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I'm working on a project and I have a question. How does the volume and pressure of the balloon affect the time the hovercraft hovers above ground?

This relates to my earlier question where I described the experiment in more detail.

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David 2000
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    A hovercraft with a balloon? It's unclear what you are talking about. – Mike Dunlavey Mar 15 '15 at 17:52
  • From what little I've thought about this, and the discussions below, I'm guessing the bottom surface of the puck could be dished somewhat. It's not necessary, or maybe even desirable, for the air layer to be the same thickness throughout. The thickness that matters is where the air escapes to the outside, at the rim. (Real hovercraft have a "skirt" around the edge.) – Mike Dunlavey Mar 16 '15 at 21:33

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I actually added this to the answer to your other question but will repeat it here...

If the balloon is bigger, the time that the toy can hover will increase - by a surprisingly large amount. Using the result from my other answer that pressure (and thus flow rate) scales with $1/r^2$, and volume scales with $r^3$, then time (which is the time it takes for the balloon to deflate) will scale with $r^5$.

Proof:

From flow rate: $$\frac{dV}{dt} \propto \frac{1}{r^2}$$ From equation for volume of balloon: $$\frac{dV}{dr}=4\pi r^2\\ $$ Combining: $$\frac{dr}{dt} = \frac{\frac{dV}{dt}}{\frac{dV}{dr}} \propto r^{-4}$$ Integrating: $$t \propto r^5$$

In other words, a bigger balloon will allow for much longer floating time, assuming that the flow rate is proportional with the pressure (and that the pressure of the fully inflated balloon is still large enough to keep the craft floating). That's an interesting result I was not expecting.

Following a conversation with Mike Dunlavey in the comments, if the flow rate goes with pressure squared (as it might for a simple aperture), then the answer changes: the time for the balloon to empty would go with the seventh (!) power of radius.

I expect that experiment will give the answer, and would urge you, once you have built the project, to report back on your findings. It is probably easiest, given modern technology, to just constrain the hovercraft from floating about (surround it with three regularly space pins into the surface below) and film it against a background of graph paper. Estimate the size for each frame (or just find the frame where the radius is "one square smaller") and plot the log of the result. The slope will help you determine the correct value of the exponent.

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Floris
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  • Pressure proportional to flow rate? Usually pressure goes as flow-rate squared (at low velocity). Like in an airplane - double the speed, quadruple the lift. – Mike Dunlavey Mar 15 '15 at 17:56
  • @MikeDunlavey - using http://hyperphysics.phy-astr.gsu.edu/hbase/pfric.html#veff flow rate scales with pressure difference. I am ignoring the small effects of density change given we are talking balloons here so that is a second order effect at best. – Floris Mar 15 '15 at 18:01
  • That's for viscous laminar flow in a tube. Flow through an orifice with a pressure drop is different. There, if you double the velocity, you double the momentum change per parcel of air, but you also double the number of parcels. (Density change does not enter, as long as the Mach number is low.) – Mike Dunlavey Mar 15 '15 at 18:08
  • @MikeDunlavey - there is a tube connecting the balloon to the rest of the system... as long as that tube provides most of the resistance to air flow it will give a flow rate proportional to pressure. If not, then the relationship gets even crazier. Of course we have all seen how balloons deflate more quickly as they get smaller... Perhaps I will do some measurements and report back. – Floris Mar 15 '15 at 18:10
  • OK, the OP made it clearer what the situation is. It looks to me like the main pressure drop is in that thin film from center to edge. I would suppose it is mainly viscous, but the velocity would have to decrease with radius (assuming a flat bottom surface), so I'm back to square 1 :) – Mike Dunlavey Mar 15 '15 at 18:13
  • Ah yes - I was assuming the pipe provided the main resistance (see my earlier answer). If that assumption is false then so is the relationship derived from it. I did build one of these 40 years ago so my memory may be hazy. – Floris Mar 15 '15 at 18:16
  • So, what would be the true answer of my question?! – David 2000 Mar 15 '15 at 18:25
  • The "true" answer can only be determined by experiment... – Floris Mar 15 '15 at 18:29
  • Sir, I posted another question. I'm really sorry but I want you to answer that question too! – David 2000 Mar 15 '15 at 18:35
  • Sir, I understood that pressure (thus the flow rate) scales with 1/r2. and volume scales with r3. So how do you determine the time scaling with r5 ??? can you prove it please? – David 2000 Mar 16 '15 at 13:08
  • Sir, my friends and I experimented the relation between Volume and Time and it doesn't match with your sayings. I want you to give me your email address, then I'll send you the data and the diagram and the chart. – David 2000 Mar 17 '15 at 11:36
  • Just add the data and graph to the question - then everyone can take a look. – Floris Mar 17 '15 at 12:05
  • OK sir. But I would be very happy to contact with you via Email. – David 2000 Mar 17 '15 at 12:08
  • Sir, I can't have your email address? – David 2000 Mar 17 '15 at 12:38
  • I'm sorry but I don't give out my email address. Please see http://meta.stackexchange.com/a/138792/212464 – Floris Mar 17 '15 at 14:47
  • OK. It seems I have no other ways, I'll add the data to the question. Please check them. – David 2000 Mar 18 '15 at 13:59
  • Ping me again when you have updated the question and I will take another look. – Floris Mar 18 '15 at 14:01
  • Floris, I want you to answer my another question " Roughness of the Surface and the time that a toy hovercraft hovers?", I've posted it recently – David 2000 Apr 01 '15 at 15:51