4
$\begingroup$

While studying some papers, I didn't understand the following sentences:

deltP the pressure drop across the turbine, evaluated using the mass average values at the inlet and outlet sections of the computational domain, respectively.

What is the exact meaning of the "mass average values"?

And is there such a option of "mass average values" in Fluent? I dont know that how to compute this parameter in Fluent.

Improve this question
$\endgroup$
  • $\begingroup$ Likely related: engineering.stackexchange.com/q/5384/16 $\endgroup$ – user16 Sep 14 '15 at 23:02
  • $\begingroup$ @GlenH7 How is this related :-) ??? $\endgroup$ – Algo Sep 15 '15 at 14:36
  • 1
    $\begingroup$ @Algo - it's background context for this question. Have a look at the OP's recent questions and you'll see a consistent theme across them. I provided the link in case it added enough context to make it easier to answer this question. $\endgroup$ – user16 Sep 15 '15 at 14:38
  • $\begingroup$ GlenH7, I dont understand your means, I am new in this site. I am grateful that explain me more. Thanks $\endgroup$ – user19061 Sep 15 '15 at 16:16
3
$\begingroup$

What is the exact meaning of the "mass average values"? And is there such a option of "mass average values" in Fluent? I don't know that how to compute this parameter in Fluent.

Yes, FLUENT has many surface integral reports like Area-weighted average and Mass average, quoting from FLUENT 6 user manual:

  • Area-weighted average: You can find the average value on a solid surface, such as the average heat flux on a heated wall with a specified temperature.
  • Mass average: You can find the average value on a surface in the flow, such as average enthalpy at a velocity inlet.

And this is an answer from cfd-online about the difference between them:

The area average of a scalar is calculated by integrating the scalar times the area divided by the total area over the region. A mass averaged quantity is obtained by integrating the scalar time mass flow divided by total mass flow over the region. $$\text{Area Average} = \frac{\int \phi \, dA}{A}$$ $$\text{Mass Average} = \frac{\int \phi \, dm^.}{m^.}$$

Where $\phi$ is an arbitary scalar property of the flow.

Improve this answer
$\endgroup$
  • $\begingroup$ Please see the following link: en.wikipedia.org/wiki/Intensive_and_extensive_properties According to your answer, I Think that first formula is for specific properties or intensive property like velocity and second formula is for extensive property like specific volume (m3/kg, the inverse of density). Am I right? Thanks $\endgroup$ – user19061 Sep 15 '15 at 16:26
  • $\begingroup$ @user19061 No, you can calculate both intensive and extensive properties through both average methods. $\endgroup$ – Algo Sep 15 '15 at 16:28
  • $\begingroup$ ok, What is major different between these tow cases? $\endgroup$ – user19061 Sep 15 '15 at 19:23
2
$\begingroup$

Averaging is a tricky business (see Cumsty 2006 for an introduction). Basically the problem is, that our thermodynamics usually talk about a pressure and a temperature etc. But in real life the pressure and temperature are fields (i.e. varying in all three dimensions). So the question is: How to come up with one representative value of (for example) the temperature.

Depending on the way the averaged value is used different averaging techniques are suited better.

For example think of the temperature of water in a big tank which is filled via two pipes from two different heat exchangers delivering two different water temperatures. Therefore not every water molecule has the same temperature. If you were supposed to calculate the temperature in the tank you would need to come up with a way to calculate a representative temperature. How high would this mixed out temperature be?

You can of course just use the mean of the two pipes. But what if one pipe pumps more water into the tank. The solution to this particular problem is to weigh every pipe-flow with the mass-flow rate. This is called a mass-flow-average.

As a rule of thumb: To validate numerical simulations the averaging method in CFD and experiment have to be the same. It might be a good idea to use area-averaging to compare basic values as temperature or pressure. If the averaged values are used to do further calculations use mass-average for all except for static pressure which is always area-averaged.

Improve this answer
$\endgroup$
1
$\begingroup$

I think they are telling you that they are considering the bulk, or macroscopic, properties of the fluid, and not the properties of the individual molecules. This is an important assumption in fluid mechanics, known as the Continuum Hypothesis.

At the molecular level, tiny, rapid fluctuations in the fluid's physical properties such as velocity and density occur as the molecules move around, bumping into each other (e.g. Brownian motion). These fluctuations would be impossible to predict and model accurately, so fluid mechanics often treats fluids as continuous media with an average velocity, average viscosity, average pressure, etc, on the assumption that the sum of the fluctuations, across millions or billions of particles, cancel out.

This greatly simplifies most fluid mechanics problems and is a valid assumption for most practical problems.

Improve this answer
$\endgroup$
  • 1
    $\begingroup$ Thanks your answer, But I do not understand the relationship between the pressure drop with mass!! $\endgroup$ – user19061 Sep 15 '15 at 13:36
1
$\begingroup$

While I haven't heard this term specifically, it may be telling you to use specific properties (those expressed in terms of mass) instead of non-mass-adjusted properties. For example, you would use the specific volume (${m^3}/{kg}$, the inverse of density) in calculating the pressure difference, because the actual volume has little meaning in a moving system. To me, this seems pretty basic, as I rarely used anything but specific values in my fluids and thermo classes, but I can see an academic paper wanting to be thorough in explaining their exact method.

Improve this answer
$\endgroup$
1
$\begingroup$

If the analysis on the turbine is two- or three-dimensional, then the pressure can vary across the transverse and/or circumferential dimensions as well as between the entrance/exit. Think about the air entering at the center of the inlet duct possibly having a different pressure than the air entering at the edge of the inlet duct. In this case, you don't have just one pressure at the inlet and one at the outlet, you have a pressure field at each. So, the author(s) take an average of the pressure at the inlet and outlet, and use those averages to calculate the pressure difference. The term "mass-average" in this context in analagous to a weighted-average.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.