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Re: Where can I buy floater balloon?
J68HZ
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. Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook!  Moderator – North American QRO Group at Groups.IO. email: bill@...
From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon? Bill,
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.
Jerry, KE7ER
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.
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Re: #pa #u3s Extreme power curve across the bands.
#pa
#u3s
Nik
Hi Allison,
I'm using the 40m lpf and the 20m lpf (positions 5 and 3 respectively). Although, I've double checked the values and the windings on the 40m filter, I've not checked the 20m one! Think I'd better take them all out and have a check, then test with just the 2 in place. (I've 80m in slot 0, 40m in 5, 30 in 4, 20 in 3, 17 in 2 and 15 in 1).
In fact I think I'll put a signal through them and have a look on the scope.
Thank you. It's nice to have someone else's thoughts.
Nik.
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Re: Where can I buy floater balloon?
Stephen Farthing G0XAR JO92ON97
Gerry,
I’m wondering what you mean by four times larger? Do you mean four times the volume? Or four times the surface area?
It’s not a trick question. I’m just trying to follow your argument. It’s 50 years since I did any academic physics so I’m a bit rusty.
Warm regards from a rainy England,
Steve G0XAR
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Re: Where can I buy floater balloon?
K9HZ wrote:
"The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion."
One has to consider the partial pressure of gasses. There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal. My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material.
Joe
On Sun, May 26, 2019 at 12:47 PM J68HZ < bill@...> wrote: 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. Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook!  Moderator – North American QRO Group at Groups.IO. email: bill@... From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon? Bill,
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.
Jerry, KE7ER
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.
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Re: Where can I buy floater balloon?
In actuality, this is
Backwards of what is happening with these small floaters.
They are in fact miniature small "Superpressure" Balloons.
When they float it is because there is it lifting gas that has
lost it's lifting power because it has become more dense that is
was at low altitudes because it is now under pressure. There is
a much larger pressure differential inside vs outside of the
balloon envelope when at float than there is when at lower
altitudes.
so the thought that at float, if it is going to leak it has less
chance is 100% totally backwards.
Joe WB9SBD
toggle quoted messageShow quoted text
On 5/26/2019 2:30 PM, Joe Street wrote:
K9HZ wrote:
"The
balloon leaks less at altitude because the pressure on the
inside and the outside come to equilibrium by diffusion."
One has to consider the partial pressure of gasses. There
is still a huge difference in partial pressure of He between
inside and outside regardless of altitude so the balloon
should continue to leak at altitude all other things being
equal. My guess is that it is what was conjectured previously
(sorry I can't remember who wrote it) something happens to
mylar at cold temperatures which densifies it at a molecular
level thereby lowering the diffusion through the material.
Joe
On Sun, May 26, 2019 at 12:47
PM J68HZ < bill@...> wrote:
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.
Dr.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ
VP5/K9HZ PJ2/K9HZ
Owner
- Operator
Big
Signal Ranch – K9ZC
Staunton,
Illinois
Owner
– Operator
Villa
Grand Piton – J68HZ
Soufriere,
St. Lucia W.I.
Rent
it: www.VillaGrandPiton.com
Like us on
Facebook! 
Moderator
– North American QRO Group at Groups.IO.
email:
bill@...
From: QRPLabs@groups.io
[mailto:QRPLabs@groups.io]
On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater
balloon?
Bill,
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.
Jerry, KE7ER
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.
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Obviously I meant 144.390 !
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Re: Where can I buy floater balloon?
J68HZ
Here’s the math. Again, excuse the bandwidth and delete this straight away if you have no interest. This will be my last post on the subject because it’s getting old. There is no reason to really talk in terms of partial pressures of gas on the balloon side. It’s a pure gas for all intent. Oh the other side, air is a mix and I suppose you could add the force from the partial pressures of N2 + O2, but they are so similar in molecular weight, size, and behavior that they can be treated as the same (or just use the reference below which gives the density of air at various temperatures, pressures, altitudes.) In general a volume V of material of density ρ immersed in a fluid of density ρf experiences a buoyant force of Fb= ρfgV and a weight of W=−gVρ. so the available lifting force is Fl=gV(ρf−ρ). Where the object is floating at the surface of a liquid the buoyant force is modified to reflect the volume of liquid displaced Fb=gVdρf where Vd is enough to cover the weight of the floating object. Similarly, buoyant force on a submerged object (e.g. a balloon submerged/ floating in air) is equal to the weight of the displaced fluid, Fb=ρfgV The physical origin of this force is actually the pressure difference between the top and bottom surfaces of the floating object. Pressure in a fluid at a certain height is related to the depth of the fluid above that height by P(height2)−P(height1)=ρfg(height2−height1), that is, density of fluid times gravitational acceleration times height difference. If you have an object whose top and bottom surfaces are parallel, then it's pretty easy to calculate the buoyant force as the pressure differential times the area of those surfaces, Fb=(ΔP)(A)=ρfg*Δheight*A=ρfgV For an irregular shape like a baloon, you'll have to do some integration (a sphere for a balloon is a good approximation). You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing some additional calculus. But, according to the US standard atmosphere model, the density of the atmosphere takes about 12 miles to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is negligible, so you're pretty safe just using a single value for the density. However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and payload), and the balloon levitates at that level. For a controlled balloon, you can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air). Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook!  Moderator – North American QRO Group at Groups.IO. email: bill@...
From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 2:37 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon? In actuality, this is Backwards of what is happening with these small floaters.
They are in fact miniature small "Superpressure" Balloons.
When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.
so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.
Joe WB9SBD On 5/26/2019 2:30 PM, Joe Street wrote: "The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion." One has to consider the partial pressure of gasses. There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal. My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material. On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote: 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. Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook! Moderator – North American QRO Group at Groups.IO. email: bill@... From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon? Bill,
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.
Jerry, KE7ER
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.
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Peter GM0EUL
Good evening everyone
Remember its the last Monday tomorrow so its the QCX party. Fun for all and conditions seem reasonable today so fingers crossed it will stay good. Hope to catch some of you on the air and if you haven't tried it yet please do, its great fun and a good opportunity to get some qcx/qcx contacts.
Don't worry if you are a beginner at cw or don't consider yourself a contester. Its not a contest and everyone is very friendly and accommodating.
Rules and info here https://www.qrp-labs.com/party.html
73
Peter GM0EUL
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Re: Where can I buy floater balloon?
On 5/26/2019 4:26 PM, J68HZ wrote:
For
an irregular shape like a baloon, you'll have to do some
integration (a sphere for a balloon is a good approximation).
You can also take into account variations in density (or
gravitational acceleration) over the size of the balloon by
doing some additional calculus. But, according to the US standard atmosphere model,
the density of the atmosphere takes about 12 miles
to drop off to near zero, which corresponds to a fraction of a
percent change over the height of a typical hot air balloon (a
few tens of meters). That fraction of a percent is negligible,
so you're pretty safe just using a single value for the
density. However, you can't neglect differences in density
between vastly different altitudes. Remember that the buoyant
force on the balloon is equal to the weight of the amount of
fluid displaced. As you go higher, the density of the air
drops, which means the balloon displaces a lower mass of air.
Therefore, as the balloon rises, the buoyant force drops.
Eventually it reaches a height at which the buoyant force
exactly balances out the weight of the balloon (and payload),
and the balloon levitates at that level. For a controlled
balloon, you can adjust the level by either heating the gas
inside the balloon (thus making it expand and displace more
air) or by letting some gas out (thus making the balloon
contract and displace less air).
You are
forgetting, that the gas in the balloon is also getting less
dense per unit of volume as it rises due to expansion.
The Balloon
expands to equalize the pressure internally vs externally.
Until the
balloon stops expanding, the buoyancy amount does not change
from 10 feet above sea level to 150,000 feet above sea level NO
DIFFERENCE!
Joe WB9SBD
Dr.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ
VP5/K9HZ PJ2/K9HZ
Owner
- Operator
Big
Signal Ranch – K9ZC
Staunton,
Illinois
Owner –
Operator
Villa Grand
Piton – J68HZ
Soufriere,
St. Lucia W.I.
Rent it: www.VillaGrandPiton.com
Like us on Facebook! 
Moderator
– North American QRO Group at Groups.IO.
email:
bill@...
In actuality,
this is Backwards of what is happening with these small
floaters.
They are in fact miniature small "Superpressure" Balloons.
When they float it is because there is it lifting gas that has
lost it's lifting power because it has become more dense that
is was at low altitudes because it is now under pressure.
There is a much larger pressure differential inside vs outside
of the balloon envelope when at float than there is when at
lower altitudes.
so the thought that at float, if it is going to leak it has
less chance is 100% totally backwards.
Joe WB9SBD
On 5/26/2019 2:30 PM, Joe Street wrote:
"The
balloon leaks less at altitude because the pressure on
the inside and the outside come to equilibrium by
diffusion."
One has to consider the partial
pressure of gasses. There is still a huge difference in
partial pressure of He between inside and outside
regardless of altitude so the balloon should continue to
leak at altitude all other things being equal. My guess
is that it is what was conjectured previously (sorry I
can't remember who wrote it) something happens to mylar at
cold temperatures which densifies it at a molecular level
thereby lowering the diffusion through the material.
On Sun, May 26, 2019 at 12:47 PM J68HZ
<bill@...> wrote:
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.
Dr.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ
PJ4/K9HZ VP5/K9HZ PJ2/K9HZ
Owner -
Operator
Big
Signal Ranch – K9ZC
Staunton,
Illinois
Owner
– Operator
Villa
Grand Piton – J68HZ
Soufriere,
St. Lucia W.I.
Rent
it: www.VillaGrandPiton.com
Like us on Facebook!
Moderator
– North American QRO Group at Groups.IO.
email: bill@...
From: QRPLabs@groups.io
[mailto:QRPLabs@groups.io]
On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy
floater balloon?
Bill,
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.
Jerry, KE7ER
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.
|
|
|
Re: Where can I buy floater balloon?
J68HZ
“The Balloon expands to equalize the pressure internally vs externally.” This is never true. The only way it could possibly be true if the balloon itself offered no containment force against the gas at any volume (e.g. was infinitely elastic and forceless as if it weren’t there). The overall force balance includes the force of the gas on the inside… against the restraint force of the balloon plus that of the atmosphere. Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook!  Moderator – North American QRO Group at Groups.IO. email: bill@...
From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Joe WB9SBD
Sent: Sunday, May 26, 2019 4:51 PM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon? On 5/26/2019 4:26 PM, J68HZ wrote: For an irregular shape like a baloon, you'll have to do some integration (a sphere for a balloon is a good approximation). You can also take into account variations in density (or gravitational acceleration) over the size of the balloon by doing some additional calculus. But, according to the US standard atmosphere model, the density of the atmosphere takes about 12 miles to drop off to near zero, which corresponds to a fraction of a percent change over the height of a typical hot air balloon (a few tens of meters). That fraction of a percent is negligible, so you're pretty safe just using a single value for the density. However, you can't neglect differences in density between vastly different altitudes. Remember that the buoyant force on the balloon is equal to the weight of the amount of fluid displaced. As you go higher, the density of the air drops, which means the balloon displaces a lower mass of air. Therefore, as the balloon rises, the buoyant force drops. Eventually it reaches a height at which the buoyant force exactly balances out the weight of the balloon (and payload), and the balloon levitates at that level. For a controlled balloon, you can adjust the level by either heating the gas inside the balloon (thus making it expand and displace more air) or by letting some gas out (thus making the balloon contract and displace less air). You are forgetting, that the gas in the balloon is also getting less dense per unit of volume as it rises due to expansion. The Balloon expands to equalize the pressure internally vs externally. Until the balloon stops expanding, the buoyancy amount does not change from 10 feet above sea level to 150,000 feet above sea level NO DIFFERENCE! Joe WB9SBD
Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook! Moderator – North American QRO Group at Groups.IO. email: bill@... In actuality, this is Backwards of what is happening with these small floaters.
They are in fact miniature small "Superpressure" Balloons.
When they float it is because there is it lifting gas that has lost it's lifting power because it has become more dense that is was at low altitudes because it is now under pressure. There is a much larger pressure differential inside vs outside of the balloon envelope when at float than there is when at lower altitudes.
so the thought that at float, if it is going to leak it has less chance is 100% totally backwards.
Joe WB9SBD On 5/26/2019 2:30 PM, Joe Street wrote: "The balloon leaks less at altitude because the pressure on the inside and the outside come to equilibrium by diffusion." One has to consider the partial pressure of gasses. There is still a huge difference in partial pressure of He between inside and outside regardless of altitude so the balloon should continue to leak at altitude all other things being equal. My guess is that it is what was conjectured previously (sorry I can't remember who wrote it) something happens to mylar at cold temperatures which densifies it at a molecular level thereby lowering the diffusion through the material. On Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...> wrote: 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. Dr. William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ VP5/K9HZ PJ2/K9HZ Owner - Operator Big Signal Ranch – K9ZC Staunton, Illinois Owner – Operator Villa Grand Piton – J68HZ Soufriere, St. Lucia W.I. Rent it: www.VillaGrandPiton.com Like us on Facebook! Moderator – North American QRO Group at Groups.IO. email: bill@... From: QRPLabs@groups.io [mailto:QRPLabs@groups.io] On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy floater balloon? Bill,
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.
Jerry, KE7ER
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.
|
|
|
Re: Where can I buy floater balloon?
In most balloons the pressure
differential between inside and outside the envelope is soo
extremely small that it took many many flights to try to measure
it.
After six flights where they did get a measurable difference it
still was soo small that noise in the sensor is in the accuracy
pressure window.
the differential is less than 0.001 Psi.
And until it reaches the point where it can not stretch any more
is there a spike in pressure differential. And then it was still
very small, like 0.004 or so psi.
I have been doing these flights for 30+ years. 70+ of them. from
small balloons like the foil pico's
to the latex medium size ones.
https://youtu.be/HJ0IT4ZwtSo
How about 118 Thousand feet?
https://youtu.be/EdAuHr-bZ1M
and done some test flights for NASA guys, some really BIG
Balloons.
Joe WB9SBD
toggle quoted messageShow quoted text
On 5/26/2019 5:00 PM, J68HZ wrote:
“The Balloon expands to equalize the
pressure internally vs externally.” This is never
true. The only way it could possibly be true if the balloon
itself offered no containment force against the gas at any
volume (e.g. was infinitely elastic and forceless as if it
weren’t there). The overall force balance includes the force
of the gas on the inside… against the restraint force of the
balloon plus that of the atmosphere.
Dr.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ
VP5/K9HZ PJ2/K9HZ
Owner
- Operator
Big
Signal Ranch – K9ZC
Staunton,
Illinois
Owner –
Operator
Villa
Grand Piton – J68HZ
Soufriere,
St. Lucia W.I.
Rent it:
www.VillaGrandPiton.com
Like us on Facebook! 
Moderator
– North American QRO Group at Groups.IO.
email:
bill@...
On 5/26/2019 4:26 PM, J68HZ wrote:
For
an irregular shape like a baloon, you'll have to do some
integration (a sphere for a balloon is a good
approximation). You can also take into account variations
in density (or gravitational acceleration) over the size
of the balloon by doing some additional calculus. But,
according to the US standard atmosphere model,
the density of the atmosphere takes about 12
miles to drop off to near zero, which corresponds
to a fraction of a percent change over the height of a
typical hot air balloon (a few tens of meters). That
fraction of a percent is negligible, so you're pretty safe
just using a single value for the density. However, you
can't neglect differences in density between vastly
different altitudes. Remember that the buoyant force on
the balloon is equal to the weight of the amount of fluid
displaced. As you go higher, the density of the air drops,
which means the balloon displaces a lower mass of air.
Therefore, as the balloon rises, the buoyant force drops.
Eventually it reaches a height at which the buoyant force
exactly balances out the weight of the balloon (and
payload), and the balloon levitates at that level. For a
controlled balloon, you can adjust the level by either
heating the gas inside the balloon (thus making it expand
and displace more air) or by letting some gas out (thus
making the balloon contract and displace less air).
You
are forgetting, that the gas in the balloon is also getting
less dense per unit of volume as it rises due to expansion.
The
Balloon expands to equalize the pressure internally vs
externally.
Until
the balloon stops expanding, the buoyancy amount does not
change from 10 feet above sea level to 150,000 feet above
sea level NO DIFFERENCE!
Joe
WB9SBD
Dr.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ PJ4/K9HZ
VP5/K9HZ PJ2/K9HZ
Owner -
Operator
Big
Signal Ranch – K9ZC
Staunton,
Illinois
Owner –
Operator
Villa
Grand Piton – J68HZ
Soufriere,
St. Lucia W.I.
Rent
it: www.VillaGrandPiton.com
Like us on
Facebook!
Moderator
– North American QRO Group at Groups.IO.
email: bill@...
In
actuality, this is Backwards of what is happening with
these small floaters.
They are in fact miniature small "Superpressure" Balloons.
When they float it is because there is it lifting gas that
has lost it's lifting power because it has become more
dense that is was at low altitudes because it is now under
pressure. There is a much larger pressure differential
inside vs outside of the balloon envelope when at float
than there is when at lower altitudes.
so the thought that at float, if it is going to leak it
has less chance is 100% totally backwards.
Joe WB9SBD
On
5/26/2019 2:30 PM, Joe Street wrote:
"The
balloon leaks less at altitude because the pressure
on the inside and the outside come to equilibrium by
diffusion."
One
has to consider the partial pressure of gasses. There
is still a huge difference in partial pressure of He
between inside and outside regardless of altitude so
the balloon should continue to leak at altitude all
other things being equal. My guess is that it is what
was conjectured previously (sorry I can't remember who
wrote it) something happens to mylar at cold
temperatures which densifies it at a molecular level
thereby lowering the diffusion through the material.
On
Sun, May 26, 2019 at 12:47 PM J68HZ <bill@...>
wrote:
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.
Dr.
William J. Schmidt - K9HZ J68HZ 8P6HK ZF2HZ
PJ4/K9HZ VP5/K9HZ PJ2/K9HZ
Owner -
Operator
Big
Signal Ranch – K9ZC
Staunton,
Illinois
Owner
– Operator
Villa
Grand Piton – J68HZ
Soufriere,
St. Lucia W.I.
Rent
it: www.VillaGrandPiton.com
Like us on
Facebook!
Moderator
– North American QRO Group at Groups.IO.
email: bill@...
From: QRPLabs@groups.io
[mailto:QRPLabs@groups.io]
On Behalf Of Jerry Gaffke via Groups.Io
Sent: Sunday, May 26, 2019 11:19 AM
To: QRPLabs@groups.io
Subject: Re: [QRPLabs] Where can I buy
floater balloon?
Bill,
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.
Jerry, KE7ER
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.
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Re: #pa #u3s Extreme power curve across the bands.
#pa
#u3s
SkipF, NT1G
I'll bet my 'ticket', your backwards coil is FINE. Leave it for last. Transformers for phasing are a different issue. Extra winding (and too few) can be repaired without unwinding. Don't make extra work for yourself. 73 es gud luck! SkipF, NT1G
On Sun, May 26, 2019, 7:45 PM < nik@...> wrote: Just spent the last few hours scrutinising the low pass filters. Found a couple of mistakes. One choke with a winding too many, and another wound backwards(!)....
I rebuilt the 20m unit just to be sure. The results are the same.
If I select the 17m filter and transmit on 20m, the result is as expected, an increase of around 300mW, and approaching 1W through the 15m filter.
I then made a bypass filter and tried that on a dummy load. That was around 2W.
I guess the filter is doing it's job well. Although the output on the fundamental is still very low at 200mW through the 5W amplifier. I'm now thinking of experimenting with something compatible with the BC170, but with a wider frequency range. Something like a BC184.
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I had this same confusion when I first started mine up! It's spelled out in the manual, but I guess it's easy to overlook.
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Re: #pa #u3s Extreme power curve across the bands.
#pa
#u3s
Nik
Just spent the last few hours scrutinising the low pass filters. Found a couple of mistakes. One choke with a winding too many, and another wound backwards(!)....
I rebuilt the 20m unit just to be sure. The results are the same.
If I select the 17m filter and transmit on 20m, the result is as expected, an increase of around 300mW, and approaching 1W through the 15m filter.
I then made a bypass filter and tried that on a dummy load. That was around 2W.
I guess the filter is doing it's job well.
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European friends, please aim your 20 meter antenna toward Japan for part of the 1300-1400z session, and I'll be listening around .060. I'll also be checking 40-meters for NA same session. JS6TMW
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Re: Where can I buy floater balloon?
Stephen, You are asking what this statement means: > At 10,000 meters the air is roughly four times less dense, so the balloon > will be 4 times larger than at sea level. By "four times larger", I mean larger in volume. The ideal gas law says that if the amount of gas and the temperature are both held constant, then the volume will be inversely proportional to any change in pressure. So if the pressure increases by a factor of four, the volume decreases by a factor of four. Likewise, if the temperature changes but the pressure is held constant, then the volume changes proportional to the change in temperature expressed in degrees Kelvin. Zero degrees Kelvin is -273 degrees Centigrade, so an increase in temperature from -50 C (-223 K) to 20 C (293 K) with pressure held constant will cause the volume of our gas to increase by a factor of 293/223 = 1.31. If we had 1 liter of gas at -50 C, it expands to 1.31 liters at 20 C. Generally, all three can change simultaneously, for example compressing a gas causes it to heat up. Both relationships can be expressed simultaneously by the formula PV = nRT, where P is pressure, V is volume, T is temperature, and you can think of the nR factor as a constant for a given amount of gas. Digging a bit deeper, the n is a count of the number of molecules of gas, and the R is a physical constant called "the ideal gas constant". https://en.wikipedia.org/wiki/Ideal_gas_lawThe first line of this passage might be a bit confusing: < 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. best to edit it as follows: > If at 10,000 meters the balloon has more helium than what fits without stretching the mylar > 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. As Joe, WB9SBD, has also stated here, the lifting power of the balloon remains the same as the balloon rises so long as the balloon can expand without exerting extra pressure on our lifting gas. Think of a big floppy weather balloon about to be set free, the hydrogen has plenty of room to expand before the balloon envelope reaches its maximum size and starts pressing back. Once it gets high enough that the balloon envelope is tight and causes the hydrogen to be at a higher pressure than the ambient air, the lifting power starts to drop. Jerry
On Sun, May 26, 2019 at 10:11 AM, Stephen Farthing G0XAR JO92ON97 wrote:
I’m wondering what you mean by four times larger? Do you mean four times the volume? Or four times the surface area?
It’s not a trick question. I’m just trying to follow your argument. It’s 50 years since I did any academic physics so I’m a bit rusty.
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Hi..I'm Graham VE3WGW residing on the shores of Lake Eire in Canada. I have just built 2 of the prog rocks. Did the change of the 39k to 27k in the filter circuit to get closer to .33v. I installed TCOX's in place of the 27mhz crystal. After setting the calibration correctly. Mine set to 27,000,025 to get it correct. They have been sitting on 10 mhz for the past number of hours. Not moved more than .3 hz...... I have a gps calibrated oscillator... so it is dead on..... One of them is going to be used in a Kendecom 220 mhz repeater as the RX crystal died... At first tests everything is excellent.. The other is going to be the crystal bank in a GE MSTR II ... for 8 chans on 6 meters. Heck... The prog rock plus the TCXO is less than the price of 1 crystal !! Love them..
Graham VE3WGW
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Re: Minimally modified QCX-17 has been heard on 80, 60, 40, 30, 20, 17, 15, 12, 10m this UTC day !
Michael, With C1, C5, C8 removed and with T1 configured for 40m RX is working well for all bands, but the lower bands (80m, 160m) are a bit deaf. So initially wind T1 for 80m would do probably better. With a LPF installed, higher band harmonics on RX are not really an issue in practice in my experience; but a LPF is recommended to prevent FM broadcast mixing products. For LPF switching, I like KD1JVs minimalistic approach by using a few series DPDT slider switches to select the appropriate LPF. Another interesting way to switch LPFs is to use 1N4007 for swiching [1], were you can use the rectified RF + 12V at C30 to bias the diodes either forward or in reverse. Guido [1] https://www.qsl.net/in3otd/electronics/PIN_diodes/1N4007.html
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Re: Where can I buy floater balloon?
Thank You Jerry!
Buoyancy can be a fun thing to play with! Check out this design
of flight train we have flown twice, and seemed to have worked,
sadly the payloads did not survive the ultra cold long night.
But the theroy is there, just to get a payload to withstand that
night time temps!
Check out 'Earth Breeze"
https://www.qsl.net/nss/earthbreeze.html
Joe WB9SBD
Near Space Sciences
KB9KHO
toggle quoted messageShow quoted text
On 5/26/2019 10:56 PM, Jerry Gaffke via
Groups.Io wrote:
Stephen,
You are asking what this statement means:
> At 10,000 meters the air is roughly four times less dense,
so the balloon
> will be 4 times larger than at sea level.
By "four times larger", I mean larger in volume.
The ideal gas law says that if the amount of gas and the
temperature
are both held constant, then the volume will be inversely
proportional
to any change in pressure. So if the pressure increases by a
factor
of four, the volume decreases by a factor of four.
Likewise, if the temperature changes but the pressure is held
constant,
then the volume changes proportional to the change in temperature
expressed in degrees Kelvin. Zero degrees Kelvin is -273 degrees
Centigrade,
so an increase in temperature from -50 C (-223 K) to 20 C (293 K)
with pressure
held constant will cause the volume of our gas to increase by a
factor of 293/223 = 1.31.
If we had 1 liter of gas at -50 C, it expands to 1.31 liters at 20
C.
Generally, all three can change simultaneously, for example
compressing a gas causes it to heat up.
Both relationships can be expressed simultaneously by the formula
PV = nRT,
where P is pressure, V is volume, T is temperature, and you can
think of
the nR factor as a constant for a given amount of gas.
Digging a bit deeper, the n is a count of the number of molecules
of gas,
and the R is a physical constant called "the ideal gas constant".
https://en.wikipedia.org/wiki/Ideal_gas_law
The first line of this passage might be a bit confusing:
< 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.
best to edit it as follows:
> If at 10,000 meters the balloon has more helium than what
fits without stretching the mylar
> 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.
As Joe, WB9SBD, has also stated here, the lifting power of the
balloon remains the same as
the balloon rises so long as the balloon can expand without
exerting extra pressure on our lifting gas.
Think of a big floppy weather balloon about to be set free, the
hydrogen has plenty of room to expand
before the balloon envelope reaches its maximum size and starts
pressing back.
Once it gets high enough that the balloon envelope is tight and
causes the hydrogen to be at a
higher pressure than the ambient air, the lifting power starts to
drop.
Jerry
On Sun, May 26, 2019 at 10:11 AM, Stephen Farthing G0XAR JO92ON97
wrote:
I’m wondering what you mean by four times larger? Do you
mean four times the volume? Or four times the surface area?
It’s not a trick question. I’m just trying to follow your
argument. It’s 50 years since I did any academic physics so
I’m a bit rusty.
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Re: Is the CLK0 pin of si5361a supposedly up to driving an HEF4013 direct?
Hello Allison, Thanks for the reply. What we are doing is driving this modified to 4013 flip flop, pair of IXD_614 fast FET driver IC's, 1kW Class D amp direct from the Si5351a output via a 220nF poly cap. So what we are seeing is the Si5351a is generating just enough to fire the 4013 but it's far from ideal or reliable? What about using a logic level flip flop like a 74HCT74? That should reliably and correctly trigger from the Si, right? The LF version of the amp runs exactly as the build notes later show, no issues. The MF version ran the FET driver mad hot, with poor gate and drain waveforms, so we modded things... The other thing we are seeing is unlike the ultra reliable, perfect FET gate and drain waveforms on the similar LF amp, the MF one shows one fairly decent gate and drain waveform on one pair of FET's, but the other pair show incredibly spiky drain waveforms and poor gate waveforms when power to the FET's is applied, into a good Bird water colled dummy load or the antenna system. We have tried changing the FET driver IC's over, tried replacing the driver IC's and the FET's even to expensive ultra low gate capacitance ones, and tried swapping the gate leads from the drivers to the FET's over but the same "side" of the output shows bad waveforms. We also swapped out the 1500pF and 10 Ohm R1, R2, C3 and C4 components, and still the same side shows bad waveforms. The transformer wiring is symmetrical, we even changed the whole transformer T1 out to a newly wound one on on a different core from the FT-240-79 one recommended, still one pair of FET's exhibit the poor gate and drain waveforms. https://www.buerklin.com/en/Products/Passive-Components/Inductors/Ferrites/Ring-core%2C-outer-%C3%98-x-inner-%C3%98-x-height-%3D-63-x-38-x-25-mm%2C-N30/p/84D258but still the same side has bad waveforms... We can't see why this should be... The circuit is as below in regard to after the FET driver. We were using an HEF4013 flip flop and a pair of IXD_614 FET driver IC's. The interconnections are pretty short. The FET driver to FET gate wires are short and symmetrical. Amp schematic link and info: http://www.w1vd.com/137-500-KWTX.htmlwe use the W1VD LPF's for LF and MF. www.ixysic.com/home/pdfs.nsf/www/IXD_614.pdf A bit off topic and definitely not QRP but I do know a few here are using U3S's to operate on LF and MF into Class D amps, so hopefully not TOO off topic :) Best regards, Chris 2E0ILY mailto:chris@... aK> BS170 and 2n7000 are not identical. aK> While the 2n7000 can handle a bit more power it has higher capacitances aK> and one has to watch the gate threshold. aK> For class D and E one has to hit the gate hard as in 5-6V to insure the device aK> is turned on hard (acts like a switch) and turned off hard. aK> As to the HEF4013 you need the correct drive voltage to get the logic to switch. aK> Generally using 3 or 3.3V logic (SI5351) to drive 5V CMOS is bad practice aK> (also unreliable) and you need a level converter to go from 3V to 5V levels. aK> Going down in frequency its generally easier to drive power MOSFETS aK> so doing class D or E at 630m or 2200m is switch mode power technology. aK> There are low cost parts that can easily hit KW power levels at 630M. aK> You still need the right drive levels. aK> Allison -- Best regards, Chris Wilson (2E0ILY)
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