10 11/06 W

1. For an ideal point vortex at the origin, the velocity field is given in cylindrical coordinates $r,\theta,z$ by
   
   $$\vec v = \frac{\Gamma}{2\pi r} {\widehat \imath}_\theta$$
   
   
   Show that the vorticity $\vec\omega=\nabla\times\vec v$ of this flow is everywhere zero. Now sketch a contour (closed curve) $C$ that loops once around the vortex at the origin, in the counter-clockwise direction. In fluid mechanics, (for any flow, not just this one), the circulation $\bar\Gamma$ of a contour is defined as
   
   $$\bar\Gamma = \oint_C \vec v \cdot { \rm d}\vec r$$
   
   
   Here the integration starts from an arbitrary point on the contour and loops back to that point in the counter-clocwise direction. Evaluate the circulation of your contour around the vortex. Do not take a circle as contour $C$; take a square or a triangle or an arbitrary curve. Of course you know that in polar coordinates an infinitesimal change ${\rm d}\vec r$ in position is given by
   
   $${\rm d}\vec r = {\widehat \imath}_r {\rm d}r + {\widehat \imath}_\theta r {\rm d}\theta$$
   
   
   (If not, you better also figure out what it is in spherical.) You should find that $\Gamma$ has a nonzero value for your contour.

2. So far so good. But the Stokes theorem of Calculus III says
   
   $$\oint \vec v \cdot { \rm d}\vec r = \int_A \nabla \times \vec v \cdot \vec n { \rm d}A$$
   
   
   where $A$ is an area bounded by contour $C$. You just showed that the left hand side in this equation is not zero, but that the right hand side is because $\nabla\times\vec v$ is. Something is horribly wrong???! To figure out what is going on, instead of using an ideal vortex, use the Oseen vortex
   
   $$\vec v = \frac{\Gamma}{2\pi r}\left(1 -e^{-r^2/4\nu t} \right) {\widehat \imath}_\theta$$
   
   
   To simplify the integrations, now take your contour C to be (the perimeter of) a circle around the origin in the $x,y$-plane, and take area $A$ to be the inside of that circle in the $x,y$-plane. Do both the contour integral and the area integral. In this case, they should indeed be equal. Now in the limit $t\downarrow 0^+$, the Oseen vortex becomes an ideal vortex (the exponential becomes zero). (The Oseen vortex is an initially ideal vortex that diffuses out in time due to viscosity.) So if you look at a very small time, you should be able to figure out what goes wrong for the ideal vortex with the Stokes theorem. You might want to plot the vorticity versus $r$ for a few times that become smaller and smaller. Based on that, explain what goes wrong for $t\downarrow 0^+$. Is the area integral of the vorticity of the ideal vortex really zero? Read up on delta functions.

3. Do bathtub vortices have opposite spin in the southern hemisphere as they have in the northern one? Derive some ballpark number for the exit speed and angular velocity of a bathtub vortex at the north pole and one at the south pole, assuming that the bath water is initially at rest compared to the rotating earth. Use Kelvin’s theorem. Note that the theorem applies to an inertial frame, not that of the rotating earth. So assume you look at the entire thing from a passing star ship. (Since you cannot see through the earth, you will either need to fly above the north pole or above the south pole, seeing different directions of rotation of the earth, counter-clockwise respectively clockwise.) What do you conclude about the starting question? In particular, how do you explain the bathtub vortices that we observe?

4. A Boeng 747 has a maximum take-off weight of about 400,000 kg and take-off speed of about 75 m/s. The wing span is 65 m. Estimate the circulation around the wing from the Kutta-Joukowski relation. This same circulation is around the trailing wingtip vortices. From that, ballpark the typical circulatory velocities around the trailing vortices, assuming that they have maybe a diameter of a quarter of the span. Compare to the typical take-off speed of a Cessna 52, 50 mph.

5. Model the two trailing vortices of a plane as two-dimensional point vortices (three-dimensional line vortices). Take them to be a distance $2\ell$ apart, and to be a height $h$ above the ground. Take the ground as the $x$-axis, and take the $y$-axis to be the symmetry axis midway between the vortices. Now:
   1. Identify the mirror vortices that represent the effect of the ground on the flow field. Make a picture of the $x,y$-plane with all vortices and their directions of circulation.
   2. Find the velocity at an arbitrary point $x$ on the ground due to all the vortices.
   3. From that, apply the Bernoulli law to find the pressure changes that the vortices cause at the ground. Sketch this pressure against $x$ for both $h$ significantly greater than $d$ and vice-versa. Ignore the fact that the flow is unsteady.

6. Continuing the previous question, Also find the velocity that the right-hand non-mirror vortex R experiences due to the other vortices. In particular find the Cartesian velocity components $u_{\rm R}$ and $v_{\rm R}$ in terms of $\Gamma$, $h$ and $\ell$.

   Now the right non-mirror vortex R moves with the velocity that the other vortices induce:
   

   $$\frac{{\rm d}\ell}{{\rm d}t} = u_{\rm R} \qquad \frac{{\rm d}h}{{\rm d}t} = v_{\rm R}$$
   
   
   If you substitute in the found velocities and take a ratio to get rid of time, you get an expression for ${\rm d}h/{\rm d}\ell$. Integrate that expression using separation of variables to find the trajectory of the vortices with time. Accurately draw these trajectories in the $x,y$-plane, indicating any asymptotes. Do the vortices end up at the ground for infinite time, or do they stay a finite distance above it?