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View Poll Results: Doe's Poisseuille's Law apply to air flowing through a tube?
No, dummy 0 0%
Air flow in a tube is not laminar 0 0%
Only applies to uncompressible fluids 0 0%
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Old Mar 7th 2013, 03:11 AM   #1
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Join Date: Mar 2013
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Question Poisseuille's Law

Hi -

I'm working on a balloon-driven helicopter toy for my daughter. You can buy these, but they waste a lot of air on an irritating whistle and I thought making our own would be more fun. The basic design is the same as commercially available ones, with a twist (sic): I add a second balloon on top for additional air and therefore prolonged flying time - hopefully...

So I am trying to ascertain the flow rate and velocity of air expelled from a deflating balloon through a tube. Applying Poisseuille's Law and using the calculator at http://hyperphysics.phy-astr.gsu.edu...ase/ppois.html I get unexpected results...

I enter the following parameters:

pressure difference: 50mmHg (I found a YouTube video where a chemistry teacher measured pressure in an inflated balloon at +-810mmHg in a lab. Difference to sea-level ambient pressure of 760mmHg - I'm at the coast - is therefore 50mmHg. Several sources online estimate this as 0.5 to 1.5 psi and 50mmHg is about 1 psi - seems to fit)

tube radius: 0.2cm (the air is expelled through a 4mm diameter drinking straw)

tube length: 20cm

air viscosity: 0.000181 Poise (online sources)

I then solve for volume flow rate (by clicking Volume Flow Rate in red above as site directs) and get the ridiculous result of 11567.43 cm3/sec - that's 11.5 litres of air per second, through a drinking straw?! I have measured the flow rate by inflating the balloon to a guesstimate 2.5 litres of air and then timing how long it takes to deflate (around 13 sec). That is more or less 200cm3/sec - orders of magnitude out. Since I am fairly sure of the other parameters, I enter the measured flow rate and solve for pressure difference and get 0.864mmHg (0.017 psi) - equally ridiculous. I also tried setting the measured flow to 200cm3/sec and 50mmHg pressure and solving for viscosity - and get 0.01 Poise - very close to the viscosity of water!? What am I doing wrong?

Is there some principle that says that the static pressure in an inflated, sealed balloon is not the same as when the balloon is allowed in the process of deflating? Or, how relevant is the tube length really? Air has such a low viscosity, I can't imagine halving the length of the straw will double the flow rate as the calculator suggests. Is Poisseuille's Law maybe not applicable to compressible fluids? Can't find anything to that effect online.

I'm well confused! Please help...

Doogli (down in South Africa)
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Poisseuille's Law-poisseuilles-law.jpeg  

Last edited by Doogli; Mar 7th 2013 at 05:02 AM. Reason: typo
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Old Mar 7th 2013, 10:33 AM   #2
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Join Date: Jun 2010
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Posts: 418
Interesting...

Hi,

Hard to respond. What does the "balloon helicopter" look like?
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Old Mar 7th 2013, 11:15 PM   #3
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Arrow The ballon helicopter

See the commercially available balloon helicopter here:



See the way the air ducts in the wings are shaped here:

https://www.google.com/search?q=ball...2F%3B450%3B450

An interesting detail of the design, is that the ducts inside the wing get MUCH wider towards the wing tip. They have a flat rectangular section which flares wider towards the tip and then narrows smoothly at the tip to a small aperture. I think that this design causes the air to be accelerated outwards by centrifugal force and then compressed by the narrowing of the duct and expelled at higher speed. This will be tough to replicate, but may prove to be the secret of its success. I estimate the section of one of its tip jets to be about 3mm2, considerably smaller than my 4 * PI = 12.6mm2.

My first attempt last night failed, but there is hope. I think perhaps the second balloon on top added too much weight, but the real problem was in the air feed - there is some kind of blockage in one of the wings. I built it by drilling some holes in a short piece of a pen tube with a hexagonal section like a pencil - it makes for neat 120 degree angles between the blades. I then inserted the bendy straws through these holes and fixed paper wings with symmetrical airfoil sections to the straws. The balloons are stretched over a bit of cork from a wine bottle fitted over the ends of the pen tube. I could seal off the top of the tube if two balloons are just too heavy. The inside diameter of the pen tube is about 6mm, so it's just too thin to carry enough air to supply all three straws' maximum capacity: (3 * PI * 2^2)) > (PI * 3^2), so I need a thicker pen or thinner straws.
One final word: the straws in my design do not all stem from the same point along the length of the tube, so I'm concerned that the upper and lower blades will bleed off a lot of pressure, leaving less airflow for the middle blade (furthest from both balloons). In the case of only one balloon at the bottom, the pressure would decrease away from the balloon, so the lowest would have highest pressure, then the middle and finally the last would get the dregs. The commercial design does not suffer from this. I'll try drilling another pen with the blades all stemming from the same point - not that easy, even with a drill press, believe me...

Thanks for your interest THERMO. All the above aside, can you tell me what's going wrong in my application of Poisseuille's Law above?

Last edited by Doogli; Mar 8th 2013 at 12:15 AM. Reason: typos
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air flow, balloon, law, poisseuille, poisseuille's law



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