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Buoyancy / air pressure

  1. Dec 19, 2006 #1
    I've been wondering a couple things:
    1. How much energy would it take to evacuate* a 1 cubic meter volume at sea level and some convenient temperature?
    2. If the structure containing the vacuum weighed, say, 1kg, how high up would it go?

    *arbitrarily close to a vacuum to avoid getting into quantum mechanical issues

    Using ideal assumptions is fine with me (e.g. no temperature gradients caused by heating from the ground, no wind, etc).

    As far as I can tell, the answer to the first question would be about 100 kilojoules. I came up with that by assuming the volume was like an accordion 1m^2, and had to be stretched to a length of 1m from an initial length of 0. Since the motion is resisted by the 100 kilopascal atmospheric pressure, and the distance is 1 meter, I get 100 kilojoules.

    I don't know what the answer to #2 is, but I would expect the answer to be less than 10km. I'll probably have follow-up questions after getting answers to these two.
  2. jcsd
  3. Dec 20, 2006 #2


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    At 1atm and 25oC, the mass of one cm of normal air is about 1.17 kg. So the object described would have only 170 grams of lift at sea level. So I would say you're right about "<10km".

    In fact, if you look at http://en.wikipedia.org/wiki/Image:Atmosphere_model.png" [Broken], it appears that atmospheric density drops to 1kg/m3 at around 2 km altitude. At this point, the lift being generated equals the wegith of the device, and climb stops.
    Last edited by a moderator: May 2, 2017
  4. Dec 20, 2006 #3


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    It depends on your assumptions.

    If you assume the volume is evacuated adiabatically (ie: no heat transfer) then the first law reduces to:
    dU = 0 = PdV - W

    Here, I'm assuming dU = 0 because there is initially no air inside your volume (volume is zero to begin with) and that never changes. Note that the control volume for the above is one which never has any air in it, as if it were "an accordian".

    The result is exactly what you calculated:

    Yes, I agree that if you do this adiabatically, it's 100 kj. This calculation can be done without integrating in a single step. However, it doesn't take into account the possibility of heat transfer. That makes this a bit more complicated.

    With heat transfer, the first law reduces to:
    dU = Q - H

    Note that now I'm assuming a constant volume which surrounds this one cubic meter volume instead of changing shape as before. I'm also assuming that energy out (enthalpy) is negative and removes energy from the volume, and energy in (heat) is positive, and that adds energy to this volume. To do this correctly, you really need to integrate the entire process. This can be done if you have the equations for enthalpy and want to do the calculus. However, if you're like me, you already have a computer program you created years ago that does this analysis iteratively.

    Interestingly, the heat transfer factor is huge. Overwhelming even. Deepthroat might swoon.

    If there was no heat transfer, the air inside the chamber would get extremely cold. I calculate that somewhere around 0.1 psia, the air left in the chamber will liquify. In fact, before the vessel is completely evatuated, it may even solidify. Unfortunately, my database doesn't do solids. <sigh>

    Regardless, adiabatic evacuation is unreasonable, since the vessel has mass and therefore the specific heat of that vessel will come into play even with perfect insulation. The result will probably be an almost isothermal expansion.

    To do this right, you need to do the integration. For an isothermal process (one in which heat transfer maintains a constant gas temperature), I come up with roughly 1.6 kj, which is a far cry from the 100 kj an adiabatic process will yield. Note that the remainder (100 - 1.6 = 98.4 kJ) is heat that was added.

    Regarding question #2, I think Lurch gave you a decent answer there.
  5. Dec 20, 2006 #4


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    Yeah, it sounds like you've stumbled across the one substance more bouyant than hydrogen: vacuum.

    If you can fill a balloon with vacuum, you have a mighty stong lifting force, AND it's not flammable.

    Trouble is, how do you keep it rigid AND extremely light? Your 1kg container will implode upon the slightest imperfection or mistreatment.

    Me, I'm envisioning fullerenes on a macro scale - beachball-sized - that are impermeable to air.
  6. Dec 21, 2006 #5
    I don't understand the heat-transfer part of your answer, Q_Goest. Are you saying that in reality, it would take very little energy to create the vacuum?
  7. Dec 22, 2006 #6


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    Hi CTho. There are various ways to create this vacuum. The one you chose to model is to create a moving wall. You used the analogy of an accordion. You could equally use the analogy of a piston in a cylinder. However, the model you use will obviously determine how you do the calculation and the analysis will reflect the model, not reality. If the model you use doesn't properly represent the actual system, it probably won't represent reality either.

    If a 1 cubic meter volume is evacuated using a vacuum pump, the model of a moving wall such as an accordion or piston in a cylinder may be incorrect, depending on what other assumptions you make. To do this properly, you need to apply the first law of thermodynamics. In the case of a vacuum pump model, we draw a control volume around the interior of the vessel and apply the first law which reduces to:

    dU = Q - H

    Where Q = heat going into the gas
    H = enthalpy of the gas coming out of the vessel.

    If we make the assumption that the walls are perfectly insulated and have no thermal mass, we can also delete the heat, Q. We are left with dU = - H

    In this case, the vessel gets very cold, because the energy out (enthalpy) contains this PdV energy. This case actually matches the accordion model you have.

    If instead we say there is heat going into the vessel during the evacuation process, we need to determine how much heat. The entire process can be calculated iteratively. We can take a small amount of mass out that has some enthalpy, calculate the difference in internal energy to determine the new internal energy, then using the new internal energy, calculate the new temperature. With that we can determine how much heat is needed to maintain a constant temperature. We can then do this again and again until all the air is removed. This makes for an easy computer program.

    That's the thermodynamics of the tank itself. The result is a pressure/temperature curve over time for the tank. It also gives us the enthalpy out over time. However, that's not what you asked for. You wanted to determine how much total energy was needed to do this. That means we need another analysis of the vacuum pump. (By the way, the program I have only had the vessel analysis, so I added the next bit in to do the analysis on the vacuum pump. More about that in a minute.)

    Next, draw a control volume around your vacuum pump. You have mass and some enthalpy going in, some mass with some enthalpy leaving, and work entering. You can also have heat entering or exiting this control volume. The first law then reduces to:

    dU = Hin - Hout + Win + Q

    Let's forget about Q, and assume the compression process (vacuum pump) is isentropic. We also note that dU = 0 since we're not storing any gas inside the vacuum pump. What we're left with is:
    Win = Hout - Hin

    In order to calculate the Hout, we need to determine the entropy of the gas going in. The entropy of the gas going into the vacuum pump is a state function, so it's an easy look up for a computer program. With that, it's easy enough to look up the state of the gas on the outlet to determine discharge enthalpy, because we have the entropy and pressure (vacuum pump discharges to atmosphere, so pressure is always 14.7 psia). Remember, you need 2 properties of the fluid to determine the state, in this case they are pressure and entropy. With enthalpy in and out, we can now calculate work in for each iteration. The total amount of work is then just the sum of the work for each iteration.

    Note also that we could assume there's some isentropic efficiency to the pump and get a slightly different Hout, and we could also assume there's some heat transfer which would affect Hout and the total work. It's actually not that hard to consider these factors if you have a computer program and a thermodynamic database and it's not unusual to do this kind of analysis in industry.

    Getting back to the question about total work, after writing this up, I realized the section I added (so I could respond to your post) with the vacuum pump work, I'd forgotten to do the entropy calculation, and had instead a constant Hout by assuming 14.7 psia and 70 F. <oops!> Ok, I'll admit it seemed a bit odd that it would take so little work compared to the adiabatic case. I should have known better! Anyway, once I put that in, of course the lower the tank pressure, the higher the discharge temperature on the pump since the heat added to the vessel affects the vacuum pump. The end result is a slight increase in total work for an isothermal case when compared to the adiabatic case. I'm now getting 135 kJ, which is probably a bit more realistic.

    The point of all this is that you need to model things as they actually occur, not just some way that makes it easy. Hope that explanation of the use of the first law was some help to you.
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