Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
National Exams — 98-Civ-A5 Hydraulic Engineering, December 2015. Full worked solutions. Closed-book format, 3 hours; six questions, candidates complete any five — all six are solved here as a study resource. All questions are of equal value (20 marks); where a question has parts, each part is of equal value.
Reference texts. Mays, Water Resources Engineering, 2nd ed. (pipe systems, equivalent pipes, valves, quasi-steady network simulation); Chow, Open-Channel Hydraulics (Manning uniform flow, gutter/triangular sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (unsteady open-channel flow, St. Venant equations, kinematic vs. dynamic waves). Exam-supplied relations: Hazen–Williams $Q = 0.278\,C\,D^{2.63}\,S^{0.54}$ with $S = h_f/L$ (SI units), Manning $Q=\tfrac{1}{n}A R^{2/3}S^{1/2}$, valve law $Q=\tau E_s\sqrt{H_{u/s}-H_{d/s}}$, and Total Dynamic Head $\mathrm{TDH}=H_S+H_f$. Unless stated, local losses and velocity head are neglected, $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
Convention. Inverting Hazen–Williams gives a single-pipe resistance $h_f = k\,Q^{1.852}$ with $k=L\big/(0.278\,C\,D^{2.63})^{1.852}$ (SI, $Q$ in m³/s). Identical pipes in parallel between two nodes share the flow equally; pipes in series add their losses at a common flow. This collapses these "N identical pipe" networks by equivalent-pipe reduction, so no Hardy–Cross iteration is needed.
