Time (h) 0 2 4 6 8 10 12 Cumulative rainfall (cm) 0 0.64 2.64 5.64 6.00 6.80 10.80
Rainfall Runoff Calculation Using Phi-Index Method
This solution details the calculation of total direct runoff for a given rainfall event using the phi-index ($\Phi$) method.
Rainfall Data Analysis
The observed cumulative rainfall over a 12-hour event is provided. First, we tabulate this data and convert it from centimeters (cm) to millimeters (mm) for consistency, as the phi-index is often applied in mm/h units, and the final answer is expected in mm.
| Time (h) | Cumulative Rainfall (cm) | Cumulative Rainfall (mm) |
|---|---|---|
| 0 | 0 | 0 |
| 2 | 0.64 | 6.4 |
| 4 | 2.64 | 26.4 |
| 6 | 5.64 | 56.4 |
| 8 | 6.00 | 60.0 |
| 10 | 6.80 | 68.0 |
| 12 | 10.80 | 108.0 |
Calculating Incremental Rainfall and Losses
The phi-index ($\Phi$) represents the average rate of loss (infiltration and other abstractions) during the storm period. It is given as 0.46 cm/h, which is equal to 4.6 mm/h.
Direct runoff is calculated for each time interval where the incremental rainfall exceeds the potential loss ($\Phi \times \Delta t$).
| Time Interval (h) | Duration ($\Delta t$) (h) | Incremental Rainfall ($P_i$) (mm) | Potential Loss ($\Phi \times \Delta t$) (mm) | Direct Runoff ($P_i - \Phi \times \Delta t$) (mm) |
|---|---|---|---|---|
| 0-2 | 2 | $6.4 - 0 = 6.4$ | $4.6 \times 2 = 9.2$ | 0 (Since $P_i < \Phi \times \Delta t$) |
| 2-4 | 2 | $26.4 - 6.4 = 20.0$ | $4.6 \times 2 = 9.2$ | $20.0 - 9.2 = 10.8$ |
| 4-6 | 2 | $56.4 - 26.4 = 30.0$ | $4.6 \times 2 = 9.2$ | $30.0 - 9.2 = 20.8$ |
| 6-8 | 2 | $60.0 - 56.4 = 3.6$ | $4.6 \times 2 = 9.2$ | 0 (Since $P_i < \Phi \times \Delta t$) |
| 8-10 | 2 | $68.0 - 60.0 = 8.0$ | $4.6 \times 2 = 9.2$ | 0 (Since $P_i < \Phi \times \Delta t$) |
| 10-12 | 2 | $108.0 - 68.0 = 40.0$ | $4.6 \times 2 = 9.2$ | $40.0 - 9.2 = 30.8$ |
Total Direct Runoff Calculation
The total direct runoff is the sum of the runoff calculated for each interval.
Total Direct Runoff = Sum of runoff from all intervals
$ \text{Total Direct Runoff} = 0 + 10.8 + 20.8 + 0 + 0 + 30.8 $
$ \text{Total Direct Runoff} = 62.4 \text{ mm} $
Therefore, the total direct runoff of the event is 62.4 mm.
Important Questions from Runoff Estimation
- A $0.80\text{ km}^2$ agricultural watershed has a slope of $0.30\%$ and the maximum length of travel of water is $1\text{ km}$. The 10-year maximum depth of rainfall is tabulated below:
Duration (minutes) 5 10 20 30 40 60 Maximum depth of rainfall (mm) 16 24 39 50 60 65
Half of the watershed has row crops (runoff coefficient = $0.40$), whilst the other half of the watershed has pasture (runoff coefficient = $0.35$). The peak flow rate for the watershed (in $\text{m}^3\text{.s}^{-1}$) for the 10-year return period is ________. (Rounded off to two decimal places) - The mass curve of a rainfall event of 100 min duration over a catchment is given in the table.
If the initial loss is 0.6 cm and $\phi$-index is 0.6 cm/hour, the total surface runoff from the catchment is ______ cm (rounded off to one decimal place).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 Rational method is used to estimate
- A $100 \text{ ha}$ watershed received rainfall at a rate of $5 \text{ cm h}^{-1}$ for $2 \text{ hours}$. If the runoff generated by the storm was at the rate of $1 \text{ m}^3 \text{ s}^{-1}$ for $10 \text{ hours}$, the runoff coefficient for the watershed would be
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 _____________
