2.019 Design of Ocean Systems
Lecture 6
Seakeeping (II)
February 21, 2011
Wave Radiation Problem
ζ3 (t) = ζ̄3 cos(ωt)
ω, λ, Vp , Vg
ζ̇3 (t) = −ωζ̄3 sin(ωt)
ζ¨3 (t) = −ω2 ζ̄3 cos(ωt)
z
x
2a
~n
Total: P (t) = −ρ ∂φ
∂t − ρgz
Hydrodynamic: Pd (t) = −ρ ∂φ
∂t = P̄d cos(ωt − ψ)
Hydrodynamic Force:
F3 (t) = −
Z Z
SB
Pd nz dS = F¯3 cos(ωt − ψ)
= F¯3 cos ψ cos(ωt) + F¯3 sin ψ sin(ωt)
F¯3 cos ψ
F¯3 sin ψ
ζ̇3 (t)
= − ¯ 2 ζ̈3 (t) − ¯
ζ3 ω
ζ3 ω
= −A33 ζ̈3 (t) − B33 ζ̇3 (t)
A33: Added mass; B33: Wave damping
Physical Meaning of Wave Damping
ω, λ, Vp , Vg
z
ζ3 (t) = ζ¯3 cos(ωt)
ζ̇3 (t) = −ωζ̄3 sin(ωt)
ζ̈3 (t) = −ω2 ζ̄3 cos(ωt)
x
2a
Energy
flux out
EVg
Control Volume
Energy
flux out
EVg
Averaged power into the fluid by the body:
Ēin
=
=
1
T
1
T
Z
Z
T
{−F3 (t)} ζ̇3 (t)dt
0
T
0
n
o
A33 ζ̈3 (t)ζ̇3 (t) + B33 ζ̇3 (t)ζ̇3 (t) dt = B33 (ζ̄3 ω)2 /2
Averaged energy flux out of the control volume:
Conservation of energy:
dĒ
dt
Ēf lux = 2Vg E ∼ 2Vg a2
≡ Ēin − Ēf lux = 0
B33 ∼ (a/ζ̄3 )2 > 0
• B33 =0 if a=0 corresponding to ω = ∞, 0
Mathematical Formulation of Heave Radiation Problem
z
ζ3 (t) = cos(ωt)
y
Φtt + gΦz = 0
η(t) = −Φt /g
x
~n
Radiation condition:
Generated waves
must propagate away
from the body
Hydrodynamic Pressure:
Radiation Force:
Radiation Moment: 2
∇ Φ(x, y, z, t) = 0
Deep water condition:
∇Φ → 0 as z → −∞
Pd (x, y, z, t) = −ρΦt
R
F~R (t) = − SB Pd~nds
R
~ R (t) = −
M
Pd (~x × ~n)ds
SB
Frequency-Domain Formulation of Heave Radiation Problem
ζ¯
3 = 1
y
−ω 2 φ3 + gφ3z = 0
η̄ = −iωφ3 /g
x
~n
Radiation condition
Deep water condition:
∇φ3 → 0
∇2 φ3 (~x) = 0
Let:
ζ3 (t) = cos ωt = <{eiωt }
Φ(~x, t) = <{φ3 (~x)eiωt }
η(x, y, t) = <{η̄(x, y)eiωt }
Pd (~x, t) = <{pd (~x)e
F~R (t) = <{f~eiωt }
~ R (t) = <{me
~ iωt }
M
iωt
}
pd = −iρωφ3 (~x)
R
f~ = − SB pd~ndS
R
m
~ = − SB pd (~x × ~n)dS
f3R = iρω
R
SB
as
z → −∞
φ3 n3 ds
F3R (t) = −A33 ζ̈3 (t) − B33 ζ̇(t) = <{[ω 2 A33 − iωB33 ]eiωt }
Thus,
A33 = <
A13 = <
n
n
iρ
ω
iρ
ω
R
SB
R
SB
o
φ3 n3 ds ,
o
φ3 n1 ds ,
A23 , B23 , ..., A63 , B63
B33
B13
n R
o
= −= iρ SB φ3 n3 ds
n R
o
= −= iρ SB φ3 n1 ds
• Aij and Bij are symmetric, i.e. Aij = Aji, Bij =Bji, i=1, …, 6; j=1, …, 6
• Aij and Bij are functions of frequency ω
|ζ3 |
A
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Examples: Added Mass at Low Frequency
At low frequencies, i.e. ω → 0:
d2 Φ
dt2
∼ ω2 → 0
as ω → 0
Thus, the free surface boundary conditoin becomes: Φz =0
(1) slender vertical circular cylinder
Surge added mass
m11 = ρπR2 h
Wave damping =0
(2) slender ship with a semi-circle
cross section
Sway added mass
πR2
m11 = ρ 2 L
Wave damping =0
Examples: Added Mass at High Frequency
At high frequencies, i.e. ω → ∞:
d2 Φ
dt2
∼ ω2 → ∞
as ω → ∞
Thus, the free surface boundary conditoin becomes: Φ=0
Slender ship with a semi-circle cross section:
Heave added mass: m33 =
Wave damping =0
πR2
ρ 2 L
Hydrostatic Restoring Effect in Body Motion
ζ3 (t)
z
x
t=0
~n
Wetted body surface: S0
SB (t) = S0 + ∆S(t)
Hydrostatic pressure: Ps = −ρgz
Hydrostatic restoring effect
R
R
R
~s = −
P ~nds = − S0 Ps~nds − ∆S(t) Ps ~nds
Hydrostatic force: F
SB (t) s
R
Balanced by other forces at equilibrium
Fs3 (t) = ρg V ol dV ol = F¯s3 − ρgSwl ζ3 (t)
Swl : Water plane surface area of the body
C33 = ρgSwl : Hydrostatic restoring coefficient (i.e. spring constant)
R
R
R
~
Hydrostatic moment:
Ms = − SB (t) Ps (~x × ~n)ds = − S0 Ps (~x × ~n)ds − ∆S(t) Ps (~x × ~n)ds
Hydrostatic restoring force/moment: Fsi3 (t) = −Ci3 ζ3 (t), i = 1, . . . , 6
P
In general Fsij (t) = − 6j=1 Cij ζj , i = 1, . . . , 6 where Cij is 6×6 restoring coef. matrix
Wave Diffraction Problem
Body is fixed
z
y
Φtt + gΦz = 0
Radiation condition:
diffracted waves must
propagate away from the
body
η(t) = −Φt /g
x
~n
Deep water condition:
∇Φ → 0 as z → −∞
∇2 Φ(x, y, z, t) = 0
~ E =???
F~E , M
Total potential:
Φ(~x, t) = ΦI (~,xt) + ΦD (~x, t)
ΦI : Incident wave potential (of a plane progressive wave)
ΦD : Diffracted (or scattered) wave potential
Total dynamic pressure: Pd = −ρΦIt − ρΦDt
~E (t) =
Total wave excitations (force/moment): F
Diffraction effect
R
ρΦIt~nds +
SB
Froude-Krylov force
~ E (t) =
M
R
SB
ρΦIt (~x × ~n)ds +
R
SB
R
SB
ρΦIt~nds = F~I + F~D
~I +M
~D
ρΦDt (~x × ~n)ds = M
Frequency-Domain Formulation of Wave Diffraction Problem
z
−ω 2 φD + gφDz = 0
Radiation condition:
diffracted waves must
propagate away from the
body
Incident wave:
ηI (x, y, t)
y
x
∂φD
∂n
~n
2
η̄D = −iωφD /g
I
= − ∂φ
∂n
∇ φD (x, y, z) = 0
Deep water condition:
∇φD → 0
as
z → −∞
= a cos(ωt − kx) = <{ae−ikx eiωt } = <{η̄I eiωt }
ΦI (~x, t) = −(ga/ω) sin(ωt − kx)ekz = <{(−iga/ω)ekz−ikx eiωt } = <{φI eiωt }
Diffraction potential:
ΦD (~x, t) = <{φD (~x)eiωt }
Total dynamic pressure: Pd (~
x, t) = <{pd (~x)eiωt },
Total wave excitations:
pd (~x) = pI + pD
pI = −iρωφI ,
f~E = f~EI + f~ED
F~E (t) = <{f~E eiωt },
R
R
~
nds = iρω SB φI ~nds
Froude-Krylov force: fEI = − SB pI ~
R
R
Diffraction force:
f~ED = −
pD ~nds = iρω
φD ~nds
SB
~ E (t) = <{m
~ E eiωt },
M
SB
m
~E =m
~ EI + m
~ ED
pD = −iρωφD
Heave Response of A Floating Body to Ambient Waves
z
Incident wave:
ηI = a cos(ωt − kx)
ζ3 (t) = <{ζ̄3 eiωt },
ζ̄3 =???
y
Φtt + gΦz = 0
x
η(t) = −Φt /g
~n
Deep water condition:
∇2 Φ(x, y, z, t) = 0
∇Φ → 0 as z → −∞
• Decompose the total problem into a sum of diffraction problem and radiation problem:
Φ(~x, t) = ΦI (~x, t) + ΦD (~x, t) + ΦR (~x, t)
Diffraction problem
Radiation problem
• From the diffraction problem:
Wave excitation force:
FE3 (t) = <{fE3 eiωt },
fE3 = f3I + f3D
• From the radiation problem:
Wave radiation force:
FR3 (t) = −A33 ζ̈3 (t) − B33 ζ̇3 (t) = <{(−ω 2 A33 − iωB33 )ζ̄3 eiωt }
Hydrostatic restoring force:
Fs3 = −C33 ζ3 (t) = <{(−C33 ζ̄3 )eiωt }
• Total hydrodynamic and hydrostatic forces:
FE3 + FR3 + Fs3 = <{[fE3 − (ω2 A33 + iωB33 + C33 )ζ̄3 ]eiωt }
• Applying Newton’s second law:
FE3 + FR3 + Fs3 = mζ¨3 (t)
<{(−ω 2 m)eiωt } = <{[fE3 − (−ω 2 A33 + iωB33 + C33 )ζ̄3 ]eiωt }
Equation of Motion:
[−ω 2 (m+ A33 ) + iωB33 + C33 ]ζ̄3 = f3I + f3D
ζ̄3 =
• Heave motion amplitude: f3I +f3D
−ω 2 (m+A33 )+iωB33 +C33
Response Amplitude Operator (RAO): ζ̄3 (ω)
a
Heave natural frequency:
(f3I + f3D )/a
−ω 2 (m + A33 ) + iωB33 + C33
¯
¯
¯
¯ iα
(f3I + f3D )/a
¯
¯ e
=
¯
2
−ω (m + A33 ) + iωB33 + C33
¯
=
−ωn3 (m + A33 ) + C33 = 0
→
ω3n =
³
C33
m+A33
´
12
Analogy to a Simple Mass-Spring-Dashpot System Body displacement
Body mass
Excitation force
Spring constant
Damping coefficient
Equation of motion:
mẍ + bẋ + cx = f (t)
For harmonic excitation,
f (t) = f0 cos ωt, we have harmonic response: x(t) = x0 cos(ω + α),
From equation of motion, we obtain:
Natural frequency:
x0 =
ωn = (c/m)1/2
f0
2
[(c−mω )2 +b2 ω 2 ]1/2
and
x0
1
= , at ω=0
f0
c
x0
1
=
, at ω=ωn
f0
bωn
x0
→ 0, as ω → ∞
f0
α = tan−1
³
x0 =??
−bω
c−mω 2
´
x0
f0
1
bωn
1
c
ωn
α
ωn
ω
ω
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2.019 Design of Ocean Systems
Spring 2011
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