Turbulent mixing

Turbulent mixing redistributes a transported scalar, such as temperature or particle concentration, through irregular fluid motions and eventual molecular diffusion. A top-hat plume model assumes this mixing keeps the scalar approximately uniform within a selected cross section; that uniformity is an additional closure, not a consequence of thin geometry alone.

Richardson pair dispersion concerns relative separation of two marked fluid elements, rather than their centre-of-mass motion. In an inertial range, scale-local relative diffusivity is of order ${ϵ}^{1/3}{r}^{4/3}$. A self-similar closure gives $d⟨r^2⟩/dt=A{ϵ}^{1/3}⟨r^2{⟩}^{2/3}$ and hence a $t^3$ regime after loss of memory of initial separation. Stationarity and inertial-range separation alone do not imply this law at arbitrarily short times.

For positive initial separation below the Kolmogorov length scale, a phenomenological pair-dispersion model has smooth-flow exponential stretching, inertial-range Richardson pair dispersion, and diffusive large-scale pair dispersion beyond the integral scale of turbulence. A positive Lyapunov exponent is needed for the first regime, scale locality and initial-memory loss for the second, and integrable relative-velocity autocorrelation for the third. The inertial-range crossing time is of order $L^{2/3}{ϵ}^{-1/3}$, one outer eddy turnover time.

At separations well above the integral scale of turbulence, two fluid velocities can decorrelate spatially. If the relative-velocity autocorrelation $R_{\mathrm{rel}}$ is stationary and integrable, the displacement identity used in Taylor turbulent dispersion gives the displayed long-time linear mean-square growth. With speed $U$ and Lagrangian integral time of order $L/U$, its relative diffusivity is of order $UL$. The characteristic separation grows as $t^{1/2}$, unlike inertial-range Richardson pair dispersion.

The dimensionless Richardson constant is the coefficient in the mean-square pair-separation law. Under the simple self-similar diffusivity closure, $g=(A/3{)}^3$. Its universality requires scale locality, adequate scale separation and loss of initial-condition memory; a finite-time fit need not isolate this asymptotic coefficient.

Taylor turbulent dispersion describes the absolute displacement of a marked fluid element in stationary homogeneous turbulence. Here $R_L$ is its Lagrangian velocity autocorrelation. Continuous correlation at zero gives ballistic spreading at short times, and an integrable correlation gives diffusive spreading at long times. This is distinct from shear-induced Taylor dispersion and from Richardson pair dispersion.

For a zero-mean stationary velocity process of variance ${\sigma }_v^2$ and normalized autocorrelation $(1+|s|{)}^{-\alpha }$, integrating velocity twice gives the displayed expression, exactly rather than just asymptotically. For $\alpha >1$ the long-time diffusion coefficient is ${\sigma }_v^2/(\alpha -1)$. At $\alpha =1$, the displacement variance is asymptotic to $2{\sigma }_v^{2}t\log t$. For $0<\alpha <1$, it is asymptotic to $2{\sigma }_v^2{t}^{2-\alpha }/[(1-\alpha )(2-\alpha )]$. Linear variance growth does not by itself imply a central limit theorem or a normal distribution for displacement.

For stationary statistics along a Lagrangian trajectory $\mathbf{X}$, the velocity autocorrelation depends on the lag $s$. Integrating velocity twice gives the mean-square displacement $2{\int }_0^t(t-s)R_L(s)ds$. If $R_L$ is integrable, the long-time displacement is diffusive. Its normalized one-sided area is the Lagrangian integral time.

The Lagrangian integral time is the area under a normalized velocity autocorrelation along a Lagrangian trajectory. If this correlation is integrable, Taylor turbulent dispersion has asymptotic mean-square displacement $2R_L(0)t_{L}t$. The integral time need not equal every short-time decorrelation scale, particularly when the autocorrelation oscillates.

For $A=U_0^2(1/L_f-1/L_{\rho })>0$, define $K_{\infty }=A{L}_v$ and $y=\sqrt{K/K_{\infty }}$. The positive-energy solution is $y(t)=[y_0+\tanh (\sqrt{K_{\infty }}t/(2L_v))]/[1+y_0\tanh (\sqrt{K_{\infty }}t/(2L_v))]$. Energies above and below $K_{\infty }$ approach it from their respective sides. The zero-energy equation is not Lipschitz and admits both permanent rest and delayed startup, so a unique positive steady limit needs a positive seed or an additional startup prescription. With negligible available and internal energy exchange, the long-time instantaneous mixing efficiency fraction is $L_f/L_{\rho }$, independent of $L_v$.

With boundary fluxes suppressed, let $B$ transfer kinetic energy to potential energy, $M$ transfer available potential energy irreversibly to background potential energy, and $W$ transfer internal energy to potential energy. Then $\dot{P}=B+W$, ${\dot{P}}_b=M+W$, $\dot{A}=B-M$, and $\dot{K}=F-B-ϵ$. The internal reservoir has $\dot{I}=ϵ-W$, conserving total energy apart from forcing $F$. The distinction between reversible buoyancy exchange and irreversible mixing is essential outside a quasi-steady negligible-$A$ regime.

Instantaneous mixing efficiency is the fraction of instantaneous irreversible mechanical-energy loss that raises background potential energy, after subtracting internal-to-potential exchange. The ratio $\Gamma =M/ϵ$ is a different quantity, related by $\eta =\Gamma /(1+\Gamma )$. Replacing $M$ by buoyancy exchange $B$ additionally assumes negligible change of available potential energy.

Cumulative mixing efficiency is the ratio of total irreversible density-mixing conversion to total irreversible mechanical-energy loss over an interval. It is a loss-rate-weighted average of instantaneous mixing efficiency, not an unweighted time average. Stored kinetic energy, changes in available potential energy and boundary work must be accounted for before replacing its denominator by integrated forcing.