Branched Distribution Network — Pressure Heads: 16-Civ-A5 Hydraulic Engineering, May 2014, Q1

Question 1: Branched Distribution 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. Reservoir R1 water level 110 m; all five demand nodes at ground elevation 45 m; every pipe ductile iron with $C=115$, $D=452\ \text{mm}=0.452\ \text{m}$, $L=200\ \text{m}$; base demand $1.0\ \text{L/s}$ at each of N1–N5; fire flow $33\ \text{L/s}$ added at N5.

Find. Steady-state pressure head at Node 4 (max day + fire) and at Node 5 (max day only).

Figure 1. Branched (tree) water-supply network. R1 feeds N1; the tree splits at N3 to N4 and to N2→N5. Arrows are nodal demands.

Approach. The network is a tree, so each pipe carries the sum of the demands downstream of it; compute each pipe head loss by Hazen–Williams and subtract cumulatively along the path from the reservoir, then pressure head = HGL − node elevation.

Assign pipe flows by downstream demand (Case a, fire at N5). Tracing the tree, $Q_{P7}=Q_{N5}=1+33=34\ \text{L/s}$, $Q_{P4}=Q_{N2}+Q_{N5}=35\ \text{L/s}$, $Q_{P5}=Q_{N4}=1\ \text{L/s}$, $Q_{P3}=Q_{N3}+Q_{P4}+Q_{P5}=37\ \text{L/s}$, and $Q_{P1}=Q_{N1}+Q_{P3}=38\ \text{L/s}$ (equal to the total draw $5\times1+33$).
Head loss per pipe. Inverting Hazen–Williams, $h_f=L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}$. With $0.278\,C\,D^{2.63}=3.86$ for this pipe, the path to Node 4 gives $h_{f,P1}=0.037\ \text{m}$ (38 L/s) and $h_{f,P3}=0.035\ \text{m}$ (37 L/s); $P5$ carries only 1 L/s so its loss is negligible ($\sim6\times10^{-5}\ \text{m}$).
HGL march to Node 4. $H_{N1}=110-0.037=109.96\ \text{m}$, $H_{N3}=109.96-0.035=109.93\ \text{m}$, $H_{N4}=H_{N3}-h_{f,P5}\approx109.93\ \text{m}$.
Pressure head at Node 4 (a). Subtract the ground elevation: $p_{N4}/\gamma=H_{N4}-45=\boxed{64.93\ \text{m}}$.
Case b flows (no fire). Now all demands are 1 L/s: $Q_{P7}=1$, $Q_{P4}=2$, $Q_{P3}=4$, $Q_{P1}=5\ \text{L/s}$. These are so small that every head loss is at the millimetre level ($\sim1\text{--}2\ \text{mm}$ total to N5).
Pressure head at Node 5 (b). $H_{N5}=110-\sum h_f\approx110.00\ \text{m}$, so $p_{N5}/\gamma=H_{N5}-45=\boxed{65.00\ \text{m}}$.

Question 1 results
Quantity	Value
a) Pressure head at Node 4 (max day + fire)	64.93 m
b) Pressure head at Node 5 (max day, no fire)	65.00 m
Governing pipe flow $Q_{P1}$ (case a / case b)	38 / 5 L/s

Check: the 452 mm mains are enormously oversized for these flows, so friction is a few centimetres at most and the pressure head is essentially the static lift $110-45=65\ \text{m}$ at every node. The fire flow at N5 only drops Node 4 by $\sim0.07\ \text{m}$.

Topic: steady-state analysis of a branched (dead-end) distribution network.

Key relations: Hazen–Williams $Q=0.278\,C\,D^{2.63}(h_f/L)^{0.54}$; conservation of mass at each node; HGL bookkeeping $p/\gamma = \text{HGL}-z$.

Why this works: a tree has no loops, so continuity alone fixes every pipe flow—each pipe carries exactly the demands lying beyond it. No Hardy–Cross iteration is needed. Once flows are known the head losses are explicit and the HGL is found by marching outward from the fixed reservoir level. Pressure head is the HGL above the local ground; on lightly loaded large mains it approaches the static difference.

Common pitfall: forgetting to add the fire flow to every pipe on the supply path to N5, or subtracting the wrong (upstream vs. node) elevation. Here the elevations are equal so only the reservoir head and friction matter.

Source: Mays, Water Resources Engineering, Ch. 3 (pipe systems and networks).

Question 1 of 6: Branched Distribution Network — Pressure Heads

Question 1: Branched Distribution Network — Pressure Heads (20 marks)