Question

A two-hour duration storm event with uniform excess rainfall of 3 cm occurred on a watershed. The ordinates of streamflow hydrograph resulting from this event are given in the table. \[ \begin{array}{|c|c|} \hline \text{Time (hours)} & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \text{Streamflow (m}^3/\text{s}) & 10 & 16 & 34 & 40 & 31 & 25 & 16 & 10 \\ \hline \end{array} \] Considering a constant baseflow of 10 m\(^3\)/s, the peak flow ordinate (in m\(^3\)/s) of the one-hour unit hydrograph for the watershed is \(\underline{\hspace{2cm}}\). (in integer)

Hint:

The peak flow ordinate for a unit hydrograph is calculated by subtracting the baseflow from the maximum streamflow observed in the hydrograph.

Answer 1

Correct Answer: 12

The peak flow ordinate of the one-hour unit hydrograph can be determined by first calculating the excess flow, which is the total streamflow minus the baseflow. The peak excess flow in the hydrograph is the maximum value reached after accounting for baseflow.
The maximum streamflow in the table is 40 m\(^3\)/s, and the baseflow is 10 m\(^3\)/s. Thus, the peak excess flow is:
\[ \text{Peak excess flow} = 40 - 10 = 30 \, \text{m}^3/\text{s}. \] Therefore, the peak flow ordinate of the one-hour unit hydrograph is \( \boxed{30} \, \text{m}^3/\text{s} \).