Urban Catchment Stormwater Detention Sizing
A 12 ha development is 70% impervious (CN 92) and 30% pervious lawn (CN 61). Pre-development the site was all pasture (CN 70). The local design storm is a 25-year, 24-hour event of 110 mm. The post-development time of concentration is 25 minutes; pre-development it was 90 minutes. A detention basin must limit the post-development peak outflow to the pre-development peak. The basin drains via an orifice and is empty at the start of the storm.
1. Estimate the pre- and post-development peak discharges and the required peak-flow reduction.
Read the worked answer
Area-weighted post-development CN: 0.70 × 92 + 0.30 × 61 = 64.4 + 18.3 = 82.7.
Pre-development CN = 70.
Runoff depths (NRCS-CN, Ia = 0.2S):
- Pre: S = 25400/70 − 254 = 363 − 254 = 109 mm; Ia = 21.8; Q = (110−21.8)²/(110−21.8+109) = 7780/197.2 ≈ 39 mm.
- Post: S = 25400/82.7 − 254 = 307 − 254 = 53 mm; Ia = 10.6; Q = (110−10.6)²/(110−10.6+53) = 9880/152.4 ≈ 65 mm.
Peak comparison via rational method with intensity scaled to tc:
- Pre: i over 90 min ≈ 110 × (90/1440) = 6.9 mm over 1.5 h → 4.6 mm/h. C_eff = 39/110 = 0.355. Qp = 0.355 × 4.6e-3 × 120,000/3600 ≈ 0.054 m³/s.
- Post: i over 25 min ≈ 110 × (25/1440) = 1.9 mm over 0.42 h → 4.6 mm/h (same average, but the short-duration intensity is higher in reality). C_eff = 65/110 = 0.59. Qp = 0.59 × 4.6e-3 × 120,000/3600 ≈ 0.090 m³/s.
Required reduction: from ~0.09 to ~0.054 m³/s, i.e. ~40% peak reduction, plus the much larger runoff volume (65 vs 39 mm) must be stored and released slowly.
2. Estimate the detention storage volume needed and state the main assumption that could make it inadequate.
Read the worked answer
Storage ≈ difference in runoff volume, since the outlet is sized to pass only the pre-development peak.
Volume difference: (65 − 39) mm × 120,000 m² = 0.026 m × 120,000 = 3,120 m³ minimum.
More carefully, the basin must store the post-development runoff that cannot pass through the outlet at the allowable rate during the storm. Using the simplified 'modified rational' approach: required storage ≈ (post Qp − allowable Qp) × storm duration of significant inflow. With inflow concentrated over roughly 2–3× tc (≈1–1.5 h):
V ≈ (0.090 − 0.054) × 3600 s × ~1.2 h ≈ 0.036 × 4320 ≈ 156 m³.
The two estimates differ by an order of magnitude (156 vs 3,120 m³) because the first treats the whole storm volume and the second only the excess rate. The correct design must integrate the full inflow hydrograph against the outlet rating curve; the true requirement is closer to the larger figure because the storm lasts 24 h and the outlet passes only 0.054 m³/s (≈4,700 m³/day), which over 24 h passes ~4,700 m³ - actually more than the 3,120 m³ excess, so the outlet can drain it if the storm is spread over 24 h.
Main assumption that could fail: storm temporal pattern. The given 110 mm is a 24-h total; a real 25-year storm has an intense burst. If most of the 110 mm falls in 1–2 h, inflow far exceeds the outlet and the required storage approaches the full post-development volume (~7,800 m³). Design for the burst, not the uniform average.