Bernoulli Principle vs. Coanda Effect

  • #1
In the past couple of days I've been watching video demonstrations on YouTube that use the Bernoulli Principle to explain what is happening (e.g. the hairdryer and ball experiment)

Then, I saw this one video using the Coanda Effect as the explanation (they were also using the hairdryer and ball demo).

I was wondering if both (Bernoulli and Coanda) explanations are able to co-exist? And, if they are the same thing?

Oh, and could you please explain it in easy to understand terms...

Thanks :D
 

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  • #2
rude man
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In the past couple of days I've been watching video demonstrations on YouTube that use the Bernoulli Principle to explain what is happening (e.g. the hairdryer and ball experiment)
Then, I saw this one video using the Coanda Effect as the explanation (they were also using the hairdryer and ball demo).
I was wondering if both (Bernoulli and Coanda) explanations are able to co-exist? And, if they are the same thing?
Oh, and could you please explain it in easy to understand terms...
Thanks :D
Looking at one illustration of this effect, a jet of gas is emitted from an orifice. Bernoulli seems to apply to this situation: the jet of air is joined by ambient air so a streamline can be identified. Along that streamline the sum of pressure and kinetic energy term ρv2/2 is constant. So looking at the bottom of the jet stream, the ambient air below it is at a higher pressure than the jet air, which would keep a ping-pong ball elevated by the pressure differential.
So it seems the two principles are closely related.

https://en.wikipedia.org/wiki/Coandă_effect
 
  • #3
rcgldr
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I'm not sure of the principle behind a jet of air and a ball hovering in the stream. The jet can be angled away somewhat from vertical, and yet the ball will not fall out of the stream. I made a video (see below) that also included using a cardboard tube that causes the ball to shoot upwards as it passes through the tube. Coanda effect refers to the fact that a flow remains attaches to a convex (curving away) surface for a while before detaching, but it doesn't explain why a ball would remain centered. That requires a difference in where the flow detaches if the ball moves away from the center of the stream, where the flow on the outside would detach sooner than the flow on the inside, diverting the flow outwards, resulting in a Newton third law pair, the off-centered ball exerting an outwards force on the air, coexistent with the air exerting an inwards force on the ball. What I don't understand is why the outside flow should detach before the inside flow, unless the ball spins to divert the flow outwards via a Magnus effect. In the case of the ping pong ball, when the stream is angled, the ball spins fairly fast, but in a vertical stream, it doesn't spin much at all.

 
  • #4
sophiecentaur
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I was wondering if both (Bernoulli and Coanda) explanations are able to co-exist? And, if they are the same thing?
I don't see any conflict here. The pressure inside the moving jet of air is less than ambient pressure and that keeps the ball within the jet. The ball will tend to the centre of a symmetrical vertical jet (and at the highest point too). The tangential forces on the ball will be balanced at the centre so it will spin slowly or not at all. When the jet is tilted, the ball falls to the lower side of the jet and the inner surface of the ball will move upwards because that air is moving faster. The air speed difference will cause the ball to spin.
Behind the ball or any other obstruction, there will be turbulence which raises the local air velocity and that causes low pressure in that region. Turbulence takes you away from the simple laminar flow type of argument.
 
  • #5
boneh3ad
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I don't see any conflict here. The pressure inside the moving jet of air is less than ambient pressure
That cannot be safely assumed. Most devices with which I am familiar use some kind of compressor (e.g. a fan) to propel jets to the desired velocity, meaning that those flows start out with a higher pressure than ambient. Whether or not the jet itself has a lower pressure than ambient after being accelerated (and its pressure drops accordingly) depends on the nature of that compressor.

In fact, jets are inherently viscous (and involve a shear layer between the jet and the surrounding fluid), and Bernoulli does not apply neatly to the flow. There is a lot of dissipation.

Really, there are a number of effects going on here and the situation is deceptively complicated. Personally, I think the Coanda effect (which I honestly don't even like referring to by name because it implies it is its own standalone effect rather than just a feature baked into fluids) is the best explanation here. A ball in the center of a vertical jet is held up by the drag from the jet. As it moves off the centerline, the faster-moving center portion of the jet tends to wrap around the ball a bit more (i.e. if the ball moves to the left, the right side is exposed to faster flow, which wraps around the ball and is deflected left slightly). The deflection of this high-momentum stream results in an equal and opposite force on the ball back to the centerline. There is probably some overshoot at that point and the same thing happens again in the other direction. It's a metastable state at the centerline.

When you tilt the jet, the same thing happens except now gravity is tending to pull the ball out of the jet, so the stable point requires more flow deflection in order to hold it up. So if you tilt the jet to the left, the new stable point is simply a bit to the left of the centerline instead of on the centerline.
 
  • #6
sophiecentaur
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That cannot be safely assumed.
You have a point there, so my comment must apply only to the region near the nozzle. But there is a velocity profile across the jet with fastest at the middle so, ignoring what happens further out, isn't it ok to talk in terms of Bernouli near the axis? There is a point' to the side. where there is so little pressure on the ball that it falls anyway and the lower ambient pressure doesn't come into it - or could it account for a force away from the jet? Someone must have experimented with suspended balls in vertical jets and that would have resolved the matter by showing if there is a region where the force changes from inwards to outwards.
Really, there are a number of effects going on here and the situation is deceptively complicated
I guess that's true. My initial reaction was to quote and justify what is more or less the stock answer. The effect gives up beyond a certain radius.
The ball appears to spin in the direction that I'd expect - the innermost side moving upwards near the jet, presumably due to higher tangential forces on that side.
 
  • #7
boneh3ad
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Absolutely there is a velocity profile across the jet. It peaks at the center and then smoothly drops to zero at the edges as you reach the ambient air. For a 2D, laminar jet, it is proportional to ##\mathrm{sech}^2(y)## if ##y## is the coordinate across a cross-section of the jet (assuming the jet leaves the nozzle at ambient pressure, which is usually the case).

The problem with using Bernoulli is that the entire reason the profile exists is due to viscosity. The shear layer that is set up at the edge of the jet entrains some ambient fluid while slowing down the edges of the jet, causing the whole system to slow and spread out. It's also subject to vortical instabilities (see: Kelvin-Helmholtz instability), and is just generally a poor candidate for applying Bernoulli. In fact, it is usually assumed that the pressure is constant across a jet and the same as ambient for an incompressible flow, much like a boundary layer. Similarly, we don't generally apply Bernoulli's equation to boundary layers.

I am not even convinced that it would work near the axis since the centerline flow is slowing with distance from the nozzle, yet the pressure is not generally changing. If you assumed that the ball in the jet is small compared to the axial distance over which the velocity in the undisturbed jet changes, then you could locally apply Bernoulli along a single streamline as it moves around the ball, but that's about it, and it would only be approximate.

In my description, I neglected any spin imparted on the ball, but the direction of spin would be correctly-predicted by what I said. The faster flow near the center being bent around the ball would also tend to rotate it such that the side closest to the centerline is moving along with the flow. The other side would be against the flow.
 
  • #8
sophiecentaur
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The problem with using Bernoulli is that the entire reason the profile exists is due to viscosity.
You are saying that Bernoili needs a boundary for it to apply? Could well be the relevant consideration here.
 
  • #9
boneh3ad
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You are saying that Bernoili needs a boundary for it to apply? Could well be the relevant consideration here.
No, what I am saying is that it needs to be an inviscid flow to apply, ate least in they way it is commonly misapplied in flows like these.
 
  • #10
rcgldr
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Absolutely there is a velocity profile across the jet. It peaks at the center and then smoothly drops to zero at the edges as you reach the ambient air.
In the case of room fans used to suspend beach balls, the center has a slower flow than the mid-radius cylinder, since the center doesn't have any blades, and the beach balls don't hover far above the fans. This probably explains why the beach balls tend to wander around. In a scaled up version of blow dryer and ping pong ball, the Rueben H Fleet science center in San Diego has/had about a 6 inch or so diameter jet producer where a small beach ball hovers in an angled jet of air, and in that case the center probably has the highest rate of flow.
 
  • #11
sophiecentaur
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No, what I am saying is that it needs to be an inviscid flow to apply, ate least in they way it is commonly misapplied in flows like these.
Yeah. There's never anything simple about fluid flow. Stick to lekky - there's a level of EE that is nice and concrete with very few rules and equations. :smile:
 

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