How do you get sigma from the 100- and 50-year floods?

sigma = (1200 - 1060)/(K100 - K50) = 223.1

What is the 95% interval for the 500-year flood?

1524 ± 1.96 x 245.3 = 1043 to 2004

What does the factor b = sqrt(1 + 1.3K + 1.1K^2) do?

Scales the standard error, increasing it for large K

2082 Baisakh · 4+2+2 marks

Gumbel 500-year flood, 95% limits, 30 years

Short answer

Subtracting equations: σ = (1200-1060)/(3.653-3.026)=223.1 m³/s, mean=385.1 m³/s. 500-yr: K=5.104, discharge=1524 m³/s. 95% confidence limits: 1524±1.96×245.3 = 1043 to 2004 m³/s.

Using 30 years data and Gumbel's method, the flood magnitude for return period of 100 and 50 years for a river are found to be 1200 and 1060 m $^3$ /s respectively.

(i) Determine the mean and standard deviation of the data used. Take values of reduced mean and reduced standard deviation in Gumbel's extreme value distribution for n = 30 as 0.5362 and 1.1124. (ii) Estimate the magnitude of the flood with a return period of 500 years. (iii) What are the 95% confidence limits for this estimate if $f(95\%) = 1.96$ .

(i) Mean and standard deviation

$\bar y_n=0.5362$ , $S_n=1.1124$ , $y_T=-\ln[\ln(T/(T-1))]$ , $K=(y_T-\bar y_n)/S_n$ , $x_T=\bar x+K\sigma$ .

T (yr)	$y_T$	$K$	$x_T$ (m³/s)
100	4.600	3.653	1200
50	3.902	3.026	1060

Subtract the two equations:

\sigma=\frac{1200-1060}{3.653-3.026}=223.1 \text{ m}^3/\text{s}

\bar x=1200-3.653\times223.1=385.1 \text{ m}^3/\text{s}

Answer: mean = 385.1 m³/s, standard deviation = 223.1 m³/s.

(ii) 500-year flood

y_{500}=-\ln[\ln(500/499)]=6.214,\qquad K=\frac{6.214-0.5362}{1.1124}=5.104

x_{500}=385.1+5.104\times223.1=1524 \text{ m}^3/\text{s}

Answer: $x_{500}$ ≈ 1524 m³/s.

(iii) 95% confidence limits of the 500-year flood

b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(5.104)+1.1(5.104)^2}=6.024

S_e=\frac{b\sigma}{\sqrt N}=\frac{6.024\times223.1}{\sqrt{30}}=245.3 \text{ m}^3/\text{s}

x_T'=x_T\pm f(c)S_e=1524\pm1.96\times245.3=1524\pm481

Answer: 95% confidence limits are 1043 m³/s and 2004 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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Gumbel 500-year flood, 95% limits, 30 years

(i) Mean and standard deviation

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