---
title: "Gumbel 30-year flood, Lothar River 1989-2009"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-lothar-1989-2009"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-lothar-1989-2009
site: BCT Engineering
---

# Gumbel 30-year flood, Lothar River 1989-2009

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2080 Baisakh, 10 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-lothar-1989-2009
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

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?

## 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.

## Answer

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.**

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Why is n = 20 used?
   A. 2006 has no record, giving 20 values matching the table for 20
   B. Outliers removed
   C. Years 1989-2009 is 20 years
   D. The table is for 21
   Answer: A. 21 years minus the missing 2006 gives 20 values. (Key point: Sample size)

2. Which divisor is used for sigma?
   A. n + 1
   B. 1
   C. n - 1 = 19
   D. n = 20
   Answer: C. sigma = sqrt(1502965/19) = 281.25. (Key point: Std deviation)

3. x30 =
   A. 281.35 + 2.692 x 281.25 = 1038 m^3/s
   B. 281.35 x 2.692
   C. 1038 x 30
   D. 281.25 + 30
   Answer: A. K = (3.384 - 0.5236)/1.0628 = 2.692. (Key point: 30-year flood)

4. How is T found for 419 m^3/s?
   A. T = 30
   B. K = (419 - mean)/sigma, y = y_n + K S_n, then T = 1/(1 - exp(-exp(-y))) = 3.4 years
   C. T = 419/281
   D. T = (N+1)/m only
   Answer: B. Reverse the Gumbel steps: K, then y, then T. (Key point: Return period of a flood)

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