Marine Hydrodynamics (Lecture 11)

2.20 - Marine Hydrodynamics, Spring 2005 Lecture 11 2.20 - Marine Hydrodynamics Lecture 11 3.11 - Method of Images • Potential for single source: φ = m  2 ln x + y 2 2π m m m • Potential for source near a wall: φ = 2π m m     2 2 ln x2 + (y − b) + ln x2 + (y + b) y dφ =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 φ (y) = φ (−y). 1 Γ • Vortex near a wall (ground effect): φ = Ux + 2π   −1 y − b −1 y + b ) − tan ( ) tan ( x x Γ b U y x b -Γ Added vortex for symmetry Verify that dφ = 0 on the wall y = 0. dy  • Circle of radius a near a wall: φ ∼ = Ux 1 + a2 a2 + x2 + (y − b)2 x2 + (y + b)2  y U b y x b ∂φ 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, φ ∼ d1 , ∇φ ∼ d12 . 2 • More than one wall: Example 1: b b U U b' b' b' Example 3: Example 2: b b b b -Γ Γ b' b' b' b' b' b' b' b' Γ b b -Γ 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’ o’ x z’ z Fixed in space Fixed in translating body x = x` + Ut 3 x’ U Reference system O’: v  , φ , p Reference system O: v , φ, p U X O S O’ ∇2 φ = 0  v · n̂ = X’ S U ∇2 φ = 0 ∂φ ∂n  · n̂ = (U, 0, 0) · (nx , ny , nz ) =U = U nx on Body  v · n̂ = ∂φ ∂n =0   φ → 0 as |x| → ∞ φ → −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) φ(x, y, z, t) = φ (x = x − U t, y, z, t) + U x ⇒ −U x + φ(x = x + U t, y, z, t) = φ (x , y, z, t) Pressure (no gravity) p∞ = − 12 ρv 2 + Co = Co = − 12 ρv 2 + Co = Co − 12 ρU 2 ∴ Co = Co − 12 ρU 2 In O: unsteady flow In O’: steady flow ∂φ 1 − 2 ρ  v 2 +Co = Co ps = −ρ ∂t  − 12 ρ  v 2 +Co ps = −ρ ∂φ ∂t U2 ∂φ ∂t ∂ ∂x =( + ∂t ∂t   0 ∴ ps = ρU 2 0 ∂ ) (φ ∂x −U − 12 ρU 2 0 + U x ) = −U 2 + Co = 12 ρU 2 + Co ps − p∞ = 12 ρU 2 stagnation pressure ps − p∞ = 12 ρU 2 stagnation pressure 4 3.12.2 Forces B n̂ Total fluid force for ideal flow (i.e., no shear stresses): F =  pˆ ndS B For potential flow, substitute for p from Bernoulli: ⎛ ⎜ ∂φ 1  ⎜ F =  −ρ ⎜ + |∇φ|2 + gy 2 ⎝∂t   B hydrodynamic force  = ↑ Gauss outward theorem normal B ∇ (−ρgy) dυ = ρg∀ĵ   (−) ↑ ⎟ ⎟ +c(t)⎟ ndS ˆ ⎠ hydrostatic force   For the hydrostatic case v ≡ φ ≡ 0 : F s =  (−ρgyn̂) dS ⎞ Archimedes principle υB We evaluate only the hydrodynamic force:    ∂φ 1 2 + |∇φ| n̂dS F d = −ρ ∂t 2 For steady motion  ∂φ ∂t  ≡0 : B  1 Fd = − ρ 2 v 2 ndS ˆ B 5 where ∀ = dυ υB 3.12.3 Example Hydrodynamic force on 2D cylinder in a steady uniform stream. n̂ S U x a B  Fd =   −ρ  B  2 Fx = F · î = = ρa 2 2π 0 |∇φ|2 nd ˆ = −ρa 2 2π 0 2π  −ρ  0 2 2 n̂adθ |∇φ|r=a dθ |∇φ|2r=a  n̂ · î − cos θ 2 |∇φ|r=a cos θdθ Velocity potential for flow past a 2D cylinder:  a2 φ = U r cos θ 1 + 2 r  Velocity vector on the 2D cylinder surface:  ∇φ|r=a = ( vr |r=a , vθ |r=a ) =   ∂φ   ∂r    0 r=a Square of the velocity vector on the 2D cylinder surface:  |∇φ|2 r=a = 4U 2 sin2 θ 6   1 ∂φ  , r ∂θ    r=a −2U sin θ Finally, the hydrodynamic force on the 2D cylinder is given by ρa Fx = 2 2π 0   1  dθ 4U 2 sin2 θ cos θ = ρU 2 (2a) 2  2  ps −p∞ diameter or projection 2π 2 dθ sin   θ cos  θ = 0 0 even   odd π 3π w.r.t , 2 2 ≡0 Therefore, Fx = 0 ⇒ no horizontal force ( symmetry fore-aft of the streamlines). Similarly, 1  Fy = ρU 2 (2a)2 2 2π dθ sin2 θ sin θ = 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. φ = Ux + Γ Γ θ = U r cos θ + θ 2π 2π Γ 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 ≡ −FV = FCS + M ΣF =  N ET (FV + FCS ) + M Where, F = Hydrodynamic force exerted on the vortex from the fluid. FV = −F = Hydrodynamic force exerted on the fluid in the control volume from the vortex. FCS = 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 θ Fy Γ U Fx Control volume  IN The hydrodynamic force on the vortex is F = FCS + 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 = Γ sin θ 2πr Γ cos θ 2πr The radial velocity on the control surface is therefore, given by ur = U  ∂x = U cos θ = V · n̂ ∂r vθ = Γ 2πr U θ a.2 The net horizontal and vertical momentum fluxes through the control surface are given by 2π (MN ET )x = − ρ 2π dθruvr = − ρ 0 2π (M N ET )y = − ρ 0 2π dθrvvr = − ρ 0 =− ρU Γ 2π   Γ dθr U − sin θ U cos θ = 0 2πr  dθr 0 2π cos2 θdθ = − 0 9 ρU Γ 2  Γ cos θ U cos θ 2πr b. Pressure force on the control surface, FCS . b.1 From Bernoulli, the pressure on the control surface is 1 p = − ρ |v |2 + C 2 b.2 The velocity |v |2 on the control surface is given by  2  2 Γ Γ 2 2 2 |v | =u + v = U − sin θ + cos θ 2πr 2πr  2 Γ Γ 2 =U − U sin θ + πr 2πr b.3 Integrate the pressure along the control surface to obtain FCS (FCS )x = (FCS )y = 2π dθrp(− cos θ) = 0 0 2π 0   ρ dθrp(− sin θ) = − 2  ΓU 2π dθ sin2 θ = − 12 ρU Γ − πr (−r) 0  π c. Finally, the force on the vortex F is given by Fx = (FCS )x + (Mx )IN = 0 Fy = (FCS )y + (My )IN = −ρU Γ i.e., the fluid exerts a downward force F = −ρU Γ on the vortex. Kutta-Joukowski Law 2D : F = −ρU Γ  × Γ 3D : F = ρU Generalized Kutta-Joukowski Law:   × F = ρU n   Γi i=1 . where F is the total force on a system of n vortices in a free stream with speed U 10