Question

The ordinates of a 6-hour S-hydrograph of a catchment are given in the table. The catchment has phi-index of 0.25 cm h\(^{-1}\) and baseflow of 10.5 m\(^3\)/s. The peak of the flood hydrograph generated from this catchment due to a storm of 45 mm received during the first 6 h in m\(^3\)/s is: \[ \begin{array}{|c|c|} \hline \text{Time (h)} & 0 6 12 18 24 30 36 42 48
\hline \text{Ordinate (m}^3/s\text{)} & 0 30 90 180 252 306 342 360 360
\hline \end{array} \]

• A. 259.5
• B. 270.0
• C. 280.5
• D. 349.5

Hint:

Always subtract losses using phi-index, compute effective rainfall, scale unit hydrograph ordinates, and then add baseflow to obtain flood hydrograph peak.

Answer 1

The Correct Option is C

Step 1: Effective rainfall. Total rainfall during 6 h = 45 mm = 4.5 cm. Loss due to phi-index = \(0.25 \times 6 = 1.5 \, cm\). \[ P_{eff} = 4.5 - 1.5 = 3.0 \, cm \]

Step 2: Unit hydrograph scaling. Given hydrograph is an S-hydrograph (due to unit depth = 1 cm of rainfall excess). For 3.0 cm effective rainfall, ordinates are scaled by factor 3.

Step 3: Flood hydrograph peak. From table, peak of S-hydrograph = 90 (at 12 h) corresponds to 1 cm rainfall. For 3 cm rainfall: \[ Q_{peak, direct} = 3 \times 90 = 270 \, m^3/s \]

Step 4: Add baseflow. \[ Q_{peak, total} = Q_{peak, direct} + Q_{baseflow} \] \[ = 270 + 10.5 = 280.5 \, m^3/s \] \[ \boxed{280.5 \, m^3/s} \]