Peak Flow Estimate for an Ungauged Culvert Site
A 3.2 km² catchment drains to a proposed culvert on a small stream. There is no stream gauge, but a rain gauge 4 km away records a 24-hour storm of 95 mm, of which the first 20 mm is lost to interception and infiltration before runoff begins. The catchment is 60% pasture on clay loam (curve number CN ≈ 75) and 40% woodland (CN ≈ 55). The time of concentration is estimated at 2.1 hours using the Kirpich method, and the storm's 24-hour depth is roughly evenly distributed so that the peak 2-hour intensity is close to the 24-hour average. The design requires a peak discharge estimate with stated assumptions.
1. Estimate the area-weighted curve number and the runoff depth using the NRCS-CN method. State whether the 20 mm initial abstraction is consistent with the method and how you handle it.
Read the worked answer
Area-weighted CN: (0.60 × 75) + (0.40 × 55) = 45 + 22 = 67.
NRCS-CN runoff: S = (25400/CN) − 254 = (25400/67) − 254 ≈ 379 − 254 = 125 mm. Standard initial abstraction Ia = 0.2S ≈ 25 mm, which is close to the stated 20 mm loss, so the method is broadly consistent. Using the standard Ia = 0.2S is the defensible choice; if you instead force Ia = 20 mm, note it as a deviation.
With P = 95 mm, Ia = 25 mm: Q = (P − Ia)² / (P − Ia + S) = (70)² / (70 + 125) = 4900 / 195 ≈ 25 mm of runoff.
Key judgement: the 20 mm figure quoted is a lumped loss, not a verified Ia. Report Q ≈ 25 mm and flag that a ±10 mm change in Ia shifts Q by roughly 30–40%, so the estimate is sensitive to that assumption.
2. Convert the runoff depth to a peak discharge using the rational-method-style approach implied by the given data, and identify the single largest source of uncertainty.
Read the worked answer
Runoff volume: Q_vol = 25 mm × 3.2 × 10⁶ m² = 0.025 m × 3.2e6 m² = 80,000 m³ over the storm.
Peak approximation: the storm's peak 2-hour intensity is close to the 24-hour average = 95/24 ≈ 3.96 mm/h. But peak runoff responds to the time of concentration (2.1 h), so use the intensity over ~2 h ≈ 95/12 ≈ 7.9 mm/h (since the 24-h depth is roughly uniform, any 2-h window carries ~1/12 of it).
Rational method: Qp = C·i·A. Effective C = Q/P = 25/95 ≈ 0.26. With i = 7.9 mm/h = 7.9e-3 m/h and A = 3.2e6 m²: Qp = 0.26 × 7.9e-3 × 3.2e6 / 3600 ≈ 1.8 m³/s.
Cross-check by volume: 80,000 m³ spread over a runoff hydrograph of roughly 2–3× tc (~5 h) gives a mean of ~4.4 m³/s, which is higher than the rational result - a sign the two methods disagree.
Largest uncertainty: the time of concentration and the storm temporal pattern. A uniform 24-h storm is unrealistic; a real burst would raise the peak substantially. Report a range (roughly 1.5–4 m³/s) rather than a single number.