FREQUENCY ANALYSIS

Siti Kamariah Md Sa’atPPK Bioprocess..2010

Flood Frequency Analysis

Statistical Methods to evaluate probability exceeding a particular outcome - P (X >20,000 m3/s) = 10%

Used to determine return periods of rainfall or flows Used to determine specific frequency flows for

floodplain mapping purposes (10, 25, 50, 100 yr) Used for datasets that have no obvious trends Used to statistically extend data sets

Probability, P

P = 1/T and P = m / N+1where P in %, T=return period/frequency Plot a graph to get relationship for

Q vs Tr or Q vs P Equation used to determine flood

probability P(X > x0)n = 1 – (1-1/T)n

where n = total number of event

Frequency distribution analysis

Gumbel’s Method Log-Pearson Type III distribution Log normal distribution

General equation

Where XT =calue of variate X of a random

hydrologic series with return period T X = mean of variate σ = standard deviation of variate K = frequency factor depend on return

period, T and the assume frequency distribution

KX TX

Gumbel’s Extreme-Value distribution Introduced by Gumbel,1941 Known as Gumbel’s distribution Most widely used for extreme values in

hidrologic studies for prediction of flood peaks, maximum rainfalls, maximum wind speed, etc.

2 method to determine discharge, Q Graph Equation

Gumbel’s distribution graph

Plotting graph Q vs Tr at special Gumbel’s graph chart.

The straight line must be intercept at coordinate (2.33years, Qav)

Gumbel Equation

Where Qav=average discharge for all flow data T=return period/frequency σ = standard deviation n=total number of event m=order number of event

)45.078.0( yQQ avT

TP

Ty 1)],11ln(ln[

)(1

22

avQnQ

nn m

Example:

Q = 650 m3/s is assume to happen again in 3 years time in Tr= 50 years.

P(X > x0)n = 1 – (1-1/T)n

P (X ≥ 50 years flood) = 1-[1-1/50]3 = 6% = 0.06Qav = 319.5 (from graph)

σ = 124.6

Tr= 2.33yr

Flowrate, cms

Using equation

For Tr = 100 years y=4.6 Q100= 319.5 + 124.6(0.78(4.6) -0.45)

= 710 m3/sFrom chart, we get Q100=718 m3/s

)45.078.0( yQQ avT

)]11ln(ln[T

y

Example 7.4

Plot graph Q vs T we have P = 1/T and P = m / N+1T = 1/P = N+1/m = 27+1/mDetermine Qav from the data and we get

Qav = 4263 cms σ = 1432.6 cms Plot graph For Tr= 100 years, y = 4.6

Q = 4263 + 1432.6(0.78(4.6)-0.45) =8758.5 cms

)45.078.0( yQQ avT