# FeaturedI Blowing between two objects -- Why is the pressure low?

1. Nov 16, 2016

### Fredrik

Staff Emeritus
Hey guys. I have a simple-looking physics problem that I don't really understand, and I remembered this site, where I used to hang out a lot in the beforetime, the long-long ago. (That's a South Park reference).

This is the problem: A person blows through a straw between two empty soda cans. Do the cans move closer together or away from each other? Explain why they move the way they do.

Partial answer: The cans will move closer together. Given that information, we can conclude that the air pressure in the region between the soda cans must be lower than normal.

What I would like to do is explain why the pressure must be lower there without using information about how the cans will move. A lot of sites discuss this, but most of them just say that Bernoulli's principle is responsible. That's not much of an explanation.

First of all, I'm not sure how to apply Bernoulli's principle here. (I have never studied fluid dynamics by the way). According to Wikipedia, Bernoulli's principle states that "an increase in the speed of a fluid occurs simultaneously with a decrease in pressure or a decrease in the fluid's potential energy".

Here's my attempt: Chop up the region in front of the straw into small pieces. If I look at pieces along the straight line defined by the straw, then the average velocity of molecules in a piece will decrease with increasing distance from the straw. Because of this, the principle tells us that the pressure in these pieces will be increasing with increasing distance from the straw. At a large enough distance, the air pressure will be "normal". This means that each of the pieces has a pressure that's lower than normal.

Is this a correct way to apply the principle? Even if it is, I don't find the argument convincing. What if I somehow produce a high pressure stream of air? Wouldn't the same argument lead to the same conclusion, that the pressure inside the stream is low?

There is clearly something I don't understand here. I would like to know how to apply Bernoulli's principle correctly, and if possible, I would like to see a simple explanation based on Newton's laws instead of Bernoulli's principle.

2. Nov 16, 2016

### Charles Kottler

In simple terms (the way I think about things...) the air molecules outside the straw move in random directions, applying equal pressure in all directions. The molecules moving inside the straw have a greater speed along the axis than across it (because momentum is lost in colliding with the walls?). As the moving air emerges from the end of the straw it is therefore exerting less lateral pressure than the surrounding air so results in lower pressure. There will presumably be an area of higher pressure which spreads out ahead of the emerging stream which would form a vortex circulating around any objects near to the source thus increasing the directional pressure difference.

3. Nov 16, 2016

### A.T.

I remember long discussions here on how to correctly explain pressure decrease in a narrowing pipe using Newton. You might try to find them.

One similar is going on right now:
https://www.physicsforums.com/threa...air-flow-characteristics.893541/#post-5621475

Last edited: Nov 16, 2016
4. Nov 17, 2016

### Fredrik

Staff Emeritus
Thanks for the input, both of you.

I believe that this part is correct, but the stuff you said before doesn't really explain the low pressure, I think. But I think I found a different argument for it. See below.

This was useful. I found a post where boneh3ad explained how to calculate the velocity in the narrow part of the pipe. I also found some links to HyperPhysics:

http://hyperphysics.phy-astr.gsu.edu/hbase/pber.html#beq
http://hyperphysics.phy-astr.gsu.edu/hbase/press.html

According to HyperPhysics, the Bernoulli equation for the narrowing pipe is just $$P_1+\frac 1 2 \rho v_1^2=P_2+\frac 1 2\rho v_2^2.$$ I chose not to include the (gravitational) potential energy terms, which should be the same on both sides if the pipe is horizontal. The 1 subscript refers to the wide part of the pipe, and the 2 subscript refers to the narrow part. Once we have concluded that $v_2>v_1$, e.g. using a mass conservation argument (that's what boneh3ead did), we can conclude that $P_2<P_1$. The equation above is basically just an energy conservation equation, so I would say that this is an explanation of the narrowing pipe that doesn't go far beyond Newton's laws.

However, I don't see how this explains the low air pressure in my problem.

I came up with something that resembles an argument for low air pressure in my problem. I'd like to hear someone's thoughts on it. I really can't tell if this is a good start or just completely wrong.

I found this page at the HyperPhysics web site. If I transfer this conclusion to my situation, I have to conclude that inside the straw, the pressure is high near the end inside the person's mouth, and equal to the outside air pressure at the other end. But the pressure inside the straw isn't just a static pressure. The pressure towards the exit end of the straw is a static pressure (which pushes in all directions) plus the pressure from the "wind" (which pushes only in direction of motion). If this sum is supposed to be equal to the (static) air pressure outside, then the static air pressure on the inside has to be less than normal air pressure.

Not sure if that made sense. Even if this argument is correct, parts of it would require further explanation. And it would still only explain the pressure at the exit. It wouldn't explain why the pressure is also low in the stream a few centimeters in front of the straw. (Why isn't air from over and under the stream pushed into the stream until the pressure has been equalized?)

5. Nov 17, 2016

### Staff: Mentor

6. Nov 17, 2016

### rcgldr

or that due to Coanda effect, the air going between the cans is diverted outwards, which would mean that the cans effectively exert an outwards force on the air, and the air exerts an inwards force on the cans (Newton third law pair).

Last edited: Nov 17, 2016
7. Nov 17, 2016

I don't have a ton of free time at the moment so let's see if I can throw in a quick thought that helps.

In a gas, pressure (thermodynamic or static pressure, they're the same thing) arises from the random motion of gas molecules whizzing about. Pressure is essentially a measure of energy per unit volume stored in a gas as a result of this movement. Additionally, a gas can have kinetic energy due to the bulk movement of the gas (flow). There are also terms such as potential energy, but that will be negligible here, so we can ignore it. So essentially, a gas, moving a not, has a pool of energy that is the sum of its random motions (static pressure) and its bulk motion (dynamic pressure). The sum of the two is often called total pressure, $p_0$. This quantity must be conserved in the absence of dissipative phenomena or the external addition of energy. This is essentially what Bernoulli's equation states:
$$p_0 = \dfrac{1}{2}\rho v^2 + p = C,$$
where $C$ is a constant. So, along any given streamline, the flow must always maintain the same $p_0$, which is why, if you take any two points along said streamline, you get
$$\dfrac{1}{2}\rho v_1^2 + p_1 = \dfrac{1}{2}\rho v_2^2 + p_2.$$
Clearly, if any two streamlines originate from points featuring the same $p_0$, then you can apply Bernoulli's equation between those two lines, not just along each one individually. This is important. In the case of, say, and airplane wing, Bernoulli's equation works because all of the air originates upstream of the wing. It originates in the atmosphere where any variations in $p_0$ are going to be over such large scales that they don't matter to that problem. In that case, you can absolutely compare the velocity above and below an airfoil to discuss how the pressure must be higher under a wing.

Now, consider blowing between two cans. The streamlines in that jet originate in your lungs, while the streamlines on the opposite sides of the cans (such that they even exist) originate in the ambient air in the room. In other words, the air on the outside edges of the cans has $p_{0,room}$ and the air being blown between them has $p_{0,lungs}$, and $p_{0,room}\neq p_{0,lungs}$. Generally, your lungs will add a few psi of pressure to the total pressure emanating from them. So, lets say that your lungs add some small additional pressure, $\Delta p_{0,lungs}$, to their air. Bernoulli's equation could then be modified to be
$$\dfrac{1}{2}\rho v_{amb}^2 + p_{amb} + \Delta p_{0,lungs} = \dfrac{1}{2}\rho v_{jet}^2 + p_{jet}.$$
Generally, the air in the room is not moving, so $v_{amb}=0$. So, in order for the pressure between the cans to be equal to the pressure outside the cans, $p_{jet} = p_{amb}$, we have
$$v_{jet} = \sqrt{ \dfrac{2\Delta p_{0,lungs}}{\rho} }.$$
In other words, saying that the stream between the cans is moving faster than the ambient air is not enough to say that the pressure is lower than ambient. In fact, the air must be moving greater than the above relation in order to result in a lower pressure in the jet than the ambient pressure. If the speed is lower than that, then that jet could actually be at a higher pressure than ambient and tend to push the cans apart. This analysis, of course, ignores the effect of viscosity or humidity or any of those factors that would complicate matters.

Finally, regarding the Coandă effect, I am not honestly sure how much of a factor that would play. On the one hand, that would tend to try and bend the air around the cans, which would tend to imply an inward force to counteract the centripetal force associated with the curve. On the other hand, the Coandă effect classically applies to a fluid jet passing by a single curved body. In the presence of two curved bodies that would essentially be competing to try and pull the flow around them, and I am not immediately sure what would actually happen in that situation. I am sure it is dependent on the space between the cans as well.

Anyway, I hope that helped. It doesn't fully answer the original question, but at the very least it should help people see how Bernoulli's equation does and does not apply to a situation like this.

8. Nov 17, 2016

### rcgldr

The wiki article about Coanda effect seems to focus on a jet, similar to the situation here:

http://en.wikipedia.org/wiki/Coandă_effect#The_Coand.C4.83_effect_mechanism

In the case of a wind where all of the air is is moving rather than just a jet, the air will still tend to follow a convex surface if the centripetal acceleration is not too great (the flow remains attached). I don't know if this is considered to be Coanda effect.

9. Nov 18, 2016

The issue isn't whether or not it is a jet. Clearly it is. The issue is that there are two surfaces. Which does it follow? I don't know the answer, and it is entirely possible that it could do something like follow neither and end up with a recirculating region where it would normally follow the cans. It's also possible it could split and follow both. I am just not really sure.

10. Nov 18, 2016

### rcgldr

I'm wondering if a combination of Coanda and Venturi effects (if the cans are close enough to each other) would allow the stream to be separated into two parts as it passes by the cans.

11. Nov 20, 2016

### Fredrik

Staff Emeritus
Thanks for the input. I haven't had a lot of free time myself these past few days. If I had, I would have responded sooner.

I'm with you this far. This is roughly what I had been able to understand on my own, but your comments helped me understand it a bit better.

I didn't understand this the first time I read it, but I think I do now.

We want to show that the $p_{jet}<p_{amb}$, or at least that this will be the case if we choose $\Delta p_{0,lungs}$ large enough. This inequality holds if and only if
$$0<p_{amb}-p_{jet}=\frac 1 2\rho_{jet}v_{jet}^2-\Delta p_{0,lungs}.$$ It seems to me that $\rho_{jet}$ and $v_{jet}$ should both be functions of $\Delta p_{0,lungs}$. This makes it difficult to see if the inequality is satisfied. The extreme possibilities that the inequality is satisfied by all values of $\Delta p_{0,lungs}$ and that the inequality isn't satisfied by any value of $p_{0,lungs}$ both seem possible to me (if I disregard the fact that I have disproved the latter by experiment).

12. Nov 22, 2016

### Zahid Iftikhar

Great answer from boneh3ad--- It helped me a lot. Thanks
Very impressive answer by boneh3ad----- It helped me a lot to understand pressure and velocity paradox. Thanks a lot.

13. Nov 22, 2016

### rcgldr

14. Nov 22, 2016

Of course the pressure in a jet can drop below ambient. It is very common in supersonic jets, for example. That's one way you end with shock diamonds behind rocket engines at sea level or afterburning jet engines. In a subsonic flow, if a pressure difference is the only driver of a flow then the jet will never be lower than ambient on its own, but if you have an example like lungs it could since pressure difference isn't the only means of moving the flow. In that case, the air would move even with no $\Delta p$ since your lungs are contracting.

15. Nov 22, 2016

### rcgldr

Why would air move outwards from a person's lungs unless the pressure in the lungs was greater than ambient? If the pressure is the same, the lungs would not contract and there would be no air flow. If the lung pressure was less, then air would be drawn inwards.

You could initiate a jet, then reduce the source pressure to below ambient, and for a period of time, the jet's momentum would allow it to continue forwards until it slowed down and reversed it's path.

16. Nov 23, 2016

If it already has momentum then a fluid can move against a pressure gradient. It will just slow down as it does so. It's like a ball rolling up a hill.

In most situations, a subsonic jet will not have a lower pressure than ambient. It will be exactly ambient in any typical situation used to create a jet and my lung example was a poor one. However, I see no reason why it would not be theoretically possible for a jet to have a pressure lower than ambient. It would just require an effort specifically to achieve that and I'm struggling for a good, concrete example at the moment.

Last edited: Nov 23, 2016
17. Nov 24, 2016

### Fredrik

Staff Emeritus
I would appreciate some help with some basic stuff. Consider this picture, from HyperPhysics.

Edit: The image looks fine when I preview. You should be able to see it if you click "Reply".

This is about water, not air. They're saying that the pressure is dropping through the horizontal pipe, as indicated by the picture, and that the pressure at the exit, further to the right, is equal to the atmospheric pressure.

I seem to have misunderstood something basic, and I'm trying to understand what. This is the source of my confusion:.It seems to me that the velocity can't change through the horizontal pipe, because the same mass of water must pass through each vertical cross section of the pipe every second. That means that $\frac 1 2 \rho v^2$ is the same everywhere in the pipe. So Bernoulli's equation implies that the static pressure is the same everywhere in the pipe. But this contradicts the HyperPhysics claim.

Last edited: Nov 24, 2016
18. Nov 24, 2016

### rcgldr

I'm wondering about an angled jet blowing over a convex surface so that a low pressure flow exits near horizontally?

Assuming density does not change, then both volume flow and mass flow remain constant. The pressure energy decreases as the flow proceeds through the pipe due to friction losses converting pressure energy into heat. Bernoulli equation needs to be modified to account for the friction losses.

19. Nov 24, 2016

### Fredrik

Staff Emeritus
This too seems to be inconsistent with what that HyperPhysics page is saying. It claims that the water pressure at the exit will be equal to the air pressure outside. That wouldn't always be the case if the pressure drop is caused by friction.

20. Nov 24, 2016

### BvU

Yes it would: the flow is pressure driven. Friction provides the resistance.

21. Nov 24, 2016

Yes it would. If viscosity is taken into account, you simply need a higher upstream pressure to achieve the same mass flow rate for a given exit pressure.

22. Nov 25, 2016

### Redbelly98

Staff Emeritus
The Newton's-laws explanation is actually pretty straightforward.

The air moving in the stream eventually slows to a stop to join the (relatively) still ambient air. Because a small volume of air within that moving stream has an acceleration opposing its direction of motion, the net force on that moving air volume is also opposing its direction of motion. (Newton's 2nd law)

For the net force to be in the "backward" direction, the ambient pressure ahead of the moving air volume must be greater than the stream pressure behind the moving volume. Or in other words, the stream pressure is lower than ambient pressure.

23. Nov 25, 2016

### rcgldr

or as the stream collides with the ambient air, the ambient air's pressure is increased, and at no time is there below ambient pressure in the jet or the air downstream of the jet affected by the jet.

This is ignoring the entrainment aspect of a jet due to viscosity as noted the wiki article linked to previously:

http://en.wikipedia.org/wiki/Coandă_effect#The_Coand.C4.83_effect_mechanism

24. Nov 26, 2016

### Fredrik

Staff Emeritus
I thought of that explanation too, and believed it to be correct for a couple of weeks. But about a week before I started this thread, I dismissed it as incorrect because if it's correct, then it's impossible to create a high pressure stream of air. (Applied to a high pressure stream, the argument leads to a contradiction).

Let's break this down to something even simpler. If we shoot a high-speed N2 molecule horizontally into the atmosphere at a location where there's no wind (i.e. at a location where the average velocity is zero), would it increase or decrease the pressure along the molecule's path? Since it increases things like density, average speed, average momentum and average kinetic energy, it seems to me that the pressure should increase, not decrease.

25. Nov 26, 2016

### Fredrik

Staff Emeritus
I don't understand this answer. How can friction alone make the pressure drop at the exact rate (Pascals per meter) required to make the water pressure equal to the air pressure at the exit? When you said "friction", did you mean the force on the water from the outside air? I would have guessed that friction in this context has something to do with turbulence caused by the fact that the insides of the pipe aren't perfectly smooth?