Question

A particular hot air balloon can be considered a sphere with diameter of 21m. The air inside the balloon is 60 celcius while the outside is 15 celcius. In Reno, the balloon launches (just barely) with 4 people on board. The next day , the balloon goes to Sacramento to launch with the same load and same temperatures (inside and outside the balloon). Will the balloon launch(will it rise up?) if yes, then how much extra load can they carry and still launch? if no, then how much load do they need to throw over board to launch? in both cases give answer in kg. Note that the ambient pressure is different in the two cities, due to the substantial difference in elevation.

Answer 1

The buoyant force acting on a hot air balloon is due to the density change between the inside and outside of the balloon.

$F_B = (\rho_{in}-\rho_{out})*volume$

Here, the volume of the hot air balloon = $V = 4\pi r^3/3 =4*3.14*10.5^3/3 = 4846.59 m^3$

The density inside the hot air balloon remains constant.

The variation of atmospheric pressure with altitude is given by the equation

$P(h) =101.325*e^{-0.00012h} kPa$

And density is given by

$\rho = p/RT =101.325*10^3*e^{-0.00012h}/(287.052*(273+15)) = 1.2256e^{-0.00012h}$

Now, Sacramento is 9 m above sea level and Reno is 1373 m above sea level.

The pressure outside is higher for Sacramento. so, here buoyant force will be less. (mass should be thrown overboard)

The change in force is

$\Delta F_B = (\rho_{in}-\rho_{B})*V - (\rho_{in}-\rho_{S})*V = (\rho_S-\rho_B)*V$

Substituting the equation of density,

$\Delta F_B = 1.2256*(e^{-0.00012h_{S}}-e^{-0.00012h_{B}})*V$

Substituting the values of h and V,

$\Delta F_B = 1.2256*(e^{-0.00012*9}-e^{-0.00012*1373})*4846.59 = 895.8878 N$

This much force must be removed

Using F = mg, The mass to be removed is

m = F/g = 895.8878/9.8 = 91.417 kg