5 Energy cascade

(Panton3 26.9,10)

In trying to understand turbulence, it helps to have some mental picture of it. Suppose you look at a typical turbulent flow in some boundary layer, jet, mixing layer, or whatever. You will perceive organized masses of fluid, eddies, in random motion, distorting while moving. The size of these eddies will be quite comparable to the thickness of the turbulent layer or jet. But if you look closer, you see that there are also smaller scale fluctuations, smaller eddies, that seem to do their own independent thing. Nonlinear motion on larger scales tends to create motion on smaller scales. (Much like squaring a $\cos x$ produces a $\cos 2x$ with half the wave length.) The idea here is that the larger eddies put some of their kinetic energy in creating smaller eddies, which in turn create still smaller eddies, and so on.

Now the motion of the smaller eddies involves less velocity fluctuations relative to their surroundings. Therefore the largest eddies have most of the turbulent kinetic energy per unit mass

$$\frac12\sqrt{\overline{u_i'u_i'}}$$

Similarly, the largest eddies will dominate the Reynolds stresses

$$\tau_{ij}^{\rm {R}}=\rho\overline{u_i'u_j'}$$

experienced by the mean flow. However, the smallest eddies, because they are small, will dominate the turbulent velocity derivatives and with it the dissipation per unit mass

$$\frac{\varepsilon}{\rho}=2 \nu s_{ij} s_{ij} \qquad s_{ij}... ..._i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \right)$$

that turns their kinetic energy irreversibly into heat. The picture is that there is an energy cascade where the largest eddies put kinetic energy into smaller ones, these smaller ones into still smaller ones, until extremely small final eddies convert that kinetic energy into heat. The typical scales of these final smallest eddies are called the Kolmogorov scales. For high Reynolds number, (small $\nu$), the difference between the largest and smallest scales can be tremendous.

To see that, consider first what governs the smallest eddies. One important factor is of course the kinetic energy $\epsilon/\rho$ that is draining through the cascade to the smallest eddies, for them to dissipate it. The other important factor is of course the kinematic viscosity $\nu$ that allows them to dissipate it in the first place. The eddies are presumably too small to see the large scale features of the flow, so $\epsilon/\rho$ and $\nu$ are the only two quantities that should govern the small eddies. Let’s do some dimensional analysis based on that. If $\eta_{\rm {K}}$ is the typical length scale of the smallest eddies, $\tau_{\rm {K}}$ the typical time scale, and $\upsilon_{\rm {K}}$ the typical velocity, the corresponding three nondimensionsl $\Pi$ groups that you can form are, noting that the viscosity $\nu$ has units ${\rm L^2/T}$ and the dissipation per unit mass $\varepsilon/\rho$ units ${\rm L^2/T^3}$:

$$\eta_{\rm {K}}\sqrt[4]{\frac{\varepsilon}{\rho\nu^3}} \qqu... ...qquad \upsilon_{\rm {K}}\sqrt[4]{\frac{\varepsilon\nu}{\rho}}$$

Since these $\Pi$ groups have nothing else to depend on, they must be finite numbers. In fact, we can define specific and meaningful scales of the smallest eddies by setting each $\Pi$ group equal to 1. The thus defined $\eta_{\rm {K}}$, $\tau_{\rm {K}}$, and $\upsilon_{\rm {K}}$ are called the Kolmogorov microscales:

$$\fbox{$\displaystyle \eta_{\rm{K}}=\sqrt[4]{\frac{\rho\nu^... ...d \upsilon_{\rm{K}}=\sqrt[4]{\frac{\rho}{\varepsilon\nu}} $}$$

To get a general idea how big those Kolmogorov scales really are, we will have to estimate $\varepsilon/\rho$, the kinetic energy dissipated into heat per unit time and unit mass. Well, in a steady state, the amount of energy that is dissipated by the small eddies must cancel the amount of energy that the largest eddies put into the cascade. Let’s try to estimate the latter. The amount of kinetic energy put into the cascade by the largest scale eddies per unit time and unit mass should presumably be proportional to the typical kinetic energy of the largest scale eddies, call it ${\textstyle\frac{1}{2}}v_{\rm {L}}^2$, with $v_{\rm {L}}$ a typical turbulent velocity of the largest eddies, and inversely proportional to the time it takes the eddies to evolve nontrivially, estimated as $v_{\rm {L}}/\ell$ where $\ell$ is the typical size of the largest eddies. So the kinetic energy put into the cascade is estimated to be of order $v_{\rm {L}}^3/\ell$. If you put that into the unit Kolmogorov $\Pi$ groups above, you find the following ratios of the smallest to the largest eddy scales:

$$\fbox{$\displaystyle \frac{\eta_{\rm{K}}}{\ell} = Re^{-3/4... ...} = Re^{-1/4} \qquad Re \equiv \frac{v_{\rm{L}}\ell}{\nu} $}$$

From this it follows that for large Reynolds number $Re$, the smallest eddies can be much, much, smaller than the largest ones. Note further that normally the turbulent velocity $v_{\rm {L}}$ of the largest eddies will be similar to the total turbulent velocity, since the largest eddies have most of the turbulent kinetic energy. And the total turbulent velocities in turn are typically quite comparable to the average flow velocity. Also the size $\ell$ of the largest eddies is typically quite comparable to the transverse thickness of the vortex layer or jet. So the Reynolds number above can be taken to be roughly one based on flow velocity and layer or jet thickness.

How about if you plot the kinetic energy in the entire cascade? In particular, assume that you take a Fourier transform and then plot the total kinetic energy per unit mass and unit wave number range $P$ versus wave number $k$? That is called a power spectrum. (It is a consequence of the orthogonality of Fourier modes that you can write the kinetic energy as a simple sum of the contributions of each individual mode. For Fourier modes, this is called Parseval’s identity.) Assume now that you look at the power spectrum for wave numbers that are so small that the large scale features of the turbulent flow are no longer visible to the eddies, but not so small that dissipation becomes a factor. In that range of wave lengths, called the inertial range, the viscosity $\nu$ is not important and the power spectrum can only depend on $\varepsilon/\rho$. That has units of ${\rm L^2/T^3}$, while the wave number $k$ has units 1/L and the power spectrum $P$ has units ${\rm L^3/T^2}$. The only $\Pi$ group you can form here is

$$\frac{Pk^{5/3}}{(\varepsilon/\rho)^{2/3}}$$

This $\Pi$ group has nothing it can depend on, so it must be a finite constant. That means that $P$ must be proportional to $k^{-5/3}$ in the inertial range. If you plot $P$ versus $k$ using logarithmic axes, the inertial range will be seen to be a straight line segment that slopes down with slope $-5/3$.