---
title: "Gumbel design flood, 21 years data, 40-year life"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-21yr-life40-rel85-b"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-21yr-life40-rel85-b
site: BCT Engineering
---

# Gumbel design flood, 21 years data, 40-year life

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2082 Bhadra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-21yr-life40-rel85-b
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Analysis of the annual flood peak of a river of 21 years yielded a mean of 8520 m$^3$/s and standard deviation of 3900 m$^3$/s. A proposed water control project on this river is to have an expected life of 40 years. The acceptable reliability by the design policy is 85%. Using Gumbel's Method recommend the flood discharge for this project. Take $y_n = 0.5252$ and $S_n = 1.0696$ for 21 years.

## Short answer

Risk=0.15 gives T=247 years. Frequency factor K=(5.506-0.5252)/1.0696=4.657. Design flood = 8520 + 4.657×3900 = 26681 m³/s.

## Answer

#### Design flood
Reliability = 0.85, so the risk $R=1-0.85=0.15$ in the life of n = 40 years.

$$
R=1-\left(1-\frac1T\right)^{n}\ \Rightarrow\ \frac1T=1-(0.85)^{1/40}\ \Rightarrow\ T=247 \text{ years}
$$

$$
y_T=-\ln\left[\ln\frac{T}{T-1}\right]=5.506, \qquad K=\frac{y_T-\bar y_n}{S_n}=\frac{5.506-0.5252}{1.0696}=4.657
$$

$$
x_T=\bar x+K\sigma=8520+4.657\times3900=26681 \text{ m}^3/\text{s}
$$

**Answer: recommended design flood ≈ 26681 m³/s (return period ≈ 247 years).**

## Check yourself

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

1. With reliability 85%, what risk is used?
   A. 0.015
   B. 0.15
   C. 0.40
   D. 0.85
   Answer: B. Risk = 1 - reliability = 0.15. (Key point: Risk)

2. Which equation gives T for n = 40?
   A. T = 1/0.15
   B. T = 40 x 0.85
   C. T = 40/0.15
   D. 1/T = 1 - 0.85^(1/40), T about 247 years
   Answer: D. Solving 0.15 = 1 - (1 - 1/T)^40 gives T of about 247 years. (Key point: Return period)

3. Which gives K with y_T = 5.506?
   A. (5.506 + 0.5252)/1.0696
   B. 5.506/1.0696
   C. (5.506 - 1.0696)/0.5252
   D. (5.506 - 0.5252)/1.0696 = 4.657
   Answer: D. K = (y_T - y_n)/S_n. (Key point: Frequency factor)

4. What is the design flood?
   A. 3900 + 4.657 x 8520
   B. 8520 + 4.657
   C. 4.657 x 8520
   D. 8520 + 4.657 x 3900 = 26681 m^3/s
   Answer: D. x_T = mean + K sigma. (Key point: Design flood)

## Close variants set in other exams

- [Gumbel design flood, 40-year life, 85% reliability](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-21yr-life40-rel85) — Engineering Hydrology, chapter 6, asked 2×
- [Gumbel design flood, 43 years data, 85% reliability](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-43yr-330-187-life50) — Engineering Hydrology, chapter 6, asked 2×

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