A watershed has 1.8 $km^2$ of cultivated area, 2.2 $km^2$ of forest land and 1.4 $km^2$ of grassed area. The runoff coefficient of cultivated area, forest land and grassed area are 0.25, 0.15 and 0.30, respectively. The main drainage channel has a fall of 25 m in the total length of 2.5 km. The Intensity-Duration-Frequency relationship for the watershed is expressed as, $I = \frac{70 T^{0.25}}{(t_c + 15)^{0.4}}$ where, I - intensity in cm $h^{-1}$, T - recurrence interval in years and $t_c$ - time of concentration in minutes. For a recurrence interval of 20 years, the peak rate of runoff for the watershed in $m^3s^{-1}$ will be __________ .
Detailed Solution
To determine the peak rate of runoff for the watershed, we will use the Rational Method. The process involves calculating the weighted runoff coefficient, the time of concentration, the rainfall intensity, and finally the peak discharge.
1. Calculate the Weighted Runoff Coefficient (\(C_{avg}\))
The watershed consists of three different land types. We calculate the total area (\(A\)) and the sum of the products of each area and its respective runoff coefficient (\(\sum C \cdot A\)).
- Cultivated Area: \(A_1 = 1.8 \text{ km}^2\), \(C_1 = 0.25\)
- Forest Land: \(A_2 = 2.2 \text{ km}^2\), \(C_2 = 0.15\)
- Grassed Area: \(A_3 = 1.4 \text{ km}^2\), \(C_3 = 0.30\)
Total Area \(A\):
$$ A = 1.8 + 2.2 + 1.4 = 5.4 \text{ km}^2 $$
Calculating \(\sum C \cdot A\):
$$ \sum (C \cdot A) = (0.25 \times 1.8) + (0.15 \times 2.2) + (0.30 \times 1.4) $$ $$ \sum (C \cdot A) = 0.45 + 0.33 + 0.42 = 1.20 \text{ km}^2 $$
Weighted Runoff Coefficient \(C_{avg}\):
$$ C_{avg} = \frac{\sum (C \cdot A)}{A} = \frac{1.20}{5.4} \approx 0.2222 $$
2. Calculate Time of Concentration (\(t_c\))
To find the time of concentration, we use the Kirpich Formula, which is standard for such hydrological problems:
$$ t_c = 0.01947 \cdot L^{0.77} \cdot S^{-0.385} $$
Where:
- \(L\) = Length of the main channel = \(2.5 \text{ km} = 2500 \text{ m}\)
- \(S\) = Average slope = \(\frac{\text{Fall}}{\text{Length}} = \frac{25}{2500} = 0.01\)
Calculation:
$$ t_c = 0.01947 \cdot (2500)^{0.77} \cdot (0.01)^{-0.385} $$ $$ t_c = 0.01947 \cdot (411.08) \cdot (5.888) \approx 47.13 \text{ minutes} $$
3. Calculate Rainfall Intensity (\(I\))
Using the provided Intensity-Duration-Frequency (IDF) relationship with a recurrence interval \(T = 20 \text{ years}\) and \(t_c = 47.13 \text{ min}\):
$$ I = \frac{70 \cdot T^{0.25}}{(t_c + 15)^{0.4}} $$ $$ I = \frac{70 \cdot (20)^{0.25}}{(47.13 + 15)^{0.4}} $$ $$ I = \frac{70 \cdot 2.1147}{(62.13)^{0.4}} = \frac{148.03}{5.2166} \approx 28.377 \text{ cm/h} $$
Converting intensity to \(\text{mm/h}\) for the standard Rational Formula:
$$ I = 28.377 \times 10 = 283.77 \text{ mm/h} $$
4. Calculate Peak Rate of Runoff (\(Q_p\))
The peak rate of runoff in \(\text{m}^3/\text{s}\) is given by the Rational Formula:
$$ Q_p = \frac{C \cdot I \cdot A}{3.6} $$
Using our calculated \(\sum (C \cdot A)\) directly:
$$ Q_p = \frac{1.20 \times 283.77}{3.6} $$ $$ Q_p = \frac{340.524}{3.6} \approx 94.59 \text{ m}^3/\text{s} $$
Summary: The peak rate of runoff for the watershed is approximately \(94.59 \text{ m}^3/\text{s}\), which falls within the expected range of 85 to 95.
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 _____________
