Question 6 of 6: Compound flood-protection channel
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper: National Exams — 98-Civ-A5 Hydraulic Engineering, May 2013 · 3 hours, closed book (one aid sheet). Six questions; any five constitute a complete paper — all six are solved here as a study resource. All parts of a question are of equal value.
Reference texts. Mays, Water Resources Engineering, 3rd ed. (pipe systems, pumps, network analysis); Chow, Open-Channel Hydraulics (Manning flow, compound sections); Crowe, Elger & Roberson, Engineering Fluid Mechanics (energy/continuity). Exam-supplied relations used throughout: Hazen–Williams $Q=0.278\,C\,D^{2.63}\,S^{0.54}$ with $S=h_f/L$ (SI), Manning $Q=\tfrac{1}{n}A\,R^{2/3}\,S^{1/2}$, and total dynamic head $\text{TDH}=H_s+H_f$. Unless stated, local losses and velocity head are neglected, and water has $\rho=1000\ \text{kg/m}^3$.
Check (Q4 data consistency): the pipe/valve data in Question 4 are internally inconsistent — the stated “initial valve flow = 400 L/s” corresponds to a node head of only 9.5 m, but the two supply pipes driven by the 96 m and 89 m tank levels deliver far more than that at 9.5 m. Continuity at the node fixes a network-consistent initial discharge of ≈693 L/s at a node head of ≈28.4 m. The simulation below is run from that physically consistent state, with the discrepancy noted (per Note 1, candidates may state assumptions).
Given. Rectangular Channel A width $b_A=7$ m; rectangular Channel B width $b_B=20$ m; separating wall height $H=2$ m (spill crest); $S_0=0.001$; $n=0.013$ for both.
Find. (a) water depth in Channel A at $Q=21$ m³/s (B dry); (b) whether Channel A alone can pass $Q=60$ m³/s, and if not, the water level in Channel B.
Figure 3. Channel A ($b=7$ m) is separated from Channel B ($b=20$ m) by a 2 m wall; once A fills to the 2 m crest, surplus spills into B.
Approach. Use Manning’s equation for uniform flow in each rectangular channel, $Q=\tfrac1n A R^{2/3}S_0^{1/2}$ with $A=by$, $P=b+2y$, $R=A/P$. For (a) solve A’s normal depth. For (b) first find A’s capacity at the 2 m crest; if the total exceeds it, the surplus is carried by Channel B, whose normal depth gives the requested level.
Manning constant. $\tfrac1n S_0^{1/2}=\tfrac{1}{0.013}\sqrt{0.001}=2.433$, so $Q=2.433\,A\,R^{2/3}$.
(a) Normal depth in Channel A at $Q=21$. With $b_A=7$: $$21=2.433\,(7y)\Big(\tfrac{7y}{7+2y}\Big)^{2/3}.$$ Solving, $$\boxed{y_A\approx 1.29\ \text{m}}.$$ Since $1.29\ \text{m}\lt 2\ \text{m}$ (the wall height), the water stays within Channel A and Channel B carries no flow — consistent with the assumption.
(b) Capacity of Channel A at the spill crest. At $y=2$ m: $A=7(2)=14$ m², $P=7+2(2)=11$ m, $R=1.273$ m, $$Q_{A,\max}=2.433(14)(1.273)^{2/3}=40.0\ \text{m}^3/\text{s}.$$
Can A carry 60? $60\ \text{m}^3/\text{s}\gt 40\ \text{m}^3/\text{s}=Q_{A,\max}$, so $$\boxed{\text{No — Channel A cannot carry all 60 m}^3/\text{s}.}$$ Channel A flows full at its 2 m crest (≈40 m³/s) and the surplus spills over the wall into Channel B.
Surplus into Channel B. $Q_B=60-40=20\ \text{m}^3/\text{s}$.
Normal depth in Channel B at $Q_B=20$. With $b_B=20$: $$20=2.433\,(20y)\Big(\tfrac{20y}{20+2y}\Big)^{2/3}\ \Rightarrow\ \boxed{y_B\approx 0.60\ \text{m}}.$$ Channel B runs about 0.60 m deep while Channel A stays full at 2 m.
Question 6 — results
Quantity
Value
(a) Water depth in Channel A at 21 m³/s
≈ 1.29 m
Channel A capacity at 2 m crest
≈ 40.0 m³/s
(b) Can A carry 60 m³/s?
No (40 m³/s max)
Surplus carried by Channel B
20 m³/s
(b) Water depth in Channel B
≈ 0.60 m
Check: Channels A and B are treated as separate uniform-flow channels sharing the same slope and roughness, with Channel A flowing full (2 m) once it spills and Channel B conveying only the surplus — the model implied by “when Channel A is full, the surplus spills onto Channel B.” Their water surfaces sit at different elevations (A at the 2 m crest plus a small spill head; B at ≈0.60 m), consistent with flow over the dividing wall acting as a side weir.
