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Vorticity Equation - Massachusetts Institute of Technology

Marine Hydrodynamics, Fall 2004. Lecture 9. c 2004 MIT - Department of Ocean Engineering, All rights reserved. Copyright . - Marine Hydrodynamics Lecture 9. Vorticity Equation Return to viscous incompressible flow. *.. v * * p *. N-S Equation : t + v v = + gy + 2 v *.. * *. *. () t + v v = 2 since = 0 for any (conservative forces). Now: * * 1 * * * *.. ( v ) v = v v v v 2 . v2 * 2 * *. v where v 2 v = v v * *. = . 2. 2 . * * v * * * * . v v = v = v 2. * * * * *. *. *. *.. = v v + v + v . | {z } | {z }. =0 = 0 since incompressible ( ~v ) = 0. fluid 1. Therefore, * * * *. *. + v = v + 2.

† Difiusion of vorticity is analogous to the heat equation: @T @t = Kr2T, where K is the heat difiusivity Also since ” ...

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Transcription of Vorticity Equation - Massachusetts Institute of Technology

1 Marine Hydrodynamics, Fall 2004. Lecture 9. c 2004 MIT - Department of Ocean Engineering, All rights reserved. Copyright . - Marine Hydrodynamics Lecture 9. Vorticity Equation Return to viscous incompressible flow. *.. v * * p *. N-S Equation : t + v v = + gy + 2 v *.. * *. *. () t + v v = 2 since = 0 for any (conservative forces). Now: * * 1 * * * *.. ( v ) v = v v v v 2 . v2 * 2 * *. v where v 2 v = v v * *. = . 2. 2 . * * v * * * * . v v = v = v 2. * * * * *. *. *. *.. = v v + v + v . | {z } | {z }. =0 = 0 since incompressible ( ~v ) = 0. fluid 1. Therefore, * * * *. *. + v = v + 2.

2 *. t or D * *. *. 2*. = v + . | {z }. Dt diffusion Kelvin's Theorem revisited. *. D . * * * *. If 0, then Dt = v, so if 0 everywhere at one time, 0 always. *. can be thought of as diffusivity of (momentum) and Vorticity , , once generated (on boundaries only) will spread/diffuse in space if is present.. Dv D . = 2v + .. = 2 + .. Dt Dt T. Diffusion of Vorticity is analogous to the heat Equation : t = K 2 T , where K is the heat diffusivity . Also since 1 or 2 mm2 /s, in 1 second, diffusion distance O t O (mm), whereas diffusion time O (L2 / ). So for a diffusion distance of L = 1cm, the necessary diffusion time needed is O(10)sec.

3 2. * * * *. For 2D, v = (u, v, 0) and z 0. So, = v is to v (parallel to z-axis). Then, . * * *. v = x + y + z v 0, |{z} x |{z} y z 0 |{z} 0. 0. so in 2D we have *. D . = 2 . *. Dt *. If = 0, DDt = 0, in 2D, following a particle, the angular velocity is conserved. Reason: in 2D, the length of a vortex tube cannot change due to continuity. For 3D, D i vi 2 i = j + . Dt x xj xj | {z j} | {z }. vortex turning and stretching dif f usion D 2 u2 u2 u2. = 1 + 2 + 3 +diffusion Dt x1 x2 x3. | {z } | {z } | {z }. vortex turning vortex stretching vortex turning 2. 1. u2. 3. x1. 3. Example: Pile on a River Scouring What really happens as length of the vortex tube L increases?

4 IFCF is no longer a valid assumption. Why? Ideal flow assumption implies that the inertia forces are much larger than the viscous effects (Reynolds number). UL. Re . Length increases diameter becomes really small Re is not that big after all. Therefore IFCF is no longer valid. 4. Potential Flow - ideal (inviscid and incompressible) and irrotational flow * *. If 0 at some time t, then 0 always for ideal flow under conservative body forces by Kelvin's theorem. * * *. Given a vector field v for which = v 0, then there exists a potential function (scalar) - the velocity potential - denoted as , for which *.

5 V = . * *. Note that = v = 0 for any , 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 (~x, t). is given by . * *. * . v x, t = x, t = , , x y z (x). u u x u=0. The velocity vector ~v is the gradient of the potential function , so it always points towards higher values of the potential function. 5. Governing Equations: Continuity: v = 0 = 2 = 0. *. Number of unknowns . Number of equations 2 = 0. Therefore the problem is closed. and p (pressure) are decoupled. can be solved independently first, and after it is obtained, the pressure p is evaluated.

6 * . p = f v = f ( ) Solve for , then find pressure. Bernoulli Equation for potential flow (steady or unsteady). Euler eq: * . v v2 * * p + v = + gy t 2 . *. Substitute v = into the Euler's Equation above, which gives: . 1 2 p + | | = + gy t 2 . or . 1 2 p + | | + + gy = 0, t 2 . which implies that 1 p + | |2 + + gy = f (t). t 2 . everywhere in the fluid for unsteady, potential flow. The Equation above can be written as . 1 2. p = + | | + gy + F (t). t 2. 6. which is the Bernoulli Equation for unsteady or steady potential flow. Summary: Bernoulli Equation for ideal flow. Steady rotational or irrotational flow along streamline.

7 1 2. p = v + gy + C( ). 2. Unsteady or steady irrotational flow everywhere in the fluid.. 1 2. p = + | | + gy + F (t). t 2. * . For hydrostatics, v 0, t = 0. p = gy + c hydrostatic pressure (Archimedes' principle).. Steady and no gravity effect ( t = 0, g 0). v 2 . p= + c = | |2 + c Venturi pressure (created by velocity). 2 2. Inertial, acceleration effect p . t + . *. p t v + . u p p p+ x x x 7. - Boundary Conditions KBC on an impervious boundary * * . v n =. |{z} u n no flux across boundary . |{z} = Un given n n Un given DBC: specify pressure at the boundary, , . 1. + | |2 + gy = given t 2.

8 Note: On a free-surface p = patm . - Stream function * * *. continuity: v = 0; irrotationality: v = = 0. * *. velocity potential: v = , then v = ( ) 0for any , irrotationality is satisfied automatically. Required for continuity: v = 2 = 0. *. *. Stream function defined by *. *. v = . *. *. *. Then v = 0 for any , satisfies continuity automatically. Required for irrotationality: *. *. *. v = 0 = 2 = 0. *. (1). | {z }. still 3 unknowns *. *. v . 8. *. For 2D and axisymmetric flows, is a scalar (so stream functions are more useful for 2D and axisymmetric flows). * . For 2D flow: v = (u, v, 0) and z 0.

9 I j k . * . *.. v = = x y z = z i + z j + y x k . y x x y x y z . Set x = y 0 and z = , then u = y ;v = . x So, for 2D: * . = x + y + z 0. x y z Then, from the irrotationality (see (1)) 2 = 0 and satisfies Laplace's Equation . * . 2D polar coordinates: v = (vr , v ) and z 0. y r r x vr v v e r re e z z }| { z }| { z }|z {. * 1 1 1 . v = = r . * z z z . = e r e + r r e z r r r r r . r z 9. 1 . Again let r = 0 and z = , then vr = r . and v = . r . *. For 3D but axisymmetric flows, also reduces to (read JNN for details). Physical Meaning of .. In 2D: u = y and v = . x . We define Z ~.

10 X Z ~. x (~x, t) = (~x0 , t) + ~v n dl = (~x0 , t) + (udy vdx). ~. x0 ~. x0. | {z }. total volume flux from left to right accross a curve C. between~x and ~x0. x C'. t xo C n . R. For to be single-valued, must be path independent. Z Z Z Z I ZZ. * *. = or = 0 v n dl = . | {z v} ds = 0. C C0 C C0 C C 0 S continuity Therefore, is unique because of continuity. 10. * * *. Let x1 , x2 be two points on a given streamline ( v n = 0 on streamline). streamline *. Zx 2. * * *. x2 = x1 + v n . |{z} dl | {z } | {z }. 2 1 *. x1 =0. along streamline Therefore, 1 = 2 , , is a constant along any streamline.


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