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. Reservoir R1 water surface $z_{R1}=70$ m; five demand nodes on the ground at
$z=20$ m. Every pipe is identical: $C=138$, $D=406\text{ mm}=0.406$ m, $L=255$ m. Base (maximum-day) demand at each node
$q=1.5$ L/s; Node 5 fire flow $q_f=33$ L/s.
Find. (a) pressure head $p/\gamma$ at Node 4 with maximum-day demand everywhere plus the fire flow
at Node 5; (b) pressure head at Node 5 with maximum-day demand only.
Figure 1. Water supply system — branched (tree) network from reservoir R1.
Approach. The network has no loops, so every pipe flow is fixed by continuity (a pipe
carries the sum of the demands lying downstream of it); apply Hazen–Williams for each pipe head loss, walk the
hydraulic grade line (HGL) down from the reservoir along the path to the node of interest, and take
$p/\gamma = \mathrm{HGL} - z_{node}$.
Trace the topology and assign pipe flows (case a). From Figure 1 the supply path is
R1 $\xrightarrow{P1}$ N1 $\xrightarrow{P3}$ N3, then N3 branches to N4 (via $P5$) and to N2 (via $P4$),
and N2 feeds N5 (via $P7$). With the fire flow, $q_5=1.5+33=34.5$ L/s. Summing downstream demands:
$$Q_{P7}=q_5=34.5,\quad Q_{P4}=q_2+q_5=36.0,\quad Q_{P5}=q_4=1.5,$$
$$Q_{P3}=q_3+q_4+q_2+q_5=39.0,\quad Q_{P1}=q_1+Q_{P3}=40.5\ \text{L/s}.$$
Hazen–Williams head loss per pipe. Inverting $Q=0.278\,C\,D^{2.63}(h_f/L)^{0.54}$ gives
$h_f = L\left[\dfrac{Q}{0.278\,C\,D^{2.63}}\right]^{1/0.54}$. With $0.278\,C\,D^{2.63}=0.278(138)(0.406)^{2.63}=3.585$:
$$h_{f,P1}=255\!\left(\tfrac{0.0405}{3.585}\right)^{1.852}=0.063\text{ m},\quad
h_{f,P3}=255\!\left(\tfrac{0.0390}{3.585}\right)^{1.852}=0.059\text{ m},\quad h_{f,P5}=0.0001\text{ m}.$$
The 406 mm PVC mains are so smooth that friction is only centimetres.
HGL and pressure head at Node 4. Path R1→N1→N3→N4:
$$\mathrm{HGL}_{N4}=70-h_{f,P1}-h_{f,P3}-h_{f,P5}=70-0.063-0.059-0.0001=69.88\text{ m}.$$
$$p/\gamma\big|_{N4}=\mathrm{HGL}_{N4}-z_{N4}=69.88-20=\boxed{49.9\ \text{m}}.$$
Case (b): remove the fire flow, reassign flows. Now every node draws only 1.5 L/s, so
$Q_{P7}=1.5,\ Q_{P4}=3.0,\ Q_{P5}=1.5,\ Q_{P3}=6.0,\ Q_{P1}=7.5$ L/s. The corresponding losses along
R1→N1→N3→N2→N5 total only $0.0028+0.0018+0.0005+0.0001=0.005$ m, so
$$p/\gamma\big|_{N5}=70-0.005-20=\boxed{50.0\ \text{m}}.$$
