Question 3 of 6: Looped distribution network — supply head and valving

Question 3: Looped distribution network — supply head and valving (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 R1 feeds nodes N1–N4 through six identical pipes ($D=0.203\text{ m}$, $C=130$, $L=1{,}500\text{ m}$). Connectivity (Figure 3): P1: R1–N1; P2: N1–N2; P3: N2–N4; P4: N1–N3; P5: N1–N4 (diagonal); P6: N3–N4.

Find. (a) the reservoir water surface elevation for 25 m of pressure head at N3; (b) the direction of change in that pressure head when P5 is closed; (c) a qualitative assessment of quality, reliability and fire protection at N4 when P5 and P6 are both closed.

Approach. Total demand ($5+10+5+30 = 50$ L/s) leaves R1 through P1. The two loops make the network statically indeterminate in the flows, so solve nodal continuity with Hazen–Williams pipe laws ($h_f=kQ^{1.852}$, $k=4610$ SI) for the unknown heads, pinning $H_{N3}$ at the required value.

1. (a) Fix the target head at N3. Required head $H_{N3}=z_{N3}+p/\gamma=20+25=45\ \text{m}$.
2. (a) Nodal continuity. For each pipe $Q_{ij}=\operatorname{sign}(H_i-H_j)\,(|H_i-H_j|/k)^{1/1.852}$. Writing $\sum Q_{\text{in}}=\text{demand}$ at N1, N2, N3, N4 gives four equations in the unknowns $H_{R1},H_{N1},H_{N2},H_{N4}$ (with $H_{N3}=45$). Solving simultaneously: $H_{N1}=46.6$, $H_{N2}=44.6$, $H_{N4}=44.3$ m.
3. (a) Reservoir level. The head required at the source to deliver 50 L/s through P1 and hold N3 at 45 m is $$\boxed{H_{R1}=64.5\ \text{m}.}$$ The resulting pipe flows are P1 = 50, P2 = 15.2, P3 = 5.2, P4 = 13.5, P5 = 16.3, P6 = 8.5 L/s, and the pressure heads are N1 31.6, N2 29.6, N3 25.0, N4 29.3 m — N3 is the controlling (lowest-pressure) node because it sits 5 m higher than the others.
4. (b) Close pipe 5, hold the reservoir at 64.5 m. Re-solving the network with $Q_{P5}=0$ gives $H_{N3}=42.8$ m, i.e. a pressure head of $$\boxed{p/\gamma\big|_{N3}=22.8\ \text{m}\ (\text{a decrease of }\sim2.2\ \text{m}).}$$ The pressure head at N3 decreases. Physically, P5 is the direct N1→N4 feed carrying 16 L/s of N4’s 30 L/s demand. Removing it forces that flow to detour through the P4–P6 path (N1→N3→N4) and the P2–P3 path. The extra discharge now pushed through P4 raises its friction loss, so the head delivered to N3 drops — less head is left over as pressure.
5. (c) Close P5 and P6 — N4 becomes a dead end. With both closed, N4 is fed only by P3 from N2 (a single branch line). The hydraulic consequences at N4 are:
   • Water quality. A dead-end branch has no through-circulation, so water age rises sharply; the disinfectant (chlorine) residual decays with detention time, and low velocities let sediment settle and biofilm grow — taste, odour and bacteriological problems follow. Looped networks avoid this by keeping water moving.
   • Reliability. N4 now has a single feed (P3). Any break, repair, or closure on P2 or P3 isolates N4 completely — there is no redundant path. The looped configuration (P3, P5, P6) normally provides two or three independent routes.
   • Fire protection. Fire flow demands are large and brief. A single 203 mm branch has high friction at fire flow ($h_f\propto Q^{1.852}$), so the residual pressure at N4 collapses and the required fire flow cannot be delivered at code-minimum residual pressure. Looping supplies fire flow from several directions simultaneously, keeping friction — and pressure loss — low.