2.20 - Marine Hydrodynamics, Spring 2005 Lecture 10 2.20 - Marine Hydrodynamics Lecture 10 3.7 Governing Equations and Boundary Conditions for P-Flow 3.7.1 Governing Equations for P-Flow (a) Continuity �2 φ = 0 � � 1 2 (b) Bernoulli for P-Flow (steady or unsteady) p = −ρ φt + |�φ| + gy + C(t) 2 3.7.2 Boundary Conditions for P-Flow Types of Boundary Conditions: (c) Kinematic Boundary Conditions - specify the ﬂow velocity �v at boundaries. ∂φ = Un ∂n (d) Dynamic Boundary Conditions - specify force F� or pressure p at ﬂow boundary. � � 1 2 p = −ρ φt + (�φ) + gy + C (t) (prescribed) 2 1 The boundary conditions in more detail: • Kinematic Boundary Condition on an impermeable boundary (no ﬂux condition) � �v ·n̂ = �� U ·n̂ = �� ﬂuid velocity �v =�φ boundary velocity Un �� = Given nornal boundary velocity �φ · n̂ = Un ⇒ (n1 ∂ ∂ ∂ + n2 + n3 )φ = Un ⇒ ∂x1 ∂x2 ∂x3 ∂φ = ∂n Un v U v n = (n1 , n 2 , n 3 ) v ( ) • Dynamic Boundary Condition: In general, pressure is prescribed � � 1 2 p = −ρ φt + (�φ) + gy + C (t) = Given 2 2 3.7.3 Summary: Boundary Value Problem for P-Flow The aforementioned governing equations with the boundary conditions formulate the Boundary Value Problem (BVP) for P-Flow. The general BVP for P-Flow is sketched in the following ﬁgure. KBC : (Lecture 19)  Free surface DBC : − ρ (φ + 1 (∇φ ) 2 + gy ) + C (t ) = GIVEN t 23  21 non − linear  ∇ 2φ = 0 1 2 p = − ρ (φt + (∇φ ) + gy ) + C (t ) 2 Solid boundary KBC : ∂φ = U n = GIVEN ∂n It must be pointed out that this BVP is satisﬁed instantaneously. 3 3.8 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 = U n =f on B ∂n • Stream function ψ. ∇2 ψ = 0 in V ψ=g on B Linear Superposition: if φ1 , φ2 , . . . are harmonic functions, i.e., �2 φi = 0, then φ = � αi φi , where αi are constants, are also harmonic, and is the solution for the boundary value problem provided the kinematic boundary conditions are satisﬁed, i.e., ∂φ ∂ = (α1 φ1 + α2 φ2 + . . .) = Un on B. ∂n ∂n The key is to combine known solution of the Laplace equation in such a way as to satisfy the kinematic boundary conditions (KBC). The same is true for the stream function ψ. The K.B.C specify the value of ψ on the boundaries. 4 3.8.1 Example �� Let φi x denote a unit-source ﬂow with source at xi , i.e., �� 1 φi x ≡ φsource x, xi = ln �x − xi � (in 2D) 2π � �� −1 = − 4π �x − xi � (in 3D), then ﬁnd mi such that φ= � � mi φi (x) satisﬁes KBC on B i Caution: φ must be regular for x ∈ V , so it is required that �x ∈ / V. v •x 2 v •x 1 ∇ 2 φ = 0 in V v v •x 3 •x 4 ∂Φ =f ∂n Figure 1: Note: �xj , j = 1, . . . , 4 are not in the ﬂuid domain V . 5 3.9 - Laplace equation in diﬀerent coordinate systems (cf Hildebrand §6.18) 3.9.1 Cartesian (x,y,z) � �v = ˆ j î ˆ k � � ∂φ ∂φ ∂φ , , ∂x ∂y ∂z u, v, w = �φ = �2 φ = ∂ 2φ ∂ 2φ ∂ 2φ + + 2 ∂x2 ∂y 2 ∂z z � êz P ( x, y , z ) y O x êx 6 ê y 3.9.2 Cylindrical (r,θ,z) r 2 = x2 + y 2 , θ = tan−1 (y/x) � ê ê ê � � ∂φ 1 ∂φ ∂φ � r z θ v = vr , vθ , vz = , , ∂r r ∂θ ∂z � ∂ 2 φ 1 ∂φ 1 ∂ 2 φ ∂ 2 φ + + 2 2 + 2 ⇔ 2 ∂z �∂r �� r ∂r� r ∂θ 1 ∂φ r ∂φ r ∂r ( ∂r ) � � 1 ∂φ ∂φ 1 ∂2φ ∂ 2φ 2 �φ = r + 2 2 + 2 r ∂r ∂r r ∂θ ∂z �2 φ = z êz P (r , θ , z ) O y θ r x êx 7 ê y 3.9.3 Spherical (r,θ,ϕ) r 2 = x2 + y 2 + z 2 , θ = cos−1 (z/r) ⇔ z = r (cos θ) ϕ = tan−1 (y/x) � � v = �φ = êr êθ êϕ vr , vθ , vϕ � � = ∂φ 1 ∂φ 1 ∂φ , , ∂r r ∂θ r(sin θ) ∂ϕ � � � ∂ 2φ ∂ 2 φ 2 ∂φ 1 ∂ 1 ∂φ �φ = + + ⇔ sin θ + 2 2 ∂θ r2 sin2 θ ∂ϕ2 �∂r �� r ∂r� r sin θ ∂θ 1 ∂ (r2 ∂φ ∂r ) r 2 ∂r � � � � 1 ∂ 1 ∂ ∂φ 1 ∂ 2φ 2 2 ∂φ �φ = 2 r + 2 sin θ + 2 2 r ∂r ∂r r sin θ ∂θ ∂θ r sin θ ∂ϕ2 2 z êz P (r , θ , φ ) θ r O y φ x êx 8 ê y 3.10 Simple Potential ﬂows 1. Uniform Stream �2 (ax + by + cz + d) = 0 1D: 2D: φ = U x + constant ψ = U y + constant; � φ = U x + V y + constant ψ = U y − V x + constant; � v = (U, 0, 0) v = (U, V, 0) � v = (U, V, W ) 3D: φ = U x + V y + W z + constant 2. Source (sink) ﬂow 2D, Polar coordinates 1 ∂ � = r ∂r 2 � � � ∂ 1 ∂2 r + 2 2 , with r = x2 + y 2 ∂r r ∂θ An axisymmetric solution: φ = a ln r + b. Verify that it satisﬁes �2 φ = 0, except at � r = x2 + y 2 = 0. Therefor, r = 0 must be excluded from the ﬂow. Deﬁne 2D source of strength m at r= 0: φ= �φ = m ln r 2π ∂φ m m êr = êr ⇐⇒ vr = , vθ = 0 ∂r 2πr 2πr y source (strength m) x 9 Net outward volume ﬂux is � � v · n̂ds = �� C � � · vds = �� S � � v · n̂ds = Sε �2π vr rε dθ = �� 0 Cε � � · vds m 2πrε m �� source strength y n̂ C Sε x ε S If m < 0 ⇒ sink. Source m at (x0 , y0 ): � m φ = ln (x − x0 )2 + (y − y0 )2 2π m m φ = ln r (Potential function) ←→ ψ = θ (Stream function) 2π 2π m θ 2π Ψ= y θ Vr = ψ=0 1 10 m 2π x 3D: Spherical coordinates 1 ∂ � = 2 r ∂r 2 � ∂ r ∂r 2 � � + � � ∂ ∂ , , · · · , where r = x2 + y 2 + z 2 ∂θ ∂ϕ A spherically symmetric solution: φ = a + b. Verify �2 φ = 0 except at r = 0. r Deﬁne a 3D source of strength m at r = 0. Then φ=− m ∂φ m ⇐⇒ vr = = , vθ = 0, vϕ = 0 4πr ∂r 4πr2 Net outward volume ﬂux is �� m � vr dS = 4πrε2 · = m (m < 0 for a sink ) 4πrε2 11 3. 2D point vortex 1 ∂ � = r ∂r 2 � � ∂ 1 ∂2 r + 2 2 ∂r r ∂θ Another particular solution: φ = aθ + b. Verify that �2 φ = 0 except at r = 0. Deﬁne the potential for a point vortex of circulation Γ at r = 0. Then Γ Γ ∂φ 1 ∂φ θ ⇐⇒ vr = = 0, vθ = = and, 2π ∂r r ∂θ 2πr 1 ∂ (rvθ ) = 0 except at r = 0 ωz = r ∂r φ= Stream function: ψ=− Γ ln r 2π Circulation: � � � � v · dx = C1 � � � v · dx + C2 �2π � � v · dx = C1 −C2 � � R R �� ωz dS=0 S 12 0 Γ rdθ = 2πr Γ �� vortex strength 4. Dipole (doublet ﬂow) A dipole is a superposition of a sink and a source with the same strength. 2D dipole: � � � � m 2 2 2 2 φ= ln (x − a) + y − ln (x + a) + y 2π � � � µ ∂ 2 2 lim φ = ln (x − ξ) + y �� a→0 2π ∂ξ ξ=0 �� µ = 2ma constant µ x µ x =− = − 2π x2 + y 2 2π r2 2D dipole (doublet) of moment µ at the origin oriented in the +x direction. ∂ NOTE: dipole = µ ∂ξ (unit source) 13 ξ unit source α x φ= −µ x cos α + y sin α −µ cos θ cos α + sin θ sin α = 2 2 2π x +y 2π r 3D dipole: ⎛ 1 ⎠ where µ = 2ma ﬁxed −� a→0 2 2 2 2 2 2 (x − a) + y + z (x + a) + y + z � � � µ ∂ 1 µ x µ x � � =− = − = − � 4π ∂ξ 4π (x2 + y 2 + z 2 )3/2 4π r3 (x − ξ)2 + y 2 + z 2 � φ = lim − m⎝ � 4π ⎞ 1 ξ=0 3D dipole (doublet) of moment µ at the origin oriented in the +x direction. 14 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 x + m � 2 ln x + y 2 2π U m U x D v stagnation point v = 0 Dividing Streamline ∂φ m x =U+ 2 ∂x 2π x + y 2 m u|y=0 = U + , v |y=0 = 0 ⇒ 2πx m V� = (u, v) = 0 at x = xs = − , y=0 2πU u= For large x, u → U , and U D = m by continuity ⇒ D = 15 m . U 3D: φ = U x − 4π � m x2 + y 2 + z 2 stagnation point div. streamlines ∂φ m x =U+ 2 2 ∂x 4π (x + y + z 2 )3/2 m x u|y=z=0 = U + , v |y=z=0 = 0, w|y=z=0 = 0 ⇒ 4π |x|3 � m � V = (u, v, w) = 0 at x = xs = − , y=z=0 4πU u= For large x, u → U and U A = m by continuity ⇒ A = 16 m . U 6. Stream + source/sink pair: Rankine closed bodies y S U +m -m S x a dividing streamline (see this with PFLOW) To have a closed body, a necessary condition is to have � min body = 0 2D Rankine ovoid: m φ = U x+ 2π � � � � � � m (x + a)2 + y 2 2 2 2 2 ln (x + a) + y − ln (x − a) + y = U x+ ln 4π (x − a)2 + y 2 3D Rankine ovoid: ⎡ φ = Ux − m⎣ � 4π ⎤ 1 2 −� (x + a) + y2 17 + z2 1 2 (x − a) + ⎦ y2 + z2 For Rankine Ovoid, � � ∂φ m x+a x−a u= =U+ − ∂x 4π �(x + a)2 + y 2 + z 2 �3/2 �(x − a)2 + y 2 + z 2 �3/2 � � m 1 1 u|y=z=0 =U + − 4π (x + a)2 (x − a)2 m (−4ax) =U + 4π (x2 − a2 )2 �2 � m � � u|y=z=0 =0 at x2 − a2 = 4ax 4πU At x = 0, u=U+ m 2a where R = y 2 + z 2 4π (a2 + R2 )3/2 Determine radius of body R0 : �R0 2π uRdR = m 0 18 7. Stream + Dipole: circles and spheres r U µx 2D: φ = U x + 2πr2 µ θ � = ↑ x=r cos θ µ � cos θ U r + 2πr The radial velocity is then ur = � ∂φ µ � = cos θ U − . ∂r 2πr2 Setting the radial velocity vr = 0 on r = a we obtain a = for a stationary circle of radius a. Therefore, for µ = 2πU a2 the potential � µ � φ = cos θ U r + 2πr is the solution to ideal ﬂow past a circle of radius a. • Flow past a circle (U, a). 19 � µ . 2πU This is the K.B.C. � φ = U cos θ r + Vθ = a2 r � � � 2 = −U sin θ 1 + ar2 � = 0 at θ = 0, π − stagnation points = −2U sin θ π 3π = �2U at θ = 2 , 2 − maximum tangential velocity 1 ∂φ r ∂θ Vθ |r=a 2U θ 2U Illustration of the points where the ﬂow reaches maximum speed around the circle. 3D: φ = U x + � µ cos θ µ � = U r cos θ 1 + 4π r2 4πr3 y µ r θ U x z The radial velocity is then vr = � ∂φ µ � = cos θ U − ∂r 2πr3 20 Setting the radial velocity vr = 0 on r = a we obtain a = for a stationary sphere of radius a. Therefore, choosing � 3 µ 2πU µ = 2πU a3 the potential � µ � φ = cos θ U r + 2πr is the solution to ideal ﬂow past a sphere of radius a. • Flow past a sphere (U, a). � � a3 φ = Ur cos θ 1 + 3 2r � � 1 ∂φ a3 vθ = = −U sin θ 1 + 3 r ∂θ 2r � 3U = 0 at θ = 0, π vθ |r=a = − sin θ at θ = π2 = − 3U 2 2 3/ 2 U θ 3/ 2 U 21 x . This is the K.B.C. 8. 2D corner ﬂow Velocity potential φ = rα cos αθ; Stream function ψ = rα sin αθ � 2 � ∂ 1 ∂ 1 ∂2 (a) �2 φ = ∂r + + φ=0 2 r ∂r r2 ∂θ2 (b) ∂φ = αrα−1 cos αθ ∂r 1 ∂φ uθ = = −αrα−1 sin αθ r ∂θ ∴ uθ = 0 { or ψ = 0} on αθ = nπ, n = 0, ±1, ±2, . . . ur = i.e., on θ = θ0 = 0, απ , 2π , . . . (θ0 ≤ 2π) α i. Interior corner ﬂow – stagnation point origin: α > 1. For example, α = 1, θ0 = 0, π, 2π, u = 1, v = 0 y x ψ=0 22 π 3π α = 2, θ0 = 0, , π, ,2π u = 2x, v = −2 y 2 2 o (90 corner) ψ=0 ψ=0 θ=2π/3, ψ = 0 α = 3 2 , θ0 = 0, (120o corner) 2π 4π , ,2π 3 3 120o 120o 120 θ=4π/3, ψ = 0 23 o θ=0, ψ = 0 θ=2π, ψ = 0 ii. Exterior corner ﬂow, |v| → ∞ at origin: α<1 π θ0 = 0, only α Since we need θ0 ≤ 2π, we therefore require π α ≤ 2π, i.e., α ≥ 1/2 only. 1/2 ≤ α < 1 π θ0 = 0, α For example, α = 1/2, θ0 = 0, 2π (1/2 inﬁnite plate, ﬂow around a tip) θ=0, ψ = 0 θ=2π, ψ = 0 α = 2/3, θ0 = 0, 3π (90o exterior corner) 2 θ=0, ψ = 0 θ=3π/2, ψ = 0 24 Appendix A1: Summary of Simple Potential Flows Cartesian Coordinate System Flow Streamlines Uniform ﬂow m 2π 2D Source/Sink (m) at (xo , yo ) Stream function ψ(x, y) U∞ x + V∞ y + W∞ z U∞ y − V∞ x ln((x − xo )2 + (y − yo )2 ) 1 m q − 4π 3D Source/Sink (m) at (xo , yo , zo ) (x−xo )2 +(y−yo )2 +(z−zo )2 Γ 2π Vortex (Γ) at (xo , yo ) α µ − 4π 25 y−yo arctan( x−x ) o m 2π y−yo arctan( x−x ) o NA Γ ln((x − x )2 + (y − y )2 ) − 2π o o (x−xo ) cos α+(y−yo ) sin α (x−xo )2 +(y−yo )2 µ (y−yo ) cos α+(x−xo ) sin α 2π (x−xo )2 +(y−yo )2 (x−xo ) ((x−xo )2 +(y−yo )2 +(z−zo )2 )3/2 NA µ − 2π 2D Dipole (µ) at (xo , yo ) at an angle α 3D Dipole (+x) (µ) at (xo , y0 , zo ) Potential φ(x, y, z) Appendix A2: Summary of Simple Potential Flows Cylindrical Coordinate System Flow Streamlines Uniform ﬂow Potential φ(r, θ, z) Stream function ψ(r, θ) U∞ r cos θ + V∞ r sin θ + W∞ z U∞ r sin θ − V∞ r cos θ 2D Source/Sink (m) at (xo , yo ) m 2π 3D Source/Sink (m) at (xo , yo , zo ) m − 4πr NA Γ θ 2π Γ ln r − 2π µ cos θ cos α+sin θ sin α − 2π r µ sin θ cos α+cos θ sin α 2π r µ cos θ − 4π 2 NA Vortex (Γ) at (xo , yo ) 2D Dipole (µ) at (xo , yo ) at an angle α α 3D Dipole (+x) (µ) at (xo , yo , zo ) ln r r 26 m θ 2π Appendix A3: Combination of Simple Potential Flows (2D) φ = U∞ x + m 2π ln r Rankine Half Body (3D) φ = U∞ x − m 4π √ Stream + Source + Sink (2D) φ = U∞ x + m 2π Rankine Closed Body (3D) φ = U∞ x + m √ 1 4π ( (x+a)2 +y 2 +z 2 Stream + Dipole (2D) φ = U∞ x + µx 2πr 2 if µ = 2πa2 U∞ φ = U∞ cos θ(r + a2 r ) Circle (Sphere) R = a (3D) φ = U∞ x + µ cos θ 4πr 2 if µ = 2πa3 U∞ φ = U∞ cos θ(r + a3 2r 2 ) 2D Corner Flow (2D) φ = Crα cos(αθ) ψ = Crα sin(αθ) θ0 = 0, nπ α Stream + Source m xs = − 2πU ∞ D= m U∞ A= m U∞ = � 1 x2 +y 2 +z 2 xs = − � m 4πU∞ ln((x + a)2 + y 2 ) − ln((x − a)2 + y 2 ) � = −√ 1 ) (x−a)2 +y 2 +z 2 = 27 Appendix B: Far Field Behavior of Simple Potential Flows Far ﬁeld behavior r >> 1 φ (2D) ∼ ln r �v = �φ ∼ 1 r Source (3D) ∼ 1 r ∼ 1 r2 (2D) ∼ 1 r ∼ 1 r2 (3D) ∼ 1 r2 ∼ 1 r3 (2D) ∼1 ∼ 1 r Dipole Vortex 28

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