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Gumbel return period and confidence limit

Short answer

Using Gumbel method with mean=140 m³/s, σ=50 m³/s, reduced mean=0.5224, Sn=1.1124 (N=30), K=(300-140)/50=3.20, yT=4.082, return period ≈60 years. 90% confidence limits: 239–361 m³/s.

A hydraulic structure is designed for a discharge of 300 m3^3/s. If the available flood data is for N years (reduced mean = 0.5224, reduced standard deviation = 1.1124) and the mean and standard deviation of the annual flood series are 140 m3^3/s and 50 m3^3/s respectively, calculate the return period for the design flood using Gumbel's method. Also estimate the 90% confidence limit, if f(90%)=1.645f(90\%) = 1.645.

The number of years N is not given; N = 30 is assumed because Sn=1.1124S_n=1.1124 is the Gumbel value for N = 30.

Return period

K=xT−xˉσ=300−14050=3.20K=\frac{x_T-\bar x}{\sigma}=\frac{300-140}{50}=3.20 yT=yˉn+KSn=0.5224+3.20×1.1124=4.082y_T = \bar y_n + K S_n = 0.5224 + 3.20\times 1.1124 = 4.082 T=11−e−e−yT=11−e−e−4.082=60 yearsT=\frac{1}{1-e^{-e^{-y_T}}} = \frac{1}{1-e^{-e^{-4.082}}} = 60 \text{ years}

Answer: return period of the design flood ≈ 60 years (probability of exceedance in any year = 0.0167).

90% confidence limit

b=1+1.3K+1.1K2=1+1.3(3.20)+1.1(3.20)2=4.053b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(3.20)+1.1(3.20)^2}=4.053 Se=b σN=4.053×5030=37.00 m3/sS_e=\frac{b\,\sigma}{\sqrt N}=\frac{4.053\times 50}{\sqrt{30}}=37.00 \text{ m}^3/\text{s} xT′=xT±f(c)Se=300±1.645×37.00=300±60.9x_T' = x_T \pm f(c)S_e = 300 \pm 1.645\times37.00 = 300 \pm 60.9

Answer: 90% confidence limits are 239 m³/s to 361 m³/s (for N = 30).

✓ Answer checked by Claude Sonnet 5.5

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