Physics, Math

A typical volume of a modern hot air balloon is 2500 cubic metres, and a typical maximum temperature of the hot air is 120 degrees Celsius. Given these figures, and an outside air temperature and density of 15 ∘C and 1.225 kg/m3 respectively, compute the maximum mass (in kilograms) of the balloon, basket and payload.

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  1. Archimedes says:
    volume of fluid displaced * density of fluid * g = upward buoyant force = weight is neutrally buoyant
    mass = weight/g = volume*density

    2500 * 1.225 = 3063 kg up

    now subtract the mass of air in the balloon
    2500 * rho
    what is rho, the density of air inside at 120 C ?
    approximate as perfect gas ?
    P V = n R T
    P is close to the same in and out or balloon would burst
    V the same
    so
    n T is about constant
    density proportional to mass in the balloon volume
    n hot T hot = n cold Tcold
    n hot /n cold = Tcold/Thot
    Tcold = 273 + 15 = 288
    T hot = 273 + 120 = 393
    so density hot/density cold = 288/393 = .733
    .733 * 1.225 = .8977 kg/m^3
    so mass of air in balloon = .8977*2500 = 2244 kg
    then finally
    mass of other stuff = mass of cold air displaced - mass of hot air = 3063 -2244 = 819 kg

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  2. The only parameters we need to convert are the temperature difference (ΔT=120−15=105) and the outside air temperature in Kelvin (T = 288.15 K).

    Now the lift of this hot air balloon is found using:

    L=ρVg(ΔTT+ΔT)
    L=1.225⋅2500⋅9.81(105288.15+105)=mg
    m⋅9.81=8023.727
    m=818kg

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  3. 818.225

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