---
title: "Return period of 350 m3/s design flood by Gumbel"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-return-period-350-25yr"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-return-period-350-25yr
site: BCT Engineering
---

# Return period of 350 m3/s design flood by Gumbel

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2080 Bhadra, 6 marks
- Video: [Gumbel's method in Hydrology for flood estimation | Flood Routing | GATE 2022 |](https://www.youtube.com/watch?v=gtUiLfn_E8U) — TechnoCivil
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-return-period-350-25yr
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

A hydraulic structure has been designed for a discharge of 350 m$^3$/s. From the 25 years annual flood series data, the mean and standard deviation are 140 and 65 m$^3$/s respectively. Calculate the return period for the design flood by using Gumbel's method. Take reduced mean variate and reduced standard deviation for this series as 0.5299 and 1.0876 respectively.

## Short answer

Frequency factor K = (350−140)/65 = 3.231. Reduced variate y_T = 0.5299 + 3.231×1.0876 = 4.044. Return period T = 1/[1−exp(−exp(−4.044))] ≈ 58 years.

## Answer

#### Frequency factor
$$
K=\frac{x_T-\bar x}{\sigma}=\frac{350-140}{65}=3.231
$$

#### Reduced variate
$$
y_T=\bar y_n+K S_n=0.5299+3.231\times1.0876=4.044
$$

#### Return period
$$
T=\frac{1}{1-e^{-e^{-y_T}}}=\frac{1}{1-e^{-e^{-4.044}}}=57.5 \text{ years}
$$

**Answer: the return period of the 350 m³/s design flood ≈ 58 years** (annual exceedance probability ≈ 0.0174).

## Check yourself

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

1. How is K found from the design flood?
   A. K = (350 + 140)/65
   B. K = 350/65
   C. K = 140/65
   D. K = (350 - 140)/65 = 3.231
   Answer: D. Rearranging x_T = mean + K sigma. (Key point: Frequency factor)

2. Then y_T =
   A. y_n + K S_n = 0.5299 + 3.231 x 1.0876 = 4.044
   B. y_n x K
   C. K - y_n
   D. K/S_n
   Answer: A. This inverts K = (y_T - y_n)/S_n. (Key point: Reduced variate)

3. T is obtained from
   A. T = exp(y_T)
   B. T = 1/y_T
   C. T = 1/(1 - exp(-exp(-y_T)))
   D. T = y_T
   Answer: C. Exceedance probability is 1 - exp(-exp(-y)), and T is its reciprocal. (Key point: Return period formula)

4. The return period is about
   A. 25 years
   B. 350 years
   C. 3 years
   D. 58 years
   Answer: D. P = 0.0174, T is about 57.5, so 58 years. (Key point: Result)

## Close variants set in other exams

- [Gumbel return period and confidence limit](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-return-period-300) — Engineering Hydrology, chapter 6, asked 2×

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