Post on 20-Jul-2020
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Waves
3. Statistics and Irregular Waves
Statistics and Irregular Waves
Statistics and Irregular Waves
β Real wave fields are not regular
β Combination of many periods, heights, directions
β Design and simulation require realistic wave statistics:
β probability distribution of heights
β energy spectrum of frequencies (and directions)
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3. STATISTICS AND IRREGULAR WAVES
3.1 Measures of height and period
3.2 Probability distribution of wave heights
3.3 Wave spectra
3.4 Reconstructing a wave field
3.5 Prediction of wave climate
Statistics and Irregular Waves
Measures of Wave Height
π»πππ₯ Largest wave height in the sample π»πππ₯ = maxπ(π»π)
π»ππ£π Mean wave height π»ππ£π =1
ππ»π
π»πππ Root-mean-square wave height π»πππ =1
ππ»π
2
π» Ξ€1 3 Average of the highest π/3 waves π» Ξ€1 3 =1
π/3
1
π/3
π»π
π»π0 Estimate based on the rms surface elevation π»π0 = 4 Ξ·2Ξ€1 2
π»π Significant wave height Either π» Ξ€1 3 or π»π0
Measures of Wave Period
ππ Significant wave period (average of highest π/3waves)
ππ Peak period (from peak frequency of energy spectrum)
ππEnergy period (period of a regular wave with same significant wave height andpower density; used in wave-energy prediction; derived from energy spectrum)
ππ§ Mean zero up-crossing period
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3. STATISTICS AND IRREGULAR WAVES
3.1 Measures of height and period
3.2 Probability distribution of wave heights
3.3 Wave spectra
3.4 Reconstructing a wave field
3.5 Prediction of wave climate
Statistics and Irregular Waves
Probability Distribution of Wave Heights
For a narrow-banded frequency spectrum the Rayleigh probability distribution is appropriate:
π height > π» = exp β Ξ€(π» )π»πππ 2
Cumulative distribution function:
πΉ π» = π height < π» = 1 β eβ Ξ€π» π»πππ 2
Probability density function:
π π» =dπΉ
dπ»= 2
π»
π»πππ 2 eβ Ξ€π» π»πππ
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Rayleigh Distribution
π»ππ£π β‘ πΈ π» = ΰΆ±0
β
π» π π» dπ»
π»πππ 2 β‘ πΈ π»2 = ΰΆ±
0
β
π»2 π π» dπ»
π»ππ£π =Ο
2π»πππ = 0.886 π»πππ
π» Ξ€1 3 = 1.416 π»πππ
π» Ξ€1 10 = 1.800 π»πππ
π» Ξ€1 100 = 2.359 π»πππ
π height > π» = exp β Ξ€(π» )π»πππ 2
Single parameter: π»πππ
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Example
Near a pier, 400 consecutive wave heights are measured.Assume that the sea state is narrow-banded.
(a) How many waves are expected to exceed 2π»πππ ?
(b) If the significant wave height is 2.5 m, what isπ»πππ ?
(c) Estimate the wave height exceeded by 80 waves.
(d) Estimate the number of waves with a height between1.0 m and 3.0 m.
Near a pier, 400 consecutive wave heights are measured. Assume that the sea state isnarrow-banded.(a) How many waves are expected to exceed 2π»πππ ?(b) If the significant wave height is 2.5 m, what is π»πππ ?(c) Estimate the wave height exceeded by 80 waves.(d) Estimate the number of waves with a height between 1.0 m and 3.0 m.
π height > π» = eβ Ξ€π» π»πππ 2
π height > 2π»πππ = eβ4(a)
= 0.01832
In 400 waves, π = 400 Γ 0.01832
π» Ξ€1 3 = 1.416 π»πππ (b)
2.5 = 1.416 π»πππ
π―πππ = π. πππ π¦
(c) π height > π» = eβ Ξ€π» π»πππ 2=
80
400= 0.2
π»/π»πππ 2 = β ln 0.2
π» = π»πππ Γ β ln0.2 = π. πππ π¦
π 1.0 < height < 3.0 = π height > 1.0 β π height > 3.0
= eβ 1.0/π»πππ 2β eβ 3.0/π»πππ
2
= 0.6699In 400 waves, π = 400 Γ 0.6699
(d)
= π. πππ
= πππ. π
3. STATISTICS AND IRREGULAR WAVES
3.1 Measures of height and period
3.2 Probability distribution of wave heights
3.3 Wave spectra
3.4 Reconstructing a wave field
3.5 Prediction of wave climate
Statistics and Irregular Waves
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Regular vs Irregular Waves
Regular wave:- single frequency
Irregular wave:- many frequencies
Wave energy depends on Ξ·2
Energy Spectrum
β βSpectralβ means βby frequencyβ
β A spectrum is usually determined by a Fourier transform
β This splits a signal up into its component frequencies
β Energy in a wave is proportional to Ξ·2, where Ξ· is surface displacement
β The energy spectrum, or power spectrum, is the Fourier transform of Ξ·2
Energy Spectrum
For regular waves (single frequency):
πΈ =1
2Οππ΄2 = Οπ )Ξ·2(π‘ =
1
8Οππ»2
For irregular waves (many frequencies):
π π dπ
ΰΆ±π1
π2
π π dπ
is the βenergyβ in a small interval dπ near frequency π
is the βenergyβ between frequencies π1 and π2
(strictly: energy/Οπ)
The energy spectrum π(π) is determined by Fourier transforming Ξ·2
Ξ· = π΄ cos ππ₯ β Οπ‘
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Model SpectraOpen ocean: Bretschneider spectrum
Fetch-limited seas: JONSWAP spectrum
Key parameters: peak frequency ππ ( =1/(peak period, ππ) )
significant wave height π»π0
Model Spectra
Bretschneider spectrum:
JONSWAP spectrum:
π π =5
16π»π02
ππ4
π5exp β
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ππ4
π4
π π = πΆ π»π02
ππ4
π5exp β
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ππ4
π4Ξ³π
π = ΰ΅0.07 π < ππ0.09 π > ππ
π = exp β1
2
Ξ€π ππ β 1
Ο
2
Ξ³ = 3.3
Significant Wave Height, π―ππ
Bretschneider spectrum: π π =5
16π»π02
ππ4
π5exp β
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ππ4
π4
Total energy:πΈ
Οπβ‘ ΰΆ±
0
β
π π dπ =1
16π»π02
π»π0 = 4 Ξ€πΈ Ο π = 4 )Ξ·2(π‘
Total energy for a regular wave:πΈ
Οπ=1
8π»πππ 2
Same energy if π»π0 = 2π»πππ = 1.414π»πππ
Rayleigh distribution: π»π = π» Ξ€1 3 = 1.416π»πππ
π―ππ and π― Ξ€π π can be used synonymously for π―π ...
β¦ but π»π0 is easier to measure!
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Finding Wave Parameters From a Spectrum
β Surface displacement Ξ·(π‘) is measured (e.g. wave buoy)
β Ξ·2(π‘) is Fourier-transformed to get energy spectrum π(π)
β Height and period parameters are deduced from the peak and the momentsof the spectrum:
ππ = ΰΆ±0
β
πππ π dπ
Peak period: ππ =1
ππ
Significant wave height: π»π = π»π0 = 4 π0
Energy period: ππ =πβ1
π0
Multi-Modal Spectra
frequency f (Hz)
spect
ral d
ensi
ty S
(m
^2 s
)
swell
wind
3. STATISTICS AND IRREGULAR WAVES
3.1 Measures of height and period
3.2 Probability distribution of wave heights
3.3 Wave spectra
3.4 Reconstructing a wave field
3.5 Prediction of wave climate
Statistics and Irregular Waves
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Using a Spectrum To Generate a Wave Field
Ξ· π‘ = )ππ cos(πππ₯ β Οππ‘ β ππ
amplitude
π ππ dπ = πΈπ =1
2ππ2 ππ = 2π ππ dπ
(angular) frequency
Οπ = 2Οππ
wavenumber
ππ2 = πππ tanhππβ
random phase
Simulated Wave Fieldππ = 8 s
π»π = 1.0 m
β = 30 mRegular waves:
Irregular waves:
Focused waves:
Example
An irregular wavefield at a deep-water location is characterisedby peak period of 8.7 s and significant wave height of 1.5 m.
(a) Provide a sketch of a Bretschneider spectrum, labelling bothaxes with variables and units and indicating the frequenciescorresponding to both the peak period and the energy period.
Note: Calculations are not needed for this part.
(b) Determine the power density (in kW mβ1) of a regular wavecomponent with frequency 0.125 Hz that represents thefrequency range 0.12 to 0.13 Hz of the irregular wave field.
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An irregular wavefield at a deep-water location is characterised by peak period of 8.7 s andsignificant wave height of 1.5 m.
(a) Provide a sketch of a Bretschneider spectrum, labelling both axes with variables and unitsand indicating the frequencies corresponding to both the peak period and the energyperiod.
frequency f (Hz)
spect
ral d
ensi
ty S
(m
^2 s
)
fp fe
An irregular wavefield at a deep-water location is characterised by peak period of 8.7 s andsignificant wave height of 1.5 m.
(b) Determine the power density (in kW mβ1) of a regular wave component with frequency0.125 Hz that represents the frequency range 0.12 to 0.13 Hz of the irregular wave field.
π = 0.125 Hz
Ξπ = 0.01 Hz
ππ = 8.7 s π»π = 1.5 m
ππ =1
ππ
π π = 1.645 m2 s
πΈ = Οπ Γ π π Ξπ
π = πΈππ = πΈ(ππ) Deep water: π =1
2
π =ππ
2Ο
π =1
π= 8 s
π· = π. πππ π€ππ¦βπ
= 0.1149 Hz
= 165.4 J mβ2
= 12.49 m sβ1
π π =5
16π»π 2ππ4
π5ex p( β
5
4
ππ4
π4)
3. STATISTICS AND IRREGULAR WAVES
3.1 Measures of height and period
3.2 Probability distribution of wave heights
3.3 Wave spectra
3.4 Reconstructing a wave field
3.5 Prediction of wave climate
Statistics and Irregular Waves
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Terminology
A wave climate or sea state is a model wave spectrum, usually defined by a representative height and period, for use in:
β forecasting (from a weather forecast in advance)
β nowcasting (from ongoing weather)
β hindcasting (reconstruction from previous event)
Prediction of Sea State
Require: significant wave height, π»π significant wave period, ππ (or peak period, ππ)
Depend on: wind speed, πfetch, πΉduration, π‘gravity, π
Dimensional analysis:
Empirical functions: JONSWAP or SMB curves
ΰ΅°ππ»π π2 = ππ’πππ‘πππ(
ππΉ
π2 ,ππ‘
π
αππππ
= ππ’πππ‘πππ(ππΉ
π2 ,ππ‘
π
Fetch-Limited vs Duration-Limited
distance travelled by wave energy greater than fetch F
F
distance travelled by wave energy less than fetch F
F
Feff
FETCH-LIMITED
DURATION-LIMITED
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JONSWAP CurvesFor fetch-limited waves:
ππ»π π2
= 0.0016ππΉ
π2
Ξ€1 2
(up to maximum 0.2433)
ππππ
= 0.286ππΉ
π2
Ξ€1 3
(up to maximum 8.134)
Waves are fetch-limited provided the storm has blown for a minimum timeπ‘πππ given by
ππ‘
π πππ= 68.8
ππΉ
π2
Ξ€2 3
up to maximum 7.15 Γ 104
Otherwise the waves are duration-limited, and the fetch used to determineheight and period is an effective fetch determined by inversion using the actualstorm duration π‘:
ππΉ
π2πππ
=1
68.8
ππ‘
π
Ξ€3 2
JONSWAP Curves (reprise)
π»π = 0.0016 πΉ Ξ€1 2
ππ = 0.286 πΉ Ξ€1 3
ΖΈπ‘πππ = 68.8 πΉ Ξ€2 3
πΉ β‘ππΉ
π2 , ΖΈπ‘ β‘ππ‘
π, π»π β‘
ππ»π π2 , ππ β‘
ππππ
JONSWAP and SMB Curves
Solid lines: JONSWAP
Dashed lines: SMB
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Example
(a) Wind has blown at a consistent π = 20m sβ1
over a fetch πΉ = 100 km for π‘ = 6 hrs .Determineπ»π and ππ using the JONSWAP curves.
(b) If the wind blows steadily for another 4 hourswhat are π»π and ππ?
(a) Wind has blown at a consistent π = 20 m sβ1 over a fetch πΉ = 100 km for π‘ = 6 hrs.Determine π»π and ππ using the JONSWAP curves.
π = 20 m sβ1
πΉ = 105 m
π‘ = 6 Γ 3600 = 21600 s
πΉ β‘ππΉ
π2
ΖΈπ‘ =ππ‘
π
ΖΈπ‘πππ = 68.8 πΉ Ξ€2 3
ΖΈπ‘πππ = 68.8 πΉ Ξ€2 3 = 12510
π»π = 0.0016 πΉ1/2
ππ = 0.286 πΉ1/3
πΉ β‘ππΉ
π2ΖΈπ‘ β‘
ππ‘
ππ»π β‘
ππ»π π2
Duration-limited
πΉπππ =ΖΈπ‘
68.8
Ξ€3 2
π»π = 0.0016 πΉπππΞ€1 2
ππ = 0.286 πΉπππΞ€1 3
π»π = π»π Γπ2
π
ππ = ππ Γπ
π
= 1910
= 0.06993
= 3.548
= π. πππ π¦
= π. πππ π¬
= 2453
= 10590
(b) If the wind blows steadily for another 4 hours what are π»π and ππ?
π = 20 m sβ1
πΉ = 105 m
π‘ = 10 Γ 3600 = 36000 s
πΉ = 2453
ΖΈπ‘ =ππ‘
π
ΖΈπ‘πππ = 68.8 πΉ Ξ€2 3
ΖΈπ‘πππ = 12510
π»π = 0.0016 πΉ1/2
ππ = 0.286 πΉ1/3
πΉ β‘ππΉ
π2ΖΈπ‘ β‘
ππ‘
ππ»π β‘
ππ»π π2
Fetch-limited
π»π = 0.0016 πΉ1/2
ππ = 0.286 πΉ1/3
π»π = π»π Γπ2
π
ππ = ππ Γπ
π
= 0.07924
= 3.857
= π. πππ π¦
= π. πππ π¬
= 17660
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Examples on Blackboard
Try examples 10-13
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