lecture11 - 2.20 Marine Hydrodynamics Spring 2005 Lecture...

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Lecture 11 - Marine Hydrodynamics Lecture 11 3.11 - Method of Images m Potential for single source: φ = ln x 2 + y 2 2 π m Potential for source near a wall: φ = m ln x 2 + ( y b ) 2 + ln x 2 + ( y + b ) 2 2 π b b Added source for 0 = φ dy d x y m m symmetry Note: Be sure to verify that the boundary conditions are satisfied by symmetry or by calculus for φ ( y ) = φ ( y ). 1 2.20 - Marine Hydrodynamics, Spring 2005 2.20
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Vortex near a wall (ground effect): φ = U x + Γ tan 1 ( y b ) tan 1 ( y + b ) 2 π x x b b Γ U x y - Γ Added vortex for symmetry Verify that = 0 on the wall y = 0. dy 2 2 φ a a Circle of radius a near a wall: = Ux 1 + + x 2 + ( y b ) 2 x 2 + ( y + b ) 2 b b U y x y This solution satisfies the boundary condition on the wall ( ∂φ = 0), and the degree it ∂n satisfies the boundary condition of no flux through the circle boundary increases as the ratio b/a >> 1, i.e., the velocity due to the image dipole small on the real circle for b >> a . For a 2D dipole, φ d 1 , φ d 1 2 . 2
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More than one wall: b' b U b' b U b' Example 1: Example 2: Example 3: b b b b - Γ Γ b' b' b' b' b' b' b' b' Γ - Γ b b b b 3.12 Forces on a body undergoing steady translation “D’Alembert’s paradox” 3.12.1 Fixed bodies & translating bodies - Galilean transformation . y y x U o x o z z Fixed in space Fixed in translating body x = x` + Ut 3
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Reference system O: v, φ, p Reference system O’: v , φ , p U S O X U S O’ X 2 φ = 0 v · ˆ n = ∂φ ∂n = U · ˆ n = ( U, 0 , 0) · ( n x , n y , n z = Un x on Body v 0 as | x | → ∞ φ 0 as | x | → ∞ ) 2 φ = 0 v · ˆ n = ∂φ ∂n = 0 v ( U, 0 , 0) as | x | → ∞ φ → − Ux as | x
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