[round off to one decimal place]
Runoff Calculation Basics
The problem asks for the percentage change in daily runoff volume after altering land use in a watershed. We are given the rainfall amount and how initial abstraction relates to potential maximum soil retention ($S$). The runoff ($R$) is calculated using the Soil Conservation Service (SCS) method's runoff equation, adapted for the given initial abstraction ($Ia$) definition.
The governing equation for runoff ($R$) is:
$ R = \frac{(P - Ia)^2}{P - Ia + S} $Where:
- $P$ is the rainfall depth (90 mm).
- $Ia$ is the initial abstraction.
- $S$ is the potential maximum soil retention.
The problem states that $Ia$ is 25% of $S$, so $Ia = 0.25 \times S$. We need to calculate runoff for two scenarios: forest (initial) and cultivated land (final).
Initial Runoff Calculation (Forest Land)
For the initial forest condition:
- $S_1 = 136$ mm
- Calculate initial abstraction: $Ia_1 = 0.25 \times S_1 = 0.25 \times 136 \text{ mm} = 34 \text{ mm}$
- Check if rainfall exceeds initial abstraction: $P = 90 \text{ mm}$, $Ia_1 = 34 \text{ mm}$. Since $P > Ia_1$, runoff occurs.
- Calculate the initial runoff volume ($R_1$): $ R_1 = \frac{(P - Ia_1)^2}{P - Ia_1 + S_1} = \frac{(90 - 34)^2}{90 - 34 + 136} = \frac{56^2}{56 + 136} = \frac{3136}{192} \approx 16.333 \text{ mm} $
Final Runoff Calculation (Cultivated Land)
After land use alteration to cultivated land:
- $S_2 = 64$ mm
- Calculate the new initial abstraction: $Ia_2 = 0.25 \times S_2 = 0.25 \times 64 \text{ mm} = 16 \text{ mm}$
- Check if rainfall exceeds initial abstraction: $P = 90 \text{ mm}$, $Ia_2 = 16 \text{ mm}$. Since $P > Ia_2$, runoff occurs.
- Calculate the final runoff volume ($R_2$): $ R_2 = \frac{(P - Ia_2)^2}{P - Ia_2 + S_2} = \frac{(90 - 16)^2}{90 - 16 + 64} = \frac{74^2}{74 + 64} = \frac{5476}{138} \approx 39.681 \text{ mm} $
Percentage Change in Runoff Volume
To find the percentage change in runoff volume:
- Calculate the absolute change in runoff: $ \Delta R = R_2 - R_1 = 39.681 \text{ mm} - 16.333 \text{ mm} = 23.348 \text{ mm} $
- Calculate the percentage change relative to the initial runoff ($R_1$): $ \text{Percentage Change} = \left( \frac{\Delta R}{R_1} \right) \times 100 $ $ \text{Percentage Change} = \left( \frac{23.348}{16.333} \right) \times 100 \approx 1.4293 \times 100 \approx 142.93\% $
- Rounding to one decimal place gives 142.9%.
This value falls within the provided correct answer range of 142.9 to 143.6.
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 _____________
