07-Str-B11 · May 2014

Question 6 of 6: Curbed Roadway Analysed as an Open Channel

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

Notes on this paper

National Examinations — May 2014 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and any non-communicating calculator are permitted. The paper prints six questions of 20 marks each and instructs the candidate to complete any five; where a question has more than one part, the parts carry equal marks. Candidates are urged to submit a clear statement of any interpretive assumptions with the answer paper. All six questions are worked below, because this set is intended as a study resource.

Reference texts: Mays, L.W., Water Resources Engineering (3rd ed., Wiley) — Hazen-Williams pipe hydraulics, branched and looped network analysis, transmission mains and valve characteristics; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American distribution practice, fire-flow criteria, minimum service pressures and gutter hydraulics; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning’s equation and the momentum treatment of the hydraulic jump; Henderson, F.M., Open Channel Flow (Macmillan) — specific force, conjugate depths and jump classification; White, F.M., Fluid Mechanics (7th ed., McGraw-Hill) — the control-volume force balance that links wall shear stress to the Darcy friction factor.

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. Local losses and velocity head are neglected, as instructed in Note 6, 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 properties are $\rho = 1000$ kg/m³ and $\nu = 1.31\times10^{-6}$ m²/s.

Question 6: Curbed Roadway Analysed as an Open Channel (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, curbed roadway carrying surface flow along its longitudinal grade, analysed as a prismatic open channel under Manning’s equation.

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 roughness, asphalt	$n$	0.013
Longitudinal roadway slope	$S_0$	0.02
Centreline (crown) elevation	—	100.00 m
Top-of-curb elevation	—	100.01 m
Present flood flow	$Q_a$	1.3 m³/s
Climate-adjusted flow, $1.15 Q_a$	$Q_b$	1.495 m³/s

Find. (a) the depth of flow at 1.3 m³/s; (b) the depth at the climate-adjusted flow, and whether the curbs contain it.

Figure 6.1 — Roadway section built from the stated geometry: a crown at 100.00 m falling 2% over each 4 m half-width to edges of pavement at 99.92 m, with curbs whose tops are at 100.01 m. The computed water surface for 1.3 m³/s lies above the top of curb, so the section is already surcharged at the present flood flow.

Approach. Build the wetted area and perimeter as explicit functions of the depth at the curb face from the stated crossfall geometry, then invert Manning’s equation numerically for the depth that passes the given discharge, and compare the resulting water surface elevation with the top of curb.

Fix the section geometry from the crossfall. Each half-width is 4 m falling at 2%, so the pavement drops $$\Delta z = \frac{W}{2}S_x = 4 \times 0.02 = 0.08\ \mathrm{m}$$ below the crown, placing the edge of pavement at $100.00 - 0.08 = \boxed{99.92\ \mathrm{m}}$ and giving a curb exposure of $$100.01 - 99.92 = \boxed{0.09\ \mathrm{m}}$$ (The elevation printed on Figure 3 of the paper is discussed in the callout below.)
Decide whether the flow spreads to the crown. If the depth at the curb were only $y$, a triangular gutter would spread $T = y/S_x = 50y$. The spread reaches the crown at $y = 0.08$ m, where each gutter carries $$Q = \frac{1}{n}A R^{2/3}S_0^{1/2} = \frac{1}{0.013}(0.32)\left(\frac{0.32}{4.08}\right)^{2/3}(0.02)^{1/2} = 0.40\ \mathrm{m^3/s}$$ for the whole section. Both design flows exceed 0.40 m³/s, so the water covers the full 8 m and the section must be treated as a single channel with an inverted-V invert, not as two independent triangular gutters.
Write area and wetted perimeter for the surcharged section. With $y$ measured at the curb face, the depth varies linearly from $y$ at each curb to $y - 0.08$ at the crown, so $$A = W y - \frac{W}{2}\Delta z = 8y - 0.32$$ $$P = 2\sqrt{\left(\tfrac{W}{2}\right)^{2} + \Delta z^{2}} + 2y = 2(4.0008) + 2y = 8.0016 + 2y$$ The two vertical curb faces contribute the $2y$ term; the sloping pavement is only 0.0008 m longer than its 4 m horizontal projection.
Establish the capacity of the section to the top of curb. Setting $y = 0.09$ m, the maximum depth the curbs can hold: $$A = 8(0.09) - 0.32 = 0.400\ \mathrm{m^2}, \qquad P = 8.0016 + 0.18 = 8.182\ \mathrm{m}, \qquad R = 0.0489\ \mathrm{m}$$ $$Q_{\mathrm{cap}} = \frac{1}{0.013}(0.400)(0.0489)^{2/3}(0.02)^{1/2} = \boxed{0.58\ \mathrm{m^3/s}}$$ This benchmark is worth computing first: it already shows that both design flows exceed what the section can carry.
Invert Manning’s equation for part (a). Solving $\frac{1}{0.013}(8y-0.32)\left(\frac{8y-0.32}{8.0016+2y}\right)^{2/3}(0.02)^{1/2} = 1.3$ numerically, $$\boxed{y_a = 0.121\ \mathrm{m}} \qquad (A = 0.650\ \mathrm{m^2},\ R = 0.0789\ \mathrm{m},\ V = 2.00\ \mathrm{m/s})$$ The corresponding water surface elevation is $$\mathrm{WSE}_a = 99.92 + 0.121 = 100.04\ \mathrm{m}$$ which stands 0.03 m above the top of curb at 100.01 m.
Repeat for the climate-adjusted flow in part (b). The design flow increases by 15%: $$Q_b = 1.15 \times 1.3 = 1.495\ \mathrm{m^3/s}$$ and inverting Manning’s equation again gives $$\boxed{y_b = 0.128\ \mathrm{m}}, \qquad \mathrm{WSE}_b = 99.92 + 0.128 = 100.05\ \mathrm{m}$$ Because the section is wide and shallow, a 15% rise in discharge lifts the depth by only 6% — roughly the $Q^{3/5}$ response expected of a broad channel.
Answer the containment question. The top of curb is at 100.01 m, so containment requires $\mathrm{WSE} \le 100.01$ m, that is $y \le 0.09$ m: $$\mathrm{WSE}_b = 100.05\ \mathrm{m} > 100.01\ \mathrm{m} \quad\Rightarrow\quad \boxed{\text{No: the road cannot contain the climate-adjusted flow}}$$ More than that, the section is already over capacity at the present flow: $Q_{\mathrm{cap}} = 0.58$ m³/s against 1.3 m³/s, so water overtops the curbs and spills onto the boulevard in both cases. The 15% climate increment is not what causes the failure — it deepens an existing one by about 7 mm. The engineering conclusion is that surface conveyance alone is insufficient here and the catchment must be relieved by catch basins and a storm sewer, or by a curb of roughly 0.13 m exposure combined with a wider section, before the climate-adjusted flow can be managed.

Check: the figure’s edge-of-pavement elevation contradicts the stated crossfall. Figure 3 of the paper labels the edge of pavement 99.02 m, but the question text fixes the geometry independently: a 2% crossfall over the 4 m half-width from a crown at 100.00 m puts the edge at 99.92 m. The printed 99.02 m would require a crossfall of 24.5%, which no road has and which contradicts the sentence that defines the problem. The self-consistent value 99.92 m is adopted here, giving a 0.09 m curb exposure, and this is the interpretive assumption a candidate should record under Note 1 of the paper. For completeness: were 99.02 m taken literally and the 2% discarded, the same Manning analysis would give depths of 0.32 m and 0.33 m at the curb and the flow would be contained by the resulting 0.99 m curb — the opposite conclusion, which is why the assumption must be stated explicitly.

Question 6 — results
Part	Quantity	Result
—	Crown-to-edge drop at 2% over 4 m	0.08 m; edge of pavement at 99.92 m
—	Curb exposure above edge of pavement	0.09 m
—	Section capacity to top of curb	0.58 m³/s
(a)	Water depth at the curb, $Q = 1.3$ m³/s	0.121 m (WSE 100.04 m)
(b)	Climate-adjusted flow, $1.15 \times 1.3$	1.495 m³/s
(b)	Water depth at the curb, $Q = 1.495$ m³/s	0.128 m (WSE 100.05 m)
(b)	Is the flow contained?	No — WSE exceeds the 100.01 m top of curb (and already does at 1.3 m³/s)

Topic: Roadway surface drainage treated as uniform flow in a shallow, wide compound channel, and the freeboard check against top of curb.

Key relations: $Q = \frac{1}{n}A R^{2/3}S_0^{1/2}$ with $R = A/P$; for a triangular gutter $T = y/S_x$, $A = y^2/(2S_x)$; for the surcharged crowned section $A = Wy - \tfrac{W}{2}\Delta z$ and $P \approx W + 2y$; containment requires $\mathrm{WSE} \le$ top of curb.

Why this works: The longitudinal slope of 2% is steep and constant and the section is prismatic, so normal-depth (uniform flow) is a sound assumption and Manning’s equation applies directly. The crown converts the problem from two independent gutters into one channel once the spread exceeds the half-width, and it is that transition — not the arithmetic — that decides which area formula is correct.

Common pitfall: Using the triangular gutter formula $Q = \frac{0.375}{n}S_x^{5/3}S_0^{1/2}T^{8/3}$ without checking that the computed spread $T$ fits within the half-width. Here it returns $T = 6.2$ m on a 4 m half-width, an impossible geometry that nonetheless yields a plausible-looking depth. The second pitfall is comparing depth against curb height measured from the crown rather than from the edge of pavement, which understates the depth of flooding by the full 0.08 m crossfall drop.

Source: Chow, Open-Channel Hydraulics, Ch. 5–6 (uniform flow, Manning’s equation, compound and irregular sections); Chin, Water-Resources Engineering, 3rd ed., §5.4 (street and gutter flow, spread criteria and inlet spacing); TAC, Geometric Design Guide for Canadian Roads (crossfall and curb practice).

← 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 roughness, asphalt	\(n\)	0.013
Longitudinal roadway slope	\(S_0\)	0.02
Centreline (crown) elevation	—	100.00 m
Top-of-curb elevation	—	100.01 m
Present flood flow	\(Q_a\)	1.3 m³/s
Climate-adjusted flow, \(1.15 Q_a\)	\(Q_b\)	1.495 m³/s