Sunday, December 4, 2016

The Physics of Hot Air Balloons

Despite the fact that I am deathly afraid of heights, one thing that has always been on my bucket list is to ride in a hot air balloon.  This got me thinking about the physics involved to keep the balloon in the air and the thermodynamics that are involved.  Charles' Law can be applied to this as it states the volume of a gas is directly related to the temperature of the gas.  In order to allow the gas to expand, the operator of the hot air balloon ignites a propane burner.


Therefore, if we say it is 70 degrees fahrenheit out (294 K), we can figure out the initial volume of the balloon if we know the heated air has an average temperature of 373 K and has an average volume of 2500 m^3 (we're pretending it's not laying on the ground).

V1=(2500 m^3/373 K)(294)
V1=1971 m^3

Thus, it makes sense because of Charles' law that the final volume is greater after the air has been heated.

We also can use the ideal gas law: PV=nRT to figure out the density of the air in the balloon.
We can change the equation around and find that P=(rho)RT

If we say that the atmosphere pressure is  Po =1.013x10^5 Pa
The gas constant for dry air is 287 J/(kg.K)
The air inside is heated to an average temperature of 372 K then:

(rho)=(1.013*10^5)/(287 J/(kg.K))(372 K)
(rho)=0.946 kg/m^3

The density of the hot air inside the balloon is less than the density of the external air (1.2 kg/m^3), therefore, it rises.


No comments:

Post a Comment

Note: Only a member of this blog may post a comment.