Buoyancy

The volume of the balloon is not the determining factor of lift, it's the volume of the hydrogen contained...

Remember you must take into account the weight of the balloon itself and subtract that first from the lift...

https://calculator.academy/hydrogen-lift-calculator/

Yup know all that,

Just wondering if the basic concept is as I think it is.

Well the volume of the balloon would increase as the balloon goes higher as a result the dropping external pressure...the added internal balloon would just add weight...

https://zmatt.net/weather-balloon-physics/

High-Altitude Zero-Pressure Balloon

https://www.mdpi.com/2226-4310/9/10/622

Agree on all that,

BUT,,,

OK balloon is 100 ctft. of H2 displacing 100 cuft of AIR.

Now we add 10 cuft of AIR into the inside smaller balloon, which also displaces 10 aditinal outside air.

BUT the 10 cuft added inside does not add any lift it's ambient air.

if anything it is adding some weight true?

Well it seems to me that the added weight would be the weight of the balloon material, plus the air would add weight according to how much the density was increased by balloon pressure...both seem minimal to me....

...."First manned flight in a hydrogen balloon, France, 1 December 1783 (1887). French aeronauts Jacques Charles (1746-1823) and Noel Robert made the first manned (free flight) ascent in a hydrogen balloon, designed by Charles, a physics professor, and constructed by Robert and his brother Jean. It took off in front of a crowd of 400,000, landing two hours later at Nesle-la-Vallee, over 27 miles away. Robert alighted there but Charles re-ascended in the balloon, as seen here, reaching an altitude of over 9000 feet. From Histoire des Ballons by Gaston Tissandier. (Paris, 1887)."...

If you are adding air inside the balloon, at what pressure will that be? If needed to have pressure to inflate another balloon, then the added air volume will be more than 100 cu ft at normal temperature and pressure. Secondary school was 70 years ago! p1v1=p2v2

Yes. Until a source of ignition changes the game.

Now the question is. If you made the Balloon "Bigger" So it is displacing more "AIR" But did NOT increase the amount of "H2" The amount of lift will decrease correct? Because of the greater amount of "AIR" being displaced.

No, the amount of lift is equal to the weight of air displaced minus the weight of the balloon. H2 doesn't have lift. It has weight, but much less than an equal volume of air.

At a fixed temperature and pressure, any gas has the same number of molecules in a given volume.

"Avogadro's hypothesis states that equal volumes of all gases at the same temperature and pressure contain equal numbers of particles."

Hydrogen has a molecular weight of 2. Air is a mixture of gases, 78% nitrogen (wt=28), 21% oxygen (wt=32), and 1% argon (wt=40). (All other gases are in minute amount.) The average air molecule weighs 28.96.

So, take a balloon full of air, replace the (wt=28.96) molecules with (wt=2) molecules, and the balloon will replace the heavier air above it just as a seesaw with a heavier weight will lift a lighter weight on the other side. The difference in weight is the "lift".

OK, think if it this way.

A balloon filled with 100 cuft of "AIR" will not float at all. Because of the weight of the balloon, But for now remove the balloon from these thinkings.

Ok Balloon has a volume of 100 cuft. so it is displacing 100 cuft of "AIR" correct?

This "AIR" has a weight of 1.291 oz per cuft, or for 100 cuft =129 oz.

If the balloon was also filled with air. the mass inside matches the air outside so neutral buoyancy correct?

But it is filled with H2.

H2 = 0.083 oz per cuft. so times 100 cuft the H2 inside the balloon weighs, 8.3 ounces correct?

So weight of displaced air 129 oz minus weight of inside balloon 8.3 oz equals 120 ounces of lift.

Now insert into the balloon enough of this amibent "AIR, so it is now displacing 110 cuft of space.

110 times 1.29 = 142 oz of displaced air.

142 -H2(8.3) =134.7 - AIR(13)= Only 4.7 of positive lift now.

If I am wrong, where am I wrong?

So you have 100 cuft of H2 and you add 10 cuft AIR. The 100 cuft H2 weighs 8.3 oz and 10 cuft AIR weighs 12.9 oz, for a total of 21.2 oz.

The 110 cuft of AIR displaced weighs 142 oz, so the lift is 142 - 21.2 or 120.8 oz.

What also makes it confusing is densities,

100 ctft H2 at 8 oz so 8.3/100=0.083 oz per cuft OK

Now the mix with the AIR, 21.2/110=0.1927 oz per cuft

So it is MUCH denser, yet has the same lifting power? Huh?

OK, we are neglecting any weight of the balloon material, so a balloon with just air is neutrally buoyant.

It makes no difference if you mix in 10 cuft air with the 100 cuft H2, or if you fasten a separate balloon containing air to the H2 balloon. The lifting power will be the same as the AIR balloon contributes nothing, being neutrally buoyant. It makes no difference how large the neutrally buoyant AIR balloon is, but if its air is mixed in the same balloon with the H2, the density of the mixture will increase with higher percentage air.

The correct way to solve for lift is to subtract the weight of the balloon(s) from the weight of the air displaced.

134.7 - AIR(13)= 121.7

How high can a helium balloon go,provided that the balloon has sufficient room for expansion?

Could it go outside of the atmosphere and be pushed along by the solar wind if the balloon is metalized with a reflective coating?

It does not have to be teardrop shape.

I think carrying a tank of excess hydrogen and a compressor would be a better way to control lift;Pump hydrogen into the tank,to decrease lift,and pump it from the tank into the balloon to increase lift.

You could use a bellows instead of a tank to save weight

Pre-calculate and allow for the weight of the compressor and pump beforehand.

Some autonomous submarines simply contract their length to decrease their displacement and increase their depth,and extend their length to rise to the surface to transmit data.No gasses required.While surfaced,they recharge their batteries via solar cells,and dive again to collect more data.

I think the added mass (weight) of the air in the internal balloon would counter the increased buoyancy. Consider heating the H2 instead.

Joe,

I will look at your questions from two different viewpoints--the first is if the balloon's envelope is able to change size so there is no pressure difference across its wall, and the second is if the balloon's envelope is rigid and holds its size regardless of pressure differences across its wall.

First: When filled to a sufficient volume with H₂, it will have displaced that volume of air and will therefore be buoyant because its total mass is lower than the displaced air, as we both know. Introducing more H₂ into it will necessitate its having a proportionally larger volume because the envelope had to expand to keep the pressure difference across its wall at zero. It will have a greater lift because the mass of displaced air is greater. Instead, introducing the same volume of air into the balloon gives an increased volume but no increased lift because the added mass of air has displaced an equal mass of air on its outside.

Second: The rigid walled balloon will have the same initial buoyancy as the above flexible walled balloon when filled with H₂ to a pressure equal to that on its outside. Adding any specified amount of additional H₂ to the interior will increase the mass but not change the volume, so its buoyancy will be lower. Similarly, any additional amount of air will result in a proportionally greater total contained mass but no change in volume, so its buoyancy will be substantially lower.

In the real world, balloons we see in parties, etc. have elastic walls in which the internal pressure increases as the volume of contained gas increases, but the balloon also expands in contained volume as the pressure increases. Therefore, adding H₂ will generally increase its buoyancy (initially) because the added H₂ is increasing its volume faster than it is increasing its pressure (and therefore its density.) However, adding air will slightly decrease its buoyancy because the added air is adding to the mass at a rate slightly faster than the increase in its volume.

To your question about how to get more lift without adding more H2, the rigid-walled "balloon" will have a lower mass if it has less H2 because it now is at a reduced pressure. It will have a greater lift. In practical terms, however, the mass of the balloon's envelope needed to maintain its rigidity will negate any increase in lift. As a test, picture a rigid sphere of specified volume and mass--if its mass is low enough it could have buoyancy. Add any gas to it and its mass will increase so at some point it will no longer be buoyant. The only way you can get more lift in a balloon is if you are filling it with a gas that has a lower density compared to its surrounding air at the same pressure.

--JMM