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2.016 Hydrodynamics Reading # 8 2.016 Hydrodynamics Prof. A.H. Techet 1. Froude Krylov Excitation Force 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 neglect diffraction effects completely. This assumption comes from the Froude-Krylov hypothesis and assures a resulting excitation force equivalent to the Froude-krylov force: t F FK () =− ρ ∫∫ φ I n d S (8.1) t Vertical Froude-Krylov Force on a Single Hull Vessel z B T x Deep water incident wave potential is: I = a ω kz i ( t kx ) eR e i (8.2) k The force in the vertical direction is found from the incident potential using eq. (8.1) along the bottom of the vessel. Here the normal in the z-direction, n , is negative: n 1, so the z z version 2.0 updated 8/30/2005 -1- © 2005, A. Techet

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2.016 Hydrodynamics Reading # 8 force per unit length in the z-direction is F FK = R e B / 2 ρω ia ω e kT e i ( t kx ) d x i (8.3) z −/ 2 B k 2 o a e kT e i t ikB / 2 e ikB / 2 = Re ⎟⎬ (8.4) k 2 e 2 = Re 2 ρ a e kT e i t s i n ( kB / 2) (8.5) k 2 Recall that sin z = e iz 2 e i iz . Using the vertical velocity we can rewrite the force in terms of the velocity. kz i ( t kx ) wt () = Re a e i (8.6) 2 kz i ( t kx ) ± = Re a e e (8.7) 2 it ± ( 0 z 0 wx =, t ) = a e (8.8) Now we can write the force in the vertical direction as a function of the vertical (heave) acceleration, 2 e kT s i n k B / 2) w ± ( 0 F z = R ( , , t ) . (8.9) 2 k 2 Let’s look at the case where 0 the wavenumber, k = g /→ 0 , also goes to zero and the following simplifications can be made: kt e ² 1 k T (8.10) sin( kB / 2) ² kB / 2 (8.11) to yield a simplified heave force. version 2.0 updated 8/30/2005 -2- © 2005, A. Techet
2.016 Hydrodynamics Reading # 8 2 F FK ω ± Re 2 ρ a (1 kT ) ( kB / 2) e i t (8.12) z k 2 2 ± Re g aB 1 T e it (8.13) g If we look at the case where 0 and consider the heave restoring coefficient, ( kx ) C 33 = g B , and the free surface elevation, η xt () ,= Re a e we can rewrite this force as FK { F ± Re C 33 ( x = 0 , t ) } (8.14) z 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 = S B φ I n dS (8.15) x x t kB = Re ρω ia 0 kz i ( t −/ 2) e i ( t + kB / 2) ⎤⎪ e dz e (8.16) i k T 2 i t = Re i a 1 e kT e 2 sin( kB / 2) (8.17) k As frequency approaches zero similar simplifications can be made like above for the vertical force: 2 a ± Re i ( KT ) e 2 kB / 2 (8.18) Ft x k kz i ( t kx ) ut = Re a e e (8.19) 2 kz e i ( t kx ) ² = Re i a e (8.20) t ² ( z 0 F ± Re { TB u x = 0 ,=, t ) } (8.21)

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