Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources...

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Chair of Hydrology and Water Resources Management Prof. Dr. Paolo Burlando HS 17, Hydrology I, Assignment 4 Page 1 Assignment 4 – Suggested solution FLOOD ANALYSIS Due: 20.12.2017 Task 1 (Flood Frequency Analysis / direct statistical analysis) 1.1 Gumbel: () exp exp x u Fx α = 6 x s α π = 0.5772 u x α = LogNormal: () ln () y y x Fx z µ σ =F =F x s V x = 2 ln(1 ) y V σ = + ln y x = 2 ln 0,5 y y x µ σ = Gumbel: 10.229 u = 2.575 α = LogNormal: 0.282 V = 0.277 y σ = 2.423 y µ =

Transcript of Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources...

Page 1: Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources Management Prof. Dr. Paolo Burlando. HS 17, Hydrology I, Assignment 4 Page 1 . Assignment 4 –

Chair of Hydrology and Water Resources Management

Prof. Dr. Paolo Burlando

HS 17, Hydrology I, Assignment 4 Page 1

Assignment 4 – Suggested solution

FLOOD ANALYSIS Due: 20.12.2017

Task 1 (Flood Frequency Analysis / direct statistical analysis) 1.1

Gumbel:

( ) exp exp x uF xα

− = − − 6

xsαπ

= 0.5772u x α= − ⋅

LogNormal:

( )ln

( ) y

y

xF x z

µσ

−= F = F

xsV

x= 2ln(1 )y Vσ = +

lny x= 2ln 0,5y yxµ σ= − ⋅

Gumbel: 10.229u = 2.575α =

LogNormal: 0.282V = 0.277yσ = 2.423yµ =

Page 2: Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources Management Prof. Dr. Paolo Burlando. HS 17, Hydrology I, Assignment 4 Page 1 . Assignment 4 –

Chair of Hydrology and Water Resources Management

Prof. Dr. Paolo Burlando

HS 17, Hydrology I, Assignment 4 Page 2

1.2

Gringorten:

0,44( )0,12

iF xn−

=+

Gringorten KS-Gumbel KS-LogNormal

0.012 0.019 0.016 0.035 0.002 0.005 0.057 0.011 0.013 0.079 0.038 0.041 0.101 0.020 0.024 0.123 0.017 0.022

… … … … … …

Goodness-of-fit-test: Criterion: 1maxn

d i theor empT F F== −

Gumbel: 0.117dT =

LogNormal: 0.103dT =

Kolmogorov-Smirnov-statistic: (0,1)1, 22 0.1818c

n= =

For Gumbel and for LogNormal is (0.1)dT c< , that means both distributions are accepted with a significance level of 10%.

0.0

0.1

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0.4

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0.6

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0.8

0.9

1.0

0 5 10 15 20 25

Q

P

empirisch nach GringortenGumbelLognormal

Page 3: Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources Management Prof. Dr. Paolo Burlando. HS 17, Hydrology I, Assignment 4 Page 1 . Assignment 4 –

Chair of Hydrology and Water Resources Management

Prof. Dr. Paolo Burlando

HS 17, Hydrology I, Assignment 4 Page 3

1.3

1 11 U O

RF F

= =−

𝐹𝐹𝑢𝑢 = 0.98

1.4

Using the LogNormal distribution: a) Return period of 93.2 years b) Return period of 1 year c) HQ100 = 21.5 m3s-1

ϕ−1(0.99) = z = �ln(HQ100) − µln

σln� = 2.33

HQ100 = eσ∙z+µ = 21.5 m3 s⁄

Task 2 (Regionalisation) 2.1

Inn Ova da Cluozza Chamuerabach

Mean 31.3 7.4 18.5

HQ* HQ* HQ* 0.8 0.7 0.4 0.6 0.9 0.6 0.7 0.8 0.5 1.0 0.7 0.8 0.9 0.7 0.6 1.0 1.1 0.7 1.4 1.0 0.8 … … … … … …

Page 4: Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources Management Prof. Dr. Paolo Burlando. HS 17, Hydrology I, Assignment 4 Page 1 . Assignment 4 –

Chair of Hydrology and Water Resources Management

Prof. Dr. Paolo Burlando

HS 17, Hydrology I, Assignment 4 Page 4

2.2

LogNormal: 0.401yσ = 0.080yµ = −

Goodness-of-fit-test:

Criterion: 0.088dT =

Kolmogorov-Smirnov-statistic: (0,1)1, 22 0.099c

n= =

For the LogNormal distribution is (0.1)dT c< , that means the distribution is accepted with a significance level of 10%.

2.3

Linear equation: 0.8303 0.3083y x= −

(Slope: a = 0.8303 Intercept: b = -0.3083) Exponent: n = 0.8303

Constant: c = 0.4917

2.4

Power law: 0.8303[ ] 0.4917E HQ A= ⋅

Mean HQ Dischmabach = 11.2 m3s-1

2.5

0.99UF = * 3 1100 2.34HQ m s−=

*100 100 [ ]HQ HQ E HQ= ⋅

3 1100 26.33HQ m s−= for Dischmabach

In task 1.4: 3 1100 21.5HQ m s−=

The HQ100 derived by regionalisation is much larger than the one derived using statistical methods. This can be explained by the fact that the annual flood series of the Dischmabach has lower peak values in comparison to the standardized annual flood series of the other 3 catchments.

Page 5: Assignment 4 – Suggested solution · PDF fileChair of Hydrology and Water Resources Management Prof. Dr. Paolo Burlando. HS 17, Hydrology I, Assignment 4 Page 1 . Assignment 4 –

Chair of Hydrology and Water Resources Management

Prof. Dr. Paolo Burlando

HS 17, Hydrology I, Assignment 4 Page 5

Task 3 (In practice) 3.1

1. Option: - Generate a DDF/IDF for the desired return period (100 years) - Derive from the IDF the precipitation depth of a 100-year rainfall event for a

duration which corresponds to the concentration time tc of the catchment - Create a synthetic hyetograph (design rainfall) with these values

Possible methods: SCS, Chicago, alternating block method - A synthetic hyetograph can be used as input into a R-R-model, which was

calibrated on beforehand only with few runoff data: o Runoff generation (Horton, SCS-CN, …) o Runoff concentration (Nash-Model, Isochrones, …)

- The model output is an estimate of the 100-year flood

2. Option – Rational Method: - Generate a DDF/IDF for the desired return period (100 years) - Derive from the IDF the precipitation depth of a 100-year rainfall event for a

duration which corresponds to the concentration time tc of the catchment - Rational method for each runoff value a prescribed precipitation value can be assigned linear relation Q=C*A*i C…runoff coefficient A…catchment area i…precipitation intensity of a 100-year precipitation event with a duration tc - This means the return periods of the precipitation and the flood event correspond (a 100-year precipitation event leads to a 100-year flood) - In reality, these processes are not linearly dependent

3. Option - For all larger precipitation events a R-R-modelling is performed (see above) - With the output data an annual flood series (AFS) is obtained - With this AFS a direct statistical analysis can be applied - Very extensive procedure