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Hot Air Balloon and Buoyancy

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data

    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.

    2. Relevant equations

    Fb=ρgV

    3. 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?
     
  2. jcsd
  3. Feb 14, 2013 #2
    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?
     
  4. Feb 14, 2013 #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.
     
  5. Feb 14, 2013 #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.
     
  6. Apr 5, 2013 #5


    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?
     
  7. Apr 5, 2013 #6

    haruspex

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    Your answer looks correct. It could be unreasonably high because the given conditions (volume, load) are unreasonable.
     
  8. Apr 6, 2013 #7
    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: Apr 6, 2013
  9. Apr 6, 2013 #8

    haruspex

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    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 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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