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Gumbel design flood, 40-year life, 85% reliability

Short answer

Return period from 40-yr life and 85% reliability: T = 1/[1 - (0.85)^(1/40)] = 247 yr. Frequency factor K = (5.506-0.5252)/1.0696 =4.657; design flood =8520+4.657×3900 =26681 m³/s. 80% confidence limits: 26681 ± 1.282×4731 → 20615 and 32746 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%.

i) 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.

ii) What would the 80% confidence limit of the above flood be if f(c)=1.282f(c) = 1.282 at 80% confidence level.

(i) 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).

(ii) 80% confidence limits

b=1+1.3K+1.1K2=5.559,Se=bσN=5.559×390021=4731 m3/sb=\sqrt{1+1.3K+1.1K^2}=5.559, \qquad S_e=\frac{b\sigma}{\sqrt N}=\frac{5.559\times3900}{\sqrt{21}}=4731 \text{ m}^3/\text{s} xT′=26681±1.282×4731=26681±6065x_T'=26681\pm1.282\times4731=26681\pm6065

Answer: 80% confidence limits are 20615 m³/s and 32746 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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