2.019 Design of Ocean Systems. Lecture 6. Seakeeping (II) February 21, 2011

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1 2.019 Design of Ocean Systems Lecture 6 Seakeeping (II) February 21, 2011

2 ω, λ,v p,v g Wave adiation Problem z ζ 3 (t) = ζ 3 cos(ωt) ζ 3 (t) = ω ζ 3 sin(ωt) ζ 3 (t) = ω 2 ζ3 cos(ωt) x 2a ~n Total: P (t) = ρ φ ρgz Hydrodynamic: P d (t) = ρ φ = P d cos(ωt ψ) t t Hydrodynamic Force: Z Z F 3 (t) = P d n z ds = F 3 cos(ωt ψ) S B = F 3 cos ψ cos(ωt)+ F 3 sin ψ sin(ωt) F 3 cos ψ F 3 sin ψ = ζ ζ 3(t) 3 ω 2 ζ 3 ω ζ 3(t) = A 33 ζ 3(t) B 33 ζ 3(t) A 33 : Added mass; B 33 : Wave damping

3 B 33 =0 if a=0 corresponding to ω =, 0 Physical Meaning of Wave Damping Averaged power into the fluid by the body: Z T Ē in = 1 { F 3 (t)} ζ 3(t)dt T 0 Z T n ζ 3 (t) = ζ 3 cos(ωt) ζ 3 (t) = ω ζ 3 sin(ωt) ω, λ,v p,v g z ζ 3(t) = ω 2 ζ 3 cos(ωt) Energy flux out EV g Control Volume Energy flux out EV g = 1 A 33 ζ 3(t)ζ 3(t)+ B 33 ζ 3(t)ζ 3(t) dt = B 33 (ζ 3ω) 2 /2 T 0 Averaged energy flux out of the control volume: Ē flux =2V g E 2V g a 2 dē Conservation of energy: Ē in Ē flux =0 dt o B 33 (a/ζ 3) 2 > 0 x 2a

4 Mathematical Formulation of Heave adiation Problem z ζ 3 (t) =cos(ωt) y adiation condition: Generated waves must propagate away from the body Φ tt + gφ z =0 η(t) = Φ t /g x ~n 2 Φ(x, y, z, t) =0 Deep water condition: Φ 0 as z Hydrodynamic Pressure: adiation Force: adiation Moment: P d (x, y, z, t) = ρφ t F~ (t) = P d ~nds S B M ~ (t) = SB P d (~x ~n)ds

5 Frequency-Domain Formulation of Heave adiation Problem ζ 3 =1 y ω 2 φ 3 + gφ 3z =0 η = iωφ 3 /g x adiation condition ~n Deep water condition: 2 φ 3 (~x) =0 φ 3 0 as z Let: ζ 3 (t) =cosωt = <{e iωt } Φ(~x, t) =<{φ 3 (~x)e iωt } η(x, y, t) =<{η (x, y)e iωt } P d (~x, t) =<{p d (~x)e iωt } F~ ( t) =<{ f ~ e iωt } ~ iω t M~ ( t) =<{ me } p d = iρωφ 3 (~x) f ~ = SB p d ~nds m~ = SB p d (~ x ~n)ds f 3 = iρω SB φ 3 n 3 ds F 3 (t) = A 33 ζ 3(t) B 33 ζ (t) =<{[ω 2 A 33 iωb 33 ]e iωt } Thus, n o n o ρ A 33 = < φ 3 n 3 ds, B 33 = = iρ φ 3 n 3 ds i ω S B n o n o iρ A 13 = < ω S B φ 3 n 1 ds, B 13 = = iρ SB φ 3 n 1 ds A 23,B 23,...,A 63,B 63 A ij and B ij are symmetric, i.e. A ij = A ji, B ij =B ji, i=1,, 6; j=1,, 6 Aij and Bij are functions of frequency ω S B

7 Examples: Added Mass at Low Frequency At low frequencies, i.e. ω 0: d 2 Φ dt ω 2 0 as ω 0 2 Thus, the free surface boundary conditoin becomes: Φ z =0 (1) slender vertical circular cylinder (2) slender ship with a semi-circle cross section Surge added mass m 11 = ρπ 2 h Wave damping =0 Sway added mass m 11 = ρ π2 2 L Wave damping =0

8 Examples: Added Mass at High Frequency At high frequencies, i.e. ω : d 2 Φ dt 2 ω 2 as ω Thus, the free surface boundary conditoin becomes: Φ=0 Slender ship with a semi-circle cross section: Heave added mass: m 33 = ρ π2 2 L Wave damping =0

9 Hydrostatic estoring Effect in Body Motion z ζ 3 (t) x t = 0 ~n Wetted body surface: S 0 S B (t) = S 0 + S(t) Hydrostatic pressure: P s = ρgz Hydrostatic force: F ~ s = SB (t) P s ~nds = S0 P s ~nds S(t) P s ~nds F s3 (t) = ρg dvol = F s3 ρgs wl ζ 3 (t) Vol S wl :Water plane surface area of the body Hydrostatic restoring effect Balanced by other forces at equilibrium C 33 = ρgs wl :Hydrostatic restoring coefficient (i.e. spring constant) Hydrostatic moment:m ~ s = SB (t) P s(~x ~n)ds = P s (~x ~n)ds S(t) P s (~x ~n)ds S 0 Hydrostatic restoring force/moment: F si3 (t) = C i3 ζ 3 (t), i =1,...,6 In general F sij (t) = P 6 j=1 C ij ζ j,i =1,...,6 where C ij is 6 6 restoring coef. matrix

10 Wave Diffraction Problem z Body is fixed y adiation condition: diffracted waves must Φ tt + gφ z =0 η(t) = Φ t /g ~n x Deep water condition: propagate away from the Φ 0 as body 2 Φ(x, y, z, t) =0 F ~ E, M ~ E =??? z Total potential: Φ(~x, t) = Φ I (~,xt)+ Φ D (~x, t) Φ I : Incident wave potential (of a plane progressive wave) Φ D :Diffracted (or scattered) wave potential Total dynamic pressure: P d = ρφ It ρφ Dt Total wave excitations (force/moment): F~ E (t) = ρφ It ~nds + ρφ It ~nds = F~ I + F~ D S B Froude-Krylov force S B Diffraction effect M ~ E(t) = SB ρφ It (~x ~n)ds + SB ρφ Dt (~x ~n)ds = M ~ I + M ~ D

11 Frequency-Domain Formulation of Wave Diffraction Problem adiation condition: z ω 2 φ D + gφ Dz =0 η D = iωφ D /g x diffracted waves must ~n n = n Deep water condition: y φ D propagate away from the φ D 0 as body 2 φ D (x, y, z) =0 φ I z Incident wave: η I (x, y, t) =a cos(ωt kx) =<{ae ikx e iωt } = <{η I e iωt } Φ I (~x, t) = (ga/ω)sin(ωt kx)e kz = <{( iga/ω)e kz ikx e iωt } = <{φ I e iωt } Diffraction potential: Φ D (~x, t) =<{φ D (~x)e iωt } Total dynamic pressure: P d (~x, t) =<{p d (~x)e iωt }, p d (~x) =p I + p D p I = iρωφ I, p D = iρωφ D Total wave excitations: F~ E (t) =<{f ~ Ee iωt }, f ~ E = f ~ EI + f ~ ED Froude-Krylov force: f ~ EI = SB p I ~nds = iρω SB φ I ~nds Diffraction force: f ~ ED = SB p D ~nds = iρω SB φ D ~nds M~ E (t) =<{m~ E e iωt }, m~ E = m~ EI + m~ ED

12 Heave esponse of A Floating Body to Ambient Waves Incident wave: η I = a cos(ωt kx) z ζ 3 (t) =<{ ζ 3 e iωt }, ζ3 =??? Φ tt + gφ z =0 x η(t) = Φ t /g y ~n 2 Φ(x, y, z, t) =0 Deep water condition: Φ 0 as z Decompose the total problem into a sum of diffraction problem and radiation problem: Φ(~x, t) =Φ I (~ x, t)+φ D (~x, t)+φ (~x, t) Diffraction problem adiation problem From the diffraction problem: Wave excitation force: F E3 (t) =<{f E3 e iωt }, f E3 = f 3I + f 3D From the radiation problem: Wave radiation force: F 3 (t) = A 33 ζ 3(t) B 33 ζ 3(t) =<{( ω 2 A 33 iωb 33 )ζ 3e iωt } Hydrostatic restoring force: F s3 = C 33 ζ 3 (t) =<{( C 33 ζ 3)e iωt }

13 Total hydrodynamic and hydrostatic forces: F E3 + F 3 + F s3 = <{[f E3 (ω 2 A 33 + iωb 33 + C 33 )ζ 3]e iωt } Applying Newton s second law: F E3 + F 3 + F s3 = mζ 3 (t) <{( ω 2 m)e iωt } = <{[f E3 ( ω 2 A 33 + iωb 33 + C 33 )ζ 3]e iωt } Equation of Motion: [ ω 2 (m + A 33 )+iωb 33 + C 33 ]ζ 3 = f 3I + f 3D Heave motion amplitude: ζ 3 = esponse Amplitude Operator (AO): ζ 3(ω) f 3I +f 3D ω 2 (m+a 33 )+iωb 33 +C 33 a (f 3I + f 3D )/a = ω 2 ( m + A 33 )+iωb 33 + C 33 (f 3I + f 3D )/a iα = e ω 2 (m + A 33 )+iωb 33 + C 33 ³ Heave natural frequency: ω n3 (m + A 33 )+C 33 =0 C ω 33 3n = m+a

14 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) = f 0 cos ωt, we have harmonic response: x(t) = x 0 cos(ω + α), f 0 and α = tan x 0 =?? ³ From equation of motion, we obtain: 1 bω x 0 = [(c mω 2 ) 2 +b 2 ω 2 ] 1/2 c mω 2 Natural frequency: ω n =(c/m) 1/2 x 0 = 1, at ω=0 f 0 c x 0 1 =, at ω=ω n f 0 bω n x 0 0, as ω f 0

15 x 0 f 0 1 bω n 1 c α ω n ω n ω ω

16 MIT OpenCourseWare Design of Ocean Systems Spring 2011 For information about citing these materials or our Terms of Use, visit:

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