Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — 16-Civ-A5 Hydraulic Engineering — December 2017. 3 hours, closed book (one 8.5×11 aid sheet permitted). Six questions of equal value; candidates answer any five. All six are solved here as a study resource. Unless stated, local losses and velocity head are neglected; water ρ = 1000 kg/m³, ν = 1.31×10−6 m²/s.
Reference texts: D. A. Chin, Water-Resources Engineering (3rd ed.), Ch. 2 (flow in closed conduits) & Ch. 7 (water-distribution systems); A. Osman Akan, Open Channel Hydraulics; M. Hanif Chaudhry, Open-Channel Flow (2nd ed.), Ch. 3 (hydraulic jump); Chadwick, Morfett & Borthwick, Hydraulics in Civil and Environmental Engineering.
Question 6: Gutter flow on a crowned road (20 marks)
Given. A crowned pavement draining both ways to curbed gutters; Manning uniform flow along the longitudinal slope.
Given data
Pavement width (edge to edge) / half-width
8 m / 4 m each side
Crossfall from centreline $S_x$
2% (0.02)
Manning n / longitudinal slope $S_L$
0.013 / 0.015
Elevations: edge / crown (CL) / top of curb
99.92 m / 100.00 m / 100.07 m
Find. (a) the flow depth at Q = 1 m³/s; (b) the depth at Q = 1.15 m³/s and whether the roadway contains it.
Figure 6. Crowned roadway cross-section (vertical scale exaggerated). Crown at CL 100.00 m, edges at 99.92 m, top of curb 100.07 m. At these flows the spread exceeds the half-width, so water covers the crown and the full 8 m section conveys the flow.
Approach. First test the usual triangular-gutter idealisation (each side carries half the flow). If the computed spread exceeds the 4 m half-width, the two gutters have merged over the crown and the section must be solved as a single full-width Manning channel bounded by the curbs.
Triangular-gutter check (part a). Each gutter would carry $Q/2 = 0.5\ \text{m}^3/\text{s}$. For a triangular gutter of depth $y$ at the curb, spread $T=y/S_x$, and Manning $Q=\tfrac{1}{n}A R^{2/3}S_L^{1/2}$. Solving gives $y = 0.119$ m and $T = 5.94$ m, which exceeds the 4 m half-width — the water has topped the crown.
Full-width flooded section. With the crown submerged, take the water-surface rise above the crown as $d = W_s-100.0$. The bed is the shallow inverted-V pavement, so over the 8 m width
$$A = 8d + 0.32\ \text{m}^2,\qquad P = 2\sqrt{4^2+0.08^2} + 2(0.08+d)\ \text{m},$$
where 0.32 m² is the area of the two triangular gutter troughs below crown level.
Depth at Q = 1 m³/s. Solving $Q=\tfrac{1}{n}A R^{2/3}S_L^{1/2}=1.0$ gives $d = 0.036$ m, so $W_s = 100.036$ m and the maximum flow depth (at the curb line) is
$$y_{\text{curb}} = W_s - 99.92 = \boxed{0.116\ \text{m}}.$$
Depth at the climate-adjusted flow (part b). With $Q = 1.15\ \text{m}^3/\text{s}$ (a 15% increase), the same section gives $d = 0.042$ m, $W_s = 100.042$ m, and
$$y_{\text{curb}} = 100.042 - 99.92 = \boxed{0.122\ \text{m}}.$$
Containment check. The section fills to the top of curb ($W_s = 100.07$ m) at a capacity of $Q_{\max} = 1.86\ \text{m}^3/\text{s}$. Since the climate-adjusted water surface (100.042 m) sits below the top of curb (100.07 m) — and $1.15 < 1.86\ \text{m}^3/\text{s}$ — the roadway does contain the new flow, with about 0.028 m of freeboard remaining.
