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A hot air balloon is heated with propane or similar hydrocarbon. One of the products of combustion is H2O with molar mass 18 so it is substantially less dense than air (78% nitrogen, molar mass 28 and 21% oxygen, molar mass 32). Another combustion product is carbon dioxide, molar mass 44. Does the balloon get much (or any) lift from the combustion products or does it all come from the higher temperature? A supplementary question: does the water condense on the envelope thereby increasing its weight and reducing lift?

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    $\begingroup$ I think that on chemistry.se you'd get a better answer $\endgroup$
    – sophit
    Commented May 2 at 16:04
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    $\begingroup$ There is no lift (aerodynamic force created by a relative wind) involved but buoyancy. Calculations use Archimedes' principle. The force comes from heat (combustion doesn't change the total mass of fuel+air, at least not in a significant quantity) $\endgroup$
    – mins
    Commented May 6 at 7:12
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    $\begingroup$ Any increase in buoyancy from the combustion products has got to be trivial. I've never heard of water condensing on the inside of a hot-air balloon envelope and that certainly didn't happen on the handful of flights that I've been on. But, I'll let someone else run through the math and get credit for a real answer. $\endgroup$ Commented May 8 at 19:44

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Burning one mol of propane yields 3 mols of CO2 and 4 mols of H2O, so a total of 192g for 7 mols of gas. Air has a mass of a little over 28g per mol so 7 mols equates to 196g. So ignoring the temperature change the buoyancy is around 2% of the mass of gas. That’s the equivalent of around a 2% change in absolute temperature, so heating the gas in the balloon by 6 or 7 degrees would have the equivalent effect. A balloon can operate at up to about 120C, so the temperature differential is much more significant but the gas buoyancy does contribute. It would make sense that water would condense on the fabric of the balloon, but a number of kg spread over the large surface would likely be absorbed by the fabric rather than raining on the occupants.

EDIT: as pointed out, the mass of combustion products is actually 204g so slightly denser than air.

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    $\begingroup$ You imply that 1 mol of reaction product replaces 1 mol of air. Avogadro's Law says we actually can make that assumption, at any fixed temperature. The depletion of oxygen (which is partially - but a large part - consumed in the flame) in the mix of hot air and combustion products will lighten things a little more, as N₂ is lighter than O₂. $\endgroup$
    – Chris H
    Commented May 9 at 12:47
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    $\begingroup$ It's also assumed here that the air in the balloon is completely replaced by combustion products. Since the air can mix freely with outside air I find this very unlikely. $\endgroup$
    – Chris
    Commented May 10 at 19:54
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    $\begingroup$ @Chris H agreed. However, combustion gases enter the balloon and force out cooler gas, so there will presumably be a tendency for air to be flushed out. That said, to lift a payload of 100kg would require around 300kg of hot gas (assuming 270K external and 360K internal) so a 40kg LPG cylinder would only displace a fraction of the air. Taking that into account the gas buoyancy becomes much less significant. $\endgroup$
    – Frog
    Commented May 10 at 21:16
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    $\begingroup$ @Frog I like your answer but am unsure of your calculation for 7 moles of combustion products which I think is 3x CO2 (44g) + 4x H2O (18g) = 204g. This reinforces your argument so I conclude that the net effect of the combustion products is negative from the lift perspective. $\endgroup$
    – Ken Mercer
    Commented May 13 at 21:07
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    $\begingroup$ @Ken Mercer you’re quite right, thanks for pointing that out $\endgroup$
    – Frog
    Commented May 15 at 5:54
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Pretty negligible. A typical balloon of volume $2800~\rm m^3$ at $100~\rm^\circ C$ at sea level pressure contains about $2600~\rm kg$ of air, or about 90000 moles.

If you burn $20~\rm kg$ (450 moles) of propane, this consumes 2250 moles of oxygen ($72~\rm kg$) and produces 1350 moles of carbon dioxide ($59.4~\rm kg$) and 1800 moles of water ($32.4~\rm kg$). Assuming all the combustion products stay in the envelope, this then displaces 900 moles of air ($26~\rm kg$).

So the new air mass inside the envelope is $$2600-72+59.4+32.4-26\approx 2594~\rm kg$$

So even with unrealistic assumptions where all the combustion products from about 10 gallons of propane all stay in the balloon, the density of air in the balloon is only decreased by about 0.2%. Assuming standard conditions, that means the $830~\rm kg$ being produced is increased to $836~\rm kg$, an increase of less than 1%.

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the ballon rises as was commented by bouyancy. it simply displaces a volume of air that weighs more than it's own weight (cold air outside, vs the hot air inside), that's all there is to it.

water vapour in air is a complex subject, beyond avation, but water has a lower mol weight but it has a very strong hydrogen bond. so molecules will bond to each other. this is what rel% measures, how effective air molecules are in preventing that, or rather how much water vapour the air can carry before molecules start bonding and creating mist. hot air is much drier (same ammount of water but lower rel%) so there's no condensation inside the balloon.

all other gases and subproducts are heavier molecules.

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There is no lift being produced. The combustion of the fuel heats the air inside the balloon envelope and it expands, but the amount of air remains the same so the density of the air inside the envelope decreases. The surrounding colder air moves around and down the envelope. A convection current is created which forces the hotter air inside the envelope up. Since the air is trapped inside the balloon envelope, it too rises with the hot air.

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  • $\begingroup$ I'm not the d'ver, but it almost sounds like you are saying that for a given net density of air within the envelope, and thus a given buoyancy, the balloon goes up faster if there's a convection current within the envelope that pushes up against the top of the envelope. $\endgroup$ Commented May 8 at 19:46
  • $\begingroup$ I did not say that, you did. $\endgroup$ Commented May 8 at 19:50
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    $\begingroup$ Ok. But it's not clear how mentioning the convection current added to the answer. PS -- re "but the amount of air remains the same" -- amount meaning volume? Certainly the number of molecules of air within the envelope does not remain the same when the air is heated. Nor does the weight of the air. Answer might be improved by clarification as to meaning of "amount". $\endgroup$ Commented May 8 at 20:11
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    $\begingroup$ Hot air balloons work because the air isn't trapped in them. When the air heats up, it pushes some of the air inside out of the mouth. The amount of air in the balloon goes down, but the volume stays the same, so the density goes down. $\endgroup$
    – Chris
    Commented May 10 at 19:57
  • $\begingroup$ "the amount of air remains the same" Hot air balloons are neither rigid nor sealed, so I see no way that could be right. $\endgroup$ Commented May 11 at 1:37

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