3.5 - Boundary Conditions for Potential Flow

Transcript

1 Marine Hydrodynamics, Fall 2004 Lecture 10 Copyright c 2004 MIT - Department of Ocean Engineering, All rights reserved Marine Hydrodynamics Lecture Boundary Conditions for Potential Flow Types of Boundary Conditions: 1. Kinematic Boundary Conditions - specify the flow velocity v at boundaries. 2. Dynamic Boundary Conditions - specify force F or pressure p at flow boundary. Kinematic Boundary Conditions on an impermeable boundary (no flu condition): n v = n U = U n (given) where v = φ n φ ( = U n φ = U n n ) = n n n n 3 3 U n = ( n,n, ) 1 2 n3 Dynamic Boundary Conditions: Pressure is prescribed in general. ( p = ρ φ t + 1 ) 2 ( φ)2 + gy + C (t) (prescribed)

2 In general: Free surface 2 φ = 0 ρ φ t 1 + ( φ) 2 + gy + C t 2 non-linear ( ) given φ given n = Linear Superposition for Potential Flow In the absence of dynamic boundary conditions, the potential flow boundary value problem is linear. Potential function φ. 2 φ = 0 in V φ n = U n = f on B Stream function ψ.

3 2 ψ = 0 in V ψ=g on B Linear Superposition: if φ 1, φ 2,... are harmonic functions, i.e. 2 φ i = 0, then φ = α i φ i, where α i are constants, are also harmonic, and is the solution for the boundary value problem provided the boundary conditions (kinematic boundary condition) are satisfied, i.e. φ n = n (α 1φ 1 + α 2 φ ) = U n on B. The key is to combine known solution of the Laplace equation in such a way as to satisfy the K.B.C. (kinematic boundary condition). The same is true for the stream function ψ. K.B.C.s specify the value of ψ on the boundaries. Eample: ( ) φ i is a unit-source flow with source at i i.e. then find m i such that: ( ) φ i φ source ( ) 1, i = 2π ln i (in 2D) = ( 4π i ) 1 (in 3D), φ = i m i φ i ( ) satisfies KBC on B Caution: φ must be regular for V, so it is required that / V.

4 2 1 2 φ = 0 in V 3 4 Φ n = f Figure 1: Note: j, j = 1,..., 4 are not in the fluid domain V. Laplace equation in different coordinate systems (cf Hildebrand 6.18) 1. Cartesian (,y,z) ( ) ĵ v = îu, v, ˆkw = φ = 2 φ = ( φ, φ y, φ ) z ( ) 2 φ + 2 φ 2 y + 2 φ 2 z 2 y (,y,z) z 2. Cylindrical (r,θ,z)

5 r 2 = 2 + y 2, θ = tan 1 (y/) ( êr ) ( φ v = vr, v êθ θ, v êz z = r, 1 φ r θ, φ ) z 2 2 φ φ = r + 1 φ φ } 2 {{ r r} r 2 θ + 2 φ 2 z 2 1 φ r r (r φ r ) y z r (r,, z) θ z 3. Spherical (r,θ,ϕ) v = φ = r 2 = 2 + y 2 + z 2, θ = cos 1 (/r) or = r (cos θ) ϕ = tan 1 (z/y) ( ) ( ê r vr, v êθ θ, v êϕ φ ϕ = r, 1 φ r θ, 1 ) φ r(sin θ) ϕ

6 2 φ = 2 φ r + 2 φ } 2 {{ r r} 1 r 2 r(r 2 φ r ) + 1 r 2 sin θ ( sin θ φ ) + θ θ 1 r 2 sin 2 θ 2 φ ϕ 2 z (r, θ, ϕ) θ ϕ r(sinθ) y Simple Potential flows 1. Uniform Stream: 2 (a + by + cz + d) = 0 1D: φ = U + constant ψ = Uy + constant 2D: φ = U + V y + constant ψ = Uy V + constant 3D: φ = U + V y + W z + constant v = (U, 0, 0) v = (U, V, 0) v = (U, V, W ) 2. Source (sink) flow: 2D: Polar coordinates 2 = 1 r An aisymmetric solution: φ = ln r (verify) ( r ) + 1r 2 r r 2 θ, with r = 2 + y 2 2

7 Define 2D source of strength m at r = 0: φ = m 2π ln r It satisfies 2 φ = 0, ecept at r = 2 + y 2 = 0 (so must eclude r = 0 from flow) φ = φ r êr = m 2πr êr, i.e. v r = m 2πr, v θ = 0 y source (strength m) Net outward volume flu is C C ε v ˆnds = S 2π v ˆnds = 0 vds = vds S ε V }{{} r m 2πrε r ε dθ = m }{{} source strength

8 nˆ y C ε S ε S If m < 0 sink. Source m at ( 0, y 0 ): φ = m 2π ln ( 0 ) 2 + (y y 0 ) 2 φ = m ln r (Potential function) 2π ψ = m θ (Stream function) 2π y θ V r m Ψ = θ 2 π m = 2 π ψ = 0 1 3D: Spherical coordinates 2 = 1 ( r 2 ) ( ) + r 2 r r θ, ϕ,, where r = 2 + y 2 + z 2

9 A spherically symmetric solution: φ = 1 r (verify 2 φ = 0 ecept at r = 0) Define 3D source of strength m at r = 0: Net outward volume flu is φ = m 4πr, then V r = φ r = V r ds = 4πr 2 ε m 4πr 2 ε m 4πr 2, V θ, V ϕ = 0 = m (m < 0 for a sink ) 3. 2D point vorte 2 = 1 r ( r ) + 1r 2 r r 2 θ 2 Another particular solution: φ = aθ + b (verify 2 φ = 0 ecept at r = 0) Define the potential for a point vorte of circulation Γ at r = 0: φ = Γ 2π θ, then V r = φ r = 0, V θ = 1 φ r θ = Stream function: Γ 2πr and ω z = 1 r r (rv θ) = 0 ecept at r = 0 Circulation: ψ = Γ 2π ln r C 1 v d = C 2 v d + v d C 1 C 2 ω z ds=0 S }{{} = 2π 0 Γ 2πr rdθ = Γ }{{} vorte strength

10 S C 1 C 2 ω z = 0 4. Dipole (doublet flow) A Dipole is a superposition of a sink and a source with the same strength. 2D dipole:

11 φ = m 2π lim φ = µ a 0 }{{} 2π µ = 2ma constant = µ 2π [ ] ln ( a) 2 + y 2 ln ( + a) 2 + y 2 ξ ln ( ξ) 2 + y 2 ξ=0 2 + y = µ 2 2π 2D dipole (doublet) of moment µ at the origin oriented in the + direction. r 2 NOTE: dipole = µ ξ (unit source) ξ unit source α 3D dipole: φ = µ cos α + y sin α 2π = µ 2 + y 2 2π cos θ cos α + sin θ sin α r φ = lim a 0 = µ 4π ξ m 1 4π ( a) 2 + y 2 + z 2 1 ( ξ) 2 + y 2 + z 2 ξ=0 1 ( + a) 2 + y 2 + z 2 = µ 4π ( 2 + y 2 + z 2 ) = µ 3/2 4π where µ = 2ma fied r 3

12 3D dipole (doublet) of moment µ at the origin oriented in the + direction. 5. Stream and source: Rankine half-body It is the superposition of a uniform stream of constant speed U and a source of strength m. U m 2D: φ = U + m 2π ln 2 + y 2 U m D U stagnation point v = 0 Dividing Streamline u = φ = U + m 2π 2 +y 2 u y=0 = U + m 2π u = 0 at = s = m 2πU

13 For large, u U, and UD = m by continuity, D = m U. 3D: m φ = U 4π 2 + y 2 + z 2 stagnation point div. streamlines u = φ = U + m 4π u y=z=0 = U + m 4π ( 2 +y 2 +z 2 ) 3/2 3 u = 0 v=w=0 at = s = m 4πU For large, u U and UA = m by continuity, A = m U. 6. Stream + source/sink pair: Rankine closed bodies

14 y U S +m -m a S dividing streamline (see this with PFLOW) To have a closed body, a necessary condition is to have m in body = 0 2D Rankine ovoid: φ = U + m 2π ( ) ln ( + a) 2 + y 2 ln ( a) 2 + y 2 3D Rankine ovoid: φ = U m 1 4π ( + a) 2 + y 2 + z 2 1 ( a) 2 + y 2 + z 2 For Rankine Ovoid, [ ] u = φ = U + m + a 4π ( ( + a) 2 + y 2 + z 2) a 3/2 ( ( a) 2 + y 2 + z 2) 3/2 u y=z=0 =U + m [ ] 1 4π ( + a) 2 1 ( a) 2 =U + m ( 4a) 4π ( 2 a 2 ) 2 u y=z=0 =0 at ( 2 a 2) 2 = ( m 4πU ) 4a

15 At = 0, Determine radius of body R 0 : u = U + m 4π 2a (a 2 + R 2 ) 3/2 where R = y2 + z 2 7. Stream + Dipole: circles and spheres R 0 2π urdr = m 0 U µ r θ 2D: φ =U + µ, where = r cos θ 2πr2 ( = cos θ Ur + µ ) then V r = φ ( 2πr r = cos θ U µ ) 2πr 2 So V r = 0 on r = a = µ 2πU } (which is the K.B.C. for a stationary circle radius a) or choose µ = 2πUa 2. Steady flow past a circle (U,a):

16 ( ) φ = U cos θ r + a2 r ( ) V θ = 1 φ = U sin θ 1 + a2 r θ r { 2 = 0 at θ = 0, π SA and S V θ r=a = 2U sin θ B stagnation points. = 2U at θ = π, 3π maimum tangential velocity 2 2 2U θ 2U Illustration of the points where the flow reaches maimum speed around the circle. 3D:

17 y µ U z r θ Steady flow past a sphere (U, a): φ = U + µ cos θ 4π r 2, = r cos θ V r = φ r = cos θ ( U µ V r = 0 on r = a }{{} K.B.C. on stationary sphere of radius a 2πr 3 ) a = 3 µ 2πU or µ = 2πUa3 ( ) φ = U cos θ r + a3 2r 2 ( ) V θ = 1 φ = U sin θ 1 + a3 r θ 2r { 3 = 0 at θ = 0, π V θ r=a = 3U sin θ 2 = 3U at θ = π 2 2 θ 3U 2

18 8. 2D corner flow Potential function: φ = r α cos αθ, and the Stream function: ψ = r α sin αθ ( ) (a) 2 φ = φ = 0 r 2 r r r 2 θ 2 (b) u r = φ = r αrα 1 cos αθ u θ = 1 φ = r θ αrα 1 sin αθ u θ = 0 { or ψ = 0} on αθ = nπ, n = 0, ±1, ±2,... i.e. on θ = θ 0 = 0, π α, 2π α,... (θ 0 2π) i. interior corner flow stagnation point origin: α > 1. e.g. α = 1, θ 0 = 0, π, 2π, u = 1, v = 0 y ψ = 0

19 π 3π α = 2, θ0 = 0,, π,,2π u = 2, v = 2y 2 2 (90 o corner) ψ = 0 ψ = 0 θ=2π/3, ψ = 0 2π 4π α = 3 2, θ0 = 0,,, 2π 3 3 (120 o corner) 120 o 120 o 120 o θ=0, ψ = 0 θ=2π, ψ = 0 θ=4π/3, ψ = 0

20 ii. Eterior corner flow, v at origin: α < 1( 1 / 2 α < 1). θ 0 = 0, π only. Since we α need θ 0 2π, we therefore require π 2π, i.e. α 1/2 only. α e.g. α = 1/2, θ 0 = 0, 2π ( 1 / 2 infinite plate, flow around a tip) θ=0, ψ = 0 θ=2π, ψ = 0 α = 2/3, θ 0 = 0, 3π 2 (90o eterior corner)

21 θ=3π/2, ψ = 0 θ=0, ψ = 0