A rectangular catchment ABCD has an area of 7 hectares. The times of concentration from the four extreme points A, B, C and D to the outlet are 10, 20, 15 and 25 minutes, respectively. The rainfall intensity-duration relationship is given by \(I = \frac{25}{t+20}\), where I = rainfall intensity in cm/hr and t = time of concentration in minutes. The runoff coefficient of the catchment is 0.4. Determine the peak discharge from the catchment. (Enter the numerical value only in m\(^3\)/s.)
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Solution and Explanation
Step 1: Understanding the Question and Method:
We need to calculate the peak discharge from a catchment using the Rational Method. The required parameters are the runoff coefficient (C), rainfall intensity (I), and catchment area (A).
Step 2: Key Formula and Parameters:
The Rational Method formula is \( Q_p = C \cdot I \cdot A \). To get the discharge in m\(^3\)/s, we must use consistent units. A common form of the formula is: \[ Q_p (\text{m}^3/\text{s}) = \frac{C \cdot I (\text{mm/hr}) \cdot A (\text{hectares})}{360} \] Let's identify the parameters from the question:
- Runoff coefficient, \(C = 0.4\)
- Catchment area, \(A = 7\) hectares
- Time of concentration (\(t_c\)): This is the longest time of travel for water to reach the outlet. Given the times 10, 20, 15, and 25 minutes, the longest is \(t_c = 25\) minutes.
Step 3: Detailed Calculation:
- Calculate Rainfall Intensity (I): The intensity is calculated for a duration equal to the time of concentration, \(t_c = 25\) min. \[ I = \frac{25}{t_c + 20} = \frac{25}{25 + 20} = \frac{25}{45} = \frac{5}{9} \text{ cm/hr} \]
- Convert Intensity Units: The standard formula uses intensity in mm/hr. \[ I = \frac{5}{9} \text{ cm/hr} = \frac{5}{9} \times 10 \text{ mm/hr} = \frac{50}{9} \text{ mm/hr} \]
- Calculate Peak Discharge (\(Q_p\)): Now we use the rational formula with the converted units. \[ Q_p = \frac{C \cdot I (\text{mm/hr}) \cdot A (\text{ha})}{360} \] \[ Q_p = \frac{0.4 \times (\frac{50}{9}) \times 7}{360} = \frac{0.4 \times 50 \times 7}{9 \times 360} = \frac{140}{3240} \] \[ Q_p \approx 0.043209 \text{ m}^3/\text{s} \]
Step 4: Final Answer:
Rounding the result, the peak discharge is 0.043 m\(^3\)/s.
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