13.01 Marine Hydrodynamics, Fall 004 Lecture 9 Copyright c 004 MIT - Department of Ocean Engineering, All rights reserved. Vorticity Equation 13.01 - Marine Hydrodynamics Lecture 9 Return to viscous incompressible flow. ( ) N-S equation: v + v p v = + gy + ν v t ρ Now: () ω + ( v v ) = ν ω since φ = 0 for any φ (conservative forces) t ( v ) v = 1 ( v v ) v ( v ) ( ) v = v ω where v v = v v ( v ) v = ( ) v = ( v ) ω ( ω ) v + ( v ω ) = ( ω v ) ω ( v ) + = 0 incompressible fluid v ( ω ) = 0 since ( v) = 0
Therefore, ω t + ( v ) ω = ( ω ) v + ν ω or Dω Dt = ( ω ) v + ν }{{ ω} diffusion Kelvin s Theorem revisited. If ν 0, then D ω Dt = ( ω ) v, so if ω 0 everywhere at one time, ω 0 always. ν can be thought of as diffusivity of (momentum) and vorticity, i.e., ω once generated (on boundaries only) will spread/diffuse in space if ν is present. ω ω D v = υ v +... Dt D ω = υ ω +... Dt Diffusion of vorticity is analogous to the heat equation: T t = K T, where K is the heat diffusivity Also since ν 1 or mm /s, in 1 second, diffusion distance O ( νt ) O (mm), whereas diffusion time O (L /ν). So for a diffusion distance of L = 1cm, the necessary diffusion time needed is O(10)sec.
For D, v = (u, v, 0) and z 0. So, ω = v is to v (parallel to z-axis). Then, so in D we have ( ω ) v = ω x 0 x + ω y 0 D ω Dt = ν ω. y + ω z z 0 v 0, If ν = 0, D ω = 0, i.e. in D, following a particle, the angular velocity is conserved. Reason: in Dt D, the length of a vortex tube cannot change due to continuity. For 3D, e.g. Dω i Dt = ω v i j x }{{ j } vortex turning and stretching + ν ω i x j x j diffusion Dω Dt = ω u 1 x 1 vortex turning + ω u x vortex stretching u + ω 3 +diffusion x }{{ 3} vortex turning 3 u x 1 1
Example: Pile on a River Scouring What really happens as length of the vortex tube L increases? Why? IFCF is no longer a valid assumption. Ideal flow assumption implies that the inertia forces are much larger than the viscous effects (Reynolds number). R e UL ν Length increases diameter becomes really small Therefore IFCF is no longer valid. R e is not that big after all.
3.3 Potential Flow - ideal (inviscid and incompressible) and irrotational flow If ω 0 at some time t, then ω 0 always for ideal flow under conservative body forces by Kelvin s theorem. Given a vector field v for which ω = v 0, then there exists a potential function (scalar) - the velocity potential - denoted as φ, for which v = φ Note that ω = v = φ 0 for any φ, so irrotational flow guaranteed automatically. At a point x and time t, the velocity vector v( x, t) in cartesian coordinates in terms of the potential function φ( x, t) is given by v ( x, t ) = φ ( x, t ) ( φ = x, φ y, φ ) z φ(x) u u u = 0 x The velocity vector v is the gradient of the potential function φ, so it always points towards higher values of the potential function.
Governing Equations: Continuity: Number of unknowns φ v = 0 = φ φ = 0 Number of equations φ = 0 Therefore the problem is closed. φ and p (pressure) are decoupled. φ can be solved independently first, and after it is obtained, the pressure p is evaluated. p = f ( v ) = f ( φ) Solve for φ, then find pressure. 3.4 Bernoulli equation for potential flow (steady or unsteady) Euler eq: ( ) v v t + ( ) p v ω = ρ + gy Substitute v = φ into the Euler s equation above, which gives: ( ) ( ) ( ) φ 1 p + t φ = ρ + gy or { φ t + 1 φ + p } ρ + gy = 0, which implies that φ t + 1 φ + p + gy = f(t) ρ everywhere in the fluid for unsteady, potential flow. The equation above can be written as [ φ p = ρ t + 1 ] φ + gy + F (t)
which is the Bernoulli equation for unsteady or steady potential flow. Summary: Bernoulli equation for ideal flow. Steady rotational or irrotational flow along streamline. ( ) 1 p = ρ v + gy + C(ψ) Unsteady or steady irrotational flow everywhere in the fluid. For hydrostatics, v 0, t = 0. ( φ p = ρ t + 1 ) φ + gy + F (t) p = ρgy + c hydrostatic pressure (Archimedes principle) Steady and no gravity effect ( t = 0, g 0) Inertial, acceleration effect p = ρv + c = ρ φ + c Venturi pressure (created by velocity) p ρ φ + t p ρ v + t u p p + p x δx δx
3.5 - Boundary Conditions KBC on an impervious boundary v ˆn ˆn φ = u ˆn U n given DBC: specify pressure at the boundary, i.e., Note: On a free-surface p = p atm. 3.6 - Stream function no flux across boundary φ n = U n given ( φ ρ t + 1 ) φ + gy = given continuity: v = 0; irrotationality: v = ω = 0 velocity potential: v = φ, then v = ( φ) 0for any φ, i.e. irrotationality is satisfied automatically. Required for continuity: Stream function ψ defined by v = φ = 0 v = ψ ( ) Then v = ψ 0 for any ψ, i.e. satisfies continuity automatically. Required for irrotationality: v = 0 ( ) ) ψ = ( ψ ψ = 0 (1) still 3 unknowns v ψ
For D and axisymmetric flows, ψ is a scalar ψ (so stream functions are more useful for D and axisymmetric flows). For D flow: v = (u, v, 0) and z 0. v = ψ = î ĵ ˆk x y z ψ x ψ y ψ z = ( ) ( y ψ z î + ) ( x ψ z ĵ + x ψ y ) y ψ x ˆk Set ψ x = ψ y 0 and ψ z = ψ, then u = ψ ; v = ψ y x So, for D: ψ = x ψ x + y ψ y + z ψ z 0 Then, from the irrotationality (see (1)) ψ = 0 and ψ satisfies Laplace s equation. D polar coordinates: v = (v r, v θ ) and z 0. y r ê ê r x v = ψ = 1 r ê r rê θ ê z r θ z ψ r ψ θ ψ z v {}} r { = 1 r ψ z θ êr v θ {}}{ ψ z r êθ + v z { ( }}{ 1 r r rψ θ θ ψ r ) ê z
Again let ψ r = ψ θ 0 and ψ z = ψ, then v r = 1 ψ r θ and v θ = ψ r. For 3D but axisymmetric flows, ψ also reduces to ψ (read JNN 4.6 for details). Physical Meaning of ψ. In D: u = ψ y We define ψ and v =. x ψ( x, t) = ψ( x 0, t) + x x 0 v ˆndl total volume flux from left to right accross a curve C between x and x 0 C = ψ( x 0, t) + x t x x 0 (udy vdx) x o C nˆ For ψ to be single-valued, must be path independent. C = C or C C = 0 Therefore, ψ is unique because of continuity. C C v ˆn dl = S }{{ v} ds = 0 continuity
Let x 1, x be two points on a given streamline ( v ˆn = 0 on streamline) streamline ψ ( ) x ψ ( x1 ) = ψ ψ 1 + x x 1 v ˆn = 0 along streamline dl Therefore, ψ 1 = ψ,i.e., ψ is a constant along any streamline. For example, on an impervious stationary body v ˆn = 0, so ψ = constant on the body is the appropriate boundary condition. If the body is moving v ˆn = U n ψ = ψ 0 + U n given dl on the boddy
u = 0 φ n = ψ = constant 0 ψ = given ψ o Flux ψ = v x = u y. Therefore, u = ψ y ψ and v = x u streamline streamline (x,y) -v (x + x, y) ψ ψ + ψ
Summary: Potential formulation vs. Stream-function formulation for ideal flows potential stream-function definition v = φ v = ψ continuity v = 0 φ = 0 automatically ( ) satisfied ) irrotationality v = 0 automatically satisfied ψ = ( ψ ψ = 0 In D : w = 0, z = 0 φ = 0 for continuity ψ ψ z : ψ = 0 for irrotationality Cauchy-Riemann equations for (φ, ψ) = (real, imaginary) part of an analytic complex function of z = x +iy Cartesian (x, y) Polar (r,θ) u = φ x v = φ y v r = φ r v θ = 1 r φ θ u = ψ y v = ψ x v r = 1 r ψ θ v θ = ψ r For irrotational flow use φ For incompressible flow use ψ For both flows use φ or ψ Given φ or ψ for D flow, use Cauchy-Riemann equations to find the other: e.g. φ = xy ψ =? φ x = y = ψ ψ = 1 y y + f 1 (x) φ y = x = ψ ψ = 1 x x + f (y) } ψ = 1 (y x ) + const