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Gumbel design flood, safety factor, confidence limits

Short answer

T =252 yr from 35-yr life and 87% reliability; Gumbel K =4.419, predicted flood =18495 m³/s. With safety factr 1.5, adopt 27743 m³/s (margin 9248 m³/s). 95% confidence limits: 18495 ± 1.96×3504 → 11627 and 25364 m³/s.

Annual flood peak flood of a river for 20 years yielded a mean value of 5460 m3^3/s and the standard deviation of 2950 m3^3/s. The proposed hydraulic project on this river has an expected life of 35 years and reliability of project is 87%.

(i) Using Gumbel's method predict the flood discharge for the project if the value of yˉn=0.5402\bar{y}_n = 0.5402 and Sn=1.1285S_n = 1.1285.

(ii) What discharge is to be adopted if the safety factor for flood magnitude is taken as 1.5 and also determine safety margin on this basis.

(iii) Calculate the confidence limits at 95% confidence probability, f(c)=1.96f(c) = 1.96.

(i) Design flood

Reliability = 0.87, so risk R=0.13R=0.13 over n = 35 years.

1T=1−(0.87)1/35 ⇒ T=252 years\frac1T=1-(0.87)^{1/35}\ \Rightarrow\ T=252 \text{ years} yT=5.527,K=5.527−0.54021.1285=4.419y_T=5.527,\qquad K=\frac{5.527-0.5402}{1.1285}=4.419 xT=5460+4.419×2950=18495 m3/sx_T=5460+4.419\times2950=18495 \text{ m}^3/\text{s}

Answer: predicted flood for the project ≈ 18495 m³/s (T ≈ 252 years).

(ii) Safety factor 1.5

Qadopted=1.5×18495=27743 m3/sQ_{adopted}=1.5\times18495=27743 \text{ m}^3/\text{s}

Safety margin = adopted − predicted flood = 27743 − 18495 = 9248 m³/s (that is 50% of the predicted flood).

Answer: adopt 27743 m³/s; safety margin = 9248 m³/s.

(iii) 95% confidence limits

b=1+1.3K+1.1K2=5.312,Se=bσN=5.312×295020=3504 m3/sb=\sqrt{1+1.3K+1.1K^2}=5.312,\qquad S_e=\frac{b\sigma}{\sqrt N}=\frac{5.312\times2950}{\sqrt{20}}=3504 \text{ m}^3/\text{s} xT′=18495±1.96×3504=18495±6868x_T'=18495\pm1.96\times3504=18495\pm6868

Answer: 95% confidence limits are 11627 m³/s and 25364 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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