I Buoyancy of Helium and Hydrogen Balloons: Calculations and STP Considerations

AI Thread Summary
The discussion centers on buoyancy calculations for balloons, particularly why volume isn't explicitly included in certain calculations. Participants clarify that buoyancy is often expressed per unit volume, which can lead to confusion about the role of volume in calculations. The conversation also explores the behavior of partially inflated hydrogen balloons as they ascend, noting that they expand due to decreasing external pressure, which can lead to bursting. It is highlighted that partially filled weather balloons are common practice to prevent premature bursting. The impact of inflation levels on maximum altitude is debated, emphasizing that fully inflated balloons experience greater pressure differences compared to partially inflated ones.
chemisthypnos
As a note, this is my first post on this site. I was looking at some bouyancy calculations with regard to balloons being in air when I came across this post. I cannot understand why the calculation for the net force of bouyancy done by Vintage (the last post on the page) does not include the volume number in the calculation. Does it have something to do with it being calculated at STP? Should the calculation not inlcude a number for the volume of the balloon? Is there some condition that the volume term can be ignored?

https://physics.stackexchange.com/questions/9204/buoyancy-helium-vs-hydrogen-balloons
 
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The numbers presented are buoyancy per volume. Look at the units.
 
Orodruin said:
The numbers presented are buoyancy per volume. Look at the units.
How did they get to buoyancy per volume from that formula, though? There is no point when you divide by the volume. I am confused.
 
You divide by the volume and so the volume cancels.
 
He's making a mathematical statement about how much buoyant force would be gained for each m^3 of volume.
 
Ok. I see. Thanks!
 
One of the favourite discussion problems of our VI form physics teacher was what happens when a hydrogen balloon is only partly inflated at ground level . Assuming it is inflated enough to rise what happens to the balloon as reaches higher altitudes ?
 
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Nidum said:
One of the favourite discussion problems of our VI form physics teacher was what happens when a hydrogen balloon is only partly inflated at ground level . Assuming it is inflated just enough to rise what happens to the balloon as reaches higher altitudes ?
The same thing that happens to a scuba diver's lungs if he holds his breath while ascending. There is a reason emergency ascent when scuba diving includes shouting "aaaaaaaaaahhhhhhhh".
 
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Anyone else ?
 
  • #10
Nidum said:
One of the favourite discussion problems of our VI form physics teacher was what happens when a hydrogen balloon is only partly inflated at ground level . Assuming it is inflated enough to rise what happens to the balloon as reaches higher altitudes ?

I concur. If it's inflated with hydrogen enough to gain positive buoyancy then it will expand as it rises to higher altitudes and eventually burst like any hydrogen balloon.

Or does the density of the atmosphere decrease faster than the pressure with respect to altitude? If that's the case then a small enough amount of hydrogen might reach neutral buoyancy before the exterior pressure decreases enough to allow it to expand.

Ok it's a conditional answer but that's my answer.
 
  • #11
Well it is actually shown in this video what happens. It grows and eventually pops. As the pressure gets lower at higher altitudes the gas expands. Density and pressure are in this case directly related via ##p =p_0 -ρgz## where ##p_0## is the pressure at ground level, ##ρ## the density, ##g## the gravitational acceleration and ##z## the height above the ground.

[EDIT]
It is actually common practice to only partially fill weather balloons with hydrogen as it would otherwise pop too early.
 
  • #12
After my last post I looked into my own question and found that indeed, the balloon would expand and burst.
[QUOTE="

[EDIT]
It is actually common practice to only partially fill weather balloons with hydrogen as it would otherwise pop too early.[/QUOTE]

That's one of the main issues with all modern airships as well. Upon gaining altitude you have to vent helium, or at least collect it back into storage, which can be a pain.
 
  • #13
Getting there .

Assuming that the balloon does not burst during ascent what effect does full or partial inflation have on the maximum altitude that can be reached by the balloon ?
 
  • #14
A balloon fully inflated at ground level would burst sooner than a partially inflated balloon, clearly, assuming the two are identical apart from the amount of gas inside them. But if we assume that neither bursts at any point during their ascent then we have to conclude they would reach the same altitude, where air density is low enough to stop providing any positive buoyancy. But in that case the balloons are clearly not identical, since one of them is withstanding a far greater pressure difference than the other without bursting.

Am I missing something?
 
  • #15
Assume that the balloon fabric is inelastic so that maximum inflation volume is fixed .

The initially fully inflated balloon is at maximum volume at launch . The initially partially inflated balloon will swell as it climbs until it too reaches maximum volume .

Once at maximum volume the buoyancy force acting on each balloon is the same at any given altitude and is reducing at the same rate as the balloons ascend further .

There is something different about the two balloons though which will affect the maximum altitudes reached by each - what do you think that difference is ?
 
  • #16
Arjan82 said:
Well it is actually shown in this video what happens. It grows and eventually pops. As the pressure gets lower at higher altitudes the gas expands. Density and pressure are in this case directly related via ##p =p_0 -ρgz## where ##p_0## is the pressure at ground level, ##ρ## the density, ##g## the gravitational acceleration and ##z## the height above the ground.
This equation is incorrect. The correct equation is $$\frac{dp}{dz}=-\rho g=-\frac{pM}{RT(z)}g$$where T is the temperature at altitude z. and M is the molecular weight of air. The relationship of the density follows from the ideal gas law. So $$p=p_0e^{-\int_0^z{\frac{Mg}{RT(z')}dz'}}$$
 
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