Physics of a Hot Air Balloon

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1. Jan 12, 2017

Ian Baughman

1. The problem statement, all variables and given/known data

A hot-air balloon stays aﬂoat because hot air at atmospheric pressure is less dense than cooler air at the same pressure.If the volume of the balloon is 500 m3 and the surrounding air is at 60◦F. What is the maximum load (including the weight of balloon, but excluding the weight of the hot air) the balloon can lift if the hot air is at 400◦F? The air density at 60◦F is 1.23 kg/m3.

2. Relevant equations

F = ma
Fbuoyant = ρDF × g × VDF
pV = nRT

3. The attempt at a solution

1) My first thought is to apply Newton's Second Law:
F = ma ⇒ FB - Fg = 0​
2) ρoutgVB = (mballoon + mload + ρinVHotAir)g
[Where ρout = air density outside of balloon and ρinVHotAir = mass of the air in the balloon]​
3) From here I can find ρin using the ideal gas law, VHotAir = VB, and I can isolate mballoon + mload but what is throwing me for a loop is the fact that it does not state this is an ideal gas and the say exclude the weight of the hot air. Would I just exclude ρinVHotAir from step two then?

2. Jan 12, 2017

Simon Bridge

You need to use a model for how air density depends on temperature.
What would be accepted depends on your course ... what do they normally do?
ie - you may do $PV = \frac{7}{2}NkT$ ... or you could look up values... or just write "modelling air as an ideal gas" and go for it.

Note: you want the weight of the balloon+load to be the same as the weight of air displaced.

3. Jan 12, 2017

@Simon Bridge A minor correction: $PV=N kT$ for an ideal gas, even a diatomic one, (to solve for $N/V$ and ultimately the density.) (The 7/2 number is a specific heat factor for diatomic molecules, etc.) $\\$ And for the OP, the buoyant force is essentially Archimedes principle.

4. Jan 12, 2017

Ian Baughman

That would mean then that I would just be ignoring the weight of the hot air correct?
In that case my equation would be:

5. Jan 12, 2017

Editing=(made a correction here)=The weight of the hot air is part of the load (of the balloon). See also my post #3. The buoyant force comes from the weight of the ambient air that is displaced.

6. Jan 12, 2017

Staff: Mentor

This is not correct. That 7/2 is for the enthalpy, not for the gas equation of state. The OP was correct in using the ideal gas law, and scaling the density in terms of the temperature. Using this, he can get the weight of the air in the balloon. The problem statement was confusing when it said to exclude the weight of the air. What they really meant to ask for was the lift force over and above the weight of the air.

Finally, at 1 atm., it is perfectly valid to use the ideal gas law for air.

7. Jan 13, 2017

Simon Bridge

The hot air is part of the balloon.
But I see the confusion ... the question puts the weight of the empty balloon as part of the load.
So weight or hot air + weight of load = weight of displaced air... which I think you got.

You decided on a model for air as a gas yet?
I see a couple of people weighing in on that 7/2 thing, but you have not commented.
Does your course have a standard you are expected to use?

8. Jan 13, 2017

Ian Baughman

I ended up treating air as an ideal gas consisting of diatomic oxygen molecules. This allowed me to find density at 400°F in the following way:

1) Using:
pV=nRT ​
2) I can get:
pV=(m/M)RT [M=molar mass=.032 kg/mol for O2]​
3) From here I was able to get density:
ρ=(m/V)=(pM/RT)​
4) I substituted this back in and was able to find the combined weight.

However, one thing I did notice that threw me off was that when I used the equation in step 3 temperature has to be in kelvin because of R. I was under the impression that Celsius and Kelvin, for the most part, were interchangeable.

9. Jan 13, 2017

Staff: Mentor

No way. In the ideal gas law, you must always use absolute temperature. Only temperature differences are the same for C and K.

Last edited: Jan 13, 2017
10. Jan 13, 2017

Staff: Mentor

With regard to that 7/5 thing, can you provide a single reference giving the equation of state for a gas (even non-ideal) in the form you presented?

Last edited: Jan 13, 2017
11. Jan 13, 2017

$PV=NkT$ and $PV=nRT$ are equivalent forms, where $N$ is the number of particles, and $k=1.381 E-23 joules/(degree \, Kelvin)$ is Boltzmann's constant. $R=.08206 \,$ liter-atm/(mol- degree Kelvin) . $n$ is number of moles. $N_A k=R$ where $N _A=6.02 E+23$ is Avogadro's number. $\\$ When working with pressure $P$ in atmospheres, it is easier to use the $PV=nRT$ form. @Chestermiller Simon incorrectly introduced a (7/2) factor in the equation (because as you pointed out, it does show up in form of the enthalpy equation), but in any case, $PV=nRT$ is the more direct route to the answer than $PV=NkT$. Otherwise, it requires a conversion of the pressure from atmospheres to units of $Newton/m^2$.

Last edited: Jan 13, 2017
12. Jan 13, 2017

Staff: Mentor

Hi Charles.

I knew all that. I was just challenging Simon because I felt that Simon still believed that the 7/2 belongs in the equation.

Last edited: Jan 13, 2017
13. Jan 13, 2017

Staff: Mentor

14. Jan 14, 2017

For the OP @Ian Baughman For doing any kind of hot air and/or helium balloon calculations, the very best you are going to do is by using the equation $PV=nRT$. (And it holds to fairly high accuracy for the pressures and temperatures of interest.) More refined equations of state such as Van der Waals $(P+\frac{an^2}{V^2})(V-nb)=nRT$ would require very precise control of the volume and temperature and pressure to be able to observe any slight corrections from any $a$ and $b$ correction factors that would be input. In any case, these corrections are quite small, and the equation $PV=nRT$ is a very reliable equation of state for the purpose at hand.

Last edited: Jan 14, 2017
15. Jan 18, 2017

llatosz

Isn't this problem impossible? Without knowing the pressure inside the balloon or the amount of gas in the balloon, there is no way to find the air density in the balloon to substitute into the buoyancy force equation

16. Jan 18, 2017

The buoyancy is determined by the gas density outside the balloon. The pressure inside the balloon is meanwhile determined by the temperature of the gas inside the balloon. The same equation $PV=nRT$ is used to solve for the density both inside and outside. The temperature is different for the two cases. Both inside and outside the balloon, the pressure is assumed to be $P$=1 atm. The pressure may be slightly higher than this inside the balloon, but to a good approximation, in a hot air balloon, the pressure is $P$= 1 atm.

17. Jan 18, 2017

Staff: Mentor

To a good approximation, the pressure inside these kinds of balloons is known to be nearly equal to the atmospheric pressure outside.

18. Jan 18, 2017

llatosz

OH!! That does make sense since the balloon is an open container, allowing excess pressure to escape. I was modelling it as a closed container. Thank you

19. Jan 18, 2017

Even for a helium balloon, to a good approximation, the pressure can be estimated to be $P$= 1 atm. You could do more conservative estimates and do the computation with $P=1.2$ atm or more, but it wouldn't change the internal weight of the helium and overall buoyancy significantly.