How is the design return period T found from R and n?

From R = 1 - (1 - 1/T)^n, so 1/T = 1 - (1 - R)^(1/n)

Asked 2/27
2070 Chaitra · 9 marks

Gumbel design flood, 43 years data, 85% reliability

Short answer

With 85% reliability over 50-year life, design return period is 308 years. Using Gumbel with sample size 43 (y_bar_n=0.5453, S_n=1.1480), frequency factor K=4.515 and design flood=1174 m³/s.

Analysis of the annual flood peak of a river for 43 years yielded a mean of 330 m $^3$ /s and a standard deviation of 187 m $^3$ /s. A proposed water control project on this river is to have an expected life of 50 years. Policy decision of the project allows an acceptable reliability of 85%. Using Gumbel's method, recommend the flood discharge for this project.

A table for reduced mean ( $\bar{y}_n$ ) and reduced standard deviation ( $S_n$ ) is given below:

N	40	41	42	43	44	45
$\bar{y}_n$	0.5436	0.5442	0.5448	0.5453	0.5458	0.5463
$S_n$	1.1413	1.1436	1.1458	1.1480	1.1499	1.1519

For N = 43 years the table gives $\bar y_n=0.5453$ and $S_n=1.1480$ .

Step 1: Design return period

Reliability 85% means risk $R=1-0.85=0.15$ during the life n = 50 years.

R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/50}\ \Rightarrow\ T=308 \text{ years}

Step 2: Reduced variate and frequency factor

y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.729

K=\frac{y_T-\bar y_n}{S_n}=\frac{5.729-0.5453}{1.1480}=4.515

Step 3: Design flood

x_T=\bar x+K\sigma=330+4.515\times187=1174 \text{ m}^3/\text{s}

Answer: recommended design flood ≈ 1174 m³/s (return period ≈ 308 years).

✓ Answer checked by Claude Sonnet 5.5

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Gumbel design flood, 43 years data, 85% reliability

Step 1: Design return period

Step 2: Reduced variate and frequency factor

Step 3: Design flood

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