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Gumbel peak flood and confidence limits, 75 years

Short answer

For P =0.4 %, T =250 yr; frequency factor K = (y_T - y_n)/S_n =4.174; peak flood =5561 +4.174×1718 =12733 m³/s. 90% confidence limits: 12733 ± 1.6×1004 → 11127 and 14339 m³/s.

In the time series data of annual peak flood for 75 years, the mean and standard deviation are found to be equal to 5561 m3^3/s and 1718 m3^3/s respectively. Using yn=0.556y_n = 0.556 and Sn=1.189S_n = 1.189 (for 75 yrs):

i) Determine the peak flood for 0.4% probability of exceedance by Gumbel's method.

ii) Compute 90% confidence limits for the above floods, using f(c)=1.6f(c) = 1.6 for 90% confidence level respectively.

(i) Peak flood for 0.4% probability of exceedance

T=1P=10.004=250 yearsT=\frac1P=\frac{1}{0.004}=250 \text{ years} yT=−ln⁡[ln⁡TT−1]=−ln⁡[ln⁡250249]=5.519y_T=-\ln\left[\ln\frac{T}{T-1}\right]=-\ln\left[\ln\frac{250}{249}\right]=5.519 K=yT−yˉnSn=5.519−0.5561.189=4.174K=\frac{y_T-\bar y_n}{S_n}=\frac{5.519-0.556}{1.189}=4.174 xT=xˉ+Kσ=5561+4.174×1718=12733 m3/sx_T=\bar x+K\sigma=5561+4.174\times1718=12733 \text{ m}^3/\text{s}

Answer: x250x_{250} ≈ 12733 m³/s.

(ii) 90% confidence limits

b=1+1.3K+1.1K2=1+1.3(4.174)+1.1(4.174)2=5.059b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(4.174)+1.1(4.174)^2}=5.059 Se=bσN=5.059×171875=1004 m3/sS_e=\frac{b\sigma}{\sqrt N}=\frac{5.059\times1718}{\sqrt{75}}=1004 \text{ m}^3/\text{s} xT′=xT±f(c)Se=12733±1.6×1004=12733±1606x_T'=x_T\pm f(c)S_e=12733\pm1.6\times1004=12733\pm1606

Answer: the 90% confidence limits are 11127 m³/s and 14339 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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