1
$\begingroup$

I don't get one thing: we say a hot air balloon rises as hot air is less dense than cooler air.

However, if the heat source continues to add heat to the air in the balloon, doesn't the hot air become so hot that pressure will be too high for the balloon to resist and burst? But it doesn't burst even if we continue to heat the air in balloon even for a long time?

Improve this question
$\endgroup$
3
  • 20
    $\begingroup$ The balloon is not sealed at the bottom. Why do you think excess pressure would not leak out? $\endgroup$
    – Bill Watts
    Nov 2, 2021 at 6:26
  • 2
    $\begingroup$ Indeed, if the air couldn't escape (and the envelope was perfectly rigid, instead of approximately so), then the balloon wouldn't rise. Ignore practical details like unfolding & filling the balloon, and assume you have a spherical balloon at ambient temperature. It has N air molecules, that weigh X kg. Heat it up, the air expands, some goes out the opening, and fewer molecules are left inside, weighing less. Seal that opening, and the weight doesn't change. $\endgroup$
    – jamesqf
    Nov 2, 2021 at 16:35
  • 3
    $\begingroup$ No mention anywhere in the answers about how the hot air balloon loses heat through its canopy? I'll say it then; the canopy may prevent hot air from mixing with the cooler outside air, preventing convective heat loss, but it does very little to prevent conductive heat loss through the balloon. Greater heat inside = greater rate of heat loss. $\endgroup$
    – PcMan
    Nov 3, 2021 at 3:41

4 Answers 4

Reset to default
27
$\begingroup$

Since the balloon isn't sealed and the bottom has an opening, air will escape through this opening thus preventing the point where the pressure inside the balloon is so high that it would explode.

Improve this answer
$\endgroup$
17
$\begingroup$

Hot air balloons take advantage of the density of hot air, like you correctly mention in your post. Pressure won't cause the balloon to suddenly burst given that the hot air balloon has an exit present at the bottom, resulting in what could be approximately equal pressure inside and out, ruling out a possible fail in the system.

Improve this answer
$\endgroup$
9
$\begingroup$

doesn't the hot air become so hot that pressure will be too high for the balloon to resist and burst?

If you have a sealed balloon, and you heat it enough, then yes.

Hot air balloons aren't sealed though. If the pressure gets too great, they can vent out of the opening where the flame goes.

Even with that though, you'd need to heat the air to the point where the balloon bursts. There's a lot of air in a balloon, and a large surface area. However much heat you put in, there's going to be an equilibrium point where you can't heat the balloon any more because all the heat you're putting in is dissipated into the surrounding (cooler) air. You can (in principle) do the sums to find where that equilibrium point is, but the main thing to note is that if that equilibrium point is less than the pressure which would explode the balloon, then the balloon stays intact.

Improve this answer
$\endgroup$
1
  • 1
    $\begingroup$ No, it's not the that the pressure in a (real world) hot air balloon keeps increasing. It's always near equilibrium with the outside air, since the inside are is free to leave. It's that air leaving that reduces the weight of air remaining nside, and creates lift. $\endgroup$
    – jamesqf
    Nov 3, 2021 at 3:52
3
$\begingroup$

The ideal gas law states that $PV=nRT$. In many cases you apply this to a closed system with a fixed quantity and volume of gas, so you often see that as temperature ($T$) increases, so does pressure ($P$). A hot air balloon is not sealed or pressurized, however - as the temperature increases, air escapes the balloon. If you consider just the air in the balloon, as $T$ increases, the number of air molecules in the balloon ($n$) decreases, resulting in a system of constant pressure and volume.

Air is forced out by a slight and transient pressure differential. Anytime the pressure in the balloon is higher than outside, air will flow through an opening in the balloon to equalize the pressure. The bigger the pressure differential, the faster air will flow to equalize the pressure, so the internal pressure is always close to (but perhaps slightly higher than) the external pressure, and in an equilibrium state, they are identical.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.