1. FREQUENCY ANALYSIS Siti Kamariah Md Sa’at PPK Bioprocess..2010

2. Flood Frequency Analysis  Statistical Methods to evaluate probability exceeding a particular outcome - P (X >20,000 m 3 /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

3. Probability, P  P = 1/T and P = m / N+1 where 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 > x 0 ) n = 1 – (1-1/T) n where n = total number of event

4. Frequency distribution analysis  Gumbel’s Method  Log-Pearson Type III distribution  Log normal distribution

5. General equation Where  X T =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

6. 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

7. 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)

9. 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

10. Example:  Q = 650 m 3 /s is assume to happen again in 3 years time in T r = 50 years. P(X > x 0 ) n = 1 – (1-1/T) n P (X ≥ 50 years flood) = 1-[1-1/50] 3 = 6% = 0.06 Q av = 319.5 (from graph) σ = 124.6

11. Tr= 2.33yr Flowrate, cms

12. Using equation  For Tr = 100 years  y=4.6  Q 100 = 319.5 + 124.6(0.78(4.6) -0.45) = 710 m 3 /s From chart, we get Q 100 =718 m 3 /s

13. Example 7.4  Plot graph Q vs T we have P = 1/T and P = m / N+1 T = 1/P = N+1/m = 27+1/m Determine 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