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March 18, 2010, 14:57 |
Continuum Validity
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#1 |
New Member
Karan Anand
Join Date: Feb 2010
Posts: 23
Rep Power: 16 |
Hi,
Out of curiosity, I would like to ask a question: for Fluid dynamics and application of Navier stokes equation, we consider that the fluid is a continuum. i.e. we assume that the fluid properties like density, etc is high enough in a given volume that sufficient number of particles exists in that volume. As a result we can take values of primitive variables like velocity or Kinetic energy for that matter as bulk or mean values. Also, when we do computational analysis, we tend to limit our grid node distance to zero for maximum accuracy. Now if the magnitude of the sides of the finite volume under consideration is so small (around the node of interest), i.e. it is of the order of or less that the mean free path of the molecules, then how can we still assume that continuum theory is valid in that finite volume? |
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March 18, 2010, 15:05 |
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#2 |
New Member
maply
Join Date: Mar 2010
Posts: 5
Rep Power: 16 |
Hi,
Interesting and still controversial question. To my knowledge, the mesh or the nodes distance does not have any physical meaning. I consider the elements as mathematical support for solving the PDE. Comments are welcome, Cheers |
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March 18, 2010, 15:10 |
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#3 |
Senior Member
Join Date: Jul 2009
Posts: 357
Rep Power: 19 |
If you had a grid cell on the order of the mean free path of the molecules the continuum assumption would be invalid. But then you would not be able to practically solve the equations on any given piece of hardware. You would be looking at using the Boltzmann equations to describe your system.
Not sure what you mean by the grid node distance being zero. A computational grid is simply a way of representing the continuous flowfield by a set of discrete values that approximate the field, much in the same way that a straight line can approximate a curve over some domain. Once we start to discretize, it's simply a mathematical process based on the continuum assumption - if we actually had the horsepower to resolve down to the molecular level we wouldn't be solving the Navier-Stokes equations. |
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March 19, 2010, 06:12 |
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#4 |
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L.Y LIN
Join Date: Mar 2010
Posts: 49
Rep Power: 16 |
adg's reply is very good.
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March 22, 2010, 10:36 |
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#5 |
Senior Member
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Once you have decided to solve numerically the Navier-Stokes Equations (instead of the motion of the single particles) there is no way to go back to the original molecular nature of the flow. The grid will influence your solution but not in the sense of a magnifiing lens (you can't see better something that is not in your model anymore, i.e. the molecules).
It is mostly the same for RANS/URANS computations, whatever is your number of cells you are not going to see any turbulent structure, because they are not in the model. If this is good or bad it is determined by your application. However, on a very conceptual basis, you can think of an entirely numerical procedure, like a FVM, which is based on a grid but, instead of solving for the NS equations it is based on performing statistical averages of properties associated to some particles distributed inside the volumes with a certain density and obeying some dynamical rules. If the method would be constructed exactly you should recover the NS equations for big enough volumes and high enough densities but also the real averages over few particles when the grid is fine enough. But, still, you would be assuming that the atoms are the smaller elements available in nature. What if, with my FV grid i also want to see sub-atomic particles (when fine enough)? Then i still have to change the model at the basis of the code and construct it on the basis of sub-atomic particles and let the code describe the formation of atoms and so on. To conclude, the mathematical continuum model used in the numerical simulations does not break-up per se on the molecular scale (if this happens or not, and why, with the NS equations is another story), it just becomes wrong but you can still use it for your simulations. |
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