Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 73 / 3 / v

The constant Reynolds stress and the logarithmic shear imply the eddy viscosity ${\nu }_T=u_{\ast }\kappa y$. On the other hand the K-epsilon model gives ${\nu }_T=C_{\mu }{k}^2/ϵ$. Insert $ϵ=u_{\ast }^3/(\kappa y)$:Canceling the positive factors yields $C_{\mu }{k}^2=u_{\ast }^4$. Since energy is nonnegative, the log-layer solution of the k-epsilon model hasindependent of $y$ to this leading overlap-layer accuracy. In particular $k_y=0$, and the $k$ diffusion term vanishes, confirming the local production–dissipation balance in the energy equation.