제2장 해양의 파_JEJU Univ.

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제2장 해양의 파

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Coastal amp Harbor Engineering Laboratory Jeju National University- 1 -

[그림 1]

2-1 서설

파속(wave celerity)rArr C =LT

(T주기(period)) 파형경사(wave steepness)rArr HL

주기에 의한 해양파의 분류

표면장력파 가장 주기가 짧은 파 주기 007초 이하 파장은 17cm이하 파고는 1～2mm

풍파(wind wives) 바람에 의해 발달한 파 주기 10～15초 파고 37m되는 파도

너울(Swell) 풍파가 풍역을 벗어나 진행하는 파 가장 긴 주기는 20sec

파랑 = 풍파 + 너울

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[그림 3]

lt 영역 gt

- h le yle η- infin le xle infin

장주기파

해면 부진동

방재또는 연안개발이나 환경보전에 중요 해일(Tsunami) 지진해일 고조(Storm surge) 폭풍해일

조석

A 해양파의 성질

2-2 파의 기본방정식

완전유체 점성무시 회전무시 속도포텐셜 φ존재

물입자의 속도성분 u( x방향) u =partφpartx

(2-2-1) 물입자의 속도성분 v( y방향) v=

partφparty

연속방정식 비압축성

partupartx

+partvparty

= 0 (2-2-2)

(2-2-1)식을 (2-2-2)식에 대입하여 정리하면

part 2φpartx2 +

part 2φparty2 = 0 ( 2차원 Laplace 방정식 ) (2-2-3)

경계조건(Boundary Condition) 수평인 해저에서 연직 속도 성분은 ldquo0rdquo

(v) y=- h = ( partφparty )

y=-h

= 0 (2-2-4)

압력방정식(Pressure Equation) 비회전

partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ

+ gη = 0 (2-2-5)

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여기서 P 압력 ρ 밀도 g 중력가속도

자유표면 y= η에서 압력은 P 0

( partφpartt )

y= η

+12 ( partφ

partx )2

+ ( partφparty )

2

y= η

+P0

ρ+ gη= 0 (2-2-6)

운동학적 조건 자유표면이 F(xyzt) = 0와 같이 표시된다면 경계상의 물입자는 항상

그 경계에 머문다고 하는 것에 의해 DFD t

= 0가 성립

DDt

=partpartt

+ upartpartx

+ vpartparty

+ wpartpartz

( Lagrange 미분 ) (2-2-7)

2차원의 경우에는 F(xyt) = η(xt) - y= 0을 편미분하면

( DFDt )

y= η

= ( partηpartt

+ upartηpartx

- v)y= η

= 0 (2-2-8)

또는

partηpartt

+partηpartx ( partφ

partx )y= η

= ( partφparty )

y= η (2-2-9)

식(2-2-4) (2-2-6) (2-2-9)의 경계조건을 만족하는 식(2-2-3)의 해를 구해야 한다

2-3 미소진폭파이론(Small Amplitude Wave Theory)

2-3-1 기초방정식의 해

기초 방정식의 해를 구하는 것은 비선형 문제의 해를 구하는 것이다

우선 파고가 파장 또는 수심에 비해서 매우 작은 경우 rArr 미소진폭파

( partφpartt )

y= η

( partφparty )

y= η

를 y = 0에 Taylor급수로 전개하면

( partφpartt )

y= η

= ( partφpartt )

y= 0

+ partparty ( partφ

partt )y= 0

η+12 part

party2 ( partφpartt )

y= 0

η 2 + (2-3-1)

( partφparty )

y= η

= ( partφparty )

y= 0


party )y= 0

η+12 part

party2 ( partφparty )

y= 0

η 2 + (2-3-1)

이 된다 2차 이상의 미소항을 생략하고 y= η에서의 대기의 압력 P 0 = 0과 같다고 생각한다

따라서 파의 기초방정식은

part 2φpartx2 +

part 2φparty2 =0 -hleyleη -infinlt xltinfin (2-3-2)

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( partφparty )

y=- h

= 0 (2-3-3)

( partφpartt )

y= 0

+ gη = 0 (2-3-4)

partηpartt

= ( partφparty )

y= 0 (2-3-5)

여기서 (2-3-4)와(2-3-5)의 관계로부터

η= -1g ( partφ

partt )y= 0

lsquo partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0

가 된다 (2-3-4)식에서 y을 소거하면

partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0

( part 2φpartt2 )

y= 0

= -g( partφparty )

y= 0

(2-3-6)

을 얻을 수 있다

지금의 해로서 변형하지 않고 x축 방향으로 파속 c로 진행하는 파를 생각하면

φ = φ(x-ct y)로 쓸 수 있다 여기서 변수분리법을 적용하고 또한 파형으로서 정현파형을

취하면

φ= f(y)sink(x- ct) = f(y) sin (kx- σ t) (2-3-7)

여기서 k=2πL

(파수 wave number) L 파장

σ =2πT

(각주파수 angular frequency) T 주기

c =σk (파속 wave celerity )

이것으로부터 (2-3-4)식을 고려하면

partpartt

f(y) sin (kx- σ t)|y= 0

+ gη= 0

을 얻을 수 있다 따라서 파형은

η=σg(f(0) cos (kx- σt))= a cos (kx- σt) (2-3-8)

여기서 파고를 H라고 하면 진폭 a =H2

가 된다

식(2-3-7 φ= f(y)sin (kx- σ t))을 식(2-2-3part 2φpartx2 +

part 2φparty2 =0)에 대입하면

part 2

partx2 (f(y) sin (kx- σt)) +part 2

party2 (f(y) sin (kx- σt)) = 0 rArr x에 관해 한번 미분하면

partpartx

(kf(y)cos (kx- σt)) +part 2

party2 (f(y) sin (kx- σt)) = 0 rArr x에 관해 다시 한 번 미분하면

-k2 f(y) sin (kx- σt) +d 2

dy2 f(y) sin (kx- σt) = 0

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( partφparty )

y=-h

= 0 larr(2-3-3)

( partφparty )

y=- h

=partparty

[ (Ae ky+ Be-ky)sin (kx- σt)] y=- h

= [ (kAe ky -kBe-ky)sin (kx- σt)] y=- h

= k(Ae-kh- Be kh) sin (kx- σt) = 0 hellip①

( part 2φpartt2 )

y= 0

larr(2-3-6)

가 된다 이것을 정리하면

d 2f(y)dy2 - k2f(y) = 0 (2-3-9)

가 된다 식(2-3-9)을 변수분리하면

d 2f(y)f(y)

-k2dy2=0

가 된다 여기서 f(y)는

f(y) = Ae ky+ Be- ky A B 적분상수 (2-3-10)

이며 따라서 φ는

φ = (Ae ky + Be-ky) sin (kx- σt) (2-3-11)

식(2-3-11)식을 경계조건식(2-3-3)과 (2-3-6)에 대입하면

이 얻어진다

( part 2φpartt2 )

y= 0

=part 2

partt2[ (Ae ky + Be- ky) sin (kx- σt)] y= 0

=part 2

partt2 (A + B)sin (kx- σt)

=partpartt

(A + B) - σ cos (kx- σt)

= (A + B) - σ 2sin(kx- σt)

-g( partφparty )

y= 0

larr(2-3-6)

-g( partφparty )

y= 0

= -gpartparty

[ (Ae ky + Be-ky) sin (kx- σt)] y= 0

= -g[ (kAe ky - kBe- ky) sin (kx- σt)] y= 0

= (-gkA + gkB)sin (kx- σt)

이것을 ( part 2φpartt2 )

y= 0

= -g( partφparty )

y= 0

에 대입하여 정리하면

-σ 2 (A+B) sin (kx-σt)= (-gkA+gkB)sin (kx-σt) hellip②

①과 ②에서 sin (kx-σt)로 나누면

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coshm =em + e-m

2

sinhm =em - e-m

2

①은 k(Ae- kh - Be kh)= 0 이고 kne0 아니므로 (2-3-12)

②는 -σ 2 (A+B)= (-gkA+gkB) rArr σ 2A-gAk+gBk+σ 2B= 0 이며

(σ 2 - gk)A + (σ 2+gk)B = 0 (2-3-13)

가 얻어진다

위에서 얻어진 두식에서 A B가 함께 0이 아닌 해를 갖기 위해서는

| |e-kh -e kh

σ 2- gk σ 2+gk ( )AB

= 0 Matrix에서[]+[]- (2-3-14)

이 되어야 한다 이것을 풀면

(σ 2 + gk)e-kh + (σ 2 - gk)e kh = 0 rArr σ 2e -kh+gke kh+σ 2e kh-gke kh=0

(e kh+e -kh)σ 2-(e kh-e -kh)gk= 0

이것을 다시 σ 2에 대해서 정리하면

σ 2 = gke kh - e-kh

e kh+ e- kh = gk2sinh (kh)2 cosh (kh)

= gk tanhkh

(2-3-15)

의 관계를 얻는다 이것을 분산관계식이라 하며 수심 h을 고정했을 때 파수 k와 각주파수 σ와는

서로 관계있음을 나타내고 있다

따라서 c =σk의 관계를 이용하면 c 2k2=gktanhkh이므로

c =gk

tanhkh =gL2π

tanh2πhL

(2-3-16)

혹은 c =LT

을 고려하여 위 식(2-3-16)을 고쳐 쓰면 c 2 =gcT2π

tanh2πhL

이 되어

c =gT2π

tanh2πhL

(2-3-17)

가 된다 또 L = cT을 고려하면

L =gT 2

2πtanh

2πhL

(2-3-18)

의 식을 얻을 수 있다

그러면 식(2-3-12)에서 Ae-kh - Be kh = 0을 Ae-kh = Be kh =D2

로 놓으면

A =D2

e kh B =D2

e- kh가 되어

속도포텐셜 φ는

φ =D2

e k( h+ y) + e-k( h+ y)sin (kx- σt)=Dcoshk(h+y)sin (kx-σt) (2-3-19)

로 된다 이것을 ( partφpartt )

y= 0

+ gη = 0에 대입하면

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[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part

partt(D coshk(h+y)sin (kx- σt))

y= 0

=σg

Dcoshkhcos (kx- σt)= a cos (kx- σt)

(2-3-20)

여기서 a =σg

Dcoshkh이므로 σ 2 = gk tanhkh을 염두에 두고 D을 구하면

D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)

로 된다 따라서 속도포텐셜 φ는

φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다

2-3-2 물입자의 속도성분과 궤적

물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)

sinhkhcos (kx- σt) = σ

coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)

sinhkhsin (kx- σt) (2-3-24)

이제부터는 Euler적인 취급으로부터 벗어

나 특정 물입자에 주목한 Lagrange적인 취

급을 해보자

시각 t에 있어서 물입자의 위치를

[x( t) y( t)]로 놓으면 수평방향 속도=

dxdt

연직방향 속도=dydt

로 표시된다

그러므로 다음의 관계가 성립된다

dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)

이 미분방정식을 만족하는 x(t) y(t)가 이 물입자의 궤적을 주게 된다

물입자의 평균위치를 x0 y 0로 하고 시각 t에서 수평 및 연직방향 변위를 δ( t) γ( t)로 하면

x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)

으로 된다 이것을 다시 윗 식에 대입하여 Taylor급수로 전개하면 다음과 같다

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[그림 5]

dxdt

=partφpartx

x( t) y( t) t = ( partφpartx )

x0 y0

+ δ( part 2φpartx2 )

x0 y0

+ γ( part 2φpartxparty )

x0 y 0

+

dydt

=partφparty

x( t) y( t) t = ( partφparty )

x0 y0

+ δ( part 2φparty2 )

x0 y0


x0 y 0

+

(2-3-27)

제1차 근사로서 위 식의 우변 제2항 이하를 생략하고 φ =aσk

coshk(h+y)sinhkh

sin (kx- σt)을

이용하면 다음의 관계식을 얻는다

dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)

sinhkhsin (kx0-σt))

= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)

sinhkhsin (kx0- σt))

= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)

이 식을 적분하면

δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)

가 된다 식(2-3-26 x(t) = x0 + δ( t) y(t) = y0 + γ( t))에서 δ = x-x0 γ = y- y0가 된다

이것을 다시 위 식에 대입하면

x- x0 = -acoshk(h+y0)

sinhkhsin (kx0- σt) (2-3-29)

y- y0 = asinhk(h+y0)

sinhkhcos (kx0- σt) (2-3-30)

가 된다 위 두식을 t을 소거하기 위해서 정리하면

x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)

가 된다 다시 양변에 제곱을 한 후 두식을 더하면

sin 2(kx0 - σt)+ cos 2(kx0 - σt) = 1이 되므로

(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)

을 얻을 수 있다 곧 물입자는 아래 그림에 표시한 것과 같은 장원궤도를 그리는 것을 알 수

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있다

2-3-3 파에 수반되는 압력의 변동

압력방정식(2-2-5 partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ

+ gy= 0)에서 2차 미소항을 생략하면

P = - ρpartφpartt

- ρgy (2-3-32)

로 된다 여기서 우변의 제2항 ρgy는 정수압을 나타내며 파의 운동에 의해서 일어나는 압력 변동을

Δp로 하면

Δp = - ρpartφpartt

= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)

coshkhcos (kx - σt) rArrη = a cos (kx - σt)

= ρgcoshk(h+y)

coshkhη

(2-3-33)

2-3-4 군속도와 에너지 전파

파고가 같고 파장에 따라서 파속도 매우 닮은 2개의 여현파를 생각한다

η = a cosk(x- ct) η = a cosk(x - ct)

선형이론에서는 해를 겹친 것도 역시 해가 되기 때문에

η = a cosk(x- ct) + a cosk(x - ct) rArrcosA+cosB=2cos (A-B

2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)

여기서 cos ( k- k2

x-kc - kc

2t)의 파에 주목하면 ( k=

2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc

2t)의 파에 대해서는

k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)

여기서 k≒k c≒c라고 생각하기 때문에 이들은 2πk

2πkc

c로 되어 파의 특성을 나타낸다

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[그림 6]

(tanh x) = sech2 x 2 sinhkh coshkh = sinh 2kh

다시 cos ( k-k2

x-kc - kc

2t)의 파로 되돌아가면 그 파장 주기는 각각의 파의 파장

주기에 비해서 크고 합성된 파의 어떤 순간에 있어서 파형의 포락선을 규정하는 것임을 알 수

있다 이와 같은 포락선 곧 파군의 전파속도를 군속도 c G라고 하는데 이는 다음과 같이 나타

낼 수 있다

cG =kc - kck- k

=δ(kc)δk

여기서 c G =d(kc)dk

=dσdk

(2-3-36)

로 고쳐 쓸 수 있다

따라서 σ 2 = gk tanhkh을 대입하여 대수를 취하고 k로 미분하면

2σdσdk

= gtanhkh + gkh sec h 2kh

dσdk

=(g tanhkh + gkh sech 2kh)

2σrArr σ 2=gk tanhkh


2gktanhkhσ=

g tanhkh2gktanhkh

σ+gkh sech 2kh2gktanhkh

σ

=σ2

1k

+ h

1

cosh 2khsinhkhcoshkh

=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)

을 얻는다 여기서 n =12 (1 +

2khsinh 2kh )이다

단위폭 1파장당의 위치 에너지(potential energy)는

E p = ⌠⌡

h

0

⌠⌡

η

0ρgy dydx= ρg⌠⌡

h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

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=ρg2

a 2⌠⌡

L

0cos 2(kx - σt) dx rArr η=a cos (kx-σt)

=ρg2

a 2 12

⌠⌡

L

01 + cos2(kx - σt) dx rArr cos 2θ=

12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)

단위폭 1파장 당 운동에너지

E k =ρ2

⌠⌡

L

0

⌠⌡

η

-h(u 2 + v2) dydx에서 연직운동은 매우 작다고 하여 생략한다 ( |η | ltlt h )

따라서 E k를 다시 쓰면

E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L

0a 2cos 2(kx - σt) dx

=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)

1파장 당 전체 에너지

E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다

단위 표면적당 평균 에너지를 각각 E E p E k로 나타내면

E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)

단위폭 당 단위시간에 물이 하는 일률 W

W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ

partx ) dy≒ ⌠⌡

0

-hp( partφ

partx ) dy (2-3-44) 여기서 p는 식(2-3-33) u는 식(2-3-23)을 이용한다

p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)

1주기에 걸쳐 평균을 취하면

=====================================================================================


=====================================================================================


sinh kh cosh kh = 12 sinh 2kh cos2x =12[1+cos 2θ]

cosh2 x =12cosh2x+12

W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)

coshkhcos (kx-σt) aσ

coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T

0cos 2(kx-σt)dt⌠⌡

0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h

12(1+cosh 2k(h+y))dy

= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1

2ksinh 2kh+h]= ρa 2σg

sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh

sinh 2kh ) rArr c= σk

n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)

2-3-5 심해파 천해파 장파

지금까지는 수심 효과가 있는 소위 천해파를 다루어 왔다 여기서 2개의 극한에 대해서

생각해보면

① 파장에 비해서 수심이 매우 큰 경우 kh rarr infin 심해파 표면파

② 파장에 비해서 수심이 매우 작은 경우 kh rarr 0 장파 극천해파

rArrhL

장파 극천해파 lt 천해파 lt 심해파 표면파

hL

120

～125

(=005～004) 12(= 05)

tanhkh≒kh tanhkh≒1

2-3-6 표면장력파

=====================================================================================


=====================================================================================


파장이 짧아지면 표면장력의 영향을 고려할 필요가 생긴다

식 (2-3-4 ( partφpartt )

y= 0

+ gη= 0)에서 표면장력 K를 고려하면

( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)

η= a cos (kx- σt)을 고려하면 part 2ηpartx2 = -k2η가 되므로 식(2-3-64)을 고쳐 쓰면

( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)

가 된다 여기서

φ =agσ

coshk(h+y)coshkh

sin(kx- σt)를 ( partφpartt )

y= 0

에 대입하여 풀면

( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)

가 된다 이것을 다시 위식에 대입하면

( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0

-ag cos (kx- σt) + (g+Kk2

ρ )a cos (kx- σt) = 0

-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )

가 된다 이것을 다시 c =gk

tanhkh에 대입하면 파속 c는

c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)

실제로 표면 장력이 영향을 받는 것은 파장이 2 cm이하의 짧은 파이므로 통상은 심해파의

조건( tanhkh≒1)으로 생각한다 따라서 이 경우의 파속은

c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)

로 된다 파속 c와 파장 L과의 관계를 나타내면 아래 그림과 같이 된다 여기서 a는 순수한

중력파의 파속 gL2π

을 b는 순수한 표면 장력파의 파속2πkρL

을 나타내고 있다

따라서 위 식에서 보이는 파속 c는 극값 cm을 가지며 이것을 주는 L을 Lm으로 하면

=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin

- sinA = sin (-A)- cosA = cos (-A)

Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다

예로서 온도 15에 있어서 물의 표면 장력 K = 735 dynecm

를 대입하면

Lm = 172cm cm = 232cms

를 얻는다 곧 실제로 물의 파에 있어서는 파속이 232cms보다

도 느린 파는 존재하지 않음을 알 수가 있다

2-3-7 중복파

파고와 주기가 같고 x의 양(+) 또는 음(-)의 방향으로 진행하는 2개의 파형을 각각 η 1 η 2로

하고 이들에 대응하는 속도포텐셜을 φ 1 φ 2로 하면

η 1 = a cos (kx- σt) φ 1 = a σk

coshk(h+y)sinhkh

sin (kx- σt)

η 2 = a cos (kx+ σt) φ 2 = -a σk

coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서

η= η 1 + η 2 = 2a coskxcos σt

(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt

이렇게 해서 얻어진 파를 정상파(Standing wave)

혹은 중복파(Clapotis)라고 한다

이 중복파에 대한 물입자의 속도 u v는

u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)

이로써 분명한 바와 같이 수평 속도 u는 x= n πk에 있어서 항상 zero이기 때문에 위의 식

=====================================================================================


=====================================================================================


[그림 8]

은 예를 들면 그림에서 도시한 바와 같이 x= 0에 직

립벽을 설치한 경우의 조건 ( partπpartx )

x= 0

=0 을 만족하는

중복파를 나타내는 것이 된다 이 x= n πk의 곳에서

는 연직 속도 v는 최대이고 또 수면변동 η도 최대이

므로 이것을 배(loop)라 한다 또 x= (n+12

)πk의

곳에서는 수평 속도 u는 최대 연직 속도 v는 zero이

고 수면 변동도 항상 zero이므로 이를 절(node)이라고 한다 배에 있어서는 수평속도 u는

항상 0이므로 배에 있어서 직립벽으로 막아도 운동 상태는 변화하지 않는다

수평유속( u) x= 0πk

2πk

hellip 에서 0 연직속도 최대 배(loop)

연직유속( v) x= π2k

3π2k

5π2k

hellip에서 0 수평속도 최대 절(node)

이 파의 단위 표면적당 평균 위치 에너지 E p와 운동 에너지 E k는

E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L

0a 2cos 2σtdxdy rArr η= 2a coskscos σt

=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0

-h(u 2 + v2)dxdydt = ρg

2a 2 (2-3-73)

이처럼 중복파의 각 에너지는 진폭 a의 진행파의 각 에너지의 2배가 된다 그러나 다시 단위

폭 1파장에 대한 위치 에너지 운동에너지를 구해보면 각각 시간의 함수로 되는 것을 알 수

가 있다 즉 sinσt = 0의 순간에는 유체내의 모든 장소에서 u v는 동시에 zero로 되고 운동

에너지는 zero이지만 그때 수면 변동은 최대이고 위치 에너지 또한 최대의 상태에 있다 이

렇게 에너지는 수송됨이 없이 운동 에너지와 위치 에너지의 사이를 시간적으로 왕복하는 것을

알 수 있다

물입자의 평균 위치 (x0 y0)에 대한 변위 (δ γ)을 구하면

=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2

cosA- cosB=-2 sinA+B

2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)

sinh khcos kx0cosσt

(2-3-74)

여기서 σt를 소거하면

γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0

=- tanhk(h+y0)coskx0

sinkx0

=- tanhk(h+y0)1

tankx0

=- tanhk(h+y0) cotkx0

(2-3-75)

를 얻는다 곧 물입자의 이동 궤적은 (x0 y 0)를 중심으로 하는 일정 경사의 직선으로 물입

자는 배에서는 연직 절에서는 수평 방향으로 왕복 운동을 한다는 것을 알 수가 있다

유선은 위의 식으로부터 다음의 미분 방정식의 해로 주어진다

vu

=dydx

=- tanhk(h+y) cotkx (2-3-76)

따라서 유선을 나타내는 곡선은

sinhk(h+y) sinkx= const (2-3-71)

이 형태는 그림에 보인 것처럼 되어 비정상 운동이지만 기본적으로는 시간적으로 변동하지

않는 유선의 형태로 되는 것을 알 수 있다

우리가 지금까지 다룬 것은 연직벽에 있어서 파가 완전히 반사할 경우에 생기는 완전 중복

파에 대한 것이다 이에 대해서 불완전 반사의 경우에는 다른 진폭을 갖는 파의 겹침이 된다

이를 부분 진폭파라 한다

x의 음의 방향으로 진행하는 입사파의 진폭을 a 1 양의 방향으로 진행하는 반사파의 진폭을

a 2로 하고 또한 a 1 gt a 2라 하면 이들 입사파의 파형은

η = a 1cos(kx+σt) + a 2cos(kx-σt)= (a 1+a 2) coskxcos σt- (a 1-a 2) sinkxsinσt

(2-3-78)

가 된다 이로부터 「그림 9」에 나타낸 것과

같이 L4마다 배와 절이 나타남을 알 수가 있다

배에서의 진폭을 a max 절에서의 진폭을 a min

라 하면

a max = a 1 + a 2 a min

= a 1 - a 2 (2-3-79)

로 주어진다 따라서

=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)

를 얻는다 이로부터 반사율(reflection coefficient) K r은

K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)

따라서 배에서의 진폭과 절에서의 진폭을 측정함으로써 미소진폭파 이론이 적용된다고 했을

때의 반사율을 구할 수 있다 이와 같은 방법을 Healy의 방법이라 한다

주기 5sec 파고 12m 수심 5m인 파에 대해서 파장과 파속을 천해파와 심해파에 대해서 계산하라

천해파

L =98times 5 2

2πtanh

10πL

을 Try and Error Method(시산법)로 풀면 L = 30289m

c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】

2-4 파에 수반되는 평균량

=====================================================================================


=====================================================================================


2-4-1 질량 수송

파가 존재하는 장(field)에서 단위폭당에 파가 진행하는 방향으로의 질량 수송량(mass

transport) 혹은 평균 운동량 (mean momentum) Mx은

Mx=⌠⌡

η

-hρudy (2-4-1)

여기서 는 파의 1주기에 걸친 평균을 나타낸다

Euler적인 속도 성분 u는미소진폭파 이론에 의해서 다음과 같이 주어진다

u = aσcoshk(h+y)

sinhkhcos (kx- σt)

따라서 평균 운동량 Mx는

Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η

0udy≒ ρ(u ) y= 0η

위의 식에서 주기 T동안의 시간 t에 대해서 적분하면

Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T

0(a cos (kx-σt)aσ coshk(h+y)

sinhkhcos (kx-σt))dt

= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k

σrArr σ 2=gktanhkh rArr

gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다

평균질량 수송속도 u L은

u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)

로 되고 이를 이용하여 질량수송 Mx를 구하면

Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ

2 sinhkh coshkhsinh 2kh

=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================


주기 128sec 파고 3m의 심해파에 대해서 미소진폭파이론으로 다음 값을 구하여라

(1)파장 파속

(2)정수면 아래 2m 20m의 해중에서 물입자 속도 u v를 구하라

ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도

u= aσe kycos(kx-σt)

v= aσe kysin(kx-σt)

a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형

진행하는 파는 해저지형의 변화 장애물의 존재 등에 의해 여러가지 변형을 받는다 파의 변

형을 다루는 경우 통상 미소진폭파 이론을 사용하지만 충분히 설명할 수 없는 경우에는 유한

진폭파 이론으로 설명한다

2-6-1 수심이 얕아짐에 따른 변형 (천수변형)

파가 얕은 수역에 들어오면 차츰 변형을 받아 파고 파장 파속이 변한다 이를 천수변형

(wave shoaling)이라 한다

우선 해저의 경사를 무시하고 각각의 위치에 있어서는 일정한 수심에 대한 미소진폭파 이

론을 적용하는 것으로 가정한다 심해파의 조건은 첨자(添字) 0에 의해서 나타내는 것으로 한

다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)

이다 여기서 k=2πL

심해파에서 tanhkh≒1 장파에서 tanhkh≒kh 이다 따라서

LL 0

=cc 0

= tanh2πhL (2-6-2)

또 음으로 파고의 변화를 구한다 단위 폭당 단위 시간에 수송되는 유속(流束) F x는

F x= Ec G = Ecn (2-6-4)

으로 주어진다 이 에너지 유속은 각 단면에서 보존되는 것으로 고려하며 하나는 심해파의 조

건 다른 하나는 수심 h의 천해파 조건을 취하여 생각하면 F x= F x0의 관계에서

18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )

를 얻는다 여기서 Ks는 천수계수라고 한다

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파고 3m 주기 12 sec의 너울성 심해파가 직선 해안에 직각으로 입사해 온다 수심

10m와 5m에서의 파형 경사를 계산하라

H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여

교재 P45의 그림 2-6-1에서 Ks값을 읽으면 104

H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272

② h = 5m의 경우

hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄

현실적으로 유체는 점성이 있기 때문에 저면에 마찰응력(friction stress)가 작용한

다 수심이 얕아지면 저면 마찰응력에 의해 파의 에너지 일부는 열로 변환되며 따라서

파는 감쇄한다

해저면에 있어서의 마찰응력 τb를 정상류의 경우에 준해서

τ b = fρu 2b (2-6-6)

여기서 ρ는 해수밀도 u b는 해저면에서의 물입자 속도 f는 마찰계수이다

「그림 1」에서와 같이 일정 수심 h의 수역을 단면 ①에서 단면②까지의 거리 Δx

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[그림 10]

을 전파한다고 하면 에너지 방정식은 다음과 같다

( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)

sinhkhcos (kx-σt) =

πHT

1sinhkh

cos (kx - σt)

단위시간 저면에 있어서 단위 면적당의 에너지 일산량 D f는

D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T

0( cos (kx-σt) 3 )dt

= π 2fρ6

H3

T 3sinh 3kh ( sin 3kx-cos 3kx)+9( sinkx+ coskx) rArrkx= 0

=4π 2fρ3

H3

T 3sinh 3kh

D f를 다시 쓰면

D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )

에너지 보존법칙d( Ecn )

dx=-D f 이므로

-43π 2 ρfH3

T 3( sinh kh) 3=ρgH8

gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3

T 4( sinh kh) 3[ tanhkh(1 +

2khsinh 2kh )]

- 1

dx

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=====================================================================================


-dHH2 =

K2s64π

3ρ

g2T 43( sinh kh) 3dx

1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)

여기서 마찰계수 f의 값으로 001 ～ 002가 적당하다

또 해저가 파장의 03배 이상인 두께의 침투층에 의해서 덮여 있는 경우에는 침투

수에 의해서 에너지가 손실된다 단위면적 단위시간당의 에너지 일산량을 D p라 하면

D p =πkρgH 2

4L1

cosh 2kh (2-6-9)

여기서 k는 투수계수이다 위의 식을 에너지 방정식에 대입하여 풀면 침투효과에 의

한 파고의 감쇠율이 구해진다

3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

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[그림 11]

2-6-3 파의 굴절

파속은 수심에 의해서 변하기 때문에

그림과 같이 수심이 h 1에서 h 2로 변

화하는 경계선에 파가 경사로 입사할

때에는 파의 굴절이 일어나는데 빛의

굴절과 마찬가지로 스넬의 법칙

(Snells Law)이 성립하며 입사각 α 1

과 굴절각 α 2의 관계는 각각의 수심

에서의 파속 c 1 c 2를 사용해서 나타낸

다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)

파향선의 변화 b 1 b 2는

b 1

cosα 1

=b 2

cosα 2

(2-6-12)

파봉선 C B는 Δt시간후에 AD에 도달 한다 파의 전파속도는

Δt=ACC 1

=BDC 2

AC=ABsin α 1 BD=ABsin α 2

ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1

굴절에 의한 파고 변화는 2개의 파향선의 사이에서 수송되는 파의 에너지는 일정하기

때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0

여기서 b 0 b는 파향선 간격 PP 0는 단위폭당 전달되는 에너지의 1주기 평균

b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8

대입하여 정리하면

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b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14

주기가 10 sec인 심해파가 파고가 5m 입사각이 55일 때 수심 7m에서의

파고와 파향을 구하여라

Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec

h = 7m에서의 상대수심 hL 0

=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c

sinθ에서 θ = sin -1(051 times sin 55) = 24 41 37

K r =cosα 0

cosα=

cos 55cos 244137

= 0795

h = 7m에서의 파고는

H = KsK rH0 = 104times 0794times 5 = 413m

【예제 5】

일정한 흐름이 있고 파가 경사로 입사하면 흐름에 의해 파속은 변화하고 파의 굴절

이 일어난다

흐름이 없는 수심 h s의 영역에 파고 Hs 파장 L s 파수 ks=2πL s

파속 c s의 파가 존재

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한다 하고 이 파가 x방향으로 유속 U를 갖는 일정한 흐름의 영역에 θs의 각도를 가

지고 입사하는 것으로 가정하면

c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs

= ( sin 2θsin 2θ s

)-

12

( nn s

)-

12

(2-6-17)

군속도와 파속의 비 n는

n =12 (1 +

2khsinh 2kh ) (2-6-18)

파가 흐름의 영역에 진입할 수 없게되는 한계의 유속 U c로서

Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절

방파제와 같은 장애물이 파가 진행하는 도중에 존재하는 경우에 파는 그 배후로

돌아와서 들어간다 이와 같은 현상을 파의 회절이라고 한다

「그림 4」과 같이 수면 위에 x y축을 취하고 연직 상향으로 z축을 취한다 수축

하지 않는 유체의 소용돌이 없는 운동에 대해서는 속도포텐셜 φ가 존재하고 연속방

정식은

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[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)

으로 된다 여기서 저면의 경계조건 ( partφpartz )

z =- h

= 0를 고려하여

φ(xyz) = AF(xy)coshk(h+z)e iσt (2-6-21)

A 진폭에 비례하는 정수 F xy의 복소수의 함수 i = -1

로 놓고 위식을 식(2-6-20)에 대입하여 정리하면

part 2

partx2(AF(xy)coshk(h+z)e iσt) +

part 2

party2(AF(xy)coshk(h+z)e iσt) +

part 2

partz 2(AF(xy)coshk(h+z)e iσt) = 0

Helmholtz의 방정식이다

part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다

φ에 대한 수면 변동량을 ζ(xyt)로 하면

ζ= -1g ( partφ

partt )z = 0

= -Aiσg

F(xy)coshkhe iσt (2-6-23)

가 된다

여기서 무한히 긴 파봉을 가지고 y축에 따라서 진행하는 입사파에 대하여

F(xy) = e - i ky (2-6-24)

로 놓으면 입사파의 수면 파형 ζ 1은

ζ 1 = -Aiσg

cosh khe i ( σt- ky) (2-6-25)

로 된다 이것에서

ζζ 1

= e ikyF(xy) (2-6-26)

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을 얻는다 그러므로 입사파의 파고를 H I 회절파의 파고를 H로 하면

HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)

이 되며 Kd를 회절 계수라 한다 따라서 Kd에 대해서 정리하면

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

r 개구부 중심에서 거리 k 파수

개구폭 10m의 방파제에 파고 20m 주기 6 sec의 파가 진입한다 개구폭

의 중점에서 50m 떨어진 항내 지점에서 입사파고는 어느 정도인가

Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사

구조물이 존재하거나 혹은 해저 지형이 급변하는 곳에서는 파 에너지의 일부는 반사

되고 일부는 전달파 에너지가 되고 나머지 에너지는 쇄파 점성 마찰 등에 의하여

소비된다

입사파 및 반사파 혹은 전달파의 파고를 각각 H I 및 HR 혹은 HT로 하면 이들 파고의

비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

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구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다

Miche는 미소진폭파 이론에 의해 사면으로부터 완전 반사되는 심해파의 한계 파형경

사를 다음과 같이 나타내고 있다

( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)

여기서 β는 수평면과 이루는 각도(radian)이다 θ는 수평면과 이루는 각(degree)이

다 파형경사가 우변의 값보다도 클 때에는 파가 사면위에서 부서져 에너지의 일부는

소모되어 부분 반사가 된다 이때에 반사율 KR은

KR =HR

H I

=γ 1 γ 2 (2-6-30)

there4 γ 1 = 사면의 조도 침투성

there4 γ 2 = 사면경사

γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it

Battijes는 사면으로부터의 반사에 대한 실험 결과에서 다음의 관계를 얻었다

KR ≒01ξ 2 (1보다 작은 경우)

= 1 (기타) (2-6-31)

여기서 ξ=tanβHL

이고 tanβ 사면경사 H 및 L 사면 끝에서의 파고 및 파장이

다

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해저경사 125

T = 8sec H0 = 35m 파가 입사할 때 모래해빈의 반사율

을 추정하라

Sol) 입사파의 심해파 파형경사 H0

L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906

한계파형경사로서

(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it

이므로 γ 2 =812times10- 5

35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파

파가 차차 얕은 수역으로 진행해 오면 파장은 짧아지고 또 파고는 증대한다 따라

서 그 파형경사가 커져 어떤 수심에서 파는 부서진다

붕괴파쇄파 파형경사가 비교적 큰 파랑이 완만한 경사의 해안에서 쇄파하는 형태로

파봉에 거품이 일기 시작하여 이것이 점차로 전면에 퍼지는 형태를 가진 쇄파

권파쇄파 수심이 비교적 낮고 해저 경사가 급한 곳에서 발생되는 쇄파 형태로 파랑

의 전면이 점점 가파러져서 결국 둥근 원호 형상으로 변하여 앞으로 구부러지면서 부서

지는 파

쇄기파쇄파 파형경사가 작은 파랑이 경사가 급한 해안에서 부서 질 때에 보이는 쇄

파의 형태로서 파봉의 전면부는 거의 연직을 이루고 후면부는 수평에 가까우며 쇄파로

인한 기포 발생이 매우 미약한 파

심해파는 파형경사 및 해저경사에 의해서 쇄파의 형태를 취한다

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쇄기파 쇄파 권파 쇄파 붕괴파 쇄파

ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0

ε b gt 20 20 gt ε b gt 04 04 gt ε b

해저경사 완만 비교적 완만 급한 경사

여기서 ε 0 =tanβH 0

L 0

ε b =tanβHb

L 0

이다

쇄파지표

Hb = 0142 L b (Michell)

Hb = 0142 L b tanh ( 2πh b

L b) (Hamada)

H b = 0827 h (Yamada)

Hb = AL 01 - exp -15πh b

L 0

(1 + 15times( tanβ)43 ) (Goda)

여기서 A = 012～ 018의 값을 가진다

T = 4sec h = 2m β =130

일 때 쇄파지표를 구하고 어떤 형

태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL

에서 L = 25 tanh2π2L

there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m

Hb = 018times251-exp -15times2π25 (1+15times( tan 1

30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029

there4붕괴파 쇄파

【예제 8】

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=====================================================================================


----------------------------------------------------------------------------------

Ex1)평형 등심선을 가지는 해저경사 130의 해안에 직각방향으로 심해파 주기 8초 파고 2m의 파가 입사

하고 있다

a 심해파의 파형경사는

=gt

b 수심 4m지점에서 파장 파수 파속 를 구하라

=gt

따라서

c 수심 4m지점에서 파고 는

=gt

d 심해에서 파향이 일 때에 수심 4m 지점에서 파향 및 굴절계수 은

=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos

Ex 2) 주기5초 파고 1m의 심해파에서 파장 파속 군속도 에너지 일율(에너지수송량)의 1주기 평균 는

=gt 파장 39m 파속 78ms 심해에서는 rarrinfin익 때문에

평균 일률

----------------------------------------------------------------

2-6-7 파에 의한 평균해면의 저하와 상승

파가 해안으로 향하여 진행해 오면 파에 의한 질량 수송은 해안에서 차단되며 따라

서 해안에서 먼 바다로 향하는 흐름이 생긴다 이렇게 해서 파와 흐름의 상호간섭이 일

어난다 이것이 radiation stress이다

평균 해면은 쇄파점으로 향함에 따라 차차 저하(wave setdown)하는 것으로 밝혀졌다

지금 그 저하량을 η b로 하면

=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)

쇄파대 내에서는 평균해면은 상승 상승량 η는

η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)

여기서 고립파 이론에 기초하면 γ=073이지만 실용적으로 γ = 08sim11 정도의 값

을 취한다

2-6-8 쇄파후의 파고변화

쇄파후의 파고 변화는 주로 해저 경사에 좌우된다 근년에 천해역 및 쇄파후의 파고

변화를 계산에 의해 구하려고 하는 시도가 행해지고 있다

2-7 해양파의 성질

불규칙한 파의 취급 통계적인 대표값 에너지 스펙트럼에 의한 값

2-7-1 파랑의 통계적 성질

Zero-up(down) cross 법 수면 파형이 상승(하강)하면서 평균수면을 지나는 시각으로부터

다음에 똑같이 지나는 시각까지의 최고수위와 최저수위와의 차를 파고로 정의하고 그 시간의

간격을 주기로 정의하는 것이다

(1) 최대파 (H max T max ) 파군중의 최대파 최대주기를 말한다

(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)

이중 가장 잘 사용되는 것은 Sverdrup-Munk에 의해 도입된 유의파

목시 관측에 의해서 보고 되어 있는 파의 파고 주기와 거의 같은 것으로 채용되었다

Ex) 최대파 110최대파 유의파(13최대파) 평균파를 구하라

답)m ax m ax sec sec sec sec

2-7-2 파고분포

Longuet-Higgins는 하나의 파군 중에서 파고가 어떤 빈도로 출현하는가 하는 문제를 다루고

파군의 주파수가 좁은 밴드폭에 들어가는 경우 출현빈도는 Rayleigh 분포에 의해 나타낸다

파고의 확률 밀도 함수를 p(H)라 하면

p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH

여기서 H는 평균파고

임의의 값 H보다도 큰 파고가 출현하는 확율(초과 확률) p(H)는

p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================


따라서 계산에 의해 다음과 같은 관계가 얻어진다

H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13

= 0706 lnN 최대치

= 0706 lnN+r

2 lnN 평균치 여기서

400파의 파랑 관측 결과에서 H rms= 45m을 얻었다

(1) 이 파군의 유의파고

(2) H = 2H rms이상인 파의 수

(3) 최대파고의 최대값과 평균값을 계산하라

H rms=1

0886H

(1) H 13 = 16 H= 160886H rms= 16times0886times45 = 6339

(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파

(3) 최대치 H max = 0706 ln 400times64 = 1728times64 = 1095m

평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결

해양의 파는 완전히 불규칙한 현상으로서 취급해 왔다 그러나 주의해서 파를 관찰하면 높

은 파가 여러 개의 파와 연결되어 나타나는 것을 알 수 있다 이와 같은 현상을 파의 연결

(Wave Grouping)이라 하는데 최근에 와서 주목을 받게 되었다

고파의 연결 길이를 나타내는 방법으로는 그림과 같은 파고의 변화가 관측됐다고 했을 때

미리 설정한 파고치 H c를 넘는 파가 연속해서 출현하는 파수 j 1 혹은 하나의 파의 연결 파고

가 H c를 넘고서부터 다음파의 연결이 H c를 넘기까지의 파수 j 2로 나타내는 일도 있다 후자

의 j 2는 고파의 반복길이를 나타내는 것으로 생각할 수 있다

=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포

파군 중의 주기의 편차는 파고의 경우보다는 작고 평균 주기 T의 05-20배의 범위에 있다

고 보면 된다 현지 데이터를 해석한 결과 의하면 다음의 관계가 성립한다

T max ≒T 110

≒T 13

≒12T

2-7-5 해양파의 에너지 스펙트럼

불규칙한 파형은 주기가 다른 무수한 여현파의 합으로서 나타낼 수 있다

각 여현파의 에너지가 각 주파수에 대해서 어떻게 분포하고 있는가를 나타낸 것을 주파수 스

펙트럼 각 파수에 대해서 나타낸 것을 파수 스펙트럼이라 하며 이들을 총칭해서 에너지 스펙

트럼이라 한다

해양파의 수면파형은 시간적으로도 공간적으로도 매우 불규칙한 변화를 나타낸다 그 수면변

동 η(xyt)는

η(xyt) = suminfin

n= 0a ncos (xkncosθ n + yknsinθ n - σ nt+ ε n)

여기서 kn 성분파의 파수 θn 성분파의 전파각도

σ n 성분파의 각 주파수 f n 성분파의 주파수

ε n 성분파의 위상 a n 성분파의 진폭

임의의 구간 (k kδkθ θ+δθ)에 있어서의 파의 방향 스펙트럼 E(k θ)와의 사이에는 다음의

관계가 성립한다

E(kθ)δkδθ= sumk+δk

ksumθ+δθ

θ

12a 2

n

수면 변위의 자승 평균을 취하면

η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ

를 파의 진행 방향에 대해서 적분하면

E(k) = ⌠⌡

2π

0E(k θ)δθ

이것이 파수 스펙트럼이다

=====================================================================================


=====================================================================================


심해파에 대해서 생각하면

σ 2n = 4π 2f2n = gkn

혹은

f2n =g4π

kn

이고 파수 k와 주파수 f는 일의적으로 대응하기 때문에

η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df

에 의해 주파수 스펙트럼 E(f)를 정의할 수가 있다

E(f)에 있어서는 파랑 관측 기록의 해석이나 이론적 고찰에 기초해서 많은 식이 제안되어져

있다

(a) Pierson-Moskowitz의 스펙트럼

외양에서 일정한 풍속의 바람이 충분히 긴거리를 불어서 파가 한껏 발달한 상태의 스펙트럼

E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)

(b) Bretschneider-Mistuyatsu 스펙트럼

대표파의 파고와 주기가 주어졌을 때

E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)

(c) Mitsuyatsu 스펙트럼

취송거리와 유한한 경우

E(f) = 858times10- 4( gFu 2

)- 0312

g2f- 5times exp -125 ( gFu 2

)- 132

( u 8f

g )-4

(m 2sdots)

여기서 는 해면에서의 바람의 마찰속도 해면 10m에서의 풍속과 관계는 이다

E(f) = 115times10- 4( gFU2

10)- 0312

g2f- 5times exp -996 ( gFU2

10)- 132

( U10f

g )- 4

(m 2sdots)

(d) JONSWAP의 스펙트럼

북해에서의 관측결과에 기초

E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p

위의 식들은 다음의 관계가 성립한다

=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5

c = 305times10 4cm 2sdotsec - 5

H 13 = 7065times10- 3U5275msdotsec

Z n에 있어서의 풍속

U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )

γ 2 = 05 U 10times10- 3 ( 3 m sec lt U 10 lt 15msec )

γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================


U10 = 30ms F = 50km인 경우의 Mitsuyatsu스펙트럼을 이용하여 H 110을 구하여라

H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면

E(f)df = k1f-mexp -k2 f

-n

= 858times10- 4( gFu 2)- 0312

g2f- 5times exp [-125 ( gFu 2

)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4

= 1716times10- 4times( gFu )1008

timesu 4

g2

Γ(m-1n

)는 Gamma의 함수 따라서 m = 5 n = 4 Γ(1) = 1

여기서 times times times

E = 1716times10- 4times( gFu

)1008

timesu 4

g2 =2891times10- 7times( gFU2

10)1008

timesU4

10

g2

【 예제 10】

H 110 = 360 E= 360 times2891times10- 7times( gFU2

10)1008

timesU4

10

g2 12

= 360times2891times10- 7times( 980621times50000

30 2 )1008

times30 4

980621 2 12

=4252m

파의 방향 스펙트럼에 관해서도 여러 가지 검토가 행하여지고 있지만 관측이 곤란하기 때문

에 아직 만족할 만한 결론에 도달하지 않았다

E(fθ) = E(f)G(θf)

여기서 G(θf)는 방향별 에너지 분포상태를 나타낸다

또 θ를 파의 탁월 방향에서 측정한다고 하면

E(f) = ⌠⌡

π

-πE(fθ)dθ = E(f)⌠⌡

π

-πg(θf)dθ

=====================================================================================


=====================================================================================


로 나타낼 수 있음을 고려하여 위의 식을 적분하면

⌠⌡

π

-πG(θ f)dθ = 1

로 된다

윗 식을 만족하는 성분파의 방향분포는 파의 진행 방향에 대해서 대팅인 것을 고려하여

G(fθ) prop cos nθ prop cos 2s( θ2

)

의 함수형을 고려할 수 있다 여기서 n s는 주파수 f의 함수

Mitsuyatsu의 방향 분포 함수

크로바 브이식 파랑계에 의해 얻어진 관측 결과에서 Mitsuyatsu는 다음의 방향 분포 함수를

제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)

S는 방향에 대한 에너지의 집중도를 나타내는 파라메타로

S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )

S max 방향 집중도 파라메타

f p 주파수 스펙트럼의 피크 주파수

f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25

S max 의 값은 다음 값을 이용하는 것을 제안하고 있다

(1) 풍파 S max = 10 (G 0 = 09033)

(2) 감쇄거리가 짧은 너울 (파형 경사가 비교적 큼) S max = 25 (G 0 = 14175)

(3) 감쇄거리가 긴 너울 (파형 경사가 작음) S max = 75 (G 0 = 24451)

2-8 풍파의 발생발달과 파랑의 추산

2-8-1 바람에 의한 파의 발생과 발달

수면위에 바람이 불면 파가 일어나는데 그 발생기구를 밝히려는 시도는 오래전부터 있어 왔

다 우선 Helmholtz는 공기와 물이라고 하는 밀도 전혀 다른 성층의 해면에 불안정이 생기기

위한 조건으로서 다음의 식을 구하고 있다

(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================


여기서 ρ는 상층에서의 밀도 (즉 공기의 밀도)

또 Kelbin은 파의 발생 단계에서는 표면장력 K가 중요한 역할을 한다고 생각하며 이 효과를

넣어서 다음 식을 얻었다

(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ

해수가 정지하고 있는 경우 U = 640cms

파의 발생 한계 풍속에 대해서의 관측 결과를 보면 Roll 40 ms

Keulegan 1200 cms

Kelvin Helmholtz에 의한 완전 유체로서의 취급에는 문제가 제기되어 점성에 의해서 경계

면에 생기는 경계층을 고려해서 논해야 한다는 관점에서 Wuest나 Lock는 이 문제를 다루었다

한편 바람의 압력 변동이 파의 발생에 대해서 중요한 역할을 한다고 하는 생각이 Eckart에 의

해 제안되었다 이상과 같은 경과를 거쳐서 1957년에 이르러 파의 발생발달에 대한 연구는

비약적으로 진보했다 그것은 Philips와 Miles가 때를 같이 하여 각가의 이론을 발표한 것에

의한다

(a) Philips이론

바람에 의한 압력변동은 바람과 함께 다양한 속도로 이동하고 해면에 발생한 미소한 파는

그 주파수에 대응한 전파속도로 진행한다 이상은 초기파의 발생기구를 설명한 것이다

(b) Miles이론

해면상에 파가 나타나면 그 파의 존재에 의해 해면 부근의 풍속의 분포는 변화를 받게 된

다 압력 성분은 파를 바람 아래쪽으로 밀어붙여 바람의 에너지를 파에 공급하며 이에 의해

파고는 증대한다 이상은 초기파의 발생 단계로부터 충분히 파고가 성장해서 파의 비선형성이

무시할 수 없게 되고 성분파 사이의 상호 간섭이 탁월해지기까지의 단계를 포괄적으로 설명하

였다

(c) 충분히 발달한 파

바람에 의한 에너지 공급과 쇄파에 의한 에너지 일산이 평형을 이루어 정상상태에 달한다

2-8-2 취송시간과 취송거리

취송시간 (Duration) 바람이 계속 불고 있는 시간

취송거리 (Fetch) 파가 바람을 받아 발달하면서 진행하는 거리

파가 충분히 발달하기 위해서는 어느 시간이상 바람이 계속 불 필요가 있으며 이 시간을 최

소 취송시간 t min이라 한다 파가 그 시간에 대응하는 한도까지 발달하는데 요하는 수역의 거

리를 최소 취송거리 F min라 한다

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파랑의 추산 방식 유의파법(SMB법 가장 널리 사용) 에너지 스펙트럼(PNJ법)

2-8-3 SMB법

SMB법은 풍역이 이동하지 않는 경우에 적용된다

여기서는 현재 신뢰도가 비교적 높은 1965년 Wilson이 관측치를 정리하여 만든 식에 대해서

만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5

(1) 풍속일정 취송거리 일정한 경우

U와 F 및 U와 t의 각각의 조합에 대하여 교재 P85 그림 2-8-7에서 H및 T을 읽어내어

어느 것이든 작은 조의 값을 취한다

F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec

따라서 U와 F의 조합 H 13 = 25m T 13 = 60 sec이 된다

【예제 11】

(2) 풍속이 변하는 경우

( ( t2 - t1) + t)와 U 2 및 F 2와 U 2의 각각의 조합에 대한 파를 구하여 작은 쪽의 값을 가지

고 그 지점의 파의 특성으로 한다

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취송거리 100km의 풍역대를 U= 10ms의 바람이 10시간 불고 나서 그 후 U= 20ms의

바람이 5시간 불었다 파고와 주기는

F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec

등에너지 U = 20ms t = 18hr

F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec

따라서 파고와 주기는 H 13 = 35m T 13 = 70 sec

【예제 12】

2-8-5 파랑의 추산법

(a) 바람의 장

U 10을 알기 위해서는 먼저 경도풍의 Ugr을 구해야 한다

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경도풍 기압경도 코리올리의 힘 원심력 등이 평행되어 부는 바람

Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr

기압경도 r 등압선의 곡률반경 w 지구의 자전각속도 φ 위도

태풍의 중심으로부터 r의 거리의 지점에서의 기압을 구하는 근사식으로서 다음의 것이 알려

져 있다

(1) Myer의 식

ρ = ρinfin - δp1 - exp ( r 0

r ) 여기서 ρinfin 태풍의 중심으로부터 충분히 떨어진 지점에서의 기압

δp 태풍의 중심에서의 기압 저하량

r 0 풍속이 거의 최대가 되는 지점의 거리에 상당하는 정수

(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장

바람의 장이 구해지면 파의파법 또는 에너지 스펙트럼법에 의해 파의 장을 산출한다 파의

방향 스펙트럼의 시간적 공간적 변화를 구하기 위해서는 다음에 나타내는 에너지 스펙트럼의

평형방정식

partEpartt

(f θ x t) = - cG ( f)sdotE(f θ x t) + F(f θ x t)

을 수치해석 한다

2-8-6 천해파의 추정

(a) Bretschneider의 법

천해에 있어서 파의 변형을 바람으로부터 에너지 공급과 해저 마찰에 의한 에너지 손실의 둘

로 나누어 생각했다

천해파의 주기를 구하기 위해서는 충분히 발달한 파에 대해서 관측한 결과에 기초한 관계로

T 13 = 386 H 13 (단위 m s)

(b) 풍역이 변동하는 경우의 천해파의 추정법

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Bretschneider의 방법은 풍역이 급격하게 변화하는 경우에는 적용하기 어려우므로 Ijima 등

은 Wilson의 심해 풍파에 대한 계산법을 천해 풍파에 확장했다

gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38

평균수심 6m 대안거리 20km의 호수에 풍속 20ms의 바람이 장시간 불었을 때 발생

하는 천해파의 파고와 주기를 추산하라

h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】

2-8-7 너울의 추정

파가 풍역을 떠나 너울로서 진행할 때에는 바람으로부터의 에너지의 공급이 없기 때문에

파고는 점차 감소한다 이 파고가 감소하는 것은 방향 분산과 속도 분산에 의해서 에너지 밀

도가 저하하기 때문이다 또 너울로써 전파해 가는 사이에 역풍을 받기도 하고 내부 마찰이

나 파의 상호 간섭 등의 영향을 받기도 해서 주로 단주기파 성분의 에너지가 손실된다 그 때

문에 너울의 파형은 풍파의 파형과는 달리 둥근 모양을 하게 된다

너울의 추정에 대해서는 여러 연구가 있지만 여기서는 Bretschneider가 제안한 관계를 알

아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

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여기서 k1 k2는 각각 k1 = 04 k2 = 20인 무차원 정수이고 F min은 풍파를 발생시킨 풍

역의 길이 D는 너울의 감쇠거리 (H 13 )F (T 13 )F는 풍역의 종단에 있어서의 유의파의 파

고 및 주기 (H 13 )D (T 13 )D는 감쇠거리 D를 진행한 뒤의 너울의 유의파 파고 및 주기이

다 또 너울이 진행거리 D을 진행하는데 요하는 시간 tD는

tD =4πD

g(T 13 )D

해안에서 2000km떨어진 해상에서 풍속 25msec의 바람에 의해 유의파고 55m의 파

가 발생했다 이 파가 너울로 되어 해안에 도달할 때의 파고와 주기 및 그 도달시간을

추정하라

U = 25msec H 13 = 55m을 가지고 교재 P85의 그림 2-8-7에서 구하면

F min = 140km T 13 = 85 sec가 되고

파고 (H 13 )D = 55times( 04times14004times140 + 2000 )

12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec

너울의 도달시간 tD =4πtimes2times10 6

98times115= 223times10 5sec ≒ 619hr

【예제 14】

2-9 불규칙한 파의 변형

불규칙한 파의 굴절과 회절에 관해서 위와 같은 취급을 한 결과를 소개하여 규칙적으로 파의

굴절과 회절이 어떻게 다른가를 보기로 한다

2-9-1 불규칙파의 굴절

Goda 해저의 등심선이 평행한 직선 모양인 해안을 고려하여 계산한다

2-9-2 불규칙파의 회절

Goda가 불규칙파의 회절도를 구하고 있다 즉 불규칙파와 규칙파의 비교

2-10 직릭벽에 작용하는 파압

2-10-1 파압현상

파고가 적으면 중복파압

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파고가 크면 중복파압 (쌍봉형)

쇄파가 한계를 넘으면 쇄파파압

쇄파가 한계를 넘어서 진행 충격쇄파압

2-10-2 중복파압

(a) 미소 진폭 중복파

파고가 작은 경우에는 미소 진폭파 이론을 적용할 수 있다

중복파에 대한 파형 η는

η 1 = a cosk(x-ct) η 2 = a cosk(x+ct)

η = η 1 + η 2 = 2a coskx coskct

또 중복파에 대한 속도포텐셜 φ는

φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)

φ 2 = -accoshk(h+y)

sinhkhsink(x+ct) = ac

coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx

여기서 압력방정식

partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0

에서 두 번째 항을 무시하면

pρ

= -gy-partφpartt

에 윗 식을 대입하면

pρg

= -y + 2ac 2kcoshk(h+y)

sinhkhcos kctcoskx

- ρgy+ Hsinhkhcoshkh

coshk(h+y)sinhkh

cos σtcos kx

로 된다 또한 coskx= 1이므로 다음과 같이 쓸 수 있다

p = - ρgy+ ρgHsinhkhcoshkh

coshk(h+y)sinhkh

cos σt

여기서 좀더 근사도를 높이기 위하여

pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)

을 사용하여 계산을 전개하여 본다 y= η 0에서 p = p0 = 0 η 0 = 2a cos σt = Hcosσt를 고려

하면

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f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh

cosh 2k(h+y) sin 2kx+ sinh 2k(h+y) cos 2kx

= H2gksin 2σtsinh 2kh

( cosh 2kch + y)sin 2kx= sinh 2k(h+y) cos 2kx

여기서

( cosh 2k(h+y) - sinh 2k(h+y))( sin 2kx+ cos 2kx)

= cosh 2k(h+y) cos 2kx- sinh 2k(h+y) sin 2kx= 0

partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0

= - ρgHcoshk(h+η 0)

coshkhcosσt

여기서 Hcosσt = η 0

( partφpartt )

y= η 0

= η 0(1 +cosh (k+y)coshkh

-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)

coshkh ) 를 얻는다 파봉이 왔을 때에는 cos σt = 1 sinσt = 0이므로

pρg

= -y + H1 +coshk(h+y)

coshkh-

coshk(h+H)coshkh

이 된다 또한 해저 y = -h에서는

pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)

coshkh≒1 + kHtanhkh +

12

k2H2를 이용하면 다음과 같다

pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식

평균위치 (x0 y0)에 있는 물입자의 입사파에 의한 궤도의 장경 γ 단경 γ는 다음과 같다

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γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2

H는 입사파의 파고이다

중복파의 물입자의 운동을 나타내면

x= x0 - 2γ sinkx0cosσt

y = y0 + ka 2sinh2k(h+y0)

sinh 2khcos 2σt + 2γ coskx0cosσt

위의 식은 Lagrange의 연속 방정식을 근사적으로 만족하고 있으며 소용돌이 있는 운동이다

여기서 Lagrange 운동 방정식의 적분을 구하면

pρ

- gy=2σ 2

k γ coskx0cosσt+k2(γ 2 + γ 2)( cos 2σt+ cos 2σt) -

ka 2

2 sinh 2khcos 2kxcos 2σt + f(t)

여기서 위식에 y식을 대입하고 γ에 대해서 kγ 2항을 무시하면

pρg

= -y0 + 2( σ2

gkγ- γ)coskx0cosσt+ f( t)

을 얻는다 여기서 σ 2 = gktanhkh 또 y 0 = 0에서 p = 0의 조건을 사용하면 f( t) = 0을 얻

을 수 있다 따라서

pρg

= -y0 + H coshk(h+y0)

coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt

로 된다 직립벽에서의 압력 분포는 x0 = 0로 놓고 구할 수 있다

중분면과 정수면과의 차 δ 0는 다음과 같이 구할 수 있다 위의 y식에서 x0 = 0으로 놓으면


sinh 2khcos 2σt+ 2γ cos σt

을 얻는다 정수시의 위치는 y = y0로 주어지고 파봉 및 파곡의 위치는 각각

ycr est = y0 + ka 2sinh2k(h+y0)

sinh 2kh+ 2γ

y tr ough = y0 + ka 2sinh2k(h+y0)

sinh 2kh- 2γ

로 된다 이것으로부터 중분면은


sinh 2kh

로 되고 정수면과 중분면과의 차 δ는

δ = ka 2 sinh2k(h+y0)

sinh 2kh

이다 해면에서는 y0 = 0이므로

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δ 0 = ka 2 sinh2kh

sinh 2kh= 2ka 2cothkh

=πH 2

Lcoth

2πhL

로 주어진다

그러면 파봉이 도달했을 때의 압력 분포는

pρg


coshkh-

sinhk(h+y0)

sinhkh

이다 따라서 해면 ( y0 = 0)에서는 pρg

= 0 해저면 (y0 = -h)에서는

pρg

= h + H1

coshkh 이다

파곡이 왔을 때의 압력 분포는

pρg

= -y0 - H coshk(h+y0)

coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다

위의 결과를 도시하면 그림과 같이 된다 방파제와 같은 구조물을 생각하면 반대쪽에는 정수

압이 작용하고 있으므로 실적으로 작용하는 파압의 분포는 그림과 같다

계산을 간단히 하고 또 안전 쪽의 값을 얻기 위하여 파압의 분포를 직선으로 나타낸 것이

다음의 간략식이며 실제 설계에는 이 식을 사용한다

(a) 벽면에 파봉이 있을 때

p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)

(b) 벽면에 파곡이 있을 때

p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

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여기서 h 0 =πH2

Lcoth

2πhL

이다

Example

수심 10m의 지점에 설치된 천단폭 5m 천단높이가 충분히 높은 직립방파제에 파고 25m 주

기 6초의 파가 작용한다 파봉 작용시에 있어서 파압의 합력과 양압력의 합력을 구하여라

Solution)

By try and error method

times

times

cos

cosh timestimes

times

there4파압의 합력

times

there4양압력의 합력

times

2-10-3 쇄파의 파압

쇄파는 매우 짧은 시간 ( 110s 이하)안에 충격적인 압력을 작용시켜 그 압력의 최대치가 일

반적으로 현저하게 변동한다는 것이 많은 실험과 현지 관측 결과에 의해서 확인되고 있다

Mitsuyasu에 의하면 충격 파압이 현저해지는 제각 수심 h 0는

hM

H0

= CM( H0

L 0)- 14

여기서 H 0는 심해파의 파고 L 0는 심해파의 파장 CM는 해저 경사 tanθ에 의해서 변하

는 계수

쇄파의 파압은 복잡한 현상이기 때문에 이것을 이론적으로 구하는 것은 불가능하다 따라서

아래의 실험식을 제시되었다

(a) Hiroi 공식

직립부의 하단으로부터 정수면위 125H의 높이까지의 파압이 일정하게 작용한다고 생각한다

p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]

이 공식에 의한 쇄파압은 실험실

이나 현지에서 관측된 국소적인 파압

강도와 반드시 일치하지는 않지만

작용 면적에 걸쳐서의 파압의 평균치

와는 잘 합치되는 것으로 인정되어

있다 그러나 많은 경우에 파압 강

도는 수면 부근에서는 강하고 수저

부근에서는 약해지고 있다 따라서

쇄파에 의한 수면 부근의 강한 충격

력을 고려한 Minikin 식이 제안되었

다

(b) Minikin 공식

혼성 방파제의 직립벽에 작용하는 쇄파의

충격압은 쇄파의 전단이 얇은 공기층을 휘

감아들이면서 직립벽에 충돌하기 때문에 일

어난다고 생각하여 다음식을 제안하였다

(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2

y는 정수면에서 연직 상향으로 측정한 거

리이며 -H2

simH2

의 값을 취한다

(2) 정압

ps= ρg( H2

- y) (정수면 위)

ps = ρgH2

(정수면 아래)

y는 정수면에서 연직 상향으로 측정한 거리이며 0simH2


2-10-4 쇄파후의 파에 의한 파압

해저 경사가 150

보다 완만할 때에는 쇄파하고 Hb를 제각 수심의 09배로 하여 Hiroi을 적

용하는 방법이 제안되어 있다

=====================================================================================


=====================================================================================


[그림 19]

해저경사가 비교적 가파르고 구조물이 어느 정

도의 전면수심 h을 가지며 게다가 쇄파후의 파가

작용할 때에는 다음 식에 의해서 파압을 구할 수

가 있다

(1) 등압

정수면상에서 pm = 16ρgh

정수면상 12h 및 제체 기부에서 0이 되는 삼각

형 분포로 한다

(2) 정압

정수면상 12h에서 0 제체 기부에서

ps= 22ρgH가 되는 삼각형 분포로 한다

2-10-5 양압력

기초가 투수성이며 압력이 전파된다고 생각되며 구조물의 천단이 높아 월파를 일으키지 않

는 경우에는 구조물의 정수면 아래의 부분에 작용하는 부력에 더하여 파에 의한 양압력을 고

려해야 한다

중복파의 경우 pu = p2

쇄파의 경우 pu = 125ρgH

2-10-6 파압에 관한 Goda식

해양의 파는 불규칙적이며 따라서 파압의 변화는 불연속이라 하는 것은 적당하지 않다 이

와 같은 생각에서 Goda는 새로운 파압의 산정식을 제안했다

파압의 분포는 그림과 같다

여기서 h는 전면 수심 d는 마운드 위의 수심 h는 직립부 저면으로부터 정수면까지의 높

이 h c은 정수면 위의 천단고이다

(a) 설계파의 파고

설계파로서는 최고파를 사용하는 것으로 하고 그 파고는 다음과 같이 정한다

쇄파대 외 H min = 18H13

쇄파대 내 H max = 방파제 의 벽면으로부터 먼 바다쪽 5H 13의 지점에 있어서의

H max

여기서 H 13은 방파제 설치 수심에서의 값

(b) 파압의 작용 높이

=====================================================================================


=====================================================================================


η = 075(1 + cotβ)H max

여기서 β는 방파제의 벽면에 대한 수선과 파의 탁월한 방향이 이루는 각도이다 그러나

파향의 불확정성을 고려하여 파의 탁월한 방향으로부터 plusmn15의 범위에서 위험한 쪽으로 기울

인 방향을 사용한다

(c) 전면의 파압 강도

p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1

cosh ( 2πhL ) min ab는 a또는 b의 어느 쪽이든 작은 값을 취하는 것을 나타내고 h b는 방파제의 벽

면으로부터 5H 13 먼 바다쪽 지점의 수심이다

(d) 양압력

양압력은 직립부 전단에서 pu 후단에서 0이 되는 삼각형 분포가 된다

pu =12(1 + cosβ)α 1α 3ρgH max

=====================================================================================


=====================================================================================


수심 8m의 지점에 설치된 직립방파제에 T 13 = p s의 파가 입사하는 경우 전체 파력을

입사파의 유의파고가 H 13 = 3m일 때 계산하시오 (Sainflou식 사용)

H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704

h 0 = πtimes3 2times1704747

= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================


수심 12m의 지점에 h = 8m d = 65m h c = 40m의 혼성 방파제에 상당 심해파 파고

H0 = 6m 유의파 주기 T 13 = 11 sec의 파가 파향 θ= 15로 입사할 때 전파력을 계산

하라 해저경사는 1100 제체폭 B = 150m

L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL

파압의 작용 높이 η = 075times(1 + cos15)times87 = 128m

LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676

sinh ( 2times0676) )2

= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872

p 1 = 05times (1 + cos15)times(0881 + 0282timescos 15)times103times87 = 1016tfm 2

p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times

times costimestimestimestimes

전 수평력

P =05 times[ (p1+p3)timesh+(p1+p4)h c]

= 05times[ (1016 +886)times80+(1016+699)times40] = 1103tfm

【 예제 16】

=====================================================================================


=====================================================================================


2-11 여러 가지 구조물에 작용하는 파력

2-11-1 사석 사면의 안정

사석 방파제는 사석이나 콘크리트 블록을 쌓아올려서 제체를 축조하고 그 사면 위에서 파를

부서지게 함으로써 에너지를 소모시켜 방파의 기능을 갖는다

사석의 중량은 Iribarren에 의해 최초에 제안되었다

사석 한 개에 작용하는 양압력 F = kρgAH

그 작용 방향에의 사석의 투영 면적 A prop ( Wρ sg

)23

여기서 W는 사석의 중량 H는 사석 전면의 파고 ρ는 해수의 밀도 ρ s는 사석의 밀도

k는 계수

따라서

F = kρgH( Wρ sg

)23

여기서 k는 새로운 계수이다

사석의 마찰 계수를 μ 사면 경사를 α로 하면 사석의 사면위에서 활동에 대한 안정 조건으

로부터

저항력 τ는 사석에 작용하는 정지 마찰력에서 사석 중량의 사면에 따른 성분을 뺀 것으로 한

다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s

)W(μ cosθ- sinθ) ge kρgH( Wρ sg

)23

(1 -ρρ s

)3

W3(μ cosθ- sinθ) 3 ge k 3ρ 3g3H3( Wρ sg

)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3

사석이 나란히 있을 때

=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3 ( Iribarren 식 )

Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )

법면 경사 12의 소파공을 설계 파고 5m에 대해서 피해율 0으로 계획하고 있다 소파블

록의 중량을 계산하라 KD = 8 wr = 23tfm 3

W=23times5 3

8times(23 - 1) 3times2= 82ton

【예제 17】

2-11-2 수중 물체에 작용하는 파력

진행파에 의하여 수중의 물체에 작용하는 수평 방향의 파력은

dF =12ρC Du |u| dS + ρCMu dV

여기서 dF는 연직거리 dy의 부분에 작용하는 수평 파력 ρ는 해수의 밀도 g는 중력 가

속도 CD는 항력 계수 CM은 질량 계수 dS는 물체의 dy부분의 흐름의 방향에 대한 투영

면적 dV는 물체의 dy부분의 체적 u는 파의 운동에 의한 수평 입자 속도 u 는 파의 운동

에 의한 수평 입자 가속도

직경 D의 직립 원주의 경우에는

dF = ( 12ρC DDu |u | + ρCM

πD 2

4u )dy

파력은 시간과 함께 변화하지만 그 최대치 dF max에 주목하는 경우에 그 값은

dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )

dFM ( 2dF D lt dFM )

여기서

dF D =12ρC Du

2max dS = C Dw0DH 2 1

4 sinh 2kh ( 2kh + sin 2kh4 )

dFM= ρCMu max dV = CM

w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================


여기서 u max u max 는 수평 속도 및 수평 가속도의 진폭 파수 k=2πL

각주파수

σ =2πT

w0는 해수의 단위중량

Hy3 = 45m T = 11 sec h = 10m D = 08m 최대파력 C D = 10 CM= 20

L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】

2-12 파의 쳐오름과 월파

2-12-1 파의 쳐오름 높이

Miche는 파형의 처오름 높이를 파형 경사가 매우 작다는 조건에서

RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0

여기서 α α c는 라디안

Takada는 파형 경사의 영향을 고려하여

RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1

4 cosh 2kh ) (Miche)

여기서 H 0는 심해파고 Ks는 천수 계수 H는 사면 선단에서의 파고 k=2πL

이다

사면의 경사각 α 쇄파를 일으키지 않는 가장 완만한 경사각 α c를 고려할 때 α lt α c일 때

의 쳐오름 높이에 대해서 다음과 같은 실험식이 있다

(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23

곧 α gt α c라면 윗 식에 의하여 α lt α c라면 아래 식에 의해 처오름 높이를 구할 수가 있으

며 최대 처오름 높이는 α = α c에서 일어난다

수심 5m의 지점에 경사각 α =13을 계획하고 있다 H0 = 25m T = 9sec 의 파가 입

사할 때 쳐올림 높이를 구하라

L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13

= 0333 there4α c gt α

L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량

제방과 호안의 천단고가 쳐오름 높이보다도 낮으면 해수는 천단을 넘어서 육지쪽으로 유입

한다 이와 같은 월파량은 파의 특성 제방의 형상 설치 위치 해저 지형 바람의 유무 등에

의해서 변화한다

따라서 월파량 Q는

Q = a(R - Hc)

2 ( x0

R- cotθ)

여기서 a는 월파계수로서

경사제일 경우 a = 76( cotθ) 073( H0

L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)

a 쇄파 영역에 있는 경우

① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18

b 중복파 영역에 있는 경우

Q = aA

A = ( 12

(1 + cot 2θ)( cotγ- cotθ)

(R - Hc)2 + 015H(R - Hc))

cotθ gt 1경우 cot γ= 67(HL

)( cotθ) 16

cotθ lt 1경우 cotγ= n +n(n-1)

2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2

해저경사= 130 수심= 3m 조위=2m T = 10 sec H 0 = 55m 허용 월파량= 005m 2ms 일 때의 천단고는

L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005

2times98times55 3= 876times10- 4

그림 2-12-12에서hH0

= 091과 q

2gH30

= 876times10- 4로 구한 값은 h c

H 0

= 105

따라서 천단고는 h c=105 times55 + 20 = 777m

만약 방파제 앞에 소파호안이 있는 경우 h c

H 0

= 06이므로 53m가 요구된다

따라서 방파제 앞에 호안이 있는 경우 천단고를 777m-53m=247m 정도 줄일 수 있다

【예제 20】

=====================================================================================


=====================================================================================


[그림 3]

lt 영역 gt

- h le yle η- infin le xle infin

장주기파

해면 부진동

방재또는 연안개발이나 환경보전에 중요 해일(Tsunami) 지진해일 고조(Storm surge) 폭풍해일

조석

A 해양파의 성질

2-2 파의 기본방정식

완전유체 점성무시 회전무시 속도포텐셜 φ존재

물입자의 속도성분 u( x방향) u =partφpartx

(2-2-1) 물입자의 속도성분 v( y방향) v=

partφparty

연속방정식 비압축성

partupartx

+partvparty

= 0 (2-2-2)

(2-2-1)식을 (2-2-2)식에 대입하여 정리하면

part 2φpartx2 +

part 2φparty2 = 0 ( 2차원 Laplace 방정식 ) (2-2-3)

경계조건(Boundary Condition) 수평인 해저에서 연직 속도 성분은 ldquo0rdquo

(v) y=- h = ( partφparty )

y=-h

= 0 (2-2-4)

압력방정식(Pressure Equation) 비회전

partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ

+ gη = 0 (2-2-5)

=====================================================================================


=====================================================================================




( partφpartt )

y= η

+12 ( partφ

partx )2

+ ( partφparty )

2

y= η

+P0

ρ+ gη= 0 (2-2-6)



= 0가 성립

DDt

=partpartt

+ upartpartx

+ vpartparty

+ wpartpartz



( DFDt )

y= η

= ( partηpartt

+ upartηpartx

- v)y= η

= 0 (2-2-8)

또는

partηpartt


partx )y= η

= ( partφparty )

y= η (2-2-9)






( partφpartt )

y= η

( partφparty )

y= η


( partφpartt )

y= η

= ( partφpartt )

y= 0


partt )y= 0

η+12 part


y= 0

η 2 + (2-3-1)

( partφparty )

y= η

= ( partφparty )

y= 0


party )y= 0

η+12 part


y= 0

η 2 + (2-3-1)



part 2φpartx2 +


=====================================================================================


=====================================================================================


( partφparty )

y=- h

= 0 (2-3-3)

( partφpartt )

y= 0

+ gη = 0 (2-3-4)

partηpartt

= ( partφparty )

y= 0 (2-3-5)

여기서 (2-3-4)와(2-3-5)의 관계로부터

η= -1g ( partφ

partt )y= 0

lsquo partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0


partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0

( part 2φpartt2 )

y= 0

= -g( partφparty )

y= 0

(2-3-6)




취하면


여기서 k=2πL


σ =2πT




partpartt


+ gη= 0




가 된다



part 2



partpartx





=====================================================================================


=====================================================================================


( partφparty )

y=-h

= 0 larr(2-3-3)

( partφparty )

y=- h

=partparty




( part 2φpartt2 )

y= 0

larr(2-3-6)


d 2f(y)dy2 - k2f(y) = 0 (2-3-9)


d 2f(y)f(y)

-k2dy2=0





식(2-3-11)식을 경계조건식(2-3-3)과 (2-3-6)에 대입하면

이 얻어진다

( part 2φpartt2 )

y= 0

=part 2


=part 2


=partpartt


= (A + B) - σ 2sin(kx- σt)

-g( partφparty )

y= 0

larr(2-3-6)

-g( partφparty )

y= 0

= -gpartparty





y= 0

= -g( partφparty )

y= 0




=====================================================================================


=====================================================================================


coshm =em + e-m

2

sinhm =em - e-m

2



(σ 2 - gk)A + (σ 2+gk)B = 0 (2-3-13)

가 얻어진다


| |e-kh -e kh


= 0 Matrix에서[]+[]- (2-3-14)







= gk tanhkh

(2-3-15)




c =gk

tanhkh =gL2π

tanh2πhL

(2-3-16)

혹은 c =LT


tanh2πhL

이 되어

c =gT2π

tanh2πhL

(2-3-17)


L =gT 2

2πtanh

2πhL

(2-3-18)



로 놓으면

A =D2

e kh B =D2

e- kh가 되어


φ =D2



y= 0


=====================================================================================


=====================================================================================


[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part


y= 0

=σg


(2-3-20)

여기서 a =σg


D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)


φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다


물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)


coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)




급을 해보자



dxdt


로 표시된다


dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)



x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)


=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




( partφpartt )

y= η

+12 ( partφ

partx )2

+ ( partφparty )

2

y= η

+P0

ρ+ gη= 0 (2-2-6)



= 0가 성립

DDt

=partpartt

+ upartpartx

+ vpartparty

+ wpartpartz



( DFDt )

y= η

= ( partηpartt

+ upartηpartx

- v)y= η

= 0 (2-2-8)

또는

partηpartt


partx )y= η

= ( partφparty )

y= η (2-2-9)






( partφpartt )

y= η

( partφparty )

y= η


( partφpartt )

y= η

= ( partφpartt )

y= 0


partt )y= 0

η+12 part


y= 0

η 2 + (2-3-1)

( partφparty )

y= η

= ( partφparty )

y= 0


party )y= 0

η+12 part


y= 0

η 2 + (2-3-1)



part 2φpartx2 +


=====================================================================================


=====================================================================================


( partφparty )

y=- h

= 0 (2-3-3)

( partφpartt )

y= 0

+ gη = 0 (2-3-4)

partηpartt

= ( partφparty )

y= 0 (2-3-5)

여기서 (2-3-4)와(2-3-5)의 관계로부터

η= -1g ( partφ

partt )y= 0

lsquo partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0


partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0

( part 2φpartt2 )

y= 0

= -g( partφparty )

y= 0

(2-3-6)




취하면


여기서 k=2πL


σ =2πT




partpartt


+ gη= 0




가 된다



part 2



partpartx





=====================================================================================


=====================================================================================


( partφparty )

y=-h

= 0 larr(2-3-3)

( partφparty )

y=- h

=partparty




( part 2φpartt2 )

y= 0

larr(2-3-6)


d 2f(y)dy2 - k2f(y) = 0 (2-3-9)


d 2f(y)f(y)

-k2dy2=0





식(2-3-11)식을 경계조건식(2-3-3)과 (2-3-6)에 대입하면

이 얻어진다

( part 2φpartt2 )

y= 0

=part 2


=part 2


=partpartt


= (A + B) - σ 2sin(kx- σt)

-g( partφparty )

y= 0

larr(2-3-6)

-g( partφparty )

y= 0

= -gpartparty





y= 0

= -g( partφparty )

y= 0




=====================================================================================


=====================================================================================


coshm =em + e-m

2

sinhm =em - e-m

2



(σ 2 - gk)A + (σ 2+gk)B = 0 (2-3-13)

가 얻어진다


| |e-kh -e kh


= 0 Matrix에서[]+[]- (2-3-14)







= gk tanhkh

(2-3-15)




c =gk

tanhkh =gL2π

tanh2πhL

(2-3-16)

혹은 c =LT


tanh2πhL

이 되어

c =gT2π

tanh2πhL

(2-3-17)


L =gT 2

2πtanh

2πhL

(2-3-18)



로 놓으면

A =D2

e kh B =D2

e- kh가 되어


φ =D2



y= 0


=====================================================================================


=====================================================================================


[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part


y= 0

=σg


(2-3-20)

여기서 a =σg


D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)


φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다


물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)


coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)




급을 해보자



dxdt


로 표시된다


dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)



x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)


=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


( partφparty )

y=- h

= 0 (2-3-3)

( partφpartt )

y= 0

+ gη = 0 (2-3-4)

partηpartt

= ( partφparty )

y= 0 (2-3-5)

여기서 (2-3-4)와(2-3-5)의 관계로부터

η= -1g ( partφ

partt )y= 0

lsquo partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0


partpartt (-

1g

partφpartt )

y= 0

= ( partφparty )

y= 0

( part 2φpartt2 )

y= 0

= -g( partφparty )

y= 0

(2-3-6)




취하면


여기서 k=2πL


σ =2πT




partpartt


+ gη= 0




가 된다



part 2



partpartx





=====================================================================================


=====================================================================================


( partφparty )

y=-h

= 0 larr(2-3-3)

( partφparty )

y=- h

=partparty




( part 2φpartt2 )

y= 0

larr(2-3-6)


d 2f(y)dy2 - k2f(y) = 0 (2-3-9)


d 2f(y)f(y)

-k2dy2=0





식(2-3-11)식을 경계조건식(2-3-3)과 (2-3-6)에 대입하면

이 얻어진다

( part 2φpartt2 )

y= 0

=part 2


=part 2


=partpartt


= (A + B) - σ 2sin(kx- σt)

-g( partφparty )

y= 0

larr(2-3-6)

-g( partφparty )

y= 0

= -gpartparty





y= 0

= -g( partφparty )

y= 0




=====================================================================================


=====================================================================================


coshm =em + e-m

2

sinhm =em - e-m

2



(σ 2 - gk)A + (σ 2+gk)B = 0 (2-3-13)

가 얻어진다


| |e-kh -e kh


= 0 Matrix에서[]+[]- (2-3-14)







= gk tanhkh

(2-3-15)




c =gk

tanhkh =gL2π

tanh2πhL

(2-3-16)

혹은 c =LT


tanh2πhL

이 되어

c =gT2π

tanh2πhL

(2-3-17)


L =gT 2

2πtanh

2πhL

(2-3-18)



로 놓으면

A =D2

e kh B =D2

e- kh가 되어


φ =D2



y= 0


=====================================================================================


=====================================================================================


[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part


y= 0

=σg


(2-3-20)

여기서 a =σg


D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)


φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다


물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)


coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)




급을 해보자



dxdt


로 표시된다


dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)



x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)


=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


( partφparty )

y=-h

= 0 larr(2-3-3)

( partφparty )

y=- h

=partparty




( part 2φpartt2 )

y= 0

larr(2-3-6)


d 2f(y)dy2 - k2f(y) = 0 (2-3-9)


d 2f(y)f(y)

-k2dy2=0





식(2-3-11)식을 경계조건식(2-3-3)과 (2-3-6)에 대입하면

이 얻어진다

( part 2φpartt2 )

y= 0

=part 2


=part 2


=partpartt


= (A + B) - σ 2sin(kx- σt)

-g( partφparty )

y= 0

larr(2-3-6)

-g( partφparty )

y= 0

= -gpartparty





y= 0

= -g( partφparty )

y= 0




=====================================================================================


=====================================================================================


coshm =em + e-m

2

sinhm =em - e-m

2



(σ 2 - gk)A + (σ 2+gk)B = 0 (2-3-13)

가 얻어진다


| |e-kh -e kh


= 0 Matrix에서[]+[]- (2-3-14)







= gk tanhkh

(2-3-15)




c =gk

tanhkh =gL2π

tanh2πhL

(2-3-16)

혹은 c =LT


tanh2πhL

이 되어

c =gT2π

tanh2πhL

(2-3-17)


L =gT 2

2πtanh

2πhL

(2-3-18)



로 놓으면

A =D2

e kh B =D2

e- kh가 되어


φ =D2



y= 0


=====================================================================================


=====================================================================================


[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part


y= 0

=σg


(2-3-20)

여기서 a =σg


D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)


φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다


물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)


coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)




급을 해보자



dxdt


로 표시된다


dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)



x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)


=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


coshm =em + e-m

2

sinhm =em - e-m

2



(σ 2 - gk)A + (σ 2+gk)B = 0 (2-3-13)

가 얻어진다


| |e-kh -e kh


= 0 Matrix에서[]+[]- (2-3-14)







= gk tanhkh

(2-3-15)




c =gk

tanhkh =gL2π

tanh2πhL

(2-3-16)

혹은 c =LT


tanh2πhL

이 되어

c =gT2π

tanh2πhL

(2-3-17)


L =gT 2

2πtanh

2πhL

(2-3-18)



로 놓으면

A =D2

e kh B =D2

e- kh가 되어


φ =D2



y= 0


=====================================================================================


=====================================================================================


[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part


y= 0

=σg


(2-3-20)

여기서 a =σg


D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)


φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다


물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)


coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)




급을 해보자



dxdt


로 표시된다


dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)



x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)


=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 4]

η = -

1g ( partφ

partt )y= 0

= -1g ( part


y= 0

=σg


(2-3-20)

여기서 a =σg


D =

agσ

1coshkh

=agσσ 2

1coshkh

=agσgk

1tanhkh

1coshkh

=aσk

1sinhkhcoshkh

1coshkh

=aσk

1sinhkh

(2-3-21)


φ =agσ

coshk(h+y)coshkh

sin (kx- σt)=aσk

coshk(h+y)sinhkh

sin (kx- σt) (2-3-22)

로 나타낸다


물입자 속도의 각 성분은 식(2-2-1) (2-3-8) (2-3-22)로 부터

u =partφpartx

= aσcoshk(h+y)


coshk(h+y)sinhkh

η (2-3-23)

v=partφparty

= aσsinhk(h+y)




급을 해보자



dxdt


로 표시된다


dxdt

=partφpartx

x(t) y( t) t

dydt

=partφparty

x(t) y( t) t

(2-3-25)



x(t) = x0 + δ( t) y(t) = y0 + γ( t) (2-3-26)


=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 5]

dxdt

=partφpartx


x0 y0


x0 y0


x0 y 0

+

dydt

=partφparty


x0 y0


x0 y0


x0 y 0

+

(2-3-27)


coshk(h+y)sinhkh

sin (kx- σt)을


dδdt

= ( partφpartx )

x0 y0

=partpartx ( aσ

k

coshk(h+y0)


= aσcoshk(h+y0)

sinhkhcos (kx0-σt)

dγdt

= ( partφparty )

x0 y0

=partparty ( aσ

k

coshk(h+y0)


= aσsinhk(h+y0)

sinhkhsin (kx0-σt)

(2-3-28)


δ = -acoshk(h+y0)

sinhkhsin (kx0-σt)

γ = asinhk(h+y0)

sinhkhcos (kx0-σt)








x- x0

acoshk(h+y0)

sinhkh

= - sin (kx0- σt)

y- y0

asinhk(h+y0)

sinhkh

= cos (kx0- σt)



(x - x0)

2

(a coshk(h+y0)

sinhkh )2 +

(y - y0)2

(a sinhk(h+y0)

sinhkh )2 = 1 (2-3-31)

(장 축) (단 축)


=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


있다



+12 ( partφ

partx )2

+ ( partφparty )

2

+ Pρ


P = - ρpartφpartt

- ρgy (2-3-32)


Δp로 하면


= - ρpartpartt (

agσ

coshk(h+y)coshkh

sin (kx- σt))

= ρagcoshk(h+y)


= ρgcoshk(h+y)

coshkhη

(2-3-33)






2)cos (

A+B2

)

= 2a cos ( k-k2

x-kc - kc

2t)cos ( k+ k

2x-

kc + kc2

t) (2-3-34)


x-kc - kc


2πL

σ=2πT

c= σk)을 이용

k- k

2=

2πL

rArr L =4π

k- k (파장)

kc - kc

2=

2πT

rArr T =4π

kc - kc (주기)

c= σk

=2πT2πL

=LT

=kc - kck- k

(파속)

또 cos ( k+ k2

x-kc + kc


k+ k

2=

2πL

rArr L =4π

k+ k (파장)

kc + kc

2=

2πT

rArr T=4π

kc + kc (주기)

c=kc + kck+ k

(파속)


2πkc


=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 6]


다시 cos ( k-k2

x-kc - kc




낼 수 있다

cG =kc - kck- k

=δ(kc)δk


=dσdk

(2-3-36)



2σdσdk


dσdk




2gktanhkhσ=

g tanhkh2gktanhkh


σ

=σ2

1k

+ h

1


=

kc2 ( 1

k+

hsinhkh coshkh )

=c2 (1 + 2kh

sinh2kh )= nc

(2-3-37)


2khsinh 2kh )이다


E p = ⌠⌡

h

0

⌠⌡

η


h

0[ y2

2 ]η

0

dx=ρg2

⌠⌡

L

0η 2 dx (2-3-39)

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


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[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


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=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


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=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


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=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

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=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

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2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


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=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



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[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

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=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

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[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


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=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



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=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

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=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

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=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

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=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

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해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





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ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

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----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

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=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








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[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


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=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









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2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


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=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


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2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


=ρg2

a 2⌠⌡

L


=ρg2

a 2 12

⌠⌡

L


12(1+ cos 2θ)

=ρg4

a 2⌠⌡

L

0dx+

ρg4

a 2⌠⌡

L

0cos2(kx- σt) dx

= ρg4

a 2L = ρg4 ( H2 )

2

L = ρgH2

16L

(2-3-40)


E k =ρ2

⌠⌡

L

0

⌠⌡

η



E k =ρ2

⌠⌡

L

0

⌠⌡

η

-hu 2 dydx =

ρ2

⌠⌡

L

0dy⌠⌡

η

-hu 2 dx =

ρ2

⌠⌡

L

0[y] η-h u

2 dydx rArr u =ghη

=ρ2

⌠⌡

L

0(η + h)u 2 dx =

ρ2

⌠⌡

L

0(h

ghη 2) dx= ρg

2⌠⌡

L


=ρg2

a 2⌠⌡

L

0 12(1 + cos 2(kx- σt)) dx= ρg

2a 2[⌠⌡

L

0dx+ ⌠

⌡

L

0cos2(kx- σt)dx]

=ρga 2

4L =

ρgH2

16L

(2-3-41)


E = E p+ E k=ρgH 2

16L +

ρgH 2

16L =

ρgH 2

8L = 2E p = 2E k (2-3-42)

로 된다


E = E p+ E k=ρgH 2

16+ρgH 2

16=ρgH 2

8= 2E p = 2E k (2-3-43)


W= ⌠⌡

η

-hpu dy = ⌠

⌡

η

-hp( partφ


0

-hp( partφ


p = aρgcoshk(h+y)

coshkhcos (kx- σt)

u = aσcoshk(h+y)

sinhkhsin (kx- σt)


=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




W =1T

⌠⌡

T

0Wdt=

1T

⌠⌡

T

0

⌠⌡

0

-hpu dy

=1T

⌠⌡

T

0

⌠⌡

0

-h(aρg coshk(h+y)


coshk(h+y)sinhkh

cos (kx-σt))dtdy

= ρa 2σgT

⌠⌡

T


0

-h( cosh 2k(h+y)

sinhkhcoshkh )dy

=ρa 2σgT

⌠⌡

T


0

-h( 2 cosh 2k(h+y)

sinh 2kh )dy

=ρa 2σgT

T2

2sinh 2kh

⌠⌡

0

-h


= ρa 2σgsinh 2kh

12 [y+ 1

2ksinh 2k(h+y)]

0

-h

=ρa 2σgsinh 2kh

12 [ 1


sinh 2kh12

12k

[ sinh 2kh+2kh]

=ρa 2σg2k

12 (1+ 2kh


n =12 (1+ 2kh

sinh 2kh )

=ρa 2g2

cn =ρg2 ( H

2 )2

c G=ρgH2

8c G= E c G

(2-3-46)



생각해보면



rArrhL


hL

120

～125

(=005～004) 12(= 05)



=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




y= 0


( partφpartt )

y= 0

+ gη-Kρ

part 2ηpartx2 = 0 (2-3-64)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0 (2-3-65)


φ =agσ

coshk(h+y)coshkh


y= 0


( partφpartt )

y= 0

=partpartt (

agσ

coshk(h+y)coshkh

sin (kx - σt))y= 0

=agσ

(-σ)coshkhcoshkh

cos (kx- σt)

= -ag cos (kx- σt)


( partφpartt )

y= 0

+ (g+Kk2

ρ )η = 0



-g + (g+Kk2

ρ ) = 0

g = (g+Kk2

ρ )



c =

gk

tanhkh =1k (g+

Kk2

ρ ) tanhkh

= ( gk

+Kkρ ) tanhkh

(2-3-66)



c = ( gk +Kkρ ) = ( gL

2π+

2πKρL ) (2-3-67)






=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 7]

cos cos cos

sin sin sin


Lm = 2πKρg

cm =4πKρLm

= 4 4gKρ

(2-3-68)

가 된다


를 대입하면




2-3-7 중복파




coshk(h+y)sinhkh

sin (kx- σt)


coshk(h+y)sinhkh

sin (kx+ σt)

(2-3-69)

따라서


(2-3-70)

φ = φ 1+φ 2 =2a σk

coshk(h+y)sinhkh

sin (-σt) cos (kx)

=-2a σk

coshk(h+y)sinhkh

coskxsinσt




u =partφpartx

= 2aσcoshk(h+y)

sinhkhsinkxsinσt

v=partφparty

= -2a σsinhk(h+y)

sinhkhcoskxsinσt

(2-3-71)


=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 8]



x= 0

=0 을 만족하는




)πk의





2πk



3π2k

5π2k



E p = ρg2TL

⌠⌡

T

0

⌠⌡

L

0η 2dxdy =

4ρg2TL

⌠⌡

T

0

⌠⌡

L


=4ρga 2

2TL⌠⌡

T

0cos 2σtdt⌠⌡

L

0cos 2kxdx

=4ρga 2

2TL⌠⌡

T

0( 12( 1+ cos 2σt))dt⌠⌡

L

0( 12( 1+ cos 2kx))dx

=4ρga 2

2TLT2

L2

=ρga 2

2

(2-3-72)

E k=ρ

2TL⌠⌡

T

0

⌠⌡

L

0

⌠⌡

0


2a 2 (2-3-73)






알 수 있다


=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


cosA+ cosB=2cosA+B

2cos

A-B2


2sin

A-B2

δ = -2a

coshk(h+y0)

sinhkhsinkx0cosσt

δ = 2asinhk(h+y0)


(2-3-74)


γδ

=2a

sinhk(h+y0)

sinh khcoskx0cosσt

-2acoshk(h+y0)

sinhkhsinkx0cosσt

=-sinhk(h+y0)

coshk(h+y0)

coskx0

sinkx0


sinkx0

=- tanhk(h+y0)1

tankx0


(2-3-75)




vu

=dydx












(2-3-78)




라 하면


= a 1 - a 2 (2-3-79)


=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 9]

a 1 =12(a max

+ a min) a 2 =

12(a max

- a min) (2-3-80)


K r =a 2

a 1

=a max

- a min

a max+ a min

(2-3-81)




천해파

L =98times 5 2

2πtanh

10πL


c =LT

=30289

5= 6058 msec

심해파

c 0 =gT2π

=98times52π

= 7799 msec

L 0 =gT 2

2π=

985 2

2π= 38992m

【예제 1】


=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


2-4-1 질량 수송



Mx=⌠⌡

η

-hρudy (2-4-1)



u = aσcoshk(h+y)

sinhkhcos (kx- σt)


Mx=⌠⌡

0

-hρudy+ ⌠

⌡

η

0ρudy = ⌠

⌡

η



Mx=ρT

⌠⌡

T

0ηudt = ρ

T⌠⌡

T



= ρT

a 2σ cothkh⌠⌡

T

0cos 2(kx-σt)dt=

12ρa 2σ cothkh

=12ρga 2 k


gk

σ 2 =1

tanhkh

=12ρg( H2 )

2 kσ

=Ec

rArr E=ρg8

H2 c= σk

(2-4-2)

가 된다


u L =12

a 2σk2cosh2k(h+y0)

sinh 2kh (2-4-4)


Mx= ρ⌠⌡

0

-hu Ldy 0 =

12

a 2σk2

sinh 2khρ⌠⌡

0

-hcosh2k(h+y0)dy 0

=12

a 2σk2ρsinh 2kh

sinh 2kh2k

=14a 2σkρ


=12ρa 2ckcoth kh =

12ρa 2σ coth kh

(2-4-5)

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



(1)파장 파속


ltSolutiongt

(1)파장 파속

C 0=gT2π

= 156T= 156128= 200ms

L 0=gT 2

2π= 156T 2=156128 2= 256m

(2)물입자 속도



a =H2

σ=2πT

k=2πL

θ=kx-σt

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2πy256cosθ

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2πy256sinθ

2m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 2)256cosθ=07011 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π (- 2)256sinθ= 07011 sinθ ms

20m일 때

u =H0

22πT

e2πyL 0cosθ=

302

2π128

e 2π ( - 20)256cosθ= 0451 cosθ ms

v=H0

22πT

e2πyL 0sinθ=

302

2π128

e 2π ( - 20)256sinθ=0451 sinθ ms y

【예제 2】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


2-6 파의 변형









다

L =

gT 2

2πtanh

2πhL

c =gT2π

tanh2πhL

L 0 =gT 2

2π c 0 =

gT2π

(2-6-1)



LL 0

=cc 0

= tanh2πhL (2-6-2)


F x= Ec G = Ecn (2-6-4)



18ρgH 2cn =

18ρgH2

0c 0n 0

로 쓸 수 있다 여기서 n 0 =12(2-3-54)이므로

HH0

=12n

c 0

c= Ks (2-6-5)

n =12 (1 +

2khsinh 2kh )


=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




H 0 = 3m T= 12 sec

L 0 = 156T 2= 15612 12 = 2246m

H0

L 0

=3

2246= 00133

① h = 10m의 경우

hL 0

=10

2246= 004452 값을 이용하여


H =KsH 0= 104times 30 = 312m

L = L 0051= 2246051 = 114546m

파형경사HL

=312

114546= 00272


hL 0

=5

2246= 00223 값을 이용하여


H= KsH0= 121times 30 = 363m

L = L 00365 = 22460365 = 81979m

파형경사HL

=363

91979= 00442

【예제 3】

2-6-2 파의 감쇄



파는 감쇄한다


τ b = fρu 2b (2-6-6)



=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 10]


( Ecn )① - ( Ecn )② = D fΔx a =H2

σ=2πT

y=-h

u b = aσcoshk(h+y)


πHT

1sinhkh

cos (kx - σt)


D f =1T

⌠⌡

T

0τ bu bdt =

fρT

⌠⌡

T

0u 3

bdt

=fρT

⌠⌡

T

0[ πHT

1sinhkh

cos (kx- σt)]3

dt

=fρH3π 3

T 4 ( 1sinhkh )

3⌠⌡

T


= π 2fρ6

H3


=4π 2fρ3

H3

T 3sinh 3kh


D f =4π 2fρ3

H 3

T 3sinh 3kh (2-6-7)

윗 식을 풀면

d(Ecn )

dx=

ddx ( ρgH

2

8 ) gT2π

tanhkh12 (1 +

2khsinh 2kh )


dx=-D f 이므로

-43π 2 ρfH3


gT2π

tanhkh12 (1 +

2khsinh 2kh ) dHdx

여기서 Ks=1

tanhkh(1 +2kh

sinh 2kh ) K2

s = [ tanhkh(1 +2kh

sinh 2kh ) ]- 1

로 가정

-dHH2 =

64

3g2

ρπ 3


2khsinh 2kh )]

- 1

dx

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


-dHH2 =

K2s64π

3ρ


1H

=K2

s 248π3ρ

3g 2T 4sinh 3khx+ C

여기서

x= 0 H = H 1 1H 1

= C

x= Δx H = H 2 1H 2

= kΔx+1H 1

H 1

H2

= H1kΔx+ 1 [H 1kΔx+ 1]- 1 =H 2

H 1

= K f

따라서

K f =H 2

H1

= [ 64π 3fK2sΔxH 1

3g 2T 4sinh 3kh+ 1]

- 1

(2-6-8)




D p =πkρgH 2

4L1

cosh 2kh (2-6-9)



3m 12 sec 25m

5km

f = 002

L 0=15612 2= 225m hL 0

=25225

= 0111 LL 0

= 074

L= 074 225= 167m hL

=25167

= 01497 kh= 0941

K s= 092 f= 002 Δx= 5000m

K f =H 2

H 1

= [ 64π 3fK2sΔxH1

3g 2T 4sinh 3kh+ 1]

- 1

= [ 643 ( 3141592 ) 3

(980621) 200235000

12 4

(092) 2

sinh 3(0941)+1]

- 1

=0938

there4 H 2=09383 0= 2814m

감소율= (30-2814)30= 0062 약 6 감소

【예제 4】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 11]

2-6-3 파의 굴절









다

c 1

sinα 1

=c 2

sinα 2

(2-6-11)


b 1

cosα 1

=b 2

cosα 2

(2-6-12)


Δt=ACC 1

=BDC 2


ABsin α 1

C 1

=ABsin α 2

C 2

C 2

sinα 2

=C 1

sinα 1


때문에

b0P0 = bP = const

P = n Ec P0= n 0E 0c 0


b0n 0E 0c 0 = bn Ec

n 0 =12

E 0 =ρgH2

0

8 E=

ρgH 2

8


=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


b0

12

ρgH20

8c 0 = bn ρ

gH 2

8c

따라서

HH0

=12n

c 0

c

b0

brArr Ks=

12n

c 0

c K r =

bo

b

=KsK r

(2-6-13)

굴절계수 K r는

K r =b1

b2

=cosα 1

cosα 2

= [1+ (1- ( cc 0

)2

) tan 2α]-

14



Sol) L 0=156T 2 = 15610 2 = 156m

c 0 = 156T = 156 10 = 156msec


=7

156= 00449이므로 45P에서 K s=104

cc 0

=LL 0

= 051

L = 051times 156 = 79546m

c = 051times 156 = 796msec

c o

sinθ 0

=c


K r =cosα 0

cosα=

cos 55cos 244137

= 0795



【예제 5】


이 일어난다



=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




c r

c s

= (1 -Uc s

cosθ s)-1 tanhkh

tanhksh s (2-6-14)

kks

= (1 -Uc s

cosθ s)2

( tanhkhtanhksh s

)-1

(2-6-15)

cosθcosθ s

= (1 -Uc s

cosθ s)-2 tanhkh

tanhksh s (2-6-16)

HHs


)-

12

( nn s

)-

12

(2-6-17)


n =12 (1 +

2khsinh 2kh ) (2-6-18)


Uc

c s

=

1 - ( cosθ stanhkh

tanhksh s)

12

cosθ s (2-6-19)

가 얻어진다

2-6-4 파의 회절





정식은

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 12]

part 2φpartx2 +

part 2φparty2 +

part 2φpartz 2 = 0 (2-6-20)


z =- h

= 0를 고려하여




part 2


part 2


part 2



part 2

partx2 F(xy)+part 2

party2 F(xy)+k2F(xy)= 0 (2-6-22)

이 된다


ζ= -1g ( partφ

partt )z = 0

= -Aiσg


가 된다


F(xy) = e - i ky (2-6-24)


ζ 1 = -Aiσg



ζζ 1

= e ikyF(xy) (2-6-26)

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



HH I

= | ζζ 1 | = |F(xy)| = Kd (2-6-27)


Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr




Sol) L 0 = 156timesT 2 = 561m k=2πL 0

= 0112

Kd =π

2 kB ( ln kB8

+ 05772)2

+π 2

4 Bπr

=3141592

2 011210 ( ln 0112108

+ 05772)2

+3141592 2

4 10

314159250

= 0179

H= KdH I = 0357m

【예제 6】

2-6-5 파의 반사



소비된다


비는

반사율 KR =HR

H I 전달율 KT =

HT

H I (2-6-28)

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


구 분 γ 1

불투과 활면 10

불투과 조면 07 ～ 09

모 래 해 빈 08

사 석 사 면 03 ～ 06

로 되며 K2R + K2

T = 1이 된다



( H0

L 0)cr it

=2βπ

sin 2βπ

( H0

L 0)cr it

=θ90

sin 2θπ

(2-6-29)




KR =HR

H I

=γ 1 γ 2 (2-6-30)



γ 2 =

( H0

L 0)cr it

( H0

L 0)

Ho

L 0

gt ( H0

L 0)cr it

1 H0

L 0

le ( H0

L 0)cr it



= 1 (기타) (2-6-31)



다

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


해저경사 125


을 추정하라


L 0

=35

(156 times 8 2)= 00351

θ = tan -1( 125 )=22906


(1) ( H0

L 0)cr it

=2times004π

times( 12501992 )

2

π= 812times10- 5

(2) ( H0

L 0)cr it

=2290690

times( sin 22906) 2

π= 812times10- 5

H0

L 0

gt ( H0

L 0)cr it


35times10- 2 = 232times10- 3

γ 1 = 08로 하면

반사율 KR = 08times 000232 = 00019 (약 02)

【예제 7】

2-6-6 쇄파







지는 파





=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



ε 0 gt 33 33 gt ε 0 gt 046 046 gt ε 0




L 0

ε b =tanβHb

L 0

이다

쇄파지표



L b) (Hamada)



L 0



T = 4sec h = 2m β =130


태의 쇄파인가

Sol) L 0 = 156T 2 = 156times4 2 = 25m

LL 0

= tanh2πhL


there4L = 1625m

Hb = 0142times1625 = 231m

Hb = 0142times1625 tanh (2π21625

) = 165m

Hb = 0827times2 = 165m


30 )43

) = 160m

ε b=tanβHbL 0

=tan (130)16025

= 029


【예제 8】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


----------------------------------------------------------------------------------


하고 있다


=gt


=gt

따라서


=gt


=gtsin

sin sin sin

sin

sin

sin

sin

13

따라서

coscos

coscos



평균 일률

----------------------------------------------------------------







=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 13]

η b = -H2

8k

sinh 2kh (2-6-33)


η= K(h b- h) + η b (2-6-34)

K=1

1 +8

3γ 2

(2-6-35)


을 취한다











(2) 110

최대파 (H 110 T 110)

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


(3) 13최대파 유의파 (H 13 T 13)

(4) 평균파 ( H T)





2-7-2 파고분포




p(H)dH=π2

H

H2exp - π

4 ( HH )

2

dH



p(H) = 1 - ⌠⌡

H

0p(H)dH= exp - π

4 ( HH )

2

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



H 13≒160H H = 0625H 13

H 110≒203H≒ 127H 13

H max

H 13


= 0706 lnN+r






H rms=1

0886H


(2) p(H) =nN

nN

= exp - π4 H

0886H rms2

= exp (-4) = 001832

n = 001832times400≒ 7파


평균치 H rms= 0706(2448 + 0118)times64 = 116m

【예제 9】

2-7-3 파의 연결








=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 15]

2-7-4 파의 주기 분

포



T max ≒T 110

≒T 13

≒12T





트럼이라 한다


동 η(xyt)는

η(xyt) = suminfin








ksumθ+δθ

θ

12a 2

n


η 2 = ⌠⌡

infin

0

⌠⌡

2π

0E(k θ)δkδθ


E(k) = ⌠⌡

2π

0E(k θ)δθ


=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




혹은

f2n =g4π

kn


η 2 = ⌠⌡

infin

0E(k)dk= ⌠

⌡

infin

0E(f)df



있다



E(f) =810times10- 3g2

(2π) 4f 5exp -074 ( g

2πU195f)4

(m 2sdots)



E(f) = 0257H213T 13(T 13 f)

- 5exp [-103(T y3 f)- 4] (m 2sdots)



E(f) = 858times10- 4( gFu 2

)- 0312


)- 132

( u 8f

g )-4

(m 2sdots)


E(f) = 115times10- 4( gFU2

10)- 0312


10)- 132

( U10f

g )- 4

(m 2sdots)



E(f) = αg2(2π)- 4 f- 5exp - 54 ( f

f p)-4

γexp -(f-f p)

2

2σ 2pf

2p


=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


H 110 = 36 E H 13 = 283 E H= 177 E

E = c 3( π2 )32

( U75

2g )5




U ( Z) = 1 + 575 γ 2 log 10(

z10

)U10 (msdotsec )


γ 2 = 26times10- 3 (U10 gt 15m sec )

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



H 110 = 360 E = 360 ⌠⌡

infin

0E(f)df

로 쓰면


-n

= 858times10- 4( gFu 2)- 0312


)( u

g )-4

f- 4]

k1 = 858times10- 4( gFu 2

)- 0312

g2 k2 = 125times( gFu 2)- 132

( u

g )- 4

E = ⌠⌡

infin

0E(f)df =

k1

n

Γ(m-1n

)

k( m- 1)

n2

=k1

4times

1k2

=14 (858times10- 4( gFu 2

)- 0312

g2)times 1

125times( gFu 2

)- 132

times( u

g )- 4


timesu 4

g2

Γ(m-1n




)1008

timesu 4


10)1008

timesU4

10

g2

【 예제 10】


10)1008

timesU4

10

g2 12


30 2 )1008

times30 4

980621 2 12

=4252m



E(fθ) = E(f)G(θf)



E(f) = ⌠⌡

π


π

-πg(θf)dθ

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



⌠⌡

π

-πG(θ f)dθ = 1

로 된다



)




제안했다

G(fθ0 = G 0cos2s( θ2 ) G 0 =

1π

2 2s- 1 T 2(S+1)T(S+1)


S = S max ( ff p

)5

( fle f p )

S max ( ff p

)- 25

( fge f p )



f p =1

105T 13

S max = 115( 2πf pU 10

g )- 25


(1) 풍파 S max = 10 (G 0 = 09033)








(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================





(U- U) 2 gtgk

(ρ 2 - ρ 2)ρρ

+ Kk(ρ + ρ)ρρ



Keulegan 1200 cms






의한다

(a) Philips이론



(b) Miles이론





였다









=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



2-8-3 SMB법



만 알아본다

gH 13

U210

= 03

1 -1

(1 + 0004( gFU2

10)

12

)2

gT y3

2πU 10

= 137

1 -1

(1 + 0008( gFU2

10)

13

)5




F = 100km U10 = 15ms t = 10hr

U와 F의 조합 H 13 = 25m T 13 = 60 sec

U와 t의 조합 H 13 = 28m T 13 = 65 sec


【예제 11】




=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




F = 100km U = 10ms H 13 = 15m T 13 = 48 sec

t = 10hr U = 10ms H 13 = 15m T 13 = 48 sec


F = 100km U= 20ms H 13 = 35m T 13 = 70 sec

t = 5+18= 68 U= 20ms H 13 = 36m T 13 = 71 sec


【예제 12】


(a) 바람의 장


=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



Ugr = γwsinφ

- 1 + 1plusmn

dpdr

ρ arΩ2sin 2φ

여기서 dpdr



져 있다

(1) Myer의 식





(2) Fujita의 식

p = pinfin -δp

1 + ( rr 0

)2

(b) 파의 장



평형방정식

partEpartt








T 13 = 386 H 13 (단위 m s)


=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




gH 13

U 2 = 03A

1 -1

1 +

0004( gFU 2 )

12

A

2

gT 13

2πU= 137B

1 -1

1 +

0008( gFU2 )

13

B

5

A = tanh 0578( ghU2 )

34

B = tanh 0520( ghU2 )

38



h = 6m F = 20km U10 = 20ms

ghU2

10

=980621times6

20 2 = 0147 gFU2

10

=980621times20000

20 2 = 490

교재 P94 그림 2-8-12에서 gH 13

U210

= 0037

H 13 = 0037times40098

= 15m T 13 = 386times( 15) 12 = 47 sec

【예제 13】








아본다

(H 13)D(H 13)F

= ( k1F min

k1F min + D )12

(T 13)D(T 13)F

= k2 + (1 - k2)(H 13 )D(H 13)F

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================






tD =4πD

g(T 13 )D



추정하라




12

= 091m

주기 (T 13 )D = 85times(20 + (10-20)times09155 )

12

= 115 sec



【예제 14】









2-10-1 파압현상


=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================





2-10-2 중복파압







φ 1 = accoshk(h+y)

sinhkhsin k(x-ct)



coshk(h+y)sinhkh

sink(-x-ct)

φ = 2ac

coshk(h+y)sinhkh

sin (-kct) cos (kx)

= -2accoshk(h+y)

sinhkhsinkctcoskx


partφpartt

+12 ( partφ

partx )2

+ ( partφparty )

2

+ pρ

+ gy= 0


pρ

= -gy-partφpartt


pρg


sinhkhcos kctcoskx


coshk(h+y)sinhkh

cos σtcos kx



coshk(h+y)sinhkh

cos σt


pρ

= -gy-12 ( partφ

partx )2

+ ( partφparty )

2

- partφpartt

+ f( t)


하면

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


f(t) = gη 0 +12 ( partφ

partx )2

+ ( partφparty )

2

y= η 0

+ ( partφpartt )

y= η 0

이다

φ = -2accoshk(h+y)

sinhkhcoskx sin σt

partφpartx

= 2ackcoshk(h+y)

sinhkhsinkx sinσt

partφparty

= -2acksinhk(h+y0)

sinhkhcoskxsin σt

( partφ

partx )2

+ ( partφparty )

2

= H2gksinhkhcoshkh

sin 2σtsinh 2kh




여기서



partφpartt

= - ρgHcoshk(h+y)

coshkhcos σt

( partφpartt )

y= η 0


coshkhcosσt


( partφpartt )

y= η 0


-coshk(h+η 0)

coshkh )

따라서

pρg

= -y + η 0(1 +cosh (k+y)

coshkh-

coshk(h+η 0)


pρg


coshkh-

coshk(h+H)coshkh


pρg

= h + H1 +1

cpshkh-

coshk(h+H)coshkh

이고 cosh k(h+H)


12


pρg

= h + H( 1cos kh

- kHtanhkh -12

k2H2)

(b) Sainflou의 식


=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


γ = acoshk(h+y0)

sinhkh γ = a

sinhk(h+y0)

sinhkh

여기서 a =H2








pρ

- gy=2σ 2


ka 2



pρg

= -y0 + 2( σ2




pρg


coshkh-

sinhk(h+y0)

sinhkh coskx0cosσt







sinh 2kh+ 2γ


sinh 2kh- 2γ



sinh 2kh



sinh 2kh


=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


δ 0 = ka 2 sinh2kh


=πH 2

Lcoth

2πhL

로 주어진다


pρg


coshkh-

sinhk(h+y0)

sinhkh



pρg

= h + H1

coshkh 이다


pρg


coshkh-

sinhk(h + y0)

sinhkh



pρg

= h + H1

coshkh 이다






p1 = (p2 + ρgh)( H+ h 0

h+ H+ h 0)

p2 =ρgH

cosh (2πhL

)


p1 = ρg(H - h 0)

p1 = p2 =ρgH

cosh (2πhL

)

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


여기서 h 0 =πH2

Lcoth

2πhL

이다

Example



Solution)


times

times

cos

cosh timestimes

times


times


times

2-10-3 쇄파의 파압




hM

H0

= CM( H0

L 0)- 14


는 계수



(a) Hiroi 공식


p = 15ρgH

p = 125ρgHcos 2θ

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 17]

[그림 18]











다

(b) Minikin 공식





(1) 충격압

pm = 1024ρgd(1 +dh ) HL

py = pm( H-2|y|H )

2


리이며 -H2

simH2


(2) 정압

ps= ρg( H2


ps = ρgH2

(정수면 아래)







=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


[그림 19]




가 있다

(1) 등압




(2) 정압



2-10-5 양압력



려해야 한다













H max



=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================







p1 =12(1 + cosβ)(α 1 + α 2cos

2β)ρgH max

p 2 =1

cosh2πhL

p 3 = α 3p 1

여기서

α 1 = 06 +12

4πhL

sinh ( 4πhL )

2

α 2 = min h b- d

3h b( H max

d )2

2d

H max

α 3 = 1 -hh 1 -

1



(d) 양압력



=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================




H 13 = 3m T = 9sec h = 8m

심해파의 파장 L 0 = 156times9 2 = 1264m

hL 0

=8

1264= 00633

LL 0

= 059 L = 059times1264 = 747m

hL

= 01071 cotkh = coth 0673 = 1704


= 0645m

p1 = (250 + 103times8)times30 + 0645

80 + 30 + 0645= 336tfm 2

p2 =103times3

cosh 0673= 250tfm 2

there4pH =

12

( p1

H0+ h 0)( H

2+ h 0)

2

+ 2p2 + (P1 - P2)

h -

H2

h

(h -

H2 )

+ 15w0H

= 05times(721 + 3704) + 1391 = 360tfm 2

【예제 15】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================





L 0 = 156times11 2 = 1888 hL 0

= 00636 H 0

L 0

= 00318

H 13≒55m (β 0 = 01038 β 1 = 0520 β max = 092 k 3 = 153)

H max≒87m (β 0=01928 β

1=0630 β max =165)

times

ks=12n

c 0

c n =12

1 +

4πhL

sinh ( 4πhL )

c 0 = 156T =gT2π

c =gT2π

tanh2πhL


LL 0

= 059 hL

= 01077

α 1 = 06 +12 ( 2times0676


= 0881

α 2 = min 123 - 653times123

times( 8765 )

2

2times6587 = min (0282 1494) = 0282

α 3 = 1 -812

times(1 -1

cosh 0676 ) = 0872


p 2 =1016

cosh 0676= 820tfm 2

p 3 = 0872times1016 = 886tm 2

times times


전 수평력



【 예제 16】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================









)23


k는 계수

따라서

F = kρgH( Wρ sg

)23



로부터


다

τ= W(1 - ρρ s

)(μ cosθ- sinθ)

τgeF

(1 -ρρ s


)23

(1 -ρρ s

)3


)2

W gek 3ρ 3gH3

ρ2s(1 - ρρ s

)3

(μ cosθ- sinθ) 3

gek 3ρ sgH

3

ρ2s( ρ s

ρ- 1)

3

(μ cosθ- sinθ) 3


=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


F = kμρg( Wρ sg

)23

W gek 3ρ sμgH

3

( ρ s

ρ- 1)

3


Wgeρ sgH

3

K D( ρ s

ρ- 1)

3

cotα

=w r H 3

K D( w r

w 0

- 1)3

cotα

( Hudson 식 )



W=23times5 3


【예제 17】










πD 2

4u )dy


dF max = dF D +dF 2

M

4dF D

( 2dF D gt dFM )


여기서

dF D =12ρC Du




w0

gπD 2

4kσ 2H2

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================



각주파수

σ =2πT



L 0 = 156times11 2 = 1888m hL 0

=10

1888= 0053

hL

= 0097 HL

= 045

F D = 405t FM= 121t

따라서 FT = 405 +1214times405

= 414t

【예제 18】




RH

=π2α

ks

2α c

π

sin 2α c

π=

H 0

L 0



RH0

= π2α

+ ( η s

H- 1)Ks

η s

H= 1 + π

HL

cothkh (Sainflou)

η s

H= 1 + π

HL

cothkh(1 +3

4 sinh 2kh-

1



이다



(1) Hunt 식

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


RH0

=101 tan α

( HL 0

)12

(2) Takada 식

RH0

= π2α c

+ ( η s

H- 1)Ks( cotα c

cotα )23





L 0 = 156times9 2 = 1264m H 0

L 0

=25

1264= 00198

hL 0

=5

1264= 00396

α c = 036r ad α =13


L = 606m K s= 1064

η s

H= 1 + π

25606

times2098times(1 +3

4times0542 2 -1

4times1138 2 ) = 09721064

RH0

= ( π2times0333

+09721064 )times1064times( 0346

0356 )23

= 328

R = 328times25 = 82m

【예제 19】

2-12-2 월파량





Q = a(R - Hc)

2 ( x0

R- cotθ)



L 0)083

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

=====================================================================================


=====================================================================================


직립제일 경우 a = 93( (R - H0)

H )12

( H0

L 0)


① 직립제 Q = 15times10- 2(R - Hc)2

② 105 경사제 Q = 78times10- 2(R - Hc)2

③ 11 경사제 Q = 21times10- 1(R - Hc)17

④ 12 경사제 Q = 12(R - H c)145

⑤ 13 경사제 Q = 55times10- 1(R - Hc)18


Q = aA

A = ( 12


(R - Hc)2 + 015H(R - Hc))


)( cotθ) 16


2cot 2θ

12

cotθ

n = -3224 log 10 1

1+(67 HL )

2


L 0 = 156 H 0

L 0

=55156

= 00353 hH0

=5

55= 0909

q

2gH30

=005



= 091과 q

2gH30


H 0

= 105



H 0



【예제 20】

제2장 해양의 파_JEJU Univ.

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