The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion. The contents will leak until equilibrium is reached (if that is, in fact what is happening). At that point then, the buoyancy balance is reduces like this: [Pgas*Mwgas/Tgas]/[Pair*Mwair/Tair]… but now Pgas=pair… so the buoyancy ratio is [Mwgas*Tair]/[Mwair*Tgas]. And futher, for the most part, the temperature of the lift gas is approximately the same as that of the air… give or take… of course, this can change in the daytime when the sun warms the balloon and causes the lift gas to get warmer… but at night, the buoyancy equation reduces to Mwgas/Mwair… (2/16) meaning you will always have lift. Again you need to do the real force balance to determine if there is enough lift gas quantity for the balloon to rise or fall… but what it does shows is the limiting case where the balloon will eventually reach a maximum and minimum buoyancy (day/night) and I think we see this in the altitude data that gets reported.
The balloon would actually leak MORE at altitude because the ambient pressure at altitude is lower.
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From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
Subject: Re: [QRPLabs] Where can I buy floater balloon?
In post 34692 Mikael said:
> If I fill a foil balloon (like the one linked to first in this thread) with helium with 6gram of lift, seal it and let it sit inside it have lost its 6 gram of lift within a week and fall to the > floor but when used as a balloon for a radio tracker at +10000m its fine for several month, I had a tracker up for 64 days last year before a storm toke it down and Dave > > have a balloon released in February with this (2) balloon stil flying, whats happening at altitude that prevents the gas from leaking out as it does at ground level?
In post 34705 he states that this is repeatable, so I assume he did not happen to have a leak
in the balloons that remained in the house.
Is there something about mylar that makes it more impermeable to helium at -50C
(as found at 10,000 meters) vs room temperature of 20C?
I don't see PV=nRT https://en.wikipedia.org/wiki/Ideal_gas_law
causing the reported behavior.
Here is my reasoning, a first attempt at thinking this through:
Let's assume the balloon has "just enough" helium to fill it out at 10,000 meters,
so the pressure inside the balloon is always equal to the ambient pressure,
whether it is at sea level or at 10,000 meters.
At 10,000 meters the air is roughly four times less dense, so the balloon
will be 4 times larger than at sea level. But at sea level, the air that the balloon displaces
will have 4 times the density, so the lift of the helium is the same as at 10,000 meters.
(The lift is equal to the weight of the air displaced by the balloon minus the weight of the helium.)
Likewise, the temperature change affects the density of helium and air in equal measure.
(Assume the helium eventually reaches the same temperature as the ambient air.)
If the mylar balloon has more helium than needed at 10,000 meters (but does not pop),
then the pressure of the helium at altitude will be greater than the "just enough" case
and thus it is more dense. The lift will be less at altitude than it is at sea level.
The balloon will rise to that altitude at which the lift is equal to the weight of the payload.
When Mikael's balloon left behind in the house sinks to the floor after a week, it has
less helium remaining in the balloon than the one that is floating at 10,000 meters.
Even though the balloon at altitude is likely overfilled, with the helium pressing on the mylar walls.
But somehow the balloon leaks less at altitude? Curious.
On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:
Some science here.
Balloons lift (are buoyant) on the basis of Archimedes principle (cite: https://en.wikipedia.org/wiki/Archimedes%27_principle) which has the corollary that less dense material will rise above more dense material. It’s the same principle that causes oil to float on water… and with gases, it causes the lightest one (less dense gas) to rise above a heavier gas. Even hot air will rise above ambient air just because the density of the air in the balloon is less than that of ambient air.
In the buoyancy calculation, the force of lift is proportional to the relative differences between the densities of the gases… and in this case, that would be H2 (or whatever the lift gas is) and air. Now, the density of a gas can be calculated from the ideal gas law… P*V=n*R*T, or rewriting, P=p*R*T/Mw where “p” is the gas density and “Mw” is the molecular weight of the gas. The ratio for lift is then [P1*Mw1/T1]/[P2*Mw2/T2]. Plug in the numbers and you can determine where the lift will cease. Of course you need to do a force balance for the real lift and include the weight of the payload and the balloon.
The point here is that the buoyancy can vary based on the ambient temperature and pressure… even of the temperature and pressure of the gas in the balloon stays constant.