation 9 m). $p_{N_3}/\gamma = 65.2-9 = 56.2$ m; $p_{N_2}/\gamma = 62.4-9 = 53.4$ m. With $P_3$ shut, the heavily-loaded $N_2$ draws part of its 200 L/s from $N_3$ through $P_5$.
6. Part (b) — open $P_3$. Re-solving with $N_2$ now fed by the parallel pair $P_3\|P_4$ (each $Q_B/2$): $$k\left(\tfrac{Q_B}{2}\right)^{1.852} = k\left(\tfrac{Q_A}{2}\right)^{1.852} + k\,Q_5^{1.852} \;\Rightarrow\; Q_5 = 51.4\ \text{L/s}.$$ Then $Q_A = 101.4$ L/s and $\text{HGL}_{N_3} = 67 - 259.3(0.0507)^{1.852} = 65.96$ m, so $p_{N_3}/\gamma = 56.96$ m.

Answer to (b): the pressure head at $N_3$ becomes higher (56.2 m → 57.0 m). The mechanism: opening $P_3$ gives node $N_2$ a second direct feeder from the tank, roughly halving the head loss on the tank–to–$N_2$ path and raising $\text{HGL}_{N_2}$. That reduces the driving head $(\text{HGL}_{N_3}-\text{HGL}_{N_2})$ across the tie pipe $P_5$, so the flow $N_3\!\to\!N_2$ falls from 86.7 to 51.4 L/s. Less throughput to $N_2$ means the parallel feeders $P_1\|P_2$ into $N_3$ now carry only 101 L/s instead of 137 L/s; their friction loss drops from 1.8 to 1.0 m, so $\text{HGL}_{N_3}$ — and the pressure head at $N_3$ — rises. In short, relieving $N_2$’s supply unloads the pipes feeding $N_3$.