Question 1 of 6: Branched supply network — pressure head under maximum-day and fire flow
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2018 — 16-Civ-A5, Hydraulic Engineering. Closed book (one aid sheet; approved calculator); 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. Each question 20 marks; parts of equal value.
Governing relations supplied on the exam cover sheet: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with slope $S=\Delta h/L$ (SI: $Q$ in m³/s, $D$ in m); Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$; Darcy–Weisbach $\Delta h = 0.0826\,\tfrac{fL}{D^{5}}Q^{2}$; total dynamic head $TDH=H_s+H_f$. Unless stated, local losses and velocity head are negligible, diameters are nominal, and the fluid is water ($\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$).
Question 1: Branched supply network — pressure head under maximum-day and fire flow (20 marks)
Given. A tree network fed by reservoir R1 at $H_{R1}=110\ \text{m}$; all five demand nodes at ground elevation $z=40\ \text{m}$; ductile-iron pipes with $C=110$, $D=0.405\ \text{m}$, $L=340\ \text{m}$ each. Node demands 1–5 each $1.0\ \text{L/s}$ (max day); Node 5 additionally $25\ \text{L/s}$ fire flow.
Given data
Quantity
Value
Reservoir head $H_{R1}$
110 m
Node ground elevation $z$
40 m
Diameter $D$ / roughness $C$
0.405 m / 110
Pipe length $L$ (each)
340 m
Max-day demand per node
1.0 L/s
Fire flow at Node 5
25 L/s
Find. (a) the pressure head at Node 4 under max-day + fire at Node 5; (b) the pressure head at Node 5 under max-day only.
Figure 1. Branched (tree) supply network. The fire path is R1→P1→N1→P3→N3→P4→N2→P7→N5; N4 hangs off N3 by P5.
Approach. The network is a tree, so nodal continuity fixes every pipe flow directly (no loops, no Hardy–Cross): each pipe carries the sum of the demands lying downstream of it. Trace the HGL from R1 to the target node using the Hazen–Williams loss on each path pipe; the pressure head is the local HGL minus the 40 m ground elevation.
Assign pipe flows by downstream demand. For case (a) the fire flow rides on every pipe of the supply path to N5. Summing demands beyond each pipe (in L/s):
$$Q_{P1}=30,\quad Q_{P3}=29,\quad Q_{P4}=27,\quad Q_{P7}=26,\quad Q_{P5}=1.$$
(Total $=5(1.0)+25=30\ \text{L/s}$ leaves R1.)
Write the Hazen–Williams loss per pipe. With $a\equiv 0.278\,C\,D^{2.63}=0.278(110)(0.405)^{2.63}=2.837$, inverting the cover-sheet formula gives
$$h_{f,i}=L\left(\frac{Q_i}{a}\right)^{1/0.54}=340\left(\frac{Q_i}{2.837}\right)^{1.852}\quad(Q_i\ \text{in m}^3/\text{s}).$$
Part (a): sum losses along R1→N4. The path to N4 is P1–P3–P5, carrying 30, 29 and 1 L/s:
$$h_f=0.0745+0.0700+0.0001=0.145\ \text{m}.$$
Then $HGL_{N4}=110-0.145=109.86\ \text{m}$ and
$$\boxed{p_{N4}/\gamma = HGL_{N4}-z = 109.86-40 \approx 69.9\ \text{m}.}$$
Part (b): max-day only (no fire). Now all demands are $1.0\ \text{L/s}$, so $Q_{P1}=5,\ Q_{P3}=4,\ Q_{P4}=2,\ Q_{P7}=1$ (L/s). The path to N5 is P1–P3–P4–P7:
$$h_f=0.00270+0.00179+0.00050+0.00014=0.0051\ \text{m}.$$
Hence $HGL_{N5}=110-0.005=109.99\ \text{m}$ and
$$\boxed{p_{N5}/\gamma = 109.99-40 \approx 70.0\ \text{m}.}$$
Check / physical sense. The 405 mm mains carry such low flows ($V_{P1}=Q/A=0.030/0.1288\approx0.23\ \text{m/s}$ even under fire) that friction is only centimetres over 340 m. The pressure head at every node is therefore essentially the static lift $110-40=70\ \text{m}$, and the fire flow lowers it by barely 0.1 m — the correct, if unremarkable, result for an over-sized ductile-iron grid.
