Peak flow by empirical method and Gumbel

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.

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.