Direction: The ordinates of a 2-h unit hydrograph at 1 hour intervals starting from time t = 0, are 0, 3, 8, 6, 3, 2 and 0 m³/s. Use trapezoidal rule for numerical integration, if required.
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A storm of 6.6 cm occurs uniformly over the catchment in 3 hours. If f-index is equal to 2 mm/h and base flow is 5 m³/s, what is the peak flow due to the storm?
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- 41.0 m³/s
- 43.4 m³/s
- 53.0 m³/s
- 56.2 m³/s
Correct Option: C
Effective rainfall = rainfall – φ index
φ index = 2 mm/hour = (2 × 3) + 10–1 (3 hours)
(10–1 to convert mm to cm)
= 0.6
Effective rainfall = 6.6 – 0.6 = 6 cm
Peak discharge = Effective rainfall × maximum ordinate of UH + Base flow
= 6 × 8 + 5 = 53 m³/s.