 Solution

The zeroth law of turbulence states that at fixed outer velocity $u$ and integral scale of turbulence $\ell$, the mean viscous dissipation per unit mass approaches a nonzero finite value as the Reynolds number $\mathrm{Re}=u\ell /\nu$ becomes large. In an established forward energy cascade, this gives $ϵ\sim C_{ϵ}{u}^3/\ell$, with $C_{ϵ}$ of order one. The small-scale velocity gradients become large enough to offset decreasing kinematic viscosity; this is the turbulent dissipation anomaly.

At the Kolmogorov microscales, the local Reynolds number is of order one, $v\eta /\nu \sim 1$, and the viscous dissipation is $ϵ\sim \nu (v/\eta {)}^2$. Solving these balances givesSubstituting the outer-scale estimate, and omitting order-one constants, givesThe associated eddy turnover time is $\eta /v=(\nu /ϵ{)}^{1/2}$.