Question

The ordinates of a 1-hour unit hydrograph (UH) are given below: \includegraphics[width=0.5\linewidth]{66image.png} These ordinates are used to derive a 3-hour UH. The peak discharge (in m$^3$/s) for the derived 3-hour UH is ______ (rounded off to the nearest integer).

Hint:

When deriving unit hydrographs for longer durations, ensure that all ordinates are accurately summed for overlapping intervals.

Answer 1

Step 1: Understand the derivation of a 3-hour UH. To derive a 3-hour UH from the given 1-hour UH, we use the principle of superposition. The 3-hour UH is obtained by summing up the ordinates of the 1-hour UH spaced 3 hours apart. Let the ordinates of the derived 3-hour UH be \( Q_3 \). Step 2: Calculate the ordinates of the 3-hour UH. The calculation for \( Q_3 \) is done as follows: \[ Q_3(t) = Q_1(t) + Q_1(t-3) + Q_1(t-6), \] where \( Q_1(t) \) represents the ordinates of the 1-hour UH. For times less than 3 hours, only \( Q_1(t) \) contributes; for times between 3 and 6 hours, \( Q_1(t) + Q_1(t-3) \) contribute; and for times greater than 6 hours, all three terms contribute. Step 3: Tabulate the ordinates of the 3-hour UH. \begin{table}[h!] \centering \begin{tabular}{|c|c|c|c|c|} \hline Time (hours) & \( Q_1(t) \) & \( Q_1(t-3) \) & \( Q_1(t-6) \) & \( Q_3(t) \)
\hline 0 & 0 & - & - & 0
\hline 1 & 13 & - & - & 13
\hline 2 & 50 & - & - & 50
\hline 3 & 80 & 0 & - & 80
\hline 4 & 95 & 13 & - & 108
\hline 5 & 85 & 50 & - & 135
\hline 6 & 55 & 80 & 0 & 135
\hline 7 & 35 & 95 & 13 & 143
\hline 8 & 15 & 85 & 50 & 150
\hline 9 & 10 & 55 & 80 & 145
\hline 10 & 3 & 35 & 95 & 133
\hline 11 & 0 & 15 & 85 & 100
\hline 12 & 0 & 10 & 55 & 65
\hline \end{tabular} \caption{Derived 3-hour UH ordinates} \end{table} Step 4: Identify the peak discharge. From the table, the peak discharge is: \[ Q_{\text{peak}} = 150 \, \text{m\(^3\)/s}. \] Conclusion: The peak discharge for the 3-hour UH is \( 150 \, \text{m\(^3\)/s} \).