teori gelombang 2 new.pdf
TRANSCRIPT
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R. TriatmadjaR. Triatmadja 20112011
This Lecture note is based on This Lecture note is based on Water Wave Mechanics for Engineer and Water Wave Mechanics for Engineer and
ScientistScientist Dean and Dalrymple , 1994Dean and Dalrymple , 1994
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LONGWAVELONGWAVE
( )( ) ( )tkxkh
zhkHgku σσ
−+= coscosh2
cosh
( )( ) ( )tkxkh
zhkHgkw σσ
−+= sincosh2
sinh
( )tkxH ση −= cos2
Kinematika gelombang secara umum
( )khgTc tanh2π
=
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LONGWAVELONGWAVE
( ) khkh ≈tanh
cgh
chgkhgTc === π
ππ2
22
ghc =
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LONGWAVELONGWAVE
( )( ) ( )tkxkh
zhkHgku σσ
−+= coscosh2
cosh
Jika kh mendekati nol (kh<0.3), maka
1)cosh( ≈kh 1)(cosh)cosh( ≈+≈ zhkkh
( )tkxgHku σσ
−= cos2 σ2max,
gHku surface =
ha
cga
cu == 2
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LONGWAVELONGWAVE
( )( ) ( )tkxkh
zhkHgkw σσ
−+= coscosh2
sinh
( )tkxzhkgHkw σσ
−+≈ cos)(2
Laakkh
cga
cw 28.6
2 ===
khgHkw surface σ2max, ≈
20<Lh
khL
huw == π2
102 ππ < ==L
huw
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Persamaan gelombang dapat diturunkan tidak dengan Persamaan gelombang dapat diturunkan tidak dengan penyederhanaan sebanyak penyederhanaan atau asumsi penyederhanaan sebanyak penyederhanaan atau asumsi pada gelombang Airypada gelombang Airy
LONGWAVELONGWAVE
0=∂
∂+∂
∂+∂
∂zw
yv
xu ρρρ Persamaan kontinuitas
0=∂∂+
∂∂+
∂∂
zw
yv
xu Persamaan kontinuitas
disederhanakan (ρ = konstan)
0),,(),,( =−−+
∂∂+
∂∂=
∂∂+
∂∂+
∂∂
∫∫∫ −−−hyxwyxwdz
yvdz
xudz
zw
yv
xu
hhhη
ηηη
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Ingat aturan Leibnitz untuk IntegrasiIngat aturan Leibnitz untuk Integrasi
( ) ( ) ( ) ( )xxxxQ
xxxxQdyyxQ
xdyyxQ
xx
x
x
x ∂∂−
∂∂+
∂∂=
∂∂
∫∫)()(,)()(,,,
)(
)(
)(
)(
ααβββ
α
β
α
LONGWAVELONGWAVE
Dalam hal ini misalnya Q(x,y) adalah u atau v, α(x)=-h, β(x)=η. Dalam hal ini dy adalah dz pada persamaan sebelumnya.
( ) ( ) ( ) ( )xxxxQ
xxxxQdyyxQ
xdyyxQ
xx
x
x
x ∂∂+
∂∂−
∂∂=
∂∂
∫∫)()(,)()(,,,
)(
)(
)(
)(
ααβββ
α
β
α
0),,(),,( =−−+
∂∂+
∂∂=
∂∂+
∂∂+
∂∂
∫∫∫ −−−hyxwyxwdz
yvdz
xudz
zw
yv
xu
hhhη
ηηη
( ) ( ) ( ) ( )xxxyxQ
xxxyxQdzzyxQ
xdzzyxQ
xx
x
x
x ∂∂+
∂∂−
∂∂=
∂∂
∫∫)()(,,)()(,,,,,,
)(
)(
)(
)(
ααβββ
α
β
α
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∫
∫
−
−
=∂∂−−
∂∂−
∂∂+
−−+∂∂−−
∂∂−
∂∂
η
η
ηη
ηηη
h
h
yhhyxv
yyxvdzv
y
hyxwyxwxhhyxu
xyxudzu
x
0),,(),,(
),,(),,(),,(),,(
Secara Keseluruhan diperoleh persamaan Kontinuitas gelombang
( ) ( ) ( ) ( )xxxyxQ
xxxyxQdzzyxQ
xdzzyxQ
xx
x
x
x ∂∂+
∂∂−
∂∂=
∂∂
∫∫)()(,,)()(,,,,,,
)(
)(
)(
)(
ααβββ
α
β
α
( ) ( ) ( ) ( )xxxxQ
xxxxQdyyxQ
xdyyxQ
xx
x
x
x ∂∂+
∂∂−
∂∂=
∂∂
∫∫)()(,)()(,,,
)(
)(
)(
)(
ααβββ
α
β
α
0),,(),,( =−−+
∂∂+
∂∂=
∂∂+
∂∂+
∂∂
∫∫∫ −−−hyxwyxwdz
yvdz
xudz
zw
yv
xu
hhhη
ηηη
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Dengan asumsi bahwa Dengan asumsi bahwa
dzuh
Uh∫ −+
=η
η 1 dzv
hV
h∫−+=
η
η 1
( )[ ] ( )[ ]
0),,(),,(),,(),,(
),,(),,(
=−−+∂∂−−
∂∂−
∂∂−−
∂∂−+
∂∂++
∂∂
hyxwyxwxhhyxu
xyxu
yhhyxv
yyxvhV
yhU
x
ηηη
ηηηη
∫
∫
−
−
=∂∂−−
∂∂−
∂∂+
−−+∂∂−−
∂∂−
∂∂
η
η
ηη
ηηη
h
h
yhhyxv
yyxvdzv
y
hywyxwxhhyxu
xyxudzu
x
0),,(),,(
),,),,(),,(),,(
Asumsi ini inclusif menganggap aliran irrotasional
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Persamaan kinematika permukaan airPersamaan kinematika permukaan air
),,(),,(),,( ηηηηηη yxwy
yxvx
yxut
=∂∂+
∂∂+
∂∂
yhhyxv
xhhyxuhyxw
∂∂−−
∂∂−−=− ),,(),,(),,(
),,( ηηηηη yxwydt
dyxdt
dxtdt
d =∂∂+
∂∂+
∂∂=
Kecepatan vertikal di permukaan
Kecepatan vertikal di dasar
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( )[ ] ( )[ ]ty
hVx
hU∂
∂−=∂
+∂+∂
+∂ ηηη
( )[ ] ( )[ ]
0),,(),,(),,(),,(
),,(),,(
=−−+∂∂−−
∂∂−
∂∂−−
∂∂−+
∂∂++
∂∂
hyxwyxwxhhyxu
xyxu
yhhyxv
yyxvhV
yhU
x
ηηη
ηηηη
yhhyxv
xhhyxuhyxw
∂∂−−
∂∂−−=− ),,(),,(),,(
),,(),,(),,( ηηηηηη yxwy
yxvx
yxut
=∂∂+
∂∂+
∂∂
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Persamaan MomentumPersamaan Momentum
∂
∂+∂
∂+
∂∂+
∂∂−=
∂∂+
∂∂+
∂∂+
∂∂=
zyxxp
zuw
yuv
xuu
tu
dtdu zxyxxx τττ
ρρ11
xg
xp
∂∂−=
∂∂− η
ρ1
( )[ ] [ ]
( )ρ
τητττηρ
ηη
ηβηβη
hyx
hx
hg
hUVy
hUxt
hU
zxzxyxxx
xxxx
−−+
∂
∂+
∂∂++
∂∂+−=
+∂∂++
∂∂+
∂+∂
)()(1)(
)()( 2
Arah x
( )
( ) ∫∫
∫
−−
−
+=
+=
+=
ηη
η
ηηβ
ηβ
hhyx
hxx
dzuh
UdzuvUVh
dzuUh
1 ; 1
1 22
Integral vertikal dengan Leibnitz rule, dan jika
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Arah YArah Y
( )[ ] ( )[ ]
( ) ( ) ( )ρ
τητττη
ρηη
ηβηβη
hyx
hy
hg
hVy
hUVxt
hV
zyzyyyxy
yyyx
−−+
∂
∂+
∂∂
++∂∂+−=
+∂∂++
∂∂+
∂+∂
1)(
)( 2
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Dengan menganggap koreksi integrasi = 1 diperolehDengan menganggap koreksi integrasi = 1 diperoleh
( ) ( ) ( )[ ]hhyxx
gyUV
xUU
tU
zxzxyxxx −−
++
∂
∂+
∂∂
+∂∂−=
∂∂+
∂∂+
∂∂ τητ
ηρττ
ρη 11
( ) ( ) ( )[ ]hhyxy
gyVV
xVU
tV
zyzyyyxy −−
++
∂
∂+
∂∂
+∂∂−=
∂∂+
∂∂+
∂∂ τητ
ηρττ
ρη 11
Arah x
Arah Y
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( ) ( )ty
Vhx
Uh∂
∂−=∂
∂+∂
∂ η
xg
tU
∂∂−=
∂∂ η
yg
tV
∂∂−=
∂∂ η
2
2
2
2
2
22
tyxC
∂∂=
∂∂+
∂∂ ηηη
Persamaan gelombang panjang linier
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Gelombang linierGelombang linier
)cos(2
tkxH ση −=
)sin(2
tkxkHgt
U σ−=∂
∂
hCtkxkHgU ησ =−= )cos(
2
Apa bedanya dengan gelombang Airy ???
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Persamaan gelombang panjang Persamaan gelombang panjang dalam bentuk differensialdalam bentuk differensial
Masa jenis konstanMasa jenis konstan Kecepatan dan momentum horizontal dianggap Kecepatan dan momentum horizontal dianggap
merata sepanjang vertikal (sehingga hanya merata sepanjang vertikal (sehingga hanya berlaku untuk gelombang panjang)berlaku untuk gelombang panjang)
Air dianggap inviscid (tak punya kekentalan) Air dianggap inviscid (tak punya kekentalan) sehingga tak ada gesekansehingga tak ada gesekan
??????
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Shoaling gelombang panjangShoaling gelombang panjang
21
2
1
41
2
112
=
bb
hhHH
F= EnCb
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Gelombang tercampur akibat Gelombang tercampur akibat refleksirefleksi
kxtH
tkxHtkxH
ri
ri
coscos
)cos(2
)cos(2
σηηη
σσηηη
=+=
++−=+=
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2nodeπ=kx (5.31a)
4nodeLx =
(5.31b)
( ) |cos|2||2 klHl =η (5.32)
( )( ) |cos|
10kll
=ηη
(5.33)
Posisi Node
l
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Gelombang panjang pada saluran Gelombang panjang pada saluran dengan tampang bervariasidengan tampang bervariasi
Permasalahan : dapatkah dihitung dengan F=EnCb ??
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Gelombang panjang pada saluran Gelombang panjang pada saluran dengan tampang bervariasidengan tampang bervariasi
( )t
bx
Uhb∂
∂−=∂
∂ η Persamaan Kontinuitas (5.34a)
xgb
tUb
∂∂−=
∂∂ η Persamaan Momentum (5.34b)
( ) konstan)lebar untuk (tx
Uh∂
∂−=∂
∂ η
xg
tU
∂∂−=
∂∂ η
Saluran ke arah X sehingga
Persamaan ini bukan terjadi karena dikalikan dengan b, tetapi diturunkan dari persamaan alinya dengan mengambil lebar b bervariasi
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2
2
tb
tUhb
x ∂∂−=
∂∂
∂∂ η
Diferensialkan ke arah t diperoleh:
Dalam hal ini h dan b tidak berubah terhadap t (apa artinya ?)
Substitusikan persamaan momentum untuk memperoleh:
2
2
txbh
dxd
bg
∂∂=
∂∂ ηη
( )t
bx
Uhb∂
∂−=∂
∂ η
xgb
tUb
∂∂−=
∂∂ η
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If
( ) txtx σηη cos)(, =
Than
( ) ( ) 02 =+
x
dxxdbh
dxd
bg ηση
2
2
txbh
dxd
bg
∂∂=
∂∂ ηη
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( ) ( ) ( ) 0)( 22
2
=+
+ x
dxxdbh
dxxd
dxbhd
bg ησηη
( ) ( ) 02 =+
x
dxxdbh
dxd
bg ηση
( ) ( ) ( ) 0)( 22
2
=+
+ x
dxxdb
dxxd
dxbd
bgh ησηη
( ) ( ) ( ) 0)( 2
22
2
=+
+ x
cdxxd
dxxd
bdxbd ησηη
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( ) ( ) ( ) 01 22
2
=++ xkdx
xdxdx
xd ηηη
( ) ( ) ( ) 0)( 2
22
2
=+
+ x
cdxxd
dxxd
bdxbd ησηη
Jika b= ax maka
Selesaian dengan Bessel Function
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( ) ( ) ( ) 01 22
2
=++ xkdx
xdxdx
xd ηηη
Selesaian dengan Bessel Function
( ) ( ) ( )[ ] tkxYCkxJCtx ση cos, 0201 +=
C : Konstanta
Untuk x=0, maka Yo= tak berhingga, solusi ini ditanggalkan (tak mungkin)
( ) ( )[ ] tkxJCtx ση cos, 01=
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( ) ( )[ ] tkxJCtx ση cos, 01=
Jika gelombang sinusoidal, dengan tinggi H di x=l, maka
( ) ( )[ ] tHtklJCtl σση cos2
cos, 01 ==Jadi:
( )klJHC0
1 2=
( ) ( )( ) tklJkxHJtx ση cos
2,
0
0=
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kx
Jo(kx)
Misal kx (yaitu lokasi yang akan dicari) = 2maka Jo(kx) =
Harga kl sudah tertentu tergantung l (panjang corong) misal kl =5
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Bessel FunctionBessel Function
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Fluktuasi Muka Air di dalam Fluktuasi Muka Air di dalam Tapper ChannelTapper Channel
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Pantai BaronPantai Baron
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( ) ( ) tklJ
tx ση cos2
1,0
=
( )klJ0max 2
1=η
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Benarkah elevasi muka air dapat mencapai Benarkah elevasi muka air dapat mencapai tak berhingga ?tak berhingga ?
Apa yang berbeda antara persamaan Apa yang berbeda antara persamaan dengan kenyataan ?dengan kenyataan ?
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Wave Height from mouth to Channel EndWave Height from mouth to Channel End
Pola Tinggi Gelombang yang Masuk Tapchan Pada Arah 175o (MSL)
0
2
4
6
8
10
12
0 200 400 600 800 1000
Jarak Model (Cm)
H m
odel
(Cm
)
T=2 dtkT=2.4 dtkT=2.77 dtkT=2.77 dtkT=2.38 dtkT=2 dtkT=2 dtkT=2.38 dtkT=2.73
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Wave Height from mouth to Channel EndWave Height from mouth to Channel End
Pola Tinggi Gelombang yang Masuk Tapchan Pada Arah 191o
(MSL)
0.00
2.00
4.00
6.00
8.00
10.00
12.00
0 200 400 600 800 1000
Jarak Model (Cm)
H m
odel
(Cm
)
T=2 dtk
T=2.4 dtk
T=2.8 dtk
T=2 dtk
T=2.4 dtk
T=2.8 dtk
T=2.8 dtk
T=2.4 dtk
T=2 dtk
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Wave Height from mouth to Channel EndWave Height from mouth to Channel EndPola Tinggi Gelombang yang Masuk Tapchan Pada Arah 210o (MSL)
0
1
2
3
4
5
6
7
8
9
10
0 200 400 600 800 1000
Jarak model (Cm)
H m
odel
(Cm
)
T=2.8 dtkT=2.4 dtkT=2 dtkT=2 dtkT=2.4 dtkT=2.8 dtkT=2.8 dtkT=2.4 dtkT=2 dtk
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QuestionsQuestions
Bandingkan tapper channel dengan Bandingkan tapper channel dengan Kontinuitas Energi Fluks Kontinuitas Energi Fluks
1. Refleksi1. Refleksi 2. Tinggi Gelombang2. Tinggi Gelombang 3. Tinggi gelombang maximum3. Tinggi gelombang maximum 4. Phase4. Phase
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Refleksi dan Transmisi Refleksi dan Transmisi Gelombang PanjangGelombang Panjang
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Refleksi dan transmisi melalui Refleksi dan transmisi melalui perubahan mendadakperubahan mendadak
( ) ( )
( )122
111
cos2
cos2
cos2
∈+−==
∈+++−=+=
txkH
txkHtxkH
tt
rri
ri
σηη
σσηηη
transimisiincoming
refleksi
1 2
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Water level fluctuation at Water level fluctuation at xx = 0 = 0
tri ηηη =+
0sin2
sin2
sincos2
cos22
cos =
∈+∈−
∈−∈+ t
tr
rt
tr
ri HHtHHHt σσ
( ) ( )
( )tt
t
rri
ri
txkH
txkHtxkH
∈+−==
∈+++−=+=
σηη
σσηηη
22
111
cos2
cos2
cos2
ttrri HHH ∈=∈+ coscos
ttrr HH ∈−=∈ sinsin
This term should be =0
This term should be =0
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Continuity EquationContinuity Equation
Based on continuity Based on continuity equation at x=0equation at x=0
hCU η=21 )()( UbhUbh =
( ) tri CbCb ηηη 2211 =−
In the direction of the wave:
Of course at X=0, ( )ri ηη − tη=
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ttrri HCbHCbHCb ∈=∈− coscos 221111
ttrr HCbHCb ∈=∈ sinsin 2211
Mengantikan harga η, dan aljabar
Dengan memisalkan Kr=Hr/Hi dan Kt= Ht/Hi diperoleh
ttrr KK ∈=∈+ coscos1
ttrr CbCbKK ∈=∈− coscos1
11
22
ttrr KK ∈−=∈ sinsin
ttrr CbCbKK ∈=∈ sinsin
11
22
ttrri HHH ∈=∈+ coscosttrr HH ∈−=∈ sinsin( ) tri CbCb ηηη 2211 =−
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ttrr KK ∈−=∈ sinsin
ttrr CbCbKK ∈=∈ sinsin
11
22
0sin111
22 =∈
+ tt Cb
CbK
5.50 Xb2C2/b1C1 tambahkan ke 5.51 dengan πε n±=
tr KK ±=±1
11
221CbCbKK tr ±=
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πε n±=tr KK ±=±1
11
221CbCbKK tr ±=±
+1122 /1
2CbCb
Kt +±=
( )( )
+−
±=1/1/
1122
1122
CbCbCbCbK r
1122
1122
/1/1
CbCbCbCbK r +
−=
-
1122 /12
CbCbK t +
=
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ExamplesExamples A long wave propagates into a converging channel. At A long wave propagates into a converging channel. At
the mouth the wave height is 0.19m, The channel the mouth the wave height is 0.19m, The channel depth is uniform at 5m, T=16s. The width of the depth is uniform at 5m, T=16s. The width of the channel is 50m at the mouth and becomes zero at channel is 50m at the mouth and becomes zero at about 340m from the channel. about 340m from the channel.
1.1. Calculate the wave height at the end of the channel Calculate the wave height at the end of the channel and and
2.2. Calculate the phase difference of the wave at the Calculate the phase difference of the wave at the mouth and at the end of the channel. mouth and at the end of the channel.
3.3. Compare the situation with Tidal wave component Compare the situation with Tidal wave component (M2)(M2)
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ExamplesExamples The channel depth is uniform at 5m, T=16s. The channel depth is uniform at 5m, T=16s. C=(5*9.8)C=(5*9.8)0.50.5 =7.0m; L =7.0*16=112 m. =7.0m; L =7.0*16=112 m. k= 2*k= 2*ππ/L= 0.0561/L= 0.0561 The width of the channel is 50m at the mouth and The width of the channel is 50m at the mouth and
becomes zero at about 340m from the channel. becomes zero at about 340m from the channel. At the mouth, kl =0.0561*340=19At the mouth, kl =0.0561*340=19 At the end of the channel, kx=0At the end of the channel, kx=0 From Bessel function, the wave height at the mouth is From Bessel function, the wave height at the mouth is
0.19 of that at the end of the channel. Hence at the 0.19 of that at the end of the channel. Hence at the channel the wave height is 1.0m. No Phase different channel the wave height is 1.0m. No Phase different between the two locationsbetween the two locations
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ExamplesExamples
Long wave of 0.1m propagates along the Long wave of 0.1m propagates along the channel of 100m wide and 20m deep. The channel of 100m wide and 20m deep. The channel abruptly changes its cross section into channel abruptly changes its cross section into 20m wide and 2m deep. 20m wide and 2m deep.
1.1. Calculate the reflected and transmitted wave.Calculate the reflected and transmitted wave.2.2. Is the energy flux conserved?Is the energy flux conserved?3.3. Compare it with Greens law ?Compare it with Greens law ?
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Bessel FunctionBessel Function
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Refleksi
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Buktikan bahwa energi pada refleksi tetap Buktikan bahwa energi pada refleksi tetap (terkonservasi)(terkonservasi)
Pada kedalaman yang kecil di hilir, terjadi Pada kedalaman yang kecil di hilir, terjadi koefisien transmisi yang tinggi sementara tinggi koefisien transmisi yang tinggi sementara tinggi gelombang yang direfleksikan mendekati 100% gelombang yang direfleksikan mendekati 100% terangkan secara fisik (tentang konservasi terangkan secara fisik (tentang konservasi energi)energi)
Gelombang datang dari kedalaman 1m ke Gelombang datang dari kedalaman 1m ke kedalaman 1000m terangkan apa yang terjadikedalaman 1000m terangkan apa yang terjadi
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LONG WAVES WITH BOTTOM LONG WAVES WITH BOTTOM FRICTIONFRICTION
Based on Dean and Dalrymple, 1984Based on Dean and Dalrymple, 1984
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LONG WAVES WITH BOTTOM FRICTIONLONG WAVES WITH BOTTOM FRICTION
gRib ρτ =
gU
DLfh f 2
2
=
8
2fUb
ρτ =
gRigDiLh
gDfU f 8222 ===
Darcy Weisbach
Correlation between U,f and Ri
Correlation between U, f and τ
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8|| UfU
bρτ =
8
2fUb
ρτ =
Is it correct ?
Why do we need such form ?
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( )∑∞
+=0
0 cos tnaaUU n σ
Suppose
tUU m σcos=
In Fourier Series it can be rewritten as
This assumption limits the solution to sinusoidal wave form
Kita sudah mengasumsi bahwa U merupakan fungsi periodik
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dtttT
UaTm |cos|cos0
2
0 σσ∫=
dt σtnσttTUa
Tmn coscoscos2
0
2
∫= σ
00 =aπ3
8 2
1mU
a = 02 =a
π158 2
3mU
a =
πρ
σπ
ρτ
3cos
3
2 UfUt
fU mmb ==
See : Fourier Series
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( )AU
xg
hh
xg
tU zx −
∂∂−=
−−
∂∂−=
∂∂ η
ρτη
2
2
2
2
xgh
tA
t ∂∂=
∂∂+
∂∂ ηηη
xg
tU
∂∂−=
∂∂ η
No Friction !!
( ) ( ) ( )[ ]hhyxx
gyUV
xUU
tU
zxzxyxxx −−
++
∂
∂+
∂∂
+∂∂−=
∂∂+
∂∂+
∂∂ τητ
ηρττ
ρη 11
( ) ( )ty
Vhx
Uh∂
∂−=∂
∂+∂
∂ η
h=constant
( )tx
Uh∂
∂−=∂
∂ η
2
22
tg
txUgh
∂∂−=
∂∂∂ η
xAUgh
xhg
xtUgh
∂∂−
∂∂−=
∂∂∂
2
22
2 η
If A is not a function of x
xUghA
xhg
xtUgh
∂∂−
∂∂−=
∂∂∂
2
22
2 η
tgA
xhg
xtUgh
∂∂+
∂∂−=
∂∂∂ ηη
2
22
2
tgA
xhg
tg
∂∂+
∂∂−=
∂∂− ηηη
2
22
2
2Long wave with bottom
friction
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2
2
2
2
xgh
tA
t ∂∂=
∂∂+
∂∂ ηηη
( ) xktfHII
I cos2
ση =
022
2
=++ fghkdtdfA
dtfd
I
( ) xktAeHI
I
tAI cos411cos
2
2
12/
−= −
σση
Suppose the solution is:
The general wave equation should be :
xkafkx II cos2
2
2
−=∂∂ η
0coscoscos 22
2
=++ xkfghakdtdfxAka
dtfdxka IIII
Total Solution :
Standing wave
( ) tAe 2/−
This solution suggests that the wave height
reduces due to energy losses depending on the
decay modulus
k1 dan σ1 merepresentasikan adanya perubahan panjang
gelombang dan perioda karena adanya gesekan
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( ) xktAeHI
II
tAI cos411cos
2
22/
−= −
σση
kxteHr
tI i coscos2
ση σ−=
2A
i ≡σ2
411
−=
IIr
Aσ
σσ
Total solution:
ghkCk IIII ==σWith
Subscript I > undamped condition
with
Refer to σI and A
Important : A is not Area,
Remember that :
kxtH coscos2
ση =
Ii σσ ≠
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∫ ∂∂−= dx
thU η1
( ) xktehk
HU Irt
irI
I i sinsin2
22 ∈++= − σσσ σ
r
i
σσ1tan −∈ =
( )( )
TATA iI eeetTt σσπ
ηη −−− ===+ )/()2/(
With :
kxteHr
tI i coscos2
ση σ−=
For partial reflection
What about the particle velocity of wave with bottom friction?
damping
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Example of wave damping due to bottom frictionExample of wave damping due to bottom frictionShiau dan Rumer (1970) conducted experiment in square Shiau dan Rumer (1970) conducted experiment in square
basin of shallow water (0.15<h<8.5 cm).basin of shallow water (0.15<h<8.5 cm).The problem is directly analogous with the case under The problem is directly analogous with the case under
consideration. If laminar flow is assumed the Stokes consideration. If laminar flow is assumed the Stokes formula is applied (Dean and Dalrymple 1984)formula is applied (Dean and Dalrymple 1984)
mbm Uσ υρτ =
From equation 5.74, 5.67 and 5.66 yield:
hA σ υ=
Ub σ υρτ =
hfUA m
π3=
πρτ
3m
bfUU=
hUA b
ρτ=
AUx
gt
U −∂∂−=
∂∂ η
πρτ
3UfUm
b =
For laminar >>General eq for laminar
and turbulence
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From previous equation the modulus decay is found ( as From previous equation the modulus decay is found ( as also suggested by Shiau dan Rumer): also suggested by Shiau dan Rumer):
( )( )
hA
eeetTt TATA iI
σ υ
ηη σσπ
=
===+ −−− )/()2/(
I
Aσπα =
4/1Pvh
πσ
πα ==
Substituting to the above equation for A, yields :
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Gelombang berjalan dengan pengaruh kekasaran dasarGelombang berjalan dengan pengaruh kekasaran dasar
( )txkeHr
xkI i ση −= − cos2
+≈
+
+=
22/1
2
81111
2 σσAkAkk I
Ir
σσAkAkk II
i 211
2
2/12
≈
−
+=
( )( )
( ) ( )σππ
ηη //2 AkkkL eee
xLx
ri −−− ≈==+
kxteHr
tI i coscos2
ση σ−=Standing WaveStanding Wave
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Gelombang berjalan dengan pengaruh kekasaran Gelombang berjalan dengan pengaruh kekasaran dasardasar
xkie−
( )txkeHr
xkI i ση −= − cos2
Perhatikan bahwa gelombang diasumsikan mengikuti persamaan dengan decay modulus
Pangkat –kx, menunjukkan bahwa gelombang akan berkurang tingginya jika x positip. Penguranganannya adalah exponensial.
rkPanjang gelombang juga berubah dan ditentukan oleh
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Harga Kr dan ki ditentukan seperti pada gelombang berdiri, dan Harga Kr dan ki ditentukan seperti pada gelombang berdiri, dan diperoleh:diperoleh:
+≈
+
+=
22/1
2
81111
2 σσAkAkk I
Ir
σσAkAkk II
i 211
2
2/12
≈
−
+=
Persamaan gelombang terhadap x diperoleh :
( )( )
( ) ( )σππ
ηη //2 AkkLk eee
xLx
Iii −−− ≈==+
σA
kk
I
i
2=
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Gelombang melalui rumpun bakauGelombang melalui rumpun bakauGelombang melalui Pemecah gelombang porousGelombang melalui Pemecah gelombang porous
Gelombang melalui Groin porousGelombang melalui Groin porous
−
−= LBζ0,2738
e1aKE
−
−= LBζm
e1aKE
( )σπ /Ae−
BAKAU
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Gelombang melalui Hutan Gelombang melalui Hutan BakauBakau
−
−= LBζ0,2738
e1aKE
( )σπ /Ae−
B
L
ζB/L= A
2738.0=σπ
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KE t = e -0.2738 ζ B/L r 2 = 0.8783
0.00
0.20
0.40
0.60
0.80
1.00
1.20
0.00 2.00 4.00 6.00 8.00 10.00 12.00
ζ B/L
KEt=
(Ht/H
i)2 0.20
0.00
0.40
0.60
0.80
1.00
-0.20
KEa=
1-KE
t
Gelombang melalui rumpun bakauGelombang melalui rumpun bakau
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Gelombang melalui Groin atau Gelombang melalui Groin atau BreakwaterBreakwater
−
= LBR ζ
eo
iηη
( )σπ /Ae−
B
L
+ Refleksi
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Kecepatan partikel pada gelombang berjalan Kecepatan partikel pada gelombang berjalan dengan pengaruh gesekan dasardengan pengaruh gesekan dasar
hCtkxkHgU ησ
σ=−= )cos(
2
( )∈−−+
=−
txkkkh
eHU r
ri
xkI
i
σσ cos2 22
r
i
kk1tan −∈ =
Apa arti ∈
)cos(2
)cos(2
)cos(2
2 tkxkh
Htkxkh
Hctkxkh
HghU σσσσ
σσ
−=−=−=
)cos(2
tkxkh
HU σσ −=
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Kecepatan partikel pada gelombang berjalan Kecepatan partikel pada gelombang berjalan dengan pengaruh gesekan dasardengan pengaruh gesekan dasar
( )∈−−+
=−
txkkkh
eHU r
ri
xkI
i
σσ cos2 22
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Efek Geostrophic pada gelombang Efek Geostrophic pada gelombang panjangpanjang Fakta:Fakta:
bumi berrotasi dengan kecepatan sudutbumi berrotasi dengan kecepatan sudut
srad /1027.7 5−×=ω Pengaruh:Pengaruh:
gaya Coriolisgaya Coriolis
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xgVf
yUV
xUU
tU
c ∂∂−=−
∂∂+
∂∂+
∂∂ η
ygUf
yVV
xVU
tV
c ∂∂−=+
∂∂+
∂∂+
∂∂ η
( ) ( ) 0=∂
+∂+∂
+∂+∂
∂y
hVx
hUt
ηηη
Pengaruhnya pada persamaan gelombang panjang tanpa kekasaran dasar adalah
srad /1027.7 5−×=ω
φω sin2=cf
Remain unchanged !!!
xgAU
tU
∂∂−=+
∂∂ η
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xgVf
yUV
xUU
tU
c ∂∂−=−
∂∂+
∂∂+
∂∂ η
ygUf
yVV
xVU
tV
c ∂∂−=+
∂∂+
∂∂+
∂∂ η
Jika gelombang berjalan ke arah X (dalam hal ini ke arah timur atau barat) maka tidak ada kecepatan ke arah Y sehingga gaya Corioli hanya
berpengaruh pada satu persamaan momentum
xg
yUV
xUU
tU
∂∂−=
∂∂+
∂∂+
∂∂ η
ygUf
yVV
xVU
tV
c ∂∂−=+
∂∂+
∂∂+
∂∂ η
Tak berubah
berubah
ygUfc ∂
∂−= ηKarena V =0, maka derivasinya =0 >>
F `V =0
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xgVf
yUV
xUU
tU
c ∂∂−=−
∂∂+
∂∂+
∂∂ η
ygUf
yVV
xVU
tV
c ∂∂−=+
∂∂+
∂∂+
∂∂ η
Jika gelombang berjalan ke arah Y (dalam hal ini ke arah utara atau selatan) maka tidak ada kecepatan ke arah X sehingga gaya Corioli hanya
berpengaruh pada satu persamaan momentum
Tak berubah
berubah
Karena U =0, maka derivasinya =0 >>
xgVf
yUV
xUU
tU
c ∂∂−=−
∂∂+
∂∂+
∂∂ η
yg
yVV
xVU
tV
∂∂−=
∂∂+
∂∂+
∂∂ η
xgVfc ∂
∂−=− η
![Page 77: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/77.jpg)
Dengan linierisasi dapat diasumsikan Dengan linierisasi dapat diasumsikan persamaan gelombang panjang ke arah Xpersamaan gelombang panjang ke arah X
( ) ( )tkxy σηη −=∧
cos
( ) ( )tkxyhCU ση −=
∧cos
dyd
hgCf c
∧∧
=− ηη
CyfceH /
2−
∧=η
( )tkxeH Cyfc ση −= − cos2
/ ( )tkxehCHU Cyfc σ−= − cos
2/
Persamaan momentum ke arah Y:
![Page 78: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/78.jpg)
Persamaan gelombang Persamaan gelombang panjang ke arah Xpanjang ke arah X
)cos(2
)cos(2
tkxhCHtkx
khHU σσσ −=−=
( )tkxeH Cyfc ση −= − cos2
/
( )tkxehCHU Cyfc σ−= − cos
2/
φω sin2=cf
The Kelvin wave
![Page 79: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/79.jpg)
( )tkxeH Cyfc ση −= − cos2
/
The Kelvin wave
Gelombang I
( )tkxeH Cyfc ση += cos2
/Gelombang II
( ) ( )tkxeHtkxeH CyfCyf cc σση +−−= − cos2
cos2
//
)cos(2
)cos(2
tkxhCHtkx
khHU σσσ −=−=
![Page 80: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/80.jpg)
kxtC
yfc tancottanh σ−=
tkxC
yf c σcot−=
0=∂
∂t
ηPada garis dengan elevasi maksimum,
( ) ( )tkxeHtkxeH CyfCyf cc σση +−−= − cos2
cos2
//
Di lokasi dekat titik (0,0), atau x dan y mendekati nol, diperoleh
tf
Ckxyc
σcot−=
![Page 81: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/81.jpg)
tf
Ckxyc
σcot−=
![Page 82: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/82.jpg)
Storm surgeStorm surge
Bisa mencapai 6 meterBisa mencapai 6 meter Membahayakan pertambakanMembahayakan pertambakan Pencemaran lingkunganPencemaran lingkungan
Kejadian oleh:Kejadian oleh: Gesekan angin terus menerus (berjam-Gesekan angin terus menerus (berjam-
jam) dengan arah ke pantai.jam) dengan arah ke pantai.
![Page 83: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/83.jpg)
angin
X
![Page 84: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/84.jpg)
|W|Wkw ρτ =
>
−×+×
≤×= −
−
cc
c
WW
Wk
W ,W
11025.2101.2
|W| ,101.22
66-
6
Gaya gesek angin
Wc=5.6 m/s
θττ coswxw =
X
Y
θPantai
![Page 85: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/85.jpg)
|W|Wkw ρτ =
( ) ( ) ( )[ ]hzxzxhxg
tU
−−+
+∂∂−=
∂∂ ττ
ηρη
η1
( ) ( ) ( )hzxzxzxn −−≡ τττ ηη
( )
( )ηττ
zx
hzxn −−= 1
θττ coswxw =
Persamaan momentum gelombang linier :
Pada keadaan sudah stabil, U konstan, karena gaya angin dapat dilawan oleh gaya hidrostatik. Ini berarti bahwa deverensi U juga nol. Karena tak mungkin menggunakan U untuk menyatakan gaya gesek,
maka
Atau
![Page 86: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/86.jpg)
( )( )ηρ
ητη+
=∂∂
hgn
xzx
Persamaan momentum, gelombang dalam kondisi steady
n=1.15 hingga 1.30 (SPM,1977)
![Page 87: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/87.jpg)
( )( )ηρ
ητη+
=∂∂
hgn
xzx
( )g
nx
h xw
ρτηη =
∂∂+
( )g
ndx
hd xw
ρτη
=+ 2
0
21
( ) Cg
xnh xw +=+
ρτ
η22
0
( ) 20
20
2h
gxn
h xw +=+ρτ
η
( ) 020
2h
gxn
hx xw −++=ρτ
η
Wind set up, jika lebar =l, kedalaman konstan
![Page 88: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/88.jpg)
( ) 12112
1 200
−+=−+=lAx
lx
ghln
hx xw
ρτη
Dalam bentuk variabel tak berdimensi
2o
w
ghln
A x
ρτ
=Dengan
![Page 89: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/89.jpg)
ContohContoh
Kedalaman laut dianggap homogen=100mKedalaman laut dianggap homogen=100m Jarak yang dipengaruhi = 600 kmJarak yang dipengaruhi = 600 km Asumsi n=1.25Asumsi n=1.25 Kecepatan angin= 20 m/sKecepatan angin= 20 m/s Berapa wind setup ??Berapa wind setup ??
![Page 90: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/90.jpg)
62
66- 1064.110
6.511025.2101.2 −− ×=
−×+×=k
0048.01010
106106.625.14
54
=⋅
⋅⋅⋅⋅=−
A
46 106.64001064.1 −− ⋅=⋅=wτ
( ) 1106
1060048.021121 5
5
0
−⋅
⋅⋅⋅+=−+=lAx
hxη
meterhl 5.00048.0 0 =⋅=η
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![Page 92: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/92.jpg)
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meterhl 5.00048.0 0 =⋅=η
Jika dasar laut miring, maka wind setup akan lebih besar. Misal untuk A=0.01, WIndsetup untuk dasar miring akan 3 kali lebih
besar. Jika A=0.05, Wind setup mencapai hampir 3 kali.
Jadi pada kondisi hitungan di atas, wind setup akan mencapai lebih dari 1.5 m jika dasar laut
miring
![Page 96: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/96.jpg)
Tekanan bergerak dengan kecepatan U, Tekanan bergerak dengan kecepatan U, atau merupakan fungsi Uatau merupakan fungsi U
Tekanan juga merupakan fungsi X atau Tekanan juga merupakan fungsi X atau lokasilokasi
Tekanan merupakan fungsi tTekanan merupakan fungsi t Pada saat t=0, x=0 Po =f(0)Pada saat t=0, x=0 Po =f(0)
( )xUtfP −=0
![Page 97: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/97.jpg)
T= 0
T= 5
T= 10
xpo
∂∂
( )xUtfP −=0
X
xph
xgh
tuh
∂∂
−∂∂−=
∂∂ 0
ρη
xg
tU
∂∂−=
∂∂ η
![Page 98: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/98.jpg)
xph
xgh
tuh
∂∂
−∂∂−=
∂∂ 0
ρη
UhhUuQ −=+−= ))(( η
hU
hUu η
ηη ≈+
=
![Page 99: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/99.jpg)
)( xUtG −=η
xU
t ∂∂−=
∂∂ ηη
Asumsi solusi
( )xUtfP −=0
![Page 100: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/100.jpg)
xU
t ∂∂−=
∂∂ ηη
( )xphghU
x ∂∂
−=−∂∂− 02
ρη
ghUp
ho
−= 2
/ ρη
gp
s ρη 0=
xph
xgh
tuh
∂∂
−∂∂−=
∂∂ 0
ρη
hU
hUu η
ηη ≈+
=
![Page 101: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/101.jpg)
ghUp
ho
−= 2
/ ρη
ghU >2 0>ηghU <2 0<η
ghU =2 ∞=η
![Page 102: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/102.jpg)
xg
tu
∂∂
−=∂∂ 1η
)(00 xUtf −=η
)(11 xUtf −=η
( )h
UhUu 01
01
01
)()( ηη
ηηηη −
≈−+
−=
![Page 103: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/103.jpg)
U
![Page 104: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/104.jpg)
hghUU
x0
2
21 ηη ∂
−
=∂
∂
ghUU
−= 2
2
01 ηη
ghU >2 01 >ηghU <2 01 <η
ghU =2 ∞=1η
![Page 105: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/105.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
∂∂+
∂∂+
∂∂−=
∂∂
2
2
2
21zu
xu
xp
tu ν
ρ
gzw
xw
zp
tw −
∂∂+
∂∂+
∂∂−=
∂∂
2
2
2
21 νρ
Navier-Stokes Linier
kxx
'
=kzz
'
=σ
'tt =
'uau σ= 'gapp ρ=
To examine the relative important of the terms, it is necessary to put them in to non dimensional variables
Non dimensional
![Page 106: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/106.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
∂∂+
∂∂+
∂∂−=
∂∂
2'
'2
2'
'22
'
'
2'
'
zu
xuk
xpgk
tu
σν
σ
Inverse square of Froude number
( ) ( )21
2
2
2
2
2 /22
)/2( CgkL
gLTTLggk −===ππ
πσ
1/ −gkC
0 to 1
![Page 107: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/107.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
∂∂+
∂∂+
∂∂−=
∂∂
2'
'2
2'
'22
'
'
2'
'
zu
xuk
xpgk
tu
σν
σ
2'
'2
22'
'22
'
'
2'
'
zu
xuk
xpgk
tu
∂∂+
∂∂+
∂∂−=
∂∂
σ δν
σν
σ
Mengingat di dekat dasar u berubah cepat terhadap vertikal. Skala vertikal di dekat dasar perlu dibedakan. Jika digunakan skala vertikal z=δz’ dengan δ ketebalan
daerah yang perubahan u sangat cepat
kzz
'
=
δ
Ingat bahwa : Khusus daerah dekat dasar
![Page 108: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/108.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
2'
'2
22'
'22
'
'
2'
'
zu
xuk
xpgk
tu
∂∂+
∂∂+
∂∂−=
∂∂
σ δν
σν
σ
Bagian akhir ruas kanan akan berkisar pada angka 1 jika
συδ ≈
Daerah dengan kecepatan u berubah sangat cepat tersebut merupakan daerah perubahan dari laminer ke turbulen, sehingga
disebut sebagai lapis batas laminer (Laminar boundary layer)
![Page 109: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/109.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
rp uuu +=
xp
tu p
∂∂−=
∂∂
ρ1
U dipisahkan menjadi dua bagian
2
2
zu
tu rr
∂∂=
∂∂ ν
Bagian irrotasional memenuhi persamaan
Euler
Bagian rotasional didekati dengan
![Page 110: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/110.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
)(
cosh)(cosh tkxi
p ekh
zhkgaku σ
σ−+=
U irrotasional sudah diperoleh
Dalam bilangan kompleks
Hanya bagian real yang digunakan
)cos(cosh
)(cosh tkxkh
zhkgaku p σσ
−+=
![Page 111: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/111.jpg)
Waves Over Real SeabedsWaves Over Real SeabedsUr diperoleh dengan metoda separasi variabel (lihat contoh Bab 3)
( ) ( )tkxihzir eAeu σνσ −+−= /
( ) ( )tkxihzir eAeu σνσ −+−−= 2/)1(
Arti dari bilangan kompleks berpangkat (z+h) menunjukkan adanya kehilangan energi (exponensial negatif). Sedang bagian terakhir ruas
kanan adalah osilasi sinusoidal. Jadi Ur berkurang terhadap (z+h)
![Page 112: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/112.jpg)
Waves Over Real SeabedsWaves Over Real Seabeds
khgakA
cosh1
σ−=
Dengan memanfaatkan kondisi batas dasar tidak bergerak (no slip boundary condition), yaitu pada z=-h, u=0 maka A dapat ditentukan
[ ] −−−+−= etkxzhkkh
gaku )cos()(coshcosh
σσ
Masukkan harga z =-h
pada persamaan sebelumnya
![Page 113: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/113.jpg)
Wave over real seabedWave over real seabed
( ) ( )tkxihzir eAeu σνσ −+−−= 2/)1(
Bisa diganti dengan
( ) ( )tkxihzir ee
khgakU σνσ
σ−+−−−= 2/)1(
cosh1
atau
( ) ( )( ))2/2/
cosh1 hztkxihz
r eekh
gakU ++−+−−= νσσνσ
σ
![Page 114: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/114.jpg)
Wave over real sea bedWave over real sea bed The real part of the horizontal velocity u is The real part of the horizontal velocity u is
thereforetherefore
( ) ( ) ( ) ( )
++−−−+= +− hztkxetkxzhk
khgaku hz
υσσσ
συσ
2coscoscosh
cosh2/
Phase shift
Contoh
![Page 115: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/115.jpg)
Water waves over a viscous mud Water waves over a viscous mud bottombottom
The bottom is not rigidThe bottom is not rigid The bottom is more like another fluid of The bottom is more like another fluid of
different densitydifferent density In the upper fluid region a solution may be In the upper fluid region a solution may be
sought :sought :
( ) ( ) ( )( ) )(1 sinhcosh,, tkxiezhkBzhkAtzx σφ −+++=
![Page 116: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/116.jpg)
Upper fluidUpper fluid( ) ( ) ( )( ) )(
1 sinhcosh,, tkxiezhkBzhkAtzx σφ −+++=
![Page 117: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/117.jpg)
Upper fluidUpper fluid( ) ( ) ( )( ) )(
1 sinhcosh,, tkxiezhkBzhkAtzx σφ −+++=
σoigakhBkhA =+ sinhcosh
It satisfy the Laplace Equation
0
1
=∂∂=
ztgφη
)( tkxioea ση −=
![Page 118: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/118.jpg)
Upper fluidUpper fluid( ) ( ) ( )( ) )(
1 sinhcosh,, tkxiezhkBzhkAtzx σφ −+++=It satisfy the Laplace Equation
kaikhBkhA oσ=+ coshsinh
tz z ∂∂=
∂∂−
=
ηφ
0
)( tkxioea ση −=
![Page 119: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/119.jpg)
Upper fluidUpper fluid
kaikhBkhA oσ=+ coshsinh
σoigakhBkhA =+ sinhcosh
khkaikhBkhkhA o coshcoshcoshsinh 2 σ=+
khigakhBkhkhA o sinhsinhsinhcosh 2
σ=+
khkaikhigakhkhB oo coshsinh)cosh(sinh 22 σ
σ−=−
khigakhkaiB oo sinhcosh
σσ −=
( )khgkk
khiaB o tanhcosh 2 −= σσ
![Page 120: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/120.jpg)
Upper fluidUpper fluid
( )khgkk
khiaB o tanhcosh 2 −= σσ
( )khgkk
khiaA o tanhcosh 2σσ
−=
If we try to impose the bottom boundary condition as in the rigid bottom that is
0=∂∂
−= hzzφ ( ) ( ) ( )( ) )(
1 sinhcosh,, tkxiezhkBzhkAtzx σφ −+++=
( ) 00cosh0sinh )( =+ − tkxieBAk σ
B should equals zero
We have 3 unknowns to solve (A,B, k)
![Page 121: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/121.jpg)
Upper fluidUpper fluid
( ) 0tanhcosh 2 =−= khgkk
khiaB o σσ
0tanh2 =− khgkσ
khgk tanh2 =σ
Which is exactly the same as before (in chapter three)
![Page 122: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/122.jpg)
Mud fluidMud fluid
( ) )()(2 ,, tkxihzk edetzx σφ −+=
The function is chosen as it has to be periodic spatially and temporally (driven by the periodic water wave)
1 at z=-h (at the boundary between the two fluids
And becoming less than 1 for deeper location in the mud.
)()(2)1(2
tkxihzi eqeu σνσ −++−=Boundary layer correction for 2φ
![Page 123: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/123.jpg)
Mud fluidMud fluid
zzt ∂∂−=
∂∂−=
∂∂ 21 φφχ
At the boundary, the vertical velocity in each region should be the same
),( txhz χ+−=On
)(),( tkxioemtx σχ −=
![Page 124: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/124.jpg)
Mud fluidMud fluid
zzt ∂∂−=
∂∂−=
∂∂ 21 φφχLinearizing the kinematic boundary condition yields
hz −=On
dkkBimo −=−=− σ
σikBmo −= dB =
![Page 125: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/125.jpg)
Mud FluidMud Fluid
Continuity of pressureContinuity of pressure21 pp = On z=-χ+h
( ) ghgzt
gzt 212
221
11 ρρρφρρφρ −−−
∂∂=−
∂∂
Requiered to accommodate densities
difference
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Mud FluidMud FluidLinearizing On z=-h
χρφρχρφρ gt
gt 2
221
11 −
∂∂=−
∂∂
Substituting for z,, 21 φφ
Yields the relation between A and B
BgkgkA
+
−= 22
1
2 1σσρ
ρ
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Mud FluidMud Fluid
BgkgkA
+
−= 22
1
2 1σσρ
ρ
( )khgkk
khiaB o tanhcosh 2 −= σσ
( )khgkk
khiaA o tanhcosh 2σσ
−=
But previously we have
![Page 128: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/128.jpg)
Mud FluidMud Fluid( ) ( ) 0tanh1tanh1tanh 2
1
22
1
24
1
2 =
−++−
+ khgkkhgkkh
ρρσ
ρρσ
ρρ
( ) 0tanh1tanh1
2
1
222 =
−−
+− khgkkhgk
ρρ
ρρσσ
gk=2σ
+
−
=kh
khgk
tanh
tanh1
1
2
1
2
2
ρρ
ρρ
σ
![Page 129: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/129.jpg)
Dispersion relationshipDispersion relationship
kh
gk
2σkhtanh
gk=2σ
51
2 =ρρ
21
2 =ρρ
Deep water wave, mud have no effect
σ2/gk=(ρ2/ρ1-1)/(ρ2/ρ1+1)
![Page 130: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/130.jpg)
Amplitudes of wavesAmplitudes of waves
kh
o
o ema =
kh
o
o ema −
−−= 1
1
2
ρρ
Surface wave
Mud wave
oo ma >
oo am >
In phase
Out of phase (180 degree)
![Page 131: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/131.jpg)
Physical understandingPhysical understanding
The first wave modeThe first wave mode)()(
21tkxizhk eAe σφφ −+==
The presence of the more dense fluid at the bottom has no effect on the wave motion.
The wave is classified as deep water wave
The mud layer should be infinitely deep
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Physical understandingPhysical understanding The second wave modeThe second wave mode
Upper layer
Lower layer
False bottom
0/ =∂∂= ozz
zφ
![Page 133: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/133.jpg)
Physical understandingPhysical understanding The second wave modeThe second wave mode
Upper layer
Lower layer
False bottom
0/ =∂∂= ozz
zφ
ozkgk tanh2 =σ
![Page 134: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/134.jpg)
DampingDampingMatching the horizontal velocities at the interface
221 u
xx+
∂∂−=
∂∂− φφ At z=-h
)(2 Adiku −=
)( 22 gkeau kho −= σ
σ
![Page 135: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/135.jpg)
At elevation ZoAt elevation Zo At the false bottom the vertical velocity is 0 At the false bottom the vertical velocity is 0
therefore therefore
0=∂∂
= zozzφ
( ) ( ) ( )( ) )(1 sinhcosh,, tkxiezhkBzhkAtzx σφ −+++=
0)(cosh =+ zhkBk0)(sinh =+ zhkAk
( ) ( ) ( )( ) 0coshsinh,, )(1 =+++=∂
∂ − tkxiezhkBkzhkAkz
tzx σφ
1)(cosh ≥+ zhk
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At elevation ZoAt elevation Zo At the false bottom the vertical velocity is 0 At the false bottom the vertical velocity is 0
therefore therefore
0=∂∂
= zozzφ
0)(cosh =+ zhkBk
1)(cosh ≥+ zhk0=B
![Page 137: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/137.jpg)
At elevation ZoAt elevation Zo0)(cosh =+ zhkBk
1)(cosh ≥+ zhk0=B
( ) 0tanhcosh 2 =−= o
oo zkgkk
zkiaB σ
σ0tanh2 =− ozkgkσ
ozkgk tanh2 =σ
![Page 138: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/138.jpg)
Wave Over Porous BedWave Over Porous Bed
Conservation of mass Conservation of mass in the porous mediain the porous media
0=•∇ u
spku ∇−=µ
Darcy Law
Where p is the pore pressure gradient in the soil
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Wave Over Porous BedWave Over Porous Bed Substituing Darcy law in the Conservation of Substituing Darcy law in the Conservation of
mass in the porous media yieldmass in the porous media yield
0=
∇−•∇ spkµ
02 =∇− spkµ
P satisfy Laplace equation as does the velocity potential φ
![Page 140: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/140.jpg)
Wave Over Porous BedWave Over Porous Bed
02 =∇− spkµ
P satisfy Laplace equation as does the velocity potential φ
[ ] )()(sinh)(cosh),,( tkxiezhkBzhkAtzx σφ −+++=
)()(),,( tkxizhk eDetzxp σ−+=
Assumed solution for both φ and p are
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Wave Over Porous BedWave Over Porous Bed[ ] )()(sinh)(cosh),,( tkxiezhkBzhkAtzx σφ −+++=
)()(),,( tkxizhk eDetzxp σ−+=
( ) ( )hxphxp s −=− ,,Dynamic Boundary condition at the interface
( )hxpt s
hz
−=∂∂
−=
,φρ
Rewriting p(x,-h) at z=-h
DAi =− σ ρ
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Wave Over Porous BedWave Over Porous Bed[ ] )()(sinh)(cosh),,( tkxiezhkBzhkAtzx σφ −+++=
)()(),,( tkxizhk eDetzxp σ−+=Kinematic Boundary condition at the interface
hz
s
hz zpK
z −=−= ∂∂−=
∂∂−
µφ
µKDB =
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Wave Over Porous BedWave Over Porous Bed[ ] )()(sinh)(cosh),,( tkxiezhkBzhkAtzx σφ −+++=
)()(),,( tkxizhk eDetzxp σ−+=Dynamic Boundary condition at the surface
( ) )(sincosh1 tkxiekhBkhAg
itg
σσφη −+−=∂∂=
( ) )()(sincosh tkxitkxi aeekhBkhAg
i σσσ −− =+
( ) akhBkhAg
i =+ sincoshσ
![Page 144: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/144.jpg)
Wave Over Porous BedWave Over Porous BedDynamic Boundary condition at the surface
( ) akhBkhAg
i =+− sincoshσ
Substituting A and B
DAi =− σ ρ
µKDB =akhKDkh
iD
gi =
− sincoshµσ ρ
σ
or
σσµ
σσσ ρ
iigakhK
gikhi
iD =
− sincosh1
![Page 145: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/145.jpg)
Wave Over Porous BedWave Over Porous Bed
DAi =− σ ρ
µKDB =
σσµ
σσσ ρ
iigakhK
gikhi
iD =
− sincosh1
gakhKg
ikhD ρµ
ρ σ =
− sincosh
gakhKg
iD ρµ
ρ σ =
− tanh1
1
tanh1cosh−
−= khKikhgaDµ
ρ σρ
![Page 146: Teori Gelombang 2 New.pdf](https://reader034.vdocuments.site/reader034/viewer/2022042508/55cf992c550346d0339bff54/html5/thumbnails/146.jpg)
Wave Over Porous BedWave Over Porous Bed
DAi =− σ ρµ
KDB =
zt ∂∂−=
∂∂ φη
khBkkhAkai coshsinh +=σ
1
tanh1cosh−
−= khKikhgaDµ
ρ σρ
Substitution for A and B in term of D yields
−=
−
υσ
υσσ KikhgkkhKi tanhtanh12
( )khgkKikhgk tanhtanh 22 συ
σσ −
−=−
Complex k
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Wave Over Porous BedWave Over Porous Bed( )khgkKikhgk tanhtanh 22 σ
νσσ −
−=−
Complex k
ri kkk +=Real wave number
Complex wave number indicates damping
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Wave Over Porous BedWave Over Porous Bed
ri kikk +=Real wave number
Complex wave number indicates damping
( ) ( ) ( ) ( )xktxkio
txikkio
irir eeaeatx −−−+ == σση )(,
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Wave Over Porous BedWave Over Porous BedIn intermediate waterdepth
hkgk rr tanh2 ≅σ
( )hkhk
kKkrr
ri 2sinh2
/2+
≅ νσ
In shallow waterdepth
( ) ( )h
Khkk oi 2
/1 νσ−=2/12
2
2
4/11
−−=
hgKgh
ghkr ν
σσ
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Wave Over Porous BedWave Over Porous BedIn shallow waterdepth
( ) ( )h
Khkk oi 2
/1 νσ−=2/12
2
2
4/11
−−=
hgKgh
ghkr ν
σσ
+
+=
σν
νσ
22sinh22
rrr
ri kK
hkhkkk
Liu (1973) incorporating laminar boundary layer to avoid velocity discontinuity at the fluid-soil interface
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Wave Over Porous BedWave Over Porous Bed
In shallow waterdepth ( ) ( )h
Khkk oi 2
/1 νσ−=2/12
2
2
4/11
−−=
hgKgh
ghkr ν
σσ
hkgk rr tanh2 ≅σ( )
hkhkkKk
rr
ri 2sinh2
/2+
≅ νσIn intermediate waterdepth
DeepShallow hko
νσ Khki
5
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Wave Over Porous BedWave Over Porous Bed
+
+=
σν
νσ
22sinh22
rrr
ri kK
hkhkkk
Liu (1973) incorporating laminar boundary layer to avoid velocity discontinuity at the fluid-soil interface
The daamping rate for a constant waerdepth and
hkKkag
r
rD 2
22
cosh2νρ
=∈
xko
iegaE 22
21 −= ρ
Porous damping alone
+=∈
σν
σνρ
22cosh2 2
22r
r
rD
kKhk
kag Including boundary layer