Cohen I.M., Kundu P.K. Fluid Mechanics
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592 Turbulence
10
23
10
28
10
25
10
23
10
21
10
10
3
10
5
10
26
10
24
10
22
1
10
22
10
2
10
21
110
Figure 13.26 Temperature and velocity spectra measured by Grant et al. (1968). The measurements were
made at a depth of 23 m in a tidal passage through islands near the coast of British Columbia, Canada.
Wavenumber K is in cm
−1
. Solid points represent in (
◦
C)
2
/cm
−1
, and open points represent S in
(cm/s)
2
/cm
−1
. Powers of K that fit the observation are indicated by straight lines. O. M. Phillips, The
Dynamics of the Upper Ocean, 1977 and reprinted with the permission of Cambridge University Press.
The average rate at which the magnitude of X
α
increases with time can be found by
finding
d(X
2
α
)/dt, where the overbar denotes ensemble average and not time average.
We can write
d
dt
(
X
2
α
) = 2X
α
dX
α
dt
, (13.75)
where we have used the commutation rule (13.3) of averaging and differentiation.
Defining
u
α
=
dX
α
dt
,
as the Lagrangian velocity component of a fluid particle at time t , equation (13.75)
becomes
15. Taylor’s Theory of Turbulent Dispersion 593
Figure 13.27 Three experimental outcomes of X(t), the current positions of particles from the origin at
time t = 0.
d
dt
(
X
2
α
) = 2X
α
u
α
= 2
t
0
u
α
(t
)dt
u
α
= 2
t
0
u
α
(t
)u
α
(t) dt
, (13.76)
where we have used the commutation rule (13.4) of averaging and integration. We
have also written
X
α
=
t
0
u
α
(t
)dt
,
which is valid as X
α
and u
α
are associated with the same particle. Because the flow is
assumed to be stationary,
u
2
α
is independent of time, and the autocorrelation of u
α
(t)
and u
α
(t
) is only a function of the time difference t − t
. Defining
r
α
(τ ) ≡
u
α
(t)u
α
(t +τ)
u
2
α
,
to be the autocorrelation of Lagrangian velocity components of a particle, equa-
tion (13.76) becomes
d
dt
(
X
2
α
) = 2u
2
α
t
0
r
α
(t
− t)dt
= 2u
2
α
t
0
r
α
(τ ) dτ, (13.77)
594 Turbulence
where we have changed the integration variable from t
to τ = t −t
. Integrating, we
obtain
X
2
α
(t) = 2u
2
α
t
0
dt
t
0
r
α
(τ ) dτ, (13.78)
which shows how the variance of the particle position changes with time.
Another useful form of equation (13.78) is obtained by integrating it by parts.
We have
t
0
dt
t
0
r
α
(τ ) dτ =
t
t
0
r
α
(τ ) dτ
t
t
=0
−
t
0
t
r
α
(t
)dt
= t
t
0
r
α
(τ ) dτ −
t
0
t
r
α
(t
)dt
= t
t
0
1 −
τ
t
r
α
(τ ) dτ.
Equation (13.78) then becomes
X
2
α
(t) = 2u
2
α
t
t
0
1 −
τ
t
r
α
(τ ) dτ. (13.79)
Two limiting cases are examined in what follows.
Behavior for small t:Ift is small compared to the correlation scale of r
α
(τ ), then
r
α
(τ ) 1 throughout the integral in equation (13.78) (Figure 13.28). This gives
X
2
α
(t) u
2
α
t
2
. (13.80)
Taking the square root of both sides, we obtain
X
rms
α
= u
rms
α
tt ᐀, (13.81)
which shows that the rms displacement increases linearly with time and is proportional
to the intensity of turbulent fluctuations in the medium.
Behavior for large t:Ift is large compared with the correlation scale of r
α
(τ ), then
τ/t in equation (13.79) is negligible, giving
X
2
α
(t) 2u
2
α
᐀t, (13.82)
where
15. Taylor’s Theory of Turbulent Dispersion 595
Figure 13.28 Small and large values of time on a plot of the correlation function.
᐀ ≡
∞
0
r
α
(τ ) dτ,
is the integral time scale determined from the Lagrangian correlation r
α
(τ ). Taking
the square root, equation (13.82) gives
X
rms
α
= u
rms
α
√
2᐀tt ᐀. (13.83)
The t
1/2
behavior of equation (13.83) at large times is similar to the behavior in a
random walk, in which the distance traveled in a series of random (i.e., uncorrelated)
steps increases as t
1/2
. This similarity is due to the fact that for large t the fluid
particles have “forgotten” their initial behavior at t = 0. In contrast, the small time
behavior X
rms
α
= u
rms
α
t is due to complete correlation, with each experiment giving
X
α
u
α
t. The concept of random walk is discussed in what follows.
Random Walk
The following discussion is adapted from Feynman et al. (1963, pp. 6–5 and 41–8).
Imagine a person walking in a random manner, by which we mean that there is
no correlation between the directions of two consecutive steps. Let the vector R
n
represent the distance from the origin after n steps, and the vector L represent the nth
step (Figure 13.29). We assume that each step has the same magnitude L. Then
R
n
= R
n−1
+ L,
596 Turbulence
Figure 13.29 Random walk.
which gives
R
2
n
= R
n
·R
n
= (R
n−1
+ L) ·(R
n−1
+ L)
= R
2
n−1
+ L
2
+ 2R
n−1
·L.
Taking the average, we get
R
2
n
= R
2
n−1
+ L
2
+ 2R
n−1
·L. (13.84)
The last term is zero because there is no correlation between the direction of the
nth step and the location reached after n − 1 steps. Using rule (13.84) successively,
we get
R
2
n
= R
2
n−1
+ L
2
= R
2
n−2
+ 2L
2
= R
2
1
+ (n − 1)L
2
= nL
2
.
The rms distance traveled after n uncorrelated steps, each of length L, is therefore
R
rms
n
= L
√
n,
(13.85)
which is called a random walk.
Behavior of a Smoke Plume in the Wind
Taylor’s analysis can be adapted to account for the presence of mean velocity. Consider
the dispersion of smoke into a wind blowing in the x-direction (Figure 13.30). Then a
photograph of the smoke plume, in which the film is exposed for a long time, would
outline the average width Y
rms
. As the x-direction in this problem is similar to time in
15. Taylor’s Theory of Turbulent Dispersion 597
Figure 13.30 Average shape of a smoke plume in a wind blowing uniformly along the x-axis. G. I. Taylor,
Proc. London Mathematical Society 20: 196–211, 1921.
Taylor’s problem, the limiting behavior in equations (13.81) and (13.83) shows that
the smoke plume is parabolic with a pointed vertex.
Effective Diffusivity
An equivalent eddy diffusivity can be estimated from Taylor’s analysis. The equiva-
lence is based on the following idea: Consider the spreading of a concentrated source,
say of heat or vorticity, in a fluid of constant diffusivity. What should the diffusiv-
ity be in order that the spreading rate equals that predicted by equation (13.77)?
The problem of the sudden introduction of a line vortex of strength , considered in
Chapter 9, Section 9, is such a problem of diffusion of a concentrated source. It was
shown there that the tangential velocity in this flow is given by
u
θ
=
2πr
e
−r
2
/4νt
.
The solution is therefore proportional to exp(−r
2
/4νt), which has a Gaussian shape in
the radial direction r, with a characteristic width (“standard deviation”) of σ =
√
2νt.
It follows that the momentum diffusivity ν in this problem is related to the variance
σ
2
as
ν =
1
2
dσ
2
dt
, (13.86)
which can be calculated if σ
2
(t) is known. Generalizing equation (13.86), we can
say that the effective diffusivity in a problem of turbulent dispersion of a patch of
598 Turbulence
particles issuing from a point is given by
κ
e
≡
1
2
d
dt
(
X
2
α
) = u
2
α
t
0
r
α
(τ ) dτ, (13.87)
where we have used equation (13.77). From equations (13.80) and (13.82), the two
limiting cases of equation (13.87) are
κ
e
u
2
α
tt ᐀, (13.88)
κ
e
u
2
α
᐀ t ᐀. (13.89)
Equation (13.88) shows the interesting fact that the eddy diffusivity initially
increases with time, a behavior different from that in molecular diffusion with con-
stant diffusivity. This can be understood as follows. The dispersion (or separation)
of particles in a patch is caused by eddies with scales less than or equal to the scale
of the patch, since the larger eddies simply advect the patch and do not cause any
separation of the particles. As the patch size becomes larger, an increasing range of
eddy sizes is able to cause dispersion, giving κ
α
∝ t. This behavior shows that it is
frequently impossible to represent turbulent diffusion by means of a large but con-
stant eddy diffusivity. Turbulent diffusion does not behave like molecular diffusion.
For large times, on the other hand, the patch size becomes larger than the largest eddies
present, in which case the diffusive behavior becomes similar to that of molecular
diffusion with a constant diffusivity given by equation (13.89).
16. Concluding Remarks
Turbulence is an area of classical fluid mechanics that is the subject of continued
intensive research. Frequent symposia are held to summarize and bring the research
community up-to-date on new results and focus on promising approaches. Some
noteworthy proceedings are listed in the Supplementary Reading section at the end
of this chapter.
Exercises
1. Let R(τ) and S(ω) be a Fourier transform pair. Show that S(ω) is real and
symmetric if R(τ) is real and symmetric.
2. Calculate the mean, standard deviation, and rms value of the periodic time
series
u(t) = U
0
cos ωt +
¯
U.
3. Show that the autocorrelation function
u(t)u(t +τ) of a periodic series
u = U cos ωt is itself periodic.
4. Calculate the zero-lag cross-correlation
u(t)v(t ) between two periodic series
u(t) = cos ωt and v(t) = cos (ωt + φ). For values of φ = 0,π/4, and π/2, plot the
Exercises 599
scatter diagrams of u vs v at different times, as in Figure 13.6. Note that the plot is
a straight line if φ = 0, an ellipse if φ = π/4, and a circle if φ = π/2; the straight
line, as well as the axes of the ellipse, are inclined at 45
◦
to the uv-axes. Argue that
the straight line signifies a perfect correlation, the ellipse a partial correlation, and the
circle a zero correlation.
5. Measurements in an atmosphere at 20
◦
C show an rms vertical velocity of
w
rms
= 1 m/s and an rms temperature fluctuation of T
rms
= 0.1
◦
C. If the correlation
coefficient is 0.5, calculate the heat flux ρC
p
wT
.
6. A mass of 10 kg of water is stirred by a mixer. After one hour of stirring, the
temperature of the water rises by 1.0
◦
C. What is the power output of the mixer in
watts? What is the size η of the dissipating eddies?
7. A horizontal smooth pipe 20 cm in diameter carries water at a temperature
of 20
◦
C. The drop of pressure is dp/dx = 8N/m
2
per meter. Assuming turbu-
lent flow, verify that the thickness of the viscous sublayer is ≈0.25 mm. [Hint: Use
dp/dx = 2τ
0
/R, as given in equation (9.12). This gives τ
0
= 0.4N/m
2
, and therefore
u
∗
= 0.02 m/s.]
8. Derive the logarithmic velocity profile for a smooth wall
U
u
∗
=
1
k
ln
yu
∗
ν
+ 5.0,
by starting from
U =
u
∗
k
ln y + const.
and matching the profile to the edge of the viscous sublayer at y = 10.7 ν/u
∗
.
9. Estimate the Monin–Obukhov length in the atmospheric boundary layer if the
surface stress is 0.1N/m
2
and the upward heat flux is 200 W/m
2
.
10. Consider a one-dimensional turbulent diffusion of particles issuing from a
point source. Assume a Gaussian Lagrangian correlation function of particle velocity
r(τ) = e
−τ
2
/t
2
c
,
where t
c
is a constant. By integrating the correlation function from τ = 0to∞, find
the integral time scale
᐀ in terms of t
c
. Using the Taylor theory, estimate the eddy
diffusivity at large times t/
᐀ 1, given that the rms fluctuating velocity is 1 m/s
and t
c
= 1s.
11. Show by dimensional reasoning as outlined in Section 10 that for
self-preserving flows far downstream, U
∞
− U
e
∼ x
−1/2
,δ ∼
√
x, for a wake,
and U
1
− U
2
= const., δ ∼ x, for a shear layer.
600 Turbulence
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Supplemental Reading 601
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