 Solution

Define the positive total shear stress by $\tau =\rho [\nu U_z-\overline{u^{\prime }{w}^{\prime }}]$, so the physical wall traction on the fluid opposes the positive flow. The steady mean momentum equation givesFor a driving pressure gradient $P_x<0$, the stress decreases upward. It is approximately constant whileAn outer boundary-layer height can impose a stricter limit. In the turbulent region outside the viscous or roughness sublayer, the only local neutral scales are $u_{\ast }$ and $z$, so $U_z=C{u}_{\ast }/z$. Conventionally write $C=1/\kappa$, where $\kappa$ is the von Kármán constant. Equivalently, a mixing length $\ell =\kappa z$ and stress closure $u_{\ast }^2={\ell }^2{U}_z^2$ give the same result. Integration yields the law of the wallThe roughness length $z_0$ is an extrapolation scale, not a statement that the logarithm is valid at the actual solid surface. For a smooth boundary, the viscous inner scales are $z^{+}=z{u}_{\ast }/\nu$, $U^{+}=U/u_{\ast }$. The viscous sublayer has $U^{+}\simeq z^{+}$; farther out $U^{+}={\kappa }^{-1}\log z^{+}+C_s$, corresponding to $z_0=(\nu /u_{\ast })e^{-\kappa C_s}$. For a fully rough boundary, $z_0$ is proportional to a geometric roughness height and viscosity no longer determines the leading profile. The logarithmic region still needs roughness height $\ll z\ll {\delta }_{\tau }$.

In local equilibrium, production of turbulent kinetic energy by the shear is $P=-\overline{u^{\prime }{w}^{\prime }}{U}_z\simeq u_{\ast }^2{U}_z$. Neglecting its transport relative to production and viscous dissipation,Dimensional similarity alone gives $ϵ\propto u_{\ast }^3/z$; the displayed coefficient additionally uses the local energy balance and the logarithmic shear.