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In 3D the capillary number $Ca$ is defined as: $$Ca=\frac{\nu \rho U}{\gamma}$$ where $\nu$ is the kinematic viscosity ($m^2/s$), $U$ is the velocity ($m/s$)and $\gamma$ is the interfacial tension ($N/m$) and $\rho$ is the density in $kg/m^3$ . However in 2D I will have $\rho$ in units of $kg/m^2$ (2D simulation), so how to find capillary number?

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Rather than leave these as comments, I guess I should answer it since this has come up before.

A 2D simulation does not mean there is no third dimension. Rather, it means we are saying there is no variation in the third dimension such that $\frac{\partial}{\partial z} = 0$. But that depth direction still exists and we typically just call it a "unit depth" or $\Delta z = 1$.

For some people, sometimes, they like to define things differently and actually adjust their units. I've come across this in papers looking at mass-spring systems where in 3D, $k = E l$ where $E$ is the Youngs Modulus and $l$ is the spring length. And in 2D, they just say that $k = E$. I don't like this approach, it's somewhat confusing, and can cause problems in a code written to handle 1, 2 and 3D without rewriting everything.

However, if you want to insist that your units of density are $kg/m^2$ you can. Just remember that your kinematic viscosity is defined as $\nu = \mu/\rho$. Since your dimensions on $\rho$ have changed, so have your dimensions on $\nu$ which is now going to be $m/s$ instead of $m^2/s$. This means your definition of $Ca$ is still the same.

But this just highlights why it's not a great idea to start dropping a length dimension from everything. It just gets really confusing and confers no benefit.

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  • $\begingroup$ The reason I use $kg/m^2$ is that in my 2D simulation,I assign mass to a certain area. In the non-dimensionalized units I assign 1 unit mass to unit area. So I only know the area mass density not the volume mass density. So how do I calculate capillary number? $\endgroup$ – zed111 Feb 7 '15 at 18:44
  • $\begingroup$ using non-dimensional units $\endgroup$ – zed111 Feb 7 '15 at 18:50
  • $\begingroup$ @zed111 As I said, if you start dropping units from one place, you have to drop them from all. So your kinematic viscosity is the molecular viscosity divided by your 2D density and so you'll end up with the right units and there is no issue. $\endgroup$ – tpg2114 Feb 7 '15 at 19:11
  • $\begingroup$ In Lattice Boltzmann method, $\nu = c_s^2(\tau-1/2)$, so it still has units of $m^2/s$ $\endgroup$ – zed111 Feb 7 '15 at 19:13
  • $\begingroup$ @zed111 I don't know how else to say it. The units must change for your viscosity otherwise the momentum equation won't have the right units either. The units of $c_s$ must change then if you want to define things in a 2D sense. There is no other way -- otherwise the units on your governing equations will not be consistent. $\endgroup$ – tpg2114 Feb 7 '15 at 19:15

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