Peak runoff rate from a design storm using a distribution graph An intense storm of uniform intensity 7.5 cm/h lasts 60 minutes over a basin of area 500 hectares. Average infiltration capacity during the storm is 1.5 cm/h. Using a 10-minute peak percentage of 16% from the basin’s distribution graph, estimate the maximum runoff rate (cumecs).

Introduction / Context:Design peak discharge can be estimated from total direct-runoff volume combined with a dimensionless distribution graph (e.g., Bernard-type), which gives the percentage of total runoff occurring in the most intense block (here, the 10-minute peak).

Given Data / Assumptions:

• Storm intensity i = 7.5 cm/h for 60 minutes.
• Average infiltration capacity f = 1.5 cm/h.
• Basin area A = 500 ha.
• 10-minute peak percentage = 16% of total runoff volume.

Concept / Approach:Total direct runoff depth equals rainfall depth minus losses by infiltration during the same duration. The peak 10-minute discharge equals the fraction of total runoff volume assigned to the 10-minute peak, divided by 600 seconds.

Step-by-Step Solution:

Verification / Alternative check:Average discharge over the full hour is V / 3600 ≈ 83.3 m³/s. A 10-min peak at 80 m³/s is reasonable given the distribution shape and chosen peak percentage.

Why Other Options Are Wrong:

• 40 and 60 m³/s underestimate the computed peak given the 16% allocation.
• 100 and 120 m³/s exceed the physically consistent result from the stated distribution percentage.

Common Pitfalls:Mixing cm and m or hectares and km²; misapplying the peak percentage to intensity rather than to total runoff volume; using 10 minutes as 10 seconds by mistake.

Final Answer:80 cumecs