lecture11 - 2.20 - Marine Hydrodynamics, Spring 2005...

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± ² ² ³ 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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± ² ± ² 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 ± ± ,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 z = Un x on Body ± v 0as | ±x |→∞ φ | ) 2 φ ± ± v ± · ˆ n ± = ± ± v ± ( U, 0 , 0) as | ± φ ± →− Ux ± as | ± Galilean transform: ± v ( x, y, z, t ) φ ( x, y, z, t ) ± + φ ( x = x ± + Ut,y,z,t ) = ± v ± ( x ± = x Ut,y, z,t )+( U, 0 , 0) = φ ± ( x ± = x )+
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This note was uploaded on 02/27/2012 for the course MECHANICAL 2.20 taught by Professor Dickk.p.yue during the Spring '05 term at MIT.

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lecture11 - 2.20 - Marine Hydrodynamics, Spring 2005...

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