Question 1 of 6: Branched pipe network — pressure heads
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 16-Civ-A5 Hydraulic Engineering — December 2017. 3 hours, closed book (one 8.5×11 aid sheet permitted). Six questions of equal value; candidates answer any five. All six are solved here as a study resource. Unless stated, local losses and velocity head are neglected; water ρ = 1000 kg/m³, ν = 1.31×10−6 m²/s.
Reference texts: D. A. Chin, Water-Resources Engineering (3rd ed.), Ch. 2 (flow in closed conduits) & Ch. 7 (water-distribution systems); A. Osman Akan, Open Channel Hydraulics; M. Hanif Chaudhry, Open-Channel Flow (2nd ed.), Ch. 3 (hydraulic jump); Chadwick, Morfett & Borthwick, Hydraulics in Civil and Environmental Engineering.
Given. A single source feeding a branched (tree) network; the reservoir level and the identical pipe properties fix every head loss.
Given data
Reservoir R1 water level
65 m
All node ground elevations
20 m
Pipe material / Hazen-Williams C
PVC / 136
Pipe diameter D, length L (all pipes)
350 mm, 155 m
Maximum-day demand at each node N1–N5
1.5 L/s
Fire flow at Node 5
60 L/s
Find. The steady-state pressure head at Node 4 (case a, with fire flow) and at Node 5 (case b, maximum day only).
Figure 1. Branched water-supply network. R1→P1→N1→P3→N3, with N3 the interior hub feeding N4 (P5) and N2 (P7→N5 through P4). Arrows are nodal demands.
The topology below follows the printed figure.
Approach. In a tree network every pipe carries the sum of the demands downstream of it (no loops, so continuity alone fixes the flows); apply the Hazen-Williams loss to each pipe on the source-to-node path, subtract from the reservoir level to get the HGL, and take pressure head = HGL − node elevation.
Head-loss law (used throughout). Rearranging Hazen-Williams $Q = 0.278\,C\,D^{2.63}S^{0.54}$ with $S=h_f/L$ gives, per pipe,
$$h_f = L\left[\dfrac{Q}{0.278\,C\,D^{2.63}}\right]^{1/0.54}.$$
With $C=136$, $D=0.350\ \text{m}$, $L=155\ \text{m}$ this is $h_f \approx 61.4\,Q^{1.852}$ ($Q$ in $\text{m}^3/\text{s}$).
Case (a): assign pipe flows (max day + 60 L/s fire at N5). Summing downstream demands along each pipe:
$$P5 = 1.5,\quad P7 = 61.5,\quad P4 = 1.5+61.5 = 63.0,\quad P3 = 1.5+1.5+63.0 = 66.0,\quad P1 = 66.0+1.5 = 67.5\ \text{L/s}.$$
The path from R1 to Node 4 is P1 → P3 → P5.
Head losses on the R1–N4 path. Substituting each flow:
$$h_{f,P1}=30.9(0.0675)^{1.852}=0.210,\quad h_{f,P3}=30.9(0.0660)^{1.852}=0.201,\quad h_{f,P5}=30.9(0.0015)^{1.852}=0.0002\ \text{m}.$$
Pressure head at Node 4. Trace the HGL from the reservoir:
$$\text{HGL}_{N4}=65-0.210-0.201-0.0002 = 64.59\ \text{m},\qquad p/\gamma\big|_{N4}=64.59-20=\boxed{44.6\ \text{m}}.$$
Case (b): re-assign flows (max day only, no fire). Now every node draws 1.5 L/s, so
$$P7=1.5,\ P4=3.0,\ P3=6.0,\ P1=7.5\ \text{L/s},$$
and the path to Node 5 is P1 → P3 → P4 → P7. The four losses sum to only $0.0036+0.0024+0.0007+0.0002 = 0.0069\ \text{m}$.
Pressure head at Node 5.
$$\text{HGL}_{N5}=65-0.007=64.99\ \text{m},\qquad p/\gamma\big|_{N5}=64.99-20=\boxed{45.0\ \text{m}}.$$
