6 Free Turbulence

(Panton3 23.5, 6, Tennekes & Lumley 4)

Maybe someday an Oracle of Delphi will arise that can actually model turbulence accurately. Do not hold your breath.

Until then, the most solid way to learn some real stuff about turbulence is without doubt dimensional analysis. Dimensional analysis does not require you to actually solve the equations. It just requires you to figure out what is important and what is not.

Some important examples follow.

Mixing layers:

$$\hbox{\epsffile{figures/mix.ps}}$$

Consider a turbulent mixing layer (AKA shear or vortex layer) as shown above. Going downstream, you would expect the velocity profile to be of the generic functional form

$$\overline{u} = F(x,y,U_{\rm ave},\Delta U,\nu,\mbox{initial conditions})$$

where $x$ is the horizontal distance from the starting point of the layer, $y$ the vertical coordinate through the layer, $U_{\rm ave}$ the average of the velocities above and below the layer, and $\Delta U$ the velocity change over the layer.

However, in a free turbulent flow, away from walls, the laminar shear stress is small compared to the turbulent one. This suggests that the laminar viscosity $\nu$ can be ignored in the relation above. (Remember from the energy cascade discussion that the laminar stress is important in the dissipation of turbulent kinetic energy. But it should not be involved in the large-scale mechanics of the turbulent mixing layer, like the instability mechanisms that sustain the big eddies. If $\nu$ changes, it should only affect at which scales the energy cascade gets rid of its energy.)

Also, if you look sufficiently far downstream, the details of the initial conditions should no longer be visible.

Dimensional analysis can now be done for the mean velocity profile $\overline{u}$ above. Selecting $x$ and $\Delta U$ to nondimensionalize the remaining variables gives

$$\fbox{$\displaystyle \mbox{Mixing layer:}\quad \frac{\ove... ... = f\left(\frac{y}{x},\frac{U_{\rm ave}}{\Delta U}\right) $}$$

But the second argument of $f$ above is a constant, independent of $x$ for a given mixing layer. So it follows that the mixing layer velocity profile is similar, and that the layer has a typical thickness $\delta$ proportional to $x$.

Exercise:

Discuss how well that seems to agree with flow visualizations.

$\bullet$

Note that while in a boundary layer approximation the mixing layer can reasonably be approximated as relatively thin compared to its streamwise extend, this approximation does not improve with streamwise distance.

Jets:

$$\hbox{\epsffile{figures/jet.ps}}$$

Now consider a turbulent jet, as shown above. Here you would expect that downstream

$$\overline{u} = F(x,y,\nu,\mbox{initial conditions})$$

Again, $\nu$ can be ignored, assuming that the turbulent Reynolds number is large enough.

Also, it can again be assumed that the details of the initial conditions become invisible sufficiently far downstream, with one exception. Integral momentum conservation between any two downstream positions of constant $x$ implies that the $x$-momentum flow integral

$$\int \rho {\overline{u}}^2 \mbox{d}A$$

must be the same at the two stations. (Pressure differences between stations sufficiently far downstream can be ignored). So the momentum flow integral above is a constant. It is determined by the strength of the jet that the initial conditions generated. Since it is constant, it cannot become invisible. Note that in the incompressible case, you can more simply assume that

$$I_0 \equiv \int {\overline{u}}^2 \mbox{d}A$$

is constant. Also note that $I_0$ has units ${\rm L^3/T^2}$ for a two dimensional jet, where $\mbox{d}A = \mbox{d}y$, but units ${\rm L^4/T^2}$ for a three-dimensional jet, for which $\mbox{d}A = r \mbox{d}r \mbox{d}\theta$.

So the functional dependence can be simplified to

$$\overline{u} = F(x,y,I_0)$$

and dimensional analysis then produces

$$\mbox{2D: } \frac{\overline{u}\sqrt{x}}{\sqrt{I_0}} = f\le... ... \frac{\overline{u}x}{\sqrt{I_0}} = f\left(\frac{y}{x}\right)$$

Cleaning this up gives:

$$\fbox{$\displaystyle \mbox{2D jet:}\quad \overline{u} = \... ...erline{u} = \frac{\sqrt{I_0}}{x} f\left(\frac{y}{x}\right) $}$$

Like for the mixing layer, the mean velocity profiles are similar, and the jet thickness is proportional to $x$. But in the two-dimensional case, the maximum jet velocity decays proportional to $1/\sqrt{x}$, slower than the $1/x$ of the three-dimensional case.

Wakes:

$$\hbox{\epsffile{figures/wake.ps}}$$

Finally consider the wake of some body, as shown above. Here you would expect that downstream

$$\Delta u = F(x,y,U,\nu,\mbox{initial conditions}) \qquad \mbox{where } \Delta u \equiv U - \overline{u}$$

Again, $\nu$ can be ignored. Also, it can again be assumed that the details of the initial conditions become invisible sufficiently far downstream, with the exception that

$$I_0 \equiv \int {\overline{u}}^2 \mbox{d}A = \int (U -\Delta u)^2 \mbox{d}A$$

cannot become invisible because it is constant. If $\Delta u$ has become small enough, we can expand the square and ignore the $(\Delta u)^2$ term to give that

$$\int U^2 \mbox{d}A - 2 \int U \Delta u \mbox{d}A$$

is constant. Since the first term is just a constant too, more simply

$$I_1 = \int \Delta u \mbox{d}A$$

must be constant. In two dimensions this has units ${\rm L^2/T}$ and in three ${\rm L^3/T}$.

The functional relationship simplifies to

$$\Delta u = F(x,y,U,I_1)$$

and dimensional analysis then gives

$$\mbox{2D: } \frac{\Delta u x}{I_1} = f_2\left(\frac{y}{x},... ...lta u x^2}{I_1} = f_2\left(\frac{y}{x},\frac{Ux^2}{I_1}\right)$$

Because of the second argument of function $f_2$, there is little useful knowledge that we can get from this.

However, we can supplement the dimensional analysis with what we believe to be true about the turbulence. Consider the two-dimensional Reynolds-averaged $x$-momentum equation in boundary layer approximation:

$$\overline{u} \frac{\partial \overline{u}}{\partial x} + \o... ...{\partial\overline{\tau_{xy}}+\tau_{xy}^{\rm {R}}}{\partial y}$$

The average laminar stress should be negligible and when $\Delta u$ has become small enough compared to $U$, we can also approximate the left hand side to give

$$- U \frac{\partial \Delta u}{\partial x} = \frac{1}{\rho} ... ... y} \qquad \frac{\tau_{xy}^{\rm {R}}}{\rho} = \overline{u'v'}$$

(Note that $\overline{v}\partial/\partial y$ should, based on continuity, be of the same order as $\overline{\Delta u}\partial/\partial x$, hence negligible compared to the retained term.)

If we ballpark the two sides in the simplified momentum equation above,

$$\frac{U\Delta u}{x} \sim \frac{(\Delta u)^2}{\delta}$$

or rearranged

$$\frac{\Delta u}{\delta} \sim \frac{U}{x}$$

In particular, there would be a problem in reasonably balancing the momentum equation if $\Delta u/\delta$ would be proportional to a different power of $x$ for large $x$ than $x^{-1}$. In addition, since $I_1$ is constant, in two dimensions we also have the constraint that $\Delta u \delta$ must stay of order $x^0$. Combining the two, in two dimensions, as far as powers of $x$ are concerned, we must have that

$$\Delta u \sim \frac{1}{\sqrt{x}} \qquad \delta \sim \sqrt{x}$$

Now we trivially rewrite the previous relationship for $\Delta u$ as

$$\frac{\Delta u x}{I_1} \sqrt{\frac{I_1}{Ux}} = \sqrt{\frac... ...t{\frac{Ux}{I_1}}\sqrt{\frac{I_1}{Ux}}, \frac{Ux}{I_1}\right)$$

Here we have cleverly formed nondimensional combinations for $\Delta u$ and $y$ that should be independent of $x$ for large $x$. But the right hand side can be written as just a different function $f$:

$$\frac{\Delta u x}{I_1} \sqrt{\frac{I_1}{Ux}} = = f\left(\frac{y}{x}\sqrt{\frac{Ux}{I_1}},\frac{Ux}{I_1}\right)$$

Function $f$ is not a new function, it is fully determined by $f_2$: if you know the arguments of $f$, you can compute those of $f_2$ and then $f$. But function $f$ cannot depend on its second argument, or else the shown nondimensional combinations for $\Delta u$ and $y$ would not be independent of $x$ as they should.

Cleaning up then gives

$$\fbox{$\displaystyle \mbox{2D wake:}\quad \frac{\Delta u}... ...c{I_1}{Ux}} f\left(\frac{y}{x}\sqrt{\frac{Ux}{I_1}}\right) $}$$

It may be noted that if substitute the above similar profile into the simplified momentum equation above, assuming some suitable constant eddy viscosity $\nu_{\rm {T}}$ for the viscous term, you can find the profile. You will get a reasonable approximation to the actual measured wake velocity profile except near the outer edges. Note that near the outer edges the flow is only part of the time truly turbulent, as near the outer edges turbulent eddies engulf regions of potential flow fluid. The intermittency $\gamma$ is defined as the fraction of time that the flow is turbulent. It turns out that if you assume that $\nu_{\rm {T}}=\nu_{\rm {T,center}}\gamma$, with $\nu_{\rm {T,center}}$ a suitable constant, you can get very good agreement with the experimental profile.

In three dimensions, accounting for the different $I_1$,

$$\fbox{$\displaystyle \mbox{3D wake:}\quad \frac{\Delta u}... ...x^2}} f\left(\frac{y}{x}\sqrt[3]{\frac{Ux^2}{I_1}}\right) $}$$

Note that in this case, the Reynolds number based on $\Delta u$ and $\delta$ is proportional to $x^{-1/3}$, so it decreases with $x$. Therefore the made assumption of high Reynolds number will eventually break down, and the wake will even become laminar. The result above assumes that you do not look that far downstream. (The two-dimensional wake above and the three-dimensional jet have Reynolds numbers that stay constant with $x$, so the approximation of high Reynolds number, while qualitatively quite reasonable, does not improve with $x$ in those cases.)