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

AI Thread Summary
To determine the temperature required for a balloon with a volume of 500 m³ to lift a 300 kg mass, the buoyant force must exceed the gravitational force. The calculations involve the difference in air densities inside and outside the balloon, using the equation (ρair - ρballoon)Vg = mg. The user initially calculated a temperature of 612 K, which seemed excessively high, prompting discussions about potential errors in understanding buoyancy and the role of air density. Clarifications were sought regarding the forces acting on the balloon, emphasizing the need to balance the buoyant force with the weight of the load and the balloon's contents. Ultimately, the conversation highlighted the complexity of buoyancy calculations and the importance of accurately applying the relevant equations.
erok81
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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...

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

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

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

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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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?
 
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.
 
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.
 
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?
 
Your answer looks correct. It could be unreasonably high because the given conditions (volume, load) are unreasonable.
 
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:
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?
 
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