13.42 Spring 2005 13.42 Design Principles for Ocean Vehicles Prof. A.H. Techet Spring 2005 Froude Krylov Excitation Force 1. Radiation and Diffraction Potentials The total potential is a linear superposition of the incident, diffraction, and radiation potentials, ωφ=(φ φ+φ) e . (1)I + D Rit The radiation potential is comprised of six components due to the motions in the six directions, 1 23456 thφj where j =, , , , , . Each function φjis the potential resulting from a unit motion in j direction for a body floating in a quiescent fluid. The resulting body boundary condition follows from lecture 15: ω 1 2 3) (2) ∂φj =inj ; ( j = , , ∂n ω( , , ∂φj =ir×n ) j −3 ; ( j = 4 5 6) (3)∂n r =(x y z ) (4),, nnj ( j =, , = 1 2 3) = (n ,ny ,n ) (5)x z In order to meet all the boundary conditions we must have waves that radiate away from the body. ikx Thus φj ∝ e∓ as x →±∞. For the diffraction problem we know that the derivative of the total potential (here the incident potential plus the diffraction potential without consideration of the radiation potential) normal to the body surface is zero on the body: ∂φT =0 on S , where φ=φφ.∂n BT I + D version 1.0 updated 3/29/2005 -1- 2005, aht13.42 Spring 2005 ∂φI ∂φD ;on S (6)∂n =− ∂nB We have so far talked primarily about the incident potential. The formulation of the incident potential is straight forward from the boundary value problem (BVP) setup in lecture 15. There exist several viable forms of this potential function each are essentially a phase shifted version of another. The diffraction potential can also be found in the same fashion using the BVP for the diffraction potential with the appropriate boundary condition on the body. This potential can be approximated for a long wave condition. This long wave approximation assumes that the incident wavelength is very long compared to the body diameter and thus the induced velocity field from the incident waves on the structure can be assumed constant over the body and approximated by the following equation: i ∂φI ∂φI ∂φI φ3 φ+ φ2 + (7) φD ω ∂x 1 ∂y ∂z  Further explanation of this approximation can be found in Newman (p. 301). Ultimately, if we assume the body to be sufficiently small as not to affect the pressure field due to an incident wave, then we can diffraction effects can be completely ignored. This assumption comes from the Froude-Krylov hypothesis and assures a resulting excitation force equivalent to the froude-krylov force: tF FK () =−ρ∫∫∂φI n dS (8)∂t version 1.0 updated 3/29/2005 -2- 2005, aht13.42 Spring 2005 2. Vertical Froude-Krylov Force on a Single Hull Vessel z B T x Deep water incident wave potential is: φI =aω kz  i (ωt −kx) eRe i e  (9) k The force in the vertical direction is found from the incident potential using eq. 8 along the bottom of the vessel. Here the normal in the z-direction, n , is negative: n =−1, so the force per unit zzlength in the z-direction is B/ 2F FK = Re ∫−/2 −ρωiaω e−kT ei (ωt −kx ) i dx  (10)z Bk  2 o =Re ρω a e −kT eiωt e−ikB/2 −eikB/2  (11)2 k   2 −kT  = Re 2ρω2 a e eiωt sin(kB /2) (12)  k Recall that sin z = eiz − 2 ei −iz . Using the vertical velocity we can rewrite the force in terms of the velocity. wt  kz i ( t −kx ) () = Re a ωe i e ω (13) wt    2 kz ei (ωt −kx ) () = Re −aω e  (14)  ω( 0 0  2 itwx =,z =,t) = Re a ω e  (15)  version 1.0 updated 3/29/2005 -3- 2005, aht13.42 Spring 2005 Now we can write the force in the vertical direction as a function of the vertical (heave) acceleration, 2ρe−kT sin kB/2) w(0 0Fz=Re  ( , ,t). (16)2k  2Let’s look at the case where ω→ 0 the wavenumber, k =ωg/→0 , also goes to zero and the following simplifications can be made: kte  1− kT (17) sin(kB/2)  kB/2 (18) to yield a simplified heave force. F  2  FK iωt   Re 2ρωa (1 − kT ) (kB/2) e  (19)z k 2     2  itRe ρg aB 1−ωT  e ω (20)  g   If we look at the case where ω→ 0 and consider the heave restoring coefficient, C33=ρg B , ( it −kx) () and the free surface elevation, ηxt,=Re a e ω we can rewrite this force as  F FK  Re C33 η(x =0,t)} (21)z{ 3. Horizontal Froude-Krylov Force on a Single Hull Vessel The horizontal force on the vessel above can be found in a similar fashion to the vertical force. F = ∫∫ρ∂φIn dS (22)x xSB ∂t  =Re ρωia 0 ekz dz  eitkB (ω+kB/2)  ω (ω−/2) −eit  (23)i ∫−T  k  2 iωt  =Re iρaω1−e−kT e 2 sin(kB/2) (24)   k version 1.0 updated 3/29/2005 -4- 2005, aht13.42 Spring 2005 As frequency approaches zero similar simplifications can be made like above for the vertical force: 2 aω ω() Re iρ (KT ) eit 2 kB /2 (25)Ftx  k  ut  kz ei (ωt −kx ) () = Re a ωe  (26)  ut  2 kz i (ωt −kx ) () = Re i a ωe e  (27)  Ft ( z 0() Re {ρTB u x =0,=, t)} (28)xρWhere ρTB =∀, and ∀ is the vessel volume such that we are left with the surge force F ρ∀u (29)xF C33η+ρ∀w (30)z4. Multi Hulled Vessel b T x z T b -B/2 B/2 Again, let’s make a few basic assumptions: ( 1b λ/ << ), ( 1B λ/ ∼ ), ( ab< ), and ( bT∼ ). Let’s look at the force in the x-direction: ( 2 0 ) ( 2 0 )FK xF bT u x B z t bT u x B z tρ ρ=−/,=, + = /,=,   (31) () cos( )x t a t kx η ω, = − (32) 2( ) sin( )kzux z t a e t kx ω ω,, =− − (33) (34) version 1.0 updated 3/29/2005 -5- 2005, aht13.42 Spring 2005 2F FK ρbT (−a ω) {sin(ωt + kB /2) + sin(ωt − kB /2)} (35)x2 −2ρbT (a ω) cos(kB /2) sin(ωt) (36) FK2 λNote that when kB/=π/2 (or B =/2 ) then F (t) =0 .x4.1. Multi Hulled Vessel with additional pontoonb T x z T b -B/2 B/2 c Use the same assumptions from above to find the x-force adjusted for the additional pontoon between the two hulls. 2F FK  − 2 ρb T (a ω) cos(kB /2) sin(ωt)x (+cp x =−B / 2 + b / 2, z = 0, t) (− cp x = B / …