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68

The spirals are used to prevent the formation of Kármán vortex sheets downwind of the chimney. They work by diverting the wind upwards on one side of the chimney and downwards on the other, creating a three-dimensional airflow pattern that disrupts the vortex sheet. Without them, the vortex shedding could cause vortex-induced vibration in the chimney, ...


13

The whirl is due to the net angular momentum the water has before it starts draining, which is pretty much random. If the circulation were due to Coriolis forces, the water would always drain in the same direction, but I did the experiment with my sink just now and observed the water to spin different directions on different trials. The Coriolis force is ...


11

Intuitive start of an answer: If you have counter rotating vortices they have zero net angular momentum (to first order). If they merged they would have to have no motion -> where did the energy go. In between the two axes of rotation the fluid moves in the same direction and has no mechanism for dissipation. By contrast for two vortices with the same ...


10

The Kutta condition is completely artificial. The potential equations are completely artificial. The potential equations are a mathematical construct we use because it's much simpler than the full Navier-Stokes set of equations. We know the Kutta condition is never actually upheld in any real flow ever. However, when we perform all of our mathematical ...


9

Because where they come close together the air in between circulates in such a way as to join them in a single path. Floris is right, but maybe this picture helps.


5

Since you want to explain it to your daughter, take a plastic bottle, cut the bottom open, turn it upside town, hold the top closed and fill it with water. Give her that bottle and have her release the top (which is on the bottom now, sorry for the bad phrasing). The water will whirl in different orientations whenever you repeat this (if it whirls at all) ...


5

The Coriolis acceleration goes like $-2\omega \times v$, which for the sake of an order of magnitude estimate we can take to be $a\sim \omega v$. But in order to get an observable effect, we don't just need an acceleration, we need a difference in acceleration between the two ends of the tub, which are separated by some distance $L\sim 1$ m. The ...


4

A fluid motion in a vortex creates a dynamic pressure that is lowest in the center increasing radially ($P \propto r^2$). The gradient of this pressure that forces the fluid to rotate around the axis. This is usually represented by a vector called vorticity, and defined by $\omega = \nabla \times v$. In simple terms, this means that the fluid is ...


4

This is a reasonable question. At the scale of a waterspout, the inertial forces of fast-moving air should be large compared to the viscous forces (i.e., very large Reynolds number). Yet the inflow along the surface of the water is laminar, where we would ordinarily expect boundary-layer vorticity (i.e., turbulence). A detailed description of the expected ...


4

So how this is done is a bit of a black art, much like how you choose what to use for other non-dimensional numbers in fluids (like Reynolds number). But you sort of have it backwards. You don't want to compute $St$ to find the shedding frequency; $St$ is good if you want to compare flows over different conditions but want to show the physics is the same, or ...


4

I think @Killercam is right, I'll try to explain the same thing a little more elaborately. Firstly. in the case considered, since the fluid and the cylinder is chosen, increase in velocity directly translates to increase in the Reynolds number as $R_e = \frac{\rho V D}{\mu}$. Before considering flow in the range $250 < R_e < 2\times 10^5$ , lets ...


4

The fly is carried away within the turbulent motion of the air the moving car generates. Therefore, it stays close to the car (for a short while) and returns without actually having to fly at 80 mph. -> Answer to your second question: No! A google search for "turbulence around car/obstacle/plane" gives colourful pictures of the wind field around moving ...


4

As best I have been able to tell, vortex air intakes are mostly a scam designed to sell useless car modifications to people, as discussed on this HowStuffWorks article. In case of the inevitable future link rot, I'll paste the article below: The internal combustion engines in cars and trucks are essentially large air pumps: The action of the pistons ...


3

A vortex ring has a finite core radius $a$. The circular line vortex is a vortex ring with an infinitesimal core radius $a\to 0$. It is unphysical and only occurs in an inviscid fluid.


3

First important thing to understand is that vortices and vorticity are not the same thing, despite the similarity of the words. A vortex is a region in a flow with spinning features (at a rather large scale if you wish), but it may be irrotational (zero vorticity). Vorticity is a local property of the fluid, the rate of rotation of an imaginary particle ...


3

The velocity of the flow divided by the diameter of the cylinder is the typical crossing time of the fluid, hence is directly related to the frequency of the observed oscillations for a specific Reynolds number. It is as simple as that. Clearly, this time scale is then correlated with observation to provide the Strouhal number for this particular phenomenon. ...


3

If you take a body of fluid having some angular momentum, like air or water in a bathtub, and draw it toward a center, such as by draining it down a hole, or draining it up with heat, it's going to concentrate its angular momentum in a smaller volume. That's how you get a vortex. Tornadoes happen when angular momentum of air is concentrated sufficiently in ...


3

Self-sustaining vortices without dissipation (energy loss) are possible in superfluids (like, e.g., liquid helium) because there is no internal friction (viscosity) for the superfluid component. Rotation goes on by inertia. This is as close a I can imagine to a "self-sustaining vortex" although admittedly has little to do with space-time.


3

The whirl happens in the draining tube, whose optimal solution to drain the bathtub is a laminar flow allowing for some rotation in the tube. What you see in the surface is the match between the solution of flow in the tube and the solution of flow in the surface. Angular momentum of the flow gets modified a lot as the tube twists and twists, sometimes even ...


3

They will equalize pressure at the entrance to the tube between them. That pressure is the density of the fluid times the height from the bottom to the lowest point in the vortex, because the fluid at the lowest point has zero velocity and so is equivalent to standing fluid.


3

The Wikipedia page on Dust Devils explains it quite well: Dust devils form when hot air near the surface rises quickly through a small pocket of cooler, low- pressure air above it. If conditions are just right, the air may begin to rotate. As the air rapidly rises, the column of hot air is stretched vertically, thereby moving mass closer to the ...


3

Yes, the velocity field inside the ellipse is really zero. To convince myself of it I have run a few numeric evaluations of the integral for various $(a,b)$ and $(x,y)$. Here is how can we obtain this result analytically. First, we introduce the elliptical coordinate system with coordinates $\chi$, $\theta$: $$x = c \cosh \chi \cos \theta, \qquad y = c ...


3

The difference between rain and water in the sink is that rain is simply falling, while water in the sink is being drawn into a center from a distance away, and the water in the sink is not perfectly still. It is rotating, if only a little bit. As it is drawn to the center, the rotation becomes more rapid. The principle is Conservation of Angular Momentum. ...


2

Turbulence isn't the same as unsteadiness - a vortex street is not necessarily a turbulent phenomenon. As an analogy that (for some reason) I find easier to understand, consider a convection experiment where we heat a fluid at the bottom and cool it at the top. Below a certain threshold value for the temperature difference, the heat is transferred only by ...


2

In basic principle, both could do the same thing. Pragmatically, water in a drain has the resistance of the sink/drain walls to influence the effect. (This is a hairpin vortex regime.) Basically, vortices differ per sink. Surface tension of a rain drop exceeds wind friction. Coriolis forces still exist within the rain drop, and could produce a ...


2

There is as far as I know no simple formula for the energy of a vortex. I am not a fluid dynamicist, but here are some problems you encounter in the theoretical computation. Hopefully a fluid dynamicist can clarify a workable practical formula: I am almost certain that such a formula will come from experiment as I hope to show by outlining the problems in ...


2

I recommend the original papers, [1]N. Read and D. Green, Phys. Rev. B 61 (2000), 10267 and [2]D. A. Ivanov, Phys. Rev. Lett. 86 (2001), 268. If you can understand the Bogoliubov-de Genns equation with the p wave order parameter, calculations are straightforward. I also recommend the paper which contains the similar calculation, [3]V. Gurarie and L. ...


2

You can think about it like this: It takes one day for the earth to perform a full rotation (about 86k seconds), on the other hand, it takes a few seconds for your sink to drain (lets say 10 seconds). So it takes 8600 times longer for the earth to do a full rotation than it takes the water to drain down the sink. It is not too hard to imagine that the ...


2

Firstly let us define what is meant by turbulent and laminar in a case such as the one you describe... The Reynolds number of a flow gives a measure of the relative importance of inertial forces (associated with convective flow) and viscos forces. From experimental observations it is seen that for values of Re below the so-called critical Reynolds number ...


2

Bernoulli's equation does not require that the flow be irrotational, just inviscid. Let's consider a vortex filament, and denote its surface by $S$ and volume by $V$. Using the identity you mentioned above, Euler's equation can be written as: $$ \rho\frac{\partial \boldsymbol{u}}{\partial t} + ...



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