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Return period of 350 m3/s design flood by Gumbel

Short answer

Frequency factor K = (350−140)/65 = 3.231. Reduced variate y_T = 0.5299 + 3.231×1.0876 = 4.044. Return period T = 1/[1−exp(−exp(−4.044))] ≈ 58 years.

A hydraulic structure has been designed for a discharge of 350 m3^3/s. From the 25 years annual flood series data, the mean and standard deviation are 140 and 65 m3^3/s respectively. Calculate the return period for the design flood by using Gumbel's method. Take reduced mean variate and reduced standard deviation for this series as 0.5299 and 1.0876 respectively.

Frequency factor

K=xT−xˉσ=350−14065=3.231K=\frac{x_T-\bar x}{\sigma}=\frac{350-140}{65}=3.231

Reduced variate

yT=yˉn+KSn=0.5299+3.231×1.0876=4.044y_T=\bar y_n+K S_n=0.5299+3.231\times1.0876=4.044

Return period

T=11−e−e−yT=11−e−e−4.044=57.5 yearsT=\frac{1}{1-e^{-e^{-y_T}}}=\frac{1}{1-e^{-e^{-4.044}}}=57.5 \text{ years}

Answer: the return period of the 350 m³/s design flood ≈ 58 years (annual exceedance probability ≈ 0.0174).

✓ Answer checked by Claude Sonnet 5.5

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