Understanding the peak discharge from a watershed is crucial in hydrology and civil engineering, especially for designing drainage systems and managing stormwater. The question asks us to determine the peak discharge from a watershed using given parameters.

Watershed Peak Discharge Calculation

To calculate the peak discharge, we will use the Rational Method. This method is widely used for estimating the peak rate of runoff from small urban and rural drainage areas. The Rational Method is applicable when the storm duration is equal to or greater than the time of concentration, which is stated in the question.

Rational Method Formula

The formula for peak discharge (\(Q_p\)) using the Rational Method is:

\[Q_p = C \cdot I \cdot A\]

Where:

• \(Q_p\) is the peak discharge (commonly in cubic meters per second, $m^3/s$).
• \(C\) is the runoff coefficient (dimensionless), representing the ratio of runoff to rainfall.
• \(I\) is the average rainfall intensity for a duration equal to the time of concentration (commonly in millimeters per hour, $mm/hr$).
• \(A\) is the watershed area (commonly in hectares, $ha$).

When \(I\) is in \(mm/hr\) and \(A\) is in \(ha\), to get \(Q_p\) in \(m^3/s\), a conversion factor of \(\frac{1}{360}\) is often applied. This is because:

• 1 hectare = 10,000 \(m^2\)
• 1 mm = 0.001 m
• 1 hour = 3600 seconds

So, the formula becomes:

\[Q_p = \frac{C \cdot I \cdot A}{360}\]

Given Parameters for Peak Discharge

Let's list the values provided in the question:

• Watershed Area (A): 90 hectares
• Runoff Coefficient (C): 0.4
• Storm Intensity (I): 4.5 cm/hr

Unit Conversion for Rainfall Intensity

The rainfall intensity is given in cm/hr. For consistency with the standard metric Rational Method formula, we need to convert it to mm/hr.

• 1 cm = 10 mm
• Therefore, 4.5 cm/hr = 4.5 \(\times\) 10 mm/hr = 45 mm/hr

So, the converted rainfall intensity \(I\) = 45 mm/hr.

Applying the Rational Method for Peak Discharge

Now, we can substitute the values into the Rational Method formula:

\[Q_p = \frac{C \cdot I \cdot A}{360}\]

\[Q_p = \frac{0.4 \cdot 45 \text{ mm/hr} \cdot 90 \text{ ha}}{360}\]

\[Q_p = \frac{1620}{360}\]

\[Q_p = 4.5 \text{ } m^3/s\]

Thus, the peak discharge created by the storm is 4.5 \(m^3/s\).

The calculated peak discharge matches one of the provided options.