How is T found for 419 m^3/s?

K = (419 - mean)/sigma, y = y_n + K S_n, then T = 1/(1 - exp(-exp(-y))) = 3.4 years

2080 Baisakh · 10 marks

Gumbel 30-year flood, Lothar River 1989-2009

Short answer

Mean = 281.35 m³/s, σ = 281.25 m³/s. For T=30 yr: y_30=3.384, K=2.692, x_30 = 281.35+2.692×281.25 = 1038 m³/s. For 1989 peak (419 m³/s): K=0.489, y_T=1.044, T ≈ 3.4 years.

The observed annual peak flood for Lothar River from 1989 to 2009 are as follows:

Year	1989	1990	1991	1992	1993	1994	1995	1996	1997	1998
Peak discharge (m $^3$ /s)	419	320	65	150	1170	392	86	74.7	219	142

Year	1999	2000	2001	2002	2003	2004	2005	2007	2008	2009
Peak discharge (m $^3$ /s)	787	245	539	69.2	245	293	58.5	77.4	67.2	208

Find the flood value with return period of 30 years using Gumbel's Method. [Use Reduced Mean = 0.5236 and Reduced Standard Variation = 1.0628 for sample size of 20]. What is the return period of peak flood occurred on 1989?

The record has no value for 2006, so the sample has n = 20 values, matching the given reduced values for a sample size of 20.

Year	x (m³/s)	x − mean	(x − mean)²
1989	419	137.7	18948
1990	320	38.7	1494
1991	65	-216.3	46807
1992	150	-131.3	17253
1993	1170	888.7	789699
1994	392	110.7	12243
1995	86	-195.3	38162
1996	74.7	-206.6	42704
1997	219	-62.3	3888
1998	142	-139.3	19418
1999	787	505.7	255682
2000	245	-36.3	1321
2001	539	257.7	66384
2002	69.2	-212.1	45008
2003	245	-36.3	1321
2004	293	11.7	136
2005	58.5	-222.8	49662
2007	77.4	-203.9	41596
2008	67.2	-214.1	45860
2009	208	-73.3	5380

\bar x=\frac{\sum x}{n}=\frac{5627.0}{20}=281.35 \text{ m}^3/\text{s}

\sigma=\sqrt{\frac{\sum(x-\bar x)^2}{n-1}}=\sqrt{\frac{1502965}{19}}=281.25 \text{ m}^3/\text{s}

Flood of 30-year return period

y_{30}=-\ln[\ln(30/29)]=3.384,\qquad K=\frac{3.384-0.5236}{1.0628}=2.692

x_{30}=\bar x+K\sigma=281.35+2.692\times281.25=1038 \text{ m}^3/\text{s}

Answer: 30-year flood ≈ 1038 m³/s.

Return period of the 1989 flood (419 m³/s)

K=\frac{419-281.35}{281.25}=0.489,\qquad y_T=0.5236+0.489\times1.0628=1.044

T=\frac{1}{1-e^{-e^{-y_T}}}=\frac{1}{1-e^{-e^{-1.044}}}=3.37 \text{ years}

Answer: the 1989 peak (419 m³/s) has a return period of ≈ 3.4 years.

✓ Answer checked by Claude Sonnet 5.5

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