07-Str-B11 · December 2015

Question 6 of 6: Roadway Cross-Section Capacity under a Climate-Adjusted Flood

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format. National Exams, December 2015 — 07-Str-B11, Hydraulic Engineering. Three hours, closed book; one 8.5 × 11 inch aid sheet (both sides) and any non-communicating calculator are permitted. Six questions of twenty marks each; candidates complete any five, and where a question has parts the parts carry equal weight. The solutions below work all six, so the paper can be used whichever five a reader chooses.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, series/parallel equivalent pipes, looped-network analysis and extended-period simulation; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, service-pressure criteria, control-valve characteristics and gutter/roadway hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and compound cross-sections; Henderson, F.M., Open Channel Flow (Macmillan) — the Saint-Venant equations, wave celerity and the kinematic/diffusion/dynamic hierarchy; Transportation Association of Canada, Geometric Design Guide for Canadian Roads — crossfall, curb reveal and roadway drainage conventions.

Notation and conventions used throughout. The paper supplies the SI Hazen-Williams form $Q = 0.278\,C\,D^{2.63}S^{0.54}$ with $Q$ in m³/s, $D$ in metres and $S = h_f/L$; it is used exactly as printed. Writing $K = 0.278\,C\,D^{2.63}$ inverts it to the head-loss form $h_f = L\,(Q/K)^{1.852}$, the exponent $n = 1.852$ being the value the paper itself quotes in its loop-correction note. Following Note 6, local losses and velocity head are neglected, so the hydraulic grade line (HGL) and the energy grade line coincide and “pressure head at a node” means $p/\gamma = \mathrm{HGL} - z$. Water is taken as $\rho = 1000$ kg/m³ with $\nu = 1.31\times10^{-6}$ m²/s.

Question 6: Roadway Cross-Section Capacity under a Climate-Adjusted Flood (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. A crowned roadway 8 m from curb to curb, draining to gutters on both sides, treated as a single open channel carrying uniform flow at the longitudinal grade.

Given data — Question 6
Quantity	Symbol	Value
Pavement width, edge to edge	$W$	8 m (half-width 4 m each side of the crown)
Crossfall from the centreline	$S_x$	2 % = 0.02
Manning’s roughness, asphalt	$n$	0.013
Longitudinal roadway slope	$S_0$	0.01
Centreline (crown) elevation	$z_{CL}$	100.00 m
Top of curb elevation	$z_{curb}$	100.01 m
Edge of pavement elevation (derived)	$z_{EP}$	100.00 − 4(0.02) = 99.92 m
Present flood flow	$Q_a$	0.8 m³/s
Climate-adjusted flood flow	$Q_b$	1.15 × 0.8 = 0.92 m³/s

Find. (a) the water depth at the curb face for 0.8 m³/s; (b) the depth for the 15 %-increased flow, and a judgement on whether the roadway section contains it.

Check: the figure’s edge-of-pavement label contradicts the stated geometry. Figure 6 of the paper annotates the edge of pavement as 99.02 m, but the question text fixes it: a crown at 100.00 m with a 2 % crossfall over a 4 m half-width puts the edge at 100.00 − 0.08 = 99.92 m. The label of 99.02 m would imply a crossfall of 24.5 % and a curb reveal of 0.99 m, neither of which is a road. The stated geometry is adopted throughout, giving a 0.09 m curb reveal. This matters: under the figure’s label the depths would be 0.32 m and 0.34 m and both flows would be contained, whereas under the question’s own geometry neither is. The opposite-reading results are given at the end of the working so a reader can see both.

Figure 6.1 — The roadway section drawn to the geometry the question states: crown at 100.00 m, edge of pavement at 99.92 m and a 0.09 m curb reveal. The water surface required to pass 0.8 m³/s stands at 100.035 m, 25 mm above the top of curb, so the flow escapes the section rather than being contained by it.

Approach. Build the stage–discharge relation for the crowned section from Manning’s equation, noting that the geometry changes character once the water rises above the crown; find the containment capacity first so the answers can be interpreted, then invert the relation for each of the two flows.

Fix the geometry. With the crown at 100.00 m and a 2 % fall over each 4 m half-width, the gutter invert at each curb face is at 99.92 m and the top of curb is at 100.01 m. Two depths therefore matter, both measured at the curb face: $$d_{\mathrm{crown}} = 4.0 \times 0.02 = 0.08\ \mathrm{m} \quad(\text{water surface level with the crown})$$ $$d_{\mathrm{curb}} = 100.01 - 99.92 = 0.09\ \mathrm{m} \quad(\text{water surface level with the top of curb})$$ Below 0.08 m the section is two independent triangular gutters; between 0.08 m and 0.09 m the crown is drowned and the section is a single channel spanning the full 8 m; above 0.09 m the water spills over the curbs.
Write the section properties. For a depth $d$ at the curb face, with $d > 0.08$ m so that the crown is submerged, the area is the two gutter triangles plus a rectangle of the full width, and the wetted perimeter is the two pavement surfaces plus the two curb faces: $$A = 2\left(\frac{d_{\mathrm{crown}}^{2}}{2S_x}\right) + W\left(d - d_{\mathrm{crown}}\right) = 0.32 + 8\left(d - 0.08\right)$$ $$P = 2\sqrt{4.0^{2} + 0.08^{2}} + 2d = 8.0016 + 2d$$ Manning’s equation then gives the discharge at that stage, $$Q = \frac{1}{n}A\left(\frac{A}{P}\right)^{2/3}\sqrt{S_0} = \frac{\sqrt{0.01}}{0.013}\,A\left(\frac{A}{P}\right)^{2/3} = 7.6923\,\frac{A^{5/3}}{P^{2/3}}$$
Establish what the roadway can actually hold. Before answering either part, evaluate the two threshold stages. At $d = 0.08$ m the two gutters are just full and the crown is on the point of drowning: $$A = 0.320\ \mathrm{m^2}, \quad P = 8.16\ \mathrm{m}, \quad R = 0.0392\ \mathrm{m} \;\Longrightarrow\; Q = 0.284\ \mathrm{m^3/s}$$ At $d = 0.09$ m the water is level with the top of curb, which is the largest flow the section can carry without spilling: $$A = 0.400\ \mathrm{m^2}, \quad P = 8.18\ \mathrm{m}, \quad R = 0.0489\ \mathrm{m}$$ $$\boxed{Q_{\mathrm{capacity}} = 0.411\ \mathrm{m^3/s}}$$ Both design flows in this question are roughly twice that figure, which frames everything that follows.
Solve part (a) for $Q$ = 0.8 m³/s. Inverting the stage–discharge relation of Step 2 by trial gives $A$ = 0.5976 m² and $P$ = 8.231 m at $$\boxed{d_a = 0.115\ \mathrm{m\ at\ the\ curb\ face}}$$ Checking: $R = 0.5976/8.231 = 0.0726$ m, $R^{2/3} = 0.1738$, and $Q = 76.923 \times 0.5976 \times 0.1738 \times 0.1 = 0.800$ m³/s as required. The corresponding water-surface elevation is $99.92 + 0.115 = 100.035$ m, which is 25 mm above the top of curb. So the 0.115 m is the normal depth the section would require, and the section cannot in fact provide it: at 0.8 m³/s the road is already overtopping its curbs.
Solve part (b) for the climate-adjusted flow. The design flow rises by 15 % to $Q_b = 1.15 \times 0.8 = 0.92$ m³/s. Repeating the inversion gives $A$ = 0.6503 m², $P$ = 8.244 m and $$\boxed{d_b = 0.121\ \mathrm{m\ at\ the\ curb\ face}}$$ a water-surface elevation of 100.041 m, now 31 mm above the top of curb.
Answer the containment question. The roadway section between the curbs holds 0.411 m³/s at most. $$\boxed{\text{No } - 0.92\ \mathrm{m^3/s\ exceeds\ the\ }0.411\ \mathrm{m^3/s\ curb\text{-}full\ capacity\ by\ a\ factor\ of\ }2.24}$$ Nor, for that matter, does the section contain the present-day 0.8 m³/s: that flow already exceeds capacity by a factor of 1.95. The climate adjustment is not what breaks the section; it makes an already-inadequate section worse. Note also how insensitive the depth is to the flow — a 15 % increase in discharge raises the depth by only 5 %, because once the crown is drowned the section is very wide and shallow, so a small rise in stage adds a great deal of area. That same insensitivity works against the designer in the other direction: there is no realistic curb height that would bring 0.92 m³/s inside the section, since containing it would need a reveal of 0.12 m and the spread would still be the entire travelled way.
State the engineering consequence. Water 0.12 m deep across the full 8 m width is not a drainage nuisance but a loss of the road: it exceeds the depth at which hydroplaning becomes likely at highway speed, it removes all usable travelled surface, and under Canadian practice a roadway would be checked so that the major-system flow leaves at least one lane passable with the spread limited to the outer lane and the gutter. The correct response is to remove the flow from the roadway rather than to deepen the section — catch basins at a much closer spacing, an intercepting swale or ditch upstream of the roadway, or a cross-culvert — and to size that system on the climate-adjusted 0.92 m³/s rather than on the historical 0.8 m³/s.
Record the alternative reading. If the figure’s 99.02 m edge-of-pavement label were taken as correct instead of the question’s 2 % crossfall, the crossfall becomes 24.5 % and the curb reveal 0.99 m. The section is then two deep triangular gutters that never drown the crown, and Manning’s equation gives $d = 0.32$ m for 0.8 m³/s and $d = 0.34$ m for 0.92 m³/s — both comfortably contained by a 0.99 m curb. The two readings therefore reverse the answer to part (b), which is why the callout above sets out the choice explicitly.

Final results — Question 6
Quantity	Result
Edge-of-pavement elevation (from the stated geometry)	99.92 m; curb reveal 0.09 m
Discharge with the gutters just full ($d$ = 0.08 m)	0.284 m³/s
Curb-full capacity of the section ($d$ = 0.09 m)	0.411 m³/s
(a) Depth at the curb for $Q$ = 0.8 m³/s	0.115 m (water surface 100.035 m, 25 mm over the curb)
(b) Climate-adjusted flow	0.92 m³/s
(b) Depth at the curb for $Q$ = 0.92 m³/s	0.121 m (water surface 100.041 m, 31 mm over the curb)
(b) Can the road contain the new flow?	No — it exceeds the 0.411 m³/s capacity by 2.24 times; the present 0.8 m³/s already exceeds it by 1.95 times
Alternative reading (edge of pavement 99.02 m as labelled)	$d$ = 0.32 m and 0.34 m; both contained by the implied 0.99 m curb

Topic. Uniform flow in a crowned roadway cross-section; stage–discharge relations for compound gutter sections and roadway containment checks.

Key relations. Manning: $Q = \frac{1}{n}AR^{2/3}S_0^{1/2}$, $R = A/P$. Triangular gutter of cross slope $S_x$ and depth $d$: $A = d^{2}/(2S_x)$ per side, spread $T = d/S_x$. Crowned section with the crown drowned: $A = 2\left(d_{\mathrm{crown}}^{2}/2S_x\right) + W(d - d_{\mathrm{crown}})$, $P = 2\sqrt{(W/2)^{2}+d_{\mathrm{crown}}^{2}} + 2d$. Containment: compare the required normal depth with the curb reveal $z_{curb} - z_{EP}$.

Why this works. Manning’s equation applies to any prismatic section provided the flow is uniform, so the roadway is treated as an open channel whose geometry happens to be set by the crossfall and the curbs. The essential step is recognising that the section changes character at the crown: below that stage there are two hydraulically independent triangular gutters and the spread grows as $d/S_x$; above it the two gutters merge into one very wide, very shallow channel. Because the top width is then the full 8 m while the depth is barely 0.1 m, the hydraulic radius is small and grows only slowly with stage, which is why doubling the discharge changes the depth so little. Working the capacity first and the required depth second turns the question from arithmetic into a design check, which is what part (b) is really asking for.

Common pitfall. Applying the triangular gutter formula at a depth greater than the crossfall drop, which silently assumes the water has spread past the crown onto the other half of the road while still computing it as one triangle. A second pitfall is answering part (b) with a depth alone: the question asks whether the road contains the flow, so the depth must be compared explicitly against the curb reveal. A third, specific to this paper, is accepting the figure’s 99.02 m label without noticing that it contradicts the 2 % crossfall stated three lines above it.

Source. Chow, Open-Channel Hydraulics, Ch. 5–6 (uniform flow, Manning’s equation, compound and irregular sections); Chin, Water-Resources Engineering, 3rd ed., §5.2 and §5.6 (gutter flow, spread, inlet capacity and roadway drainage); Transportation Association of Canada, Geometric Design Guide for Canadian Roads, Ch. 4 (crossfall and curb standards).

← Question 5 Back to the paper →

Quantity	Symbol	Value
Pavement width, edge to edge	\(W\)	8 m (half-width 4 m each side of the crown)
Crossfall from the centreline	\(S_x\)	2 % = 0.02
Manning’s roughness, asphalt	\(n\)	0.013
Longitudinal roadway slope	\(S_0\)	0.01
Centreline (crown) elevation	\(z_{CL}\)	100.00 m
Top of curb elevation	\(z_{curb}\)	100.01 m
Edge of pavement elevation (derived)	\(z_{EP}\)	100.00 − 4(0.02) = 99.92 m
Present flood flow	\(Q_a\)	0.8 m³/s
Climate-adjusted flood flow	\(Q_b\)	1.15 × 0.8 = 0.92 m³/s

Quantity	Result
Edge-of-pavement elevation (from the stated geometry)	99.92 m; curb reveal 0.09 m
Discharge with the gutters just full (\(d\) = 0.08 m)	0.284 m³/s
Curb-full capacity of the section (\(d\) = 0.09 m)	0.411 m³/s
(a) Depth at the curb for \(Q\) = 0.8 m³/s	0.115 m (water surface 100.035 m, 25 mm over the curb)
(b) Climate-adjusted flow	0.92 m³/s
(b) Depth at the curb for \(Q\) = 0.92 m³/s	0.121 m (water surface 100.041 m, 31 mm over the curb)
(b) Can the road contain the new flow?	No — it exceeds the 0.411 m³/s capacity by 2.24 times; the present 0.8 m³/s already exceeds it by 1.95 times
Alternative reading (edge of pavement 99.02 m as labelled)	\(d\) = 0.32 m and 0.34 m; both contained by the implied 0.99 m curb