---
title: "Peak flow by empirical method and Gumbel"
subject: "Hydro Power"
chapter: 2
key: "semester_6/hydro_power:peak-flow-gumbel"
source: https://www.bctengineer.com/q/semester_6/hydro_power/peak-flow-gumbel
site: BCT Engineering
---

# Peak flow by empirical method and Gumbel

Hydro Power · Chapter 2: Introduction to Hydrology

- Asked in: 2070 Magh
- Video: [Estimation of peak flow](https://www.youtube.com/watch?v=rU9H1m47c4w) — Aapka Engineer
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/hydro_power/peak-flow-gumbel
- Whole chapter: https://www.bctengineer.com/md/semester_6/hydro_power/chapter-2

## Question

How peak flow is estimated by using empirical method and Gumbel's distribution? (partly faded in the scan)

## Answer

**Peak flow** (design flood) is the maximum discharge expected at a site for a chosen return period. It is needed to design the spillway, diversion works, cofferdams and river training works of a hydropower project.

#### 1. Empirical methods
They relate peak flow mainly to catchment area $A$ (km²) with a regional constant. Simple but only valid in the region where they were developed.

| Formula | Equation | Remarks |
|---|---|---|
| Dicken's | $Q_p = C_D A^{3/4}$ | North India; $C_D$ = 6 to 30 (11–14 for hilly areas) |
| Ryves | $Q_p = C_R A^{2/3}$ | South India; $C_R$ = 6.8 to 10.1 |
| Inglis | $Q_p = \dfrac{124A}{\sqrt{A + 10.4}}$ | Fan-shaped catchments, Maharashtra |
| Rational | $Q_p = \dfrac{CIA}{3.6}$ | Small catchments; I in mm/h, A in km² |

In Nepal, regional methods such as WECS/DHM (1990) and modified Dicken's are commonly used, e.g. $Q_{100} = 14.63\,(A_{3000} + 1)^{0.7342}$, where $A_{3000}$ is the catchment area (km²) below 3000 m elevation.

Example: $A = 500$ km², $C_D = 11$: $Q_p = 11 \times 500^{0.75} = 11 \times 105.7 = 1163$ m³/s.

#### 2. Gumbel's extreme value distribution
A statistical method using the **annual maximum flood series** (one peak per year for N years).

Steps:
1. Find mean $\bar{x}$ and standard deviation $\sigma_{n-1}$ of the annual peaks.
2. Reduced variate for return period T:
$$
y_T = -\ln\left[\ln\left(\frac{T}{T-1}\right)\right]
$$
3. Frequency factor:
$$
K = \frac{y_T - \bar{y}_n}{S_n}
$$
   where $\bar{y}_n$ and $S_n$ are the reduced mean and reduced standard deviation from Gumbel's table for sample size N (for very large N, $\bar{y}_n = 0.577$, $S_n = 1.2825$).
4. Flood of return period T:
$$
x_T = \bar{x} + K\sigma_{n-1}
$$
5. Optionally plot on Gumbel probability paper and give confidence limits.

Example: $\bar{x} = 800$ m³/s, $\sigma = 250$ m³/s, large N, $T = 100$ yr: $y_T = 4.600$, $K = (4.600 - 0.577)/1.2825 = 3.137$, $x_{100} = 800 + 3.137 \times 250 = 1584$ m³/s.

| Point | Empirical | Gumbel |
|---|---|---|
| Data needed | Area only | Long flood record |
| Return period | Not explicit | Any T |
| Accuracy | Rough | Better |

## Check yourself

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

1. What do empirical formulae such as Dicken's mainly relate peak flow to?
   A. Return period alone
   B. Catchment area with a regional constant
   C. River width
   D. Rainfall duration only
   Answer: B. Qp = CD A^(3/4) uses catchment area and a region-specific constant, so it is valid only where developed. (Key point: Empirical methods)

2. For A = 500 km2 and CD = 11, what is Qp = CD A^0.75?
   A. About 55 m3/s
   B. About 5,500 m3/s
   C. About 116 m3/s
   D. About 1,163 m3/s
   Answer: D. 500^0.75 = 105.7, so Qp = 11 x 105.7 = 1163 m3/s. (Key point: Dicken's example)

3. What data does Gumbel's method use?
   A. Rainfall at one gauge
   B. Catchment area only
   C. The annual maximum flood series
   D. Monthly mean flows
   Answer: C. It fits the extreme value distribution to one peak per year for N years. (Key point: Gumbel data)

4. With mean 800, sigma 250 m3/s, large N and T = 100 yr (yT = 4.600, K = 3.137), what is x100 = mean + K sigma?
   A. About 1,584 m3/s
   B. About 1,050 m3/s
   C. About 800 m3/s
   D. About 3,937 m3/s
   Answer: A. x100 = 800 + 3.137 x 250 = 1584 m3/s. (Key point: Gumbel flood estimate)

## Related in Hydro Power

- [Head loss and flow direction in tapering oil pipe](https://www.bctengineer.com/question.md?key=semester_6%2Fhydro_power%3Atapering-pipe-oil-flow) — Hydro Power, chapter 1

## Related in other subjects

- [Gumbel peak flood and confidence limits, 75 years](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-peak-0-4pct-75yr) — Engineering Hydrology, chapter 6
- [Gumbel's distribution and frequency factor derivation](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Agumbel-distribution-frequency-factor) — Engineering Hydrology, chapter 6
- [Peak flow from 3-h UH, two successive storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Apeak-flow-3h-uh-4h-3h) — Engineering Hydrology, chapter 5
- [Derive UH from storm, peak flow from two storms](https://www.bctengineer.com/question.md?key=semester_5%2Fengineering_hydrology%3Auh-from-storm-then-peak-two) — Engineering Hydrology, chapter 5