Question 6 of 6: Gutter/crown flow on a curbed roadway

Question 6: Gutter/crown flow on a curbed roadway (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Curbed, crowned roadway 8 m edge-to-edge (half-width 4 m); design crossfall $S_x=0.02$; asphalt $n=0.013$; longitudinal slope $S_L=0.02$. From Figure 6: centreline (crown) at 100.00 m, top of curb at 100.01 m. Flows: (a) $Q=1.3\ \text{m}^3/\text{s}$; (b) $Q=1.15(1.3)=1.495\ \text{m}^3/\text{s}$.

Find. (a) the water depth at $Q=1.3\ \text{m}^3/\text{s}$; (b) the depth at the climate-adjusted flow and whether it stays within the roadway (below top of curb).

Approach. First try each side as an independent triangular gutter carrying $Q/2$ (Izzard/Manning gutter formula) and find the spread $T$. If $T$ exceeds the 4 m half-width, water tops the crown, so re-analyse the whole width as one flooded Manning section bounded by the curbs, and compare the required water-surface elevation with the top of curb.

1. Adopt the design crossfall. The figure’s edge-of-pavement label (99.02 m) implies a $\sim\!24.5\%$ crossfall, which contradicts the stated 2%. Per exam Note 1 the design value governs: $S_x=0.02$ (so the edge of pavement sits $0.02(4)=0.08\ \text{m}$ below the crown, at 99.92 m), and the spill limit is the top of curb at 100.01 m.
2. Triangular-gutter spread at $Q/2$. Using $Q=\dfrac{0.376}{n}S_x^{5/3}S_L^{1/2}T^{8/3}$ with $Q/2=0.65\ \text{m}^3/\text{s}$, $$T=\left[\frac{Q/2}{(0.376/n)S_x^{5/3}S_L^{1/2}}\right]^{3/8}=5.79\ \text{m}\;(\gt 4\ \text{m}).$$ The spread exceeds the half-width, so the gutters submerge the crown — the triangular model no longer applies.
3. Re-analyse as one flooded section. With water depth $d$ measured over the crown, integrating the shallow tent geometry across the full 8 m and adding the two curb faces: $$A=8d+0.32\ \text{m}^2,\qquad P=8.0+2(d+0.08)\ \text{m},\qquad Q=\frac{1}{n}A\,R^{2/3}S_L^{1/2}.$$
4. Part (a): depth at $Q=1.3\ \text{m}^3/\text{s}$. Solving the flooded Manning equation, $$\boxed{d\approx 0.041\ \text{m over the crown}\;(W\!s=100.041\ \text{m}).}$$ Since $100.041\gt 100.01$ (top of curb), the water would stand $\sim\!0.031\ \text{m}$ above the curb — it cannot be held within the roadway. The capacity right up to the top of curb ($d=0.01\ \text{m}$) is only $0.58\ \text{m}^3/\text{s}$, far below 1.3.
5. Part (b): climate-adjusted $Q=1.495\ \text{m}^3/\text{s}$. Solving again, $$\boxed{d\approx 0.048\ \text{m}\;(W\!s=100.048\ \text{m}),\ \text{over the curb by}\ \sim0.038\ \text{m}.}$$ The roadway cannot contain the new flow — indeed it could not contain even the base 1.3 m³/s.