18.325: Vortex Dynamics

Transcription

1 8.35: Vortex Dynamics Problem Sheet. Fluid is barotropic which means p = p(. The Euler equation, in presence of a conservative body force, is Du Dt = p χ. This can be written, on use of a vector identity, ( u t + u u ω = p χ. ( Take the curl: t (u ω = On use of a vector identity we get t ( p = p ( =. ( + u. ω ω. u u(.ω + ω(.u. (3 Now,.ω = since div curl=. Now use conservation of mass equation to substitute for.u:.u = D Dt (4 so Dω Dt ω. u ω D =. Dt (5 Dividing by gives the final result D Dt ( ω = ω. u. (6. Assume a barotropic fluid in a conservative force field. Let Γ = u.dl. (7 Take time derivative dγ dt = Du.dl + u.ddl Dt Dt But it is known that Ddl/Dt = du. Using this, together with the Euler equation, dγ ( dt = ( u p χ.dl + d (9 (8

2 But this can be written in the form ( dγ dt = dp d d χ + u.dl ( which is the integral, around a closed loop, of a total spatial differential of a single-valued function. It is therefore zero and yields Kelvin s circulation theorem for a barotropic fluid. 3. For a barotropic fluid, Euler s equation can be written in the form u t + + u + ω u = dp d d χ. ( First form of Bernoulli: Suppose the flow is steady. Taking the dot product of the Euler equation with u yields ( dp u. d d χ u ( which means that the quantity dp d d χ u (3 is constant on streamlines. Second form of Bernoulli: Suppose that the flow is irrotational. The Euler equation then says that ( dp d d + χ + u = (4 from which we deduce that dp d d + χ + u (5 is constant everywhere. Third form of Bernoulli: Suppose the flow is unsteady, but irrotational (note, by Q, we still have the persistence of irrotational flow for a barotropic fluid so this is a consistent statement. Then u = φ for some scalar φ. Then the Euler equation says that which means that ( φ t dp d d χ u = (6 φ t + dp d d + χ + u = H(t (7

3 for some function of time H(t. 4. Let u i and P be the velocity, density and pressure fields of any threedimensional steady solution of the incompressible Euler equation, i.e., u i u j + P =, (8 x j and (u i =. (9 onsider now the velocity, density and pressure fields û i, ˆ and ˆP given by where λ is assumed to be such that û i = λu i, ( ˆ = λ, ( ˆP = P ( u i λ =. (3 Note: this corresponds to the fact that λ is constant on streamlines. It must be shown that if (8, (9 and (3 hold then so do and û i û j + ˆ ˆP =, (4 x j (ˆû i =. (5 First, to show that (4 holds, note that by ( and (, û i û j + ˆ ˆP x j = λu j (λu j + λ x j ( = λ u i u j + P x j = P + λu i u j λ x j (6 where the last equality follows by (8 and provided (3 holds. 3

4 To show that (5 holds, note that (ˆû i = ( ui λ = (u i + u i λ = λ u λ i = ( λ (7 where we have used both (9 and (3. 5. Assume an ideal fluid and a flow field of the form u = (u r (r, z, t,, u z (r, z, t. (8 Taking the curl, in cylindrical polar coordinates, e r re θ e z ω = r θ z = (, ω(r, z, t, (9 u r (r, z, r u z (r, z, r where The vorticity equation is ω(r, z, t = u r z u z r. (3 t omputation of the right hand side gives = (u ω. (3 (, z (ωu z r (ωu r, (3 so only the azimuthal term gives a non-trivial equation i.e., or, equivalently, t + (ωu r + (ωu z r z ( t + ω ur r + u z + u r z r + u z z = (33 =. (34 But, the conservation of mass equation is.u = which, in cylindrical polar coordinates, takes the form u r r + u r r + u z z = (35 4

5 which can be used in the vorticity equation to reduce it to t ωu r + u r r r + u z z =. (36 But, dividing this by r, it is simply ( t + u r r + u (ω z = (37 z r which is the required result. Note that if the radius of a vortex ring increases then the vorticity equation just derived shows that the dynamics is such that the vorticity ω changes linearly with the radius, thus as a ring is stretched, the vorticity intensifies. 6. In spherical polars, the condition.u = takes the form or, multiplying by r sin θ, r r (r u r + r sin θ θ (sin θ u θ = (38 r (r sin θ u r + θ (r sin θ u θ =. (39 Therefore, introduce a streamfunction Ψ such that r sin θ u r = Ψ θ, r sin θ u θ = Ψ r. (4 omputing the vorticity field gives e r re θ r sin θe φ ω = r θ φ = (,, ω(r, θ, t (4 u r (r, θ, t u θ (r, θ, t where ω(r, θ, t = (ru θ u r r r r θ. (4 Substituting for u r and u θ in terms of Ψ then gives the final result ω = ( Ψ r sin θ r + Ψ r θ cot θ Ψ. (43 r θ For irrotational uniform flow past a sphere, we have ω = while u r U cos θ and u θ U sin θ as r. Therefore, as r, Ψ Ur sin θ/. This suggests trying a separable solution of the form Ψ(r, θ = f(r sin θ. (44 5

6 Substituting this ansatz into the vorticity equation just derived with ω = yields the ordinary differential equation which can be solved to yield = f (r f r (45 f(r = Ar + B r (46 where A and B are constants. From the far-field conditions, we must pick A = U/. On r = a (the spherical boundary, we need Ψ to be constant. Take Ψ = without loss of generality. This determines B and the final solution is Ψ = U (r a3 sin θ. (47 r Note: We will see this solution again when considering the Hill s spherical vortex. 7. Seek the solution of ψ = mδ(x x (48 that decays at infinity. Without loss of generality, take x =. Multiply this equation by e ik.x and integrate over all space (i.e. take a Fourier transform: e ik.x ψ = e ik.x mδ(x x (49 R 3 R 3 Green s identity says that R 3 ( u v v u dv = lim R S R (u v v u.ds (5 where S R is some radius-r spherical surface. The right side vanishes provided everything decays sufficiently fast at infinity. Letting u = ψ and v = e ik.x gives e ik.x ψdv = k Ψ(k (5 where Ψ(k is the Fourier transform of ψ, that is Ψ(k e ik.x ψdv. (5 R 3 Use of this in (48 gives the result Ψ(k = m k (53 6

7 Now, the easiest way to arrive at the result is to verify that the Fourier transform of m/(4πr is m/ k. But the Fourier transform of m/(4πr is π π R 4πr eik.x dv = 3 4πr ei k r cos θ r sin θdθdφdr (54 where we have adopted spherical polar coordinates to perform the integration. arrying out the φ integration gives π 4π (π dr dθe i k r cos θ r sin θdθ (55 but this allows a further integration with respect to θ yielding [ ] dr e i k r e i k r sin k r = dr = k i k k Im e i k z dz (56 where is the contour consisting of the infinite ray along the positive real z-axis. But e i k z is an analytic function of z in the first quadrant of the z-plane, moreover it decays exponentially on the contour R consisting of a large radius-r quarter-circle between the positive real and imaginary axes of the first quadrant. This means that auchy s theorem can be used to argue that the required integral is the same as k Im e i k z dz (57 where Ĉ is the ray consisting of the positive imaginary axis of the z-plane. Parametrizing this contour as z = iy for y < the integral becomes k Im Ĉ e k y idy = k (58 which verifies that the Fourier transform of /(4πr is / k as required. 8. From the Biot-Savart integral in 3d, u = 4π r (x 3 x ω(x dx dy dz (59 flow Now assume ω = everywhere off the plane z = and assume that x lies in this plane. Then u(x = (x x ω(x dx dy dz (6 4π [(x x + (y y + (z z ] 3/ flow so that, performing the z integration (using the hint, u(x = (x x ω(x dx dy 4π flow [(x x + (y y ] = (x x ω(x dx dy π ˆr flow (6 where ˆr = x x is the distance between two vectors x and x in the plane z =. This is precisely the d Biot-Savart result. 7