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
Calculating Total Surface Runoff using Phi-Index Method
The problem requires calculating the total surface runoff using a rainfall mass curve, initial loss, and the $\phi$-index. We are given:
- Total rainfall duration = 100 min
- Initial Loss (IL) = 0.6 cm
- $\phi$-index = 0.6 cm/hour
- Rainfall Mass Curve data.
Step 1: Convert $\phi$-index to cm/min
The $\phi$-index is given in cm/hour. Convert it to cm/min for consistency with the rainfall data:
$ \phi\text{-index} = \frac{0.6 \text{ cm}}{1 \text{ hour}} = \frac{0.6 \text{ cm}}{60 \text{ min}} = 0.01 \text{ cm/min} $
Step 2: Calculate Incremental Rainfall and Rates
Determine the rainfall in each 20-minute interval and the average rainfall rate during each interval.
| Time from start (min) | Cumulative Rainfall (cm) | Incremental Rainfall ($\Delta P$) (cm) | Duration ($\Delta t$) (min) | Average Rate ($\Delta P / \Delta t$) (cm/min) |
|---|---|---|---|---|
| 0-20 | 0.5 | 0.5 | 20 | $ \frac{0.5}{20} = 0.025 $ |
| 20-40 | 1.2 | $ 1.2 - 0.5 = 0.7 $ | 20 | $ \frac{0.7}{20} = 0.035 $ |
| 40-60 | 2.6 | $ 2.6 - 1.2 = 1.4 $ | 20 | $ \frac{1.4}{20} = 0.070 $ |
| 60-80 | 3.3 | $ 3.3 - 2.6 = 0.7 $ | 20 | $ \frac{0.7}{20} = 0.035 $ |
| 80-100 | 3.5 | $ 3.5 - 3.3 = 0.2 $ | 20 | $ \frac{0.2}{20} = 0.010 $ |
Step 3: Calculate Excess Rainfall for Each Interval
The $\phi$-index represents the rate of loss after the initial loss is met. Surface runoff is generated only when the rainfall rate exceeds the $\phi$-index. Calculate the excess rainfall rate ($Rate - \phi$) and the total excess rainfall depth for each interval where the rate is greater than $\phi$. Note: $\phi = 0.01$ cm/min.
| Interval (min) | Average Rate (cm/min) | $\phi$-index (cm/min) | Excess Rate (cm/min) | Duration ($\Delta t$) (min) | Excess Rainfall ($\Delta P_{excess}$) (cm) |
|---|---|---|---|---|---|
| 0-20 | 0.025 | 0.01 | $ 0.025 - 0.01 = 0.015 $ | 20 | $ 0.015 \times 20 = 0.30 $ |
| 20-40 | 0.035 | 0.01 | $ 0.035 - 0.01 = 0.025 $ | 20 | $ 0.025 \times 20 = 0.50 $ |
| 40-60 | 0.070 | 0.01 | $ 0.070 - 0.01 = 0.060 $ | 20 | $ 0.060 \times 20 = 1.20 $ |
| 60-80 | 0.035 | 0.01 | $ 0.035 - 0.01 = 0.025 $ | 20 | $ 0.025 \times 20 = 0.50 $ |
| 80-100 | 0.010 | 0.01 | $ 0.010 - 0.01 = 0.000 $ | 20 | $ 0.000 \times 20 = 0.00 $ |
Step 4: Calculate Total Surface Runoff
Sum the excess rainfall depths calculated for each interval. This sum represents the total surface runoff.
Total Surface Runoff = $ 0.30 + 0.50 + 1.20 + 0.50 + 0.00 = 2.50 \text{ cm} $
The calculation involves summing the rainfall depths occurring above the $\phi$-index rate during the periods when the rainfall intensity exceeded this rate. Rounding to one decimal place gives 2.5 cm.
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 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 _____________
- A 10 ha watershed received a 4-hour rainfall storm with rainfall intensity of 2, 1, 1.5, and 1 cm.h$^{-1}$ in each subsequent hour, respectively. Assuming phi-index ($\phi$) of 1 cm.h$^{-1}$, the volume of direct runoff, in m$^3$, is _________. (Rounded off to 2 decimal places)
