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Thermodynamics, isentropic and reversible processes

  1. Mar 16, 2008 #1
    Hi, I have some questions regarding thermo:
    1) Does an isentropic process have to be a quasistatic one? What is the relation between equilibrium and reversibilty?

    2) Naturally, there are revesible processes involving control volumes. A gas can go through an isentropic process in a reversible turbine, but as we know, turbines in a steady state have siginficant gradient of pressure inside them- the properties of the gas inside the turbine are not uniform.
    How can the gas go through a reversible process in the turbine if it is not in equilibrium? (the gas in the turbine is not in equilibrium because it's properties are not unfiorm)

    Also, it is said that the gas goes through an isentropic prcess, and let's say it goes through a turbine, what system is regarded? Is it the gas inside the control volume (always looking at the gas in the control volume- open system) ? is it a closed system that goes through the turbine, exits it and than moves on (following the gas- closed system)?

    For example: if you have an ideal gas going through a quasistatic adiabatic process, than the process can be described by the equation: (pV)^k=const
    if you have an ideal gas going through a revesible adiabatic process (isentropic)- the process can be described by the equation: (pV)^k=const
    The same equation. So does this mean any isentropic process is quasistatic? If so how can the process in the turbine can be quasistatic if the properties of the gas in the trubine are not uniform. If not any isentropic process is quasistatic ,than how can both be equations be the same?

    Generaly, how can a process in a control volume be reversible and adiabatic, if the properties of the gas are not uniform?

    I must say I am very confused about this, and it's not the first time I took this course.
    Thanks, Erez.
     
  2. jcsd
  3. Mar 16, 2008 #2
    1) Not really.

    Any real process is not isentropic. When modeling thermodynamic systems it is often convenient to use the assumption of an isentropic process as a 'first-cut' model. This is similar to ignoring friction in a mechanical model. This works well in combination with the assumption that a process is quasi-static because increases in entropy can be considered small in some situations.

    2) Being uncomfortable with the idea of an 'isentropic' turbine is well founded and that shows some insight that you have realized that, because these devices are just not isentropic. However, it is often a good place to start and is certainly a good place to start when learning the material. This is very similar to physics I when you learn about a block sliding down a frictionless inclined surface. You know perfectly well that there isn't such a thing as a frictionless surface but it is a good place to start, the same applies here.

    2a) The equation you have displayed is for a polytropic process of a gas and can actually represent a lot of different situations depending on the value of 'k'. The exponent can be one value if one is modeling the situation as isentropic (e.g. 1.4 for air). Another situation is an isothermal process, where the value of k=1.

    Just keep in mind that these are just mathematical models which represent what is actually going on and in a lot of situations it is much more complicated.
     
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