The peak of flood hydrograph due to a 3-hour duration storm in a catchment is 180 m\(^3\)/s. The total rainfall depth is 6.6 cm. It can be assumed that the average infiltration loss is 0.2 cm/h. There are no other losses. The base flow is constant at a value of 30 m\(^3\)/s.
The peak value of the 3-hour unit hydrograph for this catchment (in m\(^3\)/s) is ......... (round off to the nearest integer).
The peak value of the 3-hour unit hydrograph for this catchment (in m\(^3\)/s) is ......... (round off to the nearest integer).
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Solution and Explanation
The peak discharge is given as \( 180 \, {m}^3/{s} \), and the base flow is \( 30 \, {m}^3/{s} \).
The infiltration loss is 0.2 cm/h, and the total rainfall depth is 6.6 cm.
The peak of the 3-hour unit hydrograph \( Q_p \) is calculated using the formula: \[ Q_p = \frac{{Peak discharge} - {Base flow}}{R - \phi t} \] Where: - Peak discharge = \( 180 \, {m}^3/{s} \)
- Base flow = \( 30 \, {m}^3/{s} \)
- \( R = 6.6 \, {cm} \) (rainfall depth)
- \( \phi = 0.2 \, {cm/h} \) (infiltration loss)
- \( t = 3 \, {hours} \) (duration of the storm) Substitute the values into the formula: \[ Q_p = \frac{180 - 30}{6.6 - 0.2 \times 3} = \frac{150}{5.4} = 27.78 \, {m}^3/{s} \] Thus, the peak value of the 3-hour unit hydrograph is \( 25 \, {m}^3/{s} \).
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