---
title: "Gumbel design flood, factor of safety 1.3, 45-year life"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-25yr-life45-fos1-3"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-25yr-life45-fos1-3
site: BCT Engineering
---

# Gumbel design flood, factor of safety 1.3, 45-year life

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2079 Bhadra, 4+2 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-25yr-life45-fos1-3
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Analysis of the annual flood peak data of a certain river covering the period of 25 years has the mean of 8520 and standard deviation of 3900 m$^3$/s. The proposed water control project near this site has an expected life of 45 years. The policy decision for project provides the reliability of 85%.

(i) Using Gumbel's method recommend peak discharge. Take $y_n = 0.5309$ and $S_n = 1.0915$

(ii) Calculate the design discharge if factor of safety is 1.3 and also the corresponding safety margin.

## Short answer

With 85% reliability in 45 years, T = 277 years. y_T = 5.624, K = 4.666, peak discharge x_T = 26717 m³/s. With safety factor 1.3: design discharge = 34732 m³/s, safety margin = 8015 m³/s.

## Answer

#### (i) Peak discharge by Gumbel's method

Given: $N = 25$ years, $\bar{x} = 8520\ \text{m}^3/\text{s}$, $\sigma = 3900\ \text{m}^3/\text{s}$, $y_n = 0.5309$, $S_n = 1.0915$, design life $n = 45$ years, reliability $= 85\%$.

**Return period.** Reliability $= (1 - 1/T)^n = 0.85$, so risk $\bar{R} = 1 - 0.85 = 0.15$.

$$
\begin{aligned} 1 - \frac{1}{T} &= (0.85)^{1/45} = 0.996395\\ T &= 277.4\ \text{years} \end{aligned}
$$

**Reduced variate.**

$$
y_T = -\ln\left[\ln\frac{T}{T-1}\right] = -\ln\left[\ln\frac{277.4}{276.4}\right] = 5.6236
$$

**Frequency factor.**

$$
K = \frac{y_T - y_n}{S_n} = \frac{5.6236 - 0.5309}{1.0915} = 4.6658
$$

**Flood magnitude.**

$$
x_T = \bar{x} + K\sigma = 8520 + (4.6658)(3900) = 26717\ \text{m}^3/\text{s}
$$

**Answer (i): recommended peak discharge $\approx$ 26717 m³/s** (return period about 277 years).

#### (ii) Design discharge with factor of safety 1.3

$$
Q_{design} = 1.3 \times 26717 = 34732\ \text{m}^3/\text{s}
$$

Safety margin $= Q_{design} - x_T = 34732 - 26717 = 8015\ \text{m}^3/\text{s}$ (30 % of $x_T$).

**Answer (ii): design discharge = 34732 m³/s; safety margin = 8015 m³/s.**

## Check yourself

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

1. A project of 45 years life needs 85% reliability. How is the design return period T found?
   A. T = 1/0.15 = 6.7 years
   B. Solve (1 - 1/T)^45 = 0.85, giving T of about 277 years
   C. T = 45 years, equal to the design life
   D. T = 25 years, equal to the record length
   Answer: B. Reliability is (1 - 1/T)^n, so (1 - 1/T) = 0.85^(1/45) = 0.9964 and T = 277 years. (Key point: Return period from reliability)

2. In Gumbel's method with the given data, how is the frequency factor K obtained from y_T?
   A. K = (y_T - y_n)/S_n = 4.67
   B. K = (y_T + y_n)/S_n
   C. K = y_T/S_n
   D. K = (y_T - y_n) x S_n
   Answer: A. K = (y_T - y_n)/S_n = (5.6236 - 0.5309)/1.0915 = 4.67. (Key point: Frequency factor)

3. Which expression gives the recommended peak discharge and what is its value?
   A. x_T = mean x K = about 39750 m^3/s
   B. x_T = sigma + K x mean = about 40000 m^3/s
   C. x_T = mean - K x sigma, a negative value
   D. x_T = mean + K x sigma = 8520 + 4.6658 x 3900, about 26717 m^3/s
   Answer: D. Gumbel gives x_T = mean + K sigma = 26717 m^3/s. (Key point: Flood magnitude)

4. With a factor of safety of 1.3, what are the design discharge and safety margin?
   A. Design 26717 m^3/s; margin 0
   B. Design 20551 m^3/s; margin 6166 m^3/s
   C. Design 34732 m^3/s; margin 8015 m^3/s
   D. Design 34732 m^3/s; margin 34732 m^3/s
   Answer: C. Design = 1.3 x 26717 = 34732, and the margin is the difference from x_T: 34732 - 26717 = 8015 m^3/s. (Key point: Factor of safety and margin)

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