6.3 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

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

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

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

We can use that additional knowledge by recasting the previous nondimensional relationship for $\Delta{}u$ in terms of nondimensional ratios that do not vary with $x$ and a remaining one that does. First, we can trivially rewrite the previous nondimensional relationship as

$$\frac{x \Delta u}{I_1} \sqrt{\frac{I_1}{Ux}} = f_2\left(\f... ...{\frac{I_1}{Ux}}, \frac{Ux}{I_1}\right) \sqrt{\frac{I_1}{Ux}}$$

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 a different function $f$ whose first argument is the nondimensional combination for $y$ that does not depend on $x$:

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

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.)