How Does Temperature Affect the Lifting Capacity of a Hot Air Balloon?

In summary: No way. In the ideal gas law, you must always use absolute temperature. Only temperature differences are the same for C and K.In summary, the problem involves finding the maximum load a hot-air balloon can lift based on the volume of the balloon, the temperature and density of the surrounding air, and the temperature of the hot air inside the balloon. Using the ideal gas law and treating air as an ideal gas, the density of the hot air at 400°F can be calculated. The weight of the hot air is included in the load of the balloon, and the buoyant force is equal to the weight of the air displaced. Absolute temperature must be used in the calculations.
  • #1
Ian Baughman
36
2

Homework Statement



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.

Homework Equations



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

The Attempt at a Solution


[/B]
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?
 
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  • #2
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
Simon Bridge said:
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.
@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
Simon Bridge said:
Note: you want the weight of the balloon+load to be the same as the weight of air displaced.

That would mean then that I would just be ignoring the weight of the hot air correct?
In that case my equation would be:
ρoutgVB = (mB + mload)g​
 
  • #5
Ian Baughman said:
That would mean then that I would just be ignoring the weight of the hot air correct?
In that case my equation would be:
ρoutgVB = (mB + mload)g​
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
Simon Bridge said:
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.
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.
 
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  • #7
Ian Baughman said:
That would mean then that I would just be ignoring the weight of the hot air correct?
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
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 O[SUB]2[/SUB]]​
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
Ian Baughman said:
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 O[SUB]2[/SUB]]​
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.
No way. In the ideal gas law, you must always use absolute temperature. Only temperature differences are the same for C and K.
 
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  • #10
Simon Bridge said:
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?
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?
 
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  • #11
Chestermiller said:
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
## 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 ##.
 
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  • #12
Charles Link said:
## 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 ## Nt/m^2 ##.
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.
 
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  • #13
Chestermiller said:
With regard to that 7/2 thing, can you provide a single reference giving the equation of state for a gas (even non-ideal) in the form you presented?
 
  • #14
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.
 
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  • #15
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
llatosz said:
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
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
llatosz said:
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
To a good approximation, the pressure inside these kinds of balloons is known to be nearly equal to the atmospheric pressure outside.
 
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  • #18
Chestermiller said:
To a good approximation, the pressure inside these kinds of balloons is known to be nearly equal to the atmospheric pressure outside.
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
 
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  • #19
llatosz said:
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
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.
 

Related to How Does Temperature Affect the Lifting Capacity of a Hot Air Balloon?

What is the physics behind a hot air balloon's ability to fly?

The physics behind a hot air balloon's ability to fly is based on the principle of buoyancy. The hot air inside the balloon is less dense than the surrounding cool air, causing it to rise. This creates an upward force, or lift, that allows the balloon to fly.

How is the temperature of the hot air controlled in a hot air balloon?

The temperature of the hot air in a hot air balloon is controlled by a burner, which uses propane to heat the air inside the balloon. The pilot can adjust the temperature by controlling the amount of propane being burned. This allows the balloon to rise or descend as needed.

What is the role of the envelope in a hot air balloon?

The envelope of a hot air balloon is the large fabric bag that holds the hot air. It is made of a lightweight and heat-resistant material, such as nylon or polyester. The envelope is responsible for containing the hot air and providing the shape and structure of the balloon.

Why do hot air balloons have a wicker basket?

The wicker basket on a hot air balloon serves as the gondola for the passengers and pilot. It is made of a lightweight and sturdy material, and its open design allows for easy entry and exit from the balloon. The wicker material is also able to withstand the heat from the burner, making it a safe option for a hot air balloon basket.

How do wind conditions affect the flight of a hot air balloon?

Wind conditions can greatly impact the flight of a hot air balloon. The direction and speed of the wind can determine the direction and speed of the balloon's flight. Pilots must carefully monitor wind conditions and make adjustments to their flight plan to ensure a safe and enjoyable flight.

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