Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, December 2013 · 3 hours, closed book (one 8.5×11 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; ‘cms’ = m³/s.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, valves, network analysis); Chow, Open-Channel Hydraulics (Manning uniform flow, gutter flow); Crowe, Elger & Roberson, Engineering Fluid Mechanics (laminar film flow, 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 $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, Darcy–Weisbach, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected and water has $\rho=1000\ \text{kg/m}^3$, $\nu=1.31\times10^{-6}\ \text{m}^2/\text{s}$.
The Hazen–Williams relation is used in inverted form to get the friction loss carried by a known pipe discharge $Q$ (SI, $Q$ in m³/s):
Check (crossfall vs figure — NOTE 1 assumption). The prose gives a 2% crossfall; the figure’s edge-of-pavement elevation (99.02 m against a 100.0 m centreline over the 4 m half-width) would imply a ~24.5% crossfall, which is inconsistent. The stated 2% crossfall is adopted as the hydraulic cross-slope; the consistent crown and top-of-curb elevations (100.0 m and 100.01 m — only 0.01 m of freeboard above the crown) are used as the containment limit.
Given. A crowned road (centreline high, 2% fall to each curb); water collects in the two curb gutters. By symmetry each gutter carries half the flow.
Given data (Question 6)
Width edge-to-edge $B$
8 m (half-width 4 m)
Cross-slope $S_x$
0.02 (2%)
Manning $n$
0.013
Longitudinal slope $S_L$
0.01
Crown / top-of-curb elev.
100.00 m / 100.01 m
Flows
1.0 then 1.15 m³/s
Find. (a) water depth at $Q=1$ m³/s; (b) depth at $Q=1.15$ m³/s and whether the road contains it.
Figure 6. Crowned roadway (vertical scale exaggerated): 2% crossfall to each curb; the top of curb sits only 0.01 m above the crown.
Approach. Treat each curb gutter as a triangular Manning channel (cross-slope $S_x$, vertical curb) carrying $Q/2$. Solve for the spread $T$ and curb depth. If the required spread exceeds the 4 m half-width, the water has reached the crown and the whole pavement is analysed as one flooded section; the water level is then compared with the top of curb.
Triangular gutter geometry. For a curb depth $y$ and spread $T=y/S_x$: area $A=\tfrac{1}{2}Ty=\dfrac{y^2}{2S_x}$, wetted perimeter $P\approx T+y$, hydraulic radius $R=A/P$. Manning: $Q_{\text{gutter}}=\tfrac{1}{n}A\,R^{2/3}\,S_L^{1/2}$.
(a) Solve one gutter for $Q/2=0.5$ m³/s. Bisection gives a spread $T\approx 6.4$ m — but the half-width is only 4 m. The gutter therefore fills past the crown: the two gutters merge and the entire pavement is inundated.
Re-analyse as a flooded roadway (water surface above the crown). With crown depth $d_c$: $A=0.32+8d_c$, curb depth $y_c=d_c+S_x(4)$, $P=2\sqrt{4^2+0.08^2}+2y_c$. Solving $\tfrac{1}{n}A R^{2/3}S_L^{1/2}=1.0$ gives $d_c\approx 0.045$ m, so
$$\boxed{y_c \approx 0.125\ \text{m (depth at the curb)},\quad \text{water surface} \approx 100.045\ \text{m}.}$$
(a) Containment check. The top of curb is at 100.01 m. Since the water surface (100.045 m) already exceeds it, even $Q=1$ m³/s overtops the curbs — the road’s capacity to the top of curb is only $\approx 0.41$ m³/s.
(b) Climate-adjusted flow $Q=1.15$ m³/s. Repeating the flooded-section solution: $d_c\approx 0.053$ m, curb depth
$$\boxed{y_c \approx 0.133\ \text{m},\quad \text{water surface} \approx 100.053\ \text{m}.}$$
This is further above the 100.01 m curb top, so no — the road cannot contain the flow. It cannot even contain the present 1 m³/s; the 15% climate increase deepens the overtopping.
