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Blowing up a balloon using centrifugal force

  1. Dec 19, 2015 #1
    Hey guys,
    So I had what I thought was a cool simple idea to show the effects of the apparent centrifical force.
    I glued the mouth of a balloon onto the end of a 1.5m length of a 10mm flexible tube.
    The idea in my head was that by spinning it around the inertial forces would result in effective increase in weight of the air in the tube, and therefore pressure. I figured this would essentially compress the air at the end of the tube and fill up the balloon. A vacuum would be present at the end of the tube I am holding, and suck more air in.

    I was quite certain I would get flow into the balloon, as when spinning the balloon at 120rpm (2 revs a second) at a radius of 1.5m, this results in a centrifugal g force of approx 24g's, which I believe would give you a pressure of of 24 atmospheres (2.4MPa).

    But sadly, the balloon did not inflate at all, and I experienced no vacuum when i put my finger over the end of the tube. So I felt quite silly haha.

    Can anyone explain where my logic is failing me?

    Cheers
     
  2. jcsd
  3. Dec 19, 2015 #2

    davenn

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    Gold Member

    show the formula and figures you used ... someone should be able to see what you did/didn't do correctly :)
     
  4. Dec 19, 2015 #3
    I you had a gravity of 24g and the atmosphere of the earth above you, you'd have an air pressure of 24 atmospheres.
    But in this case there's only a tube with a length of 1.5 meters to produce the pressure, and only the air at the maximum radius feels the full 24g.
     
  5. Dec 19, 2015 #4
    Yes which would mean the balloon feels air of that full weight, and as the air outside the balloon is atmospheric, the balloon would inflate until it stretches to equilibrium (or pops) right?

    Davenn, simple calculation or online calculator:
    Fc = m v2 / r
    = m (n 2 π r / 60)2/r
    = 0.01097 m r n2
    = 0.01097m*1.5*(120)^2
    = 236*m (N)
    Therefore gravity = 236 m/s^2 or 24g (236/9.81)


    Perhaps the complication comes from air being compressible
     
  6. Dec 19, 2015 #5

    jbriggs444

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    To repeat what was said in #3 using different words...

    The 24 g's applies to the pressure gradient, not the pressure itself. Right now the air in your room is under a gravitational acceleration of 9.8 meters per second2. The effect of that is not to create one atmosphere of pressure. The effect of that is to make the pressure at floor level slightly higher than the pressure at ceiling level.

    The relevant equation is P = ρgh where ρ is the density of the air (in kilograms per cubic meter), g is the acceleration of your centrifuge (in meters per second per second and h is the height of the tube (in meters). P would then be given in Pascals. One atmosphere is approximately 100,000 Pascals.

    Work it out for yourself. Look up the density of air. Figure out the pressure required to support 1.5 meters of air at 24 g's.
     
  7. Dec 19, 2015 #6
    Oh wait I think I see now. Atmospheric pressure is related to all of the mass of the air pressing down on you.
    And all that air results in a pressure (similar to hydro static pressure) of 101.3 KPa.
    In this case, even if all of the air in the tube was at a density of 24 times atmosphere, the change in pressure is only:
    P=ρ.g.h, which gives 3/5ths of f-all...

    edit: jbriggs444, sorry missed your post but yep seems we've worked it out now, cheers
     
  8. Dec 19, 2015 #7
    I repeated the experiment (edit: without the balloon). While I was spinning the tube I simultaneously stuck my tongue end of the tube and could feel a very slight vacuum. After some thought i must conclude your analysis of the problem is incomplete. My suggestion is to actually calculate not just the acceleration but the air pressure using the acceleration along the length of the tube. This will involve some integration.
     
    Last edited: Dec 19, 2015
  9. Dec 19, 2015 #8
    I presume you had the spinning end sealed off with a balloon also. If you had both ends open you will end up the generation of a vacuum due to Bernoullis venturi effect.
    But yeah the logic of increased density at the end of the sealed tube does still hold, it's just at a much smaller magnitude due to my incorrect assumption before.
    If we presume the 24g is experienced along the full length of the tube (which it isn't, but just for a simple calculation)
    P=ρ.g.h,
    =1.225x236x1.5=433Pascals
    Atmospheric pressure = 101,300 pascals, therefore we have less than 0.5% pressure increase.
    And as the air in the tube is not all under the weight of 24g, it would actually end up being half of that, so less than a 0.25% pressure increase.
    Impressive if you could feel that!

    The thing I don't quite get about the above calculation, is that it uses the density of air of 1.225kg/m3 which is the density of air at sea level, i.e. atmospheric pressure. But the density in this case has changed, and the only way to calculate the density is to know the pressure, but that's what we're trying to figure out.. how does that work? I'm not sure you can use the hydrostatic pressure formula for a compressible fluid, hence why it's called "hydro"static..

    Hmmm, so what's the forumula for pressure of a gas?
     
  10. Dec 19, 2015 #9
    I edited my post, experiment without the balloon, and yes I could feel a very slight vacuum. I think you actually have to do the integration?
     
  11. Dec 19, 2015 #10
    Look at a small cylindrical element of gas in the tube and list the forces on the element. Consider the end of the tube closed in this calculation.
     
  12. Dec 19, 2015 #11
    It's not quite that simple for two reasons, one because the effective gravity is dependent on the position in the tube, and secondly because of the compressible nature of the fluid. So the pressure gradient would actually increase at an increasing rate because these two things working together. When the fluid compresses as you get further down the tube, whereas normally gravity is constant, in this case the effective gravity grows.

    If the tube is sealed on both ends, mass stays the same, so you could calculate the pressure from P=F.a where Force is the average weight of the air (probably somewhere between 24 and 12 g effective weight increase of normal air) acting on a 10mm diameter surface tube.

    But with the stationary end open, additional mass would be drawn into the tube, and hence the Pressure at the end would increase further.
     
    Last edited: Dec 19, 2015
  13. Dec 19, 2015 #12
    I disagree. I am pretty sure it is straightforward integration once you come up with a differential equation (who knows, maybe it must be numerically integrated).

    As Feynman would say, "Shut Up and calculate".

    Just kidding!
     
  14. Dec 19, 2015 #13
    No offence but it's no help to continually tell me to create a differential equation for the scenario. If I knew how I wouldn't be asking here.
    Everything is straight forward when you know how to do it. We shall wait!
     
  15. Dec 19, 2015 #14
    I told you to look at the forces on a small cylindrical element of gas in the tube. What did you come up with? Waiting. Edit, assume equilibrium.
     
  16. Dec 19, 2015 #15
    Stop it you're not helping anymore. I looked at it. The comprehensibility of the fluid and increasing rate of acceleration make it difficult for me to calculate, hence why I'm asking here. If you don't know yourself please stop trolling.
     
  17. Dec 19, 2015 #16
    I am pretty sure of a way to move this problem forward, look at the forces on a small element of gas.
     
  18. Dec 19, 2015 #17
    Hilarious
     
  19. Dec 19, 2015 #18
    I think you were centrifuging your tongue into the tube. ?:)
     
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