---
title: "Gumbel flood for 500-year return period"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-100-50-500"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-100-50-500
site: BCT Engineering
---

# Gumbel flood for 500-year return period

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2078 Kartik, 8 marks
- Video: [Flood discharge at various return periods using Gumbel's extreme value distribution | Hydrology](https://www.youtube.com/watch?v=RE7Q4NDVPQQ) — The Study Sphere
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-100-50-500
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

For a river the estimated flood peaks for two return periods by the use of Gumbel's method are given below:

| Return period (yrs) | Peak flood (m$^3$/s) |
|---|---|
| 100 | 1020 |
| 50 | 850 |

What flood discharge in this river will have a return period of 500 yrs?

## Short answer

Using Gumbel's reduced variate, slope b = (1020-850)/(4.6001-3.9019) = 243.48 m³/s. Then 500-year flood = 1020 + 243.48×(6.2136-4.6001) ≈ 1413 m³/s.

## Answer

In Gumbel's method $x_T=\bar x+K\sigma = a + b\,y_T$, which is a straight line in $y_T$ (the reduced variate). Using two known points gives the line, so $\bar y_n$ and $S_n$ are not needed.

$$
y_T = -\ln\left[\ln\frac{T}{T-1}\right]
$$

| T (yr) | $y_T$ | $x_T$ (m³/s) |
|---|---|---|
| 50 | 3.9019 | 850 |
| 100 | 4.6001 | 1020 |
| 500 | 6.2136 | ? |

Slope of the line (equal to $\sigma/S_n$):

$$
b=\frac{1020-850}{4.6001-3.9019}=243.48 \text{ m}^3/\text{s}
$$

$$
x_{500}=1020+243.48\,(6.2136-4.6001)=1413 \text{ m}^3/\text{s}
$$

**Answer: the 500-year flood ≈ 1413 m³/s.**

## Check yourself

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

1. x_T in Gumbel's method is a straight line in:
   A. T
   B. ln x
   C. sqrt(T)
   D. The reduced variate y_T
   Answer: D. x_T = a + b y_T, so two points define it without y_n and S_n. (Key point: Linear relation)

2. The reduced variate is:
   A. y_T = -ln[ln(T/(T-1))]
   B. y_T = ln T
   C. y_T = 1/T
   D. y_T = ln[ln(T)]
   Answer: A. This is the standard Gumbel relation. (Key point: y_T formula)

3. The slope b from the two data points is:
   A. (1020-850)/(100-50) = 3.4
   B. (1020-850)/(4.6001-3.9019) = 243.5
   C. 1020/4.6001
   D. (1020+850)/2
   Answer: B. Slope is the difference in x over the difference in y_T. (Key point: Slope)

4. The 500-year flood is about:
   A. 1020 m3/s
   B. 1130 m3/s
   C. 1413 m3/s
   D. 1700 m3/s
   Answer: C. x = 1020 + 243.48 x (6.2136-4.6001) = 1413. (Key point: 500-year flood)

## Related in Engineering Hydrology

- [Gumbel 500-year flood and 80% limits, Kamalamai River](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-kamalamai-80cl) — Engineering Hydrology, chapter 6
- [Gumbel 200-year flood from 50 and 100-year floods](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-35yr-200yr-flood) — Engineering Hydrology, chapter 6
- [Gumbel analysis from 100 and 50-year floods](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-30yr-100-50-1200-1060) — Engineering Hydrology, chapter 6
- [Gumbel 500-year flood from two risk-based floods](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-risk-63-6-4200-5800) — Engineering Hydrology, chapter 6
- [Gumbel 30-year flood, Lothar River 1989-2009](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-lothar-1989-2009) — Engineering Hydrology, chapter 6
- [Gumbel 500-year flood, 95% limits, 30 years](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-30yr-100-50-1200-1060-2) — Engineering Hydrology, chapter 6

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