frequency analysis siti kamariah md sa’at ppk bioprocess..2010
DESCRIPTION
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 eventTRANSCRIPT
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FREQUENCY ANALYSIS
Siti Kamariah Md Sa’atPPK Bioprocess..2010
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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
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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
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Frequency distribution analysis
Gumbel’s Method Log-Pearson Type III distribution Log normal distribution
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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
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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
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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)
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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
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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
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Tr= 2.33yr
Flowrate, cms
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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
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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