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This question already has an answer here:

For a project into balloon simulation I'd like to know how the force on a balloon changes with altitude:

I know that the Buoyant force on a balloon is:

$F = (\rho_{air} - \rho_{helium})gV$

Using changing air density from the international standard atmosphere and changes in gravitational acceleration I can calculate the buoyant force and add this to the (negative) force of gravity mg.

However I know that it isn't this simple, the balloon changes volume with altitude as the air pressure outside the balloon changes.

I'd like to know how the balloon inflates - this would be a great help!

I think it may be something to do with ideal gas laws - I can calculate air temperature and pressure from the international standard atmosphere but do I use pressure as the air pressure? - or air pressure minus balloon pressure?

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marked as duplicate by John Rennie, Danu, ACuriousMind, Kyle Kanos, Ali Sep 12 '14 at 15:32

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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It is reasonably easy: the balloon will want to maintain pressure equilibrium with its surroundings, i.e. $P_{in} = P_{out}$. This occurs because any pressure imbalance can be redressed on the sound-crossing time scale, i.e. the time it takes a sound wave to cross the balloon's diameter. This can easily be checked to be less than a millisecond, thus on timescales longer than this the balloon will have achieved pressure balance with its surroundings.

Since you mentioned the international standard atmosphere you probably know that it tabulates pressure as a function of altitude as well. Calling $P(h)$ the atmospheric pressure at height $h$ above ground, we find $$ n(h) k T(h) = P(h)\;\;\;\; (1) $$ where we assume $P(h)$ known from the standard atmosphere, and the left-hand side of this equation referes to quantities inside the balloon.

Now we have to determine $T(h)$ inside the balloon. A reasonable hypothesis we can make is that, at least initially, the balloon's ascent is so fast that it will be unable to exchange heat with its surroundings: thus its expansion is adiabatic, in which case we know that $T \propto n^{2/3}$. We can now write $$ T(h) = T_0 \left(\frac{n(h)}{n_0}\right)^{2/3} \;\;\;\; (2) $$ where $T_0$ and $n_0$ are the values of $T,n$ respectively at ground level.

Putting together the two previous equations we find: $$ n(h) = n_0 \left(\frac{P(h)}{P_0}\right)^{3/5} \;\;\;\; (3) $$ which is what you were searching for.

In the long run, though, the balloon will also reach thermal equilibrium with its surroundings, which means it will tend to the same temperature as the surrounding air. From the equation of pressure equilibrium we see that this implies that the balloon will also have the same particle density as its surroundings, hence the density ratio will tend to $$ \frac{\rho_{balloon}}{\rho_{atm}} \rightarrow \frac{m_{balloon}}{m_{atm}}\;\;\; (4) $$ where $m_{balloon}$ and $m_{atm}$ are the mean mass of molecules in the two mixtures.

Thus the answer depends somewhat on the time when you are asking: immediately after being released, Eq. 3 applies, later on Eq. 4 applies.

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    $\begingroup$ Note that this answer relies on pressure equilibrium between the inside of the balloon and the outside. For high altitude helium balloons this is usually a correct assumption, as the balloon starts out as a limp bag with plenty of scope for expansion: for meteorological balloons, there is a size dependent pressure difference which depends on the material of the balloon, and where it is on the stress-strain curve (see for example this earlier answer) $\endgroup$ – Floris Sep 12 '14 at 13:47
  • $\begingroup$ @Floris Thanks for pointing this out, I ignored this! $\endgroup$ – MariusMatutiae Sep 12 '14 at 14:09

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