Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 84 / 5 / d / Solution

With negligible internal-to-potential exchange and negligible change in available potential energy, the budgets give $M=B$. The model supplies $B=\sqrt{K}{U}_0^2/L_{\rho }$ and $ϵ=K^{3/2}/L_v$, so for $K(t)>0$ the instantaneous mixing efficiency isFor a positive initial seed and $L_f<L_{\rho }$, both loss rates approach positive constants. Their time integrals grow linearly at those rates, implyingIf mixing efficiency instead denotes the flux coefficient, then $\Gamma (t)=L_v{U}_0^2/[L_{\rho }K(t)]$ and ${\Gamma }_{c,\infty }=L_f/(L_{\rho }-L_f)$. During the transient, $\int (B+ϵ)dt=\int Fdt+K_0-K(t)$; equating irreversible loss with the instantaneous forcing would omit changing kinetic-energy storage. A permanently resting zero-seed branch has zero numerator and denominator, so it has no defined loss-based efficiency.