How are the mean and standard deviation found from the two given floods?

Compute K for T = 100 and 500, then sigma = difference in x divided by difference in K

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Gumbel flood, confidence limits, hydrologic risk

Short answer

From Gumbel fit using given floods, mean=4358.4 m³/s, σ=1516.8 m³/s. 200-yr flood: 10849 m³/s, 1000-yr: 13046 m³/s. 95% confidence limits: 200-yr 8044–13653, 1000-yr 9424–16668 m³/s. Risk for 12300 m³/s over 25 yr: 4.2%. Design flow for 99% reliability over 25 yr: 14289 m³/s.

A river, whose annual flood peak can be represented by Gumbel distribution, has 100-year and 500-year return period flood of magnitude 9900 m $^3$ /s and 12100 m $^3$ /s respectively. The sample size is n = 30 ( $\bar{y}_n = 0.536$ , $S_n = 1.1124$ ).

i) What is the magnitude of 200 year and 1000 year flood?

ii) What are 95% and 80% confidence limits for 200 year and 1000 year flood if $f(95\%) = 1.96$ and $f(80\%) = 1.28$ .

iii) A hydraulic structure of 25 year life was designed for 12300 m $^3$ /s peak flow. What is the hydrologic risk of the structure?

iv) What peak flow should be taken into consideration if you want the structure to be 99% reliable for a structure life of 25 years.

Gumbel's method: $x_T = \bar x + K\sigma$ , with $K=\dfrac{y_T-\bar y_n}{S_n}$ and $y_T = -\ln\!\left[\ln\dfrac{T}{T-1}\right]$ . Here $\bar y_n=0.536$ , $S_n=1.1124$ (n = 30).

(i) 200-year and 1000-year floods

First find $\bar x$ and $\sigma$ from the two given floods.

T (yr)	$y_T$	$K=(y_T-0.536)/1.1124$
100	4.600	3.653
500	6.214	5.104

\sigma = \frac{12100-9900}{5.104-3.653} = 1516.8 \text{ m}^3/\text{s}, \qquad \bar x = 9900 - 3.653\times 1516.8 = 4358.4 \text{ m}^3/\text{s}

T (yr)	$y_T$	$K$	$x_T=\bar x+K\sigma$ (m³/s)
200	5.296	4.279	10849
1000	6.907	5.727	13046

Answer: $x_{200}$ ≈ 10849 m³/s and $x_{1000}$ ≈ 13046 m³/s.

(ii) Confidence limits

x_T' = x_T \pm f(c)\,S_e, \qquad S_e = \frac{b\,\sigma}{\sqrt{N}},\quad b=\sqrt{1+1.3K+1.1K^2}

T	K	b	$S_e$ (m³/s)	95% limits (f = 1.96)	80% limits (f = 1.28)
200	4.279	5.167	1431	8044 to 13653	9017 to 12680
1000	5.727	6.673	1848	9424 to 16668	10680 to 15411

(iii) Hydrologic risk for 12300 m³/s, n = 25 years

K=\frac{12300-4358.4}{1516.8}=5.236, \quad y_T=0.536+5.236\times 1.1124=6.360

T=\frac{1}{1-e^{-e^{-y_T}}} = 579 \text{ years}

\text{Risk}=1-\left(1-\frac1T\right)^{n}=1-\left(1-\frac{1}{579}\right)^{25} = 0.0423

Answer: T ≈ 579 years and the risk ≈ 4.2%.

(iv) Flood for 99% reliability over 25 years

Reliability = 0.99 means risk = 0.01.

1-\frac1T = (0.99)^{1/25} \Rightarrow T = 2488 \text{ years}

y_T=7.819, \quad K=6.547, \quad x_T = 4358.4 + 6.547\times1516.8 = 14289 \text{ m}^3/\text{s}

Answer: design peak flow ≈ 14289 m³/s (return period ≈ 2488 years).

✓ Answer checked by Claude Sonnet 5.5

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Gumbel flood, confidence limits, hydrologic risk

(i) 200-year and 1000-year floods

(ii) Confidence limits

(iii) Hydrologic risk for 12300 m³/s, n = 25 years

(iv) Flood for 99% reliability over 25 years

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