Question 4 of 6: Three-pipe network — flows and nodal pressure heads
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 98-Civ-A5 Hydraulic Engineering — May 2016.
Closed book; one aid sheet and a non-communicating calculator permitted. Six questions of equal value (20 marks each);
candidates answer any five. All six are solved here as a study resource. Exam-supplied relations: Hazen–Williams
$Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI, $Q$ in m$^3$/s, $D,L$ in m), Manning
$Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, Darcy–Weisbach $\Delta h = 0.0826\,f\,L\,Q^2/D^5$, valve law
$Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$, and $\mathrm{TDH}=H_s+H_f$. Water: $\rho=1000$ kg/m$^3$,
$\nu=1.31\times10^{-6}$ m$^2$/s.
Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe
networks, branched systems, valves, transmission mains); Chow, Open-Channel Hydraulics (Manning uniform flow,
critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance,
laminar film flow).
Given. Three identical pipes ($D=200\text{ mm}=0.200$ m, $C=100$, $L=500$ m); a source
reservoir at 110 m feeding two demand nodes; each nodal demand $q=1$ L/s.
Find. The flow in every pipe and the pressure head at each node.
Per Note 1, the solution adopts the
standard three-pipe layout that this exam series uses (reservoir feeds two nodes N1 and N2 by pipes $P1$ and $P2$, with a
third pipe $P3$ linking N1–N2). Node elevations are taken at a common ground datum $z$; absolute pressure heads
follow by subtracting the actual $z$ once known.
Three-pipe network: reservoir feeds N1 and N2; the N1–N2 link P3 carries no flow by symmetry.
Approach. Exploit the mirror symmetry of the layout: identical pipes and identical demands
make the two node heads equal, which forces the connecting pipe to carry zero flow (the pipe-network analogue of a
zero-force truss member). Continuity then hands each feeder its own node’s demand, and Hazen–Williams gives the
(tiny) friction drop.
Symmetry argument. $P1$ and $P2$ are identical and each node draws the same 1 L/s, so by mirror
symmetry $H_{N1}=H_{N2}$. The flow in the link obeys $Q_{P3}\propto\mathrm{sign}(H_{N1}-H_{N2})$, hence
$$\boxed{Q_{P3}=0}.$$
Continuity at each node. With no flow through $P3$, each feeder alone supplies its node:
$$Q_{P1}=Q_{P2}=q=\boxed{1.0\ \text{L/s}},\qquad Q_{reservoir}=2.0\ \text{L/s}.$$
Friction drop on a feeder. Hazen–Williams at 1 L/s on a 200 mm/500 m pipe:
$$h_{f}=500\!\left[\frac{0.001}{0.278(100)(0.200)^{2.63}}\right]^{1/0.54}=0.0075\text{ m}\approx7.5\text{ mm}.$$
This is negligible, so the hydraulic grade line sits essentially at the reservoir level:
$\mathrm{HGL}_{N1}=\mathrm{HGL}_{N2}=110-0.0075\approx109.99$ m.
Pressure heads. $p/\gamma=\mathrm{HGL}-z_{node}$. Taking the nodes at the ground datum $z$:
$$p/\gamma\big|_{N1}=p/\gamma\big|_{N2}=110-z\ \text{(m)}\ \approx\ 110\ \text{m above the node datum}.$$
If the figure’s node elevations were, say, $z\approx90$–95 m (typical of these systems), the pressure heads
would be $\approx15$–20 m; the friction correction ($<1$ cm) is immaterial to that reading.
