breakwaters and closure dams: chapter 5

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April 12, 2012

Vermelding onderdeel organisatie

1

Chapter 5: use of theory

ct5308 Breakwaters and Closure Dams

H.J. Verhagen

Faculty of Civil Engineering and GeosciencesSection Hydraulic Engineering

April 12, 2012 2

Theoretical background needed• waterlevels (tides)• flow trough gaps• stability of floating objects• waves

• basics• refraction, shoaling, breaking, diffraction, reflection• wave statistics

• short term statistics (Rayleigh)• long term statistics

• Geotechnics• sliding• squeeze• liquefaction

April 12, 2012 3

Initial tidal wave by the moon and the sun

April 12, 2012 4

Adding semi-diurnal constants resulting in spring and neap tide

April 12, 2012 5

Adding diurnal to semi-diurnal constant

April 12, 2012 6

Amphidromy in the North Sea

April 12, 2012 7

typical tides

April 12, 2012 8

adding the fortnightly constant

April 12, 2012 9

flow pattern in a gap

April 12, 2012 10

Flow over a sill

subcritical flow

critical flow

2 ( )Q mBh g H h= −

2 ( )Q hu m g H hB a a

= = −

1Q = m B a 2 g H3

( ) ( )2 1Q m B H 2 g H3 3

=

( )1u m 2 g H3

=

April 12, 2012 11

modelling

xQ H+ = 0Bx t

δ δδ δ

0x2

g Q QQ ( Q u) Hg A Wt x x A RC

δ δ α δδ δ δ

+ + + − =

Solving these equation by:•physical model•mathematical model

•2 d model•1 d model•storage area approach

April 12, 2012 12

Physical model

April 12, 2012 13

two dimensional model

Korea, Gaduk port, Mike21, DHIOosterscheldewerken, Waqua, Rijkswaterstaat/WL

April 12, 2012 14

one-dimensional model

April 12, 2012 15

storage/area approach

xQ H+ = 0Bx t

δ δδ δ

April 12, 2012 16

validity of storage/area approach

length of tidal wave: L= c*T = √gh * T = √10*10 *12*3600= 432 km

basin < 0.05 L = 20 km

April 12, 2012 17

equations for storage/area approach

31 2

2 3 3 1

2 1 3 1

2 ( ) ( )

23

2 23 3

g Rdh

A g H h B Q tdt

h h for h H

h H for h H

μ − = −

= >

= <

Ag and B can be combined to one input parameter

April 12, 2012 18

parameters needed• water level in the sea• river discharge• ratio between storage area and width of closure gap• sill height• discharge coefficient of the gap

Assume for the time being that the river discharge is zero and that the tide is always semi-diurnal

Set the discharge coefficient of the gap to 1

Remaining parameters: • tidal difference• ratio storage area/gap width• sill height

April 12, 2012 19

design graph for the velocity

April 12, 2012 20

example of the use of a design graph

April 12, 2012 21

velocity as a function of the closure

April 12, 2012 22

Stability of a submerged object

April 12, 2012 23

Stability of a floating object

.5 2 3.5

112

b

b

IMCV

I yx dx LB

GVgρ

+

=

= =

=

April 12, 2012 24

Definition of a regular wave

H

H wave heightT wave periodL wave length

2 2cos2x tH

L Tπ πη ⎛ ⎞= −⎜ ⎟

⎝ ⎠

2tanh2gL hc

π⎛ ⎞= ⎜ ⎟⎝ ⎠

22

0 1.562gTL Tπ

= = c gh=

April 12, 2012 25

validity for wave theories

April 12, 2012 26

breaking

by steepness H/L< 0.14by depth H/h < 0.78 but…………….

April 12, 2012 27

Irregular wave

April 12, 2012 28

Rayleigh graph paper

2

2

( ) s

HH

P H H e

⎡ ⎤⎛ ⎞⎢ ⎥− ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦> =

April 12, 2012 29

characteristic wave heights

Name Notation H/√m0 H/Hs

Standard deviation free surface ση=√m0 1 0.250 RMS height Hrms 2√2 0.706 Mean Height H = H1 2√ln 2 0.588

Significant Height Hs= H1/3 4.005 1 Average of 1/10 highest waves H1/10 5.091 1.271 Average of 1/100 highest waves H1/100 6.672 1.666 Wave height exceeded by 2% H2% 1.4

April 12, 2012 30

characteristic wave periods

Name Notation Relation to spectral moment

T/Tp

Peak period Tp 1/fp 1 Mean period Tm √(m0/m2) 0.75 to 0.85 Significant period Ts 0.9 to 0.95

April 12, 2012 31

typical types of wave statistics patterns

April 12, 2012 32

H/T-diagram

April 12, 2012 33

waves in shallow water

shoalingrefractionbreakingdiffractionreflection

( ) ( )( )

0

1 1 4 /tanh 2 / 1

sinh 4 /

shH k

h LH h Lh L

πππ

= + =+

April 12, 2012 34

the iribarren number(surf similarity parameter)

0

tanH Lαξ =

tan α slope of the shoreline/structureH wave heightL0 wave length at deep water

April 12, 2012 35

breaker types (2)

spilling ξ < 0.5

plunging 0.5 < ξ < 3

collapsing ξ = 3

surging ξ > 3

April 12, 2012 36

breaking waves

20.142 tanhbH L hLπ⎛ ⎞= ⎜ ⎟

⎝ ⎠

0.78 ( )bH solitarywaveh≈

0.4 0.5sHh≈ −

April 12, 2012 37

change of distribution in shallow water

April 12, 2012 38

Battjes Jansen method

{ }( )

2

11

3.6

22

( ) 1 exp

Pr

1 exp

tr

tr

HF H H HH

H HHF H H HH

⎧ ⎡ ⎤⎛ ⎞⎪ ⎢ ⎥= − − ≤⎜ ⎟⎪ ⎢ ⎥⎝ ⎠⎪ ⎣ ⎦≤ = ⎨

⎡ ⎤⎪ ⎛ ⎞⎢ ⎥= − − >⎪ ⎜ ⎟⎢ ⎥⎝ ⎠⎪ ⎣ ⎦⎩

April 12, 2012 39

Influence of shallow water on the wave height

April 12, 2012 40

Wave refraction

22 1

1

sin sincc

α α⎛ ⎞

= ⎜ ⎟⎝ ⎠

2 1

1 2

H bH b

=

April 12, 2012 41

Diffraction behind a detached breakwater

April 12, 2012 42

reflection

20.1Rr

I

HKH

ξ= ≈

tot i rη η η= + = ( ) ( ) ( ) ( ) ( ) ( )2 2 2 21 cos *cos 1 sin *sin2 2

i iH Hx t x tr rL T L Tπ π π π+ + −

April 12, 2012 43

Example with Cress

run demo Cress

refractionshoaling, etcdiffraction

x(50-200;4)

y (-200,200)

April 12, 2012 44

The effect of shoaling on wave parameters

April 12, 2012 45

Typical wave record of the North Sea

( ) 212 iS a

ω

ω ωΔ

= Δ∑

( ) ( )cos 2i i it a f tη π ϕ= +∑

0 13.5%4sH m H= =

April 12, 2012

Vermelding onderdeel organisatie

46

Spectral wave periodsThe use of different wave parameters to obtain better results for wave structure interaction

ct5308 Breakwaters and closure dams

H.J. Verhagen

Faculty of Civil Engineering and GeosciencesSection Hydraulic Engineering

( )0

nnm f S f df

= ∫

April 12, 2012 47

Example wave record

28 waves, Hs = "13% wave", Hs= wave nr 4, Hs ≈ 3.828 waves in 150 seconds, so Tm = 5.3 s

April 12, 2012 48

composition of the record

H1 = 0.63 m T1= 4 sec

H2 = 1.80 m T2 = 5 sec

H3 = 1.55 m T3 = 6.67 sec

H4 = 0.90 m T4 = 10 sec Tm = 5.3 sec

April 12, 2012 49

Spectrumdiscretised spectrum

0

1

2

3

4

5

6

7

0,1 0,15 0,2 0,25

frequency (Hz)

ener

gy d

ensi

ty (m

2 s)

energy density spectrum

0

1

2

3

4

5

6

7

0 0,1 0,2 0,3 0,4

frequency (Hz)

ener

gy d

ensi

ty (m

2 s)

212

a S f= ⋅Δ

2 221.558 6 [ ]

8 8 0.05HH S f S m s

f= ⋅Δ = = =

Δ ⋅

April 12, 2012 50

Calculation of m0

0.60

0.050.05*1

0.150.05*3

0.300.05*6

0.100.05*2

04 3.1m m=

discretised spectrum

0

1

2

3

4

5

6

7

0,1 0,15 0,2 0,25

frequency (Hz)

ener

gy d

ensi

ty (m

2 s)

( )0

nnm f S f df

= ∫

April 12, 2012 51

Calculation of m1

dist * SΔf

0.098

0.0130.25*0.05

0.0300.20*0.15

0.0450.15*0.30

0.0100.10*0.10

discretised spectrum

0

1

2

3

4

5

6

7

0,1 0,15 0,2 0,25

frequency (Hz)

ener

gy d

ensi

ty (m

2 s)

( )0

nnm f S f df

= ∫

April 12, 2012 52

Calculation of m2 discretised spectrum

0

1

2

3

4

5

6

7

0,1 0,15 0,2 0,25

frequency (Hz)

ener

gy d

ensi

ty (m

2 s)

dist2 * SΔf

1.69 10-3

3.12 10-30.252*0.05

6.00 10-30.202*0.15

6.75 10-30.152*0.30

1.00 10-30.102*0.10

0

2

0.6010 5.69sec1.69

mTm

= = =

( )0

nnm f S f df

= ∫

April 12, 2012 53

Calculation of m-1 discretised spectrum

0

1

2

3

4

5

6

7

0,1 0,15 0,2 0,25

frequency (Hz)

ener

gy d

ensi

ty (m

2 s)

1/dist * SΔf

3.95

0.201/0.25*0.05

0.751/0.20*0.15

2.01/0.15*0.30

1.01/0.10*0.10

11,0

0

3.95 6.58 sec0.60m

mTm−

− = = =

( )0

nnm f S f df

= ∫

April 12, 2012 54

Overview

•Hm0 = 3.1 m (1.55+1.10+0.90+0.63=4.18)•Tm0 = 5.69 sec•Tm-1,0 = 6.58 sec•Tpeak = 6.67 sec

•Tm = 5.35 sec

− = =1,0

0

6.58 1.165.69

m

m

TT

For standard spectra:

Goda: Tp=1.1 T1/3

PM: Tp=1.15 T1/3

Jonswap: Tp=1.07 T1/3

TAW (vdMeer): Tp=1.1Tm-1,0

Old Test (vdMeer): Tp=1.04 Tm-1,0

Also: Tm-1,0=1.064T1/3

= =0 5.69 1.065.35

m

m

TT

Usual assumptions:Tm0 = TpT1/3 = Tm

April 12, 2012 55

Overview to determine shallow water wave condition

• Determine deep water wave condition, this gives wave height, peak period and spectrum shape type (e.g. Jonswap)

• Calculate shallow water condition using spectral model (e.g. with SWAN), this gives Hm0, Tm0 and Tm-1,0

• Use Battjes-Jansen method to determine H2%

April 12, 2012 56

Why these parameters ?

( )0.2

0.250.182%1,0

50

cotpl mn

H Sc P s for plunging wavesd N

α−⎛ ⎞= ⎜ ⎟Δ ⎝ ⎠

( ) ( )0.2

0.25 0.50.132%1,0 1,0

50

P

s m sn

H Sc P s for surging wavesd N

ξ− −−

− −⎛ ⎞= ⎜ ⎟Δ ⎝ ⎠

April 12, 2012 57

stress relations determined by soil testing

April 12, 2012 58

Dam profile after the slide

April 12, 2012 59

Squeeze

April 12, 2012 60

Liquefied sand

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