Branched supply network — pressure heads: 16-Civ-A5 Hydraulic Engineering, December 2013, Q1

Question 1: Branched supply network — pressure heads (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. A single reservoir R1 feeds a tree (no loops): R1 →P1→ N1 →P3→ N3, then N3 branches N3 →P5→ N4 and N3 →P4→ N2 →P7→ N5. Every pipe is identical.

Given data (Question 1)
Reservoir head $H_{R1}$	70 m	Node elevation	20 m
Pipe $C$ / $D$ / $L$	138 / 406 mm / 255 m	Node demand (each)	1.5 L/s
Fire flow at N5	33 L/s	Local losses	neglected

Find. Pressure head at N4 (case a: max-day + fire at N5) and at N5 (case b: max-day only).

Figure 1. Branched (tree) water-supply network; each node draws a demand, N5 also a fire flow.

Approach. Because the network is a tree, each pipe’s discharge is fixed by continuity (sum of the demands downstream of it); the HGL falls from 70 m by the Hazen–Williams loss along the path to the target node, and pressure head = HGL − node elevation.

Assign pipe flows by downstream continuity (case a). With fire flow, N5 draws $1.5+33=34.5$ L/s. Summing downstream demands: $Q_{P7}=34.5$, $Q_{P4}=1.5+34.5=36.0$, $Q_{P5}=1.5$ (N4 only), $Q_{P3}=1.5+1.5+36.0=39.0$, $Q_{P1}=1.5+39.0=40.5$ L/s (= total demand, check).
Head loss along the path R1→N1→N3→N4 ($P_1,P_3,P_5$), using the inverted Hazen–Williams law with $C=138$, $D=0.406$ m, $L=255$ m: $$h_{f}=L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}=23.99\,Q^{1.852}\ \text{(m, }Q\text{ in m}^3\text{/s).}$$ Then $h_{P1}=23.99(0.0405)^{1.852}=0.063$ m, $h_{P3}=23.99(0.039)^{1.852}=0.059$ m, $h_{P5}=23.99(0.0015)^{1.852}=0.0001$ m.
HGL and pressure head at N4. $\text{HGL}_{N4}=70-0.063-0.059-0.0001=69.88$ m, so $$\boxed{p_{N4}/\gamma = 69.88-20 \approx 49.9\ \text{m}.}$$
Reassign flows for case b (no fire). All demands 1.5 L/s: $Q_{P7}=1.5$, $Q_{P4}=3.0$, $Q_{P5}=1.5$, $Q_{P3}=6.0$, $Q_{P1}=7.5$ L/s. The governing path to N5 is $P_1,P_3,P_4,P_7$.
HGL and pressure head at N5. The losses are now $0.0028+0.0018+0.0005+0.0001=0.0052$ m, so $\text{HGL}_{N5}=69.995$ m and $$\boxed{p_{N5}/\gamma = 69.995-20 \approx 50.0\ \text{m}.}$$

The 406 mm mains are so oversized for these litre-per-second demands that friction is a few centimetres even with the fire flow; the pressure head everywhere is essentially the static lift $70-20=50$ m. The fire flow lowers N4 by only about 0.1 m.

Question 1 — results
Quantity	Value
Pressure head at N4 (max-day + fire at N5)	≈ 49.9 m
Pressure head at N5 (max-day only)	≈ 50.0 m

Topic. Steady flow in a branched (dead-end) distribution network.

Key relations. Continuity at a tree pipe: $Q_{\text{pipe}}=\sum(\text{downstream demands})$; Hazen–Williams $Q=0.278\,C\,D^{2.63}(h_f/L)^{0.54}$; pressure head $=\text{HGL}-z$.

Why this works. A tree has no loops, so no Hardy–Cross iteration is needed — each pipe carries exactly the water demanded beyond it. The energy grade line drops monotonically from the reservoir by the accumulated friction, and the local pressure is whatever head remains above the node’s ground elevation.

Common pitfall. Forgetting to add the fire flow into every pipe on the supply path to N5 (not just the last one), or subtracting the wrong node elevation. Here the demands are tiny, so a student who mis-assigns them still gets ~50 m and cannot tell — always trace the tree explicitly.

Source. Mays, Water Resources Engineering, 3rd ed., §3 (pipe-network hydraulics); AWWA fire-flow provisioning.

Question 1 of 6: Branched supply network — pressure heads

Question 1: Branched supply network — pressure heads (20 marks)