1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Thermodynamics - Hot-air balloon

  1. Feb 21, 2014 #1
    1. The problem statement, all variables and given/known data
    The mass of a hot-air balloon and its cargo (not including the air inside) is 200kg. The air outside is at 10C and 101kPa. The volume of the balloon is 400m^3. To what temperature must the air in the balloon be warmed before the balloon will lift off? (The density of air a 10C is 1.25kg/m^3).


    2. Relevant equations
    PV=nRT
    and maybe some other basic thermodynamics.

    3. The attempt at a solution

    [itex]\sum F = _{}\rho_{inside}Vg - _{}\rho_{outside}Vg - mg = 0[/itex]
    Simplifies to
    [itex]m= V(_{}\rho_{inside} - {}\rho_{outside})[/itex]

    Assuming the volume is the same.... But I can't really wrap my head around it. Would the volume be the same? I have no idea what to do next and I'm not even sure that I'm doing it right.
     
  2. jcsd
  3. Feb 22, 2014 #2

    ehild

    User Avatar
    Homework Helper
    Gold Member

    The volume of the balloon is given, 400m3.

    You have to sum all forces acting on the balloon and its cargo, as you did, but check the signs. Which force acts downward and which one acts upward?


    ehild
     
  4. Feb 22, 2014 #3

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    The condition for lift is net upward force > 0, which means:

    m < (ρoutside - ρinside)V

    You cannot change m (the mass of the balloon itself), V or ρoutside or P. So how do you lower ρinside? (hint: express PV= nRT in terms of density).

    AM
     
    Last edited: Feb 22, 2014
  5. Feb 22, 2014 #4
    I still don't understand how the volume can be the same on the inside and the outside... Wouldn't the volume of the gas outside be all of the air above the balloon?

    P(m/ρ)=nRT....?
     
  6. Feb 22, 2014 #5

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    V is the volume of air displaced by the balloon i.e. the volume of the balloon (you are to assume that the volume of the balloon apparatus apart from the balloon itself has negligible volume). Buoyancy requires the balloon displacing a mass of air that is greater than its mass, which is comprised of the mass of the balloon apparatus and the air inside the balloon.

    AM
     
  7. Feb 23, 2014 #6
    Well wouldn't that only make the initial volume of the balloon Vi=400
    Then using

    PiVi=PfVf I could find some final volume? Maybe?
     
  8. Feb 23, 2014 #7

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    Assume that the balloon does not expand so the volume is constant. You would you reduce the mass of the inflated balloon? (hint: you can't reduce the mass of the balloon apparatus).

    Am
     
  9. Feb 23, 2014 #8

    ehild

    User Avatar
    Homework Helper
    Gold Member

    You wrote the equation for an isothermal process.

    Have you seen a hot-air balloon launching? (watch from 15 minutes. )

    The gas in the balloon is heated by burning gas. The balloon is open, so the pressure is the same inside and outside. The air expands while heated and some of it leaves the balloon. But the volume of the balloon does not change. The mass and the density of air inside the balloon changes with the temperature - how?

    ehild
     
    Last edited by a moderator: Sep 25, 2014
  10. Feb 23, 2014 #9
    If the balloon doesn't expand... I guess heating the gas is the only option but that only means that the pressure inside increases. But if the volume can't expand I don't see how that is of any significance to the problem.

    ehild, I have no idea... It doesn't many any intuitive sense to me at all

    ρ=m/V

    PV=nRT

    ρ=mnRT/P
     
  11. Feb 23, 2014 #10

    Andrew Mason

    User Avatar
    Science Advisor
    Homework Helper

    You are trying to reduce the mass of the air inside the balloon!!! What has to happen if the pressure and volume cannot increase and the temperature increases? Hint: the balloon is open at the bottom...

    AM
     
  12. Feb 24, 2014 #11
    I guess some of the air molecules have to escape from the bottom then.
     
  13. Feb 24, 2014 #12

    ehild

    User Avatar
    Homework Helper
    Gold Member

    Very good... If some molecules escape, less stay inside the balloon. T increases, P and V are constant, so what else changes in the equation PV=nRT?

    ehild
     
  14. Feb 25, 2014 #13
    n decreases.
     
  15. Feb 25, 2014 #14

    ehild

    User Avatar
    Homework Helper
    Gold Member

    n is the number of moles in the balloon. If n decreases the mass of air inside the balloon also decreases. How much should be the mass of air inside the balloon, so that the weight of the air in the balloon + the weight of the balloon and cargo is equal or less than the buoyant force?

    ehild
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Thermodynamics - Hot-air balloon
Loading...