Density of a baloon?

Question

A recreational (open) hot air balloon (i.e., Pinside is approximately Poutside) has a volume of 2267 m3 when fully inflated. The total weight of the balloon, basket, ballast and pilot is 1989.4 N (448 lbs). By how much must the density of the air in the balloon be smaller than that of the surrounding atmosphere in order to keep the balloon floating level near the ground?

Answer 1

A balloon floats in air only when the volume of air it displaces is heavier than the weight of the balloon.

The weight of air W = Rho V g; where Rho = mass density of the outside air, V is the volume of air displace (the volume of the balloon, basket, ballast, and pilot combined), and g = 9.81 m/sec^2 close to Earth's surface.

The weight of the balloon B = rho v g + M g = 1989.4 N; where rho is the mass density of the hot air inside the balloon of combined mass M. Thus, for B < W (and the balloon floats), we need g(rho v + M) < g Rho V; so that rho v + M < Rho V, and rho < (Rho V - M)/v = Rho V/v - M/v. If we assume V = v; then this becomes rho < Rho - M/V. In terms of a ratio, rho/Rho < 1 - M/(Rho V) = 1 - M/m; where m = the mass of the displaced air.

Thus, the answer is that the hot air density must be slightly less than the density of the outside air minus the mass density of the balloon, basket, ballast, and pilot. If you know Rho, the outside air mass density, you can get a number since V and M are given.

Note that Rho varies according to temperature and some other factos; so for each ballon launch it would require recalculating. Also note, V, the volume of air displaced by the balloon includes more than just the balloon itself. The basket, ballast, and pilot also displace some volume of air. It is not clear these can be ignored in a real case by assuming V = v. That is, if V > v rho can be denser than when V = v and the balloon will still float.

Answer 2

Well, if it's floating in equilibrium, then the total weight of the balloon + the air inside must be the same as the weight of the outside air that the balloon displaces.

The air the balloon displaces weighs density times volume times gravity, or 1 kg/m^3 times 2267 m^3 times 9.8 m/s^2, that is, 22216.6 N, which is what the total weight of the balloon must be.

Since the weight of the ballast, etc., is 1989.4 N, there is 22216.6 - 1989.4 N left, that is, 20227.2 N, which is what the air inside the balloon weighs.

The density of the air inside the balloon is then 20227.2 N divided by gravity and volume, that is: 20227.2/(9.8*2267) = 0.91 kg/m^3.

Since they ask for how smaller it must be, we assumed 1 kg/m^3 for the outside air and the air inside is 0.91 kg/m^3, so the air inside must have a density which is 91% of the density of the outside air.

Answer 3

The air in the balloon must weigh 1989.4 N less than the equivalent volume of air outside the balloon. So the *difference* in density is simply 1989.4/2267 N/m^3. = 0.88 N/m^3.

BTW air at sea level weighs 11.8 N/m^3, so it's only about a 7.5% difference.