385.1 + 5.104 x 223.1 = 1524 m^3/s

Why is the interval narrower than a 95% one?

f(c) = 1.28 is smaller than 1.96

2081 Bhadra · 8 marks

Gumbel 500-year flood and 80% limits, Kamalamai River

Short answer

From 100-yr and 50-yr: σ=223.1 m³/s, mean=385.1 m³/s. 500-yr flood: K=5.104, discharge=1524 m³/s. 80% limits: 1524±1.28×245.3 = 1210 to 1838 m³/s.

30 years flood data of Kamalamai River has been used for frequency analysis and after fitting the Gumbel distribution the 100-year and 50-year floods are predicted equal to 1200 m $^3$ /s and 1060 m $^3$ /s respectively. Calculate the magnitude of flood for 500 years return period if reduced mean and reduced standard deviation in Gumbel's distribution for available data series are 0.5362 and 1.1124 respectively. Also calculate 80% confidence limits of 500-year flood if $f(c)$ for 80% is 1.28.

Statistics of the series

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

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.

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.

80% 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\pm1.28\times245.3=1524\pm314

Answer: 80% confidence limits are 1210 m³/s and 1838 m³/s.

✓ Answer checked by Claude Sonnet 5.5

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