Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams, December 2014 — 98-Civ-A5 Hydraulic Engineering, 3 hours, closed book (one aid sheet). Six questions of equal value (20 marks each); candidates complete any five. All six are solved here as a study resource. ‘cms’ = m³/s; water ρ = 1000 kg/m³, ν = 1.31×10−6 m²/s; local losses and velocity head neglected unless stated.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, valves, parallel/loop networks); Chow, Open-Channel Hydraulics (Manning uniform flow, critical depth, specific energy); Crowe, Elger & Roberson, Engineering Fluid Mechanics (pipe force balance, St-Venant equations). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI), Manning $Q = \tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach, and $\text{TDH} = H_s + H_f$. In inverted form the friction loss carried by a known discharge is $h_f = L\left(\dfrac{Q}{0.278\,C\,D^{2.63}}\right)^{1/0.54}$.
Given. A reservoir feeds two nodes N1 and N2 through pipes P1 and P2; a third pipe P3 links N1 to N2, forming a single loop. All three pipes are identical.
Given data (Question 3)
Pipe $D$ / $C$ / $L$ (all three)
200 mm / 100 / 500 m
Reservoir level
110 m
Nodal demand at N1, N2
1 L/s each
Find. The discharge in each pipe and the pressure (piezometric) head at N1 and N2.
Approach. Recognise the geometric and loading symmetry (P1 and P2 identical, equal demands), which fixes the split by inspection; confirm it satisfies both continuity and the loop head balance, then trace the HGL from the reservoir.
Total supply. The reservoir must supply both demands, so the flow leaving it is $Q_{P1} + Q_{P2} = 1 + 1 = 2\ \text{L/s}$.
Exploit symmetry. P1 and P2 are identical pipes carrying water from the same reservoir to symmetric nodes with equal demand. The mirror symmetry forces $Q_{P1} = Q_{P2} = 1\ \text{L/s}$ and, by continuity at N1 ($Q_{P1} = 1 + Q_{P3}$), $$\boxed{Q_{P3} = 0.}$$ P3 is a zero-flow member: with no head difference between the symmetric nodes, it carries nothing.
Loop check. The head balance around R–N1–N2–R requires $h_{f,P1} - h_{f,P2} - h_{f,P3} = 0$. Since $Q_{P1}=Q_{P2}$ gives $h_{f,P1}=h_{f,P2}$ and $Q_{P3}=0$, the balance is satisfied exactly — the symmetric split is the unique solution.
Friction and HGL. Each supply pipe carries 1 L/s: $$h_{f,P1} = 500\left(\dfrac{0.001}{0.278(100)(0.200)^{2.63}}\right)^{1/0.54} = 0.0075\ \text{m}\ (\approx 7.5\ \text{mm}).$$ The HGL at both nodes is $110 - 0.0075 = 109.99\ \text{m}$ — essentially the reservoir level, because 1 L/s through a 200 mm pipe is a trivially small flow.
Pressure head. With no node ground elevations given, take the nodes at datum; the pressure head equals the HGL, $$p/\gamma \big|_{N1} = p/\gamma \big|_{N2} = \boxed{\approx 110\ \text{m}.}$$ For any real node elevation $z$, subtract it: $p/\gamma = 109.99 - z$.
