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Hydrology & Water Resources Eng.

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Presentation on theme: "Hydrology & Water Resources Eng."— Presentation transcript:

1 Hydrology & Water Resources Eng.
ERT 246 Hydrology & Water Resources Eng. 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 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

3 Probability, P P = 100/T, where P in %, T=return period/frequency
Graph plot 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

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

5 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= ferquency 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 T at special Gumbel’s graph chart. The straight line must be intercept at coordinate (2.33years, Qav) T=(n+1)/m

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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

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12 Example 7.4: Annual maximum recorded floods in the river Bhima, tributary of river Krishna, for period of is given below. Verify whether the Gumbel extreme value distribution fit the recorded values. Estimate the flood discharge with recurrence interval of i) 100 years ii) 150 years by graphical extrapolation and equation.

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14 Answer: n=27 years Tp=(n+1)/m Qav=4263 σ=1432.6 y=2.25037 Q100=9600


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