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Using Enthalpy to define flow work for a non-ideal gas

  1. Sep 26, 2011 #1
    For an open system with mass flowing in and out, where mass exits the system there is work associated with the mass transport. When this work is used to say, push up a piston, this is pretty intuitive, but even when it does not lift a weight since, the force COULD be used to lift a weight, it is thermodynamic work. I'm trying to understand how we quantify that work. Since it is boundary work we say that

    Work = the integral of P dV from V1 to V2

    And then we have Enthalpy = U + PV, and from what I read the PV term represents the flow work. I don't understand how this is, since the integral of P dV from V1 to V2 is different depending on the relationship between P and V. If you look at work as the area under the P/V curve, you see how you need to more than just the end states to know the integral.

    Am I wrong to use Enthalpy to calculate the flow work in this way? Or is there something about it that I don't get, that makes it okay to do so?
  2. jcsd
  3. Sep 26, 2011 #2
    For example, if PV^k=C, we write P in terms of V as P=CV^(-k), and the integral of that with respect to V is (-C/k-1)V^(k-1), so from point a to point b the work done is

    (-C/k-1)(V2k-1 - V1k-1)

    but the the change in flow work is just

    P2V2 - P1V1

    I am confused why they are not equal.
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