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While discharging water from a short tube, I noticed that a vortex began to form on the top of the discharge tube causing the jet issuing from a short tube to contract (less than the cross sectional area of the short tube).

My question is why does the jet contract below the short tube?

When the vortex forms the velocity is at a maximum:

$$A_1v_1=A_2v_2$$

So, the area must be a minimum.

I also observed that the contracted jet maintains this state until all the water is discharged from the container. Could this be explained in terms of head loss caused by the vortex?

Jet before

Jet after vorcity formation

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  • $\begingroup$ What do A1, A2, v1, v2 refer to? And can you give an indication of the dimensions? $\endgroup$ – Han-Kwang Nienhuys Jun 4 '16 at 19:52
  • $\begingroup$ What's a "vorcity"? Do you mean "vorticity"? or maybe "vortex"? $\endgroup$ – Dave Tweed Jun 4 '16 at 21:33
  • $\begingroup$ @Han-KwangNienhuys: I think we can assume that An is cross-sectional area and vn is velocity. Their product is a constant for an incompressible fluid. $\endgroup$ – Dave Tweed Jun 4 '16 at 21:34
  • $\begingroup$ @DaveTweed - yeah, that part I guessed alright. But which locations in the flow path do '1' and '2' refer to? $\endgroup$ – Han-Kwang Nienhuys Jun 4 '16 at 21:38
  • $\begingroup$ Is that air on the second picture? If that's air than I think it answers your question right? $\endgroup$ – Lucas Franceschi Jun 5 '16 at 1:33
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On the second picture, it does seem like there is air inside the pipe. Is that correct?

If that is so, the reason there is less area outside the pipe is that the water inside the tube was only on the walls of the tube, with an empty space (air) in its core. Now when the water leaves the tube the water is not attracted to the tube's walls anymore and by surface tension cluster together on the center. Although it seems like the area is smaller that is not the case, there was air on the center of the tube and after the water leaves there's no air anymore.

I hope that helps.

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  • $\begingroup$ Hi @Lucas Franceschi, you are correct - there was air inside the pipe. Why is the water not attracted to the tube's walls when it leaves the tube? $\endgroup$ – user314409 Jun 5 '16 at 2:11
  • $\begingroup$ well, there isn't any walls anymore after it leaves the tube right? you can see a small amount of water stops at the exit of the tube, attracted to it, but the rest of the jet just clusters up to compensate for the lack of air that was present on the tubes. If you consider that the tube isn't completely filled but has air in it than the 'jet contraction' is more understandable. $\endgroup$ – Lucas Franceschi Jun 5 '16 at 2:16
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I think flow rate and surface tension are the two, main ingredients here. You can see the same effect at your kitchen tap when you vary from high flow to low flow. The contraction is readily visible.

I would suspect that low velocities allow surface tension more time to act and contract the water boundary. Higher flows may tend to dominate the surface tension.

Perhaps a good analogy would be how the velocity profile within a pipe during laminar flow is decidedly different from the velocity profile during turbulent flow. In turbulent flow, it's decidedly more "square".

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