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Adiabatic or Isothermal? Compressed air energy storage

  1. Jan 31, 2016 #1
    Hi,

    I'm doing a high school physics project and am trying to figure out if a certain setup that I'm using is adiabatic or isothermal, in order to determine what equation I can use to calculate the work that my setup does-- the threads I've come across so far only explain the difference, but not how to tell whether a certain setup is one or the other.

    Setup: I have an ordinary empty water bottle, which has a small hole on its bottom made by a heated paperclip. This hole is small enough such that a thumb tack can fit tightly through it as a 'seal' of sorts. Additionally, the bottle has a valve on its top, through which compressed air can be sent and contained with very marginal loss over relatively short periods of time such as a few minutes or less (which is what I'm dealing with for my setup).

    I pump a certain pressure of compressed air into the bottle (say, 30 psi). I then attach it to a toy car and remove the thumb tack, and a stream of pressurized air is expelled, propelling the car. This experiment is performed for an ordinary water bottle (0.5 L), a 1 L bottle, and a 2 L bottle that typically carries pop.

    I understand the definitions of 'isothermal' and 'adiabatic', but I still don't understand how to tell which (if either) my project is? This is (unfortunately) far above anything I have or likely will be covering in my high school physics class.

    Isothermal indicates that the temperature of the gas in the system is constant throughout.
    Adiabatic indicates that there is no heat transfer between the fluid (air, in this case), and the surroundings, which may or may not be isentropic if the process is internally reversible.

    If someone could clarify this for me, along with a citable source such as a university or formal research paper (Wikipedia doesn't count), that would be much appreciated. Thank you.
     
  2. jcsd
  3. Jan 31, 2016 #2
    Also! I may have to account for air humidity in my calculations, as I will have to calculate the density of the air as per the project's requirements. I found something here (see 'pressure dependence') assuming an isothermal process, but it seemed to apply to an air-water system, which my setup most likely is not. Does this apply, assuming the process is in fact isothermal? If not, are there other assumptions underlying this? I had trouble relations on air humidity based on pressure apart from that of the atmosphere or the aforementioned setup.

    Once again, if possible, a citable source would be much appreciated. Thank you.
     
  4. Jan 31, 2016 #3
    Which process are you interested in, (a) filling the bottle, or (b) bottle emptying while accelerating the car?

    Regarding correcting for humidity in determining the density: it is not significant, because there is so little water vapor in the air at room temperature.
     
  5. Jan 31, 2016 #4
    Definitely the bottle emptying; I'm trying to relate the loss of pressure (flow rate) to the overall acceleration of the car at specific time increments using F = ma, then compare a rough calculation of the 'theoretical' under ideal conditions to the experimental values that I actually obtain. This is obviously a bit more complicated by the fact that the forces acting on the car (force of thrust from the stream of gas and the force of friction) and the mass are both changing, due to the loss of pressurized air from the bottle. I recognize that my calculations won't be exactly the same as my experimental values obtained due to the force of friction etc. being difficult to calculate exactly, but my teacher specifically noted that the size of the % error doesn't matter as long as the process used to arrive at the values is sound and we demonstrate that we've learned how to apply our physics & laboratory concepts with regards to error analysis.

    About the density of the air: assuming correcting for humidity is insignificant, does this mean I can use a variation of ideal gas law to determine density? (I found the equation density = pressure/(R*temperatuere).) Additionally, do you have any idea where I might be able to find a citable academic source that states such a correction is insignificant?
     
  6. Jan 31, 2016 #5
    The most inexact part of your calculation is going to be that of estimating the discharge rate of gas through the hole. This is going to determine the force that the gas exerts on the bottle/car. You need to focus on that. You have an open hole discharging from a container in which there is a gas under pressure.
    Actually, the equation you gave is for the molar density. The actual mass density is equal to the molar mass of air (29 g) times the molar density. If you are worried about water vapor, look up the equilibrium vapor pressure of water at room temperature. If you multiply this by the relative humidity of the air (say 70%), you get the pressure of the water vapor in the air. You can compare this with atmospheric pressure to decide whether you think the water vapor is significant.
     
  7. Jan 31, 2016 #6
    Okay. Would emptying the bottle be isothermal or adiabatic? I'm uncertain as to whether the above principles apply (i.e. temperature remains constant throughout process, or no heat is lost during process). Depending on which process it is the equations that I can use also differ, it seems.

    Does this mean that the density of the air in the bottle is changing as the air is released due to changing pressure? Is this going to be a problem in calculating values for force of thrust, force of friction, mass, etc., some of which are dependent on the air's density or pressure, and will I have to factor this change into any equations I used that depend on these values?
     
  8. Jan 31, 2016 #7
    I think it is going to be closer to adiabatic than isothermal, because the time interval is going to be pretty short, so not much heat can enter. But, do it both ways, and see how the results compare.
    Sure.
    It will have to be taken into account. Whether this is a problem or not depends on your perspective.
     
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