Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 84 / 4

The K-epsilon turbulence model evolves turbulent kinetic energy $k$ and its turbulent kinetic energy dissipation rate $ϵ$, rather than resolving turbulent fluctuations. It closes the Reynolds stress with an isotropic eddy viscosity, ${\nu }_T=C_{\mu }{k}^2/ϵ$, and models transport of $k$ and $ϵ$ by gradient diffusion with coefficients ${\nu }_T/{\sigma }_k$ and ${\nu }_T/{\sigma }_{ϵ}$. Production is tied to mean strain; dissipation-equation production and destruction are modeled terms with empirical constants. The characteristic velocity and length are $\sqrt{k}$ and $k^{3/2}/ϵ$. These assumptions are primarily for high Reynolds number, and strongly anisotropic flow, sharp interfaces and molecular-viscosity-dominated regions need additional treatment. The dissipation equation is a closure, not a second exact energy-conservation law.

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.

The K-epsilon turbulence front hasIts characteristic integral length and the specified frequency scale areThe ratio $ϵ/k$ is an inverse eddy-turnover time. The PDF calls it turbulent vorticity, but it should not be identified with the exact root-mean-square vorticity at dissipative scales, which involves molecular viscosity as well.

The dominant balance is convection of small turbulent energy and dissipation by the steady mean flow against their modeled turbulent diffusion. Diffusion supplies turbulent quantities from the turbulent side to the front; there is no local turbulent production in this problem. For $q=2p-1$, the energy-destruction term divided by its advection term is proportional to ${\xi }^{q-(p-1)}={\xi }^p$. In the dissipation equation, the destruction term has exponent $2q-p$, so its ratio to advection is likewise ${\xi }^{(2q-p)-(q-1)}={\xi }^p$. Both tend to zero for $p>0$. Thus advection balances turbulent diffusion, while both modeled destruction terms are asymptotically smaller.

A continuous non-turbulent edge should have vanishing $k$, $ϵ$, characteristic turbulent length $L$ and turnover frequency $\omega$. With $p,q>0$ and $\sigma <2$, the length in part (c) vanishes exactly when $\sigma <3/2$, while the frequency vanishes exactly when $\sigma >1$. Hence the requested formal consistency interval isAt the endpoints one of these characteristic quantities remains finite rather than vanishing. This is consistency of the ideal turbulent closure and its edge limits, not a claim of validity arbitrarily close to $x=0$ in a real fluid: ${\nu }_T\to 0$ eventually makes neglected molecular viscosity important. Nor does it establish the stronger local scale separation $L\ll x$; the displayed solution has $L/x\to \infty$, another reason to treat the exact edge as a formal model limit.

Since $2p-q=1$, the leading turbulent viscosity is linear:Using $U_0=C_{\mu }p\sqrt{k_0}/{\sigma }_k$ and $1/p=2-\sigma$ gives