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-   -   How to calculate path-dependent thermodynamic integrals? (https://www.cfd-online.com/Forums/main/235537-how-calculate-path-dependent-thermodynamic-integrals.html)

Procyon April 19, 2021 04:19
How to calculate path-dependent thermodynamic integrals?
 
Hello everybody,


I would like to calculate path-dependent integrals like \int_1^2 v \, \mathrm{d}p (the "specific amount of useful work") using results of a CFD simulation. However, I am unsure about how to transform this integral into something that can be used in CFD post-processing (I am anlysing turbo compressors). This is what I derived so far:
  1. v \, \mathrm{d}p
    • is specific to one infinitesimal fluid parcel (mass element \mathrm{d}m
    • uses a Lagrange'ian view/description (follow the mass element)
    • does not contain any parametrisation of the infinitesimal changes, i.e. no definition of an infinitesimal change of pressure per time? distance travelled?
  2. Choose time t as the parametrisation of the infinitesimal changes, because time is "equal" in Eulerian and Lagrangian formulations and is an independent variable. Thus, v \, \frac{\mathrm{d}p}{\mathrm{d}t} (the "specific amount of useful power")
  3. Use the material derivative to transform the time derivative into an Eulerian formulation:
    \frac{\mathrm{d}p}{\mathrm{d}t} = \frac{\partial p}{\partial t}+\mathbf{u}\cdot\mathrm{grad}\left(p\right)
  4. Multiply by \mathrm{d}m = \rho \, \mathrm{d}V to get an extensive quantity (specific volume v and density \rho cancel out) and assume \frac{\partial p}{\partial t} = 0:
    \mathbf{u}\cdot\mathrm{grad}\left(p\right) \, \mathrm{d}V
  5. Since \mathrm{d}V is constant/independent in an Eulerian formulation, we can integrate over the whole volume \mathcal{V}:
    \dot{Y} = \int_{\mathcal{V}} \mathbf{u}\cdot\mathrm{grad}\left(p\right) \, \mathrm{d}V
  6. Divide by the massflow rate \dot{m} to get "the volume-averaged specific useful work" \bar{y}:
    \bar{y} = \frac{\dot{Y}}{\dot{m}} = \frac{1}{\dot{m}} \int_{\mathcal{V}} \mathbf{u}\cdot\mathrm{grad}\left(p\right) \, \mathrm{d}V
The "volume-averaged specific useful work" \bar{y} should be pretty close to the polytropic head of the compressor, right?



Is that the correct way to do it? Or did I make any mistakes?

Sincerely,
Procyon

digger April 21, 2021 01:12
Hi,


I would extract the interesting path (i.e., variables along the path) and calculate energy differentials along it (de=e_b-e_a, where a, b are two points placed close to each other on the path). The sum of those differentials would be the amount of energy change while passing from start to end point along the path. The e is defined as the internal energy e(T), while other forms of energy like the enthalpy (p/rho) and kinetic energy (v^2) can also be added if needed.


https://en.wikipedia.org/wiki/Bernou...thermodynamics


The above expression yields the specific energy at points, for example, along the path.


Another way would be probably to calculate the integral of the substatial derivative of the total energy D(e+v^2)/Dt or enthalpy for an open system.


What would be the practical relevance of that information?


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