 Solution

Put $\xi =x/{\delta }_0$ and $n=2p-q$. The eddy viscosity is ${\nu }_T=C_{\mu }\sqrt{k_0}{\delta }_0{\xi }^n$, since ${\delta }_0=k_0^{3/2}/{ϵ}_0$. In the energy equation, advection scales as ${\xi }^{p-1}$, turbulent diffusion as ${\xi }^{n+p-2}$, and destruction as ${\xi }^q$. A leading advection–diffusion K-epsilon turbulence front requires $n=1$, henceComparing the leading coefficients then gives $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$. Similarly, the leading coefficients of the dissipation equation give $U_0=C_{\mu }q\sqrt{k_0}/{\sigma }_{ϵ}$. Their equality implies $q=\sigma p$, where $\sigma ={\sigma }_{ϵ}/{\sigma }_k$. Combining this with $q=2p-1$ yieldsThe assumed positive exponents require $0<\sigma <2$. Under these values, the omitted destruction terms are indeed higher order, as checked in part (d), so the leading balance is not circular.