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as seen by an observer moving with the fluid at the point). The contribution to the right-hand
side is positive, so that
2
p
tends to become negative near the point considered, and
p
tends
to be amaximum there. Crudely, the rate-of-strain contribution to fluctuating pressure results
from eddy collisions. The vorticity contribution is negative, implying pressure minima,
and results from eddy rotation. The two may be succinctly labeled the “splat” and “spin”
contributions, respectively.
∇
We'll see in
Chapter 5
that turbulent pressure fluctuations play a critical role in
the maintenance of turbulent fluxes, and we'll revisit the form
(2.74)
of the pressure
equation in
Chapter 7
.
2.8 Eddy diffusivity
Molecular diffusion
, a macroscopic effect of molecular collisions in a medium,
produces a flux of a property (e.g., thermal energy, momentum, constituent concen-
tration) in the direction opposite to that of its gradient. This
down-gradient
diffusion
occurs in both fluids and solids. A familiar form is conduction heat transfer, the
molecular diffusion of thermal energy down the temperature gradient.
Although
turbulent diffusion
differs in many ways frommolecular diffusion, it is
convenient and tempting to treat it like molecular diffusion but with a much larger,
“eddy” diffusivity. We can explore this eddy-diffusivity representation through a
simple thought problem.
†
Figure 2.4
shows a tank filled to depth
d
, halfway with dyed water (with dye con-
centration
c
0). The horizontal dimen-
sions of the tank are much greater than
d
, making the problem one-dimensional.
The fluids are motionless and separated by a thin membrane so the boundary
between them is initially perfectly flat and distinct. We imagine that at time zero
the membrane is dissolved and the dye begins to diffuse into the clear fluid.
Here the dye conservation
equation (1.31)
reduces to
=
c
initial
) and the rest with clear water (
c
=
γ
∂
2
c
∂c
∂t
=
∂z
2
.
(2.75)
We can remove the dependence on
c
initial
by defining a new dependent variable
c
∗
c/c
initial
which still satisfies
Eq. (2.75)
.
c
∗
=
is initially 1 in the lower half of
thetankand0above.
One expects that for large time
c
∗
(z, t)
0
.
5. We can infer the
time required to approach this final state as follows.
‡
The only important physical
parameters in the problem are the tank depth
d
and the molecular diffusivity
γ
,so
→
constant
=
†
This problem is motivated by a discussion in
Tennekes and Lumley
(
1972
).
In this simple example of
dimensional analysis
the answer follows immediately on dimensional grounds. In
problems with more physical parameters and dimensions a systematic approach is necessary. We'll discuss it
in
Chapter 10
.
‡
