6.2 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.