# For an aerodynamics simulation, how accurate is the assumption that air molecules possess only translational energy?

Assume that I am running an aerodynamics simulation for air with the following properties:

• 20 degrees celsius
• Wind speed of 5 metres per second
• Pressure of one atmosphere

For any other important property I have left out, assume average back yard conditions in a temperate geographical zone.

I know that air is a mixture of molecules that have translational, rotational and vibrational energy components.
My question is, for the conditions described above, how big is the translational energy component, compared to the other two?
An answer which calculates all three components for the conditions described and gives numbers would be ideal.
How does this change if the wind speed is zero?

My reason for asking is that I wish to write my own simulation which models gas as a collection of particles that have purely translational energy, ie. an xyz velocity to keep things simple.

I'm well aware of the challenges of doing this from a computer science point of view and I do not wish this to turn into a discussion of the technicalities around that. My question is purely to do with the physics.

This has been edited as I realise my original question was too programming oriented and therefore misleading.

• Simulating gas flows with many particles is the domain of Molecular Dynamics and is limited to relatively small number of particles (there are on the order of $10^{25}$ in a $m^3$ of air, imagine solving that many equations). Simulating industrial scale gas flows (over complex boundaries) is the domain of CFD through the use of the Navier-Stokes equations, either directly solving them or a indirectly (LBM, SPH, etc.) – nluigi Aug 14 '16 at 16:28
• You should spend time looking into the advantages and disadvantages of both these methods before taking on this task. Especially since you seem to just be starting out in the field. – nluigi Aug 14 '16 at 16:29
• I know it would be infeasible to simulate the amount of particles that exist in reality at human sized scales. I'd like to simulate around a million as an approximation and just for fun. I simply wanted to know how much of a difference the inclusion of rotational energy would make. – Karl Aug 14 '16 at 18:43
• @Karl , unless you can run your program on a massively parallel set of processors, it will take a LONG time to get results. Unless you have a degree in computer science, don't jump into this program without doing some serious research regarding the size of the code and the run time of the code. – David White Aug 14 '16 at 19:08
• "some translational energy is turned into rotational energy and therefore the collision is inelastic." If you don't want to worry about that, model a monatomic gas. You can then ignore the rotational energy and momentum of a "point-sized" atom compared with its translational energy - assuming you are going to make the model using classical mechanics, not quantum mechanics. Of course, you can't ignore the rotational energy and momentum of polyatomic molecules, or the energy of their internal vibration modes. – alephzero Aug 14 '16 at 19:20

The conversion from translational to rotational energy is not an inelastic process - it just means that when you consider the mean energy of the molecule (and the heat capacity) you have to take account of the additional degrees of freedom.

Note also that by definition the volume occupied by an "ideal" gas molecule is zero - from which it follows that collisions can't occur (mean free path scales with inverse cross sectional area).

For the purpose of your proposed calculation, considering the gas as "real" (mean free path, heat capacity) is essential. But you can probably get away with not modeling explicitly the attractive (Van der Waals) forces; and the pressures won't be high enough that you would have to take account of the volume of the atoms for the PV curve.

• I disagree that collisions cannot occur. Ideal gases (thermally perfect for example) still have varying heat capacity and chemical reactions, which require collisions. Wikipedia also disagrees for what it's worth... collisions do occur, but are elastic and there are no forces at a distance, only collisions transfer energy between molecules. The zero volume assumption just means the collisions are point-like and not sphere-like. – tpg2114 Aug 14 '16 at 20:17
• @tpg2114 True points have no volume, and therefore the probability of a collision goes to zero. Meaning that an "ideal gas with collisions" is a contradictio in terminis. – Floris Aug 14 '16 at 21:22

For an aerodynamics simulation, the details of the microscopic model are irrelevant as long as they yield the correct macroscopic behavior.

Since you are interested in the properties of a fluid, and the motion of an macroscopic body through a fluid, starting from the molecular scale would be inefficient and ultimately unnecessary.

My argument is that, assuming the computational resources are available (at least $10^{23}$ molecules otherwise is barely different than brownian motion) and the model completely accurate, such simulation should behave according to the fluids' mechanics equations (which describe the behavior of macroscopic fluids) in which case an enormous amount of time, energy and computation was spent to reproduce something that has already a much simpler representation.

However, if your goal is to establish the connection between microscopic behavior and fluid behavior, this is much more relevant. Our current simulations establish connection between microscopic models and macroscopic thermodynamic behavior (as is claimed, I'm not of the same opinion). But I think this has not been done for mechanical properties of fluids. I mean, specifically your questions, as what is the relevance on the macroscopic behavior of the fluid, of the microscopic details (fraction of rotational versus translational motion of molecules). I think this does have research value and would be of interest.