Discussion
The ordinates of a 6-hour S-hydrograph of a catchment are given in Table below. The catchment has phi-index of 0.25 cm h-1 and baseflow of 10.5 m3 s-1. The peak of the flood hydrograph generated from this catchment due to a storm of 45 mm received during the first 6 h in m3 s-1 is
| Time (h) | 0 | 6 | 12 | 18 | 24 | 30 | 36 | 42 | 48 |
| Ordinate (m3 s-1) | 0 | 30 | 90 | 180 | 252 | 306 | 342 | 360 | 360 |
Answer: C
The peak of the flood hydrograph is 280.5 m3/s.
The 6-hour unit hydrograph is the S-curve minus itself lagged by 6 hours: 0, 30, 60, 90, 72, 54, 36, 18, 0 m3/s at 0, 6, ..., 48 h, with a peak of 90 m3/s at 18 h.
\[ \begin{aligned} \text{loss} &= 0.25 \times 6 = 1.5\ \text{cm} \\ \text{excess} &= 4.5 - 1.5 = 3.0\ \text{cm} \\ \text{peak} &= 3 \times 90 + 10.5 = 280.5 \end{aligned} \]
270.0 is the peak direct runoff without the baseflow, and 259.5 subtracts the baseflow instead of adding it. 349.5 would need 339 m3/s of direct runoff, far more than 3 cm of rainfall excess can produce from a unit hydrograph whose peak is 90 m3/s per cm.
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