. - Marine Hdrodnamics, Spring 5 Lecture 11. - Marine Hdrodnamics Lecture 11 3.11 - Method of Images m Potential for single source: φ = ln + π m ( ) Potential for source near a wall: φ = m ln +( ) +ln +( + ) π m d φ d = m Added source for smmetr Note: e sure to verif that the oundar conditions are satisfied smmetr or calculus for φ () = φ ( ). 1
Vorte near a wall (ground effect): φ = + tan 1 ( ) tan 1 ( + ) π - Added vorte for smmetr dφ Verif that = on the wall =. d φ a a Circle of radius a near a wall: = 1+ + +( ) +( + ) This solution satisfies the oundar condition on the wall ( φ = ), and the degree it n satisfies the oundar condition of no flu through the circle oundar increases as the ratio /a >> 1, i.e., the velocit due to the image dipole small on the real circle 1 1 for >>a. For a D dipole, φ d, φ d.
More than one wall: Eample 1: Eample : Eample 3: - - 3.1 Forces on a od undergoing stead translation D Alemert s parado 3.1.1 Fied odies & translating odies - Galilean transformation. z o z o Fied in space Fied in translating od = ` + t 3
Reference sstem O: v, φ, p Reference sstem O : v,φ,p O X S O X S φ = v ˆn = φ = ˆn = (,, ) (n n,n,n z ) = n on od v as φ as Galilean transform: v(,, z, t) φ(,, z, t) + φ( = + t,,z,t) Pressure (no gravit) p = 1 ρv + C o = C o = φ = v ˆn = φ n = v (,, ) as φ as v ( = t,, z,t)+(,, ) = φ ( = t,, z,t)+ = φ (,,z,t) = 1 ρv + C o = C o 1 ρ In O: unstead flow C o p s = ρ φ 1 ρ t }{{} v +C o φ =( + )(φ + )= t }{{} t }{{} t p s = ρ 1 ρ + C o = 1 ρ + C o = C o 1 ρ In O : stead flow p s = ρ φ t }{{} 1 ρ }{{} v +C o = C o p s p = 1 ρ stagnation pressure p s p = 1 ρ stagnation pressure 4
3.1. Forces nˆ Total fluid force for ideal flow (i.e., no shear stresses): F = pˆ nds For potential flow, sustitute for p from ernoulli: For the hdrostatic case v φ : φ 1 F = ρ + φ + g +c(t) nds ˆ } t {{ }}{{} F s = ( ρgnˆ) ds = ( ) ( ρg) dυ = ρg ĵ where = dυ }{{} We evaluate onl the hdrodnamic force: φ 1 F d = ρ + φ ˆndS t φ For stead motion : t hdrodnamic force 1 F d = ρ hdrostatic force Gauss outward υ Archimedes υ theorem normal principle v nds ˆ 5
3.1.3 Eample Hdrodnamic force on D clinder in a stead uniform stream. S nˆ a ( F d = ) ρ φ π ρ ndl ˆ = φ î = ρa F =F dθ φ nˆ î }{{} = ρa π φ Velocit potential for flow past a D clinder: a φ = r cos θ 1+ r Velocit vector on the D clinder surface: φ 1 φ φ =(v r,v θ )=, }{{} r r θ } {{ } φ π cos θdθ cos θ Square of the velocit vector on the D clinder surface: =4 sin θ ˆnadθ sin θ 6
Finall, the hdrodnamic force on the D clinder is given π π ρa (1 ) F = dθ 4 sin θ cos θ = ρ (a) dθ sin θ cos θ = }{{}}{{}}{{}}{{} odd diameter even p s p or π (1 ) F = ρ (a) dθ sin θ sin θ = π 3π w.r.t, projection }{{} Therefore, F = no horizontal force ( smmetr fore-aft of the streamlines). Similarl, In fact, in general we find that F, on an D or 3D od. D Alemert s parado : No hdrodnamic force acts on a od moving with stead translational (no circulation) velocit in an infinite, inviscid, irrotational fluid. The moment as measured in a local frame is not necessaril zero. 7
3.13 Lift due to Circulation 3.13.1 Eample Hdrodnamic force on a vorte in a uniform stream. φ = + θ = rcos θ + θ π π Consider a control surface in the form of a circle of radius r centered at the point vorte. Then according to Newton s law: Where, F d stead flow ΣF = L CV dt (F V + F CS)+ M NET = F F V = F CS + M NET = Hdrodnamic force eerted on the vorte from the fluid. F V = F = Hdrodnamic force eerted on the fluid in the control volume from the vorte. F CS M NET d L CV dt = Surface force (i.e., pressure) on the fluid control surface. = Net linear momentum flu in the control volume through the control surface. = Rate of change of the total linear momentum in the control volume. F F θ Control volume The hdrodnamic force on the vorte is F = F CS + M IN 8
a. Net linear momentum flu in the control volume through the control surfaces, M NET. Recall that the control surface has the form of a circle of radius r centered at the point vorte. a.1 The velocit components on the control surface are u = sin θ πr v = cos θ πr The radial velocit on the control surface is therefore, given u r = r = cos θ = V nˆ v θ = πr θ a. The net horizontal and vertical momentum flues through the control surface are given π π (M NET ) = ρ dθruv r = ρ dθr sin θ cos θ = πr π π (MNET) = ρ dθrvv r = ρ dθr cos θ cos θ πr π ρ ρ = cos θdθ = π 9
. Pressure force on the control surface, F CS..1 From ernoulli, the pressure on the control surface is p = 1 ρ v + C. The velocit v on the control surface is given v =u + v = sin θ + cos θ πr πr = sin θ + πr πr.3 Integrate the pressure along the control surface to otain F CS π (F CS ) = dθrp( cos θ) = π π ρ (F CS ) = dθrp( sin θ) = ( r) dθ sin 1 θ = πr ρ c. Finall, the force on the vorte F is given }{{} π F =(F CS ) +(M ) IN = F =(F CS ) +(M ) IN = ρ i.e., the fluid eerts a downward force F = ρ on the vorte. Kutta-Joukowski Law D : F = ρ 3D : F = ρ Generalized Kutta-Joukowski Law: n F = ρ i where F is the total force on a sstem of n vortices in a free stream with speed. 1 i=1