5
$\begingroup$

During the scaling of the Navier-Stokes equations it is often made use the viscous time scale $L^2/\nu$, where $L$ is the characteristic length and $\nu$ the kinematic viscosity. What is the physical interpretation of the definition term $L^2/\nu$?

Like $U/L$ might be the "the large eddy turnover time" with $U$ being the characteristic velocity etc.

$\endgroup$
4
$\begingroup$

It is the timescale required by viscosity $\nu$ to diffuse momentum significantly over a characteristic length scale $L$.

Analogous to how mass and heat can be transfered by molecular diffusion through collisions between particles, momentum can also be transfered through similar mechanisms. For mass and heat transfer the respective time scales are $L^2/\mathcal{D}$ and $L^2/\alpha$ where $\mathcal{D}$ is the mass diffusion coefficient and $\alpha=\frac{\kappa}{\rho c_p}$ is the thermal diffusion coefficient.

The presence of the square of the characteristic length is explained by the fact that this type of molecular diffusion is usually stochastic and anisotropic (see random walk model) and diffuses mass, heat and momentum in all direction equally. It therefore spreads out over a characteristic area $L^2$.

$\endgroup$
  • $\begingroup$ Nice, thanks! Just to be clear, "to diffuse momentum significantly" is like $1/e$ reduction? I know it can't be put that exact but qualitatively is that so? $\endgroup$ – Victor Pira Feb 7 '18 at 10:10
  • 1
    $\begingroup$ It can be quantified somewhat using penetration theory; the penetration distance as a function of time is $x_p=\sqrt{\pi\nu t}$ (only valid for about $x_p/L<0.5$). So for $t=0.1L^2/\nu$, we have $x_p/L = \sqrt{0.1\pi} = 0.56 $, i.e. for only a tenth of a diffusion time scale, it has penetrated halfway through the domain. $\endgroup$ – nluigi Feb 7 '18 at 10:25

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.