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Thermodynamics : Open and close control systems

  1. Feb 26, 2014 #1
    Hi all,

    Q: How do you find out the boundary work using control volume analysis for a tire that is punctured all of a sudden?

    To proceed,

    I started off by using the tire as a control volume with one outlet.

    Using first Law,

    E(in) - E(out) = dE/dt

    Final Conditions: P(final) = Patm. , T (final) = T(atm)

    Can someone help me out with this?


    Thanks,
    JT
     
  2. jcsd
  3. May 4, 2015 #2
    I'm not 100% sure on this, but I think it's not isometric (the tire decreases in volume), it's not isothermal (though it could be depending on the size of the puncture and the tire it takes to relieve the pressure). It's not isobaric since the pressure is changing. So, I would say it's going to be a polytropic or adiabatic process, assuming a somewhat large hole the the tire loses full pressure over the course of a few seconds.

    So, rather than going through the energy balance in this case (there's probably a way to use that, it just isn't coming to me at the moment),and since it just asks for boundary work anyway, I would assume adiabatic if they didn't give you a coefficient for the polytropic process. n is some number greater than 1, for air close to room temperature use 1.4 typically. R is the gas constant specific to that gas (not the universal), 287 J/kg K for air.

    [tex]W_{boundary} = \frac{P_2 V_2 - P_1 V_1}{n-1} = \frac{R(m_2 T_2 - m_1 T_1)}{n-1}[/tex]
     
  4. May 4, 2015 #3

    Randy Beikmann

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    Gold Member

    Is there more information to your question? For example, if the tire were supporting a vehicle, and the vehicle drops as the air leaks out, you find the work done from the force/displacement curve. To produce work, the boundary must apply a force to something, and also move.
     
  5. May 5, 2015 #4
    This sounds like a homework problem. Is it?

    Since mass is leaving the tire, you need to use the open system version of the first law. I think they expect you to assume that the volume of the tire doesn't change during the part of the process that you are examining. So no "shaft work" is being done. They also expect you to assume that the process is adiabatic.

    There is no enthalpy entering, but there is enthalpy leaving in the exit stream. There rate of mass flow out time the enthalpy per unit mass of the exit stream is equal to the rate of internal energy change within the tire.

    Chet
     
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