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 put some of their kinetic energy in creating 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.

Now in a steady state, the total kinetic energy in eddies of a given size must remain constant. So the kinetic energy that the largest eddies give to smaller eddies must be compensated for by kinetic energy that these eddies pick up from the large-scale instability of the flow. Indeed, since turbulent flows are at high Reynolds numbers, laminar viscous dissipation of kinetic energy into heat by large eddies should be negligible. For somewhat smaller eddies, the kinetic energy energy that they pick up from the somewhat larger eddies must similarly be the same as the kinetic energy that they put in the somewhat smaller eddies that they create. The net result is that a constant amount of kinetic energy is transmitted down the eddie scales towards smaller and smaller eddies. We have what is called an energy cascade down the eddy scales.

But eventually, the eddies become so small that viscous dissipation does become important. These smallest eddies then turn the kinetic energy that they get from the larger eddies, not into still smaller eddies, but into heat. In a steady state, the amount of energy cascading down the eddies must then be equal to the dissipation $\varepsilon/\rho$ produced by the smallest eddies. The typical scales of the 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 more precisely, dimensional analysis can be used. Consider 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 these assumptions. 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.

Dimensional analysis is also useful in describing eddies significantly larger than the Kolmogorov scale, but still significantly smaller than the largest eddies. Again the assumption is that these eddies, too, only see the amount of kinetic energy $\varepsilon/\rho$ that the large eddies put in the cascade. They do not see the large scale features of the flow that gives rise to the large eddies. Now suppose we want to look at the kinetic energy that the eddies of a given size have? Now the “kinetic energy for intuitive eddies of a given size” is not a quantity that is easy to translate into mathematics. It is more convenient to look at the kinetic energy that is found in Fourier modes, like say $\sin(kx)$ and $\cos(kx)$, in a given range of period lengths. Now the length of the period is inversely proportional to the wave number $k$, so a range of period lengths corresponds to a range of wave numbers $k$. So we can look at the kinetic energy per unit wave number range as a function of $k$. This function is called the “power spectrum” $P$. Fourier modes are convenient because they are orthogonal; that implies that you can get the total kinetic energy by just summing the kinetic energies of the individual Fourier modes. (For Fourier modes, this is called the Parseval identity.)

Assume now that you look at the power spectrum for wave lengths 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 $k$, being inversely proportional to the wave length, 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 $-5/3$.