This solution calculates the direct runoff volume for a given watershed using the phi-index method based on the provided rainfall data.
Calculate the excess rainfall for each hour where intensity ($I$) exceeds the phi-index ($\phi$). The excess rainfall depth is $I - \phi$.
The total depth of direct runoff ($R$) is the sum of the hourly excesses: $R = 1 + 0 + 0.5 + 0 = 1.5$ cm.
Convert the watershed area and runoff depth to compatible units (meters).
Calculate the volume ($V$) using the formula $V = A \times R$.
$V = 100,000 \text{ m}^2 \times 0.015 \text{ m}$
$V = 1500 \text{ m}^3$.
The volume of direct runoff is 1500 m$^3$.
| Duration (minutes) | 5 | 10 | 20 | 30 | 40 | 60 |
| Maximum depth of rainfall (mm) | 16 | 24 | 39 | 50 | 60 | 65 |
| Time from start of rainfall (min) | 0 | 20 | 40 | 60 | 80 | 100 |
|---|---|---|---|---|---|---|
| Cumulative rainfall (cm) | 0 | 0.5 | 1.2 | 2.6 | 3.3 | 3.5 |
The Curve Number (CN) of a watershed of $40$ ha area under given hydrologic soil group, land use and management practices, and Antecedent Moisture Condition (AMC)-II is $80$. The initial abstraction is $20$% of maximum retention. For the rainfall event of $40$ mm, the direct runoff in mm (round off to $2$ decimal places) is _____________