# Tag Info

## Hot answers tagged vortex

4

### Frequency of Vortex Shedding

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 ...

3

### Can a vortex be self-sustaining? [closed]

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

### Frequency of Vortex Shedding

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

### Why water in the sink follow a curved path?

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

### Why water in the sink follow a curved path?

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 ...

1

### Explanation for the next steps of chaplygin dipole

Equation (2.5) expresses the velocity field in function of the stream function. It's not clear to me it really should be presented at this stage in the process, I guess it's useful to impose the conditions at infinity. Equation (2.6) expresses that the two pieces of the total solution, the one inside the disk $\psi$ and the one outside $\psi_1$, have to ...

1

### how to determine if a vortex is laminar or turbulent

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 ...

1

### Explanation for the next steps of lamb-chaplygin dipole

When introducing the stream function, the steps that you usually take are as follows. Replace $u$ and $v$ by the streamfunction. Derive the horizontal momentum equation (for $u$) with respect to $y$ and the other with respect to $x$. Eliminate the pressure term, to end op with a single equation in $\psi$.

1

### Vortex street and Reynolds number

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 ...

1

### Vortex street and Reynolds number

The point at which the flow becomes turbulent is very sensitive to the flow geometry. The value of 2,300 you quote applies to flow in smooth pipes. To take an example, the flow round a sphere becomes ceases to be laminar at an Re of about 1. The flow becomes increasing turbulent as you raise the Reynold's number until vortex shedding starts around Re = 50.

1

### Effect of rotation on turbulence threshold for Reynolds number?

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 ...

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