The ordinate of a 4-hour unit hydrograph at 2 hour time intervals starting from time t = 0 are as follows : $0, 20, 80, 128, 152, 135, 88, 55, 24, 7, 1 and 0 m^{3} / s$ . The peak of 2-hour unit hydrograph would be_______ $m^{3} / s$ ,

168

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345

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Solution

The correct option is A 168
$\begin{matrix} Time & 4h UH & Lag 4h & S_{A} - curve & S_{B} - curve & S_{A} - S_{B} & 2h UH S-Curve & = \frac{S_{A} - S_{B} \times 4}{2} Addition 0 & 0 & - & 0 & - & 0 & 0 2 & 20 & - & 20 & 0 & 20 & 40 4 & 80 & 0 & 80 & 20 & 60 & 120 6 & 128 & 20 & 148 & 80 & 68 & 136 8 & 152 & 80 & 232 & 148 & 84 & 168 10 & 135 & 148 & 283 & 232 & 51 & 102 12 & 88 & 232 & 320 & 283 & 37 & 74 14 & 55 & 283 & 338 & 320 & 18 & 36 16 & 24 & 320 & 344 & 338 & 6 & 12 18 & 7 & 338 & 345 & 344 & 1 & 2 20 & 1 & 344 & 345 & 345 & 0 & 0 22 & 0 & 345 & 345 & 345 & 0 & 0 \end{matrix}$

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