Buoyancy-driven turbulent density diffusion ($F_{\rho }=K{W}^2\sqrt{g/{\rho }_0}({\partial }_z\bar{\rho }{)}^{3/2}$)

In a narrow tube with unstable mean density gradient $q={\partial }_z\bar{\rho }>0$, width-limited turbulent eddies have velocity scale $u_{\ast }\sim W\sqrt{gq/{\rho }_0}$. A mixing-length diffusivity gives downward density flux $F_{\rho }=K{W}^2\sqrt{g/{\rho }_0}{q}^{3/2}$ per unit area. With upward coordinate $z$, conservation is ${\bar{\rho }}_t={\partial }_z{F}_{\rho }$ and $q_t=K{W}^2\sqrt{g/{\rho }_0}{\partial }_z^2(q^{3/2})$. The closure presupposes established high-Reynolds number turbulence and requires boundary fluxes and finite available potential energy to be respected.

Ignoring end boundaries, the buoyancy-driven turbulent density diffusion model admits $\bar{\rho }={\rho }_0[1+A(t)C{z}^5]$ for $C>0$. Since $({\partial }_z\bar{\rho }{)}^{3/2}\propto A^{3/2}{z}^6$, the amplitude obeys $A^{\prime }=65^{3/2}K{W}^2\sqrt{gC}{A}^{3/2}$. With $A(0)=1$, $A=(1-t/t_{\ast }{)}^{-2}$ and $t_{\ast }=[35^{3/2}K{W}^2\sqrt{gC}{]}^{-1}$. Its formal growth is not a physical finite-time singularity of a closed mixing tube: ignored end fluxes, loss of the Boussinesq approximation and the finite energy budget invalidate the continuation.