Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
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.
Given. Crowned road, 8 m edge-to-edge (half-width $B=4$ m each side), crossfall $S_x=2\%=0.02$ from the centreline crown, longitudinal slope $S_0=0.01$, $n=0.013$. Crown at elevation 100.00 m and top-of-curb at 100.01 m (0.01 m freeboard); adopting $S_x=2\%$, each edge of pavement sits at $100.00-0.02(4)=99.92$ m. Flows: (a) $Q=0.8\ \text{m}^3/\text{s}$; (b) $Q=1.15(0.8)=0.92\ \text{m}^3/\text{s}$.
Find. the water depth at each flow, and whether the flow stays below the top of curb.
Check (figure vs. stated slope): The figure's edge-of-pavement elevation (99.02 m against a 100.00 m crown) implies a $\approx24.5\%$ crossfall, which contradicts the stated 2%. The problem statement governs, so $S_x=0.02$ is used and the top of curb (100.01 m) is taken as the spill limit.
Figure 6. Crowned section: water gathers in both gutters and, once the spread reaches the crown, floods across the whole 8 m as one section bounded by the two curbs.
Approach. Check whether each triangular gutter's spread reaches the 4 m half-width. It does, so the crown is submerged and the section is treated as one crowned channel bounded by the two curbs; Manning's equation on that section, with depth-over-crown $h$ as the unknown, gives the water level, which is compared with the top of curb.
Gutter check. A single triangular gutter reaches spread $T=4$ m (crown) at only $Q_{\text{side}}=\tfrac{0.375}{n}S_x^{5/3}S_0^{1/2}T^{8/3}=0.17\ \text{m}^3/\text{s}$ (0.34 m³/s for both sides). Both design flows exceed this, so the crown is overtopped and the whole width conveys water.
Full crowned section. With depth-over-crown $h$, the flow area (both sides) is $A=8h+0.32$ and the wetted perimeter $P=8.0+2(0.08+h)$; Manning gives $Q=\tfrac{1}{n}A\left(\tfrac{A}{P}\right)^{2/3}S_0^{1/2}$.
Capacity to top of curb. At $h=0.01$ m (water surface at 100.01 m), $A=0.40\ \text{m}^2$, $P=8.18\ \text{m}$, $R=0.049\ \text{m}$, so $$Q_{\text{cap}}=\frac{1}{0.013}(0.40)(0.049)^{2/3}(0.01)^{1/2}=0.41\ \text{m}^3/\text{s}.$$ The roadway can hold only about 0.41 m³/s before spilling over the curb.
(a) Depth at $Q=0.8$ m³/s. Solving $Q(h)=0.8$ gives $$\boxed{h=0.035\text{ m over the crown, depth at curb }\approx0.115\text{ m, water surface }100.035\text{ m}.}$$ Since $0.8\gt0.41$, the water surface already sits above the top of curb.
(b) Depth at $Q=0.92$ m³/s and containment. Solving $Q(h)=0.92$ gives $$\boxed{h=0.041\text{ m, depth at curb }\approx0.121\text{ m, water surface }100.041\text{ m}.}$$ Because the required capacity ($0.92\ \text{m}^3/\text{s}$) far exceeds the top-of-curb capacity ($0.41\ \text{m}^3/\text{s}$), $$\boxed{\text{the road cannot contain the climate-adjusted flow — water overtops the curbs.}}$$
