07-Str-B11 · May 2013

Question 6 of 6: Compound Flood-Protection Channel with a Spill Wall

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

Notes on this paper

National Examinations — May 2013 — 07-Str-B11 Hydraulic Engineering. Three-hour, CLOSED-BOOK examination; one 8.5 × 11 in aid sheet (both sides) and one approved Casio or Sharp 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 state any interpretive assumptions with their answer. 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, pipe networks, pump-system curves and quasi-steady reservoir routing; Chow, V.T., Open-Channel Hydraulics (McGraw-Hill, 1959) — uniform flow, Manning's equation, compound and divided channel sections, and the specific-energy and momentum treatment of channel obstructions; Chin, D.A., Water-Resources Engineering (3rd ed., Pearson) — North-American design practice for transmission mains, minimum service pressures and pump selection; Henderson, F.M., Open Channel Flow (Macmillan) — surges, hydraulic jumps and the unsteady response of a channel to a sudden blockage.

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. 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: Compound Flood-Protection Channel with a Spill Wall (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. Two rectangular channels share a longitudinal slope and roughness and are separated by a 2 m spill wall; the narrow channel carries minor flows and overspills into the wide channel during floods.

Given data — Question 6
Quantity	Symbol	Value
Bed width, Channel A	$B_A$	7 m
Bed width, Channel B	$B_B$	20 m
Height of the spill wall above the common bed	$H_w$	2 m
Longitudinal slope, both channels	$S_0$	0.001
Manning roughness, both channels	$n$	0.013
Minor storm discharge, part (a)	$Q_a$	21 m³/s
Flood discharge, part (b)	$Q_b$	60 m³/s

Find. (a) the uniform-flow depth in Channel A at 21 m³/s with Channel B dry, and (b) whether Channel A alone can pass 60 m³/s and, if not, the resulting water level in Channel B.

Figure 6-S. Compound section with the computed water levels. At 21 m³/s Channel A runs 1.285 m deep and the wall is not overtopped; at 60 m³/s Channel A is limited to 40.0 m³/s at its 2 m capacity and the 20.0 m³/s surplus spills into Channel B, which fills to 0.601 m.

Approach. Treat each channel as an independent rectangular section in uniform flow and apply Manning's equation, solving the implicit depth relation for part (a); for part (b), first test the capacity of Channel A at the spill-wall crest, then route the surplus through Channel B by the same method.

Assemble the constant part of Manning's equation. Both channels share $n$ and $S_0$, so $$Q = \frac{1}{n}A R^{2/3}S_0^{1/2} = C_m\,A\left(\frac{A}{P}\right)^{2/3}, \qquad C_m = \frac{\sqrt{0.001}}{0.013} = 2.4325\ \text{m}^{1/3}\text{/s}$$ For a rectangular section of width $B$ and depth $y$ the geometry is $A = By$ and $P = B + 2y$, the two vertical walls being wetted.
Solve part (a) for the depth in Channel A. With $B_A = 7$ m and $Q_a = 21$ m³/s, $$21 = 2.4325\,(7y)\left(\frac{7y}{7+2y}\right)^{2/3}$$ This is implicit in $y$ and is solved by iteration or bisection, giving $$\boxed{y_A = 1.285\ \text{m}}$$ Checking the result, $A = 8.996$ m², $P = 9.570$ m, $R = 0.940$ m and $Q = 2.4325(8.996)(0.940)^{2/3} = 21.00$ m³/s, which recovers the given discharge.
Confirm that the wall is not overtopped in part (a). Since $1.285\ \text{m} \lt 2.000\ \text{m}$, the water surface stands 0.715 m below the crest of the spill wall and Channel B does indeed remain dry, which is consistent with the assumption the question instructs us to make. The mean velocity is $V = 21/8.996 = 2.33$ m/s and the Froude number is $\mathrm{Fr} = 2.33/\sqrt{9.81(1.285)} = 0.66$, so the flow is subcritical, as uniform flow on a mild slope of 0.001 must be.
Establish the capacity of Channel A at the crest. Setting $y = 2.000$ m, the depth at which spilling begins, $$A = 7(2) = 14.0\ \text{m}^2, \qquad P = 7 + 2(2) = 11.0\ \text{m}, \qquad R = 1.2727\ \text{m}$$ $$Q_{A,\max} = 2.4325(14.0)(1.2727)^{2/3} = \boxed{40.0\ \text{m}^3\text{/s}}$$
Answer the first half of part (b). Because $60\ \text{m}^3\text{/s} \gt 40.0\ \text{m}^3\text{/s}$, Channel A cannot carry the whole flood. It fills to the 2 m crest, passes its maximum 40.0 m³/s, and the surplus $$Q_B = 60.0 - 40.0 = 20.0\ \text{m}^3\text{/s}$$ spills laterally over the wall into Channel B, which is precisely the behaviour the compound section was designed for.
Solve for the depth in Channel B. Applying Manning's equation to the 20 m wide channel carrying the surplus, $$20.0 = 2.4325\,(20y)\left(\frac{20y}{20+2y}\right)^{2/3} \quad\Rightarrow\quad \boxed{y_B = 0.601\ \text{m}}$$ Checking, $A = 12.01$ m², $P = 21.20$ m, $R = 0.567$ m and $Q = 2.4325(12.01)(0.567)^{2/3} = 20.0$ m³/s. Since the two beds are at the same level, the water level in Channel B stands 0.601 m above the common bed, which is 1.399 m below the crest of the spill wall.
Confirm that the two channels remain hydraulically separate. The check that closes the solution is that Channel B's surface at 0.601 m lies well below the 2.000 m crest, so the wall acts as a free overfall weir rather than a drowned one. Had $y_B$ come out above the crest, the spill would have been submerged, the two channels would have had to be solved together as a single compound section with a common water surface, and the simple sequential calculation used here would not have been valid. Channel B's mean velocity of 1.67 m/s and Froude number of 0.69 confirm subcritical flow there as well.

Final results — Question 6
Case	Quantity	Value
(a) Q = 21 m³/s	Depth in Channel A	1.285 m above the bed
	Freeboard to the spill wall crest	0.715 m
	Mean velocity / Froude number in A	2.33 m/s / 0.66 (subcritical)
(b) Q = 60 m³/s	Capacity of Channel A at the 2 m crest	40.0 m³/s
	Can Channel A carry all the flow?	No — it is 20.0 m³/s short
	Surplus spilled into Channel B	20.0 m³/s
	Water level in Channel B	0.601 m above the bed
	Mean velocity / Froude number in B	1.67 m/s / 0.69 (subcritical)

Topic. Uniform flow in rectangular channels by Manning's equation, and the divided-channel treatment of a compound flood-protection section with a spill wall.

Key relations. Manning's equation $Q = \tfrac{1}{n}AR^{2/3}S_0^{1/2}$ with $R = A/P$; for a rectangle $A = By$ and $P = B + 2y$; channel capacity at a fixed crest elevation $Q_{\max} = Q(y = H_w)$; flow split $Q_B = Q_{\text{total}} - Q_{A,\max}$; and the Froude number $\mathrm{Fr} = V/\sqrt{gy}$ as the subcritical check.

Why this works. Manning's equation describes uniform flow, in which the gravitational component of weight along the channel is exactly balanced by boundary friction so the water surface runs parallel to the bed. That balance is what allows the depth to be found from the discharge alone, without reference to any control section. The relation is implicit in depth because both the area and the hydraulic radius depend on it, but the right-hand side increases monotonically with depth, so the root is unique and any bracketing method converges. The divided-channel treatment used in part (b) is valid here for a specific reason: with the wall standing 1.4 m above Channel B's surface the spill is a free overfall, so the two channels do not share a water surface and each may be given its own uniform-flow calculation. The interface between them is a vertical shear layer that is conventionally excluded from both wetted perimeters, since it exerts far less resistance than a solid boundary.

Common pitfall. Including the divider wall in Channel B's wetted perimeter, or the free water surface in either, both of which understate the hydraulic radius and the capacity. A more serious error in part (b) is assuming that the two channels share a common water surface and solving the whole compound section for a single depth; that treatment is correct only once Channel B's level rises above the crest and drowns the spill, which is not the case at 60 m³/s. It is also worth remembering that the answer to part (a) must be checked against the wall height, since a depth returned above 2 m would invalidate the very assumption the question makes.

Source. Chow, Open-Channel Hydraulics, Ch. 5 and Ch. 6 (uniform flow, the Manning formula, computation of normal depth) and Ch. 7 (compound and divided channel sections); Mays, Water Resources Engineering, 3rd ed., Ch. 5 (open-channel flow and channel design).

← Question 5 Back to the paper →

Quantity	Symbol	Value
Bed width, Channel A	\(B_A\)	7 m
Bed width, Channel B	\(B_B\)	20 m
Height of the spill wall above the common bed	\(H_w\)	2 m
Longitudinal slope, both channels	\(S_0\)	0.001
Manning roughness, both channels	\(n\)	0.013
Minor storm discharge, part (a)	\(Q_a\)	21 m³/s
Flood discharge, part (b)	\(Q_b\)	60 m³/s