# MIT 2 20 - Marine Hydrodynamics (10 pages)

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## Marine Hydrodynamics

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- 2 20 - Marine Hydrodynamics

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2 20 Marine Hydrodynamics Fall 2005 Lecture 11 Copyright c 2005 MIT Department of Mechanical Engineering All rights reserved 2 20 Marine Hydrodynamics Lecture 11 3 11 Method of Images Potential for single source m p 2 ln x y 2 2 m m m Potential for source near a wall 2 m m q q 2 2 ln x2 y b ln x2 y b y d 0 dy b x b m Added source for symmetry Note Be sure to verify that the boundary conditions are satisfied by symmetry or by calculus for y y 1 Vortex near a wall ground effect Ux 2 y b 1 y b tan tan x x 1 b U y x b Added vortex for symmetry Verify that d 0 on the wall y 0 dy a2 a2 Circle of radius a near a wall U x 1 x2 y b 2 x2 y b 2 y U b y x b This solution satisfies the boundary condition on the wall n 0 and the degree it 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 d1 d12 2 More than one wall Example 1 b b U U b b b Example 3 Example 2 b b b b b b b b b b b b 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 o y x o z z Fixed in space Fixed in translating body x x Ut 3 x U Reference system O v p Reference system O v p U X O S O v n S U 2 0 2 0 X n n U 0 0 nx ny nz U U nx on Body v n n 0 0 as x U x as x v U 0 0 as x v 0 as x Galilean transform v x y z t v x x U t y z t U 0 0 x y z t x x U t y z t U x U x x x U t y z t x y z t Pressure no gravity p 12 v 2 Co Co 12 v 2 Co Co 12 U 2 Co Co 12 U 2 In O unsteady flow ps 12 z v 2 Co t U2 t 0 x U x U 2 x t t z z 0 ps U 2 U 12 U 2 In O steady flow 1 ps 2 z v 2 Co Co t z 0 Co 21 U 2 Co ps p 21 U 2 stagnation pressure ps p 21 U 2 stagnation pressure 4 3 12 2 Forces B n Total fluid force for ideal flow i e no shear stresses F ZZ pn dS B For potential flow substitute for p from Bernoulli F ZZ 1 2 gy c t n dS 2 t z z B hydrostatic force hydrodynamic force For the hydrostatic case v 0 Fs ZZ gyn dS Gauss outward theorem normal B ZZZ B gy d g j z Archimedes principle We evaluate only the hydrodynamic force ZZ 1 2 F d n dS t 2 B For steady motion t 0 1 Fd 2 ZZ B 5 v 2 n dS where ZZZ B d 3 12 3 Example Hydrodynamic force on 2D cylinder in a steady uniform stream n S U x a B Fd R B 2 Fx F i 2 n d a 2 2 0 a 2 R2 0 R2 R2 2r a n ad d 2r a z n i cos 2r a cos d 0 Velocity potential for flow past a 2D cylinder a2 U r cos 1 2 r Velocity vector on the 2D cylinder surface r a vr r a v r a r z 0 r a Square of the velocity vector on the 2D cylinder surface 2 r a 4U 2 sin2 6 1 r z r a 2U sin Finally the hydrodynamic force on the 2D cylinder is given by a Fx 2 Z2 0 1 d 4U 2 sin2 cos U 2 2a 2 2 z z ps p diameter or projection Z2 0 2 d sin z 0 z cos even z 0 odd 3 w r t 2 2 Therefore Fx 0 no horizontal force symmetry fore aft of the streamlines Similarly 1 Fy U 2 2a 2 2 Z2 d sin2 sin 0 0 In fact in general we find that F 0 on any 2D or 3D body D Alembert s paradox No hydrodynamic force acts on a body moving with steady translational no circulation velocity in an infinite inviscid irrotational fluid The moment as measured in a local frame is not necessarily zero 7 3 13 Lift due to Circulation 3 13 1 Example Hydrodynamic force on a vortex in a uniform stream Ux U r cos 2 2 U Consider a control surface in the form of a circle of radius r centered at the point vortex Then according to Newton s law d steady flow LCV dt N ET 0 F F V F CS M F N ET F V F CS M Where F Hydrodynamic force exerted on the vortex from the fluid F V F Hydrodynamic force exerted on the fluid in the control volume from the vortex F CS Surface force i e pressure on the fluid control surface N ET M Net linear momentum flux in the control volume through the control surface d L dt CV Rate of change of the total linear momentum in the control volume y U Fy Fx Control volume IN The hydrodynamic force on the vortex is F F CS M 8 x N ET a Net linear momentum flux in the control volume through the control surfaces M Recall that the control surface has the form of a circle of radius r centered at the point vortex a 1 The velocity components on the control surface are u U v sin 2 r cos 2 r The radial velocity on the control surface is therefore given by ur U x U cos V n r v 2 r U a 2 The net horizontal and vertical momentum fluxes through the control surface are given by d ruvr 0 Z2 Z2 d rvvr Z2 MN ET x Z2 M N ET y 0 U 2 d r U sin U cos 0 2 r d r 0 0 Z2 cos2 d 0 9 U 2 cos U cos 2 r b Pressure force on the control surface F CS b 1 From Bernoulli the pressure on the control surface is 1 p v 2 C 2 b 2 The velocity v 2 on the control surface is given by 2 2 2 2 2 v u v U sin cos 2 r 2 r 2 2 U U sin r 2 r b 3 …

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