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Steam Turbine enthalpy drops

  1. Nov 4, 2009 #1
    Can someone explain something to me:

    I believe when throttling h1=h2.
    When calculating enthalpy drop across a FIXED converging nozzle then kinetic energy is caluclated using the following formula:
    h1-h2 = v2^2/2 - v1^2/2
    If the nozzle is converging then there will be an increase in kinetic energy, but, if assuming the nozzle is throttling with a respective pressure drop then how can h1=h2 as this would mean v1=v2?

    Thanks
     
  2. jcsd
  3. Nov 6, 2009 #2

    stewartcs

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    The typical assumption for a throttling device is that the kinetic and potential energies are neglected. Hence the enthalpy is constant.

    CS
     
  4. Nov 6, 2009 #3

    Q_Goest

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    Hi leon
    Note that this equation can be rewritten:
    h1 + v1^2/2 = h2 + v2^2/2
    This is the first law of thermo with the velocity added in to account for kinetic energy. It says that the total energy prior to expansion is equal to the total energy after expansion. This obviously assumes no heat or work was done on or by the fluid.

    As stewart mentions, kinetic energy is normally very small and can be neglected.
     
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