Question 6 of 7

Problem 6 (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.

(i) Stormwater wet pond design approach for TSS/nutrient reduction (10 marks)

A stormwater wet pond (permanent-pool detention basin) is designed around a permanent pool sized to hold the target water-quality storm volume (in Canadian practice, typically the runoff volume from the 90th-percentile storm, or a stated depth over the contributing catchment) for a target hydraulic residence time — commonly on the order of 24–48 h — long enough for the settleable and semi-settleable fraction of total suspended solids (and the particulate-bound fraction of phosphorus/nutrients, which tends to sorb onto fine sediment) to settle out under quiescent conditions before the next storm's inflow displaces the treated water. Key design elements are: a sediment forebay at the inlet to capture coarse sediment separately (concentrating maintenance dredging in one accessible cell rather than across the whole pond); a length-to-width ratio of at least about 3:1 (with baffles/berms if the pond is more compact) to maximize the flow path and minimize hydraulic short-circuiting between inlet and outlet, which otherwise lets a portion of each storm's inflow bypass the design residence time; a permanent pool depth deep enough (typically 1–2 m) to resist re-suspension of settled sediment by wind-driven mixing and to limit algal growth in the photic zone; and an outlet control structure that releases the water-quality volume slowly (via a small orifice/reverse-slope pipe drawing from just below the permanent pool surface) while still passing larger design storms through a higher-capacity overflow, so the pond provides quality treatment for frequent small storms without becoming a flood-control liability for rare large ones.

Two maintenance issues. (1) Sediment accumulation in the forebay and main pool — as TSS is trapped by design, the forebay progressively fills and must be dredged/cleaned out on a scheduled interval (often every 5–10 years, sooner if the contributing catchment is under active construction), or the pond's effective treatment volume and residence time shrink and performance degrades; (2) vegetation, inlet/outlet and embankment upkeep — littoral (shallow marginal) vegetation must be maintained to take up dissolved nutrients and stabilize banks against erosion, while the inlet and outlet structures must be kept clear of debris/trash so that neither short-circuiting (blocked baffles) nor uncontrolled bypass (blocked low-flow orifice forcing everything over the emergency spillway) develops.

(ii) Rational Method design peak runoff for two catchments in series (10 marks)

Given.

100-year IDF curve (read from the supplied chart, Tr = 100 years): $i\approx116$ mm/h at duration 40 min; $i\approx88$ mm/h at duration 60 min (values read at the two durations of interest; chart-derived, ±10% engineering tolerance).

Find. The governing 100-year design peak runoff at the outlet, using the Rational Formula $Q=CiA/360$ ($Q$ in m³/s, $i$ in mm/h, $A$ in ha).

Approach. With two sub-catchments in series, the design storm duration that maximizes the OUTLET peak is not obvious a priori: a short, intense storm (duration = A1's own $t_c$) only lets A1 contribute at full intensity (A2 has not yet reached its own outlet-response time), while a longer storm (duration = A2's $t_c$, the time for the combined system to fully respond) lets the FULL combined area contribute, but at a lower intensity. Both candidate peaks are computed and the larger governs.

1. Candidate 1 — storm duration = $t_1=40$ min (only A1 fully contributing). $$Q_1=\frac{C_1A_1i(t_1)}{360}=\frac{(0.6)(30)(116)}{360}=\boxed{5.80\ \text{m}^3/\text{s}}$$
2. Candidate 2 — storm duration = $t_2=60$ min (combined A1+A2 area contributing). $$Q_2=\frac{(C_1A_1+C_2A_2)\,i(t_2)}{360}=\frac{[(0.6)(30)+(0.7)(50)]\,(88)}{360}=\frac{(53)(88)}{360}=\boxed{12.96\ \text{m}^3/\text{s}}$$
3. Governing peak. Since $Q_2=12.96\ \text{m}^3/\text{s} > Q_1=5.80\ \text{m}^3/\text{s}$, the longer storm duration (letting the full composite area contribute) governs, even though its intensity is lower: $$\boxed{Q_{100}=12.96\ \text{m}^3/\text{s}}$$