turbulence physics and modeling lectures

Bharathram GanapathisubramaniRoom 5059, Bldg 13Ph: 22305Email: [email protected]

Turbulence

courtesy: S. Lardeau

Turbulence: Physics and Modelling

Richard SandbergRoom 5007, Bldg 13

Ph: 27386Email: [email protected]

Lectures:

Tuesdays at 2PM: Bldg 13, Room 3019Wednesdays at 10AM: Bldg 7, Room 3021Fridays at 10AM: Bldg 29, Room 2075

mailto:[email protected]

mailto:[email protected]?subject=

Turbulence


Focus: Physics of incompressible turbulent flows and strategies available to address turbulence closure problem

Key outcome:-Insight in to the nature of turbulence-Knowledge of tools and concepts used for its analysis

Useful texts:- Turbulence for the 21st Century, W. K. George

(http://turbulence-online.com) - Pope, Cambridge University Press - Tennekes and Lumley, MIT press - Davidson, Oxford University Press - D C Wilcox: Turbulence Modeling for CFD, 2nd ed., DCW Industries,

1998- R. Pletcher, J. Tannehill, D.Anderson, Computational fluid

mechanics and heat transfer, CRC Press, 2012

http://turbulence-online.com

Turbulence


House-keeping

70% from exam15% coursework on processing turbulence data15% coursework on numerical modelling project

Three lectures a week throughout (36 lectures in all)

First 12 lectures by me on turbulence tools and theoryNext 16 lectures from Richard on numerical modellingFinal 6 lectures from me on miscellaneous/recent advances

Turbulence


The nature of turbulence Its origin - stability Mean and fluctuation quantities General features and major effects

Tools for studying turbulence Definitions: stationarity and ergodicity

Statistical tools: Amplitude statistics: probability density function and moments Time-domain statistics: correlation and spectral functions.

Equations and scales of motion Reynolds averaging; momentum and Reynolds stress equations Turbulence energy equation

Transport, Production and Dissipation of kinetic energy including Kolmogorov’s hypothesis

Turbulence


Computational strategies: hierarchy of turbulence closures Engineering/CFD models: RANS schemes 'Numerical experiments': DNS and LES Hybrid RANS/LES: DES Survey of relevance and limitations

Reynolds-Averaged Navier-Stokes (RANS) models Motivation, philosophy, and classification Modelling canonical flows: Homogeneous isotropic turbulence, 2D wall layers Industrial models: Complex geometry, pressure gradients/separation, surface roughness, mean three-dimensionality, compressibility Eddy-viscosity/Boussinesq closures: Algebraic/zero-, one- and two-equation models Example: anatomy of the Spalart-Allmaras one-equation scheme Limitations of eddy-viscosity closures Guidelines for CFD users

Turbulence


Direct Numerical Simulations (DNS) Numerical issues:

Domain size, resolution requirements Spectral and high-order finite-difference methods FFTs, aliasing, modified wavenumber analysis Examples of classical and recent simulations: plane-channel, 2D and 3D turbulent boundary layers separation bubbles, shock-boundary layer interaction Recent developments: Inflow treatments, and immersed-boundary techniques

Large Eddy Simulations (LES) Implicit and explicit filtering; subgrid-scale (SGS) modelling Recent developments: dynamic modelling, approximate deconvolution Inherent limitations of LES

New frontiers Unsteady RANS Hybrid RANS/LES methods

Turbulence


Equations and scales of motion Reynolds averaging; momentum and Reynolds stress equations Turbulence energy equation

Transport, Production and Dissipation of kinetic energy including Kolmogorov’s hypothesis

Vorticity and enstrophy; vortex stretching Invariants of velocity gradient tensor

Canonical turbulent flows Free flows:

Homogeneous isotropic turbulence and homogeneous shear flow Self-preserving jets, wakes and mixing layers Wall flows: 2D channel flow smooth- and rough-wall boundary layers

Turbulence


Mean-flow similarity Self-preserving jets, wakes and mixing layers

Fine-scale features Vorticity and enstrophy; vortex stretching

Invariants of velocity gradient tensor

Revisions (2 lectures)

Turbulence

“Violent or unsteady movement of air, water or any other fluid”

-Oxford Dictionary

Turbulence

The Great Wave off Kanagawa The Dragon of Smoke Escaping from Mt Fuji.

Hokusai ca 1830

“Violent or unsteady movement of air, water or any other fluid”

-Oxford Dictionary

Wednesday, 23 October 13

Examples of Turbulence

•Shear-generated turbulent flow


•Shear-generated turbulent flowFast stream adjacent to a slow stream

Courtesy: S. Lardeau



•Wall-bounded turbulent flowThe presence of solid wall creates shear


•Wall-bounded turbulent flowThe presence of solid wall creates shear



•Wake-generated turbulent flow

Guadalupe IslandSelkirk Island

Courtesy: G. Fishpool & M. Leschziner


•Wake-generated turbulent flow

Guadalupe IslandSelkirk Island

Courtesy: G. Fishpool & M. Leschziner



•Buoyancy-generated turbulent flow


•Atmospheric turbulence

A combination of shear, wall-bounded, buoyant and wake turbulence


•Atmospheric turbulence

A combination of shear, wall-bounded, buoyant and wake turbulence



•Galactic, Planetary or Stellar turbulence


•Galactic, Planetary or Stellar turbulence


What is Turbulence?

“Turbulenza” - Leonardo DaVinci

“Observe the motion of the surface of the water, which resembles that of hair, which has two motions, of which one is caused by the weight of the hair, the other by the direction of the curls; thus the water has eddying motion, one part of which is due to the principal current, the other to random and reverse motion”

Early View

What makes the flow turbulent?

translated by U. Piomelli

What determines the state of the flow?

Laminar Transitional Turbulent

Forces acting on the fluid particles

•Inertial Forces •Viscous Forces

the flow becomes turbulentAs Reynolds number ,


Inertial ForcesViscous Forces

=Reynolds Number (Re) =

What determines the state of the flow?

Laminar Transitional Turbulent

Forces acting on the fluid particles

•Inertial Forces •Viscous Forces

Inertial ForcesViscous Forces

ULµ=Reynolds Number (Re) =

the flow becomes turbulentAs Reynolds number ,



First carried out by O. Reynolds circa 1895

Source: Google videos

Laminar

Low Re

Instabilities

Moderate Re

Turbulent

High Re


Pipe Flow

Properties of Turbulence

However, there is some degree of organisation“Big whorls have little whorls that feed on their velocity, and little whorls have lesser whorls and so on to viscosity”

-L.F. Richardson, 1922

“Chaotic” and unsteady with wide range of interacting eddies











Reynolds number also indicates ratio of largest to smallest scales


Higher Re, greater the range of scales

Most processes are highly non-linear and non-local with multiscale interactions

Turbulence

“Last unsolved problem in classical physics”

A wide variety of engineering applications• Drag, energy loss fuel consumption, range, economy• Maximum lift, stall landing, take-off speed, weight• Aircraft, cars, trucks, trains, ships • Noise – from aircraft, road vehicle and engine• Combustion, engine emissions• Vibrations, buffet, flutter• Mixing, dispersion• Erosion• Heat transfer, heating, cooling

Understanding, Predicting and Controlling Turbulence is of enormous importance

Turbulence: Engineering Applications

Higher Skin Friction Drag in most transport systems

Aircrafts 50% Ships 70% Gas Pipelines 90%

Source: Fishpool & Leschziner


Laminar Turbulent

y

U

Much higher drag






Laminar Turbulent

y

U

Much higher drag






Separation leads to higher form drag


Turbulence and Acoustics

ONERA

ONERA

ONERA

Ali Mani, Meng Wang and Parviz Moin, CTR, Stanford University


Turbulence and Mixing

ONERA

ONERA

Kewcharoenwong, Rossi & Vassilicos

Turbulence: lecture

In this module, we limit ourselves to the following conditions

We consider only continuum flows of an incompressible, single-phase, Newtonian fluid

Continuum: Model fluid from a macroscopic view rather than the microscopic viewpoint

We assume that the fluid flow scales are much larger than the mean free path of the molecular motions


Turbulence: lecture



Continuum: Model fluid from a macroscopic view rather than the microscopic viewpoint

We assume that the fluid flow scales are much larger than the mean free path of the molecular motions

Kn =

L<<1

Characterised by Knudsen number,




Turbulence: lecture


Incompressible: If entropy is constant along streamline then: dρ/ρ=-M2du/u (from dp +ρudu =0 and ∂ρ/∂p|s = a2) so that fractional changes in density are small compared to fractional changes in velocity, provided M2«1

Relative changes in density is small - Incompressible

Turbulence: lecture


Two-phase flows are what they seem – mixtures of two phases (steam, oil & water, &c). Many engineering and environmental flows are genuinely two-phase, but the complications are considerable and we avoid them here.

Turbulence in mono-phase flows will be complex enough!


Turbulence: lecture


Newtonian fluids are those with linear relation between stress τij and rate of strain Sij, with

Viscous stress Rate of strain

Note: We will try to use tensor notation (also known as Einstein summation convention most times)


Turbulence: lecture

source: Turbulence for the 21st century, W. K. George

Turbulence: lecture

source: Turbulence for the 21st century, W. K. GeorgeWednesday, 23 October 13

Turbulence: lectureTurbulence: lecture

42 CHAPTER 3. REYNOLDS AVERAGED EQUATIONS

term in brackets) is zero. Thus for incompressible flows, the mass conservationequation reduces to:

Dρ

Dt=

∂ρ

∂t+ uj

∂ρ

∂xj= 0 (3.3)

From equation 3.2 it follows that for incompressible flows,

∂uj

∂xj= 0 (3.4)

The viscous stresses (the stress minus the mean normal stress) are represented

by the tensor T (v)ij . From its definition, T (v)

kk = 0. In many flows of interest, thefluid behaves as a Newtonian fluid in which the viscous stress can be related tothe fluid motion by a constitutive relation of the form

T (v)ij = 2µ

!sij −

1

3skkδij

"(3.5)

The viscosity, µ, is a property of the fluid that can be measured in an independentexperiment. sij is the instantaneous strain rate tensor defined by

sij ≡1

2

#∂ui

∂xj+

∂uj

∂xi

$

(3.6)

From its definition, skk = ∂uk/∂xk. If the flow is incompressible, skk = 0 and theNewtonian constitutive equation reduces to

T (v)ij = 2µsij (3.7)

Throughout this text, unless explicity stated otherwise, the density, ρ = ρ andthe viscosity µ will be assumed constant. With these assumptions, the instanta-neous momentum equations for a Newtonian fluid reduce to:

#∂ui

∂t+ uj

∂ui

∂xj

$

= −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

(3.8)

where the kinematic viscosity, ν, has been defined as:

ν ≡ µ

ρ(3.9)

Note that since the density is assumed constant, the tilde is no longer necessary.Sometimes it will be more instructive and convenient to not explicitly include

incompressibility in the stress term, but to refer to the incompressible momentumequation in the following form:

ρ

#∂ui

∂t+ uj

∂ui

∂xj

$

= − ∂p

∂xi+

∂T (v)ij

∂xj(3.10)

This form has the advantage that it is easier to keep track of the exact role of theviscous stresses.

Conservation of mass for incompressible flows

@u

@x

+@v

@y

+@w

@z

= 0

In tensor notation, you sum over repeated indicies,

@u1

@x1+

@u2

@x2+

@u3

@x3= 0


Turbulence: lecture


Requirements for turbulence

P

As P increases, the deflection in the beam with increase

What happens when P is excessive?

The beam breaks!

Cannot predict this: Because we use linearised equations

Same thing happens repeatedly in fluids

Turbulence: lectureTurbulence: lecture


(a) The nonlinear terms in the N-S equations are what lead to the energy cascade to higher

frequency (smaller scale) components and thus instabilities ⇒ turbulence. We shall have

much more to say on this topic in what follows.

Example (to show effect of nonlinear terms):

Consider this simple nonlinear equation (1D Burgers equation):

0=+xuu

tu

∂∂

∂∂ ,

and suppose that at t=to, u(x,to)=Acos(kx). For small t-to the solution can be written

Notice the smaller spatial scales (i.e. higher wavenumbers) introduced by the nonlinear term.

(Recall sin(kx)cos(kx) = (1/2)sin(2kx).)

Exercise: Repeat this example for a 2D case (i.e. including v∂u/∂y in the equation plus extra

equation for v) and assume u=Acos(kx)cos(ly), v=Bcos(k’x)cos(l’y), to show that in general

quadratic nonlinearities generate sums and differences of the original wavenumbers (k,l,k’,l’).

Hence the velocity field’s Fourier components interact.

(b) Generally, a high enough Reynolds number (Re) or, equivalently for convective flows, a

high enough Raleigh number (Ra), is required to make the viscous term in the N-S equations

small enough compared with the nonlinear terms to allow instabilities to grow. Further

details are given in accompanying Handout.

. . . )cos()sin()()cos(

. . . )(),(

. . . )(),(),(

2 +−+=

+"#

$%&

'−−=

+−+=

kxkxkAttkxA

xu

utttxu

tutttxutxu

o

too

too

o

o

∂∂

∂∂



Turbulence: lecture

Requirements for turbulenceTurbulence: lecture




Dρ

Dt=

∂ρ

∂t+ uj

∂ρ

∂xj= 0 (3.3)


∂uj

∂xj= 0 (3.4)




T (v)ij = 2µ

!sij −

1

3skkδij

"(3.5)


sij ≡1

2

#∂ui

∂xj+

∂uj

∂xi

$

(3.6)


T (v)ij = 2µsij (3.7)


#∂ui

∂t+ uj

∂ui

∂xj

$

= −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

(3.8)


ν ≡ µ

ρ(3.9)



ρ

#∂ui

∂t+ uj

∂ui

∂xj

$

= − ∂p

∂xi+

∂T (v)ij

∂xj(3.10)


Non-dimensionalise this equation, using a representative U , , L and P

"@u

i

@t

+ u

j@u

i

@x

j

#= @p

@x

i

+

UL

@

2u

i

@x

i2

* quantities are non-dimensional"@u

i

@t

+ u

j@u

i

@x

j

#= @p

@x

i

+1

Re

@

2u

i

@x

i2

High Reynolds number prevents viscous damping of instabilitiesWednesday, 23 October 13

Turbulence: lecture

General features of turbulence

•Chaotic - seemingly random but statistics are reproducible

•Wide range of spatial and time scales. The range increases with increasing Reynolds numbers.

•It is still a continuum phenomena - The smallest scale of motion is still much larger than the mean-free path of molecular motions.

•Nonlinearity ensures that long term evolution is sensitive to very small changes.

Turbulence: lecture


Turbulence: lecture


•It is rotational -i.e. contains vorticity

! = r u

We can take the curl of the NS equations and arrive at the vorticity transport equation

1.3 General features of turbulence

1. It is a ‘random’ process – time and space dependent with an infinite (or, at least,

very large) number of degrees of freedom. Unpredictable in detail but its statistical

properties are usually reproducible (more later). Note that its inherent unsteadiness

does not mean that the mean flow cannot be steady – it often is. By a steady flow,

we generally imply that the flow’s statistical properties are independent of time and

can be found by suitable averaging over sufficiently large times.

2. Contains a wide range of spatial and time scales – smaller ones ‘living inside’ bigger

ones. And, as we will see, this range increases with Re. Hence it is extremely

difficult to calculate by direct application of (1.1). Quantify later.

3. It is rotational – i.e. it contains vorticity. Recall that ω = ∇×U and that taking a

derivative ‘amplifies’ the small scale variations. Note that the ‘vorticity transport’

equation (found by taking the curl of the N-S eq.) is:

[1] [2] [3]

where [1] is convection of vorticity, [2] is stretching or tilting of vorticity (often referred

to as ‘vortex stretching’) and [3] is viscous diffusion. (We have ignored the vorticity

generation term that exists in stratified flow.) Velocity derivatives are dominated by

the smallest scales (more later).

Note that (1.4) can be written in vector form:

ωνωω 2∇+∇•= UDtD

)4.1(2

jj

i

j

ij

j

ij

i

xxxu

xu

t ∂∂ω∂

ν∂∂

ω∂∂ω

∂∂ω

+=+


Turbulence: lecture


Turbulence: lecture


•It is rotational -i.e. contains vorticity

! = r u















[1] [2] [3]







)4.1(2

jj

i

j

ij

j

ij

i

xxxu

xu

t ∂∂ω∂

ν∂∂

ω∂∂ω

∂∂ω

+=+


Turbulence: lecture















[1] [2] [3]







)4.1(2

jj

i

j

ij

j

ij

i

xxxu

xu

t ∂∂ω∂

ν∂∂

ω∂∂ω

∂∂ω

+=+

[1] - convection of vorticity[3] - diffusion of vorticity

[2] - Tilting or stretching of vorticityAlso known as vortex stretching. This happens when the rate of strain has a component along vorticity.This is crucial as this is responsible for production of enstrophy and hence sustaining turbulence (more later)


Turbulence: lecture


•It dissipates energy - via the cascade process, which transfers energy from the large- to the small-scales (more later)

•It is intermittent.

•It is intrinsically 3D. This becomes obvious when we examine the vorticity transport equation. For 2D flows, there is no vortex stretching. Therefore, no cascade and hence no production of enstrophy (exercise - show to self)

Turbulence: lecture

Elements of statistical analysis

Ensemble average


The key is the events must be independent

Turbulence: lecture


Ensemble average

20 CHAPTER 2. THE ELEMENTS OF STATISTICAL ANALYSIS

2.2 The Ensemble and Ensemble Averages

2.2.1 The mean (or ensemble) average

The concept of an ensemble average is based upon the existence of independentstatistical events. For example, consider a number of individuals who are simul-taneously flipping unbiased coins. If a value of one is assigned to a head and thevalue of zero to a tail, then the arithmetic average of the numbers generated isdefined as:

XN =1

NΣxn (2.1)

where our nth flip is denoted as xn and N is the total number of flips.Now if all the coins are the same, it doesn’t really matter whether we flip

one coin N times, or N coins a single time. The key is that they must all beindependent events — meaning the probability of achieving a head or tail in agiven flip must be completely independent of what happens in all the other flips.Obviously we can’t just flip one coin once and count it N times; these clearlywould not be independent events.

Exercise Carry out an experiment where you flip a coin 100 times in groups of10 flips each. Compare the values you get for X10 for each of the 10 groups, andnote how they differ from the value of X100.

Unless you had a very unusual experimental result, you probably noticed thatthe value of the X10’s was also a random variable and differed from ensemble toensemble. Also the greater the number of flips in the ensemble, the closer you gotto XN = 1/2. Obviously the bigger N , the less fluctuation there is in XN .

Now imagine that we are trying to establish the nature of a random variable,x. The nth realization of x is denoted as xn. The ensemble average of x is denotedas X (or ⟨x⟩), and is defined as

X = ⟨x⟩ ≡ limN→∞

1

N

N!

n=1

xn (2.2)

Obviously it is impossible to obtain the ensemble average experimentally, sincewe can never have an infinite number of independent realizations. The most wecan ever obtain is the arithmetic mean for the number of realizations we have.For this reason the arithmetic mean can also referred to as the estimator for thetrue mean or ensemble average.

Even though the true mean (or ensemble average) is unobtainable, nonetheless,the idea is still very useful. Most importantly, we can almost always be sure theensemble average exists, even if we can only estimate what it really is. Note thatin some particularly difficult cases this may require imagining that we can sum ourrandom variable at a given instant and time across an infinite number of universeswhich are governed by the same statistical rules. So the fact of the existence ofthe ensemble average does not always mean that it is easy to obtain in practice.


The key is the events must be independent


Turbulence: lecture


Ensemble average


Turbulence: lecture


Ensemble average






XN =1

NΣxn (2.1)






X = ⟨x⟩ ≡ limN→∞

1

N

N!

n=1

xn (2.2)







XN =1

NΣxn (2.1)






X = ⟨x⟩ ≡ limN→∞

1

N

N!

n=1

xn (2.2)




Turbulence: lecture


Ensemble average


Unless stated otherwise, all analyses will utilise the concept of ensemble average. This means that we need to be aware of or take in to account the “statistical differences” between the true mean and the estimates of mean

Turbulence: lecture


Ensemble average


Turbulence: lecture


Ensemble average


2.2. THE ENSEMBLE AND ENSEMBLE AVERAGES 21

Nonetheless, unless stated otherwise, all of the theoretical deductions in this bookwill use this ensemble average; and therefore are completely general. Obviouslythis will mean we have to account for these “statistical differences” between truemeans and estimates of means when comparing our theoretical results to actualmeasurements or computations.

In general, the xn could be realizations of any random variable. The X definedby equation 2.2 represents the ensemble average of it. The quantityX is sometimesreferred to as the expected value of the random variable x, or even simply its mean.

For example, the velocity vector at a given point in space and time, x, t, in agiven turbulent flow can be considered to be a random variable, say ui(x, t). If

there were a large number of identical experiments so that the u(n)i (x, t) in each of

them were identically distributed, then the ensemble average of u(n)i (x, t) would

be given by

⟨ui(x, t)⟩ = Ui(x, t) ≡ limN→∞

1

N

N!

n=1

u(n)i (x, t) (2.3)

Note that this ensemble average, Ui(x, t), will, in general, vary with the inde-pendent variables x and t. It will be seen later that under certain conditionsthe ensemble average is the same as the average which would be generated byaveraging in time, or even space. But even when a time (or space) average isnot meaningful, however, the ensemble average can still be defined; e.g., as in annon-stationary or periodic flow. Only ensemble averages will be used in the de-velopment of the turbulence equations in this book unless otherwise stated. Thusthe equations derived will be completely general, and quite independent of theparticular nature of the flow, or even its statistical character.

2.2.2 Fluctuations about the mean

It is often important to know how a random variable is distributed about themean. For example, Figure 2.1 illustrates portions of two random functions oftime which have identical means, but are obviously members of different ensemblessince the amplitudes of their fluctuations are not distributed the same. It ispossible to distinguish between them by examining the statistical properties ofthe fluctuations about the mean (or simply the fluctuations) defined by:

x′ = x−X (2.4)

It is easy to see that the average of the fluctuation is zero, i.e.,

⟨x′⟩ = 0 (2.5)

On the other hand, the ensemble average of the square of the fluctuation isnot zero. In fact, it is such an important statistical measure we give it a specialname, the variance, and represent it symbolically by either var[x] or ⟨(x′)2⟩. The


Turbulence: lecture


Fluctuations about the mean

Possible to distinguish between these two signals by looking at

the fluctuations


Turbulence: lecture


Fluctuations about the mean22 CHAPTER 2. THE ELEMENTS OF STATISTICAL ANALYSIS

Figure 2.1: A typical random function of time with non-zero mean value.

variance is defined as:

var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)

Note that the variance, like the ensemble average itself, can never really be mea-sured, since it would require an infinite number of members of the ensemble.

It is straightforward to show from equation 2.2 that the variance in equation 2.6can be written as:

var[x] = ⟨x2⟩ −X2 (2.8)

Thus the variance is the second-moment minus the square of the first-moment (ormean). In this naming convention, the ensemble mean is the first moment.

Exercise Use the definitions of equations 2.2 and 2.7 to derive equation 2.8.

The variance can also referred to as the second central moment of x. Theword central implies that the mean has been subtracted off before squaring andaveraging. The reasons for this will be clear below. If two random variables areidentically distributed, then they must have the same mean and variance.

The variance is closely related to another statistical quantity called the stan-dard deviation or root mean square (rms) value of the random variable x,which is


the fluctuations







be given by


1

N

N!

n=1

u(n)i (x, t) (2.3)




x′ = x−X (2.4)


⟨x′⟩ = 0 (2.5)



The signals can be distinguished by calculating the variance


Turbulence: lecture





var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)






the fluctuations







be given by


1

N

N!

n=1

u(n)i (x, t) (2.3)




x′ = x−X (2.4)


⟨x′⟩ = 0 (2.5)





Turbulence: lecture





var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)






the fluctuations







be given by


1

N

N!

n=1

u(n)i (x, t) (2.3)




x′ = x−X (2.4)


⟨x′⟩ = 0 (2.5)





Turbulence: lecture





var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)






the fluctuations







be given by


1

N

N!

n=1

u(n)i (x, t) (2.3)




x′ = x−X (2.4)


⟨x′⟩ = 0 (2.5)






Turbulence: lecture




Turbulence: lecture







var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)






Turbulence: lecture




Turbulence: lecture







var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)








var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)






Figure 2.2: Two random functions of time having the same mean and variance,but very different higher moments.

denoted by the symbol, σx. Thus,

σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].

2.2.3 Higher moments

Figure 2.2 illustrates two random variables of time which have the same meanand also the same variances, but clearly they are still quite different. It is useful,therefore, to define higher moments of the distribution to assist in distinguishingthese differences.

The m-th moment of the random variable is defined as:

⟨xm⟩ = limN→∞

1

N

N!

n=1

xmn (2.10)

It is usually more convenient to work with the central moments defined by:

⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)


Turbulence: lecture







var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)








var[x] ≡ ⟨(x′)2⟩ = ⟨[x−X]2⟩ (2.6)

= limN→∞

1

N

N!

n=1

[xn −X]2 (2.7)



var[x] = ⟨x2⟩ −X2 (2.8)








σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)


Turbulence: lecture


Higher moments


Two signals with the same mean and variance

Turbulence: lecture


Higher moments





σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)




σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)





σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)


Turbulence: lecture


Higher moments





σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)




σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)





σx ≡ (var[x])1/2 (2.9)

or σ2x = var[x].





1

N

N!

n=1

xmn (2.10)


⟨(x′)m⟩ = ⟨(x−X)m⟩ = limN→∞

1

N

N!

n=1

[xn −X]m (2.11)


Turbulence: lecture


Probability


Turbulence: lecture


Probability



The central moments give direct information on the distribution of the values ofthe random variable about the mean. It is easy to see that the variance is thesecond central moment (i.e., m = 2).

2.3 Probability

2.3.1 The histogram and probability density function

The frequency of occurrence of a given amplitude (or value) from a finite num-ber of realizations of a random variable can be displayed by dividing the rangeof possible values of the random variables into a number of slots (or windows).Since all possible values are covered, each realization fits into only one window.For every realization a count is entered into the appropriate window. When allthe realizations have been considered, the number of counts in each window isdivided by the total number of realizations. The result is called the histogram(or frequency of occurrence diagram). From the definition it follows immediatelythat the sum of the values of all the windows is exactly one.

The shape of a histogram depends on the statistical distribution of the randomvariable, but it also depends on the total number of realizations, N , and the sizeof the slots, ∆c. The histogram can be represented symbolically by the functionHx(c,∆c,N) where c ≤ x < c + ∆c, ∆c is the slot width, and N is the numberof realizations of the random variable. Thus the histogram shows the relativefrequency of occurrence of a given value range in a given ensemble. Figure 2.3illustrates a typical histogram. If the size of the sample is increased so that thenumber of realizations in each window increases, the diagram will become lesserratic and will be more representative of the actual probability of occurrence ofthe amplitudes of the signal itself, as long as the window size is sufficiently small.

If the number of realizations, N , increases without bound as the window size,∆c, goes to zero, the histogram divided by the window size goes to a limitingcurve called the probability density function, Bx(c). That is,

Bx(c) ≡ limN → ∞∆c → 0

H(c,∆c,N)/∆c (2.12)

Note that as the window width goes to zero, so does the number of realizationswhich fall into it, NH. Thus it is only when this number (or relative number) isdivided by the slot width that a meaningful limit is achieved.

The probability density function (or pdf) has the following properties:

• Property 1:Bx(c) > 0 (2.13)

always.


Turbulence: lecture


Probability


2.3. PROBABILITY 25

Figure 2.3: Histogram, together with it’s limiting probability density function.

• Property 2:Probc < x < c+ dc = Bx(c)dc (2.14)

where Prob is read “the probability that”.

• Property 3:

Probc < x =! x

−∞Bx(c)dc (2.15)

• Property 4: ! ∞

−∞Bx(x)dx = 1 (2.16)

The condition imposed by property (1) simply states that negative probabilitiesare impossible, while property (4) assures that the probability is unity that arealization takes on some value. Property (2) gives the probability of finding therealization in a interval around a certain value, while property (3) provides theprobability that the realization is less than a prescribed value. Note the necessityof distinguishing between the running variable, x, and the integration variable, c,in equations 2.14 and 2.15.

Since Bx(c)dc gives the probability of the random variable x assuming a valuebetween c and c + dc, any moment of the distribution can be computed by inte-grating the appropriate power of x over all possible values. Thus the n-th momentis given by:

⟨xn⟩ =! ∞

−∞cnBx(c)dc (2.17)

Turbulence: lecture


Probability




2.3 Probability





Bx(c) ≡ limN → ∞∆c → 0

H(c,∆c,N)/∆c (2.12)



• Property 1:Bx(c) > 0 (2.13)

always.


2.3. PROBABILITY 25




• Property 3:

Probc < x =! x

−∞Bx(c)dc (2.15)


−∞Bx(x)dx = 1 (2.16)



⟨xn⟩ =! ∞

−∞cnBx(c)dc (2.17)

Turbulence: lecture


Probability




2.3 Probability





Bx(c) ≡ limN → ∞∆c → 0

H(c,∆c,N)/∆c (2.12)



• Property 1:Bx(c) > 0 (2.13)

always.


Turbulence: lecture


Probability


Turbulence: lecture


Probability




2.3 Probability





Bx(c) ≡ limN → ∞∆c → 0

H(c,∆c,N)/∆c (2.12)



• Property 1:Bx(c) > 0 (2.13)

always.Wednesday, 23 October 13

Turbulence: lecture


Probability


Turbulence: lecture


Probability




2.3 Probability





Bx(c) ≡ limN → ∞∆c → 0

H(c,∆c,N)/∆c (2.12)



• Property 1:Bx(c) > 0 (2.13)

always.

2.3. PROBABILITY 25




• Property 3:

Probc < x =! x

−∞Bx(c)dc (2.15)


−∞Bx(x)dx = 1 (2.16)



⟨xn⟩ =! ∞

−∞cnBx(c)dc (2.17)


Turbulence: lecture


Probability


2.3. PROBABILITY 25




• Property 3:

Probc < x =! x

−∞Bx(c)dc (2.15)


−∞Bx(x)dx = 1 (2.16)



⟨xn⟩ =! ∞

−∞cnBx(c)dc (2.17)

Turbulence: lecture


Probability


2.3. PROBABILITY 25




• Property 3:

Probc < x =! x

−∞Bx(c)dc (2.15)


−∞Bx(x)dx = 1 (2.16)



⟨xn⟩ =! ∞

−∞cnBx(c)dc (2.17)


Exercise: Show (by returning to the definitions) that the value of the momentdetermined in this manner is exactly equal to the ensemble average defined earlierin equation 2.10. (Hint: use the definition of an integral as a limiting sum.)

If the probability density is given, the moments of all orders can be determined.For example, the variance can be determined by:

varx = ⟨(x−X)2⟩ =! ∞

−∞(c−X)2Bx(c)dc (2.18)

The central moments give information about the shape of the probability den-sity function, and vice versa. Figure 2.4 shows three distributions which have thesame mean and standard deviation, but are clearly quite different. Beneath themare shown random functions of time which might have generated them. Distribu-tion (b) has a higher value of the fourth central moment than does distribution(a). This can be easily seen from the definition

⟨(x−X)4⟩ =! ∞

−∞(c−X)4Bx(c)dc (2.19)

since the fourth power emphasizes the fact that distribution (b) has more weightin the tails than does distribution (a).

It is also easy to see that because of the symmetry of pdf’s in (a) and (b),all the odd central moments will be zero. Distributions (c) and (d), on the otherhand, have non-zero values for the odd moments, because of their asymmetry. Forexample,

⟨(x−X)3⟩ =! ∞

−∞(c−X)3Bx(c)dc (2.20)

is equal to zero if Bx is an even function.

2.3.2 The probability distribution

Sometimes it is convenient to work with the probability distribution insteadof with the probability density function. The probability distribution is definedas the probability that the random variable has a value less than or equal to agiven value. Thus from equation 2.15, the probability distribution is given by

Fx(c) = Probx < c =! c

−∞Bx(c

′)dc′ (2.21)

Note that we had to introduce the integration variable, c′, since c occurred in thelimits.

Equation 2.21 can be inverted by differentiating by c to obtain

Bx(c) =dFx

dc(2.22)


Turbulence: lecture


Probability


What are the distinguishing moments for these distributions?

Turbulence: lecture


Probability


2.3. PROBABILITY 27

Figure 2.4: Relation of skewness to the shape of the pdf and nature of the signal.

What are the distinguishing moments for these distributions?


Turbulence: lecture


Skewness and Kurtosis


Turbulence: lecture


Skewness and Kurtosis



2.3.3 Gaussian (or normal) distributions

One of the most important pdf’s in turbulence is the Gaussian or Normal distri-bution defined by

BxG(c) =1√2πσx

e−(c−X)2/2σ2x (2.23)

where X is the mean and σx is the standard derivation. The factor of 1/√2πσx

insures that the integral of the pdf over all values is unity as required. It is easyto prove that this is the case by completing the squares in the integration of theexponential (see problem 2.2).

The Gaussian distribution is unusual in that it is completely determined by itsfirst two moments, X and σ. This is not typical of most turbulence distributions.Nonetheless, it is sometimes useful to approximate turbulence as being Gaussian,often because of the absence of simple alternatives.

It is straightforward to show by integrating by parts that all the even centralmoments above the second are given by the following recursive relationship,

⟨(x−X)n⟩ = (n− 1)(n− 3)...3.1σnx (2.24)

Thus the fourth central moment is 3σ4x, the sixth is 15σ6

x, and so forth.

Exercise: Prove this.

The probability distribution corresponding to the Gaussian distribution canbe obtained by integrating the Gaussian pdf from −∞ to x = c; i.e.,

FxG(c) =1√2πσx

! c

−∞e−(c′−X)2/2σ2

xdc′ (2.25)

The integral is related to the erf-function tabulated in many standard tables andfunction subroutines, but usually with the independent variable normalized byσx; i.e., c′ = c/σx. Alternatively we can subtract FxG(c) from unity to obtain theprobability that x ≥ c as 1− FxG(c). This is related to the complementary errorfunction, erfc(c), also usually easily available.

2.3.4 Skewness and kurtosis

Because of their importance in characterizing the shape of the pdf, it is usefulto define scaled (or normalized) versions of third and fourth central moments:the skewness and kurtosis respectively. The skewness is defined as third centralmoment divided by the three-halves power of the second; i.e.,

S =⟨(x−X)3⟩

⟨(x−X)2⟩3/2 (2.26)

The kurtosis is defined as the fourth central moment divided by the square of thesecond; i.e.,

K =⟨(x−X)4⟩⟨(x−X)2⟩2 (2.27)


Turbulence: lecture


Probability distribution


Turbulence: lecture


Probability distribution



Exercise: Show (by returning to the definitions) that the value of the momentdetermined in this manner is exactly equal to the ensemble average defined earlierin equation 2.10. (Hint: use the definition of an integral as a limiting sum.)

If the probability density is given, the moments of all orders can be determined.For example, the variance can be determined by:

varx = ⟨(x−X)2⟩ =! ∞

−∞(c−X)2Bx(c)dc (2.18)

The central moments give information about the shape of the probability den-sity function, and vice versa. Figure 2.4 shows three distributions which have thesame mean and standard deviation, but are clearly quite different. Beneath themare shown random functions of time which might have generated them. Distribu-tion (b) has a higher value of the fourth central moment than does distribution(a). This can be easily seen from the definition

⟨(x−X)4⟩ =! ∞

−∞(c−X)4Bx(c)dc (2.19)

since the fourth power emphasizes the fact that distribution (b) has more weightin the tails than does distribution (a).

It is also easy to see that because of the symmetry of pdf’s in (a) and (b),all the odd central moments will be zero. Distributions (c) and (d), on the otherhand, have non-zero values for the odd moments, because of their asymmetry. Forexample,

⟨(x−X)3⟩ =! ∞

−∞(c−X)3Bx(c)dc (2.20)

is equal to zero if Bx is an even function.

2.3.2 The probability distribution

Sometimes it is convenient to work with the probability distribution insteadof with the probability density function. The probability distribution is definedas the probability that the random variable has a value less than or equal to agiven value. Thus from equation 2.15, the probability distribution is given by

Fx(c) = Probx < c =! c

−∞Bx(c

′)dc′ (2.21)

Note that we had to introduce the integration variable, c′, since c occurred in thelimits.

Equation 2.21 can be inverted by differentiating by c to obtain

Bx(c) =dFx

dc(2.22)


Turbulence: lecture


Normal distribution


Distribution completely determined by first two moments

Turbulence: lecture


Normal distribution





BxG(c) =1√2πσx

e−(c−X)2/2σ2x (2.23)





⟨(x−X)n⟩ = (n− 1)(n− 3)...3.1σnx (2.24)


x, and so forth.



FxG(c) =1√2πσx

! c

−∞e−(c′−X)2/2σ2

xdc′ (2.25)




S =⟨(x−X)3⟩

⟨(x−X)2⟩3/2 (2.26)


K =⟨(x−X)4⟩⟨(x−X)2⟩2 (2.27)





BxG(c) =1√2πσx

e−(c−X)2/2σ2x (2.23)





⟨(x−X)n⟩ = (n− 1)(n− 3)...3.1σnx (2.24)


x, and so forth.



FxG(c) =1√2πσx

! c

−∞e−(c′−X)2/2σ2

xdc′ (2.25)




S =⟨(x−X)3⟩

⟨(x−X)2⟩3/2 (2.26)


K =⟨(x−X)4⟩⟨(x−X)2⟩2 (2.27)


Turbulence: lecture


Normal distribution





BxG(c) =1√2πσx

e−(c−X)2/2σ2x (2.23)





⟨(x−X)n⟩ = (n− 1)(n− 3)...3.1σnx (2.24)


x, and so forth.



FxG(c) =1√2πσx

! c

−∞e−(c′−X)2/2σ2

xdc′ (2.25)




S =⟨(x−X)3⟩

⟨(x−X)2⟩3/2 (2.26)


K =⟨(x−X)4⟩⟨(x−X)2⟩2 (2.27)





BxG(c) =1√2πσx

e−(c−X)2/2σ2x (2.23)





⟨(x−X)n⟩ = (n− 1)(n− 3)...3.1σnx (2.24)


x, and so forth.



FxG(c) =1√2πσx

! c

−∞e−(c′−X)2/2σ2

xdc′ (2.25)




S =⟨(x−X)3⟩

⟨(x−X)2⟩3/2 (2.26)


K =⟨(x−X)4⟩⟨(x−X)2⟩2 (2.27)


Turbulence: lecture


Joint pdfs and joint moments


It is usually important in turbulence to consider joint statistics of flow variables

The joint probability density function can be constructed using the joint histogram following a procedure described

before, but, now, we need to do this with two variables

Turbulence: lecture




It is usually important in turbulence to consider joint statistics of flow variables

2.4. MULTIVARIATE RANDOM VARIABLES 29

Both these are easy to remember if you note the S and K must be dimensionless.The pdf’s in Figure 2.4 can be distinguished by means of their skewness and

kurtosis. The random variable shown in (b) has a higher kurtosis than that in (a).Thus the kurtosis can be used as an indication of the tails of a pdf, a higher kurtosisindicating that relatively larger excursions from the mean are more probable. Theskewnesses of (a) and (b) are zero, whereas those for (c) and (d) are non-zero.Thus, as its name implies, a non-zero skewness indicates a skewed or asymmetricpdf, which in turn means that larger excursions in one direction are more probablethan in the other. For a Gaussian pdf, the skewness is zero and the kurtosis isequal to three (see problem 2.4). The flatness factor, defined as (K − 3), issometimes used to indicate deviations from Gaussian behavior.

Exercise: Prove that the skewness and kurtosis of a Gaussian distributed randomvariable are 0 and 3 respectively.

2.4 Multivariate Random Variables

2.4.1 Joint pdfs and joint moments

Often it is important to consider more than one random variable at a time. Forexample, in turbulence the three components of the velocity vector are interre-lated and must be considered together. In addition to the marginal (or singlevariable) statistical moments already considered, it is necessary to consider thejoint statistical moments.

For example if u and v are two random variables, there are three second-ordermoments which can be defined ⟨u2⟩, ⟨v2⟩, and ⟨uv⟩. The product moment ⟨uv⟩ iscalled the cross-correlation or cross-covariance. The moments ⟨u2⟩ and ⟨v2⟩ arereferred to as the covariances, or just simply the variances. Sometimes ⟨uv⟩ isalso referred to as the correlation.

In a manner similar to that used to build-up the probability density functionfrom its measurable counterpart, the histogram, a joint probability densityfunction (or jpdf), Buv, can be built-up from the joint histogram. Figure 2.5illustrates several examples of jpdf’s which have different cross-correlations. Forconvenience the fluctuating variables u′ and v′ can be defined as

u′ = u− U (2.28)

v′ = v − V (2.29)

where as before capital letters are used to represent the mean values. Clearly thefluctuating quantities u′ and v′ are random variables with zero mean.

A positive value of ⟨u′v′⟩ indicates that u′ and v′ tend to vary together. Anegative value indicates that when one variable is increasing the other tends to bedecreasing. A zero value of ⟨u′v′⟩ indicates that there is no correlation between

The joint probability density function can be constructed using the joint histogram following a procedure described

before, but, now, we need to do this with two variables


Turbulence: lecture




u’ and v’ are random variables about the mean

Turbulence: lecture





Figure 2.5: Contours of constant probability for four different joint probabilitydensity functions. Try to figure out what the moments would be for each and howthey would differ.










u′ = u− U (2.28)

v′ = v − V (2.29)




hu0v0i > 0 (positive correlation)

hu0v0i < 0 (negative correlation)

hu0v0i = 0 (uncorrelated)


Turbulence: lecture















u′ = u− U (2.28)

v′ = v − V (2.29)








Turbulence: lecture















u′ = u− U (2.28)

v′ = v − V (2.29)








Turbulence: lecture




correlation coefficient (-1 to 1)

1 : perfect correlation-1: perfect anti-correlation

Turbulence: lecture















u′ = u− U (2.28)

v′ = v − V (2.29)








Turbulence: lecture







u′ and v′. As will be seen below, it does not mean that they are statisticallyindependent.

It is sometimes more convenient to deal with values of the cross-varianceswhich have been normalized by the appropriate variances. Thus the correlationcoefficient is defined as:

ρuv ≡⟨u′v′⟩

[⟨u′2⟩⟨v′2⟩]1/2(2.30)

The correlation coefficient is bounded by plus or minus one, the former represent-ing perfect correlation and the latter perfect anti-correlation.

As with the single-variable pdf, there are certain conditions the joint proba-bility density function must satisfy. If Buv(c1, c2) indicates the jpdf of the randomvariables u and v, then:

• Property 1:

Buv(c1, c2) > 0 (2.31)

always.

• Property 2:

Probc1 < u < c1 + dc1, c2 < v < c2 + dc2 = Buv(c1, c2)dc1, dc2 (2.32)


−∞

! ∞

−∞Buv(c1, c2)dc1dc2 = 1 (2.33)


−∞Buv(c1, c2)dc2 = Bu(c1) (2.34)

where Bu is a function of c1 only.


−∞Buv(c1, c2)dc1 = Bv(c2) (2.35)

where Bv is a function of c2 only.

The functions Bu and Bv are called the marginal probability density functions,and they are simply the single variable pdf’s defined earlier. The subscript is usedto indicate which variable is left after the others are integrated out. Note thatBu(c1) is not the same as Buv(c1, 0). The latter is only a slice through the c2-axis,while the marginal distribution is weighted by the integral of the distribution ofthe other variable. Figure 2.6 illustrates these differences.

correlation coefficient (-1 to 1)

1 : perfect correlation-1: perfect anti-correlation


Turbulence: lecture








[⟨u′2⟩⟨v′2⟩]1/2(2.30)



• Property 1:

Buv(c1, c2) > 0 (2.31)

always.

• Property 2:



−∞

! ∞

−∞Buv(c1, c2)dc1dc2 = 1 (2.33)


−∞Buv(c1, c2)dc2 = Bu(c1) (2.34)



−∞Buv(c1, c2)dc1 = Bv(c2) (2.35)



Turbulence: lecture








[⟨u′2⟩⟨v′2⟩]1/2(2.30)



• Property 1:

Buv(c1, c2) > 0 (2.31)

always.

• Property 2:



−∞

! ∞

−∞Buv(c1, c2)dc1dc2 = 1 (2.33)


−∞Buv(c1, c2)dc2 = Bu(c1) (2.34)



−∞Buv(c1, c2)dc1 = Bv(c2) (2.35)







[⟨u′2⟩⟨v′2⟩]1/2(2.30)



• Property 1:

Buv(c1, c2) > 0 (2.31)

always.

• Property 2:



−∞

! ∞

−∞Buv(c1, c2)dc1dc2 = 1 (2.33)


−∞Buv(c1, c2)dc2 = Bu(c1) (2.34)



−∞Buv(c1, c2)dc1 = Bv(c2) (2.35)




Turbulence: lecture




Turbulence: lecture








[⟨u′2⟩⟨v′2⟩]1/2(2.30)



• Property 1:

Buv(c1, c2) > 0 (2.31)

always.

• Property 2:



−∞

! ∞

−∞Buv(c1, c2)dc1dc2 = 1 (2.33)


−∞Buv(c1, c2)dc2 = Bu(c1) (2.34)



−∞Buv(c1, c2)dc1 = Bv(c2) (2.35)




Turbulence: lecture




Turbulence: lecture





Figure 2.6: Surface representation of a joint probability density function.

If the joint probability density function is known, the joint moments of allorders can be determined. Thus the m,n-th joint moment is

⟨umvn⟩ =! ∞

−∞

! ∞

−∞cm1 c

n2Buv(c1, c2)dc1dc2 (2.36)

where m and n can take any value. The corresponding central-moment is:

⟨(u− U)m(v − V )n⟩ =! ∞

−∞

! ∞

−∞(c1 − U)m(c2 − V )nBuv(c1, c2)dc1dc2 (2.37)

In the preceding discussions, only two random variables have been considered.The definitions, however, can easily be generalized to accommodate any numberof random variables. In addition, the joint statistics of a single random vari-able at different times or at different points in space could be considered. Thiswill be discussed later when stationary and homogeneous random processes areconsidered.

2.4.2 The bi-variate normal (or Gaussian) distribution

If u and v are normally distributed random variables with standard deviationsgiven by σu and σv, respectively, with correlation coefficient ρuv, then their joint


Turbulence: lecture


Statistical independence and lack of correlation


Turbulence: lecture





probability density function is given by

BuvG(c1, c2) =1

2πσuσvexp

!(c1 − U)2

2σ2u

+(c2 − V )2

2σ2v

− ρuvc1c2σuσv

"

(2.38)

This distribution is plotted in Figure 2.7 for several values of ρuv where u and vare assumed to be identically distributed (i.e., ⟨u2⟩ = ⟨v2⟩).

It is straightforward to show (by completing the square and integrating) thatthis yields the single variable Gaussian distribution for the marginal distributions(see problem 2.5). It is also possible to write a multivariate Gaussian probabilitydensity function for any number of random variables.

Exercise: Prove that equation 2.23 results from integrating out the dependenceof either variable using equations 2.34 or 2.35.

2.4.3 Statistical independence and lack of correlation

Definition: Statistical Independence Two random variables are said to bestatistically independent if their joint probability density is equal to the productof their marginal probability density functions. That is,

Buv(c1, c2) = Bu(c1)Bv(c2) (2.39)

It is easy to see that statistical independence implies a complete lack of corre-lation; i.e., ρuv ≡ 0. From the definition of the cross-correlation,

⟨(u− U)(v − V )⟩ =# ∞

−∞

# ∞

−∞(c1 − U)(c2 − V )Buv(c1, c2)dc1dc2

=# ∞

−∞

# ∞

−∞(c1 − U)(c2 − V )Bu(c1)Bv(c2)dc1dc2

=# ∞

−∞(c1 − U)Bu(c1)dc1

# ∞

−∞(c2 − V )Bv(c2)dc2

= 0 (2.40)

where we have used equation 2.39 since the first central moments are zero bydefinition.

It is important to note that the inverse is not true — lack of correlation doesnot imply statistical independence! To see this consider two identically distributedrandom variables, u′ and v′, which have zero means and a non-zero correlation⟨u′v′⟩. From these two correlated random variables two other random variables,x and y, can be formed as:

x = u′ + v′ (2.41)

y = u′ − v′ (2.42)


Turbulence: lecture





probability density function is given by

BuvG(c1, c2) =1

2πσuσvexp

!(c1 − U)2

2σ2u

+(c2 − V )2

2σ2v

− ρuvc1c2σuσv

"

(2.38)

This distribution is plotted in Figure 2.7 for several values of ρuv where u and vare assumed to be identically distributed (i.e., ⟨u2⟩ = ⟨v2⟩).

It is straightforward to show (by completing the square and integrating) thatthis yields the single variable Gaussian distribution for the marginal distributions(see problem 2.5). It is also possible to write a multivariate Gaussian probabilitydensity function for any number of random variables.

Exercise: Prove that equation 2.23 results from integrating out the dependenceof either variable using equations 2.34 or 2.35.

2.4.3 Statistical independence and lack of correlation

Definition: Statistical Independence Two random variables are said to bestatistically independent if their joint probability density is equal to the productof their marginal probability density functions. That is,

Buv(c1, c2) = Bu(c1)Bv(c2) (2.39)

It is easy to see that statistical independence implies a complete lack of corre-lation; i.e., ρuv ≡ 0. From the definition of the cross-correlation,

⟨(u− U)(v − V )⟩ =# ∞

−∞

# ∞

−∞(c1 − U)(c2 − V )Buv(c1, c2)dc1dc2

=# ∞

−∞

# ∞

−∞(c1 − U)(c2 − V )Bu(c1)Bv(c2)dc1dc2

=# ∞

−∞(c1 − U)Bu(c1)dc1

# ∞

−∞(c2 − V )Bv(c2)dc2

= 0 (2.40)

where we have used equation 2.39 since the first central moments are zero bydefinition.

It is important to note that the inverse is not true — lack of correlation doesnot imply statistical independence! To see this consider two identically distributedrandom variables, u′ and v′, which have zero means and a non-zero correlation⟨u′v′⟩. From these two correlated random variables two other random variables,x and y, can be formed as:

x = u′ + v′ (2.41)

y = u′ − v′ (2.42)


Turbulence: lecture


Stationarity and Ergodicity

Turbulence: lecture



2. Statistical & Experimental Tools

2.1 Analysis tools

Examples of u(t) measured with a hot wire anemometer at various points in a boundary layer

(shown in class) demonstrate again the range of time scales, intermittency, etc., and the fact

that fully turbulent flow signals appear random.

We decompose the ‘signal’ by defining a time mean, U, and a fluctuating part, u’, thus: u(t) =

U + u’. (Note that overbars signify a time average, but will not generally be used for the

mean velocities, e.g. uU ≡ ). The total velocity u(x,t) will normally be denoted just by u.

We have to treat the fluctuations statistically (since we normally cannot compute them),

which means having to characterise them in terms of their amplitude and time-domain

features.

2.1.1 Stationarity & Ergodicity

Take two realisations of all possible time histories of some property of a random (or

turbulent) flow:

x1(t)

x2(t)

t=t1

These samples could be produced by doing the experiment twice. The collection of all

possible realisations is the ‘random process’. The mean value at time t1 of all the samples is:

€

x(t1) = limN →∞

1N

xn(t1)n=1

∞∑ ,

and similarly for powers of x(t). This is called an ensemble average.


Turbulence: lecture




2.1 Analysis tools









features.



turbulent) flow:

x1(t)

x2(t)

t=t1



€

x(t1) = limN →∞

1N

xn(t1)n=1

∞∑ ,


Turbulence: lecture




2.1 Analysis tools









features.



turbulent) flow:

x1(t)

x2(t)

t=t1



€

x(t1) = limN →∞

1N

xn(t1)n=1

∞∑ ,



2.1 Analysis tools









features.



turbulent) flow:

x1(t)

x2(t)

t=t1



€

x(t1) = limN →∞

1N

xn(t1)n=1

∞∑ ,



Turbulence: lecture



Turbulence: lecture



If all these ensemble averages do not vary with t1 the process is stationary. In other words, a

stationary process is one in which all possible moments and joint-moments are time-

invariant.

But it is also possible to describe the properties of x(t) by computing time averages over

specific samples (realisations). Consider the n’th sample, xn(t). It has a mean value given by:

If x(t) is stationary AND xn does not depend on n, the process is ergodic. So, for ergodic

processes, the time-averaged mean values are equal to the corresponding ensemble averages,

i.e. x(t) = xn . Note that only stationary processes can be ergodic. Fortunately, in practice,

random data representing stationary physical phenomena are generally ergodic, so we can

analyse them properly on the basis of a single observed time history – provided it is long

enough. For non-stationary data – e.g. from an unsteady ‘mean’ flow like, for instance,

within the earth’s planetary boundary layer (being affected by the diurnal cycle and the

passage of large-scale weather fronts) or within an internal combustion engine – the only way

of formally obtaining its properties is via ensemble averaging. In the latter case, however,

one could also consider a phase average, by viewing each full cycle of the piston stroke as a

separate experiment, such that the ensemble-averaged statistics depended solely on the

crankshaft angle.

2.1.2 Amplitude domain statistics

2.1.2.1 Probability density function (PDF)

This describes the probability that the data (e.g. velocity u) will lie within some defined

amplitude range, at any particular instant.

.d)(1lim0

ttxT

xT

nTn ∫∞→=


Turbulence: lecture



If all these ensemble averages do not vary with t1 the process is stationary. In other words, a

stationary process is one in which all possible moments and joint-moments are time-

invariant.

But it is also possible to describe the properties of x(t) by computing time averages over

specific samples (realisations). Consider the n’th sample, xn(t). It has a mean value given by:

If x(t) is stationary AND xn does not depend on n, the process is ergodic. So, for ergodic

processes, the time-averaged mean values are equal to the corresponding ensemble averages,

i.e. x(t) = xn . Note that only stationary processes can be ergodic. Fortunately, in practice,

random data representing stationary physical phenomena are generally ergodic, so we can

analyse them properly on the basis of a single observed time history – provided it is long

enough. For non-stationary data – e.g. from an unsteady ‘mean’ flow like, for instance,

within the earth’s planetary boundary layer (being affected by the diurnal cycle and the

passage of large-scale weather fronts) or within an internal combustion engine – the only way

of formally obtaining its properties is via ensemble averaging. In the latter case, however,

one could also consider a phase average, by viewing each full cycle of the piston stroke as a

separate experiment, such that the ensemble-averaged statistics depended solely on the

crankshaft angle.

2.1.2 Amplitude domain statistics

2.1.2.1 Probability density function (PDF)

This describes the probability that the data (e.g. velocity u) will lie within some defined

amplitude range, at any particular instant.

.d)(1lim0

ttxT

xT

nTn ∫∞→=


Turbulence: lecture


Stationarity and ErgodicityFor ergodic processes, the time averages are equal to the corresponding ensemble averages

Only stationary processes are ergodic

In practice, random data representing physical phenomena are ergodic. Therefore, we can analyse them based on single observed time history, provided that the data record is long enough for the quantity of interest

Turbulence: lecture


Stationarity and ErgodicityFor ergodic processes, the time averages are equal to the corresponding ensemble averages

x(t) = xn

Only stationary processes are ergodic

In practice, random data representing physical phenomena are ergodic. Therefore, we can analyse them based on single observed time history, provided that the data record is long enough for the quantity of interest


Turbulence: lecture


Estimation from finite number of realisations


Turbulence: lecture



2.5. ESTIMATION FROM A FINITE NUMBER OF REALIZATIONS 35

many realizations are enough? The answer to this question must be sought bylooking at the statistical properties of estimators based on a finite number ofrealizations. There are two questions which must be answered. The first one is:

• Is the expected value (or mean value) of the estimator equal to the trueensemble mean? Or in other words, is the estimator unbiased?

The second question is:

• Does the difference between the value of the estimator and that of the truemean decrease as the number of realizations increases? Or in other words,does the estimator converge in a statistical sense (or converge in probability).Figure 2.8 illustrates the problems which can arise.

2.5.2 Bias and convergence of estimators

A procedure for answering these questions will be illustrated by considering asimple estimator for the mean, the arithmetic mean considered above, XN . ForN independent realizations, xn, n = 1, 2, · · · , N where N is finite, XN is given by:

XN =1

N

N!

n=1

xn (2.44)

Now, as we observed in our simple coin-flipping experiment, since the xn arerandom, so must be the value of the estimator XN . For the estimator to beunbiased, the mean value of XN must be the true ensemble mean, X; i.e.,

limN→∞XN = X (2.45)

It is easy to see that since the operations of averaging and adding commute,

⟨XN⟩ = ⟨ 1N

N!

n=1

xn⟩ (2.46)

=1

N

N!

n=1

⟨xn⟩ (2.47)

=1

NNX = X (2.48)

(Note that the expected value of each xn is just X since the xn are assumedidentically distributed). Thus xN is, in fact, an unbiased estimator for the mean.

The question of convergence of the estimator can be addressed by defining thesquare of variability of the estimator, say ϵ2XN

, to be:

ϵ2XN≡ varXN

X2=

⟨(XN −X)2⟩X2

(2.49)2.5. ESTIMATION FROM A FINITE NUMBER OF REALIZATIONS 37

Now we want to examine what happens to ϵXN as the number of realizationsincreases. For the estimator to converge it is clear that ϵx should decrease as thenumber of samples increases. Obviously, we need to examine the variance of XN

first. It is given by:

varXN = ⟨(XN −X)2⟩

= ⟨!1

N

N"

n=1

xn − X)

#2⟩ (2.50)

= ⟨!1

N

N"

n=1

xn −1

N

N"

n=1

X)

#2⟩ (2.51)

= ⟨!1

N

N"

n=1

(xn −X)

#2⟩ (2.52)

since ⟨XN⟩ = X from equation 2.46. Using the fact that the operations of av-eraging and summation commute, the squared summation can be expanded asfollows:

⟨!

N"

n=1

(xn −X)

#2⟩ =

1

N2

N"

n=1

N"

m=1

⟨(xn −X)(xm −X)⟩

=1

N2

N"

n=1

⟨(xn −X)2⟩

=1

Nvarx, (2.53)

where the next to last step follows from the fact that the xn are assumed tobe statistically independent samples (and hence uncorrelated), and the last stepfrom the definition of the variance. It follows immediately by substitution intoequation 2.49 that the square of the variability of the estimator, XN , is given by:

ϵ2XN=

1

N

varxX2

=1

N

$σx

X

%2(2.54)

Thus the variability of the estimator depends inversely on the number of in-dependent realizations, N , and linearly on the relative fluctuation level of therandom variable itself, σx/X. Obviously if the relative fluctuation level is zero(either because there the quantity being measured is constant and there are nomeasurement errors), then a single measurement will suffice. On the other hand,as soon as there is any fluctuation in the x itself, the greater the fluctuation (rel-ative to the mean of x, ⟨x⟩ = X), then the more independent samples it will taketo achieve a specified accuracy.

Example: In a given ensemble the relative fluctuation level is 12% (i.e.,σx/X = 0.12). What is the fewest number of independent samples that mustbe acquired to measure the mean value to within 1%?


Turbulence: lecture




many realizations are enough? The answer to this question must be sought bylooking at the statistical properties of estimators based on a finite number ofrealizations. There are two questions which must be answered. The first one is:

• Is the expected value (or mean value) of the estimator equal to the trueensemble mean? Or in other words, is the estimator unbiased?

The second question is:

• Does the difference between the value of the estimator and that of the truemean decrease as the number of realizations increases? Or in other words,does the estimator converge in a statistical sense (or converge in probability).Figure 2.8 illustrates the problems which can arise.

2.5.2 Bias and convergence of estimators

A procedure for answering these questions will be illustrated by considering asimple estimator for the mean, the arithmetic mean considered above, XN . ForN independent realizations, xn, n = 1, 2, · · · , N where N is finite, XN is given by:

XN =1

N

N!

n=1

xn (2.44)

Now, as we observed in our simple coin-flipping experiment, since the xn arerandom, so must be the value of the estimator XN . For the estimator to beunbiased, the mean value of XN must be the true ensemble mean, X; i.e.,

limN→∞XN = X (2.45)

It is easy to see that since the operations of averaging and adding commute,

⟨XN⟩ = ⟨ 1N

N!

n=1

xn⟩ (2.46)

=1

N

N!

n=1

⟨xn⟩ (2.47)

=1

NNX = X (2.48)

(Note that the expected value of each xn is just X since the xn are assumedidentically distributed). Thus xN is, in fact, an unbiased estimator for the mean.

The question of convergence of the estimator can be addressed by defining thesquare of variability of the estimator, say ϵ2XN

, to be:

ϵ2XN≡ varXN

X2=

⟨(XN −X)2⟩X2

(2.49)2.5. ESTIMATION FROM A FINITE NUMBER OF REALIZATIONS 37




= ⟨!1

N

N"

n=1

xn − X)

#2⟩ (2.50)

= ⟨!1

N

N"

n=1

xn −1

N

N"

n=1

X)

#2⟩ (2.51)

= ⟨!1

N

N"

n=1

(xn −X)

#2⟩ (2.52)


⟨!

N"

n=1

(xn −X)

#2⟩ =

1

N2

N"

n=1

N"

m=1

⟨(xn −X)(xm −X)⟩

=1

N2

N"

n=1

⟨(xn −X)2⟩

=1

Nvarx, (2.53)


ϵ2XN=

1

N

varxX2

=1

N

$σx

X

%2(2.54)




Turbulence: lecture




We examine the variance of XN,

Turbulence: lecture








= ⟨!1

N

N"

n=1

xn − X)

#2⟩ (2.50)

= ⟨!1

N

N"

n=1

xn −1

N

N"

n=1

X)

#2⟩ (2.51)

= ⟨!1

N

N"

n=1

(xn −X)

#2⟩ (2.52)


⟨!

N"

n=1

(xn −X)

#2⟩ =

1

N2

N"

n=1

N"

m=1

⟨(xn −X)(xm −X)⟩

=1

N2

N"

n=1

⟨(xn −X)2⟩

=1

Nvarx, (2.53)


ϵ2XN=

1

N

varxX2

=1

N

$σx

X

%2(2.54)



We examine the variance of XN,


Turbulence: lecture




Using the fact that summation and averaging commute,

Turbulence: lecture




Using the fact that summation and averaging commute,





= ⟨!1

N

N"

n=1

xn − X)

#2⟩ (2.50)

= ⟨!1

N

N"

n=1

xn −1

N

N"

n=1

X)

#2⟩ (2.51)

= ⟨!1

N

N"

n=1

(xn −X)

#2⟩ (2.52)


⟨!

N"

n=1

(xn −X)

#2⟩ =

1

N2

N"

n=1

N"

m=1

⟨(xn −X)(xm −X)⟩

=1

N2

N"

n=1

⟨(xn −X)2⟩

=1

Nvarx, (2.53)


ϵ2XN=

1

N

varxX2

=1

N

$σx

X

%2(2.54)




Turbulence: lecture




The square of the variability of XN is given by

Turbulence: lecture




The square of the variability of XN is given by





= ⟨!1

N

N"

n=1

xn − X)

#2⟩ (2.50)

= ⟨!1

N

N"

n=1

xn −1

N

N"

n=1

X)

#2⟩ (2.51)

= ⟨!1

N

N"

n=1

(xn −X)

#2⟩ (2.52)


⟨!

N"

n=1

(xn −X)

#2⟩ =

1

N2

N"

n=1

N"

m=1

⟨(xn −X)(xm −X)⟩

=1

N2

N"

n=1

⟨(xn −X)2⟩

=1

Nvarx, (2.53)


ϵ2XN=

1

N

varxX2

=1

N

$σx

X

%2(2.54)




Turbulence: lecture




Convergence to true mean

Convergence to wrong mean

No convergence

Turbulence: lecture





Figure 2.8: Three different estimators for the mean: The first of which convergesto the mean with increasing number of samples, the middle converges to the wrongmean, and the bottom does not converge at all.

Convergence to true mean

Convergence to wrong mean

No convergence


Turbulence: lecture

Time/space domain analysis

So far, we have looked at amplitude statistics. These do not tell you anything about temporal/spatial structures

Turbulent flow Synthetic signal

We need additional tools to tell these apart

Turbulence: lecture


So far, we have looked at amplitude statistics. These do not tell you anything about temporal/spatial structures

Amplitude and frequency modulation in wall turbulence 87

Original signal Synthetic signal

–5

0

5

5 10 15 20–10

–5

0

5

10

5 10 15 20

0 1 2 3 4 5 6 7 8

10–3

10–2

10–1

100

–3 –2 –1 0 1 2 3

104

101

102

103

10–3

10–2

10–1

100

–3 –2 –1 0 1 2 3

104

101

102

103

0.04 0.06 0.08 0.10 0.12 0.14

–30 –20 –10 0 10 20 30

10–2

100

–2.0 –1.5 –1.0 –0.5 0 0.5 1.0 1.5 2.0

10–2

100

–1.5 –1.0 –0.5 0 0.5 1.0 1.5

–10

10

–2.0 2.0

(a)

(b)

(d )

(c)

FIGURE 15. (a) Instantaneous sample of original and synthetic signals at y

+ = 15. Plotsof (b) hu2

S

+i and (c) h f

+m

i as a function of u

+L

and wall-normal location. (d) Plots of 1u

2S

for u

+L

= 2 and 1f

+m

for u

+L

= 2 as a function of phase lag ( ) and wall-normal location.The data presented in panels (b–d) are calculated using the synthetic signal. The large- andsmall-scale fluctuations are calculated from the synthetic signal based on a filter time scale of

t

U1/ = 2 (this filter scale is identical to the one used to analyse the original data in thispaper).

Turbulent flow Synthetic signal

We need additional tools to tell these apart


Turbulence: lecture


Autocorrelation

Measure of memory of the process

You can replace time with space and it will tell you about longevity of the process


Turbulence: lecture


Autocorrelation

150 CHAPTER 8. STATIONARITY RANDOM PROCESSES

to represent the instantaneous value (i.e., u(t)), a capital letter to represent itsaverage (i.e., U = ⟨u(t)), and we define the fluctuation by a lower case letter (i.e.,u = u(t)−U). Also since the process is assumed stationary, the mean velocity andvelocity moments are time-independent. In fact, the probability density itself istime-independent, as should be obvious from the fact that the moments are timeindependent.

An alternative way of looking at stationarity is to note that the statisticsof the process are independent of the origin in time. It is obvious from theabove, for example, that if the statistics of a process are time independent, then⟨un(t)⟩ = ⟨un(t+ T )⟩, etc., where T is some arbitrary translation of the origin intime. Less obvious, but equally true, is that the product ⟨u(t)u(t′)⟩ depends onlyon the time difference t′ − t and not on t (or t′) directly. This consequence ofstationarity can be extended to any product moment. For example, ⟨u(t)v(t′)⟩can depend only on the time difference t′ − t. And ⟨u(t)v(t′)w(t′′)⟩ can dependonly on the two time differences t′ − t and t′′ − t (or t′′ − t′) and not t, t′ or t′′

directly.

8.2 The autocorrelation

One of the most useful statistical moments in the study of stationary random pro-cesses (and turbulence, in particular) is the autocorrelation defined as the aver-age of the product of the random variable evaluated at two times, i.e. ⟨u(t)u(t′)⟩.Since the process is assumed stationary, this product can depend only on the timedifference τ = t′ − t. Therefore the autocorrelation can be written as:

C(τ) ≡ ⟨u(t)u(t+ τ)⟩ (8.1)

The importance of the autocorrelation lies in the fact that it indicates the“memory” of the process; that is, the time over which a process is correlated withitself. Contrast the two autocorrelations shown in Figure 8.1. The autocorrelationof a deterministic sine wave is simply a cosine as can be easily proven. Note thatthere is no time beyond which it can be guaranteed to be arbitrarily small sinceit always “remembers” when it began, and thus always remains correlated withitself. By contrast, a stationary random process like the one illustrated in thefigure will eventually lose all correlation and go to zero. In other words it has a“finite memory” and “forgets” how it was. Note that one must be careful to makesure that a correlation really both goes to zero and stays down before drawingconclusions, since even the sine wave was zero at some points. Stationary randomprocesses always have two-time correlation functions which eventually go to zeroand stay there.

Example 1.Consider the motion of an automobile responding to the movement of the

wheels over a rough surface. In the usual case where the road roughness is ran-

Measure of memory of the process

You can replace time with space and it will tell you about longevity of the process


Turbulence: lecture


Autocorrelation


Turbulence: lecture


Autocorrelation


8.2. THE AUTOCORRELATION 151

Figure 8.1: Autocorrelations for two random processes and a periodic one.

domly distributed, the motion of the car will be a weighted history of the road’sroughness with the most recent bumps having the most influence and with dis-tant bumps eventually forgotten. On the other hand if the car is traveling down arailroad track, the periodic crossing of the railroad ties represents a deterministicinput and the motion will remain correlated with itself indefinitely. This can bea very bad thing if the tie crossing rate corresponds to a natural resonance of thesuspension system of the vehicle.

Since a random process can never be more than perfectly correlated, it cannever achieve a correlation greater than is value at the origin. Thus

|C(τ)| ≤ C(0) (8.2)

An important consequence of stationarity is that the autocorrelation is sym-metric in the time difference, τ = t′ − t. To see this simply shift the origin in timebackwards by an amount τ and note that independence of origin implies:

⟨u(t)u(t+ τ)⟩ = ⟨u(t− τ)u(t)⟩ = ⟨u(t)u(t− τ)⟩ (8.3)

Since the right hand side is simply C(−τ), it follows immediately that:

C(τ) = C(−τ) (8.4)


Turbulence: lecture


Autocorrelation


8.2. THE AUTOCORRELATION 151

Figure 8.1: Autocorrelations for two random processes and a periodic one.

domly distributed, the motion of the car will be a weighted history of the road’sroughness with the most recent bumps having the most influence and with dis-tant bumps eventually forgotten. On the other hand if the car is traveling down arailroad track, the periodic crossing of the railroad ties represents a deterministicinput and the motion will remain correlated with itself indefinitely. This can bea very bad thing if the tie crossing rate corresponds to a natural resonance of thesuspension system of the vehicle.

Since a random process can never be more than perfectly correlated, it cannever achieve a correlation greater than is value at the origin. Thus

|C(τ)| ≤ C(0) (8.2)

An important consequence of stationarity is that the autocorrelation is sym-metric in the time difference, τ = t′ − t. To see this simply shift the origin in timebackwards by an amount τ and note that independence of origin implies:

⟨u(t)u(t+ τ)⟩ = ⟨u(t− τ)u(t)⟩ = ⟨u(t)u(t− τ)⟩ (8.3)

Since the right hand side is simply C(−τ), it follows immediately that:

C(τ) = C(−τ) (8.4)

Turbulence: lecture


Autocorrelation


Turbulence: lecture


Autocorrelation



8.3 The autocorrelation coefficient

It is convenient to define the autocorrelation coefficient as:

ρ(τ) ≡ C(τ)

C(0)=

⟨u(t)u(t+ τ)⟩⟨u2⟩ (8.5)

where

⟨u2⟩ = ⟨u(t)u(t)⟩ = C(0) = var[u] (8.6)

Since the autocorrelation is symmetric, so is its coefficient, i.e.,

ρ(τ) = ρ(−τ) (8.7)

It is also obvious from the fact that the autocorrelation is maximal at the originthat the autocorrelation coefficient must also be maximal there. In fact from thedefinition it follows that

ρ(0) = 1 (8.8)

andρ(τ) ≤ 1 (8.9)

for all values of τ .

8.4 The integral scale

One of the most useful measures of the length of time a process is correlated withitself is the integral scale defined by

Tint ≡! ∞

0ρ(τ)dτ (8.10)

It is easy to see why this works by looking at Figure 8.2. In effect we have replacedthe area under the correlation coefficient by a rectangle of height unity and widthTint.

8.5 The temporal Taylor microscale

The autocorrelation can be expanded about the origin in a MacClaurin series; i.e.,

C(τ) = C(0) + τdC

dτ

"""""τ=0

+1

2τ 2

d2C

dτ 2

"""""τ=0

+1

3!τ 3

d3C

dt

"""""τ=0

(8.11)

But we know the autocorrelation is symmetric in τ , hence the odd terms in τmust be identically zero (i.e., dC/dτ |τ=0 = 0, d3/dτ 3|τ=0, etc.). Therefore theexpansion of the autocorrelation near the origin reduces to:

Sometimes, R =


Turbulence: lecture


Integral length scale


8.5. THE TEMPORAL TAYLOR MICROSCALE 153

Figure 8.2: The autocorrelation coefficient showing relation of the integral scaleto the area under the autocorrelation coefficient curve.

C(τ) = C(0) +1

2τ 2

d2C

dτ 2

!!!!!τ=0

+ · · · (8.12)

Similarly, the autocorrelation coefficient near the origin can be expanded as:

ρ(τ) = 1 +1

2

d2ρ

dτ 2

!!!!!τ=0

τ 2 + · · · (8.13)

where we have used the fact that ρ(0) = 1. If we define ′ = d/dτ we can writethis compactly as:

ρ(τ) = 1 +1

2ρ′′(0)τ 2 + · · · (8.14)

Since ρ(τ) has its maximum at the origin, obviously ρ′′(0) must be negative.We can use the correlation and its second derivative at the origin to define a

special time scale, λτ (called the Taylor microscale 1) by:

λ2τ ≡ − 2

ρ′′(0)(8.15)

Using this in equation 8.14 yields the expansion for the correlation coefficientnear the origin as:

ρ(τ) = 1− τ 2

λ2τ

+ · · · (8.16)

1The Taylor microscale is named after the famous English scientist G.I. Taylor who inventedit in the 1930’s. Among his many other accomplishments he designed the CQR anchor which isstill found on many boats today.

Spatial equivalant: L

Turbulence: lecture


Integral length scale





ρ(τ) ≡ C(τ)

C(0)=

⟨u(t)u(t+ τ)⟩⟨u2⟩ (8.5)

where

⟨u2⟩ = ⟨u(t)u(t)⟩ = C(0) = var[u] (8.6)


ρ(τ) = ρ(−τ) (8.7)


ρ(0) = 1 (8.8)

andρ(τ) ≤ 1 (8.9)




Tint ≡! ∞

0ρ(τ)dτ (8.10)




C(τ) = C(0) + τdC

dτ

"""""τ=0

+1

2τ 2

d2C

dτ 2

"""""τ=0

+1

3!τ 3

d3C

dt

"""""τ=0

(8.11)




C(τ) = C(0) +1

2τ 2

d2C

dτ 2

!!!!!τ=0

+ · · · (8.12)


ρ(τ) = 1 +1

2

d2ρ

dτ 2

!!!!!τ=0

τ 2 + · · · (8.13)


ρ(τ) = 1 +1

2ρ′′(0)τ 2 + · · · (8.14)



λ2τ ≡ − 2

ρ′′(0)(8.15)


ρ(τ) = 1− τ 2

λ2τ

+ · · · (8.16)


Spatial equivalant: L


Turbulence: lecture


Taylor microscale




C(τ) = C(0) +1

2τ 2

d2C

dτ 2

!!!!!τ=0

+ · · · (8.12)


ρ(τ) = 1 +1

2

d2ρ

dτ 2

!!!!!τ=0

τ 2 + · · · (8.13)


ρ(τ) = 1 +1

2ρ′′(0)τ 2 + · · · (8.14)



λ2τ ≡ − 2

ρ′′(0)(8.15)


ρ(τ) = 1− τ 2

λ2τ

+ · · · (8.16)


Turbulence: lecture


Taylor microscale





ρ(τ) ≡ C(τ)

C(0)=

⟨u(t)u(t+ τ)⟩⟨u2⟩ (8.5)

where

⟨u2⟩ = ⟨u(t)u(t)⟩ = C(0) = var[u] (8.6)


ρ(τ) = ρ(−τ) (8.7)


ρ(0) = 1 (8.8)

andρ(τ) ≤ 1 (8.9)




Tint ≡! ∞

0ρ(τ)dτ (8.10)




C(τ) = C(0) + τdC

dτ

"""""τ=0

+1

2τ 2

d2C

dτ 2

"""""τ=0

+1

3!τ 3

d3C

dt

"""""τ=0

(8.11)



Turbulence: lecture


Taylor microscale


Turbulence: lecture


Taylor microscale




C(τ) = C(0) +1

2τ 2

d2C

dτ 2

!!!!!τ=0

+ · · · (8.12)


ρ(τ) = 1 +1

2

d2ρ

dτ 2

!!!!!τ=0

τ 2 + · · · (8.13)


ρ(τ) = 1 +1

2ρ′′(0)τ 2 + · · · (8.14)



λ2τ ≡ − 2

ρ′′(0)(8.15)


ρ(τ) = 1− τ 2

λ2τ

+ · · · (8.16)



Turbulence: lecture


Taylor microscale


Turbulence: lecture


Taylor microscale



Figure 8.3: The autocorrelation coefficient for positive time lags together with itsoscullating parabola showing the Taylor microscale.

Thus very near the origin the correlation coefficient (and the autocorrelation aswell) simply rolls off parabolically; i.e.,

ρ(τ) ≈ 1− τ 2

λ2τ

(8.17)

This parabolic curve is shown in Figure 8.3 as the osculating (or ‘kissing’) parabolawhich approaches zero exactly as the autocorrelation coefficient does. The inter-cept of this osculating parabola with the τ -axis is the Taylor microscale, λτ .

The Taylor microscale is significant for a number of reasons. First, for manyrandom processes (e.g., Gaussian), the Taylor microscale can be proven to bethe average distance between zero-crossing of a random variable in time. Thisis approximately true for turbulence as well. Thus one can quickly estimate theTaylor microscale by simply observing the zero-crossings using an oscilloscopetrace.

The Taylor microscale also has a special relationship to the mean square timederivative of the signal, ⟨[du/dt]2⟩. This is easiest to derive if we consider twostationary random signals, say u and u′, we obtain by evaluating the same signalat two different times, say u = u(t) and u′ = u(t′). The first is only a functionof t, and the second is only a function of t′. The derivative of the first signal isdu/dt and the second du′/dt′. Now lets multiply these together and rewrite themas:

du′

dt′du

dt=

d2

dtdt′u(t)u′(t′) (8.18)

where the right-hand side follows from our assumption that u is not a function oft′ nor u′ a function of t.




ρ(τ) ≈ 1− τ 2

λ2τ

(8.17)




du′

dt′du

dt=

d2

dtdt′u(t)u′(t′) (8.18)


For many random processes, Taylor microscale can be shown to be equal to the average distance between zero-crossings


Turbulence: lecture


Taylor microscale





ρ(τ) ≈ 1− τ 2

λ2τ

(8.17)




du′

dt′du

dt=

d2

dtdt′u(t)u′(t′) (8.18)





ρ(τ) ≈ 1− τ 2

λ2τ

(8.17)




du′

dt′du

dt=

d2

dtdt′u(t)u′(t′) (8.18)





Turbulence: lecture


Taylor microscale


is related to the mean squared of the derivative of the signal

Using standard calculus and some manipulation,

8.6. TIME AVERAGES OF STATIONARY PROCESSES 155

Now if we average and interchange the operations of differentiation and aver-aging we obtain:

⟨du′

dt′du

dt⟩ = d2

dtdt′⟨u u′⟩ (8.19)

Here comes the first trick: u u′ is the same as u(t)u(t′), so its average is justthe autocorrelation, C(τ). Thus we are left with:

⟨du′

dt′du

dt⟩ = d2

dtdt′C(t′ − t) (8.20)

Now we simply need to use the chain-rule. We have already defined τ = t′− t.Let’s also define ξ = t′ + t and transform the derivatives involving t and t′ toderivatives involving τ and ξ. The result is:

d2

dtdt′=

d2

dξ2− d2

dτ 2(8.21)

So equation 8.23 becomes:

⟨du′

dt′du

dt⟩ = d2

dξ2C(τ)− d2

dτ 2C(τ) (8.22)

But since C is a function only of τ , the derivative of it with respect to ξ isidentically zero. Thus we are left with:

⟨du′

dt′du

dt⟩ = − d2

dτ 2C(τ) (8.23)

And finally we need the second trick. Let’s evaluate both sides at t = t′ (orτ = 0 to obtain the mean square derivative as:

⟨!du

dt

"2

⟩ = − d2

dτ 2C(τ)

#####τ=0

(8.24)

But from our definition of the Taylor microscale and the facts that C(0) = ⟨u2⟩and C(τ) = ⟨u2⟩ρ(τ), this is exactly the same as:

⟨!du

dt

"2

⟩ = 2⟨u2⟩λ2τ

(8.25)

This amazingly simple result is very important in the study of turbulence, espe-cially after we extend it to spatial derivatives.

8.6 Time averages of stationary processes

It is common practice in many scientific disciplines to define a time average byintegrating the random variable over a fixed time interval, i.e.,

Turbulence: lecture


Taylor microscale


is related to the mean squared of the derivative of the signal

Using standard calculus and some manipulation,




ρ(τ) ≈ 1− τ 2

λ2τ

(8.17)




du′

dt′du

dt=

d2

dtdt′u(t)u′(t′) (8.18)


8.6. TIME AVERAGES OF STATIONARY PROCESSES 155

Now if we average and interchange the operations of differentiation and aver-aging we obtain:

⟨du′

dt′du

dt⟩ = d2

dtdt′⟨u u′⟩ (8.19)

Here comes the first trick: u u′ is the same as u(t)u(t′), so its average is justthe autocorrelation, C(τ). Thus we are left with:

⟨du′

dt′du

dt⟩ = d2

dtdt′C(t′ − t) (8.20)

Now we simply need to use the chain-rule. We have already defined τ = t′− t.Let’s also define ξ = t′ + t and transform the derivatives involving t and t′ toderivatives involving τ and ξ. The result is:

d2

dtdt′=

d2

dξ2− d2

dτ 2(8.21)

So equation 8.23 becomes:

⟨du′

dt′du

dt⟩ = d2

dξ2C(τ)− d2

dτ 2C(τ) (8.22)

But since C is a function only of τ , the derivative of it with respect to ξ isidentically zero. Thus we are left with:

⟨du′

dt′du

dt⟩ = − d2

dτ 2C(τ) (8.23)

And finally we need the second trick. Let’s evaluate both sides at t = t′ (orτ = 0 to obtain the mean square derivative as:

⟨!du

dt

"2

⟩ = − d2

dτ 2C(τ)

#####τ=0

(8.24)

But from our definition of the Taylor microscale and the facts that C(0) = ⟨u2⟩and C(τ) = ⟨u2⟩ρ(τ), this is exactly the same as:

⟨!du

dt

"2

⟩ = 2⟨u2⟩λ2τ

(8.25)

This amazingly simple result is very important in the study of turbulence, espe-cially after we extend it to spatial derivatives.

8.6 Time averages of stationary processes

It is common practice in many scientific disciplines to define a time average byintegrating the random variable over a fixed time interval, i.e.,


Turbulence: lecture


Cross correlation


What if the second signal is not “u”, but another variable?

AB() =hA(t)B(t+ )ip

hA2ip

hB2iCross correlation between A & B

Gives you information about how one signal is related to another - We will come back to this in Reynolds shear stress

Turbulence: lecture


Cross correlation





ρ(τ) ≡ C(τ)

C(0)=

⟨u(t)u(t+ τ)⟩⟨u2⟩ (8.5)

where

⟨u2⟩ = ⟨u(t)u(t)⟩ = C(0) = var[u] (8.6)


ρ(τ) = ρ(−τ) (8.7)


ρ(0) = 1 (8.8)

andρ(τ) ≤ 1 (8.9)




Tint ≡! ∞

0ρ(τ)dτ (8.10)




C(τ) = C(0) + τdC

dτ

"""""τ=0

+1

2τ 2

d2C

dτ 2

"""""τ=0

+1

3!τ 3

d3C

dt

"""""τ=0

(8.11)


What if the second signal is not “u”, but another variable?

AB() =hA(t)B(t+ )ip

hA2ip

hB2iCross correlation between A & B

Gives you information about how one signal is related to another - We will come back to this in Reynolds shear stress


Turbulence: lecture


Matlab implementation


>>[R,lags] = xcorr(A,B,‘unbiased’); >> R = R./(std(A).*std(B)); >> plot(lags.*dt,R);

If A = B, then we have autocorrelation

`Unbiased’ is crucial since we need to account for the limited number of samples as we go farther towards the edge of the signals

Ideally, we want to compute correlations within the central part of the signal

Turbulence: lecture


Fourier analysis


This provides a frequency analysis of a signal - amplitude or energy content in different frequencies

Turbulence: lecture


Fourier analysis


This provides a frequency analysis of a signal - amplitude or energy content in different frequencies

318 APPENDIX C. FOURIER ANALYSIS OF TIME VARYING SIGNALS

C.1 Fourier series

If we have a periodic signal (i.e., a signal that repeats itself every time interval T )it can be developed in a Fourier series as:

u(t) = A0 +∞!

n=1

An cos"2πn

t

T

#+

∞!

n=1

Bn sin"2πn

t

T

#(C.1)

The frequencies present in this decomposition, fn = n/T , are harmonics (or in-teger multiples) of the fundamental frequency 1/T . The Fourier coefficients, An

and Bn, are given by:

An =1

T

$ T/2

−T/2u(t) cos

"2πn

t

T

#dt (C.2)

Bn =1

T

$ T/2

−T/2u(t) sin

"2πn

t

T

#dt (C.3)

Example: Square wave

Consider the square wave signal shown in figure C.1. Show that it can bereconstructed using the following expression:

u(t) =4

π

∞!

n=1,3,5...

1

nsin

"2πn

t

T

#(C.4)

Figure C.1 illustrates that for n = 1 (dashed curve) the result is just a sine curve.As the Fourier components for increasing values of n are added, however, theresulting curve approaches the square wave signal.

It is sometimes convenient to use a complex notation in the formulation ofFourier series. We can define a complex coefficient as Cn = An − iBn and rewriteequations (C.2–C.3) as:

Cn =1

T

$ T/2

−T/2u(t)e−i2πnt/Tdt (C.5)

The ratio between the real and imaginary values can provide phase informationabout the signal.

It is also convenient to introduce negative values of n. This corresponds tonegative frequencies (which can be thought of as waves going backwards in time).Then the values are symmetric in the sense that A(−n) = A(n) and B(−n) =B(n). The use of negative values of n means that the reconstruction formula nowcan be written very compactly as

u(t) =∞!

n=−∞Cne

+i2πnt/T (C.6)


Turbulence: lecture


Fourier analysis



C.1 Fourier series


u(t) = A0 +∞!

n=1

An cos"2πn

t

T

#+

∞!

n=1

Bn sin"2πn

t

T

#(C.1)



An =1

T

$ T/2

−T/2u(t) cos

"2πn

t

T

#dt (C.2)

Bn =1

T

$ T/2

−T/2u(t) sin

"2πn

t

T

#dt (C.3)



u(t) =4

π

∞!

n=1,3,5...

1

nsin

"2πn

t

T

#(C.4)



Cn =1

T

$ T/2




u(t) =∞!

n=−∞Cne

+i2πnt/T (C.6)

Turbulence: lecture


Fourier analysis



C.1 Fourier series


u(t) = A0 +∞!

n=1

An cos"2πn

t

T

#+

∞!

n=1

Bn sin"2πn

t

T

#(C.1)



An =1

T

$ T/2

−T/2u(t) cos

"2πn

t

T

#dt (C.2)

Bn =1

T

$ T/2

−T/2u(t) sin

"2πn

t

T

#dt (C.3)



u(t) =4

π

∞!

n=1,3,5...

1

nsin

"2πn

t

T

#(C.4)



Cn =1

T

$ T/2




u(t) =∞!

n=−∞Cne

+i2πnt/T (C.6)


C.1 Fourier series


u(t) = A0 +∞!

n=1

An cos"2πn

t

T

#+

∞!

n=1

Bn sin"2πn

t

T

#(C.1)



An =1

T

$ T/2

−T/2u(t) cos

"2πn

t

T

#dt (C.2)

Bn =1

T

$ T/2

−T/2u(t) sin

"2πn

t

T

#dt (C.3)



u(t) =4

π

∞!

n=1,3,5...

1

nsin

"2πn

t

T

#(C.4)



Cn =1

T

$ T/2




u(t) =∞!

n=−∞Cne

+i2πnt/T (C.6)


Turbulence: lecture


Fourier analysis


Fourier series is only valid for time-periodic signals.

We need to deal with signals that result from random processes


C.2 Fourier transform

The Fourier series discussed in the previous section only applies for a periodicdeterministic signal; i.e., a signal that is exactly the same from period to period.But sometimes we must deal with a single pulse or even a random processes. Wetherefore need to be able to decompose a signal that is not repeatable. We canovercome this by using the Fourier transform. This can be viewed as a Fourierseries in the limit for which the period, T becomes infinite, and the correspond-ing values of n/T become a continuous range of frequencies f , meaning that allfrequencies are now possible.

We define then, the Fourier transform of the function u(t) as:

u(f) =! ∞

−∞e−i2πftu(t)dt (C.7)

These are really the continuous counterpart to the Fourier series coefficients of aperiodic signal, and can similarly be used to reconstruct the original signal. Wecall this reconstruction the inverse Fourier transform and define it as:

u(t) =! ∞

−∞e+i2πftu(f)df (C.8)

As in the complex Fourier series, we use negative values of the frequency f .An implicit assumption is that the integrals converge. This is, of course, never

true for a stationary random process, so we need to use the idea of generalizedfunctions to insure that they do. These are discussed in detail in section E, and willbe important when we consider stationary random processes. In practice, however,these Fourier integrals exist for all experimental signals, sincethe time domain overwhich the integral can be performed is finite. Some of the consequences of thistruncation in time are discussed under the sections about filtering and windowsbelow.

C.3 Convolution

A convolution is an operation on two functions f and g defined as:

f ⊗ g =! ∞

−∞f(τ)g(t− τ)dτ (C.9)

A convolution is an integral that expresses the amount of overlap of one functiong as it is shifted over another function f . It therefore “blends” one function withanother. We have already seen one example of a convolution in the autocorrelationfunction defined in section 8.2. The use of windows discussed in the next sectioncan also conveniently be expressed in terms of convolutions.

For the Fourier transform, some important relations for convolutions exist.Here, we will let F denote the Fourier transform and F−1 denote the inverse

We assume that these integrals converge!

Turbulence: lecture


Fourier analysis


But, we cannot collect data for infinite time

The length of the time-record becomes really important

Turbulence: lecture


Fourier analysis


But, we cannot collect data for infinite time

C.4. THE FINITE FOURIER TRANSFORM 321

Fourier transform. The most important results are:

F [f ⊗ g] = F [f ]F [g] (C.10)

F(f g) = F [f ]⊗ F(g) (C.11)

F−1[F(f)F(g)] = f ⊗ g (C.12)

F−1[F(f)⊗ F(g)] = f g (C.13)

In words, this means that the convolution of two functions in the time spacecorresponds to the product of the transformed functions in the frequency space –and vice versa.

C.4 The finite Fourier transform

In any application of Fourier analysis we are always limited by the length of thetime record, T . This means that the most we can expect to be able to transformis the finite time transform given by:

uiT (f) =! T/2

−T/2e−i2πftu(t)dt (C.14)

where for convenience we have written it over the symmetric interval in time(−T/2, T/2).

Now with a little thought, it is clear that we are actually taking the Fouriertransform of the product of two functions, the correlation (the part we want) plusthe window function; i.e.,

uiT (f) = F [u(t)wT (t)] =! ∞

−∞e−i2πftu(t)wT (t)dt (C.15)

where wT (τ) is defined by:

1, −T/2 ≤ τ ≤ T/2

wT (τ) = (C.16)

0, |τ | > T/2

From the results of the preceding section we immediately recognize that theFourier transform we seek is the convolution of the true Fourier transform withthe Fourier transform of the window function; i.e,,

uiT (f) = ui(f)⊗ wT (f)

=! ∞

−∞u(f − f ′)wT (f

′)df ′ (C.17)

=! ∞

−∞u(f ′)wT (f − f ′)df ′

The length of the time-record becomes really important


Turbulence: lecture


Fourier analysis


What does this window function do?

Turbulence: lecture


Fourier analysis


C.4. THE FINITE FOURIER TRANSFORM 321

Fourier transform. The most important results are:

F [f ⊗ g] = F [f ]F [g] (C.10)

F(f g) = F [f ]⊗ F(g) (C.11)

F−1[F(f)F(g)] = f ⊗ g (C.12)

F−1[F(f)⊗ F(g)] = f g (C.13)

In words, this means that the convolution of two functions in the time spacecorresponds to the product of the transformed functions in the frequency space –and vice versa.

C.4 The finite Fourier transform

In any application of Fourier analysis we are always limited by the length of thetime record, T . This means that the most we can expect to be able to transformis the finite time transform given by:

uiT (f) =! T/2

−T/2e−i2πftu(t)dt (C.14)

where for convenience we have written it over the symmetric interval in time(−T/2, T/2).

Now with a little thought, it is clear that we are actually taking the Fouriertransform of the product of two functions, the correlation (the part we want) plusthe window function; i.e.,

uiT (f) = F [u(t)wT (t)] =! ∞

−∞e−i2πftu(t)wT (t)dt (C.15)

where wT (τ) is defined by:

1, −T/2 ≤ τ ≤ T/2

wT (τ) = (C.16)

0, |τ | > T/2

From the results of the preceding section we immediately recognize that theFourier transform we seek is the convolution of the true Fourier transform withthe Fourier transform of the window function; i.e,,

uiT (f) = ui(f)⊗ wT (f)

=! ∞

−∞u(f − f ′)wT (f

′)df ′ (C.17)

=! ∞

−∞u(f ′)wT (f − f ′)df ′

What does this window function do?


Turbulence: lecture


Fourier analysis


The window contaminates the amplitude at a given frequency by leaking information from other frequencies

Most people use a top-hat window without actually realising that they are using a top-hat window

Need to ensure the record length is much much larger than the largest scale to be considered

Conservatively, keep the record length at least one order of magnitude larger than the integral scale

Turbulence: lecture


Fourier analysis


Usually, we sample signals digitally for a length of time at a given frequency of acquisition

The frequency of acquisition is very important

We will incur ALIASING if sampling frequency is too low

Satisfy “Nyquist criterion”

Sample at least twice the frequency as the maximum frequency

Turbulence: lecture


Fourier analysis


Usually, we sample signals digitally for a length of time at a given frequency of acquisition

The frequency of acquisition is very important

D.1. ALIASING OF PERIODICALLY SAMPLED DATA 327

fs-fs 0

0

0

2fs-2fs

fs-fs 2fs-2fs

fs-fs 4fs2fs 3fs-4fs -3fs -2fs

|û f( )|

|û f( )|

|û f( )|

(a)Fouriertransformoftruesignal

(b)Fouriertransformofproperlysampledsignal

(c)Fouriertransformofaliasedsignal

Figure D.1: Relation between aliasing and sample frequency fs = 1/∆t

It follows immediately that the Fourier transform of our sampled signal is givenby:

F [us(t)] =! ∞

−∞u(f − f ′)g(f ′)df ′

=∞"

m=−∞

! ∞

−∞u(f − f ′)δ(f ′ −m/∆t)df

=∞"

m=−∞u(f −m/∆t) (D.13)

Thus the Fourier transform of our sampled signal is an infinitely repeated versionof the Fourier transform of the original signal.

This is illustrated in figure D.1. We will assume that figure D.1(a) shows thereal Fourier transform of our signal. Note that the signal is symmetric aroundf = 0 as we have just discussed. The result of a discrete Fourier transformwith data sampled with a sample frequency fs that is three times the maximumfrequency present in the signal is shown in figure D.1(b). We see that the result

We will incur ALIASING if sampling frequency is too low

Satisfy “Nyquist criterion”

Sample at least twice the frequency as the maximum frequency


Turbulence: lecture


Fourier analysis


328 APPENDIX D. DIGITAL FOURIER TRANSFORMS

is repeated with a period of fs. If we instead use a sample frequency that isonly 1.5 times the maximum frequency in the signal, we get the result shown infigure D.1(c). The result from the true distribution is still repeated with a periodof fs, but there now is an overlap between the periods. The result (shown as athick line) is an incorrect (or aliased) distribution in the regions of overlap.

This is an important problem called aliasing. If we sample too slowly, theresults get corrupted – lower frequencies get a contribution from higher frequenciesand vice versa. This is also illustrated in figure D.3 of section D.3. The sinesignal has period frequency of 0.7 the sampling frequency. However, two othersine signals with period time of 0.3 and 1.3 of sampling frequency, respectively,also match the data samples. After sampling, there is no way of determining whatthe real signal really was.

The key to avoiding aliasing is to satisfy the so-called “Nyquist criterion”:A signal must be sampled with a sample rate that is at least twice the maximumfrequency present in the signal. In figure D.3 this means that only the slowestsignal (period frequency of 0.3 time the sampling rate) would be correctly sampledwith the illustrated sampling; i.e., there is no sine function with a frequencylower than 0.3 that matches the sampled data. You may think that you areonly interested in the lower frequencies and therefore do not have to worry abouthigher frequencies. This is wrong. If you proceed this way it is quite likely thatthe lower frequencies will be corrupted by the higher frequencies! Worse, you willhave little chance of detecting the problem and you will have no way of fixingthe problem after the data is taken. You may argue that you don’t know whatthe highest frequencies are in your signal. There is always some noise that wecannot control. Well, it is your task to ensure this is not a problem! This is thereason why most A/D cards have a low pass filter before the sampling and theA/D conversion. To be safe, the filter frequency should be 3–5 times lower thanthe sampling frequency. We will have a closer look on filters in section G.2.

D.2 The Discrete Fourier transform

As in section B.1, the problem is both that the signal is only known at discretepoints and that we only will have data for a limited time period. Now we will lookat how a discrete formulation can be made. First we have to change eqs. (C.7) toa finite time estimate,

uT (f) =! T

0e−i2πftu(t)dt (D.14)

Note that we have deliberately shifted the time axis by +T/2. This introducesa linear phase shift equivalent to multiplying the Fourier coefficients by e+iπfT

compared to the symmetric finite transform of the previous chapter. Generallythis will not present a problem, but should not be ignored.

This is not exactly what we want, since we have sampled the velocity at discrete

Continuous fourier transform

What we have is discrete data: Discrete fourier transform

D.2. THE DISCRETE FOURIER TRANSFORM 329

times. We really have

un = u(n∆t) n = 0, 1, 2, . . . , N − 1 (D.15)

Using the basic definitions of integral calculus, we can discretize the integral ofequation (D.14) as:

uT (f) =N−1!

n=0

e−i2πfn∆tun∆t (D.16)

N is the total number of samples and T = N∆t is the total sample time. The timebetween samples, ∆t, is given by the sampling frequency, ∆t = 1/fs = T/N . Thisis, of course, an approximation which becomes exact in the limit as the numberof points, N , becomes infinite and as the interval between them, ∆t, goes to zero.

Now since we only have N data points, we can only calculate N independentFourier coefficients. In fact, since we are in the complex domain, we can onlycalculate N/2, since the real and imaginary parts are independent of each other.So we might as well pick the frequencies for which we will evaluate the sum ofequation D.16 for maximum convenience. For almost all applications this turnsout to be integer multiples of the inverse record length; i.e.,

fm =m

T=

m

N∆tm = 0, 1, 2, . . . , N − 1 (D.17)

Substituting this into equation D.16 yields our discretized Fourier transform as:

uT (fm) = T

"1

N

N−1!

n=0

e−i2πmn/Nun

#

m = 0, 1, 2, . . . , N − 1 (D.18)

This equation can be evaluated numerically for each of the frequencies fm definedin eq. (D.17). The Fourier coefficients, uT (fm), found from eq. (D.18) are complexnumbers. For future reference note they negative frequencies are mapped intofm = N − 1, N − 2, N − 3, N −m instead of at negative values of m, as illustratedin Figure D.1.

It is easy to show the original time series data points can be recovered by usingthe inverse discrete Fourier transform:

un =1

T

"N−1!

m=0

e+i2πmn/N uT (fm)

#

(D.19)

The real and imaginary value of at frequency fm corresponds to the coefficientsof the cosine and sine parts respectively, in a reconstruction of the signal. In fact ifwe divide uT (fm) by T, then it is exactly the Fourier series coefficient that wouldrepresent a periodic signal of period T , but equal to our piece of the record overthe interval 0, T . Thus the algorithms for treating a discretized periodic signal anda discretized piece of an infinitely long signal are the same. Because of this the

Turbulence: lecture


Fourier analysis


Turbulence: lecture


Fourier analysis




un = u(n∆t) n = 0, 1, 2, . . . , N − 1 (D.15)


uT (f) =N−1!

n=0




fm =m

T=

m

N∆tm = 0, 1, 2, . . . , N − 1 (D.17)


uT (fm) = T

"1

N

N−1!

n=0

e−i2πmn/Nun

#

m = 0, 1, 2, . . . , N − 1 (D.18)



un =1

T

"N−1!

m=0

e+i2πmn/N uT (fm)

#

(D.19)



Turbulence: lecture


Fourier analysis


Evaluating these discrete coefficients is numerically taxing

Fast Fourier Transform (FFT) allows inexpensive computation

>> uh = fft(A);

Matlab implementation

Turbulence: lecture


Fourier analysis




un = u(n∆t) n = 0, 1, 2, . . . , N − 1 (D.15)


uT (f) =N−1!

n=0




fm =m

T=

m

N∆tm = 0, 1, 2, . . . , N − 1 (D.17)


uT (fm) = T

"1

N

N−1!

n=0

e−i2πmn/Nun

#

m = 0, 1, 2, . . . , N − 1 (D.18)



un =1

T

"N−1!

m=0

e+i2πmn/N uT (fm)

#

(D.19)


Evaluating these discrete coefficients is numerically taxing

Fast Fourier Transform (FFT) allows inexpensive computation

>> uh = fft(A);Matlab implementation


Turbulence: lecture


Fourier analysis

T = 101dt = 1

a = sin(2*pi*t/5);

Peak at m = 21

uh = fft(a); ah = abs(uh); plot(ah);

In matlab, zero index does not existSo, m=1 is in fact zero frequency

f = (m-1)/T

f = 0.2 (= 1/5 that we put in)

You only need to consider till N/2

Turbulence: lecture


Power spectral density function

Now that we know how to compute fourier transforms, we can use this to see the energy/power content within signals

R (in dimensional form) and S form perfect fourier transform pairs

Different people use different symbols of S

Obviously, the symbol or letter does not matter

RAB() =

Z 1

1ei2fSAB(f)df

Turbulence: lecture




Define: SAB(f) = A(f) B(f)




SAB(f) =

Z 1

1ei2fRAB(f)d

RAB() =

Z 1

1ei2fSAB(f)df


Turbulence: lecture




Define: SAB(f) = A(f) B(f)




SAB(f) =

Z 1

1ei2fRAB(f)d

RAB() =

Z 1

1ei2fSAB(f)df


Turbulence: lecture




RAB() =

Z 1

1ei2fSAB(f)df

For = 0, if we evaluate this integral,

This tells us how the cross correlation at zero time lag comes from the entire distribution of frequencies

Turbulence: lecture




RAB() =

Z 1

1ei2fSAB(f)df

For = 0, if we evaluate this integral,

RAB(0) =

Z 1

1SAB(f)df

This tells us how the cross correlation at zero time lag comes from the entire distribution of frequencies


Turbulence: lecture




Experimentally, negative frequencies do not make sense

Suu is an even function and hence we can write this as,

Turbulence: lecture




If A = B = u

Ruu(0) =

Z 1

1Suu(f)df



Ruu(0) = 2

Z 1

0Suu(f)df

Ruu(0) =

Z 1

0Euu(f)df


Turbulence: lecture




If A = B = u

Ruu(0) =

Z 1

1Suu(f)df



Ruu(0) = 2

Z 1

0Suu(f)df

Ruu(0) =

Z 1

0Euu(f)df


Turbulence: lecture




Ruu(0) has units of energy per unit mass (u2)

df has units of frequency

Euu(f) has units of energy per unit mass per frequency

Turbulence: lecture




Ruu(0) has units of energy per unit mass (u2)

df has units of frequency

Euu(f) has units of energy per unit mass per frequency

Ruu(0) =

Z 1

0Euu(f)dfhu2i =


Turbulence: lecture


Matlab implementation: Power spectrum


>> u = load(‘hwdata.dat’); % load data >> N = length(u) % number of samples >> uh = fft(u); %Do FFT >> f = [0:N-1]./(N*dt) “dt is the time separation between successive time records”

>> eh = (conj(uh).*uh)/(N*dt); % calculate spectrum. Dividing by N*dt ensure %that the units are correct

>>plot(f(2:N/2),eh(2:N/2)); % The area under the curve here should be equal to the variance of the signal

Turbulence: lecture



F.4. AN EXAMPLE: DIGITAL SPECTRAL ANALYSIS OF RANDOMDATA347

102 103 104

102

103

104

105

106

107

108

f [Hz]

S(f)

First record128 records

Figure F.2: Power spectrum of the data presented in figure A.1.

Turbulence: lecture




Take lots of blocks of data and average


Turbulence: lecture


Taylor’s hypothesis

Most of the time, we obtained temporal dataAlmost always, the theories in turbulence are related to space

Need a way to convert time to space

Turbulence: lecture



Most of the time, we obtained temporal dataAlmost always, the theories in turbulence are related to space

Need a way to convert time to space

x = Ut or l = f/U

Proc. R. Soc. Lond. A February 18, 1938, Vol 164 No. 919 pp 476-490; doi:10.1098/rspa.1938.0032


Turbulence: lecture



At a first glance, it looks pretty frozen

But, if you start peeling the layers, then not really frozenThere is an entire range of convection velocities

It is not really correct, but, we are limited at this time

Turbulence: lecture

Equations for average velocities

It is convenient to analyse turbulent flows by decomposing it in to the mean and the fluctuation

Thus, the instantaneous velocities and stresses can be written as,


3.2. EQUATIONS FOR THE AVERAGE VELOCITY 43

3.2 Equations for the Average Velocity

Turbulence is that chaotic state of motion characteristic of solutions to the equa-tions of motion at high Reynolds number. Although laminar solutions to theequations often exist that are consistent with the boundary conditions, perturba-tions to these solutions (sometimes even infinitesimal) can cause them to becometurbulent. To see how this can happen, it is convenient to analyze the flow in twoparts, a mean (or average) component and a fluctuating component. Thus theinstantaneous velocity and stresses can be written as:

ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)

where Ui, p, and T (v)ij represent the mean motion, and ui, p, and τij the fluctuating

motions. This technique for decomposing the instantaneous motion is referred toas the Reynolds decomposition. Note that if the averages are defined as ensemblemeans, they are, in general, time-dependent. For the remainder of this book,unless otherwise stated, the density will be assumed constant so ρ ≡ ρ and itsfluctuation is zero.

Substitution of equations 3.11 into equations 3.10 yields

ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)

This equation can now be averaged to yield an equation expressing momentumconservation for the averaged motion. Note that the operations of averagingand differentiation commute; i.e., the average of a derivative is the same as thederivative of the average. Also, the average of a fluctuating quantity is zero.2

Thus the equation for the averaged motion reduces to:

ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)

where the remaining fluctuating product term has been moved to the right-handside of the equation. Whether or not this last term is zero like the other fluctu-ating terms depends on the correlation of terms in the product. In general, thesecorrelations are not zero.

The mass conservation equation can be similarly decomposed. In incompress-ible form, substitution of equations 3.11 into equation 3.4 yields:

∂(Uj + uj)

∂xj= 0 (3.14)

2These are easily proven from the definitions of both.

Turbulence: lecture






ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)





ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)


Turbulence: lecture





Dρ

Dt=

∂ρ

∂t+ uj

∂ρ

∂xj= 0 (3.3)


∂uj

∂xj= 0 (3.4)




T (v)ij = 2µ

!sij −

1

3skkδij

"(3.5)


sij ≡1

2

#∂ui

∂xj+

∂uj

∂xi

$

(3.6)


T (v)ij = 2µsij (3.7)


#∂ui

∂t+ uj

∂ui

∂xj

$

= −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

(3.8)


ν ≡ µ

ρ(3.9)



ρ

#∂ui

∂t+ uj

∂ui

∂xj

$

= − ∂p

∂xi+

∂T (v)ij

∂xj(3.10)





ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)





ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)





ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)



Turbulence: lecture






ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)





ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)


Turbulence: lecture






ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)





ui = Ui + ui

p = P + p

T (v)ij = T (v)

ij + τ (v)ij (3.11)




ρ

!∂(Ui + ui)

∂t+ (Uj + uj)

∂(Ui + ui)

∂xj

"

= −∂(P + p)

∂xi+

∂(T (v)ij + τ (v)ij )

∂xj(3.12)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ρ⟨uj

∂ui

∂xj⟩ (3.13)



∂(Uj + uj)

∂xj= 0 (3.14)



of which the average is:∂Uj

∂xj= 0 (3.15)

It is clear from equation 3.15 that the averaged motion satisfies the same formof the mass conservation equation as does the instantaneous motion, at leastfor incompressible flows. How much simpler the turbulence problem would beif the same were true for the momentum! Unfortunately, as is easily seen fromequation 3.13, such is not the case.

Equation 3.15 can be subtracted from equation 3.14 to yield an equation forthe instantaneous motion alone; i.e.,

∂uj

∂xj= 0 (3.16)

Again, like the mean, the form of the original instantaneous equation is seen tobe preserved. The reason, of course, is obvious: the continuity equation is linear.The momentum equation, on the other hand, is not; hence the difference.

Equation 3.16 can be used to rewrite the last term in equation 3.13 for themean momentum. Multiplying equation 3.16 by ui and averaging yields:

⟨ui∂uj

∂xj⟩ = 0 (3.17)

This can be added to ⟨uj∂ui/∂xj⟩ to obtain:

⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)

where again the fact that arithmetic and averaging operations commute has beenused.

The equation for the averaged momentum, equation 3.13 can now be rewrittenas:

ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xjρ⟨uiuj⟩ (3.19)

The last two terms on the right-hand side are both divergence terms and can becombined; the result is:

ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj

#T (v)ij − ρ⟨uiuj⟩

$(3.20)

Now the terms in square brackets on the right have the dimensions of stress. Thefirst term is, in fact, the viscous stress. The second term, on the other hand, is nota stress at all, but simply a re-worked version of the fluctuating contribution to thenon-linear acceleration terms. The fact that it can be written this way, however,indicates that at least as far as the mean motion is concerned, it acts as though itwere a stress — hence its name, the Reynolds stress. In the succeeding sectionsthe consequences of this difference will be examined.


Turbulence: lecture





∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)


The entire problem of turbulence would be so much easier if the same thing were true for momentum equation

Turbulence: lecture





∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)


Turbulence: lecture



Turbulence: lecture





∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)




∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)



Turbulence: lecture





∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)




∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)



Turbulence: lecture

The problem with turbulence


Appearance of Reynolds stress is the problem

They are created by the flow that we want to study/understand/predict in the first place!

•The equations are not closed•Simple models do not work (more later)•Compiling tables/handbooks carry substantial risk

Turbulence: lecture

The problem with turbulence




∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)


Appearance of Reynolds stress is the problem

They are created by the flow that we want to study/understand/predict in the first place!

•The equations are not closed•Simple models do not work (more later)•Compiling tables/handbooks carry substantial risk


Turbulence: lecture

Closure and eddy viscosity


These have to be related to the mean motion in order to be able to solve for the mean motions

The fact that we do not have these relations is referred to as the closure problem

Turbulence: lecture





∂xj= 0 (3.15)



∂uj

∂xj= 0 (3.16)



⟨ui∂uj

∂xj⟩ = 0 (3.17)


⟨uj∂ui

∂xj⟩+ 0 = ⟨uj

∂ui

∂xj⟩+ ⟨ui

∂uj

∂xj⟩ = ∂

∂xj⟨uiuj⟩ (3.18)



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂



ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂

∂xj


$(3.20)


There are 6 unknowns for the 3 equations

hu21i, hu2

2i, hu23i, hu1u2i, hu1u3i&hu2u3i

These have to be related to the mean motion in order to be able to solve for the mean motions

The fact that we do not have these relations is referred to as the closure problem


Turbulence: lecture



Analogy with viscous stress

We wrote down a constitutive relationship,

The viscosity only depends on the fluid and not on its motions

Turbulence: lecture



Analogy with viscous stress

We wrote down a constitutive relationship,



Dρ

Dt=

∂ρ

∂t+ uj

∂ρ

∂xj= 0 (3.3)


∂uj

∂xj= 0 (3.4)




T (v)ij = 2µ

!sij −

1

3skkδij

"(3.5)


sij ≡1

2

#∂ui

∂xj+

∂uj

∂xi

$

(3.6)


T (v)ij = 2µsij (3.7)


#∂ui

∂t+ uj

∂ui

∂xj

$

= −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

(3.8)


ν ≡ µ

ρ(3.9)



ρ

#∂ui

∂t+ uj

∂ui

∂xj

$

= − ∂p

∂xi+

∂T (v)ij

∂xj(3.10)


The viscosity only depends on the fluid and not on its motions


Turbulence: lecture



We can write a similar equations for the Reynolds stress

The second term on the right hand side is zero (continuity)

Turbulence: lecture



We can write a similar equations for the Reynolds stress

3.6. THE EDDY VISCOSITY 51

the so-called Reynolds stress, into the averaged equations. There are six newindividual Reynolds stress components we must deal with to be exact: ⟨u2

1⟩, ⟩u22⟩,

⟨u23⟩, ⟨u1u2⟩, ⟨u1u3⟩, and ⟨u2u3⟩. These have to be related to the mean motion itself

before the equations can be solved, since the number of unknowns and numberof equations must be equal. The absence of these additional equations is oftenreferred to as the Turbulence Closure Problem.

A similar problem arose when the instantaneous equations were written (equa-tions 3.1 and 3.2), since relations had to be introduced to relate the stresses (inparticular, the viscous stresses) to the motion itself. These relations (or consti-tutive equations) depended only on the properties of the fluid material, and noton the flow itself. Because of this fact, it is possible to carry out independentexperiments, called viscometric experiments, in which these fluid properties canbe determined once and for all. Equation 3.5 provides an example of just sucha constitutive relation, the viscosity, µ, depending only in the choice of fluid.For example, once the viscosity of water at given temperature is determined, thisvalue can be used in all flows at that temperature, not just the one in which theevaluation was made. Or for another example, if we are working on a problemof air flow, we only need to go to reference book somewhere and we can find acomplete specification of how the viscosity of air depends on temperature andpressure. Someone somewhere else has already compiled this information fromindependent experiments.

It is tempting to try such an approach for the turbulence Reynolds stresses(even though we know the underlying requirements of scale separation are notsatisfied). For example, a Newtonian type closure for the Reynolds stresses, oftenreferred to as an “eddy” or “turbulent” viscosity model, looks like:

−ρ⟨uiuj⟩+1

3⟨uiui⟩ = µt

!Sij −

1

3Skkδij

"(3.23)

where µt is the turbulence “viscosity” (also called the eddy viscosity), and Sij isthe mean strain rate defined by:

Sij =1

2

#∂Ui

∂xj+

∂Uj

∂xi

$

(3.24)

The second term vanishes identically for incompressible flow. For the simplecase of a two-dimensional shear flow, equation 3.23 for the Reynolds shear stressreduces to

−ρ⟨u1u2⟩ = µt∂U1

∂x2(3.25)

Note this “model” is the direct analogy to the Newtonian model for viscousstress in a fluid. The Reynolds stresses, ⟨−uiuj⟩ replaces the viscous stress, τ (v)ij .The counterpart to the mechanical pressure is the mean normal Reynolds stress,⟨uiui⟩/3. And like it’s fluid counterpart it, the Reynolds stress can depend onlyon the mean strain rate at a single instant and single location in the flow, so has

The second term on the right hand side is zero (continuity)


Turbulence: lecture



For a simple case, 2D shear flow

Direct analogy to the Newtonian model for viscous stress

The counterpart to the mechanical pressure is the the normal stress

Reynolds stress depends on mean strain at a single location /instant in the flow and has no history or non-local dependence

This is fatal!!!

Turbulence: lecture



For a simple case, 2D shear flow


the so-called Reynolds stress, into the averaged equations. There are six newindividual Reynolds stress components we must deal with to be exact: ⟨u2

1⟩, ⟩u22⟩,

⟨u23⟩, ⟨u1u2⟩, ⟨u1u3⟩, and ⟨u2u3⟩. These have to be related to the mean motion itself

before the equations can be solved, since the number of unknowns and numberof equations must be equal. The absence of these additional equations is oftenreferred to as the Turbulence Closure Problem.

A similar problem arose when the instantaneous equations were written (equa-tions 3.1 and 3.2), since relations had to be introduced to relate the stresses (inparticular, the viscous stresses) to the motion itself. These relations (or consti-tutive equations) depended only on the properties of the fluid material, and noton the flow itself. Because of this fact, it is possible to carry out independentexperiments, called viscometric experiments, in which these fluid properties canbe determined once and for all. Equation 3.5 provides an example of just sucha constitutive relation, the viscosity, µ, depending only in the choice of fluid.For example, once the viscosity of water at given temperature is determined, thisvalue can be used in all flows at that temperature, not just the one in which theevaluation was made. Or for another example, if we are working on a problemof air flow, we only need to go to reference book somewhere and we can find acomplete specification of how the viscosity of air depends on temperature andpressure. Someone somewhere else has already compiled this information fromindependent experiments.

It is tempting to try such an approach for the turbulence Reynolds stresses(even though we know the underlying requirements of scale separation are notsatisfied). For example, a Newtonian type closure for the Reynolds stresses, oftenreferred to as an “eddy” or “turbulent” viscosity model, looks like:

−ρ⟨uiuj⟩+1

3⟨uiui⟩ = µt

!Sij −

1

3Skkδij

"(3.23)

where µt is the turbulence “viscosity” (also called the eddy viscosity), and Sij isthe mean strain rate defined by:

Sij =1

2

#∂Ui

∂xj+

∂Uj

∂xi

$

(3.24)

The second term vanishes identically for incompressible flow. For the simplecase of a two-dimensional shear flow, equation 3.23 for the Reynolds shear stressreduces to

−ρ⟨u1u2⟩ = µt∂U1

∂x2(3.25)

Note this “model” is the direct analogy to the Newtonian model for viscousstress in a fluid. The Reynolds stresses, ⟨−uiuj⟩ replaces the viscous stress, τ (v)ij .The counterpart to the mechanical pressure is the mean normal Reynolds stress,⟨uiui⟩/3. And like it’s fluid counterpart it, the Reynolds stress can depend onlyon the mean strain rate at a single instant and single location in the flow, so has

Direct analogy to the Newtonian model for viscous stress

The counterpart to the mechanical pressure is the the normal stress

Reynolds stress depends on mean strain at a single location /instant in the flow and has no history or non-local dependence

This is fatal!!!


Turbulence: lecture



Example, buoyant plume

Want to “predict” the mean flow at different downstream locations

Turbulence: lecture



Example, buoyant plume52 CHAPTER 3. REYNOLDS AVERAGED EQUATIONS

Figure 3.1: Schematic of axisymmetric plume

no history or non-local dependence. This absence will turn out to be fatal in mostapplications. Moreover, unlike like the viscosity, µ, which depends only on thefluid and not the motion itself, the “turbulence viscosity”, µt, depends entirely onthe motion.

That such a simple model can adequately describe the mean motion in at leastone flow is illustrated by the axisymmetric buoyant plume sketched in Figure 3.1.Figures 3.2 and 3.3 show the calculation of the mean velocity and temperatureprofiles respectively. Obviously the mean velocity and temperature profiles arereasonably accurately computed, as are the Reynolds shear stress and lateralturbulent heat flux shown in Figures 3.4 and 3.5.

The success of the eddy viscosity in the preceding example is more apparentthan real, however, since the value of the eddy viscosity and eddy diffusivity (forthe turbulent heat flux) have been chosen to give the best possible agreement withthe data. This, in itself, would not be a problem if that chosen values could havebeen obtained in advance of the computation, or even if they could be used tosuccessfully predict other flows. In fact, the values used work only for this flow,thus the computation is not a prediction at all, but a postdiction or hindcastfrom which no extrapolation to the future can be made. In other words, ourturbulence “model” is about as useful as having a program to predict yesterday’sweather. Thus the closure problem still very much remains.

Another problem with the eddy viscosity in the example above is that it failsto calculate the vertical components of the Reynolds stress and turbulent heatflux. An attempt at such a computation is shown in Figure 3.6 where the verticalturbulent heat flux is shown to be severely underestimated. Clearly the value ofthe eddy viscosity in the vertical direction must be different than in the radial



Figure 3.2: Mean velocity profiles for axisymmetric plume

Figure 3.3: Mean temperature profiles for axisymmetric plume


Turbulence: lecture














Turbulence: lecture




Model generated from measurements of Reynolds stresses

As good as predicting yesterday’s weather!

Only good for this particular set of conditions

Turbulence: lecture














Turbulence: lecture



Example, buoyant plumeDoes not really work for other quantities of interest such as vertical components of Reynolds stress


Figure 3.6: Vertical turbulent heat flux for axisymmetric plume

Turbulence: lecture



Example, buoyant plumeDoes not really work for other quantities of interest such as vertical components of Reynolds stress


Figure 3.6: Vertical turbulent heat flux for axisymmetric plume


Figure 3.4: Reynolds shear stress profiles for axisymmetric plume

Figure 3.5: Radial turbulent heat flux for axisymmetric plumeWednesday, 23 October 13

Turbulence: lecture




This means that turbulence is not an isotropic medium


direction. In other words, the turbulence for which a constitutive equation is beingwritten is not an isotropic “medium”. In fact, in this specific example the problemis that the vertical component of the heat flux is produced more by the interactionof buoyancy and the turbulence, than it is by the working of turbulence againstmean gradients in the flow. We will discuss this in more detail in the next chapterwhen we consider the turbulence energy balances, but note for now that simplegradient closure models never work unless gradient production dominates. Thisrules out many flows involving buoyancy, and also many involving recirculationsor separation where the local turbulence is convected in from somewhere else.

A more general form of constitutive equation which would allow for the non-isotropic nature of the “medium” (in this case the turbulence itself) would be

−ρ⟨uiuj⟩+1

3⟨ukuk⟩δij = µijkl

!Skl −

1

3Smmδkl

"(3.26)

This closure relation allows each component of the Reynolds stress to have its ownunique value of the eddy viscosity. It is easy to see that it is unlikely this will solvethe closure problem since the original six unknowns, the ⟨uiuj⟩, have been tradedfor eighty-one new ones, µijkl. Even if some can be removed by symmetries,the remaining number is still formidable. More important than the number ofunknowns, however, is that there is no independent or general means for selectingthem without considering a particular flow. This is because turbulence is indeeda property of the flow, not of the fluid.

3.7 The Reynolds Stress Equations

It is clear from the preceding section that the simple idea of an eddy viscositymight not be the best way to approach the problem of relating the Reynoldsstress to the mean motion. An alternative approach is to try to derive dynamicalequations for the Reynolds stresses from the equations governing the fluctuationsthemselves. Such an approach recognizes that the Reynolds stress is really afunctional6 of the velocity; that is, the stress at a point depends on the velocityeverywhere and for all past times, not just at the point in question and at aparticular instant in time.

The analysis begins with the equation for the instantaneous fluctuating ve-locity, equation 3.21. This can be rewritten for a Newtonian fluid with constantviscosity as:

ρ

#∂ui

∂t+ Uj

∂ui

∂xj

$

= − ∂p

∂xi+

∂τ (v)ij

∂xj− ρ

#

uj∂Ui

∂xj

$

− ρ

%

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩&

(3.27)

Note that the free index in this equation is i. Also, since we are now talking aboutturbulence again, the capital letters represent mean or averaged quantities.

6A functional is a function of a function

We have traded 6 unknowns for 81 unknowns!!!

Closure problem remains!

There are no general solutions to this problemFlow is turbulent and the turbulence depends on circumstances

Turbulence: lecture

Reynolds stress equations


It is clear that eddy viscosity approach is flawed.

Equation for instantaneous fluctuating velocity,


direction. In other words, the turbulence for which a constitutive equation is beingwritten is not an isotropic “medium”. In fact, in this specific example the problemis that the vertical component of the heat flux is produced more by the interactionof buoyancy and the turbulence, than it is by the working of turbulence againstmean gradients in the flow. We will discuss this in more detail in the next chapterwhen we consider the turbulence energy balances, but note for now that simplegradient closure models never work unless gradient production dominates. Thisrules out many flows involving buoyancy, and also many involving recirculationsor separation where the local turbulence is convected in from somewhere else.

A more general form of constitutive equation which would allow for the non-isotropic nature of the “medium” (in this case the turbulence itself) would be

−ρ⟨uiuj⟩+1

3⟨ukuk⟩δij = µijkl

!Skl −

1

3Smmδkl

"(3.26)

This closure relation allows each component of the Reynolds stress to have its ownunique value of the eddy viscosity. It is easy to see that it is unlikely this will solvethe closure problem since the original six unknowns, the ⟨uiuj⟩, have been tradedfor eighty-one new ones, µijkl. Even if some can be removed by symmetries,the remaining number is still formidable. More important than the number ofunknowns, however, is that there is no independent or general means for selectingthem without considering a particular flow. This is because turbulence is indeeda property of the flow, not of the fluid.

3.7 The Reynolds Stress Equations

It is clear from the preceding section that the simple idea of an eddy viscositymight not be the best way to approach the problem of relating the Reynoldsstress to the mean motion. An alternative approach is to try to derive dynamicalequations for the Reynolds stresses from the equations governing the fluctuationsthemselves. Such an approach recognizes that the Reynolds stress is really afunctional6 of the velocity; that is, the stress at a point depends on the velocityeverywhere and for all past times, not just at the point in question and at aparticular instant in time.

The analysis begins with the equation for the instantaneous fluctuating ve-locity, equation 3.21. This can be rewritten for a Newtonian fluid with constantviscosity as:

ρ

#∂ui

∂t+ Uj

∂ui

∂xj

$

= − ∂p

∂xi+

∂τ (v)ij

∂xj− ρ

#

uj∂Ui

∂xj

$

− ρ

%

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩&

(3.27)

Note that the free index in this equation is i. Also, since we are now talking aboutturbulence again, the capital letters represent mean or averaged quantities.

6A functional is a function of a functionTake NS equation and subtract mean momentum equation

So, we need dynamical equations for the Reynolds stresses that recognises that these stresses depend on velocities everywhere now and in past times

Turbulence: lecture



3.7. THE REYNOLDS STRESS EQUATIONS 57

Multiplying equation 3.27 by uk and averaging yields:

ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩

Now since both i and k are free indices they can be interchanged to yield a secondequation given by7:

ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩

Equations 3.28 and 3.29 can be added together to yield an equation for theReynolds stress,

∂⟨uiuk⟩∂t

+ Uj∂⟨uiuk⟩∂xj

= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]

It is customary to rearrange the first term on the right hand side in the fol-lowing way:

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂

∂xj[⟨pui⟩δkj + ⟨puk⟩δij]

The first term on the right is generally referred to as the pressure strain-rate term.The second term is written as a divergence term, and is generally referred to asthe pressure diffusion term. We shall see later that divergence terms can nevercreate nor destroy anything; they can simply move it around from one place toanother.

7Alternatively equation 3.21 can be rewritten with free index k, then multiplied by ui andaveraged



ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂






ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂






ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂




Turbulence: lecture





ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂




Let’s look at this equation term-by-term



ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂




Turbulence: lecture



The first term on the right is pressure-strain term

The second term, written in divergence form, is pressure-diffusion term

Turbulence: lecture





ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂




The first term on the right is pressure-strain term

The second term, written in divergence form, is pressure-diffusion term


Turbulence: lecture





ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂







ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂




Turbulence: lecture



The first term on the right hand side is,“Dissipation of Reynolds stress” by turbulent viscous stress

If we substitute the Newtonian constitutive relation,

The second term is a divergence term

Turbulence: lecture




The third term on the right-hand side of equation 3.30 can similarly be re-written as:

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦ = −[

⟨τ (v)ij

∂uk

∂xj⟩+ ⟨τ (v)kj

∂ui

∂xj⟩]

(3.32)

+∂

∂xj[⟨uiτ

(v)kj ⟩+ ⟨ukτ

(v)ij ⟩]

The first of these is also a divergence term. For a Newtonian fluid, the last is theso-called “dissipation of Reynolds stress” by the turbulence viscous stresses. Thisis easily seen by substituting the Newtonian constitutive relation to obtain:

1

ρ

[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

= 2ν

[

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩]

(3.33)

It is not at all obvious what this has to do with dissipation, but it will becomeclear later on when we consider the trace of the Reynolds stress equation, whichis the kinetic energy equation for the turbulence.

Now if we use the same trick from before using the continuity equation, wecan rewrite the second term on the right-hand side of equation 3.30 to obtain:

[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

=∂

∂xj⟨uiukuj⟩ (3.34)

This is also a divergence term.We can use all of the pieces we have developed above to rewrite equation 3.30

as:





⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦ = −[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

(3.32)

+∂

∂xj[⟨uiτ

(v)kj ⟩+ ⟨ukτ

(v)ij ⟩]


1

ρ

[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

= 2ν

[

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩]

(3.33)



[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

=∂



as:



Turbulence: lecture





⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦ = −[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

(3.32)

+∂

∂xj[⟨uiτ

(v)kj ⟩+ ⟨ukτ

(v)ij ⟩]


1

ρ

[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

= 2ν

[

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩]

(3.33)



[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

=∂



as:





⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦ = −[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

(3.32)

+∂

∂xj[⟨uiτ

(v)kj ⟩+ ⟨ukτ

(v)ij ⟩]


1

ρ

[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

= 2ν

[

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩]

(3.33)



[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

=∂



as:



Turbulence: lecture





ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂







ρ

[

⟨uk∂ui

∂t⟩+ Uj⟨uk

∂ui

∂xj⟩]

= − ⟨uk∂p

∂xi⟩+ ⟨uk

∂τ (v)ij

∂xj⟩ (3.28)

−ρ

[

⟨ukuj⟩∂Ui

∂xj

]

− ρ

⟨ukuj∂ui

∂xj⟩


ρ

[

⟨ui∂uk

∂t⟩+ Uj⟨ui

∂uk

∂xj⟩]

= − ⟨ui∂p

∂xk⟩+ ⟨ui

∂τ (v)kj

∂xj⟩ (3.29)

− ρ

[

⟨uiuj⟩∂Uk

∂xj

]

− ρ

⟨uiuj∂uk

∂xj⟩


∂⟨uiuk⟩∂t


= −1

ρ

[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

−[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

(3.30)

+1

ρ

⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦

−[

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

]


[

⟨ui∂p

∂xk⟩+ ⟨uk

∂p

∂xi⟩]

= ⟨p[∂ui

∂xk+

∂uk

∂xi

]

⟩ (3.31)

+∂




Turbulence: lecture





⎡

⎣⟨ui

∂τ (v)kj

∂xj⟩+ ⟨uk

∂τ (v)ij

∂xj⟩⎤

⎦ = −[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

(3.32)

+∂

∂xj[⟨uiτ

(v)kj ⟩+ ⟨ukτ

(v)ij ⟩]


1

ρ

[

⟨τ (v)ij

∂uk


∂ui

∂xj⟩]

= 2ν

[

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩]

(3.33)



[

⟨uiuj∂uk

∂xj⟩+ ⟨ukuj

∂ui

∂xj⟩]

=∂



as:Note that in each of the three terms on the RHS of

equation 3.30, we have now identified a divergence term

Now, lets group terms that has this divergence form and rewrite equation 3.30.

Turbulence: lecture




∂

∂t⟨uiuk⟩ + Uj

∂

∂xj⟨uiuk⟩ = ⟨p

ρ

!∂ui

∂xk+

∂uk

∂xi

"

⟩

+∂

∂xj

#

−1

ρ[⟨puk⟩δij + ⟨pui⟩δkj ]− ⟨uiukuj⟩

+2ν[⟨sijuk⟩+ ⟨skjui⟩]

−!

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

"

− 2ν

!

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩"

(3.35)

This is the so-called Reynolds Stress Equation which has been the primaryvehicle for much of the turbulence modeling efforts of the past few decades.

The left hand side of the Reynolds Stress Equation can easily be recognizedas the rate of change of Reynolds stress following the mean motion. It seemsto provide exactly what we need: nine new equations for the nine unknowns wecannot account for. The problems are all on the right-hand side. These terms arereferred to respectively as

1. the pressure-strain rate term

2. the turbulence transport (or divergence) term

3. the “production” term, and

4. the “dissipation” term.

Obviously these equations do not involve only Ui and ⟨uiuj⟩, but depend on manymore new unknowns.

It is clear that, contrary to our hopes, we have not derived a single equationrelating the Reynolds stress to the mean motion. Instead, our Reynolds stresstransport equation is exceedingly complex. Whereas the process of averaging theequation for the mean motion introduced only six new independent unknowns,the Reynolds stress, ⟨uiuj⟩, the search for a transport equation which will relatethese to the mean motion has produced many more unknowns. They are:

⟨pui⟩ − 3 unknowns (3.36)

⟨uisjk⟩ − 27 (3.37)

⟨sijsjk⟩ − 9 (3.38)

⟨uiukuj⟩ − 27 (3.39)

⟨p∂ui

∂xj⟩ − 9 (3.40)

TOTAL − 75 (3.41)

Pressure-strain term

dissipation

production

Transport

Rate of change of Reynolds stress following mean motion

Turbulence: lecture



Turbulence: lecture




∂

∂t⟨uiuk⟩ + Uj

∂


ρ

!∂ui

∂xk+

∂uk

∂xi

"

⟩

+∂

∂xj

#

−1



−!

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

"

− 2ν

!

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩"

(3.35)










⟨uisjk⟩ − 27 (3.37)

⟨sijsjk⟩ − 9 (3.38)

⟨uiukuj⟩ − 27 (3.39)

⟨p∂ui

∂xj⟩ − 9 (3.40)

TOTAL − 75 (3.41)


Turbulence: lecture



Closure problem cannot be solved

By deriving new dynamical equations,

or by using “eddy” viscosity models

We need to understand how the turbulence behaves

This may allow us to develop new constitutive models for turbulence and also understands its limitations

Turbulence is a topic that we are still studying!So, this module by definition is incomplete

Turbulence: lecture

Kinetic energy of turbulent fluctuations


We can start with the Reynolds stress equations and contract the free indices (i.e. i = k)


∂

∂t⟨uiuk⟩ + Uj

∂


ρ

!∂ui

∂xk+

∂uk

∂xi

"

⟩

+∂

∂xj

#

−1



−!

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

"

− 2ν

!

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩"

(3.35)










⟨uisjk⟩ − 27 (3.37)

⟨sijsjk⟩ − 9 (3.38)

⟨uiukuj⟩ − 27 (3.39)

⟨p∂ui

∂xj⟩ − 9 (3.40)

TOTAL − 75 (3.41)

Turbulence: lecture



We can start with the Reynolds stress equations and contract the free indices


∂

∂t⟨uiuk⟩ + Uj

∂


ρ

!∂ui

∂xk+

∂uk

∂xi

"

⟩

+∂

∂xj

#

−1



−!

⟨uiuj⟩∂Uk

∂xj+ ⟨ukuj⟩

∂Ui

∂xj

"

− 2ν

!

⟨sij∂uk

∂xj⟩+ ⟨skj

∂ui

∂xj⟩"

(3.35)










⟨uisjk⟩ − 27 (3.37)

⟨sijsjk⟩ − 9 (3.38)

⟨uiukuj⟩ − 27 (3.39)

⟨p∂ui

∂xj⟩ − 9 (3.40)

TOTAL − 75 (3.41)

Turbulence: lecture




We can simplify this equation further

Turbulence: lecture




Chapter 4

The Turbulence Kinetic Energy

4.1 The Kinetic Energy of the Fluctuations

It is clear from the previous chapter that the straightforward application of ideasthat worked well for viscous stresses do not work too well for turbulence Reynoldsstresses. Moreover, even the attempt to directly derive equations for the Reynoldsstresses using the Navier-Stokes equations as a starting point has left us with farmore equations than unknowns. Unfortunately this means that the turbulenceproblem for engineers is not going to have a simple solution: we simply cannotproduce a set of reasonably universal equations. Obviously we are going to haveto study the turbulence fluctuations in more detail and learn how they get theirenergy (usually from the mean flow somehow), and what they ultimately do withit. Our hope is that by understanding more about turbulence itself, we will gaininsight into how we might make closure approximations that will work, at leastsometimes. Hopefully, we will also gain an understanding of when and why theywill not work.

An equation for the fluctuating kinetic energy for constant density flow can beobtained directly from the Reynolds stress equation derived earlier, equation 3.35,by contracting the free indices. The result is:

!∂

∂t⟨uiui⟩ + Uj

∂

∂xj⟨uiui⟩

"

=∂

∂xj

#

−2

ρ⟨pui⟩δij − ⟨q2uj⟩+ 4ν⟨sijui⟩

$

−2⟨uiuj⟩∂Ui

∂xj− 4ν⟨sij

∂ui

∂xj⟩ (4.1)

where the incompressibility condition (∂uj/∂xj = 0) has been used to eliminatethe pressure-strain rate term, and q2 ≡ uiui.

The last term can be simplified by recalling that the velocity deformation ratetensor, ∂ui/∂xj, can be decomposed into symmetric and anti-symmetric parts;

61

We can simplify this equation further


Turbulence: lecture



62 CHAPTER 4. THE TURBULENCE KINETIC ENERGY

i.e.,∂ui

∂xj= sij + ωij (4.2)

where the symmetric part is the strain-rate tensor, sij, and the anti-symmetricpart is the rotation-rate tensor, ωij, defined by:

ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)

Since the double contraction of a symmetric tensor with an anti-symmetric tensoris identically zero, it follows immediately that:

⟨sij∂ui

∂xj⟩ = ⟨sijsij⟩+ ⟨sijωij⟩

= ⟨sijsij⟩ (4.4)

Now it is customary to define a new variable k, the average fluctuating kineticenergy per unit mass, by:

k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)

By dividing equation 4.1 by 2 and inserting this definition, the equation for theaverage kinetic energy per unit mass of the fluctuating motion can be re-writtenas:

!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ 2ν⟨sijui⟩

$

−⟨uiuj⟩∂Ui

∂xj− 2ν⟨sijsij⟩ (4.6)

The role of each of these terms will be examined in detail later. First note thatan alternative form of this equation can be derived by leaving the viscous stress interms of the strain rate. We can obtain the appropriate form of the equation forthe fluctuating momentum from equation 3.21 by substituting the incompressibleNewtonian constitutive equation into it to obtain:

!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)

If we take the scalar product of this with the fluctuating velocity itself and average,it follows (after some rearrangement) that:

!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)

Turbulence: lecture




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)


i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)


Turbulence: lecture



Another way to arrive at this equation is by taking dot product of momentum equation with velocity fluctuation

Turbulence: lecture




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)


i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)



Turbulence: lecture




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)


i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)



Turbulence: lecture




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)

Both equations represent the same thing

Used for modelling

Used for understanding

Turbulence: lecture




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)


i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)

Both equations represent the same thing

Used for modelling

Used for understanding


Turbulence: lecture



4.1. THE KINETIC ENERGY OF THE FLUCTUATIONS 63

Both equations 4.6 and 4.8 play an important role in the study of turbulence.The first form given by equation 4.6 will provide the framework for understandingthe dynamics of turbulent motion. The second form, equation 4.8 forms the basisfor most of the second-order closure attempts at turbulence modelling; e.g., the so-called k-ϵ models (usually referred to as the “k-epsilon models”). This becauseit has fewer unknowns to be modelled, although this comes at the expense of someextra assumptions about the last term. It is only the last term in equation 4.6that can be identified as the true rate of dissipation of turbulence kinetic energy,unlike the last term in equation 4.8 which is only the dissipation when the flowis homogeneous. We will talk about homogeniety below, but suffice it to saynow that it never occurs in nature. Nonetheless, many flows can be assumedto be homogeneous at the scales of turbulence which are important to this term,so-called local homogeniety.

Each term in the equation for the kinetic energy of the turbulence has a distinctrole to play in the overall kinetic energy balance. Briefly these are:

• Rate of change of kinetic energy per unit mass due to non-stationarity; i.e.,time dependence of the mean:

∂k

∂t(4.9)

• Rate of change of kinetic energy per unit mass due to convection (or advec-tion) by the mean flow through an inhomogenous field :

Uj∂k

∂xj(4.10)

• Transport of kinetic energy in an inhomogeneous field due respectively tothe pressure fluctuations, the turbulence itself, and the viscous stresses:

∂

∂xj

!

−1

ρ⟨pui⟩δij −

1


"

(4.11)

• Rate of production of turbulence kinetic energy from the mean flow (gradi-ent):

−⟨uiuj⟩∂Ui

∂xj(4.12)

• Rate of dissipation of turbulence kinetic energy per unit mass due to viscousstresses:

ϵ ≡ 2ν⟨sijsij⟩ (4.13)

These terms will be discussed in detail in the succeeding sections, and the role ofeach examined carefully.Wednesday, 23 October 13

Turbulence: lecture



Turbulence: lecture

KE transport


This transport term is best understood by considering a turbulent flow confined in a control volume70 CHAPTER 4. THE TURBULENCE KINETIC ENERGY

Figure 4.1: Turbulence confined by rigid walls.

side of equation 4.6 for the fluctuating kinetic energy can be integrated over thevolume to yield:

! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν⟨sijui⟩

#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)

where we have used the divergence theorem — again!We assumed our enclosure to have rigid walls; therefore the normal component

of the mean velocity (un = ujnj) must be zero on the surface since there can be noflow through it (the kinematic boundary condition). This immediately eliminatesthe contributions to the surface integral from the ⟨pujnj⟩ and ⟨q2ujnj⟩ terms. Butthe last term is zero on the surface also. This can be seen in two ways: eitherby invoking the no-slip condition which together with the kinematic boundarycondition insures that ui is zero on the boundary, or by noting from Cauchy’stheorem that νsijnj is the viscous contribution to the normal contact force perunit area on the surface (i.e., t(v)n ) whose scalar product with ui must be identicallyzero since un is zero. Therefore the entire integral is identically zero and its netcontribution to the rate of change of kinetic energy is zero.

Thus the only effect of the turbulence transport terms (in a fixed volume atleast) can be to move energy from one place to another, neither creating nordestroying it in the process. This is, of course, why they are collectively calledthe transport terms. This spatial transport of kinetic energy is accomplished bythe acceleration of adjacent fluid due to pressure and viscous stresses (the first

Integrate this within this control volume

Apply divergence theorem, convert this to surface integration

Turbulence: lectureKE transport





! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)





i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)





! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)








! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)










! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)





i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)





! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)








! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)










! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)





i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)





! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)








! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)










! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)





i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)





! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)








! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)





Turbulence: lecture

KE transport





! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)




Now consider the boundary conditionsNormal velocity component is zero




! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)




are eliminated

Either no-slip (ui = 0) or slip (sij = 0) Therefore, the last term is also zero

Role of this transport term is to move energy from one place to another

The transport happens by acceleration of adjacent fluid elements by pressure or viscous stress or by the fluctuation itself

Turbulence: lecture

KE transport


The fact that this term only transports energy is exploited in turbulence modelling

The rationale: This term will transport energy from regions of high energy to regions of low energy

We took 9 unknowns and lumped them together



The fact that this term only transports energy is exploited in turbulence modelling

The rationale: This term will transport energy from regions of high energy to regions of low energy

4.4. THE TRANSPORT (OR DIVERGENCE) TERMS 71

and last terms respectively), and by the physical transport of fluctuating kineticenergy by the turbulence itself (the middle term).

This role of these turbulence transport terms in moving kinetic energy aroundis often exploited by turbulence modellers. It is argued, that on the average, theseterms will only act to move energy from regions of higher kinetic energy to lower.Thus a plausible first-order hypothesis is that this “diffusion” of kinetic energyshould be proportioned to gradients of the kinetic energy itself. That is,

−1

ρ⟨puj⟩ −

1

2⟨q2uj⟩+ ν⟨sijui⟩ = νke

∂k

∂xj(4.32)

where νke is an effective diffusivity like the eddy viscosity discussed earlier. If weuse the alternative form of the kinetic energy equation (equation 4.8), there is noneed to model the viscous term (since it involves only k itself). Therefore ourmodel might be:

−1

ρ⟨puj⟩ −

1

2⟨q2uj⟩ = νkealt

∂k

∂xj(4.33)

These, of course, look much more complicated in a real model because of the needto insure proper tensorial invariance, etc., but the physics is basically the same.

If you think about it, that such a simple closure is worth mentioning at all ispretty amazing. We took 9 unknowns, lumped them together, and replaced theirnet effect by the simple gradient of something we did know (or at least wanted tocalculate), k. And surprisingly, this simple idea works pretty well in many flows,especially if the value of the turbulent viscosity is itself related to other quantitieslike k and ϵ. In fact this simple gradient hypothesis for the turbulence transportterms is at the root of all engineering turbulence models.

There are a couple of things to note about such simple closures though, beforegetting too enthused about them. First such an assumption rules out a counter-gradient diffusion of kinetic energy which is known to exist in some flows. Insuch situations the energy appears to flow up the gradient. While this may seemunphysical, remember we only assumed it flowed down the gradient in the firstplace. This is the whole problem with a plausibility argument. Typically energydoes tend to be transported from regions of high kinetic energy to low kineticenergy, but there is really no reason for it always to do so, especially if there areother mechanisms at work. And certainly there is no reason for it to always betrue locally, and the gradient of anything is a local quantity.

Let me illustrate this by a simple example. Let’s apply a gradient hypothesisto the economy — a plausibility hypothesis if you will. By this simple model,money would always flow from the rich who have the most, to the poor whohave the least. In fact, as history has shown, in the absence of other forces(like revolutions, beheadings, and taxes) this almost never happens. The richwill always get richer, and the poor poorer. And the reason is quite simple, thepoor are usually borrowing (and paying interest), while the rich are doing theloaning (and collecting interest). Naturally there are individual exceptions and

We took 9 unknowns and lumped them together


Turbulence: lecture

KE transport


We replaced complex physical mechanisms by a simple gradient diffusion process

Surprisingly this works pretty well

But, by definition we do not allow counter-gradient diffusion

In fact, this is at the root of almost all engineering turbulence modelling

Although maybe true on average, there is no reason for gradient diffusion to be true locally

Turbulence: lecture







∂k

∂t(4.9)


Uj∂k

∂xj(4.10)


∂

∂xj

!

−1

ρ⟨pui⟩δij −

1


"

(4.11)


−⟨uiuj⟩∂Ui

∂xj(4.12)


ϵ ≡ 2ν⟨sijsij⟩ (4.13)


Turbulence: lecture



Turbulence: lecture

KE production


Turbulence: lectureKE production


4.3. THE PRODUCTION 67

4.3 The Kinetic Energy of the Mean Motion andthe “Production” of Turbulence

An equation for the kinetic energy of the mean motion can be derived by a pro-cedure exactly analogous to that applied to the fluctuating motion. The meanmotion was shown in equation 3.19 to be given by:

ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)

By taking the scalar product of this equation with the mean velocity, Ui, we canobtain an equation for the kinetic energy of the mean motion as:

Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)

Unlike the fluctuating equations, there is no need to average here, since all theterms are already averages.

In exactly the same manner that we rearranged the terms in the equationfor the kinetic energy of the fluctuations, we can rearrange the equation for thekinetic energy of the mean flow to obtain:

!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1

2⟨uiuj⟩Ui + 2ν⟨SijUi⟩

$

+⟨uiuj⟩∂Ui

∂xj− 2ν⟨SijSij⟩ (4.23)

where

K ≡ 1

2Q2 =

1

2UiUi (4.24)

The role of all of the terms can immediately be recognized since each term has itscounterpart in the equation for the average fluctuating kinetic energy.

Comparison of equations 4.23 and 4.6 reveals that the term −⟨uiuj⟩∂Ui/∂xj

appears in the equations for the kinetic energy of BOTH the mean and the fluc-tuations. There is, however, one VERY important difference. This “production”term has the opposite sign in the equation for the mean kinetic energy than inthat for the mean fluctuating kinetic energy! Therefore, whatever its effect onthe kinetic energy of the mean, its effect on the kinetic energy of the fluctuationswill be the opposite. Thus kinetic energy can be interchanged between the meanand fluctuating motions. In fact, the only other term involving fluctuations in theequation for the kinetic energy of the mean motion is a divergence term; thereforeit can only move the kinetic energy of the mean flow from one place to another.




ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)










ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)







ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)





Turbulence: lecture

KE production







ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)





Turbulence: lecture

KE production





i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)




ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)




Positive here

Negative here





i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)




ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)




Positive here

Negative here


Positive

Negative

Turbulence: lecture

KE production


This term is the only way by which energy can be transferred from the mean to the fluctuations

Overall, this exchange can be understood by treating the Reynolds stress as a “stress”






Therefore this “production ” term provides the only means by which energy canbe interchanged between the mean flow and the fluctuations.

Understanding the manner in which this energy exchange between mean andfluctuating motions is accomplished represents one of the most challenging prob-lems in turbulence. The overall exchange can be understood by exploiting theanalogy which treats −ρ⟨uiuj⟩ as a stress, the Reynolds stress. The term:

−ρ⟨uiuj⟩∂Ui/∂xj (4.25)

can be thought of as the working of the Reynolds stress against the mean ve-locity gradient of the flow, exactly as the viscous stresses resist deformation bythe instantaneous velocity gradients. This energy expended against the Reynoldsstress during deformation by the mean motion ends up in the fluctuating motions,however, while that expended against viscous stresses goes directly to internal en-ergy. As we have already seen, the viscous deformation work from the fluctuatingmotions (or dissipation) will eventually send this fluctuating kinetic energy on tointernal energy as well.

Now, just in case you are not all that clear exactly how the dissipation termsreally accomplish this for the instantaneous motion, it might be useful to examineexactly how the above works. We begin by decomposing the mean deformationrate tensor ∂Ui/∂xj into its symmetric and antisymmetric parts, exactly as wedid for the instantaneous deformation rate tensor in Chapter 3; i.e.,

∂Ui

∂xj= Sij + Ωij (4.26)

where the mean strain rate Sij is defined by

Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)

and the mean rotation rate is defined by

Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)

Since Ωij is antisymmetric and −⟨uiuj⟩ is symmetric, their contraction is zero soit follows that:

−⟨uiuj⟩∂Ui

∂xj= −⟨uiuj⟩Sij (4.29)

Equation 4.29 is an analog to the mean viscous dissipation term given forincompressible flow by:

T (v)ij

∂Ui

∂xj= T (v)

ij Sij = 2µSijSij (4.30)

It is easy to show that this term transfers (or dissipates) the mean kinetic energydirectly to internal energy, since exactly the same term appears with the opposite







∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)














∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)









∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)




Turbulence: lecture

KE production








∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)









∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)











∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)









∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)









∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)




Turbulence: lecture

KE production








∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)



The product of two terms here can be positive or negativeHow do we know that this is production?









∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)



The product of two terms here can be positive or negativeHow do we know that this is production?

4.4. THE TRANSPORT (OR DIVERGENCE) TERMS 69

sign in the internal energy equations. Moreover, since SijSij ≥ 0 always, this is aone-way process and kinetic energy is decreased while internal energy is increased.Hence it can be referred to either as “dissipation” of kinetic energy, or as “pro-duction” of internal energy. As surprising as it may seem, this direct dissipationof energy by the mean flow is usually negligible compared to the energy lost tothe turbulence through the Reynolds stress terms. (Remember, there is a termexactly like this in the kinetic energy equation for the fluctuating motion, but in-volving only fluctuating quantities; namely, 2µ⟨sijsij⟩.) We shall show later thatfor almost always in turbulent flow, ⟨sijsij⟩ >> SijSij. What this means is thatthe energy dissipation in a turbulent flow is almost entirely due to the turbulence.

There is a very important difference between equations 4.29 and 4.30. Whereasthe effect of the viscous stress working against the deformation (in a Newtonianfluid) is always to remove energy from the flow (since SijSij > 0 always), the effectof the Reynolds stress working against the mean gradient can be of either sign, atleast in principle. That is, it can either transfer energy from the mean motion tothe fluctuating motion, or vice versa.

Almost always (and especially in situations of engineering importance), −⟨uiuj⟩and Sij have the opposite sign. Therefore, −⟨uiuj⟩Sij > 0 almost always, so ki-netic energy is removed from the mean motion and added to the fluctuations.Since the term −⟨uiuj⟩∂Ui/∂xj usually acts to increase the turbulence kineticenergy, it is usually referred to as the “rate of turbulence energy production”, orsimply the “production”.

Now that we have identified how the averaged equations account for the ‘pro-duction’ of turbulence energy from the mean motion, it is tempting to think wehave understood the problem. In fact, ‘labeling’ phenomena is not the same as‘understanding’ them. The manner in which the turbulence motions cause thisexchange of kinetic energy between the mean and fluctuating motions varies fromflow to flow, and is really very poorly understood. Saying that it is the Reynoldsstress working against the mean velocity gradient is true, but like saying thatmoney comes from a bank. If we want to examine the energy transfer mechanismin detail we must look beyond the single point statistics, so this will have to be astory for another time.

Example: Consider how the production term looks if the Reynolds stress ismodelled by an turbulent viscosity.

4.4 The Transport (or Divergence) Terms

The overall role of the transport terms is best understood by considering a turbu-lent flow which is completely confined by rigid walls as in Figure 4.1. First consideronly the turbulence transport term. If the volume within the confinement is de-noted by Vo and its bounding surface is So, then first term on the right-hand


Turbulence: lecture

KE production


Simple homogeneous shear flow (jet, mixing-layer, pipe, channel)




4.5. THE INTERCOMPONENT TRANSFER OF ENERGY 75

The role of the pressure strain rate terms can best be illustrated by looking at asimple example. Consider a simple homogeneous shear flow in which Ui = U(x2)δ1iand in which the turbulence is homogeneous. For this flow, the assumption ofhomogeneity insures that all terms involving gradients of average quantities vanish(except for dU1/dx2). This leaves only the pressure-strain rate, production anddissipation terms; therefore equations 4.36, 4.37 and 4.40 reduce to:

1-component:

∂⟨u21⟩

∂t= −

!

⟨p∂u2

∂x2⟩+ ⟨p∂u3

∂x3⟩"

− ⟨u1u2⟩∂U1

∂x2− ϵ1 (4.41)

2-component:

∂⟨u22⟩

∂t= +⟨p∂u2

∂x2⟩ − ϵ2 (4.42)

3-component:

∂⟨u23⟩

∂t= +⟨p∂u3

∂x3⟩ − ϵ3 (4.43)

where

ϵ1 ≡ 2ν⟨s1js1j⟩ (4.44)

ϵ2 ≡ 2ν⟨s2js2j⟩ (4.45)

ϵ3 ≡ 2ν⟨s3js3j⟩ (4.46)

It is immediately apparent that only ⟨u21⟩ can directly receive energy from the

mean flow because only the first equation has a non-zero production term.Now let’s further assume that the smallest scales of the turbulence can be

assumed to be locally isotropic. While not always true, this is a pretty goodapproximation for high Reynolds number flows. (Note that it might be exactlytrue in many flows in the limit of infinite Reynolds number, at least away fromwalls.) Local isotropy implies that the component dissipation rates are equal;i.e., ϵ1 = ϵ2 = ϵ3. But where does the energy in the 2 and 3-components comefrom? Obviously the pressure-strain-rate terms must act to remove energy fromthe 1-component and redistribute it to the others.

As the preceding example makes clear, the role of the pressure-strain-rateterms is to attempt to distribute the energy among the various components of theturbulence. An easy way to remember this is to think of the pressure strain rateterms as the ‘Robin Hood’ terms: they steal from the rich and give to the poor.In the absence of other influences, they are so successful that the dissipation byeach component is almost equal, at least at high turbulence Reynolds numbers. Infact, because of the energy re-distribution by the pressure strain rate terms, it isuncommon to find a turbulent shear flow away from boundaries where the kineticenergy of the turbulence components differ by more than 30-40%, no matter whichcomponent gets the energy from the mean flow.


Turbulence: lecture

KE production



Pressure redistributes energy between the components

Production only happens in streamwise component






1-component:

∂⟨u21⟩

∂t= −

!

⟨p∂u2

∂x2⟩+ ⟨p∂u3

∂x3⟩"

− ⟨u1u2⟩∂U1

∂x2− ϵ1 (4.41)

2-component:

∂⟨u22⟩

∂t= +⟨p∂u2

∂x2⟩ − ϵ2 (4.42)

3-component:

∂⟨u23⟩

∂t= +⟨p∂u3

∂x3⟩ − ϵ3 (4.43)

where

ϵ1 ≡ 2ν⟨s1js1j⟩ (4.44)

ϵ2 ≡ 2ν⟨s2js2j⟩ (4.45)

ϵ3 ≡ 2ν⟨s3js3j⟩ (4.46)













1-component:

∂⟨u21⟩

∂t= −

!

⟨p∂u2

∂x2⟩+ ⟨p∂u3

∂x3⟩"

− ⟨u1u2⟩∂U1

∂x2− ϵ1 (4.41)

2-component:

∂⟨u22⟩

∂t= +⟨p∂u2

∂x2⟩ − ϵ2 (4.42)

3-component:

∂⟨u23⟩

∂t= +⟨p∂u3

∂x3⟩ − ϵ3 (4.43)

where

ϵ1 ≡ 2ν⟨s1js1j⟩ (4.44)

ϵ2 ≡ 2ν⟨s2js2j⟩ (4.45)

ϵ3 ≡ 2ν⟨s3js3j⟩ (4.46)








Turbulence: lecture

KE production






! ! !

Vo

∂

∂xj

"

−1

ρ⟨pui⟩δij −

1


#

dV

=! !

So

"

−1

ρ⟨pui⟩δij −

1


#

njdS (4.31)




positivehas to be negative

That means, on average, when u1 is negative u2 is positive and vice-versa

Without mean shear, we have no production

Recall from signal analysis,

hu1u2i =Z 1

1u1(f) u2(f) df

Contribution by fluctuations across different frequencies to the mean shear stress




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)




ρ

!∂Ui

∂t+ Uj

∂Ui

∂xj

"

= −∂P

∂xi+

∂T (v)ij

∂xj− ∂

∂xj(ρ⟨uiuj⟩) (4.21)


Ui

!∂

∂t+ Uj

∂

∂xj

"

Ui = −Ui

ρ

∂P

∂xi+

Ui

ρ

∂T (v)ij

∂xj− Ui

∂⟨uiuj⟩∂xj

(4.22)



!∂

∂t+ Uj

∂

∂xj

"

K =

∂

∂xj

#

−1

ρ⟨PUi⟩δij −

1


$

+⟨uiuj⟩∂Ui


where

K ≡ 1

2Q2 =

1

2UiUi (4.24)




Positive here

Negative here


Turbulence: lecture

KE dissipation


The viscous dissipation in the mean is usually very small compared to the viscous dissipation in the fluctuations

Almost all of the dissipation in turbulence is via the fluctuations

hsijsiji >> hSijSiji

Turbulence: lecture







∂k

∂t(4.9)


Uj∂k

∂xj(4.10)


∂

∂xj

!

−1

ρ⟨pui⟩δij −

1


"

(4.11)


−⟨uiuj⟩∂Ui

∂xj(4.12)


ϵ ≡ 2ν⟨sijsij⟩ (4.13)


Turbulence: lecture



Turbulence: lecture

KE dissipation


This term is always greater than zero

Since it is preceded by negative sign, it is always a sink

Energy is dissipated because the turbulent viscous stress is doing work to resist the deformation of the

fluid element by turbulent strain rate

Turbulence: lectureKE dissipation



4.2 The Rate of Dissipation of the TurbulenceKinetic Energy.

The last term in the equation for the kinetic energy of the turbulence has beenidentified as the rate of dissipation of the turbulence energy per unit mass; i.e.,

ϵ = 2ν⟨sijsij⟩ = ν

!

⟨∂ui

∂xj

∂ui

∂xj⟩+ ⟨∂ui

∂xj

∂uj

∂xi⟩"

(4.14)

It is easy to see that ϵ ≥ 0 always, since it is a sum of the average of squaredquantities only (i.e., ⟨sijsij ≥ 0). Also, since it occurs on the right hand side ofthe kinetic energy equation for the fluctuating motions preceded by a minus sign,it is clear that it can act only to reduce the kinetic energy of the flow. Thereforeit causes a negative rate of change of kinetic energy; hence the name dissipation.

Physically, energy is dissipated because of the work done by the fluctuatingviscous stresses in resisting deformation of the fluid material by the fluctuatingstrain rates; i.e.,

ϵ = ⟨τ (v)ij sij⟩ (4.15)

This reduces to equation 4.14 only for a Newtonian fluid. In non-Newtonianfluids, portions of this product may not be negative implying that it may not allrepresent an irrecoverable loss of fluctuating kinetic energy.

It will be shown in Chapter 5 that the dissipation of turbulence energy mostlytakes place at the smallest turbulence scales, and that those scales can be char-acterized by the so-called Kolmogorov microscale defined by:

ηK ≡#ν3

ϵ

$1/4

(4.16)

In atmospheric motions where the length scale for those eddies having the mostturbulence energy (and most responsible for the Reynolds stress) can be measuredin kilometers, typical values of the Kolmogorov microscale range from 0.1−10 mil-limeters. In laboratory flows where the overall scale of the flow is greatly reduced,much smaller values of ηK are not uncommon. The small size of these dissipativescales greatly complicates measurement of energy balances, since the largest mea-suring dimension must be about equal to twice the Kolmogorov microscale. Andit is the range of scales, L/η, which makes direct numerical simulation of mostinteresting flows impossible, since the required number of computational cells isseveral orders of magnitude greater than (L/η)3. This same limitation also affectsexperiments as well, which must often be quite large to be useful.

One of the consequences of this great separation of scales between those con-taining the bulk of the turbulence energy and those dissipating it is that thedissipation rate is primarily determined by the large scales and not the small.This is because the viscous scales (which operate on a time scale of tK = (ν/ϵ)1/2)dissipate rapidly any energy sent down to them by the non-linear processes of

This term is always greater than zero

Since it is preceded by negative sign, it is always a sink

Energy is dissipated because the turbulent viscous stress is doing work to resist the deformation of the

fluid element by turbulent strain rate


Turbulence: lecture

KE dissipation


mostly in small scalesTurbulence: lecture

KE dissipation


mostly in small scales

of vorticity. Note that the former must therefore be positive – vortex stretching

amplifies turbulent vorticity in the mean, as expected.

DEFINITIONS of Isotropy and homogeneity.

1. A turbulent flow is deemed isotropic if all the statistics of the fluctuating

variables (e.g. ui’) are independent of coordinate system rotations or

reflections – i.e. there is no preferred direction.

2. A turbulent flow is said to be homogeneous if all the fluctuating statistics are

independent of spatial position (and, strictly, all joint probability density

functions remain unchanged if distances are displaced by a constant

amount). So, for example, homogeneity implies that

€

∂∂xj

(ui 'uj ') = 0.











€

∂∂xj

(ui 'uj ') = 0.

Notice that as the Reynolds number increases, so does the separation between the

different scales.

4.2. The energy cascade & Kolmogorov hypotheses Richardson (1922) first introduced the idea that energy enters the turbulence

(through the production mechanism) at the largest scales of motion and this energy

is then transferred (by inviscid, non-linear processes) to smaller and smaller scales

until, at the smallest scales, it is dissipated by viscosity.

Recall (Section 4.1) that the dissipation rate scales with velocity and length scales

appropriate to the large motions – ε ∝ u'3 / L. Kolmogorov (1941) – K41 – proposed

that:

1. Hypothesis of local isotropy: At sufficiently high Re, the small-scale motions are

statistically isotropic (say for length scales l<lEI).

2. 1st Similarity hypothesis: In every turbulent flow at sufficiently high Re, the statistics

of the small-scale motions have a universal form, uniquely determined by ν and ε.

Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2

as given in Section 4.1

The region l<lEI is referred to as the universal equilibrium range.

Recall that the Reynolds number based on these scales is unity and note that

velocity gradients of the dissipative eddies are O( 1/τη ). Recall also the ratios of

smallest to largest scales (eqs. (4.3) and (4.4)).

3. Since, at sufficiently large Re, there must be a range of scales which are much

larger than η but much smaller than L, K41 also proposed his:

2nd similarity hypothesis, that in every turbulent flow at sufficiently large Re, the

statistics of the motions of scale l in the range η<<l<<L have a universal form which is

uniquely determined by ε. This is normally called the ‘inertial sub-range’.

Thus we can define various ranges as in figure below.















€

∂∂xj

(ui 'uj ') = 0.











€

∂∂xj

(ui 'uj ') = 0.


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2


























€

∂∂xj

(ui 'uj ') = 0.











€

∂∂xj

(ui 'uj ') = 0.


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2













Turbulence: lecture

KE dissipation


mostly in small scalesTurbulence: lecture

KE dissipation













€

∂∂xj

(ui 'uj ') = 0.











€

∂∂xj

(ui 'uj ') = 0.


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2

















different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2












9.8. ISOTROPY 179

where A(r) and B(r) are two functions which depend on r only, and δij is theisotropic tensor (or Kronecker delta tensor). We will find considerable use for thespectral equivalent of this result in the next chapter.

It is customary (and convenient) to express the functions A and B in terms ofthe two-point velocity correlations we commonly measure (see Figure 9.5). Thefirst of these is denoted as BLL(r) and called the longitudinal correlation becausethe two-components of velocity are aligned with the separation vector. And thesecond is denoted as BNN(r) and called the normal or transverse correlation,since the two components of velocity are chosen to be in the same plane butperpendicular to the separation variable.

To see how these determine A and B simply follow the directions above. Forexample BLL(r) would have to be the same as B1,1(r, 0, 0), but B2,2(0, r, 0) wouldalso suffice (as would any choice for which the vectors are collinear). Substitutinginto equation 9.70 yields BLL(r) as:

BLL(r) = B1,1(r, 0, 0) = A(r)r2 +B(r) (9.71)

Similarly, BNN(r) can be determined from:

BNN(r) = B2,2(r, 0, 0) = B(r) (9.72)

since r2r2 = 0 if r = (r, 0, 0). We now have two equations in two unknowns, so wecan solve for A(r) and B(r) to obtain the general form:

Bi,j(r) = [BLL(r)− BNN(r)]rirjr2

+BNN(r)δij (9.73)

The number of unknown functions can be reduced to one (either BLL or BNN) byemploying the continuity equations derived in above in Section 9.5

Exercise: Find BLL in terms of BNN Use the isotropic relation of equation 9.73together with the incompressible continuity equations 9.43 and 9.44 to obtain:

BLL = BNN + rdBLL

dr(9.74)

(Hint: Note that ∂BLL/∂ri = dBLL/dr (∂r/∂ri) and that ∂r/∂rj = rj/r. Showthis by differentiating r2 = r21 + r22 + r23) term by term.)

9.8.4 Derivative moments in isotropic turbulence

Isotropy has a powerful influence on the velocity derivative moments. In fact,only one of them is independent, so all of the others can be expressed in terms ofit. For example, if we choose to express the others in term of ⟨[∂u1/∂x1]2⟩, theresults are:

⟨!∂u1

∂x1

"2⟩ = ⟨

!∂u2

∂x2

"2⟩ = ⟨

!∂u3

∂x3

"2⟩, (9.75)

180 CHAPTER 9. HOMOGENEOUS RANDOM PROCESSES

⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)

It follows immediately the dissipation for isotropic turbulence can be quitesimply expressed as:

ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)

But any of the other derivatives could have been used as well; e.g.,

ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)

9.8.5 Isotropic integral scale and Taylor microscales

It is straightforward to show from the form of the isotropic correlation functionthat the integral scales based on the longitudinal correlation, BLL(r), is doublethat of the normal correlation, BNN(r).

Exercise: Prove that L(1)2,2 and L(2)

1,1 are twice L(1)1.1. (Hint: Use equation 9.74 and

integrate by parts.)

It is similarly quite easy to show from the derivative relations and the defi-nitions of the Taylor microscales that λf =

√2λg. It follows immediately that

the dissipation can be expressed in terms of the Taylor microscales for isotropicturbulence as:

ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)






different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2












9.8. ISOTROPY 179




BLL(r) = B1,1(r, 0, 0) = A(r)r2 +B(r) (9.71)


BNN(r) = B2,2(r, 0, 0) = B(r) (9.72)



+BNN(r)δij (9.73)



BLL = BNN + rdBLL

dr(9.74)




⟨!∂u1

∂x1

"2⟩ = ⟨

!∂u2

∂x2

"2⟩ = ⟨

!∂u3

∂x3

"2⟩, (9.75)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)






different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2












9.8. ISOTROPY 179




BLL(r) = B1,1(r, 0, 0) = A(r)r2 +B(r) (9.71)


BNN(r) = B2,2(r, 0, 0) = B(r) (9.72)



+BNN(r)δij (9.73)



BLL = BNN + rdBLL

dr(9.74)




⟨!∂u1

∂x1

"2⟩ = ⟨

!∂u2

∂x2

"2⟩ = ⟨

!∂u3

∂x3

"2⟩, (9.75)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)






different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2












9.8. ISOTROPY 179




BLL(r) = B1,1(r, 0, 0) = A(r)r2 +B(r) (9.71)


BNN(r) = B2,2(r, 0, 0) = B(r) (9.72)



+BNN(r)δij (9.73)



BLL = BNN + rdBLL

dr(9.74)




⟨!∂u1

∂x1

"2⟩ = ⟨

!∂u2

∂x2

"2⟩ = ⟨

!∂u3

∂x3

"2⟩, (9.75)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)


⟨!∂u1

∂x1

"2⟩ =

1

2⟨!∂u1

∂x2

"2⟩ = 1

2⟨!∂u2

∂x1

"2⟩

=1

2⟨!∂u1

∂x3

"2⟩ = 1

2⟨!∂u3

∂x1

"2⟩

=1

2⟨!∂u2

∂x3

"2⟩ = 1

2⟨!∂u3

∂x2

"2⟩ (9.76)

and

⟨!∂u1

∂x1

"2⟩ = −⟨∂u1

∂x2

∂u2

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x1⟩

= −⟨∂u2

∂x3

∂u3

∂x2⟩ (9.77)


ϵ = 15ν⟨!∂u1

∂x1

"2⟩ (9.78)


ϵ =15

2ν⟨!∂u2

∂x1

"2⟩. (9.79)









ε = 30νu2

λ2f

= 15νu2

λ2g

(9.80)




Scales of turbulent motion4. Scales of Motion in Turbulent Flow

4.1 General Comments

We have noted the wide range of time scales that exist in typical turbulent flows.

Consider a turbulent shear flow:

The following length and velocity scales apply:

x, U global length and velocity scales

Lx, u’ length and velocity scales of the large motions

λ, u’ length and velocity scales for dissipation process

η, v length and velocity scales for the smallest (viscous) motions – Kolmogorov

scales – see later

ξ, a length and velocity scales of the molecular motions (mean free path and

speed of sound)

The smallest scales are created by vortex stretching. Imagine a vortex tube inside a

turbulent flow:

U

U=0

x

δ ∼ Lx

λ

Γ=ω 1 A

A

Γ=ω2a

a

t=t1 t=t2

If a<A, ω2<ω 1

Remember the power spectrum

Supposed to give us frequency by frequency distribution of energy

content (or scale-by-scale)


Turbulence: lecture

KE dissipation


Scales of turbulent motion




Scales of turbulent motion4. Scales of Motion in Turbulent Flow

4.1 General Comments

We have noted the wide range of time scales that exist in typical turbulent flows.

Consider a turbulent shear flow:

The following length and velocity scales apply:

x, U global length and velocity scales

Lx, u’ length and velocity scales of the large motions

λ, u’ length and velocity scales for dissipation process

η, v length and velocity scales for the smallest (viscous) motions – Kolmogorov

scales – see later

ξ, a length and velocity scales of the molecular motions (mean free path and

speed of sound)

The smallest scales are created by vortex stretching. Imagine a vortex tube inside a

turbulent flow:

U

U=0

x

δ ∼ Lx

λ

Γ=ω 1 A

A

Γ=ω2a

a

t=t1 t=t2

If a<A, ω2<ω 1







Turbulence: lecture

KE dissipation


Scales of turbulent motionA lot of the energy is

in the large-scales

hu2i =Z 1

0Euu(f)df

Apply Taylor’s hypothesis

This log-log representation shows that energy distrbiution



Scales of turbulent motionA lot of the energy is

in the large-scales

hu2i =Z 1

0Euu(f)df

Apply Taylor’s hypothesis

hu2i =Z 1

0Euu(k)dk

This log-log representation shows that energy distrbiution


Turbulence: lecture

KE dissipationScales of turbulent motion

Under homogeneous isotropic assumptions

Let’s see where the kinetic energy and dissipation are distrbuted across the scales

Turbulence: lectureKE dissipationScales of turbulent motion


KE =

Z 1

0Euu dk =

Z 1

0k Euu d(ln(k))

hsijsiji = 15

*@u1

@x1

2+

= 7.5

*@u1

@x2

2+

Dissipation = 2hsijsiji

In fourier space, @u1/@x1 is just premultiplication with wavenumber

15

*@u1

@x1

2+

= 15

Z 1

0k

2Euu(k)dk = 15

Z 1

0k

3Euu(k)d(ln(k))





KE =

Z 1

0Euu dk =

Z 1

0k Euu d(ln(k))

hsijsiji = 15

*@u1

@x1

2+

= 7.5

*@u1

@x2

2+



15

*@u1

@x1

2+

= 15

Z 1

0k

2Euu(k)dk = 15

Z 1

0k

3Euu(k)d(ln(k))





KE =

Z 1

0Euu dk =

Z 1

0k Euu d(ln(k))

hsijsiji = 15

*@u1

@x1

2+

= 7.5

*@u1

@x2

2+



15

*@u1

@x1

2+

= 15

Z 1

0k

2Euu(k)dk = 15

Z 1

0k

3Euu(k)d(ln(k))



Turbulence: lecture


kE

ln(k)

k3E

spread increases with Reynolds number

This shows that the dissipation is in the small-scales

How does it get there?

Turbulence: lecture


Integrate this over the volume, and apply divergence theorem,

!A = constant

The fluctuating field stretches the intial vortex and through this process transfers energy to the smaller scales

This process continues until it is limited by viscosity

Turbulence: lectureKE dissipationScales of turbulent motionconsider the continuity equation, r · u = 0 and take the curl, r · ! = 0


!A = constant

t = t1

!1, A1

t = t2 > t1

!2 > !1, A2 < A1


This process continues until it is limited by viscosityWednesday, 23 October 13

t = t2 > t1



!A = constant

t = t1

!1, A1

t = t2 > t1

!2 > !1, A2 < A1



t = t1



!A = constant

t = t1

!1, A1

t = t2 > t1

!2 > !1, A2 < A1



Turbulence: lecture


“Big whorls have little whorls that feed on their velocity, and little whorls have lesser whorls and so on to viscosity”

-L.F. Richardson, 1922This describes the cascade process

"Great fleas have little fleas upon their backs to bite 'em, And little fleas have lesser fleas, and so ad infinitum."

- Jonathan Swift, 1915

Turbulence: lecture

KE dissipation

It turns out that the dissipation rate is determined by the large-scales and not the small-scales even though the

dissipation is happening at the small-scales







∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)



scales as u03

Lx

Then, if dissipation rate is controlled by this process, dimensionally, dissipation rate should also scale the same way

Cornerstone of turbulence theory


It turns out that the dissipation rate is determined by the large-scales and not the small-scales even though the

dissipation is happening at the small-scales







∂Ui

∂xj= Sij + Ωij (4.26)


Sij =1

2

!∂Ui

∂xj+

∂Uj

∂xi

"

(4.27)


Ωij =1

2

!∂Ui

∂xj− ∂Uj

∂xi

"

(4.28)


−⟨uiuj⟩∂Ui



T (v)ij

∂Ui

∂xj= T (v)



scales as u03

Lx

Then, if dissipation rate is controlled by this process, dimensionally, dissipation rate should also scale the same way

2hsij

sij

i = u03

Lx

Cornerstone of turbulence theory

Au03

Lx


Turbulence: lecture

KE dissipation

Lets look at the scaling for dissipation itself

What other scaling law can we think of?

Not the smallest scale of motion.Characteristic scale for dissipation



hsij

sij

i u03

Lx

u02

2 u03

Lx

Lets say that the velocity scale is the same, but, the length scale is different, (sort of obvious from the spectrum)

*@u1

@x1

2+

Recall, from the signal analysis

*@u1

@x1

2+

hu2i

2

: Taylor microscale

: Taylor microscale

Not the smallest scale of motion.Charactersistic scale for dissipationWednesday, 23 October 13



hsij

sij

i u03

Lx

u02

2 u03

Lx


*@u1

@x1

2+


*@u1

@x1

2+

hu2i

2

: Taylor microscale

: Taylor microscale




hsij

sij

i u03

Lx

u02

2 u03

Lx


*@u1

@x1

2+


*@u1

@x1

2+

hu2i

2

: Taylor microscale

: Taylor microscale




hsij

sij

i u03

Lx

u02

2 u03

Lx


*@u1

@x1

2+


*@u1

@x1

2+

hu2i

2

: Taylor microscale

: Taylor microscale


Turbulence: lecture

KE dissipation

If we rearrange this equation,

This shows that the ratio between the characteristic length scale of dissipation and the characteristic length scale of

energy containing motions increases with Reynolds number

ln(k)

E D


What are the smallest scales of motion?


u02

2 u03

Lx


Lx

= Re1/2L

x



ln(k)

E D


What are the smallest scales of motion?Wednesday, 23 October 13


u02

2 u03

Lx


Lx

= Re1/2L

x



ln(k)

E D


What are the smallest scales of motion?Wednesday, 23 October 13

Turbulence: lecture

KE dissipationSmallest scales of motionIf we scale dissipation rate with these scales,

u2

2

But, at this scale the energy is immediately dissipated by viscosity and converted to internal energy

The local Reynolds number has to be close to 1 (whatever comes from large scales is immediately lost)

Turbulence: lectureKE dissipationSmallest scales of motion

Length scale: Velocity scale: u

Time scale: tIf we scale dissipation rate with these scales,

u2

2



Re =u

1





u2

2



Re =u

1


Turbulence: lecture

KE dissipationSmallest scales of motion

We can rearrange these two equations and solve,

u'3

Lx

≈ ν u'2

λ2 .

Thus:

λLx

≈ ReLx

−1 / 2 ; ReLx = u' Lx

ν (4.1)

λ is called the dissipation, or Taylor, microscale. It is rather like δ in a laminar

boundary layer and is thus a sort of viscous scale for the turbulent flow. Note that it

is NOT a measure of the very smallest scales, but rather a measure of the order of

magnitude of velocity derivatives (

2

2 ~ ¸¸¹

·¨¨©

§

j

iji

xuuu

∂∂

λ).

For the very smallest scales, we note that no matter what the Reynolds number,

there is a range of scales where the Reynolds number based on local velocity and

length scales is of order unity. Thus we introduce a new length, η, and velocity

scale, v, such that:

u'3

Lx≈ ε ≈ ν v2

η2 and ηvν = 1.

Solving for η and v yields:

η ≈ (ν3 / ε)1 / 4 , v ≈ (νε)1/ 4 . (4.2)

So, effectively, once the large scales (u’ and Lx) are fixed, multiple instabilities occur

until there are fluctuations with scales v and η such that P~ε. Using the above

relations we can estimate useful ratios of scales:

ηLx

≈ ν 3

Lx4ε

§

© ¨ ¨

·

¹ ¸ ¸

1/ 4

= ν 3Lx

Lx4u'3

§

© ¨ ¨

·

¹ ¸ ¸ = Re L

x

−3/ 4 (4.3)

vu'

≈ νεu'4§ © ¨

· ¹ ¸

1/ 4

= νu'3

Lx u'4§

© ¨

·

¹ ¸

1/ 4

= Re Lx

−1/ 4 (4.4)

Note that if we wanted to compute all the scales in the turbulent flow, we’d need a

number of grid points of order N = (Lx /η)3 = ReLx 9/4, which is 3 x 1013 for RLx=106!!

We can also easily show that the vorticity at small scales is much larger (O(ReLx1/2))

than at large scales, as we anticipated – velocity gradients are large cf. velocities.

Kolmogrov scales of motion



Time scale: t u2

2Re =

u

1and


3

14

u ()14 t

12

u'3

Lx

≈ ν u'2

λ2 .

Thus:

λLx

≈ ReLx

−1 / 2 ; ReLx = u' Lx

ν (4.1)





2

2 ~ ¸¸¹

·¨¨©

§

j

iji

xuuu

∂∂

λ).





u'3

Lx≈ ε ≈ ν v2

η2 and ηvν = 1.


η ≈ (ν3 / ε)1 / 4 , v ≈ (νε)1/ 4 . (4.2)




ηLx

≈ ν 3

Lx4ε

§

© ¨ ¨

·

¹ ¸ ¸

1/ 4

= ν 3Lx

Lx4u'3

§

© ¨ ¨

·

¹ ¸ ¸ = Re L

x

−3/ 4 (4.3)

vu'

≈ νεu'4§ © ¨

· ¹ ¸

1/ 4

= νu'3

Lx u'4§

© ¨

·

¹ ¸

1/ 4

= Re Lx

−1/ 4 (4.4)






t

tL

x

=

p/

Lx

/u0 Re1/2L





u2

2



Re =u

1




Time scale: t u2

2Re =

u

1and


3

14

u ()14 t

12

u'3

Lx

≈ ν u'2

λ2 .

Thus:

λLx

≈ ReLx

−1 / 2 ; ReLx = u' Lx

ν (4.1)





2

2 ~ ¸¸¹

·¨¨©

§

j

iji

xuuu

∂∂

λ).





u'3

Lx≈ ε ≈ ν v2

η2 and ηvν = 1.


η ≈ (ν3 / ε)1 / 4 , v ≈ (νε)1/ 4 . (4.2)




ηLx

≈ ν 3

Lx4ε

§

© ¨ ¨

·

¹ ¸ ¸

1/ 4

= ν 3Lx

Lx4u'3

§

© ¨ ¨

·

¹ ¸ ¸ = Re L

x

−3/ 4 (4.3)

vu'

≈ νεu'4§ © ¨

· ¹ ¸

1/ 4

= νu'3

Lx u'4§

© ¨

·

¹ ¸

1/ 4

= Re Lx

−1/ 4 (4.4)






t

tL

x

=

p/

Lx

/u0 Re1/2L




Time scale: t u2

2Re =

u

1and


3

14

u ()14 t

12

u'3

Lx

≈ ν u'2

λ2 .

Thus:

λLx

≈ ReLx

−1 / 2 ; ReLx = u' Lx

ν (4.1)





2

2 ~ ¸¸¹

·¨¨©

§

j

iji

xuuu

∂∂

λ).





u'3

Lx≈ ε ≈ ν v2

η2 and ηvν = 1.


η ≈ (ν3 / ε)1 / 4 , v ≈ (νε)1/ 4 . (4.2)




ηLx

≈ ν 3

Lx4ε

§

© ¨ ¨

·

¹ ¸ ¸

1/ 4

= ν 3Lx

Lx4u'3

§

© ¨ ¨

·

¹ ¸ ¸ = Re L

x

−3/ 4 (4.3)

vu'

≈ νεu'4§ © ¨

· ¹ ¸

1/ 4

= νu'3

Lx u'4§

© ¨

·

¹ ¸

1/ 4

= Re Lx

−1/ 4 (4.4)






t

tL

x

=

p/

Lx

/u0 Re1/2L


Turbulence: lecture

KE dissipationKolmogrov hypothesis

From dimensional analysis,


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2













Kolmogrov hypothesis


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2













= f(, ) u = f(, ) t = f(, ) : L u : LT1 t : T : L2T1 : L2T3

3

14

u ()14 t

12


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2
















different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2














3

14

u ()14 t

12


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2
















different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2














3

14

u ()14 t

12


different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2













Turbulence: lecture



different scales.







that:





Thus:

η ≡ (ν 3 / ε)1 / 4

v ≡ (εν )1/ 4

τη = (ν / ε)1 / 2












What is this range and how do we get to this?

Turbulence: lecture


Let’s define velocity and time-scales that depend on local length scale ‘l’ and dissipation

Rate of energy transfer T ~ O(u2/t) Independent of local length scale and viscosity

Kinetic energy trickles down the scale at the same rate no matter what the length scale is




Given a length scale in this inertial sub-range, l, corresponding velocity and time

scales can be expressed as:

3/23/23/12

3/13/13/1

)/()/()/()()/(’)/()()(

xo

x

Lllll

Llulvllu

τητετηε

η ∝==∝==

(where το is time scale of large eddies – Lx/u’). This suggests that the rate, T(l), at

which energy is transferred from eddies larger than l to eddies smaller that l, which

might be expected to be O(u2/τ) if it’s accomplished primarily by eddies of scale l, is

in fact independent of l. This turns out to be true.

The energy Spectrum

In terms of the wavenumber spectrum (wavenumber κ=2π/l), total turbulence energy

is ³∞

=0

)( κκ dEkt . Using ε and κ ( wavenumber κ=2π/l) to non-dimensionalise the

spectrum, we thus have:

E(κ ) = ε 2 / 3κ −5/ 3Ψ(κη) ,

where Ψ(κη) is the (non-dimensional) ‘compensated Kolmogorov spectrum

function’. In the inertial sub-range, the second similarity hypothesis implies that for

κη<<1, this function must become a constant (since it contains η). Hence:

E(κ ) =Cε 2 / 3κ −5/ 3 .

(This is the famous –5/3 spectrum).

Lx lEI lDI

η

Energy-containing range

Dissipation range

Inertialsub-range

Dissipation = εProduction = PEnergy transfer T(l)

Decreasing length scale


This is large time-scale








3/23/23/12

3/13/13/1

)/()/()/()()/(’)/()()(

xo

x

Lllll

Llulvllu

τητετηε

η ∝==∝==





The energy Spectrum


is ³∞

=0



E(κ ) = ε 2 / 3κ −5/ 3Ψ(κη) ,




E(κ ) =Cε 2 / 3κ −5/ 3 .


Lx lEI lDI

η


Dissipation range

Inertialsub-range




This is large time-scale



Turbulence: lecture


Using dimensional analysis, k : L1, : L2T3, E : L3T2

E(k) = 2/3k5/3g(kL, k)




k = 2/l, where l is the length scaleE(k) = E(k, Lx

, , ),In non-dimensional form, E(k) = E(, k, kL

x

, k)

And saying that rate of energy transfer should be independent of lengthscale and viscosity

E(k) = 2/3k5/3g(kL, k)





k = 2/l, where l is the length scaleE(k) = E(k, Lx

, , ),In non-dimensional form, E(k) = E(, k, kL

x

, k)

And saying that rate of energy transfer should be independent of lengthscale and viscosity

E(k) = 2/3k5/3g(kL, k)


Turbulence: lecture


E(k) = 2/3k5/3g(kL, k)

Lecture 11

KPhysical picture: energy trickles down the scales till it gets to a length-scale where

the energy dissipates by viscous forces. This length-scale must depend on the rate of energy transfer which

K

H

1

3 4

K K

determines how quickly energy is dissipated and viscosity :

, is the Kolmogorov micro-scale.

ª º« »« »« »Q« »« »§ ·Q« »K K H Q ¨ ¸H« »© ¹¬ ¼

Hence: ,

523 3

Kindependentof L and

E k E k,L, , k g kL,k

Q

H K H K K by dimensional analysis and carry on

with argument very similar to that which leads to the log law. This argument works with KE k,L, , and not with e.g. E k,E,L, .H K Q

(a) Inner scaling: at wavenumbers k too large compared to 2 ,LS we can assume that the

energy spectrum does not know how large the experimental setup is i.e. in the limit where kL>>2S we assume that K i Kg kL,k g k .K o K

(b) Outer scaling: at wavenumbers k too small to resolve KK we can assume that the

energy spectrum does not know what the value of Q is since the energy is simply transferred and not dissipated, equally for all values of small enough Q. Hence, in the limit where Kk 2 , we assume thatK S

K 0g kL,k g kLK o .

(c) Overlap region exists if

43

L LK

LRe 1 as Re .§ ·

!! ¨ ¸K© ¹ This region is called the inertial

range and is where

K

2 2kLS S

K

In the inertial range, 5 5

2 23 33 3

i K 0E k k g k k g kL

H K H

i K 0 i 0 g k g kL g g constant K (independent of L and This constant is denoted and is called the Kolmogorov Constant. It follows that

K ).K

KC

5

2 33

KK

2 2E k C k in inertial range k .L

S SH

K

Lecture 11



K

H

1

3 4

K K



ª º« »« »« »Q« »« »§ ·Q« »K K H Q ¨ ¸H« »© ¹¬ ¼

Hence: ,

523 3



Q







K 0g kL,k g kLK o .


43

L LK

LRe 1 as Re .§ ·


range and is where

K

2 2kLS S

K


2 23 33 3


H K H


K ).K

KC

5

2 33

KK


S SH

K

Turbulence: lecture

KE dissipation

E(k) = 2/3k5/3g(kL, k)

31

The energy spectrum as a function of Rλ

2( )Euηκ

η

κ η

Rλ=30

100

300

1000Inner scaling

Lecture 11



K

H

1

3 4

K K



ª º« »« »« »Q« »« »§ ·Q« »K K H Q ¨ ¸H« »© ¹¬ ¼

Hence: ,

523 3



Q







K 0g kL,k g kLK o .


43

L LK

LRe 1 as Re .§ ·


range and is where

K

2 2kLS S

K


2 23 33 3


H K H


K ).K

KC

5

2 33

KK


S SH

K


E(k) = 2/3k5/3g(kL, k)

31


2( )Euηκ

η

κ η

Rλ=30

100

300

1000Inner scaling

32


0

( )Ek lκ

0lκ

Rλ=30 100300

1000

Outer scaling

Lecture 11



K

H

1

3 4

K K



ª º« »« »« »Q« »« »§ ·Q« »K K H Q ¨ ¸H« »© ¹¬ ¼

Hence: ,

523 3



Q







K 0g kL,k g kLK o .


43

L LK

LRe 1 as Re .§ ·


range and is where

K

2 2kLS S

K


2 23 33 3


H K H


K ).K

KC

5

2 33

KK


S SH

K


Turbulence: lecture

KE dissipation

E(k) = C2/3k5/3

The famous -5/3 spectrum

The value of C from experiments is about 1.5

31


2( )Euηκ

η

κ η

Rλ=30

100

300

1000Inner scaling


E(k) = 2/3k5/3g(kL, k)

31


2( )Euηκ

η

κ η

Rλ=30

100

300

1000Inner scaling

32


0

( )Ek lκ

0lκ

Rλ=30 100300

1000

Outer scaling

Lecture 11



K

H

1

3 4

K K



ª º« »« »« »Q« »« »§ ·Q« »K K H Q ¨ ¸H« »© ¹¬ ¼

Hence: ,

523 3



Q







K 0g kL,k g kLK o .


43

L LK

LRe 1 as Re .§ ·


range and is where

K

2 2kLS S

K


2 23 33 3


H K H


K ).K

KC

5

2 33

KK


S SH

K


Turbulence: lecture

KE dissipation

Thursday, 22 November 12

-5/3 spectrum for various flows across different

Reynolds numbers

The plot is in “inner” scaling and hence collapses at small

scales

Very fast roll-off after the inertial range due to viscous dissipation

Turbulence: lecture




3/23/23/12

3/13/13/1

)/()/()/()()/(’)/()()(

xo

x

Lllll

Llulvllu

τητετηε

η ∝==∝==





The energy Spectrum


is ³∞

=0



E(κ ) = ε 2 / 3κ −5/ 3Ψ(κη) ,




E(κ ) =Cε 2 / 3κ −5/ 3 .


Lx lEI lDI

η


Dissipation range

Inertialsub-range



Turbulence: lecture


This one-way cascade is in fact incorrect

There is some evidence for energy being transferred in both directions

However, the “net” transfer is from large to small

There is a whole research industry trying to disprove Kolmogrov hypothesis

Turbulence: lecture

Vorticity and enstrophy equations



[2] - Tilting or stretching of vorticity

Also known as vortex stretching. This happens when the rate of strain has a component along vorticity.This is crucial as this is responsible for production of enstrophy and hence sustaining turbulence

Turbulence: lectureVorticity and enstrophy equations



[2] - Tilting or stretching of vorticityAlso known as vortex stretching. This happens when the rate of strain has a component along vorticity.This is crucial as this is responsible for production of enstrophy and hence sustaining turbulence














[1] [2] [3]







)4.1(2

jj

i

j

ij

j

ij

i

xxxu

xu

t ∂∂ω∂

ν∂∂

ω∂∂ω

∂∂ω

+=+

94

From tensorial algebra, the following identities hold:

r (r) 0, r · (r u) 0, ru2/2 u ·ru + ur u, (A.10)

r (AB) (B ·r+ B r)A + (A ·r+ Ar)B, (A.11)

where is a scalar field, u is a vector field with modulus u kuk, and A and B are second-order

tensor fields. Use these tensor identities and the incompressibility condition, r · u = 0 to rewrite

equation A.9 as:

@!

@t+ u ·r! = ! ·ru + r2

!. (A.12)

The first term of the right-hand side of equation A.12 is responsible for the vortex-stretching.

Dot product equation A.12 with !, decompose the velocity gradient tensor as ru = S + in

the first term of the right-hand side, and use the antisymmetry of

1 to obtain:

DD t

12

!i

!i

= !i

Sik

!k

+ !i

@2!i

@xk

@xk

. (A.13)

To express equation A.13 in terms of , particularize

ij

@2kj

@xp

@xp

=14

ijl

!l

kjm

@2!m

@xp

@xp

=14

!m

@2!m

@xp

@xp

ik

!k

@2!i

@xp

@xp

(A.14)

for k = i, obtaining:

ij

@2ij

@xp

@xp

=12

!i

@2!i

@xp

@xp

. (A.15)

Then, from equation A.6 and the antisymmetry of it results !i

!i

= 2 ij

ij

, !i

!j

= 4 ik

kj

+

2 ij

mn

mn

, which can be substituted, along with the relation A.15, into equation A.13 to obtain:

DD t

12

ij

ij

= 2ik

kj

Sji

+ ij

@2ij

@xk

@xk

, (A.16)

1If A is an antisymmetric second-order tensor, then a · A · a = 0, for any vector field a.


Turbulence: lecture

Vorticity and enstrophy equations

It is more useful to look at kinetic energy equivalent for vorticity: Enstrophy

94

From tensorial algebra, the following identities hold:

r (r) 0, r · (r u) 0, ru2/2 u ·ru + ur u, (A.10)

r (AB) (B ·r+ B r)A + (A ·r+ Ar)B, (A.11)

where is a scalar field, u is a vector field with modulus u kuk, and A and B are second-order

tensor fields. Use these tensor identities and the incompressibility condition, r · u = 0 to rewrite

equation A.9 as:

@!

@t+ u ·r! = ! ·ru + r2

!. (A.12)

The first term of the right-hand side of equation A.12 is responsible for the vortex-stretching.

Dot product equation A.12 with !, decompose the velocity gradient tensor as ru = S + in

the first term of the right-hand side, and use the antisymmetry of

1 to obtain:

DD t

12

!i

!i

= !i

Sik

!k

+ !i

@2!i

@xk

@xk

. (A.13)

To express equation A.13 in terms of , particularize

ij

@2kj

@xp

@xp

=14

ijl

!l

kjm

@2!m

@xp

@xp

=14

!m

@2!m

@xp

@xp

ik

!k

@2!i

@xp

@xp

(A.14)

for k = i, obtaining:

ij

@2ij

@xp

@xp

=12

!i

@2!i

@xp

@xp

. (A.15)

Then, from equation A.6 and the antisymmetry of it results !i

!i

= 2 ij

ij

, !i

!j

= 4 ik

kj

+

2 ij

mn

mn

, which can be substituted, along with the relation A.15, into equation A.13 to obtain:

DD t

12

ij

ij

= 2ik

kj

Sji

+ ij

@2ij

@xk

@xk

, (A.16)

1If A is an antisymmetric second-order tensor, then a · A · a = 0, for any vector field a.

Dot product of this equation with ! and decompose ru = S +

Also, use the fact that is anti-symmetric,

Rate of change of enstrophy = Interaction + diffusion

12

D!2

Dt = !iSij!j + !ir2!i

Turbulence: lecture

Enstrophy and dissipation equations

We can do the same sort of thing for dissipation

12

D!2

Dt = !iSij!j + !ir2!i

12

D

2

Dt

= Sij

Sjk

Ski

14!

i

Sij

!j

Sij

@

2p

@xi@xj+ S

ij

r2Sij

Selfinteraction

vorticitystrain

interaction

pressurestrain

interaction

Viscous diffusion

vorticitystrain

interaction

Viscous diffusion

Turbulence: lectureStrain-rotation interaction

!iSij!j = !2i(ei · !)i are the principal strain rates (i.e. eigen values of the strain rate tensor)

1 > 2 > 31: Extensive strain rate

2: Intermediate strain rate

3: Compressive strain rate

1 + 2 + 3 = 0

Irrespective of the eigen value, a principal strain rate will only contribute to vortex stretching if there is component

of strain along the vorticity

So, we can examine alignments between principal strain rates and vorticity


Turbulence: lecture

Strain-rotation interaction




1 + 2 + 3 = 0




Turbulence: lectureStrain-rotation interaction

!iSij!j = !2i(ei · !)i are the principal strain rates (i.e. eigen values of the strain rate tensor)




1 + 2 + 3 = 0





Turbulence: lecture

Strain-rotation interaction

0 0.2 0.4 0.6 0.8 10.5

1

1.5

2

|ei · ω|

pdf

e1 · ωe2 · ωe3 · ω

No preference for extensiveParallel alignment with intermediatePerpendicular with compressive

This is a universal feature

Observed in most turbulent flows

Turbulence: lecture


@u1

@x1

@u1

@x2

@u1

@x3

@u2

@x1

@u2

@x2

@u2

@x3

@u3

@x1

@u3

@x2

@u3

@x3

Chakraborty et al. (2005), J. Fluid Mech.

P = r · u

Q =1

2

kk2 kSk2

R = det(ru)

P, Q and R are the invariants of the velocity gradient tensor

P = 0 for incompressible flows

Turbulence: lectureInvariants of velocity gradient tensor

@u1

@x1

@u1

@x2

@u1

@x3

@u2

@x1

@u2

@x2

@u2

@x3

@u3

@x1

@u3

@x2

@u3

@x3

ru =

Chakraborty et al. (2005), J. Fluid Mech.

ru

P = r · u

Q =1

2

kk2 kSk2

R = det(ru)

P, Q and R are the invariants of the velocity gradient tensor

P = 0 for incompressible flows


-0.02 -0.01 0 0.01 0.02 0.03

-0.3

-0.2

-0.1

0

0.1

0.2

Q/(

/

2)2

Turbulence: lecture


Tear-drop shape in Q-R space is considered universal feature

Q/(

/

2)2

R/(/2)3-0.02 -0.01 0 0.01 0.02 0.03

-0.3

-0.2

-0.1

0

0.1

0.2

Q/(

/

2)2

R/(/2)3

Discriminant

-0.02 -0.01 0 0.01 0.02 0.03

-0.3

-0.2

-0.1

0

0.1

0.2

> 0, complex eigenvalues

> 0 implies swirling regions (nodes)

< 0 implies straining regions (saddles)

Q criterion more restrictive in vortex

identification

P = r · u

Q =1

2

kk2 kSk2

R = det(ru)

Turbulence: lectureDecaying turbulence

5. Some Simple Turbulent Flows

5.1 Homogeneous, isotropic (‘grid’) turbulence

Bi-planar grids mounted across a wind tunnel working section, produce turbulence

which is approximately isotropic and homogeneous in the frame moving with the mean

velocity, U. No mean velocity gradients exist (some way downstream of the grid), so

the production term (1) in eq. (3.3) is zero and the turbulence must thus decay with x.

M

xU

Define u’ as the rms turbulent velocity. The characteristic time-scale for the turbulence

decay is −u' 2 /

du'2

dt. A characteristic time for the energy containing eddies is Lx/u’ and

since the overall decay can only occur as these large eddies break up to smaller ones,

these two time-scales must be comparable. Thus:

−Au' 2

ddt

(u' 2 )= Lx

u' (with A=O(1) by expt.)

so that

du' 2

dt= −A

u' 3

Lx

.

But (3.4) reduces to: dkdt

≡ 32

du' 2

dt= −ε (5.1)

Consider the turbulent flow downstream of a bi-planar grid.

It is one of the simplest flows we can consider.


i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)

Let’s look at the kinetic energy equation for this flow,

This equation simplifies to,source: Turbulence for the 21st century, W. K. George


Turbulence: lecture

Decaying turbulence











M

xU


decay is −u' 2 /

du'2




−Au' 2

ddt

(u' 2 )= Lx


so that

du' 2

dt= −A

u' 3

Lx

.


≡ 32

du' 2

dt= −ε (5.1)




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)












M

xU


decay is −u' 2 /

du'2




−Au' 2

ddt

(u' 2 )= Lx


so that

du' 2

dt= −A

u' 3

Lx

.


≡ 32

du' 2

dt= −ε (5.1)




i.e.,∂ui



ωij =1

2

!∂ui

∂xj− ∂uj

∂xi

"

(4.3)


⟨sij∂ui


= ⟨sijsij⟩ (4.4)


k ≡ 1

2⟨uiui⟩ =

1

2⟨q2⟩ = 1

2[⟨u2

1⟩+ ⟨u22⟩+ ⟨u2

3⟩] (4.5)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1


$

−⟨uiuj⟩∂Ui



!∂

∂t+ Uj

∂

∂xj

"

ui = −1

ρ

∂p

∂xi+ ν

∂2ui

∂x2j

−!

uj∂Ui

∂xj

"

−#

uj∂ui

∂xj− ⟨uj

∂ui

∂xj⟩$

(4.7)


!∂

∂t+ Uj

∂

∂xj

"

k =∂

∂xj

#

−1

ρ⟨pui⟩δij −

1

2⟨q2uj⟩+ ν

∂

∂xjk

$

− ⟨uiuj⟩∂Ui

∂xj− ν⟨∂ui

∂xj

∂ui

∂xj⟩ (4.8)




Turbulence: lecture

Decaying turbulence

For isotropic turbulence, this written as

80 CHAPTER 5. A FIRST LOOK AT HOMOGENEOUS TURBULENCE:

these experiments to be in error. The resolution of these two anomalies gave usquantum mechanics and relativity, both of which completely changed our pictureof the world around us and had dramatic influences on engineering and life onthis planet.

Will such be the case with our view of turbulence when we resolve the remain-ing dilemmas of homogeneous flows in turbulence? Or will we simply find outthat most of the conflicting experiments have been wrong. The search for answerswill be exhilarating for some, and frustrating for others, but necessary for all —especially for those like you entering the field now. If we find the answers, theymay completely change how we do things, like quantum mechanics or relativitydid. Or we may find they have little effect at all and our old way of doing thingsis the best we can do. A good example of this was the recognition by Copernicusthat the earth was not the center of the universe. The new world view completelychanged the way we think about the universe, but had no effect on navigation.Sailors still find their way across great oceans using the principles of Ptolemaicnavigation which is based on the idea that the earth is at the center.4

My personal suspicion is that things will be changed a lot as our understandingof these anomalies grows, simply because our present turbulence models reallydon’t extrapolate too well to new problems. Unlike the example of Ptolemaicnavigation which works perfectly well on the surface of the earth, our turbulencemodels are really very reliable, especially at predicting new things we haven’t builtinto them. This probably means we still have much to learn about turbulence,and as we learn our ideas will change and our models improve. So we might aswell begin this learning process with the problem which is at least in principle theeasiest: homogeneous turbulent flows.

5.3 Decaying turbulence: a brief overview

Look, for example, at the decay of turbulence which has already been gener-ated. If this turbulence is homogeneous and there is no mean velocity gradient togenerate new turbulence, the kinetic energy equation reduces to simply:

d

dtk = −ϵ (5.1)

Note that the time derivative is just an ordinary derivative, since there is no de-pendence of any single point quantity on position. This is often written (especiallyfor isotropic turbulence) as:

d

dt

!3

2u2"= −ϵ (5.2)

where

k ≡ 3

2u2 (5.3)

4I can personally testify that this works, since I arrived in Kinsale, Ireland after an arduousvoyage across the Atlantic from America in my 42 foot sailboat Wings by exactly this method.

where,

This is the simplest equation we can write, yet, we cannot solve this

Rearranging, Lx

pRe

L

x








M

xU


decay is −u' 2 /

du'2




−Au' 2

ddt

(u' 2 )= Lx


so that

du' 2

dt= −A

u' 3

Lx

.


≡ 32

du' 2

dt= −ε (5.1)







d

dtk = −ϵ (5.1)


d

dt

!3

2u2"= −ϵ (5.2)

where

k ≡ 3

2u2 (5.3)









d

dtk = −ϵ (5.1)


d

dt

!3

2u2"= −ϵ (5.2)

where

k ≡ 3

2u2 (5.3)


where,


= Au3

Lx

= 15u2

2Rearranging, L

x

pRe

L

x









M

xU


decay is −u' 2 /

du'2




−Au' 2

ddt

(u' 2 )= Lx


so that

du' 2

dt= −A

u' 3

Lx

.


≡ 32

du' 2

dt= −ε (5.1)







d

dtk = −ϵ (5.1)


d

dt

!3

2u2"= −ϵ (5.2)

where

k ≡ 3

2u2 (5.3)









d

dtk = −ϵ (5.1)


d

dt

!3

2u2"= −ϵ (5.2)

where

k ≡ 3

2u2 (5.3)


where,


= Au3

Lx

= 15u2

2Rearranging, L

x

pRe

L

x


Turbulence: lecture

Decaying turbulence


Now for homogeneous, isotropic turbulence it can be shown that

ε = 15νu' 2

λ2 (5.2)

so we deduce that λ = 10νLx

u'§ © ¨

· ¹ ¸ (for A=1). Turbulence Reynolds numbers can then

be defined by, as in §4: ReLx

= u' Lx

ν and

Reλ = u' λ

ν and we also see that

Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.

(Note again the increasing separation of scales with increasing Re).

Experimentally, it is found thatn

ot

MxxA

Uk −

¸¹·

¨©§ −=2 where xo is a virtual origin – see figure

below.

In the moving frame (x-xo=U(t-to)) this can be written as kt=kt,o(t/to)-n and thus (using

(5.1) ε=εo(t/to)-(n+1). The exponent n is typically found to be around 1.3. The constant A

depends on grid geometry.

STUDENT exercise. Show that the length scale defined by L=kt3/2/ε increases like t(1-n/2)

and the Reynolds number based on kt1/2 and L decreases with time.

Since Re decreases with time (or distance downstream), eventually the effects of

viscosity will dominate and the turbulence enters a final period of decay. For genuine

isotropic conditions, n=1 so that Re=constant. The smaller eddies break up more

rapidly, leaving the larger eddies behind, as it were.

u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100


ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100


Turbulence: lecture

Decaying turbulence


ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100



ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100


ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100


ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100

Research continues even today to determine what “n” and A are...

There are different results depending on

experiment




ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100


ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100


ε = 15νu' 2

λ2 (5.2)


u'§ © ¨



= u' Lx

ν and

Reλ = u' λ


Lx

λ=

ReLx

10§ © ¨

· ¹ ¸

1/ 2

.



ot

MxxA

Uk −

¸¹·


below.










u' 2 /U 2

(x-xo)/M

log scale

log scale

v'2 /U 2 kt/U2

10 -3

10-4

10 100



experiment




experiment

Turbulence: lecture

Simple shear flows

Turbulence: lectureSimple shear flows

98 CHAPTER 6. TURBULENT FREE SHEAR FLOWS

layers. By introducing a different length scale for changes perpendicular to thewall than for changes in the flow direction, he was able to explain how viscousstresses could survive near the wall at high Reynolds number. These allowed theno-slip condition at a surface to be satisfied, and resolved the paradox of howthere could be drag in the limit of zero viscosity.

It may seem strange to be talking about Prandtl’s boundary layer idea in asection about free shear flows, but as we shall see below, the basic approximationscan be applied to all “thin” (or slowly growing) shear flows with or without asurface. In this section, we shall show that free shear flows, for the most part,satisfy the conditions for these “boundary layer approximations”. Hence theybelong to the general class of flows referred as “boundary layer flows”.

One important difference will lie in whether momentum is being added to theflow at the source (as in jets) or taken out (by drag, as in wakes). A relatedinfluence is the presence (or absence) of a free stream in which our free shear flowis imbedded. We shall see that stationary free shear flows fall into two generalclasses, those with external flow and those without. One easy way to see whythis makes a difference is to remember that these flows all spread by sharing theirmomentum with the surrounding flow, or by stealing momentum from it, almostalways entraining mass from the surrounding fluid at the same time. (You don’thave to think very hard to see that as mass is entrained, it is carries its ownmomentum with it into the shear flow.) You should expect (and find) that evena small free stream velocity (or pressure gradient or even free stream turbulence)can make a significant difference, since the momentum carried in is mixed in withthat of the fluid particles which are already part of the turbulence. The longerthe flow develops (or the farther downstream one looks), the more these simpledifferences can make a difference in how the flow spreads. In view of this, itshould be no surprise that the presence or absence of an external stream plays amajor role in determining which mean convection terms which must be retainedin the governing equations. And also in determining whether an experiment orsimulation is a valid approximation to a flow in an infinite environment, sinceboth most be performed in a finite box or windtunnel.

We will consider here only flows which are plane (or two-dimensional) in themean (although similar considerations can be applied to flows that are axisymmet-ric in the mean, see the exercises). In effect, this is exactly the same as assumingthe flow is homogeneous in the third direction. Also we shall restrict our atten-tion to flows which are statistically stationary, so that time derivatives of averagedquantities can be neglected. And, of course, we have already agreed to confineour attention to Newtonian flows at constant density.

It will be easier to abandon tensor notation for the moment, and use thesymbols x, U, u for the streamwise direction, mean and fluctuating velocities re-spectively, and y, V, v for the cross-stream. Given all this, the mean momentumequations reduce to:

6.2. THE AVERAGED EQUATIONS 99

x-component:

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2(6.2)

y-component:

U∂V

∂x+ V

∂V

∂y= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2(6.3)

In addition, we have the two-dimensional mean continuity equation which reducesto:

∂U

∂x+

∂V

∂y= 0 (6.4)

6.2.2 Order of magnitude estimates

Now let’s make an order of magnitude estimate for each of the terms. Thisprocedure may seem trivial to some, or hopeless hand-waving to others. Thefact is that if you fall into either of these groups, you have missed somethingimportant. Learning to make good order-of-magnitude arguments and knowingwhen to use them (and when not to use them) are two of the most importantskills in fluid mechanics, and especially turbulence. To do this right we will haveto be very careful to make sure our estimates accurately characterize the termswe are making them for.

Naturally we should not expect changes of anything in the x-direction to scalethe same as changes in the y-direction, especially in view of the above. (This,after all, is the whole idea of “thin” shear flow.) So let’s agree that we will pick alength scale, say L, characteristic of changes or mean quantities in the x-direction;i.e.

∂

∂x∼ 1

L(6.5)

where for now ‘∼ means “of the order of magnitude of”. And we can do the samething for changes of mean quantities in the y-direction by defining a second lengthscale, say δ, to mean:

∂

∂y∼ 1

δ(6.6)

Note that both these scales will vary with the streamwise position where we eval-uate them. A good choice for δ might be proportional to the local lateral extentof the flow (or its “width”), while L is related to the distance from the source.

Consider the mean velocity in equation 7.10. It occurs in five different terms: Ualone; twice with x-derivatives, ∂U/∂x and ∂2U/∂x2; and twice with y-derivatives,∂U/∂y and ∂2U/∂y2. Now it would be tempting to simply pick a scale velocityfor U , say Us, and use it to estimate all five terms, say as: Us, Us/L, Us/L2, Us/δ,and Us/δ2. But this is much too naıve, and fails to appreciate the true role of theterms we are evaluating.

Consider 2D flows that have a planar mean. The flows are homogenous in

the third direction and are stationary (i.e. time derivatives are zero).










x-component:

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2(6.2)

y-component:

U∂V

∂x+ V

∂V

∂y= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2(6.3)


∂U

∂x+

∂V

∂y= 0 (6.4)




∂

∂x∼ 1

L(6.5)


∂

∂y∼ 1

δ(6.6)














x-component:

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2(6.2)

y-component:

U∂V

∂x+ V

∂V

∂y= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2(6.3)


∂U

∂x+

∂V

∂y= 0 (6.4)




∂

∂x∼ 1

L(6.5)


∂

∂y∼ 1

δ(6.6)






Turbulence: lecture

Simple shear flows: Order of magnitude estimates


considering specific problems and applying similarity techniques, the seeminglyarbitrary choices here become quite precise constraints (as we shall see).

We still haven’t talked about how to estimate the velocity scale for V , thecross-stream mean velocity component. From the continuity equation we knowthat:

∂V

∂y= −∂U

∂x(6.10)

From our considerations above, we know that:

∂U

∂x∼ ∆Us

L(6.11)

If there is no mean cross flow in the external stream, then the scale for V is thesame as the scale for changes in V . Therefore,

∂V

∂y∼ Vs

δ(6.12)

It follows immediately that the order of magnitude of the cross-stream velocity is:

Vs ∼ ∆Usδ

L(6.13)

We might have expected something like this if we had thought about it. If the V -velocity were of the same order as the U -velocity, how could the flow in any sensebe a “thin shear flow”. On the other hand, it also makes sense that Vs/Us ∝ δ/L,since both are some measure of how the flow spreads. Note that equation 6.13would not be the correct estimate for Vs if there were an imposed cross-flow, sincethen we would have to consider V and changes in V separately (exactly as for U).

The mean pressure gradient term is always a problem to estimate at the outset.Therefore it is better to simply leave this term alone, and see what is left at theend. In the estimates below you will see a question mark, which simply meanswe are postponing judgement until we have more information. Sometimes we willhave to keep the term simply because we don’t know enough to throw it away.Other times it will be obvious that it must remain because there is only one termleft that must be balanced by something.

Now we have figured out how to estimate the order of magnitude of all theterms except the turbulence terms. For most problems this turns out to be prettystraightforward if you remember our discussion of the pressure strain-rate terms.They so effectively distribute the energy that even that the three components ofvelocity are usually about the same order of magnitude. So if we chose a turbulencescale as simply u, then ⟨u2⟩ ∼ u2, ⟨v2⟩ ∼ u2, and ⟨w2⟩ ∼ u2. But what aboutthe Reynolds shear stress components like ⟨uv⟩? When acting against the meanshear to produce energy, they tend to be well-correlated and the maximum valueof ⟨uv⟩/urmsvrms < 0.5. Obviously the right choice for the order of magnitude is:⟨uv⟩ ∼ u2. This is not always the right choice though since some cross-stress like

• Streamwise length scale: L

• Streamwise velocity scale: Uo

or Us

(depending on the situation)

• Cross-stream length scale: << L

• Cross-stream velocity scale: Vs

Us

(/L)

This completes order of magnitudes for mean flows

How about turbulence terms?


Turbulence: lectureSimple shear flows: Order of magnitude estimates


considering specific problems and applying similarity techniques, the seeminglyarbitrary choices here become quite precise constraints (as we shall see).

We still haven’t talked about how to estimate the velocity scale for V , thecross-stream mean velocity component. From the continuity equation we knowthat:

∂V

∂y= −∂U

∂x(6.10)

From our considerations above, we know that:

∂U

∂x∼ ∆Us

L(6.11)

If there is no mean cross flow in the external stream, then the scale for V is thesame as the scale for changes in V . Therefore,

∂V

∂y∼ Vs

δ(6.12)

It follows immediately that the order of magnitude of the cross-stream velocity is:

Vs ∼ ∆Usδ

L(6.13)

We might have expected something like this if we had thought about it. If the V -velocity were of the same order as the U -velocity, how could the flow in any sensebe a “thin shear flow”. On the other hand, it also makes sense that Vs/Us ∝ δ/L,since both are some measure of how the flow spreads. Note that equation 6.13would not be the correct estimate for Vs if there were an imposed cross-flow, sincethen we would have to consider V and changes in V separately (exactly as for U).

The mean pressure gradient term is always a problem to estimate at the outset.Therefore it is better to simply leave this term alone, and see what is left at theend. In the estimates below you will see a question mark, which simply meanswe are postponing judgement until we have more information. Sometimes we willhave to keep the term simply because we don’t know enough to throw it away.Other times it will be obvious that it must remain because there is only one termleft that must be balanced by something.

Now we have figured out how to estimate the order of magnitude of all theterms except the turbulence terms. For most problems this turns out to be prettystraightforward if you remember our discussion of the pressure strain-rate terms.They so effectively distribute the energy that even that the three components ofvelocity are usually about the same order of magnitude. So if we chose a turbulencescale as simply u, then ⟨u2⟩ ∼ u2, ⟨v2⟩ ∼ u2, and ⟨w2⟩ ∼ u2. But what aboutthe Reynolds shear stress components like ⟨uv⟩? When acting against the meanshear to produce energy, they tend to be well-correlated and the maximum valueof ⟨uv⟩/urmsvrms < 0.5. Obviously the right choice for the order of magnitude is:⟨uv⟩ ∼ u2. This is not always the right choice though since some cross-stress like

• Streamwise length scale: L

• Streamwise velocity scale: Uo

or Us

(depending on the situation)

• Cross-stream length scale: << L

• Cross-stream velocity scale: Vs

Us

(/L)

This completes order of magnitudes for mean flows

How about turbulence terms?


It turns out they all scale as u2, sometimes, q2

o


Turbulence: lecture





⟨uw⟩ are identically zero in plane flows because of homogeniety. Sometimes theReynolds shear stress will be identically zero at some points in the flow (like whenit is changing sign from one side to the other of a symmetry plane). When thishappens terms you neglected can sneak back into the problem. The importantpoint to remember is that you only estimating which terms might be the mostimportant most of the time, and must look again after you have analyzed or solvedthe equations to see if your estimates were correct

6.2.3 The streamwise momentum equation

Let’s look now at the x-component of the mean momentum equation and writebelow each term its order of magnitude.

U∂U

∂x+ V

∂U

∂y

Us∆Us

L

!

∆Usδ

L

"∆Us

δ

= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2

?u2

L

u2

δν∆Us

L2ν∆Us

δ2

Now the survival of at least one of the terms on the left-hand side of theequation is the essence of free shear flow, since the flow is either being speededup or slowed down by the external flow or surroundings. Since we have chosenthe primary flow direction to be x, then the largest of these acceleration (ordeceleration) terms is the first. Therefore to see the relative importance of theremaining terms, we need to re-scale the others by dividing all the estimates byUs∆Us/L. Doing this we have:

U∂U

∂x+ V

∂U

∂y

1∆Us

Us

= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2

?u2

Us∆Us

u2

Us∆Us

#L

δ

$ν

UsL

ν

Usδ

#L

δ

$

So what do these mean? And how do we decide whether they are of the sameorder as our leading term, or much less? (Note that if any are bigger than our firstterm, it either means we have scaled it wrong, or that we guessed wrong aboutwhich term was the largest.) The beginning of the answer lies in remembering

Divide through out by Us(Us/L)







U∂U

∂x+ V

∂U

∂y

Us∆Us

L

!

∆Usδ

L

"∆Us

δ

= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2

?u2

L

u2

δν∆Us

L2ν∆Us

δ2


U∂U

∂x+ V

∂U

∂y

1∆Us

Us

= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2

?u2

Us∆Us

u2

Us∆Us

#L

δ

$ν

UsL

ν

Usδ

#L

δ

$


Divide through out by Us(Us/L)


Turbulence: lecture







U∂U

∂x+ V

∂U

∂y

Us∆Us

L

!

∆Usδ

L

"∆Us

δ

= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2

?u2

L

u2

δν∆Us

L2ν∆Us

δ2


U∂U

∂x+ V

∂U

∂y

1∆Us

Us

= −1

ρ

∂P

∂x− ∂⟨u2⟩

∂x− ∂⟨uv⟩

∂y+ ν

∂2U

∂x2+ ν

∂2U

∂y2

?u2

Us∆Us

u2

Us∆Us

#L

δ

$ν

UsL

ν

Usδ

#L

δ

$


How do we decide what to keep and what to throw out?

We need some information about Reynolds number and turbulence intensities and length scales

Turbulence: lecture



Almost all flows of interest are at high Reynolds numbers


that this is a book about turbulence. Therefore almost all the interesting flows areat high Reynolds number. But what does this mean? Now we can say. It couldmean the viscous terms in the mean momentum equation are negligible, which inturn means that:

UsL

ν>> 1

and

Usδ

ν>>

L

δ

Obviously the second criterion is much more stringent, since L/δ is typically oforder 10 or less for our “thin” shear layers. When Usδ/ν > 1000, the contributionsof the viscous stresses are certainly negligible, at least as far as the x-componentof the mean momentum equation is concerned. But if Usδ/ν ∼ 100 only, this is apretty marginal assumption, and you might want to retain the last term in youranalysis. Such is unfortunately the case in many experiments that are often atquite low Reynolds numbers.

So, what then, you ask, do we do about the turbulence terms? To repeat: thisis a book about TURBULENCE! Which means there is no way we are going tothrow away all the turbulence terms on the right-hand side of the equation. Thereis no magic or hocus-pocus about this; you simply can’t have turbulence withoutat least one turbulence term. And if there is no turbulence, we really aren’t toointerested.

So, what does this tell us? It tells us about δ! Surprised? I bet you thoughtit would tell us about u, right? Not so. Look at the right-hand side of the secondequation on the previous page and the orders of magnitude below it. Almostalways, u < Us, sometimes much less and almost never larger. Then the biggestturbulence term is the one involving the Reynolds shear stress, ∂⟨−uv⟩/∂y, whichwe have estimated as [u2/(Us∆Us)](L/δ). Hence there can be no turbulence termsat all unless:

δ

L∼ u2

Us∆Us

Wow! Look what we have learned about turbulence without ever solving a singleequation. We know how the growth of our free shear flow relates to the turbu-lence intensity – at least in terms of order of magnitude. The similarity theoriesdescribed below will even be able to actually deduce the x-dependence of δ, againwithout actually solving the equations.

Does the argument above mean the other turbulence term is negligible since itis only of order u2/Us∆Us? Well, no question we should expect it to be smaller,typically less than 5%. But unlike the viscous terms, this term often does notget smaller the farther we go in the streamwise direction. So the answer dependson the question we are asking and how accurate we want our answer to be. If



UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us



There can be no turbulence at all unless,

We still don’t know what to do with pressure...






UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us





UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us



If

u2

UsUs<< 1 then,

The only turbulence term that will matter is the production term



UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us











UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us





UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us



If

u2

UsUs<< 1 then,

The only turbulence term that will matter is the production term



UsL

ν>> 1

and

Usδ

ν>>

L

δ




δ

L∼ u2

Us∆Us






Turbulence: lecture



Lets look at the transverse momentum equation,


using the continuity equation together with l’Hopital’s rule show that to leadingorder at the centerline (assume symmetry around this line) the turbulence kineticenergy equation reduces to:

U∂k

∂x= −[⟨u2⟩ − ⟨v2⟩]∂U

∂x(6.14)

Assuming that upstream ⟨u2⟩ > ⟨v2⟩, re-examine your assessment of which termsare important near the points where k is maximal and where ⟨u2⟩ = ⟨v2⟩. (Notethat these points occur nearly, but not quite, in the same place.)

6.2.4 The transverse momentum equation

Now we must also consider the transverse or y-momentum equation. The mostimportant thing we have to remember is that the y-equation can NOT be con-sidered independently from the x-momentum equation. We would never considertrying to simplify a vector like (a, b, c) by dividing only one component by S sayto produce (Sa, b, c), since this would destroy the whole idea of a vector as havinga direction. Instead you would re-scale as (Sa, Sb, Sc) to preserve the direction.We must do the same for a vector equation. We have already decided that thefirst term on the left-hand side of the x-momentum equation was the term wehad to keep, and we divided by its order of magnitude, Us∆Us/L, to make sureit was of order one. Thus we have already decided how we are going to scale allthe components of the vector equation, and so we must do exactly the same thinghere. But first we must estimate the order of magnitude of each term, exactly asbefore.

Using our previous results, here’s what we get:

U∂V

∂x+ V

∂V

∂y

Us∆Usδ/L

L

!

∆Usδ

L

"∆Usδ/L

δ

= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

L

u2

δν∆Usδ/L

L2ν∆Usδ/L

δ2

Now dividing each term by Us∆Us/L, exactly as before, we obtain:

U∂V

∂x+ V

∂V

∂y

δ

L

∆Us

Us

!δ

L

"






U∂k

∂x= −[⟨u2⟩ − ⟨v2⟩]∂U

∂x(6.14)





U∂V

∂x+ V

∂V

∂y

Us∆Usδ/L

L

!

∆Usδ

L

"∆Usδ/L

δ

= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

L

u2

δν∆Usδ/L

L2ν∆Usδ/L

δ2


U∂V

∂x+ V

∂V

∂y

δ

L

∆Us

Us

!δ

L

"



U∂k

∂x= −[⟨u2⟩ − ⟨v2⟩]∂U

∂x(6.14)





U∂V

∂x+ V

∂V

∂y

Us∆Usδ/L

L

!

∆Usδ

L

"∆Usδ/L

δ

= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

L

u2

δν∆Usδ/L

L2ν∆Usδ/L

δ2


U∂V

∂x+ V

∂V

∂y

δ

L

∆Us

Us

!δ

L

"Wednesday, 23 October 13

Turbulence: lecture






U∂k

∂x= −[⟨u2⟩ − ⟨v2⟩]∂U

∂x(6.14)





U∂V

∂x+ V

∂V

∂y

Us∆Usδ/L

L

!

∆Usδ

L

"∆Usδ/L

δ

= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

L

u2

δν∆Usδ/L

L2ν∆Usδ/L

δ2


U∂V

∂x+ V

∂V

∂y

δ

L

∆Us

Us

!δ

L

"


= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

Us∆Us

u2

Us∆Us

!L

δ

"ν

UsL

#δ

L

$ν

Usδ

Unless you have seen this all before, the left-hand side is probably a surprise:none of the mean convection terms are of order one! In fact the only estimatedterm that is of order one in the whole equation is ∂⟨v2⟩/∂y, and only becausewe have already agreed that (u2/Us∆Us)(L/δ) had to be of order one to keepa turbulence term in the x-momentum equation. Of course, there cannot be anequation with only a single term equal to zero, unless we have badly over-estimatedits order of magnitude. Fortunately there is the pressure term left to balance it, soto first order in δ/L ∼ u2/Us∆Us, the y-momentum equation reduces to simply:

0 ≈ −1

ρ

∂P

∂y− ∂⟨v2⟩

∂y(6.15)

This simple equation has an equally simple interpretation. It says that the changein the mean pressure is only due to the radial gradient of the transverse componentof the Reynolds normal stress, ⟨v2⟩.

We can integrate equation 6.15 across the shear layer from a given value of yto infinity (or anywhere else for that matter) to obtain:

P (x, y) = P (x,∞)− ρ⟨v2⟩, (6.16)

assuming of course the free stream value of ⟨v2⟩ to be zero. If the free stream isat constant (or zero) mean velocity, then P (x,∞) = P∞ = constant, so we canwrite simply:

P (x, y) = P∞ − ρ⟨v2⟩ (6.17)

Either of these can be substituted into the x-momentum equation to eliminatethe pressure entirely, as we shall show below.

Now since ⟨v2⟩ << ∆U2s typically, P ≈ P∞, at least to first order in u2/∆U2

s .Therefore one might be tempted to conclude that the small pressure changes acrossthe flow are not important. And they are not, of course, if only the streamwisemomentum is considered. But without this small mean pressure gradient acrossthe flow, there would be no entrainment and no growth of the shear layer. In otherwords, the flow would remain parallel. Clearly whether an effect is negligible ornot depends on which question is being asked.

6.2.5 The free shear layer equations

If we assume the Reynolds number is always large enough that the viscous termscan be neglected, then as noted above, the pressure term in the x-momentum

Everything drops out and the only terms remaining are,


= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

Us∆Us

u2

Us∆Us

!L

δ

"ν

UsL

#δ

L

$ν

Usδ


0 ≈ −1

ρ

∂P

∂y− ∂⟨v2⟩

∂y(6.15)



P (x, y) = P (x,∞)− ρ⟨v2⟩, (6.16)


P (x, y) = P∞ − ρ⟨v2⟩ (6.17)







= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

Us∆Us

u2

Us∆Us

!L

δ

"ν

UsL

#δ

L

$ν

Usδ


0 ≈ −1

ρ

∂P

∂y− ∂⟨v2⟩

∂y(6.15)



P (x, y) = P (x,∞)− ρ⟨v2⟩, (6.16)


P (x, y) = P∞ − ρ⟨v2⟩ (6.17)






Turbulence: lecture




= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

Us∆Us

u2

Us∆Us

!L

δ

"ν

UsL

#δ

L

$ν

Usδ


0 ≈ −1

ρ

∂P

∂y− ∂⟨v2⟩

∂y(6.15)



P (x, y) = P (x,∞)− ρ⟨v2⟩, (6.16)


P (x, y) = P∞ − ρ⟨v2⟩ (6.17)






6.3. TWO-DIMENSIONAL TURBULENT JETS 107

equation can be evaluated in terms of P∞ and ⟨v2⟩. Thus, to second-order inu2/Us∆Us or δ/L, the momentum equations for a free shear flow reduce to asingle equation:

U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)

As we have seen above, the second term on the left-hand side (in curly brackets)may or may not be important, depending on whether Us = ∆Us or ∆Us/Us <<1. Clearly this depends on the velocity deficit (or excess) relative to the freestream. The second term in brackets on the right-hand side is also second-order(in u2/Us∆Us ∼ δ/L) compared to the others, and so could rightly be neglected.It has been retained for now since in some flows it does not vanish with increasingdistance from the source. Moreover, it can be quite important when consideringintegrals of the momentum equation, since the profiles of ⟨u2⟩ and ⟨v2⟩ do notvanish as rapidly with increasing y as do the other terms.

6.3 Two-dimensional Turbulent Jets

Turbulent jets are generated by a concentrated source of momentum issuing intoan ambient environment. The state of the environment and the external boundaryconditions are very important in determining how the turbulent flow evolves.The simplest case to consider is that in which the environment is at rest andunbounded, so all the boundary conditions are homogeneous. The jet itself canbe assumed to issue from a line source or a slot as shown in Figure 6.5. The jetthen evolves and spreads downstream by entraining mass from the surroundingswhich are at most in irrotational motion induced by the vortical fluid within thejet. But no new momentum is added to the flow downstream of the source, andit is this fact that distinguishes the jet from all other flows. As we shall seebelow, however, that the rate at which momentum crosses any x-plane is notquite constant at the source value due to the small streamwise pressure gradientarising from the turbulence normal stresses.

The averaged continuity equation can be integrated from the centerline (whereit is zero by symmetry) to obtain V , i.e.,

V = −% y

0

∂U

∂xdy (6.19)

It immediately follows that the V velocity at ±∞ is given by:

V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)

Twice the integral represents the total volume flow rate crossing a given x-plane.Therefore it makes sense that the rate of increase of this integral is the entrainmentvelocity, V∞. Obviously, V∞ cannot be zero if the jet is to spread, at least in a




= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

Us∆Us

u2

Us∆Us

!L

δ

"ν

UsL

#δ

L

$ν

Usδ


0 ≈ −1

ρ

∂P

∂y− ∂⟨v2⟩

∂y(6.15)



P (x, y) = P (x,∞)− ρ⟨v2⟩, (6.16)


P (x, y) = P∞ − ρ⟨v2⟩ (6.17)








U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)



= −1

ρ

∂P

∂y− ∂⟨uv⟩

∂x− ∂⟨v2⟩

∂y+ ν

∂2V

∂x2+ ν

∂2V

∂y2

?u2

Us∆Us

u2

Us∆Us

!L

δ

"ν

UsL

#δ

L

$ν

Usδ


0 ≈ −1

ρ

∂P

∂y− ∂⟨v2⟩

∂y(6.15)



P (x, y) = P (x,∞)− ρ⟨v2⟩, (6.16)


P (x, y) = P∞ − ρ⟨v2⟩ (6.17)








U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)


1


Turbulence: lecture




Figure 6.5: Sketch of plane jet showing jet source, coordinate system and typicalmean velocity profile.

two-dimensional flow. It is easy to see that the entrainment velocity calculatedfrom the integral is consistent with our order of magnitude estimate above (i.e.,V ∼ ∆Usδ/L). Note that the integral makes no sense if the mean velocity U doesnot go to zero with increasing |y|.

If the free stream is assumed to have zero streamwise velocity, then the scalesfor the velocity and gradients of it are the same, so the equations reduce to simplyequation 7.16 and the mean continuity equation. The latter can be multiplied byU to yield:

U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)

When this is added to equation 7.16 the terms can be combined to obtain:

∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)

This can be integrated across the flow for any given value of x to obtain:

d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)

where we have assumed that U , ⟨u2⟩, and ⟨v2⟩ vanish as |y| → ∞. (Rememberthese assumptions the next time someone tries to tell you that a small co-flowingstream and modest background turbulence level in the external flow are not im-portant.)

Equation 6.23 can in turn be integrated from the source, say x = 0, to obtain:

Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)

where Mo is the rate at which momentum per unit mass per unit length is addedat the source. The first two terms of the integrand are the flux of streamwisemomentum due to the mean flow and the turbulence. The last term is due to thestreamwise pressure gradient obtained by integrating the y-momentum equation.



U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)


These two equations come in handy when we look for self similarity







U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


Add this to the momentum equation,



U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


1



U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)


These two equations come in handy when we look for self similarity


Turbulence: lecture

Self-similarity

This idea is not the same as ‘self-similarity’ – or ‘self-preservation’ in Townsend’s

terminology – which we now discuss.

Consider a quantity Q(x,y) – dependent on two independent variables (x & y). Define

characteristic scales for the dependent and independent variables, Qo(x) and δ(x),

respectively. So scaled variables can be defined by:

η ≡ y/δ(x) and ˜ f (η, x) ≡ Q(x, y)/ Qo (x) .

If the scaled dependent variable is independent of x (i.e. there is a function f(η) such that ˜ f (η, x) ≡ f (η) ) then Q(x,y) is self-similar. Q(x,y) can be expressed as functions of single

independent variables – Qo(x), δ(x) and f(η).

Of course, the scales must be chosen appropriately. And, usually, self-similarity only

occurs over limited ranges of x. Also, if Q(x,y) is governed by a pde then Qo(x), δ(x) and

f(η) are all governed by odes.

So, in general, we look for solutions of our governing equations in which:

U = U1 + uo f(η), (η=y/δ)

uv = qo2g12(η) , u

2 = qo2g1(η) and v

2 = qo2g2(η) .

U1 is a translational velocity (almost always the free stream) and uo and δ are

characteristic velocity and length scales.

We may therefore write: ∂U∂y

= uo

δf ' (η) (dashes denote differentiation w.r.t. η.

∂U∂x

= dU1

dx+ duo

dxf (η) − uo

δdδdx

ηf ' (η) and

∂ 2U∂y2 = uo

δ 2 f ' ' (η) .

The continuity equation yields

With this, we can predict what will happen at different downstream distances by obtaining information at only one location

Turbulence: lecture

Self-similarity













U = U1 + uo f(η), (η=y/δ)

uv = qo2g12(η) , u

2 = qo2g1(η) and v

2 = qo2g2(η) .




= uo


∂U∂x

= dU1

dx+ duo

dxf (η) − uo

δdδdx

ηf ' (η) and

∂ 2U∂y2 = uo

δ 2 f ' ' (η) .


Turbulence: lecture

Plane jets

U

o

(x) = U(x, 0) Centerline velocity of the jet

(x) - Jet half width - U

o

(x)/2 = U(x, )

Look at the momentum equation,



U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


@U

2

@x

+@(UV )

@y

= @huvi@y

Integrate with respect to y,d

dx

Z 1

1U

2dy = 0 M =

Z 1

1U

2dy = constant

Turbulence: lecturePlane jets





U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


U

o



o

(x)/2 = U(x, )




U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


1

@U

2

@x

+@(UV )

@y

= @huvi@y


dx

Z 1

1U

2dy = 0 M =

Z 1

1U

2dy = constant







U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


U

o



o

(x)/2 = U(x, )




U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


1

@U

2

@x

+@(UV )

@y

= @huvi@y


dx

Z 1

1U

2dy = 0 M =

Z 1

1U

2dy = constant


Turbulence: lecture

U

o



o

(x)/2 = U(x, )

In the self-similar region,

M = U

2o

(x)(x)

Z 1

1f()2d Since M is a conserved,

U

2o

(x)(x) = constant

U

o

dU

o

dx

= 1

2

d

dx

U = U

o

(x)f(), where = y/(x)

huvi = U2o

g()

Substitute all self-similar profiles in to the momentum and cros-stream velocity equations

Plane jets

1

2

d

dx

f

2 + f

0Z

0fd

= g

0






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


U

o



o

(x)/2 = U(x, )




U∂U

∂x+

!

V∂U

∂y

"

= −dP∞

dx− ∂

∂y⟨uv⟩ −

!∂

∂x

#⟨u2⟩ − ⟨v2⟩

$"

(6.18)





V = −% y

0

∂U

∂xdy (6.19)


V∞ = −V−∞ = − d

dx

% ∞

0U(x, y)dy (6.20)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)






U

!∂U

∂x+

∂V

∂y

"

= 0 (6.21)


∂

∂xU2 +

∂

∂yUV +

∂

∂y⟨uv⟩+ ∂

∂x

#⟨u2⟩ − ⟨v2⟩

$= 0 (6.22)


d

dx

% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy = 0 (6.23)



Mo =% ∞

−∞[U2 + (⟨u2⟩ − ⟨v2⟩)]dy (6.24)


1

@U

2

@x

+@(UV )

@y

= @huvi@y


dx

Z 1

1U

2dy = 0 M =

Z 1

1U

2dy = constant


Turbulence: lecture

Plane jets1

2

d

dx

f

2 + f

0Z

0fd

= g

0

For self-similarity, d

dx

= S

(x) x

U

o

(x) x

1/2

Self-similarity in mean and shear stress implies that eddy viscosity should also be self similar

T

(x) = U

o

(x)(x)T

() T (x) px

Local Reynolds number, Re

o

(x) =

U

o

px, increases with distance

Turbulent Reynolds number, ReT (x) =U

o

T

is independent of x

Turbulence: lecture

Wall-bounded flow

Finally, note that although this flow is the most fundamental turbulent flow possible

(about which much has been written), it is in a sense pathological, for it rarely occurs in

practice (no turbulence production by shear), except perhaps when turbulence is

convected in a uniform free-stream.

5.2 Two-dimensional Channel Flow

The laminar (fully developed) equivalent yields the well-known parabolic – Poiseuille – profile and in that case the shear stress is linearly distributed. Define the bulk velocity as U and assume the top and bottom walls are no-slip boundaries. We discuss the corresponding fully developed turbulent (high Reynolds number) flow.

δ y

x

Consider the various elements of u' i u' j . Using cartesian x,y,z coordinates for the (1,2,3)

directions and (u,v,w) for the corresponding velocities, we note first that u' w' and v ' w '

are identically zero by symmetry about the z=0 plane (w is as likely to be negative as

positive). The mean flow is in the x-direction so V=W=0 and U depends only on y.

Thus the mean momentum equations simplify to:

0 = − 1ρ

∂P∂x

− ∂uv∂y

+ ν ∂ 2U∂y 2 , (5.3a)

0 = − 1ρ

∂P∂y

− ∂v 2

∂y, (5.3b)

0 = − 1ρ

∂P∂z

(5.3c)

(Mass continuity is automatically satisfied). Notice that these differ from the laminar

case only by the turbulent stress terms in (5.1a) and (5.1b). Clearly P is independent of z

and, from (5.3b)







δ y

x






0 = − 1ρ

∂P∂x

− ∂uv∂y

+ ν ∂ 2U∂y 2 , (5.3a)

0 = − 1ρ

∂P∂y

− ∂v 2

∂y, (5.3b)

0 = − 1ρ

∂P∂z

(5.3c)



and, from (5.3b)







δ y

x






0 = − 1ρ

∂P∂x

− ∂uv∂y

+ ν ∂ 2U∂y 2 , (5.3a)

0 = − 1ρ

∂P∂y

− ∂v 2

∂y, (5.3b)

0 = − 1ρ

∂P∂z

(5.3c)



and, from (5.3b)

: Channel

Turbulence: lecture

Channel flow







δ y

x






0 = − 1ρ

∂P∂x

− ∂uv∂y

+ ν ∂ 2U∂y 2 , (5.3a)

0 = − 1ρ

∂P∂y

− ∂v 2

∂y, (5.3b)

0 = − 1ρ

∂P∂z

(5.3c)



and, from (5.3b)

Identical to the laminar flow equation you have seen before, except now we have the turbulent stresses

P is independent of z

Integrate equation 2

P = Pw (x ) − ρv2 (y) ,

where Pw is the mean pressure at the walls and depends only on x. Substituting this

expression into (5.3a) yields:

∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU

dy− ρuv is the total mean shear stress. In (5.4), since the LHS is a

function only of x and the RHS only of y both terms must be constant. It follows that

τ = τ w y / δ , where τw = µ dU

dyy =0

, just as for the laminar flow equivalent. In this case,

however, the total stress includes a Reynolds stress contribution. The skin friction

coefficient is defined as usual by: Cf = τw

12 ρU

2 .

We can define viscous scales: the friction velocity, uτ = τ w

ρ and viscous length scale,

δν = ν

uτ. Note that the Reynolds number based on these is identically unity, whereas

the friction Reynolds number is defined by: Reτ = u τδ

ν. And we can measure the

distance from the wall in viscous lengths – wall units – denoted by:

y+ ≡ y

δν= uτ y

ν

Since the latter is like a Reynolds number it will provide a measure of how large the

viscous stress is compared with the total stress. The channel Reynolds number is often

defined as Re = U (2δ) /ν .

_____________________________________________________________________________

STUDENT Exercise

An experiment is performed on fully developed turbulent channel flow at Re=105. The

fluid is water (ν=1.14 x 10-6 m2s-1) and δ=2cm. The skin friction coefficient is found to be

Substitute in equation 1

P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



Turbulence: lecture

Channel flow

P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



LHS depends on x and RHS depends on y

Must be a constant

P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



But, total stress includes a Reynolds shear stress contribution

Skin friction coefficient:

P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



Reynolds number:

Turbulence: lecture

Wall-bounded flows

P = Pw (x ) − ρv2 (y) ,



∂Pw

∂x=

§ © ¨

· ¹ ¸

dPdx

= dτdy

, (5.4)

where τ(y) = µ dU




dyy =0




12 ρU

2 .



δν = ν





y+ ≡ y

δν= uτ y

ν




_____________________________________________________________________________

STUDENT Exercise



When non-dimensionalised using the viscous scales, it is referred to as “inner” scaling since the scaling represents the view point with

respect to the wall

“Outer” length scale: “Outer” velocity scale: Ucl, Ue, U1

Turbulence: lecture

There are two independent non-dimensional groups: y+ and y/ (y/ = y+/Re )

Mean velocity can be written as, U = u

Fo

y, Re

Fo

is a universal function that is yet to be determined

Instead of this form, lets examine the gradient of mean velocity as it is more dynamically relevant

The gradient of mean velocity depends on,

dU

dy=

u

y

y+,

y

where, is a non-dimensional universal function to be determined

Wall-bounded flows

Turbulence: lecture

Fully developed turbulent channel flow can be specified by , , and u .

There are two independent non-dimensional groups: y+ and y/ (y/ = y+/Re )

Mean velocity can be written as, U = u

Fo

y, Re

Fo

is a universal function that is yet to be determined

Instead of this form, lets examine the gradient of mean velocity as it is more dynamically relevant

The gradient of mean velocity depends on,

dU

dy=

u

y

y+,

y

where, is a non-dimensional universal function to be determined

Wall-bounded flows


Turbulence: lecture

y+,

y

! (y+), for

y

<< 1

dU

dy=

u

y1(y

+) In non-dimensional form,

dU+

dy+=

1

y+1(y

+)

The integral of this equation is the Law of the wall

U+= fw(y

+), where, fw(y

+) =

Z y+

0

1

y01(y

0)dy0

The critical point is not the integral but the

hypothesis that U+depends only on y+.

Wall-bounded flows: Mean velocity profilesTurbulence: lecture

Prandtl postulated that high Reynolds numbers, close to the wall (y/ << 1)

that there is an “inner” layer in which the mean velocity profile is determined

by the viscous scales and is independent of and U1

y+,

y

! (y+), for

y

<< 1

dU

dy=

u

y1(y

+) In non-dimensional form,

dU+

dy+=

1

y+1(y

+)

The integral of this equation is the Law of the wall

U+= fw(y

+), where, fw(y

+) =

Z y+

0

1

y01(y

0)dy0

The critical point is not the integral but the

hypothesis that U+depends only on y+.

Wall-bounded flows: Mean velocity profiles


Turbulence: lecture

Viscous sub-layerWall-bounded flows: Mean velocity profiles

Turbulence: lecture

Viscous sub-layerNo-slip condition gives fw(0) = 0, while the viscous stress law gives f 0

w(0) = 1

Taylor’s expansion for small y+

fw(y+) = y+ +O(y+

2)

• For y+ < 5, the departure is minimal.

• For y+ > 10 the deviation is 25%

• “Bu↵er” layer beyond viscous sublayer



Turbulence: lecture


w(0) = 1


fw(y+) = y+ +O(y+

2)






Turbulence: lecture


w(0) = 1


fw(y+) = y+ +O(y+

2)






Turbulence: lecture

The log law

The outer part of this layer corresponds to large y+

Analogous to Kolmogorov’s hypothesis, in this outer part of the inner layer,

Suppose the viscosity has little e↵ect

Therefore, 1(y+) ! constant =

1

dU+

dy+=

1

y+for y+ >> 1 and y/ << 1

integrate, U+ =1

ln y+ +B

- von Karman constant = 0.39 (latest “universal” number).

B = 4.3

Marusic et al. (2013), J. Fluid Mech.


Turbulence: lecture

The log law

The outer part of this layer corresponds to large y+

Analogous to Kolmogorov’s hypothesis, in this outer part of the inner layer,

Suppose the viscosity has little e↵ect

Therefore, 1(y+) ! constant =

1

dU+

dy+=

1

y+for y+ >> 1 and y/ << 1

integrate, U+ =1

ln y+ +B

- von Karman constant = 0.39 (latest “universal” number).

B = 4.3

Di↵erent layers in wall flows,

• Inner layer y/ < 0.15

• Viscous sublayer y+ < 5

• Outer part of inner layer y+ > 3pRe - beginning of log region

• Bu↵er layer 5 < y+ < 3pRe

• Log layer 3pRe < y+ < 0.15Re (y/ = 0.15)

• Outer layer y+ > 0.15Re (also called wake region in BLs)

Marusic et al. (2013), J. Fluid Mech.



y+

y

y+ = 100y

= 0.15

Turbulence: lecture

The log lawWall-bounded flows: Mean velocity profiles

y+

U+

y

y+ = 100y

= 0.15

Turbulence: lecture

The log lawWall-bounded flows: Mean velocity profiles


Turbulence: lecture

These numbers are constantly changing

Di↵erent layers in wall flows,

• Inner layer y/ < 0.15

• Viscous sublayer y+ < 5

• Outer layer y+ > 3pRe - beginning of log region

• Bu↵er layer 5 < y+ < 3pRe

• Log layer 3pRe < y+ < 0.15Re (y/ = 0.15)


Turbulence: lecture

In the outer layer, (

+, y/) is independent of .

This implies, tends to a function of y/

(y+, y/) !

o

(y/)

Substituting this in to the law of the wall,

Uo

U

u

= FD

(y/) where, FD

=

Z 1

y/

1

y0

o

(y0)dy0

At suciently high Reynolds number,

There is an overlap between inner (y/ < 0.15) and outer layers (y+ > 3

pRe ).

y

u

dU

dy= 1(y

+) =

o

(y/) for

<< y <<

Wall-bounded flows: Mean velocity profiles (defect form)

Turbulence: lecture

y

u

dU

dy= 1(y

+) =

o

(y/) for

<< y <<

This argument was put forward by Millikan (1938) and is an alternative derivation of the log law

More importantly, it establishes the defect law as well

Wall-bounded flows: Mean velocity profiles (defect form)

Uo

U

u

= lny

+B1

Turbulence: lecture


Law of the wake

8. Wall-bounded flows

8.1 Regular boundary layers

Since the viscous term is always significant near the wall, strict self-preservation is not

possible unless (using notation of (6.12) δuo/ν is constant. This requires a=-1 – i.e. a

decreasing free-stream velocity and thus an adverse pressure gradient (converging

duct, for example). (Townsend, 1976, goes into great detail about self-preserving forms;

we do not consider such flows here).

Recall that the inner and outer layer profiles (5.9) and (5.11) are given by:

u+ ≡ U

uτ= 1

κln y +( )+ B (5.9)

and

U − U1

uτ= 1

κln( y / δ ) − B1 , (5.12)

where y+ = yuτ

ν. Matching these two layers leads to:

U1

uτ+ 1

κln U1

uτ

§ © ¨ ·

¹ ¸ = 1

κln(Re δ ) + B + B1 (8.1)

so that, since Reδ is a large quantity, uτ/U1 must be a small one and thus the surface

friction falls only slowly with distance downstream. uτ/U1 is typically 0.03-0.05.

Note that the outer layer profile can be expressed in terms of the law of the wake – i.e.

U+ = f y+( )+ Πκ

w yδ§ © ¨

· ¹ ¸ (8.2)

where w is the wake function and is often assumed to be universal; it is defined to

satisfy the conditions w(0)=0 and w(1)=2. Π is called the wake strength parameter. In

terms of (8.1), B1=2Π/κ and is approximately 2.75, but much lower in pipes and channels

(typically 0.2 – 0.7). Figure 8.1 shows the boundary layer profile plotted as u+ vs. ln(y+).

w(y/) = 2sin2y2

In terms of defect profile, B1 = 2/Strong for boundary layers, but, weak for pipes and channels

Turbulence: lecture

Skin friction and Reynolds number

integrate, U+ =1

ln y+ +B

Adding these two together

For a given Reo

, can be solved for Uo

/u

cf

=2

w

U2o

= 2

u

Uo

2

Uo

u

=1

ln

U

+B +B1

=

2

Uo

u

=1

ln

"Re

o

Uo

u

1#+B +B1

=

2

Uo

U

u

= lny

+B1

Turbulence: lecture

Effect of surface roughness

8.3 Effects of surface roughness

If the surface is rough, with protuberances large enough to completely dominate the

viscous wall region, one might expect closer self-preservation throughout the flow –

since rough surfaces augment the drag so that a large enough Re is more easily

obtainable . This would require, setting uτ=uo in the self-preserving forms (6.12):

dδdx

= const., uτ ≡U1

Cf

2= const.

A rough surface changes the universal log-law region (see Section 5.2) to:

u+ = 1

κln(y+) +C(h+ , geometry)

where h is the typical element height and the constant must now depend on the nature

of the surface. In terms of the usual smooth wall log-law, we can therefore write:

u+ = 1

κln(y+) +Co − ∆U +(h+ , geometry) (8.4)

where ∆U is the roughness function given by ∆U + = C(h+ , geom.) − Co and Co=B in the

previous notation. The figure (from Raupach et al) contains experimental data showing

how ∆U varies with h.

Assuming that, for large enough h+, viscosity becomes irrelevant – i.e. the flow has

‘Reynolds number similarity’ – ν/uτ is an inappropriate length scale and so we write the

law-of-the-wall as:

u+ = 1

κln y

h§ © ¨ · ¹ ¸ + c(h

+ ,geom.) .

We can re-express this in terms of yo, a roughness length, so that:

u+ = 1

κln y

yo

§

© ¨ ¨

·

¹ ¸ ¸ . (8.5)

(This is the way that meteorologists often express the log law.) Note that yo contains all

the information about the surface morphology – changes in nature, shape, spacing, etc.,

of the roughness elements will change yo.

factor for a fully rough case with fixed h/d would remain independent of Reynolds

number (as for the fixed x/ks lines in fig.8.4) and, as in boundary layers, the particular

value would depend on the size of the roughness.

Note also that, often, the rough surface effectively displaces the flow vertically a little

(or a lot!), so one would normally used (y-d), where d is a ‘zero-plane displacement’,

rather than y for the vertical distance.

Note, finally, that if the elements are large enough, there may be no log-law region at all

– as in many urban situations, for example.

Turbulence: lecture


8.3 Effects of surface roughness

If the surface is rough, with protuberances large enough to completely dominate the

viscous wall region, one might expect closer self-preservation throughout the flow –

since rough surfaces augment the drag so that a large enough Re is more easily

obtainable . This would require, setting uτ=uo in the self-preserving forms (6.12):

dδdx

= const., uτ ≡U1

Cf

2= const.

A rough surface changes the universal log-law region (see Section 5.2) to:

u+ = 1

κln(y+) +C(h+ , geometry)

where h is the typical element height and the constant must now depend on the nature

of the surface. In terms of the usual smooth wall log-law, we can therefore write:

u+ = 1

κln(y+) +Co − ∆U +(h+ , geometry) (8.4)

where ∆U is the roughness function given by ∆U + = C(h+ , geom.) − Co and Co=B in the

previous notation. The figure (from Raupach et al) contains experimental data showing

how ∆U varies with h.

Assuming that, for large enough h+, viscosity becomes irrelevant – i.e. the flow has

‘Reynolds number similarity’ – ν/uτ is an inappropriate length scale and so we write the

law-of-the-wall as:

u+ = 1

κln y

h§ © ¨ · ¹ ¸ + c(h

+ ,geom.) .

We can re-express this in terms of yo, a roughness length, so that:

u+ = 1

κln y

yo

§

© ¨ ¨

·

¹ ¸ ¸ . (8.5)

(This is the way that meteorologists often express the log law.) Note that yo contains all

the information about the surface morphology – changes in nature, shape, spacing, etc.,

of the roughness elements will change yo.

Turbulence: lecture

Effect of surface roughnessUsing (8.4) and (8.5) it can be shown that

∆U+ ≈ ln(h+)

Figure 8.3 Variation of the roughness function, ∆U+, with h+=huτ/ν.

which is emphasised by the data in figure 8.3; this shows laboratory and atmospheric

data over a wide range of h+. Note that in the fully rough regime, h+>100, say, ∆U+ (and

hence yo) at any fixed h+ depends on the nature of the roughness.

In contrast, for a smooth wall – i.e. h+ → 0 , ∆U+ → 0 and youτ/ν=0.14 (using standard

values for the von Karman constant, κ, and B (i.e. Co)). So yo remains defined for a

smooth surface but is flow-dependent, unlike in the fully rough limit.

Typically, the wall is aerodynamically smooth if 0 < h+<5

and fully rough flow for h+>70 (with yo/h=30)

according to classical sand-grain experiments in pipe flows. There is therefore an

‘admissable roughness’ height; roughness of a smaller size (than h+=5, say) will not

affect the flow.

Schlichting expresses this as: hadm<100ν/U1 or, more precisely,

U1hadmν

< 7Cf '

.

Problem is same h+ can have different drag

Turbulence: lecture


Using (8.4) and (8.5) it can be shown that

∆U+ ≈ ln(h+)

Figure 8.3 Variation of the roughness function, ∆U+, with h+=huτ/ν.

which is emphasised by the data in figure 8.3; this shows laboratory and atmospheric

data over a wide range of h+. Note that in the fully rough regime, h+>100, say, ∆U+ (and

hence yo) at any fixed h+ depends on the nature of the roughness.

In contrast, for a smooth wall – i.e. h+ → 0 , ∆U+ → 0 and youτ/ν=0.14 (using standard

values for the von Karman constant, κ, and B (i.e. Co)). So yo remains defined for a

smooth surface but is flow-dependent, unlike in the fully rough limit.

Typically, the wall is aerodynamically smooth if 0 < h+<5

and fully rough flow for h+>70 (with yo/h=30)

according to classical sand-grain experiments in pipe flows. There is therefore an

‘admissable roughness’ height; roughness of a smaller size (than h+=5, say) will not

affect the flow.

Schlichting expresses this as: hadm<100ν/U1 or, more precisely,

U1hadmν

< 7Cf '

.

Turbulence: lecture


(or ks): Equivalent sandgrain roughness

We don’t know its value from geometry!

So, we develop correlations between surface parameters and equivalant sandgrain roughness that will allow us to predict drag

Turbulence: lectureEffect of surface roughness

CD varies with L/ for the same ReL

(or ks): Equivalent sandgrain roughness

We don’t know its value from geometry!

So, we develop correlations between surface parameters and equivalant sandgrain roughness that will allow us to predict drag


Turbulence: lecture

Boundary layers: Effects of pressure gradient

221 fe

e

Cdxdu

udxd

=δ+θ+θ *)(

Momentum integral

equation

the viscous stress is of course the only contribution). A standard, eddy viscosity model

is often used for the shear stress, with the eddy viscosity written in terms of k and ε (the

‘k-ε’ model), i.e.

−u ' v' = νt

dUdy

= Cµk2

εdUdy

where Cµ is a constant. Now recall that the friction velocity, uτ, defined via the wall

stress by uτ = τ w / ρ , is also given by uτ2 = u ' v' . So on substituting for the velocity

gradient (8.4) can be written

−u ' v' u ' v'

Cµk2 / ε

ª

¬ « «

º

¼ » »

= ε i.e. Cµ = u ' v'k2

§

© ¨

·

¹ ¸

2

which shows that for consistency with regular log-law wall conditions the constant Cµ

must be chosen appropriately to match the (square of the) ratio of the measured shear

stress and kinetic energy in the wall region. It is straightforward to show that the point

where the energy production is a maximum must be in a region where the shear stress

−u ' v' is increasing, so must be nearer the wall than the log-law region, because −u ' v' is

constant in the latter but zero at the wall.

8.2 Effects of pressure gradient.

The momentum integral equation can be rewritten:

dθdx

= uτ2

U12 1 + Ψ 1+ 2θ

δ *

§ © ¨

· ¹ ¸

® ¯

½ ¾ ¿

where the quantity Ψ = δ *

τ w

dP1

dx= − δ *U1

uτ2

dU1

dx is a parameter which measures the

importance of the pressure gradient. If it’s positive (pressure increasing, velocity

decreasing) the layer grows more quickly so that boundary layer assumptions become

less and less valid and much of the above does not apply. Note that for constant

pressure gradient, increasing δ∗ and decreasing uτ (because of falling velocities) will

dP1

dx

= U1dU1

dx

Turbulence: lecture

Boundary layers: Effects of pressure gradient

both increase the growth rate. So the influence increases – a kind of positive feedback!

The outer layer is affected first and has the effect merely of changing surface stress, with

the inner layer practically unchanged, ‘though with a smaller log-law region (in terms

of y/δ). Quantitative theory is on much less solid ground for non-zero pressure gradient

cases.

Note what happens to the shape factor, H=θ/δ. For zero pressure-gradient this is around

1.3 – but depending on Reynolds number of course. Favourable pressure gradient

causes the velocity profile to steepen and thus decreases H and increases Cf. and vice

versa for adverse gradients.

Note also what happens to the shear stress profile – from the momentum equation, at

the wall, we must have:

∂τ∂y

= − dP1

dx

so the shear stress increases with y in an adverse pressure gradient, as shown in Figure

8.2 As the pressure gradient (Ψ) increases from zero the velocity profile becomes less

full and the shear stress profile develops a peak, with a lower value at the wall. This

value gradually decreases with increasing until separation is reached, when the velocity

profile has zero slope at the wall and the surface stress is zero.

both increase the growth rate. So the influence increases – a kind of positive feedback!

The outer layer is affected first and has the effect merely of changing surface stress, with

the inner layer practically unchanged, ‘though with a smaller log-law region (in terms

of y/δ). Quantitative theory is on much less solid ground for non-zero pressure gradient

cases.

Note what happens to the shape factor, H=θ/δ. For zero pressure-gradient this is around

1.3 – but depending on Reynolds number of course. Favourable pressure gradient

causes the velocity profile to steepen and thus decreases H and increases Cf. and vice

versa for adverse gradients.

Note also what happens to the shear stress profile – from the momentum equation, at

the wall, we must have:

∂τ∂y

= − dP1

dx

so the shear stress increases with y in an adverse pressure gradient, as shown in Figure

8.2 As the pressure gradient (Ψ) increases from zero the velocity profile becomes less

full and the shear stress profile develops a peak, with a lower value at the wall. This

value gradually decreases with increasing until separation is reached, when the velocity

profile has zero slope at the wall and the surface stress is zero.

PhD opportunities

• Back up slides

Turbulence: lecture

Self-similarity













U = U1 + uo f(η), (η=y/δ)

uv = qo2g12(η) , u

2 = qo2g1(η) and v

2 = qo2g2(η) .




= uo


∂U∂x

= dU1

dx+ duo

dxf (η) − uo

δdδdx

ηf ' (η) and

∂ 2U∂y2 = uo

δ 2 f ' ' (η) .


Turbulence: lecture

Self-similarity













U = U1 + uo f(η), (η=y/δ)

uv = qo2g12(η) , u

2 = qo2g1(η) and v

2 = qo2g2(η) .




= uo


∂U∂x

= dU1

dx+ duo

dxf (η) − uo

δdδdx

ηf ' (η) and

∂ 2U∂y2 = uo

δ 2 f ' ' (η) .


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − δ duo

dxfdη

0

η

³ + uodδdx

ηf ' dη0

η

³ .

Since ηf ' = ddη

(ηf ) − f this reduces to:

V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − d(δuo)

dxfdη

0

η

³ + uodδdx

ηf .

Substituting these into the boundary layer equation (6.6a) yields:

uodU1

dx( f − ηf ' ) + U1

duo

dxf − uo

δdδdx

.ηf '§ ©

· ¹ + uo

duo

dxf 2 − uo

δd(δuo)

dx. f ' fdy³

+ qo2 dδ

dx(g1 ' −g2 ' ) +

qo2

δg12 ' = 0

(6.9)

To be consistent with self-preservation, the coefficients must be zero or proportional to

one another, since the equation must be a function of η only. Since we know that, for

example, the shear stress term is not zero, we can multiply all coefficients by δ/qo2 so

that the following parameters must all be constants:

δqo

2 uodU1

dx, δ

qo2 U1

duo

dx, uoU1

qo2

dδdx

, δqo

2 uoduo

dx, uo

qo2

d(δuo )dx

. (6.10)

Similar inspection of the t.k.e. equation (6.7), recognising that ε cannot be zero and

dividing the factors multiplying the production and dissipation terms, shows that qo/uo

must be a constant. Thus we can replace qo by uo in (6.10) and the necessary

requirements for self-preservation become that all the quantities:

U1δuo

2duo

dx, U1

uo

dδdx

, δuo

duo

dx, dδ

dx (6.11)

should be constant.

V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − δ duo

dxfdη

0

η

³ + uodδdx

ηf ' dη0

η

³ .

Since ηf ' = ddη


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − d(δuo)

dxfdη

0

η

³ + uodδdx

ηf .


uodU1

dx( f − ηf ' ) + U1

duo

dxf − uo

δdδdx

.ηf '§ ©

· ¹ + uo

duo

dxf 2 − uo

δd(δuo)

dx. f ' fdy³

+ qo2 dδ

dx(g1 ' −g2 ' ) +

qo2

δg12 ' = 0

(6.9)





δqo

2 uodU1

dx, δ

qo2 U1

duo

dx, uoU1

qo2

dδdx

, δqo

2 uoduo

dx, uo

qo2

d(δuo )dx

. (6.10)





U1δuo

2duo

dx, U1

uo

dδdx

, δuo

duo

dx, dδ

dx (6.11)

should be constant.

Turbulence: lectureSelf-similarity













U = U1 + uo f(η), (η=y/δ)

uv = qo2g12(η) , u

2 = qo2g1(η) and v

2 = qo2g2(η) .




= uo


∂U∂x

= dU1

dx+ duo

dxf (η) − uo

δdδdx

ηf ' (η) and

∂ 2U∂y2 = uo

δ 2 f ' ' (η) .














U = U1 + uo f(η), (η=y/δ)

uv = qo2g12(η) , u

2 = qo2g1(η) and v

2 = qo2g2(η) .




= uo


∂U∂x

= dU1

dx+ duo

dxf (η) − uo

δdδdx

ηf ' (η) and

∂ 2U∂y2 = uo

δ 2 f ' ' (η) .


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − δ duo

dxfdη

0

η

³ + uodδdx

ηf ' dη0

η

³ .

Since ηf ' = ddη


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − d(δuo)

dxfdη

0

η

³ + uodδdx

ηf .


uodU1

dx( f − ηf ' ) + U1

duo

dxf − uo

δdδdx

.ηf '§ ©

· ¹ + uo

duo

dxf 2 − uo

δd(δuo)

dx. f ' fdy³

+ qo2 dδ

dx(g1 ' −g2 ' ) +

qo2

δg12 ' = 0

(6.9)





δqo

2 uodU1

dx, δ

qo2 U1

duo

dx, uoU1

qo2

dδdx

, δqo

2 uoduo

dx, uo

qo2

d(δuo )dx

. (6.10)





U1δuo

2duo

dx, U1

uo

dδdx

, δuo

duo

dx, dδ

dx (6.11)

should be constant.

V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − δ duo

dxfdη

0

η

³ + uodδdx

ηf ' dη0

η

³ .

Since ηf ' = ddη


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − d(δuo)

dxfdη

0

η

³ + uodδdx

ηf .


uodU1

dx( f − ηf ' ) + U1

duo

dxf − uo

δdδdx

.ηf '§ ©

· ¹ + uo

duo

dxf 2 − uo

δd(δuo)

dx. f ' fdy³

+ qo2 dδ

dx(g1 ' −g2 ' ) +

qo2

δg12 ' = 0

(6.9)





δqo

2 uodU1

dx, δ

qo2 U1

duo

dx, uoU1

qo2

dδdx

, δqo

2 uoduo

dx, uo

qo2

d(δuo )dx

. (6.10)





U1δuo

2duo

dx, U1

uo

dδdx

, δuo

duo

dx, dδ

dx (6.11)

should be constant.


Turbulence: lecture

Self-similarity

V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − δ duo

dxfdη

0

η

³ + uodδdx

ηf ' dη0

η

³ .

Since ηf ' = ddη


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − d(δuo)

dxfdη

0

η

³ + uodδdx

ηf .


uodU1

dx( f − ηf ' ) + U1

duo

dxf − uo

δdδdx

.ηf '§ ©

· ¹ + uo

duo

dxf 2 − uo

δd(δuo)

dx. f ' fdy³

+ qo2 dδ

dx(g1 ' −g2 ' ) +

qo2

δg12 ' = 0

(6.9)





δqo

2 uodU1

dx, δ

qo2 U1

duo

dx, uoU1

qo2

dδdx

, δqo

2 uoduo

dx, uo

qo2

d(δuo )dx

. (6.10)





U1δuo

2duo

dx, U1

uo

dδdx

, δuo

duo

dx, dδ

dx (6.11)

should be constant.

Substituting these in the momentum equation,

V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − δ duo

dxfdη

0

η

³ + uodδdx

ηf ' dη0

η

³ .

Since ηf ' = ddη


V = − ∂U

∂x0

y

³ dy = −δ dU1

dxη − d(δuo)

dxfdη

0

η

³ + uodδdx

ηf .


uodU1

dx( f − ηf ' ) + U1

duo

dxf − uo

δdδdx

.ηf '§ ©

· ¹ + uo

duo

dxf 2 − uo

δd(δuo)

dx. f ' fdy³

+ qo2 dδ

dx(g1 ' −g2 ' ) +

qo2

δg12 ' = 0

(6.9)





δqo

2 uodU1

dx, δ

qo2 U1

duo

dx, uoU1

qo2

dδdx

, δqo

2 uoduo

dx, uo

qo2

d(δuo )dx

. (6.10)





U1δuo

2duo

dx, U1

uo

dδdx

, δuo

duo

dx, dδ

dx (6.11)

should be constant.

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