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# Breakwaters and closure dams: chapter 5

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### Breakwaters and closure dams: chapter 5

1. 1. Chapter 5: use of theoryct5308 Breakwaters and Closure DamsH.J. VerhagenApril 12, 2012 1Faculty of Civil Engineering and GeosciencesVermelding onderdeel organisatieSection Hydraulic Engineering
2. 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 • liquefactionApril 12, 2012 2
3. 3. Initial tidal wave by the moon andthe sunApril 12, 2012 3
4. 4. Adding semi-diurnal constantsresulting in spring and neap tideApril 12, 2012 4
5. 5. Adding diurnal to semi-diurnalconstantApril 12, 2012 5
6. 6. Amphidromy in the North SeaApril 12, 2012 6
7. 7. typical tidesApril 12, 2012 7
8. 8. adding the fortnightly constantApril 12, 2012 8
9. 9. flow pattern in a gapApril 12, 2012 9
10. 10. Flow over a sill subcritical flow Q = mBh 2g (H − h) Q h u= =m 2g (H − h) Ba a critical flow 1 Q = mBa 2g H 3 2 1 Q = m B( H ) 2 g ( H ) 3 3 1 u = m 2 g( H ) 3April 12, 2012 10
11. 11. modellingδQ δH + Bx =0δx δtδ Q δ ( α Q u) δH g Q Q + +g A + 2 −W x = 0δt δx δx C AR Solving these equation by: •physical model •mathematical model •2 d model •1 d model •storage area approachApril 12, 2012 11
12. 12. Physical modelApril 12, 2012 12
13. 13. two dimensional modelOosterscheldewerken, Waqua, Rijkswaterstaat/WL Korea, Gaduk port, Mike21, DHI April 12, 2012 13
14. 14. one-dimensional modelApril 12, 2012 14
15. 15. storage/area approach δQ δH + Bx =0 δx δtApril 12, 2012 15
16. 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 kmApril 12, 2012 16
17. 17. equations for storage/area approach dh3 μ Ag 2g (H1 − h2 ) = B − QR (t ) dt 2 h2 = h3 for h3 > H1 3 2 2 h2 = H1 for h3 < H1 3 3 Ag and B can be combined to one input parameterApril 12, 2012 17
18. 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 gapAssume for the time being that the river discharge is zero and that the tide is always semi-diurnalSet the discharge coefficient of the gap to 1Remaining parameters:• tidal difference• ratio storage area/gap width• sill heightApril 12, 2012 18
19. 19. design graph for the velocityApril 12, 2012 19
20. 20. example of the use of a design graphApril 12, 2012 20
21. 21. velocity as a function of the closureApril 12, 2012 21
22. 22. Stability of a submerged objectApril 12, 2012 22
23. 23. Stability of a floating object I MC = V +.5b 1 3 I =∫ yx2dx = LB −.5b 12 G V= ρgApril 12, 2012 23
24. 24. Definition of a regular wave H H wave height H cos ⎛ 2π x − 2π t ⎞ η= 2 ⎜ T wave period ⎟ ⎝ L T ⎠ L wave length gL ⎛ 2π h ⎞ c= tanh ⎜ ⎟ 2π ⎝ L ⎠ gT 2 c = ghL0 = = 1.56T 2 2πApril 12, 2012 24
25. 25. validity for wave theoriesApril 12, 2012 25
26. 26. breakingby steepness H/L< 0.14by depth H/h < 0.78 but…………….April 12, 2012 26
27. 27. Irregular waveApril 12, 2012 27
28. 28. Rayleigh graph paper ⎡ ⎛ H ⎞2 ⎤ ⎢−2⎜ ⎟ ⎥ ⎢ ⎝ Hs ⎠ ⎥ P(H > H ) = e ⎣ ⎦April 12, 2012 28
29. 29. characteristic wave heightsName Notation H/√m0 H/HsStandard deviation free surface ση=√m0 1 0.250RMS height Hrms 2√2 0.706Mean Height H = H1 2√ln 2 0.588Significant Height Hs= H1/3 4.005 1Average of 1/10 highest waves H1/10 5.091 1.271Average of 1/100 highest waves H1/100 6.672 1.666Wave height exceeded by 2% H2% 1.4April 12, 2012 29
30. 30. characteristic wave periods Name Notation Relation to spectral T/Tp moment Peak period Tp 1/fp 1 Mean period Tm √(m0/m2) 0.75 to 0.85 Significant period Ts 0.9 to 0.95April 12, 2012 30
31. 31. typical types of wave statisticspatternsApril 12, 2012 31
32. 32. H/T-diagramApril 12, 2012 32
33. 33. waves in shallow water shoaling H 1 1 = + = ksh refraction H0 tanh ( 2π h / L) 1+ ( 4π h / L) breaking diffraction sinh ( 4π h / L) reflectionApril 12, 2012 33
34. 34. the iribarren number(surf similarity parameter) tan α ξ= H L0 tan α slope of the shoreline/structure H wave height L0 wave length at deep waterApril 12, 2012 34
35. 35. breaker types (2) spilling ξ < 0.5 plunging 0.5 < ξ < 3 collapsing ξ = 3 surging ξ > 3April 12, 2012 35
36. 36. breaking waves ⎛ 2π ⎞ Hb = 0.142 L tanh ⎜ h⎟ ⎝ L ⎠ Hb ≈ 0.78 (solitarywave) h Hs ≈ 0.4 − 0.5 hApril 12, 2012 36
37. 37. change of distribution in shallowwaterApril 12, 2012 37
38. 38. Battjes Jansen method ⎧ ⎡ ⎛ H ⎞2 ⎤ ⎪ F1 ( H ) = 1 − exp ⎢ − ⎜ ⎟ ⎥ H ≤ H tr ⎪ ⎪ ⎢ H ⎥ ⎣ ⎝ 1⎠ ⎦Pr { H ≤ H } = ⎨ ⎪ ⎡ ⎛ H ⎞3.6 ⎤ ⎪ F2 ( H ) = 1 − exp ⎢ − ⎜ ⎟ ⎥ H > H tr ⎪ ⎩ ⎢ ⎝ H2 ⎠ ⎥ ⎣ ⎦April 12, 2012 38
39. 39. Influence of shallow water on thewave heightApril 12, 2012 39
40. 40. Wave refraction ⎛ c2 ⎞ sin α2 = ⎜ ⎟ sin α1 ⎝ c1 ⎠ H2 b1 = H1 b2April 12, 2012 40
41. 41. Diffraction behind a detachedbreakwaterApril 12, 2012 41
42. 42. reflection HRKr = ≈ 0.1ξ 2 HI η = η +η tot i r = (1+ r ) Hi 2 ( L) ( T ) H 2 ( L) ( cos 2π x *cos 2π t + (1 − r ) i sin 2π x *sin 2π t T ) April 12, 2012 42
43. 43. Example with Cress run demo Cress refraction shoaling, etc diffraction y (-200,200) x(50-200;4)April 12, 2012 43
44. 44. The effect of shoaling on waveparametersApril 12, 2012 44
45. 45. Typical wave record of the North Sea η (t ) = ∑ a cos ( 2π f t + ϕ ) i i i Δω S (ω ) = 1 ∑ ai2 Δω 2 Hs = 4 m0 = H13.5%April 12, 2012 45
46. 46. Spectral wave periodsThe use of different wave parameters to obtain better results for wavestructure interaction ∞ mn = ∫ f S ( f ) df n 0ct5308 Breakwaters and closure damsH.J. VerhagenApril 12, 2012 46Faculty of Civil Engineering and GeosciencesVermelding onderdeel organisatieSection Hydraulic Engineering
47. 47. Example wave record 28 waves, Hs = "13% wave", Hs= wave nr 4, Hs ≈ 3.8 28 waves in 150 seconds, so Tm = 5.3 sApril 12, 2012 47
48. 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 secApril 12, 2012 48
49. 49. Spectrum discretised spectrum energy density spectrum1 2 a = S ⋅ Δf 7 72 6 6 energy density (m 2s) energy density (m 2s) 5 5 4 4 3 3 2 2 1 1 0 0 0,1 0,15 0,2 0,25 0 0,1 0,2 0,3 0,4 frequency (Hz) frequency (Hz) H2 1.552H = 8S ⋅ Δf S= = = 6 [m2 s] 8Δf 8 ⋅ 0.05 April 12, 2012 49
50. 50. Calculation of m0 discretised spectrum 7 6 energy density (m 2s) 5 4 3 0.05*2 0.10 2 1 0.05*6 0.30 0 0,1 0,15 0,2 0,25 frequency (Hz) 0.05*3 0.15 0.05*1 0.05 0.60 4 m0 = 3.1 m ∞mn = ∫ f n S ( f ) df 0 April 12, 2012 50
51. 51. Calculation of m1 discretised spectrum 7 6 energy density (m 2s) 5 4 dist * SΔf 3 2 0.10*0.10 0.010 1 0 0.15*0.30 0.045 0,1 0,15 0,2 0,25 frequency (Hz) 0.20*0.15 0.030 0.25*0.05 0.013 0.098 ∞mn = ∫ f n S ( f ) df 0 April 12, 2012 51
52. 52. Calculation of m2 discretised spectrum 7 6 energy density (m 2s) 5 4 dist2 * SΔf 3 2 0.102*0.10 1.00 10-3 1 0 0.152*0.30 6.75 10-3 0,1 0,15 0,2 0,25 frequency (Hz) 0.202*0.15 6.00 10-3 0.252*0.05 3.12 10-3 m0 0.60 T= = 10 = 5.69sec 1.69 10-3 m2 1.69 ∞mn = ∫ f n S ( f ) df 0 April 12, 2012 52
53. 53. Calculation of m-1 discretised spectrum 7 6 energy density (m 2s) 5 4 1/dist * SΔf 3 2 1/0.10*0.10 1.0 1 0 1/0.15*0.30 2.0 0,1 0,15 0,2 0,25 frequency (Hz) 1/0.20*0.15 0.75 1/0.25*0.05 0.20 m−1 3.95 Tm −1,0 = = = 6.58 sec 3.95 m0 0.60 ∞mn = ∫ f n S ( f ) df 0 April 12, 2012 53
54. 54. Overview Usual assumptions: Tm0 = Tp•Hm0 = 3.1 m(1.55+1.10+0.90+0.63=4.18) T1/3 = Tm•Tm0 = 5.69 sec•Tm-1,0 = 6.58 sec•Tpeak = 6.67 sec For standard spectra: Tm −1,0 6.58 Goda: Tp=1.1 T1/3• = = 1.16 PM: Tp=1.15 T1/3 T m0 5.69•Tm = 5.35 sec Jonswap: Tp=1.07 T1/3 Tm 0 5.69 TAW (vdMeer): Tp=1.1Tm-1,0• = = 1.06 Tm 5.35 Old Test (vdMeer): Tp=1.04 Tm-1,0 Also: Tm-1,0=1.064T1/3April 12, 2012 54
55. 55. Overview to determine shallow water wavecondition• 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 55
56. 56. Why these parameters ? 0.2 0.18 ⎛ S ⎞ (s ) H 2% 0.25 = c pl P ⎜ ⎟ m −1,0 cot α for plunging waves Δd n 50 ⎝ N⎠ 0.2 −0.13 ⎛ S ⎞ (s ) (ξ ) H 2% −0.25 P − 0.5 = cs P ⎜ ⎟ m −1,0 s −1,0 for surging waves Δd n 50 ⎝ N⎠April 12, 2012 56
57. 57. stress relations determined by soiltestingApril 12, 2012 57
58. 58. Dam profile after the slideApril 12, 2012 58
59. 59. SqueezeApril 12, 2012 59
60. 60. Liquefied sandApril 12, 2012 60