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Thermodynamics Energy balance equation for Non - Steady Flow

  1. Sep 24, 2016 #1

    SWJ

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    Hi,
    I am currently studying Thermodynamics and stumbled upon this equation and is slightly confused as to how this works. Hopefully someone can help me with the understanding.

    thermo.png

    According to this text here, if I am not wrong e can be h + ke + pe or u + ke + pe depending on where i am looking the system at. So does that mean the final equation can be equal to (m1u1 - m2u2) or (m1h1 - m2h2) depending on where i am looking the system at? Or i should just ask when would it be mh rather than mu?

    Thank you very much.
     
  2. jcsd
  3. Sep 25, 2016 #2
    Most developments of the open system (control volume) version of the first law of thermodynamics separate the work into two parts:

    1. The work needed to force material (in the inlet and exit streams) into and out of the control volume

    2. All other work (which is usually referred to as "shaft work.")

    In these typical developments, the work in the equations is the shaft work, which is usually signified by a subscript s. The work needed to force the material into and out of the control volume is usually included in the energy term, and the term would then involve enthalpy h rather than internal energy u.

    In the particularly development that you have cited, this separation of the work into two parts is not done. So the work in your equations includes the work needed to force material into and out of the control volume. As a result, the energy term involves internal energy u, and not enthalpy h. This approach is very unconventional.
     
  4. Sep 25, 2016 #3

    SWJ

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    Ah I see what u meant. Then are these "Shaft Work" included when say the flow is turning a turbine before exiting?
     
  5. Sep 25, 2016 #4
    Yes. Shaft work includes that.
     
  6. Sep 25, 2016 #5

    SWJ

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    Alright! Thank you very much for the clarification now I have cleared up the confusion I had. :biggrin:
     
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