= Solution
Keep the forcing ratio $r=L_f/L_\rho$ fixed in $(0,1)$. Then
$$
\frac{K_\infty}{U_0^2}=\frac{L_v}{L_\rho}\frac{1-r}{r},
\qquad \frac{B}{\epsilon}=\frac{L_v}{L_\rho}\frac{U_0^2}{K}.
$$
For $L_v\ll L_\rho$, dissipation is strong at a given $K$ and $K_\infty\ll U_0^2$ when $r$ is not extreme. If the initial energy is of order $U_0^2$, it decays toward a small steady value and starts with a small <instantaneous mixing efficiency>. This is a regime of short velocity-fluctuation scales and efficient viscous loss relative to the modeled buoyancy conversion at that initial energy.
For $L_v\gg L_\rho$, dissipation is weak at that same energy and $K_\infty\gg U_0^2$. A moderate initial energy grows and initially has a large <instantaneous mixing efficiency>; increasing stored energy eventually makes dissipation comparable with the remaining forcing. The energy-relaxation time is of order $L_v/\sqrt{K_\infty}$, which grows as $\sqrt{L_v}$ at fixed forcing and density lengths.
Within this closure, \b[both limits have the same long-time loss fraction $\eta_{c,\infty}=L_f/L_\rho$]. The steady energy changes with $L_v$ so as to compensate for its explicit effect on dissipation. Different initial conditions and finite observation windows give different transient efficiencies; one cannot infer a universal long-time efficiency from $L_v/L_\rho$ alone. In a real experiment the length scales need not remain independent of the changing flow, so this conclusion belongs to the stated model.
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