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 _____________
Hydrology Solution: Direct Runoff Calculation using CN Method
Given Data Analysis
The question requires calculating the direct runoff ($Q$) for a watershed using the rainfall data and the Curve Number (CN) method. Key parameters are provided.
- Watershed Area: $40$ ha (This is contextual information and not directly used in the runoff calculation formula itself).
- Curve Number (CN): $80$ (specified for Antecedent Moisture Condition AMC-II).
- Initial Abstraction ($I_a$): Given as $20$% of the Maximum Retention ($S$).
- Rainfall Event ($P$): $40$ mm.
- Requirement: Calculate Direct Runoff ($Q$) in mm, rounded to two decimal places.
Key Formulas for CN Method
The calculation relies on the following standard formulas:
- Maximum Soil Moisture Retention ($S$): This relates to the CN value. Using the standard SCS formula, $S$ (in inches) is calculated as $S = \frac{1000}{CN} - 10$. To use metric units consistently with the given rainfall ($P$), we convert $S$ to millimeters (mm), knowing $1$ inch $= 25.4$ mm. Thus, $S_{mm} = 25.4 \times (\frac{1000}{CN} - 10)$.
- Initial Abstraction ($I_a$): The problem states $I_a = 0.20 \times S$.
- Direct Runoff ($Q$): The amount of direct runoff is calculated using $Q = \frac{(P - I_a)^2}{(P - I_a + S)}$, provided that the rainfall $P$ exceeds the initial abstraction $I_a$.
Step-by-Step Calculation of Runoff
Step 1: Calculate Maximum Retention ($S$) in mm
First, calculate $S$ in inches using $CN = 80$:
$S_{inches} = \frac{1000}{80} - 10 = 12.5 - 10 = 2.5$ inches.
Now, convert $S$ to millimeters:
$S = 2.5 \text{ inches} \times 25.4 \text{ mm/inch} = 63.5$ mm.
Step 2: Calculate Initial Abstraction ($I_a$) in mm
Using the given relationship $I_a = 0.20 \times S$:
$I_a = 0.20 \times 63.5 \text{ mm} = 12.7$ mm.
Step 3: Verify Rainfall Condition for Runoff Occurrence
Compare the rainfall depth ($P$) with the initial abstraction ($I_a$):
Given $P = 40$ mm and calculated $I_a = 12.7$ mm.
Since $P (40 \text{ mm}) > I_a (12.7 \text{ mm})$, direct runoff will occur.
Step 4: Calculate Direct Runoff ($Q$) in mm
Apply the direct runoff formula $Q = \frac{(P - I_a)^2}{(P - I_a + S)}$:
First, calculate the difference $P - I_a$:
$P - I_a = 40 \text{ mm} - 12.7 \text{ mm} = 27.3$ mm.
Next, calculate the denominator $P - I_a + S$:
$P - I_a + S = 27.3 \text{ mm} + 63.5 \text{ mm} = 90.8$ mm.
Now, calculate $Q$:
$Q = \frac{(27.3)^2}{90.8} = \frac{745.29}{90.8}$
$Q \approx 8.208589...$ mm.
Step 5: Round the Result
The question asks to round the result to two decimal places.
$Q \approx 8.21$ mm.
Final Answer Check
The computed direct runoff value of $8.21$ mm is consistent with the provided correct answer range (between 8.1 and 8.3).
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
- 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)
