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

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