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.
balloon would actually leak MORE at altitude because
the ambient pressure at altitude is lower.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ
Signal Ranch – K9ZC
Grand Piton – J68HZ
St. Lucia W.I.
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– North American QRO Group at 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
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
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
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
But somehow the balloon leaks less at altitude?
On Sun, May 26, 2019 at 12:21 AM, J68HZ wrote:
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
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
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