Question 1 of 6: Two-reservoir supply to a demand node
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, May 2013 · 3 hours, closed book (one aid sheet). Six questions; any five constitute a complete paper — all six are solved here as a study resource. All parts of a question are of equal value.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, pumps, network analysis); Chow, Open-Channel Hydraulics (Manning flow, compound sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (energy/continuity). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected, and water has $\rho=1000\ \text{kg/m}^3$.
Check (Q4 data consistency): the pipe/valve data in Question 4 are internally inconsistent — the stated “initial valve flow = 400 L/s” corresponds to a node head of only 9.5 m, but the two supply pipes driven by the 96 m and 89 m tank levels deliver far more than that at 9.5 m. Continuity at the node fixes a network-consistent initial discharge of ≈693 L/s at a node head of ≈28.4 m. The simulation below is run from that physically consistent state, with the discrepancy noted (per Note 1, candidates may state assumptions).
Question 1: Two-reservoir supply to a demand node (20 marks)
Given. Two reservoirs feed one demand node through separate pipes; the node draws 20 L/s.
Given data
Quantity
Pipe 1 (Res 1→node)
Pipe 2 (Res 2→node)
Reservoir water level
96 m
91 m
Length $L$
200 m
300 m
Diameter $D$
200 mm
200 mm
Hazen–Williams $C$
120
120
Node demand
20 L/s
Find. The flow in each pipe and the pressure head (HGL elevation, node taken as datum) at the demand node.
Figure 1. Two reservoirs feeding a common demand node; the HGL sits between the two water levels.
Approach. Let $h$ be the HGL elevation at the node. Each pipe’s flow follows from Hazen–Williams with $h_f$ = (reservoir level $-\,h$); node continuity closes the system for the single unknown $h$.
Set the pipe conductance. With $C=120$, $D=0.200\ \text{m}$, the exam form gives a common coefficient $$0.278\,C\,D^{2.63}=0.278(120)(0.200)^{2.63}=0.4844,$$ so each pipe carries $Q_i=0.4844\,(h_{f,i}/L_i)^{0.54}$ (SI, $Q$ in m$^3$/s).
Identify the flow regime. If both reservoirs supplied the node, Pipe 1 alone would deliver >60 L/s, far exceeding the 20 L/s demand. Hence the node HGL lies between the two levels: Reservoir 1 feeds the node, and the node feeds Reservoir 2 (i.e. $91\ \text{m}\lt h\lt 96\ \text{m}$).
Write node continuity. Inflow from Res 1 = demand + outflow to Res 2: $$Q_1=Q_{\text{demand}}+Q_2\ \Rightarrow\ 0.4844\Big(\tfrac{96-h}{200}\Big)^{0.54}=0.020+0.4844\Big(\tfrac{h-91}{300}\Big)^{0.54}.$$
Solve for the node HGL. Iterating (Newton/bisection) gives $$\boxed{h\approx 92.9\ \text{m}}.$$
Back-substitute the pipe flows. $$Q_1=0.4844\Big(\tfrac{96-92.9}{200}\Big)^{0.54}=0.0512\ \text{m}^3/\text{s}=51.2\ \text{L/s},$$ $$Q_2=0.4844\Big(\tfrac{92.9-91}{300}\Big)^{0.54}=0.0312\ \text{m}^3/\text{s}=31.2\ \text{L/s (node}\rightarrow\text{Res 2)}.$$ Check: $Q_1-Q_2=51.2-31.2=20.0\ \text{L/s}$, matching the demand.
State the pressure head. Taking the node as datum ($z=0$), the pressure head equals the node HGL, $h\approx 92.9\ \text{m}$.
