• 검색 결과가 없습니다.

Chapter 6. Wind-Generated waves

N/A
N/A
Protected

Academic year: 2022

Share "Chapter 6. Wind-Generated waves"

Copied!
20
0
0

로드 중.... (전체 텍스트 보기)

전체 글

(1)

Chapter 6. Wind-Generated waves

6.1 Waves at Sea

local wind wave (short-crested) + swell (long-crested)

fully developed sea: unlimited fetch and duration of wind; wind energy input is balanced by energy dissipation due to wave breaking.

growing sea: limited fetch and duration of wind (most cases in nature)

(2)

6.2 Wind-Wave Generation and Decay

• Phillips (1957): initial stage

Turbulent wind energy is transferred to water by pressure fluctuation.

t t

H( )∝ (linear growth of waves)

• Miles (1957): developing stage

After some waves are developed, eddies are formed at troughs.

e t

t

H( )∝ α (exponential growth of waves)

Assume that wave growth = sum of linear + exponential growth, and find coefficients using field dat

(3)

• Hasselmann (1962)

wave interactions → energy transfer to lower frequencies

Growth of wind waves depends on 1) fetch length, F

2) wind speed, W 3) duration of wind, td 4) fetch width, B 5) water depth, d

M

(4)

If td >F/Cg, fetch-limited, H,T = f(W,F)

If td = xA /Cg <F/Cg, duration-limited, H,T = f(W,td)

6.3 Wave Record Analysis for Height and Period

Zero-crossing method:

H = average of the highest n n% of the wave heights

(5)

e.g.

=

=

= /10

1 10

/ 1

10 /10

1 N

i

Hi

H N H

↑ ↑

in text more common expression

s N

i

i H

N H H

H = =

= 3 /

1 3

/ 1

33 /3

1 (significant wave height)

=

=

= N

i

Hi

H N H

1 100

1

=

= N

i i

rms H

H N

1

1 2

(root-mean-squared wave height)

=

=

= /3

3 1

/ 1

3 / 1

N

i H H

s T i

T N

T ; = period of the wave of height of

Hi

T Hi

Wave height distribution

Find probability density function, p(H), by plotting histogram of wave height:

m waves out of N have

2 2

H H H H

H −Δ < i < +Δ

N H m H p( )Δ =

1 )

0 ( =

p H dH for probability density function

(6)

Longuet-Higgins: p(H) is given by a Rayleigh distribution

( / )2

2

) 2

( H Hrms

rms

H e H H

p =

cumulative probability, ( / )2

0 ( ) 1

) ' ( y probabilit )

(H H H H p H dH e H Hrms

P = ≤ =

= −

exceedance probability = probability(H'>H)=1−P(H)=e(H/Hrms)2

Hrms

dH H Hp

H ( ) 0.886

0 =

=

=? Hs

( )

[ ]

3 / 1

exp− h1/3 Hrms 2 = → h1/3

rms rms

h

h h

s H H

dH H Hp dH

H p

dH H Hp

H 1.416 2

3 / 1

) ( )

( ) (

3 / 1

3 / 1

3 /

1 = = ≅

=

(7)

Fig. 6.5: line a → exceedance prob. ⎟⎟

⎜⎜ ⎞

⎛ >

rms

rms H

H H

H '

Note: P in the y-axis of Fig. 6.5 must be 1−P

line b → average of the highest % waves n

s

s N H

H

Hmax =0.707 ln ≅2.0 : used for design of offshore structures

Wave period distribution

Joint probability of H and T

(8)

p

p f

T 1

= ; Ts ≅0.95Tp

where Ts = significant wave period from zero-crossing method Tp = peak period from spectral analysis.

(9)

6.4 Wave Spectral Characteristics

Irregular waves = superposition of many sinusoidal waves of different frequency, amplitude, phase, and direction

Using directional spectrum analysis, we obtain directional spectrum, S(f,θ). Assuming single wave direction, we obtain frequency spectrum, S( f).

(10)

For a sinusoidal wave, energy per unit surface area or energy density is

2

8 1 gH E = ρ

Since ρ = constant, we can write g

∑ ∑

+ +

= f df

f

d H

dfd f S

θ θ

θ

θ

θ) 8

, (

2

; H =H(f,θ)

+

= f df

f

df H f

S( ) 8

2

; H =H( f)

+

=

+ +

=

+

=

=

* 0

* 0

* 0

* 0

2

* 2

2 2

*

2

*

*

2 2

) 2

2 (

) 2 (

1 1

T T

T T p p

T dt d g

g

dt d

T d g

dt g d

T T dE E

ρ η ρ

η ρ η

ρ η

where = length of wave measurement. Considering the potential energy due to waves,

T*

(11)

= *

0

2

2 * T

p dt

T E ρg η

Since Ep =Ek, the total energy density is

2 0

2

*

2 ρ *η ρ η

g T dt

E g

E= p =

T =

where the over-bar of η2 denotes time-average. Note that η2 is the variance of η . The definition of variance is Var(x)=E

[

(xμ)2

]

. In our case, the mean, μ is zero.

For a discretely sampled η ,

=

= N

i

N i

E g

1

η2

ρ

where N = number of samples in T*. Recalling

=

= N

i i

rms H

H N

1

1 2

we get

=

= N

i i

rms H

H N

1 2

2 1

E N E

g H H N

g N

i i N

i

i

rms =

=

=

=

=1 1

2

2 1

8 1 8

ρ ρ

2 2

16 8 rms g Hs

g H

E = ρ = ρ

From Eq. (6.8)

(12)

+

= f df

f

df H f

S( ) 8

2

0

all 0

2 ( )

8

gH g S f df gm

E

f

ρ

ρ = ρ

=

∑ ∫

where m0 = zeroth moment of spectrum. The nth moment is given by

= 0 f S(f)df

mn n

Since E= ρgη2 , η2 =m0. Also, since 2 16g Hs E= ρ

, Hs =4 m0Hm0

In deep water, HsHm0. As kd decreases, Hs >Hm0 (see Fig. 6.7)

6.5 Wave Spectral Models

General form of frequency spectrum: ( ) 5 e B/f4 f

f A

S =

where A,B = empirical constants

Bretschneider spectrum

)4

/ ( 675 . 0 4

3 2

44 . ) 3

( e T T

T H T T

S =

Using S(f)=S(T)T2 and T =1/ f ,

4 4 4

/ 675 . 0 4 5

2

) / ( 675 . 0 4

5 2

44 . 3

44 . ) 3 (

f T

T T

e f T

H e T

H f T

S

=

=

(13)

Bretschneider-Mitsuyasu spectrum (applicable to finite depth)

( )

5

[ ( )

4

]

2 exp 0.75

205 . 0 )

(f = H T T f T f

S s s s s

Pierson-Moskowitz spectrum (for fully developed seas)

)4

2 / ( 74 . 0 5 4

2

) 2 ) (

( e g Wf

f f g

S π

π

α

=

where α =8.1×103 and W = wind speed at 19.5 m above SWL ≅(1.05~1.1)W10

JONSWAP spectrum (growing seas in deep water) Based on data of JOint North Sea WAve Project

[ 2 2 2]

4 exp ( ) /2

) / ( 25 . 1 5 4

2

) 2 ) (

( e fp f f fp fp

f f g

S γ σ

π

α

=

where γ = peak enhancement factor (typical value = 3.3), and

⎩⎨

= <

p p

f f

f f for 09 . 0

for 07 . σ 0

(14)

JONSWAP spectrum with γ =1.0 = Pierson-Moskowitz spectrum

TMA spectrum (includes effect of finite water depth) )

, (f d S

STMA = JΦ

where Φ(f,d) = Kitaigordskii shape function for finite depth effect:

( )

⎪⎩

⎪⎨

>

<

= Φ

2 for 1

2 1

for 2

5 . 0 1

1 for 5

. 0 ) ,

( 2

2

d d d

d d

d f

ω ω ω

ω ω

where ωd =2πf(d/g)1/2

(15)

Directional wave spectra

Frequency spectrum assumes waves with many different frequencies but a single direction. The real waves consist of many component waves with different frequencies and directions. Therefore, we need directional wave spectra.

)

; ( ) ( ) ,

(f θ S f G f θ

S =

where G(f;θ) = directional spreading function, which represents directional distribution of wave energy. In general, G(f;θ) varies with frequency, f :

small f → long-period waves → narrow spreading large f → short-period waves → wide spreading

We take

1 )

;

( =

π

πG f θ dθ

so that G(f;θ) represents relative magnitude of directional spreading of wave energy.

∫ ∫

∫ ∫

=

=

=

0 0 0

) (

)

; ( ) (

) , ( energy

total

df f S

df d f G f S

df d f S

π π π π

θ θ θ θ

Cosine square function:

θ

θ π

θ 2cos2

) ( )

;

(f = G =

G ← independent of f

(16)

Mitsuyasu-type function:

) 1 2 (

) 1 ( ) 2

( 2 ; cos ) ( )

; (

2 1 2 2

+ Γ

+

= Γ

⎟⎠

⎜ ⎞

= ⎛

s s s

G s

G f

G

s s

π θ θ

⎪⎩

⎪⎨

>

= <

p p

p p

f f f

f s

f f f

f s s

for )

/ (

for )

/ (

5 . 2 max

5 max

smax

s= at f = fp, and s decreases as ffp increases.

Swell → larger smax → narrow spreading Wind wave → smaller smax → wide spreading

(17)

6.6 Wave Prediction – SMB Method

Sverdrup, Munk, Bretschneider Consider a box storm:

By dimensional analysis,

⎟⎠

⎜ ⎞

= ⎛

W gt W f gF W gT W

gHs s d

2 ,

, 2

2 π

Using Fig. 6.10,

W2

gFH ,s Ts (1)

W gtd

H ,s Ts (2)

Choose the smaller values of (1) and (2). If (1) is smaller, it is fetch-limited condition. If (2) is smaller, duration-limited condition.

For a typhoon, use Eqs. (6.37) and (6.38):

R = radius to maximum wind speed (WR)

e

a p

p p= −

Δ = strength of typhoon V = forward speed of typhoon F

≅1

α for slow moving typhoon

(18)

6.7 Wave Prediction – Spectral Models

/ 4

) 5

( e B f

f f A

S = ← general form B

A, = empirical constants = f(W,F,td)

Given W,F,tdS( f) → Hs Hm0 =4 m0; m0 =

0S(f)df =0

f

S at f = fp → find fpTpTs =0.95Tp

SPM: W,F,td →(w/o calculation of S( f))→ Hm0,Tp for JONSWAP spectrum

23 .

71 1

.

0 W

WA =

5 . 0

2 2

0 0.0016 ⎟⎟⎠

⎜⎜ ⎞

= ⎛

A A

m

W gF W

gH (6.40)

33 . 0

286 2

.

0 ⎟⎟⎠

⎜⎜ ⎞

= ⎛

A A

p

W gF W

gT (6.41)

66 . 0

8 2

. ' 68

⎟⎟⎠

⎜⎜ ⎞

= ⎛

A A

d

W gF W

gt (6.42)

where = minimum duration for fetch-limited condition, whereas = actual duration.

d'

t td

If , fetch-limited → Use Eqs. (6.40), (6.41) tdtd'

If , duration-limited → Calculate td <td' F using td'=td with Eq. (6.42)

→ Use Eqs. (6.40), (6.41) with new F.

6.8 Numerical Wave Prediction Models

(read text)

(19)

6.9 Extreme Wave Analysis

Return period (再現期間)?

Ex) of 50 year return period = significant wave height which can occur once in every 50 years on the average.

Hs

How can we estimate Hs of 50 year return period with limited data (e.g. 2 year data)?

Return period analysis using Gumbel distribution 1) Hs was measured every hour for 2 years.

2) Select daily maximum Hs.

number of data, N =365×2=730,

r = time interval in years = 0.00274 3651 =

3) Rearrange Hs in descending order from Hs(1) to Hs(N). 4) Compute cumulative probability, P(Hs) for each Hs.

5) Plot Hs vs −ln

{

−ln

[

P(Hs)

] }

→ Find β,γ for best fit.

Gumbel distribution:

⎪⎭

⎪⎬

⎪⎩

⎪⎨

⎧ ⎥

⎢ ⎤

⎡ ⎟⎟

⎜⎜ ⎞

⎛ −

= β

γ H H

P( ) exp exp

H =−βln

{

−ln

[

P(H)

] }

+γ 6) Calculate

Tr

Hs

P( ) by using )

(

1 s

r

H T P

r = − → 0.9999452

50 00274 . 1 0 1

)

( =50 = − = − =

r T

s T

H r

P r

(20)

7) Calculate Hs (Tr = 50 yr) by

{ }

γ

β − +

=50 = ln ln(0.9999452)

Tr

Hs

Use similar procedures for other distributions (see Table 6.1) Encounter probability:

E =1−eT/Tr ← 재현기간 Tr인 사건이 기간 T 동안에 발생할 확률

참조

관련 문서

Daily mean of air pressure, air temperature, dew-point temperature, wind direction and speed, relative humidity and cloud amount is the average of hourly

Daily mean of air pressure, air temperature, dew-point temperature, wind direction and speed, relative humidity and cloud amount is the average of hourly

Daily mean of air pressure, air temperature, dew-point temperature, wind direction and speed, relative humidity and cloud amount is the average of hourly

Daily mean of air pressure, air temperature, dew-point temperature, wind direction and speed, relative humidity and cloud amount is the average of hourly

- Renewable energy makes certain of energy security and is a clean energy without contaminants. -

Students with family obligations and the revival of individual success is complicated on sail.. Second image of the sea is energy of young and

Temperature, Temperature difference between 10m and 300m, Humidity, Wind speed, Wind speed difference between 10m and 300m, Wind direction, Vertical Wind average,

▶ One example where the Rayleigh distribution naturally arises is when wind speed is analyzed into its orthogonal 2-dimensional vector components. Assuming that the magnitude