Question 1 of 6: Branched supply network — pipe length and pressure under fire flow

Question 1: Branched supply network — pipe length and pressure under fire flow (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 water level $H_{R1}=65\ \text{m}$; all demand nodes at ground elevation $z=20\ \text{m}$; PVC pipes $C=140$, $D=0.502\ \text{m}$, all of one common length $L$. Demands: nodes 1–5 each $1.5\ \text{L/s}$ (max day); node 5 additionally $20\ \text{L/s}$ fire flow.

Find. (a) the common pipe length $L$ that yields a 40 m pressure head at Node 5 under max-day + fire; (b) the pressure head at Node 4 when $C$ is lowered to 130 (same $L$, same flows).

Approach. The network is a tree, so each pipe carries the sum of the demands downstream of it (no loops, no Hardy–Cross). Trace the hydraulic grade line (HGL) from R1 along the path to the target node; the total Hazen–Williams friction loss along that path fixes the unknown length in (a), and re-evaluating the same losses with the lower $C$ gives the Node-4 pressure in (b).

1. Assign pipe flows by downstream demand. With total demand $\sum=5(1.5)+20=27.5\ \text{L/s}$, each pipe carries what lies beyond it toward the leaves: $$Q_{P1}=27.5,\quad Q_{P3}=26.0,\quad Q_{P4}=23.0,\quad Q_{P7}=21.5,\quad Q_{P5}=1.5\ \ (\text{L/s}).$$ The fire flow rides on every pipe of the supply path P1–P3–P4–P7 to N5.
2. Write the Hazen–Williams loss per pipe. Inverting $Q=0.278\,C\,D^{2.63}S^{0.54}$ with $a\equiv 0.278\,C\,D^{2.63}=0.278(140)(0.502)^{2.63}=6.354$, $$h_{f,i}=\left(\frac{Q_i}{a}\right)^{1/0.54} L .$$
3. Impose the Node-5 head constraint (part a). The pressure head at N5 is 40 m, so $HGL_{N5}=z+40=60\ \text{m}$ and the friction available from R1 is $\Delta H = 65-60 = 5\ \text{m}$ over the four path pipes: $$L\sum_{P1,P3,P4,P7}\!\left(\frac{Q_i}{a}\right)^{1/0.54}=5 .$$ Evaluating the sum $=1.3651\times10^{-4}\ \text{m}^{-1}$ gives $$\boxed{L=\frac{5}{1.3651\times10^{-4}}\approx 3.66\times10^{4}\ \text{m}\;(\approx 36.6\ \text{km per pipe}).}$$
4. Re-evaluate to Node 4 with $C=130$ (part b). The demands (hence flows) are unchanged; only friction rises. The path to N4 is P1–P3–P5, so $$\Delta H_{N4}=\sum_{P1,P3,P5} \left(\frac{Q_i}{0.278(130)D^{2.63}}\right)^{1/0.54}L = 3.36\ \text{m}.$$ Then $HGL_{N4}=65-3.36=61.64\ \text{m}$ and $$\boxed{p_{N4}/\gamma = HGL_{N4}-z = 61.64-20 \approx 41.6\ \text{m}.}$$ (For reference, the same $C=130$ drops Node 5 to $65-5.74-20=39.3\ \text{m}$, just below its design 40 m.)