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Hot air balloon releasing sand

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    A hot air balloon with a total mass of 7000 kg carries a gondola (or basket) which includes 1000 kg of sand bags fixed to the outside. While hovering at a fixed altitude, sand starts leaking out of thebags at a rate of 50 kgs-1. Determine the velocity of the balloon at the moment when all the sand has leaked out. You may assume that the acceleration due to gravity and the upward force from the hot air remains constant. You may also neglect air resistance.

    2. Relevant equations
    Going from first principles,
    Δp=p(t+dt)-p(t)
    F=dp/dt

    3. The attempt at a solution
    Let the balloon be moving upwards at speed v at time t with mass m. Then at time t+dt it moves upwards with speed v+dv, mass m-dm, whilst the 'sand packet' has mass dM and moves upwards still at speed v (because it is not propelled out, only dropped).

    Then dp(t)=mv and dp(t+dt)=(m-dm)(v+dv)+vdM
    dp=mdv-vdm+vdM to first order
    dp/dt=mdv/dt-vdm/dt+vdM/dt
    Then dM/dt=-dm/dt so dp/dt=mdv/dt-2vdm/dt. I believe this step is wrong. I can't see a flaw in my reasoning though.

    Then the next issue is deciding what dp/dt is in terms of the forces. I have an expression for dP/dt for the whole system so I guess I need the force that acts on the whole system. But I'm not sure what to do... Thanks for any help :biggrin:
     
  2. jcsd
  3. Apr 29, 2014 #2

    mfb

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    It is not a rocket - the leaking sand won't give any velocity change. The (broken) force balance between hot air and the sand is the important point here.
     
  4. Apr 29, 2014 #3
    Hmm I'm lost.

    I have let the velocity of the balloon be v+dv at time t+dt because it must be changing... I don't really know where to go.
     
  5. Apr 29, 2014 #4

    haruspex

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    I think you mean
    p(t)=mv and p(t+dt)=(m-dm)(v+dv)+vdM
    No, you did -dm and +dM, so in fact dm and dM are the same.
    No, your equation did not include sand that had previously leaked out.
     
  6. Apr 30, 2014 #5
    Yes I did mean that!

    Ah I realise now my system isn't the whole mass is it, it's the mass of the balloon plus the differential sand mass released at any time. Thus this system experiences a force F-mg with F the thrust which is constant. I'm still a little bit unsure about the fact that this thrust force F acts on the m-dm rather than the m, whilst my system is the whole m

    Thanks for your help - I think If I can get things to be consistent with the rocket equation derivation that will be the best way of understanding everything.
     
    Last edited: Apr 30, 2014
  7. Apr 30, 2014 #6

    haruspex

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    In the equation F-mg = a(m-dm), the dm is insignificant.
     
  8. Apr 30, 2014 #7
    I have an answer of
    v=(F/a)ln[m0/(m0-aT)]-gT
    where m0 is the initial mass, a the rate of decrease of the sand on board, T the time for the sand supply to be exhausted, F the thrust force which is constant.

    F=7000g from the initial condition, a=50kgs-1 T=20s and m0=7000kg. This gives v=15.5ms-1. Not sure if this is correct.
     
  9. Apr 30, 2014 #8

    haruspex

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    I concur with all that.
     
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