Kolmogorov scale why $d\sim \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$ and not $d\propto \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$? I am looking at the Kolmogorov scale and in numerous sources (e.g. this one). I have seen the following:
$$d\sim \left( \frac{\nu^3}{\epsilon} \right)^{1/4}$$
for the Kolmogorov scale. I can see why we must have:
$$d\propto \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$$
on dimension grounds. But to replace $\propto$ with $\sim$ we would have to be sure that the constant of proportionality is of order unity. I have not been able to find justification for this assumption. Does anyone now such a justification?
 A: I think the most intuitive but maybe not the most mathematically correct interpretation is the implicit assumption that viscous forces start dominating over inertial forces at the Kolmogorov scale. This is easily seen when defining the kolmogorov length and velocity scales:
$$l\sim\left(\frac{\nu^{3}}{\epsilon}\right)^{1/4}\quad v\sim\left(\nu\epsilon\right)^{1/4}$$
The Reynolds number at the Kolmogorov scale is then evaluated as:
$$\mathrm{Re}=\frac{vl}{\nu}\sim 1$$
For a proportionality sign, the proportionality constant may be much larger than order unity. The Reynolds number would become much larger than order unity as well which would contradict the assumption of viscous forces being of the same order as inertial forces. Hence it suffices to make the constant of order unity and introduce the 'on the order of' sign.
More mathematically, according to the wiki:

The definition of the Kolmogorov time scale can be obtained from the inverse of the mean square strain rate tensor and the definition of the energy dissipation rate per unit mass.

Perhaps from these definitions it follows automatically that the proportionality constant is of order unity, however, i have not investigated this further.
A: Throughout this answer $\sim$ means 'of the order of'.
The viscous force takes the form:
$$F_V\sim \nu\frac{v}{l^2}$$
So the rate of energy desperation (per unit mass) is:
$$\varepsilon \sim \nu\frac{l}{t} \frac{v}{l^2}$$
$$\sim \nu\frac{v^2}{l^2} \tag{1}$$
Now the inertial force is given by: 
$$ F_I \sim \frac{v^2}{l}$$
On the assumption that $F_V \sim F_I$ we have:
$$ \nu \frac{v}{l^2} \sim \frac{v^2}{l}$$
$$\nu \frac{v^2}{l^2} \sim \frac{v^3}{l}$$
So:
$$\varepsilon  \sim \frac{v^3}{l} \tag{2}$$
From $(1)^3/(2)^2$ We have:
$$\varepsilon \sim \frac{\nu^3}{l^4}$$
Giving us:
$$l \sim \left( \frac{\nu^3}{\varepsilon} \right)^{1/4}$$
References
1) Fluid Mechanics: An Introduction to the Theory of Fluid Flows By F.Durst (page 544, link to Google Books).
2) nluigi's answer to this question.
