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I believe that there is big misconception about Bernoulli principle. I was curious about how liquid atomizers work and all those explanations just don't make sense for me. People say that its pressure reduction because of air acceleration. But the key point is pressure drop is a reason, not a result. Pressure that built up in sphere right before point 1, drops to atmospheric pressure meanwhile air gains velocity at the exit of point 1. So there are no further pressure drops. The drop had already happened from some positive effective pressure to atm pressure when air traveled through point 1. So after that air travels with atm pressure and velocity V which was gained by excessive pressure in the spherical pump. That's it. So how liquid is being sucked up at point 3? What is the actual cause? Maybe it is something related to 'turbulence suck'?

enter image description here

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  • $\begingroup$ Re, "pressure drop is a reason,not a result," Isn't a reason exactly what you are asking for when you ask, "why...?" $\endgroup$
    – Solomon Slow
    Commented Feb 14, 2021 at 19:41
  • $\begingroup$ reason is pressure drop,which in result gives acceleration in flow.But it happens at point 1.I was asking reason for pressure drop at top of vertical tube.. $\endgroup$
    – Tym
    Commented Feb 14, 2021 at 20:42
  • $\begingroup$ I don't know that the pressure drop happens entirely at point 1. But what matters is that by the time the air gets to point 3, the pressure has dropped below atmospheric. $\endgroup$
    – Solomon Slow
    Commented Feb 14, 2021 at 20:45
  • $\begingroup$ yeah it seems odd.so it means flow deccelerates as moves to exit as atm pressure from outside shouldnt let the flow go out. $\endgroup$
    – Tym
    Commented Feb 14, 2021 at 20:49

3 Answers 3

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The Bernoulli equation contains the math, but it doesn't furnish much physical insight. Here is a simple way to think about this.

When the air zooming through the horizontal tube encounters the opening of the vertical tube coming up from the bottom, that moving air has a tendency to try and drag extra air along with it as it crosses the opening, and carry that air away. The faster the air is moving in the horizontal tube across the opening, the harder it pulls on the air in the vertical tube, urging it to join in.

If you can get the air in the horizontal tube moving fast enough, the suction it applies to the end of the vertical tube is great enough to pull fluid in the vertical tube upwards against gravity and it then mixes with the moving air, which carries it off as a spray of fine droplets.

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  • $\begingroup$ that does make sense..i was looking for something like this.its obvious that there is nothing to do with bernoulli after flow passes point 1.Air travels at horizontal tube at atm pressure and drags along air from vertical tube.Everywhere people tend to associate it with bernoulli equation..there is btw a lot videos on youtube about bernoulli misconseptions and airlift. $\endgroup$
    – Tym
    Commented Feb 14, 2021 at 20:40
  • $\begingroup$ where in turkey do you live? $\endgroup$
    – niels nielsen
    Commented Feb 14, 2021 at 20:47
  • $\begingroup$ In istanbul.Im studying here $\endgroup$
    – Tym
    Commented Feb 14, 2021 at 20:50
  • $\begingroup$ best friend of mine from my engineering days, Ozgur Yildirim, came from a very small village in the east $\endgroup$
    – niels nielsen
    Commented Feb 14, 2021 at 22:04
  • $\begingroup$ yeah name is turkish , unfortunately there is a lot Ozgur's here :D $\endgroup$
    – Tym
    Commented Feb 14, 2021 at 22:30
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Bernoulli is applicable to this case, unlike many situations where it’s incorrectly cited. For steady incompressible inviscid flow along a streamline, it just says that the total energy density is constant—where there’s more kinetic energy due to motion there’s less potential energy due to pressure and vice versa. To see that the pressure at the T intersection (your point 3) is low, ignore the bulb—you’re right that that introduces complication because the pressure there is higher than atmospheric by an unknown amount. Instead, consider a streamline running from the T to the exit. Imagine instead of just ending there, the tube flares out to a larger diameter (streamlines will be similar either way, but this makes it easier to visualize). The pressure at the outlet of this larger tube is atmospheric. Following the streamline back to the T, you see that the velocity is higher (same mass flux through a smaller cross section) so pressure is lower.

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enter image description here

I have to make some changes to your diagram by moving the straw near the outlet of the horizontal tube. And the outlet of the suction pipe moves upward, slightly higher than the wall of the horizontal pipe.

enter image description here

When air flows through the horizontal tube, the air will move in a curved motion in the outlet area of the straw, as shown in the blue curve of my graph.

At the outlet of the straw (indicated by the green line segment), a pressure lower than atmospheric pressure will be generated.

Why does this result in a pressure lower than atmospheric pressure? Because in this design, the outlet of the straw resembles the upper surface of a flat wing. The curved motion of air is similar to having an angle of attack on a wing. If, like the diagram of your atomizer, the outlet of the straw is aligned with the horizontal pipe wall, then the air will not move along the curve, and the pressure at the outlet of the straw will not be lower than atmospheric pressure. So it is impossible to suck up the liquid in the straw.

Regarding the low pressure on the surface of the wing, you can refer to this link:

https://physics.stackexchange.com/a/516003/176092

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