10 11/15 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 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. (But define the direction of rotation as the one someone on earth looking at the bathtub sees.) 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.