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Relationship between work, internal energy and enthalpy.

  1. Nov 28, 2011 #1
    Hi,
    I'm taking a thermodynamics class and I'm stuck on how work relates to enthalpy and internal energy.
    Does work done by a system equal change in internal energy, change in enthalpy or does it depend on the situation?

    The question I'm stuck on asks for the work produced by a steam turbine and gives initial and final properties for the steam. I feel like the work produced by the turbine is just equal to the ΔU for the steam, but I could also see an argument for using ΔH. Any thoughts would be appreciated.
     
  2. jcsd
  3. Nov 28, 2011 #2
    Mass flowing across a control surface brings energy (neglecting potential and kinetic energy changes):

    u + P [itex]\nu[/itex]

    where,

    u = fluid specific internal energy
    P = fluid static pressure
    [itex]\nu[/itex] = fluid specific volume

    But this is the definition of enthalpy:

    h = u + P [itex]\nu[/itex]

    So enthalpy includes the internal energy and the "flow work" term P [itex]\nu[/itex].

    So for your turbine you will use Δh as you mentioned.
     
  4. Nov 28, 2011 #3

    I like Serena

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    The work done by a system depends on the situation.
    The general formulas are:
    dU=dQ+dW
    dU=TdS-PdV
    dH=TdS+VdP

    I think you would use ΔH=ΔQ for a quasi-static, constant-pressure process.
    And you would use ΔU=-ΔW for an adiabatic process.

    I believe both processes apply in a steam turbine cycle.
     
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