3.7 Governing Equations and Boundary Conditions for P-Flow

Transcript

1 .0 - Maine Hydodynaics, Sping 005 Lectue Maine Hydodynaics Lectue Govening Equations and Bounday Conditions fo P-Flow Govening Equations fo P-Flow (a Continuity φ = 0 ( 1 (b Benoulli fo P-Flow (steady o unsteady p = ρ φ t + + gy + C(t φ 3.7. Bounday Conditions fo P-Flow Types of Bounday Conditions: φ (c Kineatic Bounday Conditions - specify the flow velocity v at boundaies. = U n n (d Dynaic Bounday Conditions - specify foce F o pessue p at flow bounday. ( 1 p = ρ φ t + ( φ + gy + C (t (pescibed 1

2 The bounday conditions in oe detail: Kineatic Bounday Condition on an ipeeable bounday (no flux condition v nˆ = U nˆ = U n = Given }{{} }{{} }{{} fluid velocity bounday velocity nonal bounday velocity v= φ φ nˆ = U n (n 1 x 1 + n x + n 3 x 3 φ = U n φ n = U n U v v n = v ( n,n, 1 n3 Dynaic Bounday Condition: In geneal, pessue is pescibed ( 1 p = ρ φ t + ( φ + gy + C (t = Given

3 3.7.3 Suay: Bounday Value Poble fo P-Flow The afoeentioned govening equations with the bounday conditions foulate the Bounday Value Poble (BVP fo P-Flow. The geneal BVP fo P-Flow is sketched in the following figue. KBC : (Lectue19 Fee suface 1 DBC : ( + ( ρ φ 1 φ non t 3 linea + gy + C( t = GIVEN φ = 0 1 p = ρ ( φt + + ( φ + gy C( t φ Solid bounday KBC : U n = n = GIVEN It ust be pointed out that this BVP is satisfied instantaneously. 3

4 3.8 Linea Supeposition fo Potential Flow In the absence of dynaic bounday conditions, the potential flow bounday value poble is linea. Potential function φ. φ = 0 in V φ U n = n = f on B Stea function ψ. ψ = 0 in V ψ=g on B Linea Supeposition: if φ 1, φ,... ae haonic functions, i.e., φ i = 0, then φ = αi φ i, whee α i ae constants, ae also haonic, and is the solution fo the bounday value poble povided the kineatic bounday conditions ae satisfied, i.e., φ = (α 1 φ 1 + α φ +... = U n on B. n n The key is to cobine known solution of the Laplace equation in such a way as to satisfy the kineatic bounday conditions (KBC. The sae is tue fo the stea function ψ. The K.B.C specify the value of ψ on the boundaies. 4

5 3.8.1 Exaple Let φ i ( x denote a unit-souce flow with souce at xi, i.e., then find i such that ( ( 1 φ i x φ souce x,x i = ln x x i (in D π ( = 4π x xi 1 (in 3D, φ = i i φ i ( x satisfies KBC on B Caution: φ ust be egula fo x V, so it is equied that x / V. x v x v 1 φ = 0 in V x v 3 x v 4 Φ n = f Figue 1: Note: x j, j = 1,..., 4 ae not in the fluid doain V. 5

6 3.9 - Laplace equation in diffeent coodinate systes (cf Hildeband Catesian (x,y,z ( ( î ˆ j k ˆ φ φ φ v = u, v, w = φ =,, x y z φ φ φ φ = + x y + z z ê z P( x, y, z O y ê y x ê x 6

7 3.9. Cylindical (,θ,z = x + y, θ = tan 1 (y/x ( ê ( ê θ ê z φ v = v, v θ, v z =, 1 φ θ, φ z φ = φ = φ 1 φ φ φ + θ z }{{} 1 φ φ ( 1 φ ( φ 1 φ φ + + θ z z ê z P (, θ, z x O θ y ê y ê x 7

8 3.9.3 Spheical (,θ,ϕ = x + y + z, θ = cos 1 (z/ z = (cos θ ϕ = tan 1 (y/x ( ( ê ê θ ê ϕ φ 1 φ v = φ = v, v θ, v ϕ =, θ, 1 (sin θ φ ϕ φ = φ = φ φ 1 ( + + sin θ φ 1 φ + sin θ θ θ sin θ ϕ }{{} 1 φ ( 1 ( φ 1 ( φ 1 φ + sin θ + sin θ θ θ sin θ ϕ z ê z P (, θ, φ θ x O φ y ê y ê x 8

9 3.10 Siple Potential flows 1. Unifo Stea (ax + by + cz + d = 0 1D: φ = Ux + constant ψ = Uy + constant; v = (U, 0, 0 D: φ = Ux + V y + constant ψ = Uy V x + constant; v = (U, V, 0 3D: φ = Ux + V y + W z + constant v = (U, V, W. Souce (sink flow D, Pola coodinates = ( 1 1 +, with = x + y θ An axisyetic solution: φ = a ln + b. Veify that it satisfies φ = 0, except at = x + y = 0. Theefo, = 0 ust be excluded fo the flow. Define D souce of stength at = 0: φ = π ln φ = φ ê = π ê v = π, v θ = 0 y souce (stength x 9

10 Net outwad volue flux is C C ε v ˆnds = v ˆnds = nˆ S π 0 y vds = v }{{} πε ε dθ = C S ε vds }{{} souce stength ε S ε x S If < 0 sink. Souce at (x 0, y 0 : φ = ln (x x 0 + (y y 0 π φ = ln (Potential function ψ = θ π π (Stea function y θ V Ψ = θ π = π ψ = 0 1 x 10

11 3D: Spheical coodinates 1 ( ( = + θ, ϕ,, whee = x + y + z a A spheically syetic solution: φ = + b. Veify φ = 0 except at = 0. Define a 3D souce of stength at = 0. Then φ φ = 4π v = = 4π, v θ = 0, v ϕ = 0 Net outwad volue flux is v ds = 4π ε 4πε = ( < 0 fo a sink 11

12 3. D point votex = 1 ( + 1 θ Anothe paticula solution: φ = aθ + b. Veify that φ = 0 except at = 0. Define the potential fo a point votex of ciculation Γ at = 0. Then Stea function: φ = Γ π θ v = φ ω z = 1 = 0, v θ = 1 (v θ = 0 except at = 0 φ θ = Γ π and, Ciculation: C 1 v d x = C v d x + ψ = Γ π ln C 1 C v d x } R R {{} ω z ds=0 S = π 0 Γ π dθ = Γ }{{} votex stength 1

13 4. Dipole (doublet flow A dipole is a supeposition of a sink and a souce with the sae stength. D dipole: [ ] φ = ln (x a + y ln (x + a + y π µ li φ = ln (x ξ + y a 0 }{{} π ξ ξ=0 µ = a constant µ x µ x = π x + y = π D dipole (doublet of oent µ at the oigin oiented in the +x diection. NOTE: dipole = µ ξ (unit souce 13

14 ξ unit souce α x φ = µ x cos α + y sin α µ cos θ cos α + sin θ sin α = π x + y π 3D dipole: 1 1 φ = li whee µ = a fixed a 0 4π (x a + y + z (x + a + y + z µ 1 µ x µ x = 4π ξ = ( x ξ + y + z 4π (x + y + z = 3/ 4π 3 ξ=0 3D dipole (doublet of oent µ at the oigin oiented in the +x diection. 14

15 5. Stea and souce: Rankine half-body It is the supeposition of a unifo stea of constant speed U and a souce of stength. U D: φ = Ux + π ln x + y U x D U v stagnation point = 0 Dividing Stealine φ x u = = U + x π x + y u y=0 = U +, v y=0 = 0 πx V = (u, v = 0 at x = x s =, y = 0 πu Fo lage x, u U, and UD = by continuity D =. U 15

16 3D: φ = Ux 4π x + y + z stagnation point div. stealines φ x u = = U + x 4π (x + y + z 3/ x u y=z=0 = U + 3, v y=z=0 = 0, w y=z=0 4π x = 0 V = (u, v, w = 0 at x = x s =, y = z = 0 4πU Fo lage x, u U and UA = by continuity A =. U 16

17 6. Stea + souce/sink pai: Rankine closed bodies y U S + - a S x dividing stealine (see this with PFLOW To have a closed body, a necessay condition is to have in body = 0 D Rankine ovoid: ( ( (x + a + y φ = Ux+ ln (x + a + y ln (x a + y = Ux+ ln π 4π (x a + y 3D Rankine ovoid: 1 1 φ = Ux 4π (x + a + y + z (x a + y + z 17

18 Fo Rankine Ovoid, [ ] φ x + a u = = U + x 4π ( (x + a + y + z x a 3/ ( (x a + y + z 3/ [ ] 1 1 u y=z=0 =U + 4π (x + a (x a ( 4ax =U + 4π (x u y=z=0 =0 at ( x a a ( = 4ax 4πU At x = 0, Deteine adius of body R 0 : a u = U + whee R = y + z 4π (a + R 3/ R 0 π urdr = 0 18

19 7. Stea + Dipole: cicles and sphees U µ θ µx ( µ D: φ = Ux + = cos θ U + π π x= cos θ The adial velocity is then u = φ ( = cos θ U µ π Setting the adial velocity v = 0 on = a we obtain a = µ fo a stationay cicle of adius a. Theefoe, fo µ = πua the potential ( µ φ = cos θ U + π is the solution to ideal flow past a cicle of adius a. Flow past a cicle (U, a.. πu. This is the K.B.C. 19

20 ( φ = U cos θ + a (1 V θ = 1 φ = U sin θ + a θ { = 0 at θ = 0, π stagnation points V θ =a = U sin θ = U at θ = π axiu tangential velocity, 3π U θ U Illustation of the points whee the flow eaches axiu speed aound the cicle. µ cos θ 3D: φ = Ux + 4π ( µ = U cos θ 1 + 4π 3 y µ U z θ x The adial velocity is then v = φ ( µ = cos θ U π 3 0

21 Setting the adial velocity v = 0 on = a we obtain a = πu fo a stationay sphee of adius a. Theefoe, choosing 3 µ. This is the K.B.C. µ = πua 3 the potential ( µ φ = cos θ U + π is the solution to ideal flow past a sphee of adius a. Flow past a sphee (U, a. ( 3 a φ = U cos θ v θ = 1 φ = U sin θ (1 + a3 θ 3 v θ =a = 3U { = 0 at θ = 0, π sin θ = 3U at θ = π 3 / U θ x 3 / U 1

22 8. D cone flow Velocity potential φ = α cos αθ; Stea function ψ = α sin αθ ( (a φ = φ = 0 θ (b φ u = = α α 1 cos αθ 1 φ u θ = = α α 1 sin αθ θ u θ = 0 { o ψ = 0} on αθ = nπ, n = 0, ±1, ±,... i.e., on θ = θ 0 = 0, π α, π α,... (θ 0 π i. Inteio cone flow stagnation point oigin: α > 1. Fo exaple, α = 1, θ 0 = 0, π, π, u = 1, v = 0 y x ψ = 0

23 π 3π α =, θ0 = 0,, π,,π u = x, v = y (90 o cone ψ = 0 ψ = 0 θ=π/3, ψ = 0 π 4π α = 3, θ0 = 0,,, π 3 3 (10 o cone 10 o 10 o 10 o θ=0, ψ = 0 θ=π, ψ = 0 θ=4π/3, ψ = 0 3

24 ii. Exteio cone flow, v at oigin: α < 1 π θ 0 = 0, only α Since we need θ 0 π, we theefoe equie π α π, i.e., α 1/ only. Fo exaple, 1/ α < 1 θ 0 = 0, π α α = 1/, θ 0 = 0, π ( 1 / infinite plate, flow aound a tip θ=0, ψ = 0 θ=π, ψ = 0 α = /3, θ 0 = 0, 3π (90 o exteio cone θ=0, ψ = 0 θ=3π/, ψ = 0 4

25 Appendix A1: Suay of Siple Potential Flows Catesian Coodinate Syste Flow Stealines Potential Stea function φ(x, y, z ψ(x, y Unifo flow U x + V y + W z U y V x D Souce/Sink ( at (x o, y o π ln((x x o + (y y o π actan( y y o x xo 3D Souce/Sink ( at (x o, y o, z o 4π 1 NA q(x xo +(y yo +(z zo Votex (Γ at (x o, y o Γ y yo actan( π x xo Γ π ln((x x o + (y y o D Dipole (µ at (x o, y o at an angle α α µ (x xo cos α+(y yo sin α µ (y yo cos α+(x xo sin α π (x xo +(y yo π (x xo +(y yo 3D Dipole (+x (µ at (x o, y 0, z o µ (x xo 4π ((x xo +(y yo +(z zo 3/ NA 5

26 Appendix A: Suay of Siple Potential Flows Cylindical Coodinate Syste Flow Stealines Potential Stea function φ(, θ, z ψ(, θ Unifo flow U cos θ + V sin θ + W z U sin θ V cos θ D Souce/Sink ( at (x o, y o π ln π θ 3D Souce/Sink ( at (x o, y o, z o 4π NA Votex (Γ at (x o, y o Γ π θ Γ π ln D Dipole (µ at (x o, y o at an angle α α µ cos θ cos α+sin θ sin α π µ sin θ cos α+cos θ sin α π 3D Dipole (+x (µ at (x o, y o, z o µ 4π cos θ NA 6

27 Appendix A3: Cobination of Siple Potential Flows Stea + Souce (D φ = U x + π ln x s = πu D = U = Rankine Half Body (3D φ = U x 4π 1 x +y +z x s = 4πU A = U Stea + Souce + Sink = Rankine Closed Body (D (3D [ φ = U x + π ln((x + a + y ln((x a + y ] φ = U x + 4π ( 1 1 (x+a +y +z (x a +y +z Stea + Dipole = Cicle (Sphee R = a (D (3D φ = U x + µx π if µ = πa U φ = U cos θ( + a φ = U x + µ cos θ 4π if µ = πa 3 U φ = U cos θ( + a 3 D Cone Flow (D φ = C α cos(αθ ψ = C α sin(αθ θ 0 = 0, nπ α 7

28 Appendix B: Fa Field Behavio of Siple Potential Flows Fa field behavio >> 1 φ v = φ Souce (D (3D ln Dipole (D (3D Votex (D 1 1 8