2.20 – Marine Hydrodynamics, Fall 2005Lecture 11Copyrightc 2005 MIT - Department of Mechanical Engineering, All rights reserved.2.20 - Marine HydrodynamicsLecture 113.11 - Method of Images• Potential for single source: φ =m2πlnpx2+ y2mm• Potential for source near a wall: φ =m2πlnqx2+ (y − b)2+ lnqx2+ (y + b)2bbAddedsourceforsymmetry0=φdydmxymmNote: Be sure to verify that the boundary conditions are satisfied by symmetry orby calculus for φ (y) = φ (−y).1• Vortex near a wall (ground effect): φ = Ux+Γ2πtan−1(y − bx) − tan−1(y + bx)bbAddedvortexforsymmetryΓUxy-ΓVerify thatdφdy= 0 on the wall y = 0.• Circle of radius a near a wall: φ∼=Ux1 +a2x2+ (y − b)2+a2x2+ (y + b)2bbUyxyThis solution satisfies the boundary condition on the wall (∂φ∂n= 0), and the degree itsatisfies the boundary condition of no flux through the circle boundary increases asthe ratio b/a >> 1, i.e., the velocity due to the image dipole small on the real circlefor b >> a. For a 2D dipole, φ ∼1d, ∇φ ∼1d2.2• More than one wall:b'bUb'bUb'b'bbbbb'b'b'Γ-Γ-ΓΓbbb'b'b'b'bbExample2:Example1:Example3:3.12 Forces on a body undergoing steady translation“D’Alembert’s paradox”3.12.1 Fixed bodies & translating bodies - Galilean transformation.x’Uo’y’z’xoyzFixedinspace Fixedintranslatingbodyx=x`+ Ut3Reference system O: ~v, φ, p Reference system O’: ~v′, φ′, p′USOXUSO’X’∇2φ = 0 ∇2φ′= 0⇀v · ˆn =∂φ∂n=~U · ˆn = (U, 0, 0) · (nx, ny, nz)= Unxon Body⇀v′· ˆn′=∂φ′∂n= 0⇀v → 0 as |~x| → ∞⇀v′→ (−U, 0, 0) as |~x′| → ∞φ → 0 as |~x| → ∞ φ′→ −Ux′as |~x′| → ∞Galilean transform:⇀v(x, y, z, t) =⇀v′(x′= x − Ut, y, z, t) + (U, 0, 0)φ(x, y, z, t) = φ′(x′= x − Ut, y, z, t) + Ux′⇒−Ux′+ φ(x = x′+ Ut, y, z, t) = φ′(x′, y, z, t)Pressure (no gravity)p∞= −12ρv2+ Co= Co= −12ρv′2+ C′o= C′o−12ρU2∴ Co= C′o−12ρU2In O: unsteady flow In O’: steady flowps= −ρ∂φ∂t−12ρ v2|{z}U2+Cops= −ρ∂φ′∂t|{z}0−12ρ v′2|{z}0+C′o= C′o∂φ∂t= (∂∂t|{z}0+∂x′∂t|{z}−U∂∂x′) (φ′+ Ux′) = −U2∴ ps= ρU2−12ρU2+ Co=12ρU2+ Cops− p∞=12ρU2stagnation pressureps− p∞=12ρU2stagnation pressure43.12.2 ForcesBnˆTotal fluid force for ideal flow (i.e., no shear stresses):~F =ZZBpˆndSFor potential flow, substitute for p from Bernoulli:⇀F =ZZB−ρ∂φ∂t+12|∇φ|2|{z }hydrodynamicforce+ gy|{z}hydrostaticforce+c(t)ˆndSFor the hydrostatic case⇀v ≡ φ ≡ 0:⇀Fs=ZZB(−ρgyˆn) dS =↑Gausstheorem(−)↑outwardnormalZZZυB∇ (−ρgy) dυ = ρg∀ˆj|{z}Archimedesprinciplewhere ∀ =ZZZυBdυWe evaluate only the hydrodynamic force:⇀Fd= −ρZZB∂φ∂t+12|∇φ|2ˆndSFor steady motion∂φ∂t≡ 0:⇀Fd= −12ρZZBv2ˆndS53.12.3 Example Hydrodynamic force on 2D cylinder in a steady uniform stream.BSUax   nˆ⇀Fd=RB−ρ2|∇φ|2ˆndℓ =2πR0−ρ2|∇φ|2r=aˆnadθFx=⇀F ·ˆi =−ρa22πR0dθ |∇φ|2r=aˆn ·ˆi|{z}− cos θ=ρa22πR0|∇φ|2r=acos θdθVelo city potential for flow past a 2D cylinder:φ = Ur cos θ1 +a2r2Velo city vector on the 2D cylinder surface:∇φ|r=a= (vr|r=a, vθ|r=a) =∂φ∂r|{z}0r=a,1r∂φ∂θr=a|{z }−2U sin θSquare of the velocity vector on the 2D cylinder surface:|∇φ|2r=a= 4U2sin2θ6Finally, the hydrodynamic force on the 2D cylinder is given byFx=ρa22πZ0dθ4U2sin2θ cos θ=12ρU2|{z}ps−p∞(2a)|{z}diameterorprojection22πZ0dθ sin2θ|{z}evencos θ|{z}oddw.r.tπ2,3π2|{z }≡0= 0Therefore, Fx= 0 ⇒ no horizontal force ( symmetry fore-aft of the streamlines). Similarly,Fy=12ρU2(2a)22πZ0dθ sin2θ sin θ = 0In 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.73.13 Lift due to Circulation3.13.1 Example Hydrodynamic force on a vortex in a uniform stream.φ = Ux +Γ2πθ = Ur cos θ +Γ2πθUΓConsider a control surface in the form of a circle of radius r centered at the point vortex.Then according to Newton’s law:Σ~F =ddt~LCVsteady flow−→(~FV+~FCS) +~MNET= 0 ⇔~F ≡ −~FV=~FCS+~MNETWhere,~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.~MNET= Net linear momentum flux in the control volume through the control surface.ddt~LCV= Rate of change of the total linear momentum in the control volume.ControlvolumexUyθFyFxΓThe hydrodynamic force on the vortex is~F =~FCS+~MIN8a. Net linear momentum flux in the control volume through the control surfaces,~MNET.Recall that the control surface has the form of a circle of radius r centered at the pointvortex.a.1 The velocity components on the control surface areu = U −Γ2πrsin θv =Γ2πrcos θThe radial velocity on the control surface is therefore, given byur= U∂x∂r= U cos θ =⇀V · ˆnr2vπΓ=θθUa.2 The net horizontal and vertical momentum fluxes through the control surface aregiven by(MNET)x= − ρ2πZ0dθruvr= − ρ2πZ0dθrU −Γ2πrsin θU cos θ = 0(MNET )y= − ρ2πZ0dθrvvr= − ρ2πZ0dθrΓ2πrcos θU cos θ= −ρUΓ2π2πZ0cos2θdθ = −ρUΓ29b. Pressure force on the control surface,~FCS.b.1 From Bernoulli, the pressure on the control surface isp = −12ρ |~v|2+ Cb.2 The velocity |~v|2on the control surface is given by|~v|2=u2+ v2=U −Γ2πrsin θ2+Γ2πrcos θ2=U2−ΓπrU sin θ +Γ2πr2b.3 Integrate the pressure along the control surface to obtain~FCS(FCS)x=2πR0dθrp(− cos θ) = 0(FCS)y=2πR0dθrp(− sin θ) =−ρ2−ΓUπr(−r)2πZ0dθ sin2θ|{z }π= −12ρUΓc. Finally, the force on the vortex~F is given byFx= (FCS)x+ (Mx)IN= 0Fy= (FCS)y+ (My)IN= −ρUΓi.e., the fluid exerts a downward force F = −ρUΓ on the vortex.Kutta-Joukowski Law2D : F = −ρUΓ3D :~F = ρ~U ×~ΓGeneralized