---
title: "Gumbel 500-year flood, 95% limits, 30 years"
subject: "Engineering Hydrology"
chapter: 6
key: "semester_5/engineering_hydrology:gumbel-30yr-100-50-1200-1060-2"
source: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-30yr-100-50-1200-1060-2
site: BCT Engineering
---

# Gumbel 500-year flood, 95% limits, 30 years

Engineering Hydrology · Chapter 6: Flood Hydrology

- Asked in: 2082 Baisakh, 4+2+2 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/engineering_hydrology/gumbel-30yr-100-50-1200-1060-2
- Whole chapter: https://www.bctengineer.com/md/semester_5/engineering_hydrology/chapter-6

## Question

Using 30 years data and Gumbel's method, the flood magnitude for return period of 100 and 50 years for a river are found to be 1200 and 1060 m$^3$/s respectively.

(i) Determine the mean and standard deviation of the data used. Take values of reduced mean and reduced standard deviation in Gumbel's extreme value distribution for n = 30 as 0.5362 and 1.1124.
(ii) Estimate the magnitude of the flood with a return period of 500 years.
(iii) What are the 95% confidence limits for this estimate if $f(95\%) = 1.96$.

## Short answer

Subtracting equations: σ = (1200-1060)/(3.653-3.026)=223.1 m³/s, mean=385.1 m³/s. 500-yr: K=5.104, discharge=1524 m³/s. 95% confidence limits: 1524±1.96×245.3 = 1043 to 2004 m³/s.

## Answer

#### (i) Mean and standard deviation
$\bar y_n=0.5362$, $S_n=1.1124$, $y_T=-\ln[\ln(T/(T-1))]$, $K=(y_T-\bar y_n)/S_n$, $x_T=\bar x+K\sigma$.

| T (yr) | $y_T$ | $K$ | $x_T$ (m³/s) |
|---|---|---|---|
| 100 | 4.600 | 3.653 | 1200 |
| 50 | 3.902 | 3.026 | 1060 |

Subtract the two equations:

$$
\sigma=\frac{1200-1060}{3.653-3.026}=223.1 \text{ m}^3/\text{s}
$$

$$
\bar x=1200-3.653\times223.1=385.1 \text{ m}^3/\text{s}
$$

**Answer: mean = 385.1 m³/s, standard deviation = 223.1 m³/s.**

#### (ii) 500-year flood
$$
y_{500}=-\ln[\ln(500/499)]=6.214,\qquad K=\frac{6.214-0.5362}{1.1124}=5.104
$$

$$
x_{500}=385.1+5.104\times223.1=1524 \text{ m}^3/\text{s}
$$

**Answer: $x_{500}$ ≈ 1524 m³/s.**

#### (iii) 95% confidence limits of the 500-year flood
$$
b=\sqrt{1+1.3K+1.1K^2}=\sqrt{1+1.3(5.104)+1.1(5.104)^2}=6.024
$$

$$
S_e=\frac{b\sigma}{\sqrt N}=\frac{6.024\times223.1}{\sqrt{30}}=245.3 \text{ m}^3/\text{s}
$$

$$
x_T'=x_T\pm f(c)S_e=1524\pm1.96\times245.3=1524\pm481
$$

**Answer: 95% confidence limits are 1043 m³/s and 2004 m³/s.**

## Check yourself

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

1. How do you get sigma from the 100- and 50-year floods?
   A. sigma = 1200 - 1060
   B. sigma = 1200/K100
   C. sigma = (1200 + 1060)/2
   D. sigma = (1200 - 1060)/(K100 - K50) = 223.1
   Answer: D. Subtracting x_T = mean + K sigma for the two T eliminates the mean. (Key point: Mean and sigma)

2. How is K computed for T = 500?
   A. 0.5362/6.214
   B. (6.214 - 0.5362)/1.1124 = 5.104
   C. (6.214 + 0.5362)/1.1124
   D. 6.214/1.1124
   Answer: B. K = (y_T - y_n)/S_n. (Key point: Frequency factor)

3. What is the 95% interval for the 500-year flood?
   A. 1524 ± 223.1
   B. 1524 ± 1.96 x 245.3 = 1043 to 2004
   C. 1524 ± 245.3
   D. 1524 ± 1.96 x 223.1
   Answer: B. The standard error S_e = b sigma/sqrt(N) = 245.3 is multiplied by f(c) = 1.96. (Key point: Confidence limits)

4. What does the factor b = sqrt(1 + 1.3K + 1.1K^2) do?
   A. Scales the standard error, increasing it for large K
   B. Gives the mean
   C. Is the number of years
   D. Is the skewness
   Answer: A. S_e = b sigma/sqrt(N); b grows with K, so extreme floods are less certain. (Key point: Role of b)

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