Search:
D68 /

# ScalingOfDiffusion

One has to differ two regimes in the diffusion process of gases. The basic diffusion is the molecular diffusion which is solely based on the stochastic motion and interference of the molecules. This process might be supported by micro-vortices generally associated with turbulence in the gas. This turbulent diffusion is typically orders of magnitude more efficient.

Diffusion may transport mass, impulse and heat. For each transport there is a coefficient characterising the corresponding transport properties for the given thermodynamic state. Disregarding the actual transported quantity the 2nd Fick's law allows to identify the units of the diffusion coefficients $$[D] = [\nu] = [a] = {m^2 \over s}$$ Evidently ratios of these coefficients deliver dimensionless parameters characterising the gas itself: the Prandtl number $$Pr = {\nu \over a}$$ relates the impulse transport to the heat transport and the Schmidt number $$Sc = {\nu \over D}$$ the impulse transport to the mass transport. As the transport properties mainly depend on the degrees of freedom of the gas species, the related dimensionless numbers nearly the same for two-atomic gases (like H2, O2 and N2) for instance.

After switching off turbulence generating sources the turbulence connected with the contained gases will decay. Most experiments on decaying turbulence have been performed in wind tunnels up to now, where the distance from the grid gives the decay time t, if the mean velocity is known. In the garage experiment the mean velocity after the release phase is zero.

Applying the Pi-theorem to the momentum decay in a turbulent gas layer we will receive the following matrix:

$$\epsilon$$ $$u$$ $$L$$ $$t_\ell$$ $$\nu$$ $$2$$ $$1$$ $$1$$ $$0$$ $$2$$ $$0$$ $$0$$ $$0$$ $$0$$ $$0$$ $$-3$$ $$-1$$ $$0$$ $$1$$ $$-1$$

It is evident that the problem, the rank of the dimension matrix respectively, is reduced as the dynamic row represented by the kg is zero. So there must be 5-2=3 dimensionless parameters for this problem. The experienced reader easily identifies the Reynolds number, the Taylor-Reynolds respectively, $$Re = {{u L} \over \nu}$$

$$Re_\lambda = {{u \lambda} \over \nu} = {{u^2} \over {{{\partial u} \over {\partial x}} \nu}}$$

the dimensionless decay time $$\Pi_t = {{t_\ell} \over \tau} ={{t_\ell \nu} \over {L^2}}$$ and - maybe more difficult to see immediately - the ratio of dissipation times in a characteristic time step and of the specific kinetic energy of the flow field: $$\Pi_e = {{\epsilon L} \over {u^3}}$$

As for this problem the Re_\lambda depends implicitly on the time of the observation Re_\lambda, the whole might be summarised to $$\Pi_e = f(Re_\lambda(t))$$

In Fig.2 we can identify the turbulence lifetime t_\ell where the Reynolds number starts to drop close to t / \tau \approx 10^{-1} where Re_\lambda \approx 16. Figure 1. Time dependence of the Taylor-Reynolds number for freely decaying turbulence

From further considerations and from experience one may derive that also for the dimensionless decay time t_\ell / \tau there should be for high Reynolds numbers a Reynolds number independent limit, which is according to [Lohse]: $$\Pi_{t, max} = 0.18$$ For the garage benchmark this would yield approximately one minute after the release end for the turbulence lifetime.

Again, these findings are based on the assumptions of free and isotropic turbulence. Deviations for the garage example are the wall effects of the roof and the influence of the gravitation.

After the turbulent regime only the molecular diffusion remains.