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Thermodynamics and liftforce

  1. Apr 23, 2014 #1
    I'm doing some physics questions from an old test but struggle with a few questions.

    Problem:

    An air balloon with the volume 5000m^3 should be able to lift a total mass of 1000kg, including the cargo and everything. The surrounding air keeps a constant temperature of 12C and has the density of 1.2kg/m^3. How hot does the air in the balloon need to be in order to lift this?

    I've tried using some different gas law formulas, in particular pV = nRT but to no success, always seems like I'm missing some information.

    All help and thoughts is much appreciated!
     
  2. jcsd
  3. Apr 23, 2014 #2
    This is a buoyancy problem. Do you remember Archimedes Principle?

    Chet
     
  4. Apr 23, 2014 #3
    Any object, wholly or partially immersed in a fluid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
     
  5. Apr 23, 2014 #4
    Excellent. So now you know how to do the problem, correct?

    Chet
     
  6. Apr 23, 2014 #5
    Well,we have 5000m^3 of air displaced, but I can't know the weight of this since I don't have the density nor temperature.
     
  7. Apr 23, 2014 #6

    SteamKing

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    Perhaps if you would re-read the OP very carefully, you would find the necessary info. Hint: the balloon rises because the air inside (not outside) is heated.
     
  8. Apr 23, 2014 #7
    You are trying to solve for the temperature, and you know that the pressure inside the balloon is going to be about equal to the pressure outside. You can use the ideal gas law to determine the number of moles of air inside the balloon as a function of the temperature. Once you know the number of moles of air inside the balloon, you know the mass of air inside the balloon, and the weight of the air inside the balloon (in terms of the temperature).

    Chet
     
  9. Apr 24, 2014 #8
    One of the "laws" that led to the ideal gas law will allow a scaling of volume of a given mass of a gas for temperature. That will allow you to use the density given at 12 C without worrying about ideal gas constants or moles.

    Don't forget to do calculations in Kelvin!
     
  10. Apr 24, 2014 #9
    I haven't done this in a very long time, I think I'm confusing the units. Here is an attempt using both of your information.

    If we use the same density given at 12 C, we get that the density for air is about [tex]\rho = 1.2kg/m^3[/tex] hence [tex] m = \rho V = 1.2kg/m^3 \cdot 5000m^3 = 6000 kg.[/tex]

    Now calculating for the substance mass n we obtain [tex] n = \frac{m}{M} = \frac{6000000g}{28.8 \ \text{moles}/g} \approx 2.083\cdot 10^{5} \ \ \text{moles}.[/tex]

    So solving for T we get (using standard atmospheric pressure at sea level)

    [tex]T = \frac{pV}{nR} = \frac{1.013\cdot 10^5 \cdot 5000}{2.083\cdot 10^{5}\cdot 8.314} = 292.47 \ K = 19.47 \ C.[/tex]

    So how much do I need to increase the temperature so that the balloon can lift what it's supposed to?

    EDIT: On a completely unrelated note, is there a shortcut to write the latex tags [tex][/tex] and how do I use this to have latex formating "in line" and not having it auto displayed?
     
    Last edited: Apr 24, 2014
  11. Apr 24, 2014 #10
    I don't follow what you are doing. The mass you are calculating is based on a density of 1.2 kg/m^3. We know the density has to be lower.

    I suggest that you check out Charle's Law. Because it applies to a fixed mass, volume is then inversely related to density. You already have a density at a specific temperature. This could give a relation between density and temperature.

    Try setting up this relationship and test it out with a couple of known air density / temperatures.
     
  12. Apr 24, 2014 #11
    Well, that was the value given in the test as a collection of constans at the bottom of the page. As a help to remember constants during the test. I've attached an image, it's in Swedish though.

    This assignment shouldn't be this complex, it doesn't give many points on the test.

    And for the record: this is no homework, I'm just practicing for an admittance test which I'm going to have 10th of may. I tend to understand the problem quite well once I see a complete solution or at least a start to it if anyone doesn't mind. It's not that I'm lazy but it's getting tedious to struggle with one problem for days and making no progress, I need to move on I have a lot of other concepts to grasp.
     

    Attached Files:

  13. Apr 24, 2014 #12
    From Charles Law

    V1/T1 = V2/T2

    Since V = m/density and Charles Law pertains to a fixed mass you can restate Charles Law in terms of density giving you a relation between density and temperature.

    Give this approach a try.
     
  14. Apr 29, 2014 #13
    I'm sorry I can't make sense of this. What is V1, V2, T1 and T2 in this instance? Say this

    T1 = x

    V1 = 5000m^3, which is the volume inside the balloon. But Since I don't have neither the density nor the mass of this, I cant get the temperature of it.

    How should I restate it in terms of density if I don't have the mass but only the volume?

    V2 = the outside air. But How do I know how much outside air it is? Entire globe? Or is it the mass of air displaced needed? which is 1000kg at density 1.2, giving me 833.3 m^3 displaced air at temperature 12 C.

    T2 = 12

    So by charles law we get

    [tex] \frac{5000}{T_1} = \frac{?}{285}.[/tex]

    Not really sure what to do with this.
     
  15. Apr 29, 2014 #14
    The volume of outside air displaced is 5000 m3, and its density is 1.2 kg/m3, so the mass of displaced air is 6000 kg. The volume of air inside the balloon is also 5000 m3. From the ideal gas law, the density of the air inside the balloon is ρ=pM/(RT), where p is the pressure (1 atm), M is the molecular weight of air, R is the gas constant, and T is the temperature inside the balloon. Since the pressures inside and outside the balloon are the same,

    [tex]\frac{ρ_{inside}}{ρ_{outside}}=\frac{T_{outside}}{T_{inside}}[/tex]

    where ρoutside=1.2 kg/m3, Toutside=(273+12)K. So, from this equation (and the fact that the volume of air in the balloon is 5000 m3), in terms of the temperature Tinside, what is the mass of air inside the balloon? So, from Archimedes Principle, in terms of Tinside, what is the net buoyant force on the balloon. This must be equal to the weight of the 1000 kg cargo.

    Chet
     
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