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Gumbel design flood, 21 years data, 40-year life

Short answer

Risk=0.15 gives T=247 years. Frequency factor K=(5.506-0.5252)/1.0696=4.657. Design flood = 8520 + 4.657×3900 = 26681 m³/s.

Analysis of the annual flood peak of a river of 21 years yielded a mean of 8520 m3^3/s and standard deviation of 3900 m3^3/s. A proposed water control project on this river is to have an expected life of 40 years. The acceptable reliability by the design policy is 85%. Using Gumbel's Method recommend the flood discharge for this project. Take yn=0.5252y_n = 0.5252 and Sn=1.0696S_n = 1.0696 for 21 years.

Design flood

Reliability = 0.85, so the risk R=1−0.85=0.15R=1-0.85=0.15 in the life of n = 40 years.

R=1−(1−1T)n ⇒ 1T=1−(0.85)1/40 ⇒ T=247 yearsR=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/40}\ \Rightarrow\ T=247 \text{ years} yT=−ln⁡[ln⁡TT−1]=5.506,K=yT−yˉnSn=5.506−0.52521.0696=4.657y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.506, \qquad K=\frac{y_T-\bar y_n}{S_n}=\frac{5.506-0.5252}{1.0696}=4.657 xT=xˉ+Kσ=8520+4.657×3900=26681 m3/sx_T=\bar x+K\sigma=8520+4.657\times3900=26681 \text{ m}^3/\text{s}

Answer: recommended design flood ≈ 26681 m³/s (return period ≈ 247 years).

✓ Answer checked by Claude Sonnet 5.5

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