How Hot Must the Air in a Balloon Be to Lift a Given Mass?

In summary: It's a statics question: what temperature will generate enough lift to balance the other forces. What do you find strange?I don't see anything strange. Seems like a reasonable calculation.
  • #1
erok81
464
0

Homework Statement



I have a balloon with a volume of 500m3
Outside air temp of 300K
Mass to lift of 300kg
Molar mass of air is 28 g/mol (I didn't end up using this)

I am to find the temperature inside the balloon to barely lift the given mass. I have apparently forgotten everything buoyancy related.

Homework Equations



Fb=ρgV

The Attempt at a Solution



Okay...I know that in order for the balloon to lift Fb > Fg (subscript b = buoyancy and subscript g is gravity force)

Using Fb=ρgV I tried taking the differences of the the two air densities and setting it equal to my mass I need to lift. As seen here.

airballon)Vg=mg

Since I need to find the temp required, I found this relationship.

ρ=P/RspecT where P=pressure, Rspec is a known value, and of course T=temp. Rspec = 287.058 J/kg*K for dry air

Pballon and Pair are equal.

Subbing that stuff in I get...

[itex]\left(\frac{P}{R_{spec}T_{air}}-\frac{P}{R_{spec}T_{balloon}}\right) Vg=m_{cargo}g[/itex]

I enter my values into the above equation...the units work out, but I am leaving them off for brevity.

[itex]\left(\frac{1.013e5}{(287.058)(300)}-\frac{1.013e5}{287.058T}\right)(500)(9.81)=(300)(9.81)[/itex]

After some rearraging etc I end up with T=612K.

Looking up hot air balloon temps I see my answer is waaay off.

What I am not understanding here that is causing my incorrect answer?
 
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  • #2
erok81 said:
Rspec is a known value, and of course T=temp. Rspec = 287.058 J/kg*K for dry air
Ok... but shouldn't it be different for dry air at 300K and air at a higher temperature? Consider that and try solving the problem once again. I don't see any other errors.

Could you give the answer as well?
 
  • #3
Hmm...I'm not sure that it depends on temp.

Rspec=R/M where R is gas constant and M is molar mass.

With that said, I don't see how it can vary with temp.
 
  • #4
Oh yeah... sorry! That said, I have no idea on how to help you. I really see no error in your working. You could go through your calculations once again! 600 K does seem to be too much to me as well.
 
  • #5
erok81 said:
1.
Using Fb=ρgV I tried taking the differences of the the two air densities and setting it equal to my mass I need to lift. As seen here.

airballon)Vg=mg



I have a very similar problem so I'd rather just bump this thread.

I don't understand the step above. It seems that there are two upward forces(one outside of the balloon) but from my intuition I can only count one(the one inside). So can someone clarify this step?
 
  • #6
Your answer looks correct. It could be unreasonably high because the given conditions (volume, load) are unreasonable.
 
  • #7
haruspex said:
Your answer looks correct. It could be unreasonably high because the given conditions (volume, load) are unreasonable.

I think you have misunderstood my concern. This is not my thread but I have a very similar exercise which I solved using the formulas above(indeed correct), without understanding one of the steps. This one:

(ρair-ρballon)Vg=mg

Since ρVg=F_{b} I understand this as there are THREE acting forces, one inside the balloon and one outside the balloon(which is in the NEGATIVE y-direction) and mg. I find this strange since the balloon is not moving.
 
Last edited:
  • #8
zeralda21 said:
I think you have misunderstood my concern.
I was responding to the OP. Didn't notice it was a bit old. That's a risk with piggy-backing on an existing thread.
I solved using the formulas above(indeed correct), without understanding one of the steps. This one:

(ρair-ρballon)Vg=mg

Since ρVg=F_{b} I understand this as there are THREE acting forces, one inside the balloon and one outside the balloon(which is in the NEGATIVE y-direction) and mg. I find this strange since the balloon is not moving.
I can't follow your reasoning. Which density are you using in ρVg=Fb, and what exactly is Fb?
There are forces all over the place, but you can group them into:
- gravity on the balloon contents
- net force from surrounding air
- suspended load
It's a statics question: what temperature will generate enough lift to balance the other forces. What do you find strange?
 

What is a hot air balloon?

A hot air balloon is a type of aircraft that uses heated air to create lift and fly. It consists of a large fabric envelope, called the envelope or the balloon, with a wicker basket attached to the bottom.

How does a hot air balloon work?

A hot air balloon works on the principle of buoyancy, which states that an object will float if it is less dense than the air around it. In a hot air balloon, the air inside the envelope is heated using a burner, making it less dense than the surrounding air and causing the balloon to rise.

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

Buoyancy is the force that allows a hot air balloon to float in the air. It is created when the hot air inside the balloon is less dense than the cooler air outside. This difference in density causes the balloon to rise and stay afloat.

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

The temperature inside a hot air balloon is controlled by using a burner, which is a device that burns liquid propane gas to heat the air. The pilot can control the amount of heat released by adjusting the burner, which allows for precise control of the balloon's altitude.

What are the safety precautions for hot air balloon flights?

Hot air balloon flights are generally considered safe, but there are some precautions that should be taken to ensure a safe flight. These include checking weather conditions, performing regular maintenance on the balloon, having a trained pilot, and following safety protocols during takeoff and landing.

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