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Conditions are adiabatic and reversible about a turbine

  1. Nov 14, 2006 #1
    Why is it when the conditions are adiabatic and reversible about a turbine, the assumption is its isentropic?
     
  2. jcsd
  3. Nov 14, 2006 #2

    Astronuc

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  4. Nov 16, 2006 #3

    Andrew Mason

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    If dQ = 0, then dS = dQ/T = 0.

    This would seem to imply that all adiabatic processes are isentropic (constant entropy - ie dS = 0) which is not true. The relation: dS = dQ/T assumes a quasi static process in which the system is always at equilibrium. If the process is quasi-static and adiabatic, the process is isentropic.

    I don't see how this would apply to a turbine, however. The expanding gas is necessarily dynamic (in order to drive the turbine), not quasi-static/reversible.

    AM
     
  5. Nov 17, 2006 #4
    you are missing a few terms in your entropy equation. You can't simply assume that dS=dQ/T.
     
  6. Nov 18, 2006 #5

    Andrew Mason

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    I am not assuming that dS = dQ/T. That is the thermodynamic definition of dS.

    AM
     
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