# I don't understand Bernoulli's principle applications

1. Dec 3, 2011

### Curl

I'm extremely raged right now, I see people applying Bernoulli's equation at fans, atmonizers etc. Let's look at the classic example where you blow through a straw and make liquid be drawn up from a can (through a second straw):
http://parallax.sci.csupomona.edu/demo/index.php?option=com_content&view=article&id=361&Itemid=397 [Broken]

The reason you got the air to flow through your straw is because you are pressurizing the air in your mouth (say to 1.001atm) and it starts flowing down the straw, creating a negative gradient across the straw (from 1.001 atm in your mouth to 1.000atm at the other end).

How the ************ does that make the liquid in the canister rise up the pipe?

Last edited by a moderator: May 5, 2017
2. Dec 3, 2011

### gsal

Refer to Bernoulli's[/PLAIN] [Broken] Principle.

You can think of it like this...when the fluid is moving faster in one direction, it has less time to worry about producing pressure in the perpendicular direction and so, it is almost like it is producing a bit of a vacuum respect to stationary 1atm air...this little dP drives the fluid up and away

...I remember growing up with a hand sprayer like this...an air pump and a container under, to combat insects I would spray DDT...back when it was legal.

Last edited by a moderator: May 5, 2017
3. Dec 3, 2011

### rcgldr

That's not Bernoulli effect, at least not directly. When air flows across the end of an open tube, the air is directed away from and then across the end of the tube creating a vortice, like a small horizontal tornado. The vortice has reduced pressure and that is what draws the fluid upwards. Some atomizers operate using only this vortice effect, while others direct some of the pressure into the chamber in addition to the vortice effect.

Many carberators make use of the vortice effect in addition to a single or dual stage venturi. The venturi chambers reduce pressure due to Bernoulli (and somewhat due to friction of the walls and viscosity), while the protuding fuel tube adds an additional vortice effect. Often the end of the fuel tube is angled away from the direction of air flow as seen in the image in the second link below.

http://www.secondchancegarage.com/articles/images/carbtheory/drawing2.gif

http://www.motorera.com/dictionary/pics/c/carb.gif

In the case of a fan or propeller, Bernoulli applies fore and aft of the fan, but in the imediate vicinty of the fan, there's a pressure jump (with little change in air speed) that violates Bernoulli since it increases the energy of the air. Because of the pressure jump, the air continues to acclerate away from the fan into the downwind ambient air. Link to NASA article:

http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

Getting back to the vortice effect, it can be eliminated if the end of a tube is flush mounted with a surface parallel to the relative wind flow so that the end of the tube "hides" in the boundary layer near the parallel surfce. The pressure at the end of the tube will equal the static pressure of the air flowing across the tube, regardless of the air's speed (within reason, at sub-sonic speeds). This is how flush mounted static ports operate on some aircraft.

http://en.wikipedia.org/wiki/Pitot-static_system

Link to image of a pair of flush mounted static port from the wiki article above:

http://en.wikipedia.org/wiki/File:StaticPort.jpg

If you tried blowing air through a straw across a static port with the other end of the tube in a chamber filled with fluid, there would be almost no effect, due to the very small difference between the static pressure of the air you blow and the ambient pressure.

Last edited: Dec 3, 2011
4. Dec 4, 2011

### klimatos

I believe they were called "Flit" guns.

5. Dec 4, 2011

### AlephZero

Bernouilli's principle is basically Newton's laws of motion applied to a fluid.

You have a stream of air flowing over the top of the vertical tube. After it goes past the tube, it decelerates to zero velocity.

What makes it decelerate? A force.
Where does the force comes from? A change in pressure in the air.
What is the pressure after the air has stopped moving? Atmospheric.
So what is the pressure when the air is moving past the vertical tube? Less than atmospheric.

How the work is done to accelerate the air (by blowing, heating the air, a chemical reaction to produce gas, or whatever) doesn't matter.

6. Dec 4, 2011

### rcgldr

Why would the air stop when there's nothing to stop it? The stagnation zones on the leading and trailing sides of a vertical tube are tiny compared to the diameter of the tube. Most of the air flows around the sides or over the end of the tube. At the end of the vertical tube, the air gets diverted upwards, then across the open end of a tube, generating a vortice, which reduces the pressure. Perhaps you were referring to the horizontal component of velocity of the air or fluid inside the vertical tube, which would be zero?

It does matter, the exhaust just aft of a jet engine is very high presure and high velocity. The jet exhaust continues to accelerate aftward as its pressure decreases to ambient. The same is true for a fan or propeller as mentioned in the NASA article I linked to above.

In the case of air being decelerated to zero, such as a pitot tube, Bernoulli total pressure = static + dynamic pressure is a first order estimate that ignores compression effects on the air. Pitot tube based instruments instead use the more accurate impact pressure to determine air speed.

http://en.wikipedia.org/wiki/Impact_pressure

Last edited: Dec 4, 2011
7. Dec 4, 2011

### Studiot

There is a secondary, non bernoulli effect which is founded in the same reason as washing drying better on a windy day.

The reduced pressure in the vertical tube promotes increased vapour pressure (evaporation)from the liquid, which is then carried up and away by the passing air stream in a continuous process.

8. Dec 4, 2011

He is talking about the fact that if you blow straight with no tube, that stream of air will stop. This means that there has to be a force slowing it down and therefore the pressure in the jet you created is lower than atmosphere. With the tube there, that means that you have less than atmosphere at one end and atmosphere at the other, so fluid is pushed through the tube towards the air flow.

It doesn't matter in the context discussed here, which is incompressible.

9. Dec 4, 2011

### Curl

This doesn't explain anything. You are assuming the air "slows" down to v=1 at p=1atm. The only reason the flow is slowed is by viscosity, which is not treated by bernoulil's equation. Bernoulil's equation is for an INVISCID fluid.
The pressure gradient must be negative in the straw, from start to end. It must be HIGHER THAN 1 ATM anywhere in the straw, and it is 1.0000ATM at the opening. Thus, anywhere else in the pipe the pressure will be GREATER than 1ATM according to Bernoulli's equation, meaning the liquid is pushed down rather than pulled up.

10. Dec 4, 2011

### Studiot

Why do you say this?

11. Dec 4, 2011

### Curl

OMG, because you are blowing it from 1.01ATM (in your mouth) to 1.00 ATM outside.
It will move from ~0 velocity in your mouth to some final velocity when it comes out the straw. Thus, the pressure decreases continuously from 1.01 to 1.00 across the straw.

12. Dec 4, 2011

For someone who once made a thread about how easy all this mechanical engineering stuff is, you certainly are confused. Even in an inviscid fluid the jet would eventually slow to zero, unless of course you don't think the jet loses any energy when "pushing" the stagnant air out of the way. Viscosity certainly adds to the effect, but it is not the only reason the jet slows.

Your reasoning aside, you can do this experiment in your kitchen and observe the liquid moving up the straw, so your explanation clearly cannot be correct.

Your original claim that the stream of air moving over the straw has a static pressure greater than $1 atm$ is simply wrong. The static pressure of the stream of air is below atmospheric pressure. The total pressure of the jet would be $1 atm$, but that includes static and dynamic pressure.

13. Dec 4, 2011

### Studiot

Oh the straw you are blowing into!

But that one is of little consequence and could be a hair drier or just a passing zephyr.

No problem

I was referring to the straw in the tub of liquid, and I thought you were too since you refer to liquid being pushed down. I don't see how liquid could be pushed down in a horizontal straw?

14. Dec 4, 2011

### rcgldr

The fact that a stream eventually stops doesn't provide any information about the initial pressure of the stream. The stream could be intially accelerating due to higher than ambient pressure, then transition into decleration as it's pressure decreases. This is what happens to the stream aft of a fan or propeller.

Why would incompressability negate the existance higher than 1 atm pressure from the source of a stream of air?

- - -

The main point I'm making is blowing air across the end of an open tube protuding perpendicular into the stream involves a vortice effect, making it near useless for determining the static pressure of the stream. If you want to know the static pressure of a stream, use a static port or some equivalent device. In the case of a static port, you could connect one end of a clear tube or straw to the static port and have the other end in a cup of soapy water (to reduce skin effect), then note if the water in the tube is raised, indicating stream pressure less than 1 atm, or if the water is lowered, indicating stream pressure greater than 1 atm. This would be a crude instrument, and in the case of a person blowing a straw, I doubt there'd be much effect; you'd need a more precise instrument.

If you want a demonstration of Bernoulli effect, the classic spool and card demonstration, sometimes called the Bernoulli levitator, is much better method:

http://www.seykota.com/rm/spool_card/spool_card.htm

There's also a device used to drain water that combines venturi and diffuser effects on a stream of water from a tap to produce the low pressure used to drain water. (Note in spite of the patents, this type of device dates back to 1930's. There are also other brands of virtually identical devices, but this was the first good image that I found of such a device).

http://andysworld.org.uk/aquablog/?postid=247

Figure 4 in this image shows the internals of this device in venturi + diffuser (drain) mode:

internal_diagram.gif

Last edited: Dec 4, 2011
15. Dec 4, 2011

Sure it does. There is only one way that a free jet could speed up without external intervention, and that is in the case of an underexpanded jet, which only occurs in compressible flows.

For a propeller, after leaving the small region of direct influence of the propeller the flow absolutely does not accelerate. If it did without some sort of additional external "encouragement" it would be violating the second law of thermodynamics.

Here is a pretty concise CFD simulation I found on YouTube showing axial velocity contours. It dies down pretty quickly after the propeller.

The incompressibility matters because you were talking about a jet engine. If the flow accelerates after it leaves the engine it is only because the flow was underexpanded. That flow is moving much more quickly than the present case so it is compressible while the present case is not.

It also matters in reference to the comment on Pitot tubes. Since compressibility effects are not important here, a Pitot tube simply using Bernoulli would be sufficient to get the speed.

The reason I say there can't be any air outside your mouth whose pressure greater than $1\; \textrm{atm}$ is because if there was the flow would still be accelerating and it isn't. It can't be.

Do you have a picture of this "vortex effect"? I can't picture any vortices that would form that would drastically change the flow field in a way that they could overcome a pressure gradient. The pressure would already be pushing fluid out of the tube and any vortices forming would take the form of those found on a circular cylinder with a free end (link to example) likely with some horseshoe vortex structure around the edges of the tip as a result of the mass flow coming out of it.

Regardless, any vortex effect will only effect a very short distance upstream of the tube and therefore will leave formulae like Bernoulli as perfectly reasonable estimates of the pressure in the jet up to that point assuming that you are close enough to the center of the jet that viscous effects are still minimal. You only need to think about the pressure just before the tube, so the vortices will have no effect on the present discussion.

Sure that would work, though the water would go down if the static port was measuring a pressure less than the ambient, not up. It would be easier just to use a differential pressure transducer though, as is commonly done.

Indeed. This concept is used again and again in many applications across many industries. The wind tunnel I work at uses this effect to draw the diffuser pressure down to near vacuum ($60\; \textrm{torr}$ or so).

Last edited by a moderator: Sep 25, 2014
16. Dec 4, 2011

### Curl

Ok thanks rcgldr. So I was right, this has nothing to do with bernouli's principle.

Last edited: Dec 4, 2011
17. Dec 4, 2011

### rcgldr

Look at that NASA article about propellers (again):

http://www.grc.nasa.gov/WWW/K-12/airplane/propanl.html

As the stream speeds up, it's cross-sectional area decreases, so compressability is not required. In an idealized incompressable case, the surrounding air will have to fill in any voids cause by the constricting stream (or the air was there all along), but again compressability isn't required to fill in those voids (the air could simply spill into the voids by falling inwards from above, or it was already there to begin with).

The NASA article implies that only 1/2 of the increase in velocity of the air by the propeller occurs at the propeller, and the rest of the increase occurs as the air continues to accelerate (and constrict) while it's pressure decreases to ambient. The actual wording is that the velocity at the propeller is the average of the free stream and exit velocities. I assume this is somewhat idealized, but the general concept is true.

Impact pressure = dynamic pressure x (1 + M2 /4 + ... ), where M is the mach speed of an aircraft. For a civilian airliner traveling at mach .8, the actual impact pressure is about 16% greater than what dynamic pressure would predict for total pressure at mach .8. That 16% difference would greatly affect the accuracy of the indicated air speed gauge.

No, but the tubes used to atomize the fuel in many carburetors take advantage of this effect. Avoiding this vortex effect is the reason why static port tubes need to be flush mounted as opposed to protuding into the air stream.

It's enough of an effect for atomizers like the old flit gun to operate. If the pumped air was flowing across a static port, it wouldn't draw any liquid out of the tube.

You could conduct an experiment using a leaf blower to generate a high speed flow across a protuding tube and across a static port, both mounted on a flat surface, with the other end of the tubes setting in a pail of water. In the case of the protuding tube, water will be drawn upwards and atomized into the stream, if the tube isn't too high above the water level in the pail. The effect on the static port will be minimal by comparason.

The pressure on the fluid at it's surface and within the straw at fluid surface level is 1 atm. If the pressure of the open end of the straw is less than 1 atm, then fluid is drawn upwards.

Last edited: Dec 4, 2011
18. Dec 5, 2011

Fair enough, but it doesn't include a length scale over which the acceleration occurs. This length scale is actually fairly tiny (think on the order of millimeters, http://psfvip4.univ-fcomte.fr/Fpsfvip4/sources/F4035.pdf [Broken]). This occurs because even though the acceleration of the air out of your mouth is modeled as being instant, it obviously can't be, so there does have to be some sort of accelerated region. I concede that point to you.

I am not convinced about the case of blowing out of your mouth though, as the method of generating flow is entirely different.

I never said this wasn't the case. What I did say is that the Mach number of air being blown out of your mouth is, what, maybe Mach 0.01, so there is no reason to concern yourself with impact pressure here.

Right, but they take advantage of it because it creates a vortex downstream of the tip of the fuel tube, enhancing mixing so you have a more homogeneous fuel-air mixture when you get into the combustion chamber. It does very little if anything to the flow field over the tip of the tube and upstream.

Do you have a source for this? I maintain that it would draw air out of the tube. The vortices would atomize the particles downstream as a result of the mixing they induce. They wouldn't draw the fluid out.

Again, in the absence of me owning a leaf blower, I will need to find some source that shows this before I am a believer.

Simple miscommunication. I thought you were talking about the water level in whatever reservoir of water you were using as is typical in a manometer. It seems we are both correct.

Last edited by a moderator: May 5, 2017
19. Dec 5, 2011

### rcgldr

You could try a hair dryer, or the output port of a vacuum cleaner. You could drill a hole in a flat piece of wood (or reasonably stiff cardboard) so that a straw or flexible tube like the tubes used for air pumps on aquariums has a tight fit. The other end of the straw would rest in a pail or pan of water. You could then compare the effect of the straw or tube protuding from the wood (or cardboard) versus the end of the straw or tube flush with the surface of the wood (or cardboard). I did try this years ago and there was a huge difference.

Although probably not needed for such an experiment, you can buy an actual static port for \$20 (USA) or less, but then you'd need to customize a piece of wood to mount the port on, and you'd need to find a flexible tube that would fit the inside end of the static port.