Density has units of $kg/m^3$. In simulations often scaling is done to deal with non-dimensional values, so the physical density is converted to the non-dimensional density so that it can be used in simulations. How is this done in 2D simulations? One may need to assign a value of mass to certain area.

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    $\begingroup$ One way would be to pick a "depth" for your simulation (eg, 1m). This isn't a bad method since 2D simulations are often done under the assumption that the behavior in the 3rd dimension is uniform. $\endgroup$ Feb 6, 2015 at 17:04
  • $\begingroup$ @TylerOlsen So then I can find the "area"-mass density as $\rho\times 1m$ where $\rho$ is in $kg/m^3$? Then the ratio of this area mass density and the non-dimensional density will give the scaling factors? $\endgroup$
    – zed111
    Feb 6, 2015 at 17:41
  • $\begingroup$ I guess the short answer to that question is "yes". My suggestion is to do all of your scaling in 3D, then convert to 2D at the very end. Do all of your reasoning with the full 3D problem in mind (that's the way the world is, after all), and then make an appropriate conversion to 2D. To me, this is easier than trying to re-derive all of physics in an imaginary 2D world. $\endgroup$ Feb 14, 2015 at 7:01

2 Answers 2


When the Navier-Stokes equations are non-dimensionalized, one of the coefficients that comes out of the non-dimensionalization process is the Reynolds number. The Reynolds number parameter is defined as:

$$ Re = \frac{\rho VL}{\nu} $$

What is important is not so much that you are using the same density as what you are trying to simulate, but that you are using the same Reynolds number. This concept is called similarity, i.e. two problems that have the same Reynolds numbers (even though the densities may be different) produce a similar flowfield. So, in most CFD codes you wouldn't specify the density as an input, but rather the Reynolds number.

You basically need to compute this for the simulation you are running. $L$ here is some characteristic length of your object. For an airplane this might be the wing span, for a river it might be the width of the river, for a submarine it is usually the length of the submarine. $\nu$ is the kinematic viscosity. $V$ is the freestream (unperturbed) velocity.


In some cases one can integrate properties like the density across the thickness of the 2D area. The 2D sheet does not have to have a uniform thickness, but it can. For example, say I wanted to calculate the mass of a thin sheet, but the material is not homogeneous. Its density is $\rho (x,y,z)$. Say I also have a function, $f$, that will calculate the mass of a sheet, but $f$ only takes the mass per unit area as an argument.

Since I have a thin sheet, I might want to assume the density varies linearly across the thickness, in the $z$-direction. This allows me to assume the density is the average of the density at the top and the bottom of the sheet, or is the density at the middle of the sheet. Then the density of thin sheet becomes $\rho (x,y,z_{mid}) \times t(x,y)$, where $t(x,y)$ is the thickness of the sheet and $z_{mid}$ is the coordinate of the middle of the sheet. I would define a density function $\sigma (x,y) = \rho (x,y,z_{mid}) \times t(x,y)$ which is the mass per unit area of the sheet. Defining $\sigma$ does not change anything, but it helps me visualize what the terms mean and keep my units straight. $\sigma(x,y)$ is what I would use in my function $f$. If the sheet was too thick to assume the density varies linearly, then I would use some other numerical method to integrate across the thickness.

Note that this method depends on being able to integrate, at least in principle, across the thickness. If I had an equation in the $x,y,z$-coordinates, I would want to integrate each expression in the equation across the thickness of the sheet. The result would be an equation in just the $x,y$-coordinates.


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