lecture14

lecture14 - 2.20 Marine Hydrodynamics Spring 2005 Lecture...

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Lecture 14 - Marine Hydrodynamics Lecture 14 3.20 Some Properties of Added-Mass Coeﬃcients 1. m ij = ρ · [function of geometry only] F, M = U, Ω] [linear function of m ij ] × [function of instantaneous U, ˙ not of motion history 2. Relationship to ﬂuid momentum. F(t) where we define Φ to denote the velocity potential that corresponds to unit velocity U = 1. In this case the velocity potential φ for an arbitrary velocity U is φ = U Φ. The linear momentum L in the ﬂuid is given by L = vdV = ρ φdV = + ρφ ˆ ρ ndS V V Green’s B theorem φ 0 at L x ( t = T ) = ρU Φ n x dS = U ρ Φ n x dS B B The force exerted on the ﬂuid from the body is F ( t ) = ( m A U ˙ ) = m A U ˙ . 1 2.20 - Marine Hydrodynamics, Spring 2005 2.20

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T T Newton’s Law ˙ T dt [ F ( t )] = m A Udt = m A U ] 0 = L x ( t = T ) L x ( t = 0) = U ρ Φ n x dS 0 0 m A U B Therefore, m A = total ﬂuid momentum for a body moving at U = 1 (regardless of how we get there from rest) = ﬂuid momentum per unit velocity of body. Φ K.B.C. ∂φ = φ · n ˆ = ( U, 0 , 0) · n ˆ = Un x , ∂φ = Un x ∂U Φ = Un x = n x ∂n ∂n ∂n ∂n Φ m A = ρ Φ dS ∂n B For general 6 DOF: Φ j m ji = ρ Φ i n j dS = ρ Φ i dS = j ﬂuid momentum due to ∂n i body motion j force/moment B potential due to body B i direction of motion moving with U i =1 3. Symmetry of added mass matrix m ij = m ji . Φ j m ji = ρ Φ i dS = ρ Φ i ( Φ j · n ˆ) dS = ρ ∇ · i Φ j ) dV ∂n B B Divergence V Theorem = ρ Φ i · ∇ Φ j + Φ i 2 Φ j dV V =0 Therefore, m ji = ρ Φ i · ∇ Φ j dV = m ij V 2
4. Relationship to the kinetic energy of the ﬂuid. For a general 6 DoF body motion U i = ( U 1 , U 2 , . . . , U 6 ), φ = U i Φ i ; Φ i = potential for U i = 1 notation 1 K.E. = ρ φ · ∇ φdV = 2 1 ρ U i Φ i · U j Φ j dV 2 V V 1 = 2 ρU i U j Φ i · ∇ Φ j dV = 1 2 m ij U i U j V K.E. depends only on m ij and instantaneous U i . 5. Symmetry simplifies m ij . From 36 21 ‘?’. Choose such coordinate system symmetry that some m ij = 0 by symmetry. Example 1 Port-starboard symmetry.

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