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I am having a few questions about 'flow work' and the First law of thermodynamics when applying to a closed non-stationary system.

I know for a non-stationary close system, say an element(Lagrangian) of fluid, the First law can be written as

$Q=U+W+KE+PE$

i)If the kinetic energy ($KE$) and the potential energy ($PE$) of this element of fluid are negligible then

$Q=U+W$

If we consider the element to undergo an adiabatic process then

$U=-W$

Usually, the work W here is taken to be the $PdV$ work (boundary work). But what if the element is incompressible? Then can we substitute the $Vdp$ work here? i.e.

$U = -VdP$

ii) I have read multiple posts that state that Vdp is the flow work. But for a control volume through which mass flow is taking place, we usually call $PV$ the flow-work. What is the difference between the $PV$ flow work and the $VdP$ flow work and why do they have the same name?

Ref: How to apply first law of thermodynamics closed non-stationary incompressible element of fluid passing a nozzle?

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  • $\begingroup$ vdP is the shaft work per unit mass passing through the control volume, d(Pv) is the net amount of work required for force fluid into and out of the control volume per unit mass; the latter is usually lumped together with the internal energy change per unit mass. to give the enthalpy change per unit mass. You need to review the derivation of the open system version of the 1st law of thermodynamics to understand the distinction between total work and shaft work. $\endgroup$ Commented Jul 12, 2022 at 1:42

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Usually, the work W here is taken to be the $PdV$ work (boundary work). But what if the element is incompressible? Then can we substitute the $Vdp$ work here? i.e.

$U = -VdP$

No.

$VdP$ is differential flow work. Flow work is associated with pushing fluid across the boundaries into and out of an open system. It does not apply to a closed system where, by definition, mass does not cross the boundary. For an adiabatic closed system consisting of an incompressible fluid, boundary work is zero and therefore the change in internal energy is zero.

What is the difference between the $PV$ flow work and the $VdP$ flow work and why do they have the same name?

$Pv$ work refers to the flow work at the inlet or outlet of the open system control volume. The net $Pv$ work in moving the fluid both into and out of the control volume is $\Delta (Pv)$ or $P_{e}v_{e}-P_{i}v_{i}$ where $i$ and $e$ are the inlet and exit pressure and specific volume.

In general, the net reversible flow work, in the absence of changes in kinetic and potential energy, is given by

$$w_{rev}=-\int_i^e vdP$$

For an incompressible fluid ($v$ = constant)

$$w_{rev}=-v(P_{e}-P_i)$$

For reversible adiabatic flow the net flow work equals the net output work of the system, typically shaft work.

Regarding your follow up question, yes you can get the same result tracking an element of compressible fluid and treating it as a closed system. But in that case $du=-Pdv$, where $Pdv$ is the boundary work on the element. The reversible adiabatic turbine work $dw_{T}$ is based on the change in enthalpy $dh$ of the element as it passes through the control volume. In general $dh=Tds +vdP$. For an adiabatic reversible turbine, $ds=0$ and therefore $dw_{T}=dh=vdP$.

Hope this helps.

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  • $\begingroup$ Thank you! I am trying to understand the energy equation by tracking an element of fluid (closed system) in a flow, say through a turbine. When every flow is involved the Open system/control volume approach is applied. But the same result should be obtained by following the energy interaction with a 'closed' element of fluid too. How do i write the energy equation for an incompressible element of fluid passing through a turbine? I know the element loses (internal) energy and also the (static)pressure drops (ignoring KE and PE changes)? $\endgroup$
    – GRANZER
    Commented Jul 12, 2022 at 4:21
  • $\begingroup$ I know that the change in internal energy will be equal to the work done dU=W. $\endgroup$
    – GRANZER
    Commented Jul 12, 2022 at 4:34
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    $\begingroup$ @GRANZER You can get the same result by tracking an element of fluid and treating it as a closed system. But $W$ is boundary work on the element, not turbine work produced by the element. I will update my answer to elaborate. $\endgroup$
    – Bob D
    Commented Jul 12, 2022 at 13:34
  • $\begingroup$ Thank you, that will be very helpful! $\endgroup$
    – GRANZER
    Commented Jul 12, 2022 at 13:37
  • $\begingroup$ @GRANZER See my update. Hope it helps. $\endgroup$
    – Bob D
    Commented Jul 12, 2022 at 13:49

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