Question

A 100-hectare watershed with uniform land use has a time of concentration of 2 hours. What will be the peak runoff rate at its outlet due to a storm with uniform rainfall intensity of 6 cm/h for a fixed return period? The runoff coefficient is 0.60.

• A. 5 m\(^3\)/sec
• B. 8 m\(^3\)/sec
• C. 10 m\(^3\)/sec
• D. 12 m\(^3\)/sec

Hint:

The Rational Method formula \(Q = CIA\) is simple, but unit conversion is the most common source of errors. Using the formula \( Q = \frac{CIA}{360} \) with I in mm/hr and A in hectares directly gives Q in m\(^3\)/s, which is a useful shortcut.

Answer 1

The Correct Option is C

Step 1: Identify the Rational Method formula. The Rational Method is used to estimate the peak runoff rate from a small watershed. The formula is: \[ Q = C \cdot I \cdot A \] where \(Q\) is the peak discharge, \(C\) is the runoff coefficient, \(I\) is the rainfall intensity for a duration equal to the time of concentration, and \(A\) is the watershed area.
Step 2: Ensure consistent units.
To get the discharge \(Q\) in m\(^3\)/s, the other parameters must be in compatible units. A common form of the equation is: \[ Q (\text{m}^3/\text{s}) = \frac{C \cdot I (\text{mm/h}) \cdot A (\text{ha})}{360} \] Let's convert the intensity to mm/h: \(I = 6\) cm/h \( = 60\) mm/h.
Step 3: Substitute the given values into the formula.
- Runoff coefficient, \(C = 0.60\)
- Rainfall intensity, \(I = 60\) mm/h
- Area, \(A = 100\) ha \[ Q = \frac{0.60 \times 60 \times 100}{360} = \frac{3600}{360} = 10 \text{ m}^3/\text{s} \]