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Chapter 6

Several forms of the equations ofmotion

6.1 The Navier-Stokes equations

Under the assumption of a Newtonian stress-rate-of-strain constitutive equation and alinear, thermally conductive medium, the equations of motion for compressible flow becomethe famous Navier-Stokes equations. In Cartesian coordinates,

@⇢

@t+

@

@xi(⇢Ui) = 0

@⇢Ui

@t+

@

@xj

✓⇢UiUj + P �ij � 2µSij +

✓2

3µ� µv

◆�ijSkk

◆� ⇢Gi = 0

@⇢ (e+ k)

@t+

@

@xi

✓⇢Uiht � (@T/@xi)� 2µUjSij +

✓2

3µ� µv

◆�ijUjSkk

◆� ⇢UiGi = 0

(6.1)

The Navier-Stokes equations are the foundation of the science of fluid mechanics. With theinclusion of an equation of state, virtually all flow solving revolves around finding solutionsof the Navier-Stokes equations. Most exceptions involve fluids where the relation betweenstress and rate-of-strain is nonlinear such as polymers, or where the equation of state isnot very well understood (for example supersonic flow in water) or rarefied flows wherethe Boltzmann equation must be used to explicitly account for particle collisions. Theequations can take on many forms depending on what approximations or assumptions may

6-1

CHAPTER 6. SEVERAL FORMS OF THE EQUATIONS OF MOTION 6-2

be appropriate to a given flow. In addition, transforming the equations to di↵erent formsmay enable one to gain insight into the nature of the solutions. It is essential to learn themany di↵erent forms of the equations and to become practiced in the manipulations usedto transform them.

6.1.1 Incompressible Navier-Stokes equations

If there are no body forces and the flow is incompressible, r · U = 0, the Navier-Stokesequations reduce to what is probably their most familiar form.

@Ui

@xi= 0

@Ui

@t+

@

@xj

✓UiUj +

P

⇢�ij � 2

✓µ

⇢

◆Sij

◆= 0

@T

@t+

@

@xi

✓UiT �

✓

⇢C

◆@T

@xi

◆� 2

✓µ

⇢C

◆SijSij = 0

(6.2)

where the internal energy is assumed to be e = CT and C is the heat capacity of the mate-rial. The equation of state in this case is just ⇢ = constant. Notice that for incompressibleflow the continuity and momentum equations are completely decoupled from the energyequation and can be solved separately. Once the velocity is known the energy equationcan be solved for the temperature. The last term in the energy equation is always positiveand represents a source of internal energy due to dissipation of kinetic energy by viscousfriction. This will be discussed in much more detail in Chapter 7.

6.2 The momentum equation expressed in terms of vortic-ity

If the transport coe�cients, µ and µv are assumed to be constant (a reasonable assump-tion if the Mach number is not too large), the compressible momentum equation can bewritten,

@⇢U

@t+r ·

�⇢U U

�+rP � µr2U �

✓1

3µ+ µv

◆r�r · U

�� ⇢G = 0 (6.3)

where the body force is the gradient of a potential function G = �r The vorticity isdefined as the curl of the velocity.


⌦ = r⇥ U (6.4)

If we use the vector identities

U ·rU =�r⇥ U

�⇥ U +r

✓U · U2

◆

r⇥�r⇥ U

�= r

�r · U

��r2U

(6.5)

together with the continuity equation, the momentum equation can be written in theform,

⇢@U

@t+⇢�⌦⇥ U

�+⇢r

✓U · U2

◆+rP �

✓4

3µ+ µv

◆r�r · U

�+µr⇥ ⌦+⇢r = 0 (6.6)

If the flow is irrotational ⌦ = 0 equation (6.6) reduces to

⇢@U

@t+ ⇢r

✓U · U2

◆+rP �

✓4

3µ+ µv

◆r�r · U

�+ ⇢r = 0 (6.7)

According to (6.7) viscous forces play very little role in momentum transport when theflow is irrotational. If the flow is compressible the e↵ect of viscosity is only through aterm that depends on r · U which is usually relatively small in flows at moderate Machnumber. Remarkably, if the flow is incompressible the viscous term disappears altogetherand viscosity plays no role whatsoever in momentum transport. Viscous stresses do act inthe fluid but they generate no net momentum transfer in an irrotational flow. An examplewould be the flow over a wing outside the boundary layer near the surface of the wing andoutside the viscous wake.

6.3 The momentum equation expressed in terms of the en-tropy and vorticity

In Chapter 2 we noted that when an equation of state is combined with Gibbs equationany thermodynamic variable can be expressed in terms of any two others. In an unsteady,three-dimensional flow of a continuous medium


s = s (h (x, y, z, t) , P (x, y, z, t)) (6.8)

Taking the gradient of (6.8) leads to

Trs = rh� rP

⇢(6.9)

which we derived in Chapter 2.

Equation (6.9) can be used to replace the gradient of the pressure in (6.6). Thus themomentum equation in terms of the entropy is

@U

@t+�⌦⇥ U

�+r

✓h+

U · U2

+

◆� Trs� 1

⇢

✓4

3µ+ µv

◆r�r · U

�+

✓µ

⇢

◆r⇥ ⌦ = 0

(6.10)

6.3.1 Crocco’s theorem

The inviscid form of the above equation with µ = 0, µv = 0 is

@U

@t+�⌦⇥ U

�+r

✓h+

U · U2

+

◆� Trs = 0 (6.11)

Equation (6.11) is called Croccos theorem and demonstrates the close relationship betweenvorticity and the gradient of the entropy in a compressible flow.

The relation

⌦⇥ U = 0 (6.12)

is satisfied if U = 0 (a trivial situation), if the vorticity and velocity are parallel (this iscalled a Beltrami flow) or if the flow is irrotational,r⇥ U = 0.

6.3.2 The energy equation for inviscid, non-heat conducting flow

With the viscosities and thermal conductivity, , set to zero, the energy equation in (6.1)becomes


@⇢ (e+ k)

@t+

@

@xi(⇢Uiht)� ⇢GiUi = 0 (6.13)

If we are going to assume that the fluid is inviscid then it is consistent to also assumethat the heat conductivity is zero since the underlying molecular collision mechanisms forboth fluid properties are fundamentally the same in a gas. In Chapter 7 we will cometo recognize that assuming the gas is inviscid and non-heat-conducting is equivalent toassuming that the entropy of a fluid element cannot change.

In terms of the stagnation enthalpy ht = h+ k (6.13) can be written as

@⇢ht@t

+@

@xi(⇢Uiht)� ⇢GiUi =

@P

@t(6.14)

Use the continuity equation in (6.14) and Gi = �@ /@xi to produce

@ht@t

+ Ui@

@xi(ht + ) =

1

⇢

@P

@t(6.15)

We developed the viscous, heat-conducting version of (6.15) previously in Chapter 5.

6.3.3 Steady flow

If the flow is steady, the inviscid energy equation (6.15) reduces to

Ui@

@xi

✓h+

U2

2+

◆= U ·r

✓h+

U2

2+

◆= 0 (6.16)

The quantity

h+U2

2+ (6.17)

is called the steady Bernoulli integral or Bernoulli function. In a general, inviscid, non-heat-conducting, steady flow the energy equation reduces to the statement (6.16) that thevelocity field is normal to the gradient of the Bernoulli function. In the absence of bodyforces, Dht/Dt = 0 in such a flow. The flow is necessarily adiabatic and the stagnationenthalpy of a fluid element is preserved. If the flow is fed from a reservoir where theenthalpy is everywhere uniform then, since ht is preserved for each fluid element, the


enthalpy remains uniform and rht = 0 everywhere. In this case the momentum equation(Croccos theorem) reduces to

⌦⇥ U = Trs (6.18)

If the flow is steady, inviscid and irrotational then both the stagnation enthalpy and the en-tropy are constant everywhere. A flow where rs = 0 is called homentropic. In this instanceboth the momentum equation and energy equations reduce to the same result

h+U2

2+ = Const (6.19)

6.4 Inviscid, irrotational, homentropic flow

It is instructive to examine this case further. Again, assume there are no body forces = 0, set µ = µv = 0, = 0 in the energy equation and r ⇥ U = 0 in the momentumequation (6.6). The governing equations become

@⇢

@t+

@

@xk(⇢Uk) = 0

⇢@Ui

@t+ ⇢

@

@xi

✓UkUk

2

◆+@P

@xi= 0

P

P0

=

✓⇢

⇢0

◆�

(6.20)

In this approximation the energy equation reduces to the isentropic relation between pres-sure and density. These equations are the starting point for the development of inviscidflow theory as well as the theory for the propagation of sound.

6.4.1 Steady, inviscid, irrotational flow

If the flow is steady the equations (6.20) reduce to


@

@xk(⇢Uk) = 0

@

@xi

✓UkUk

2

◆+

1

⇢

@P

@xi= 0

P

P0

=

✓⇢

⇢0

◆�

(6.21)

Take the gradient of the isentropic relation. The result is

rP = a2r⇢ (6.22)

For an ideal gas

r✓P

⇢

◆=

rP

⇢� P

⇢2r⇢ =

✓� � 1

�

◆rP

⇢(6.23)

where (6.22) is used. Using (6.23) the momentum equation now becomes

r✓✓

� � 1

�

◆rP

⇢+

U · U2

◆= 0 (6.24)

The term in parentheses in (6.24) is the stagnation enthalpy. For this class of flows themomentum equation reduces to rht = 0 and the entropy is the same everywhere as justnoted in the previous section. Substitute (6.22) into the continuity equation.

U ·✓rP

⇢

◆+ a2r · U = 0 (6.25)

Using (6.23) the continuity equation becomes

The energy per unit mass of a moving fluid element is e+ k where e is the internal energyper unit mass of the medium and

U ·r�a2�+ (� � 1) a2r · U = 0 (6.26)

Using (6.24) we can write

The rate of change of the energy inside the control volume in Figure is


✓a2

� � 1

◆= ht �

U · U2

(6.27)

The continuity equation now becomes

(� � 1)

✓ht �

U · U2

◆r · U � U ·r

✓U · U2

◆= 0 (6.28)

The equations governing steady, inviscid, irrotational motion reduce to a single equationfor the velocity vector U .

6.5 The velocity potential

The condition r ⇥ U = 0 implies that the velocity can be expressed in terms of a scalarpotential.

U = r� (6.29)

Substitute (6.29) into (6.28). The result is the full potential equation for steady, irrotationalflow.

(� � 1)

✓ht �

r� ·r�2

◆r2��r� ·r

✓r� ·r�

2

◆= 0 (6.30)

The range of flows where (6.30) applies includes subsonic flow over bodies at Mach numbersbelow the critical Mach number at which shocks begin to form and supersonic flows that in-volve smooth expansion and compression such as the flow in a nozzle without shocks.

6.5.1 Unsteady potential flow, the unsteady Bernoulli integral

The irrotational, unsteady momentum equation is, from (6.10),

@U

@t+r

✓h+

U2

2+

◆� Trs� 1

⇢

✓4

3µ+ µv

◆r�r · U

�= 0 (6.31)

Insert (6.29) into (6.31). The equation becomes


r✓@�

@t+ h+

U2

2+

◆� Trs� 1

⇢

✓4

3µ+ µv

◆r�r2�

�= 0 (6.32)

If the flow is inviscid,

r✓@�

@t+ h+

U2

2+

◆� Trs = 0 (6.33)

If the flow is inviscid and homentropic (homogeneously isentropic) with rs = 0, the un-steady momentum equation reduces to

r✓@�

@t+ h+

U2

2+

◆= 0 (6.34)

The quantity

@�

@t+ h+

U2

2+ = F (t) (6.35)

which can be at most a function of time is called the unsteady Bernoulli integral and iswidely applied in the analysis of unsteady, inviscid, compressible flows.

If the flow is a calorically perfect gas with h = CpT , then (6.35) can be written

@�

@t+

✓�

� � 1

◆P

⇢+

U2

2+ = F (t) (6.36)

Generally F (t) can be taken to be a constant independent of time. The sort of unusualsituation where it could be time dependent might occur in a closed wind tunnel where thepressure throughout the system was forced to change due to some overall volume changeof the tunnel. Such changes can occur but they are usually negligible unless there is someintention to change the volume for a particular purpose, perhaps to exert some form offlow control.

In summary, the equations of inviscid, homentropic flow in terms of the velocity potentialare


1

⇢

@⇢

@t+

r� ·r⇢⇢

+r2� = 0

r✓@�

@t+

✓�

� � 1

◆P

⇢+

r� ·r�2

+

◆= 0

P

Pref=

✓⇢

⇢ref

◆�

(6.37)

Note that the isentropic relation provides the needed equation to close the system in lieuof the energy equation. Equation (6.30) is the steady version of (6.37) all boiled down toone equation for the potential. We can reduce (6.37) to a single equation for the velocitypotential as follows.

@�

@t+

✓�

� � 1

◆P

⇢+

r� ·r�2

+ = F (t) (6.38)

Replace the pressure in (6.38) with the density using the isentropic relation.

⇢ =

✓⇢ref �

Pref

✓� � 1

�

◆✓F (t)� @�

@t� r� ·r�

2�

◆◆⇣1

��1

⌘

1

⇢

@⇢

@t=

1

� � 1

✓dF (t)

dt� �tt �r�t ·r�

◆✓F (t)� @�

@t� r� ·r�

2�

◆�1

1

⇢r⇢ =

1

� � 1

✓�r�t �r

✓r� ·r�

2

◆�r

◆✓F (t)� @�

@t� r� ·r�

2�

◆�1

(6.39)

where the body force potential is assumed to be only a function of space. Substitute (6.39)into the continuity equation. The full unsteady potential equation becomes

✓dF (t)

dt� �tt �r�t ·r�

◆�r� ·

✓r�t +r

✓r� ·r�

2

◆+r

◆+

(� � 1)

✓F (t)� @�

@t� r� ·r�

2�

◆r2� = 0

(6.40)


6.5.2 Incompressible, irrotational flow

If the flow is incompressible, r · U = 0 , and irrotational, ⌦ = 0, the momentum equationin the absence of gravity , = 0, reduces to,

@U

@t+r

✓U · U2

◆+r

✓P

⇢

◆= 0 (6.41)

Note again that viscous forces, though present, have no e↵ect on the momentum in anincompressible, irrotational flow. Since r ⇥ U = 0 we can write U = r� and in termsof the velocity potential, the momentum equation for incompressible, irrotational flowbecomes

r✓@�

@t+

U2

2+

P

⇢

◆= 0 (6.42)

The quantity in parentheses is the incompressible form of the Bernoulli integral and, as inthe compressible case, is at most a function of time.

@�

@t+

U2

2+

P

⇢= F (t) (6.43)

An incompressible, irrotational flow (steady or unsteady) is governed by Laplace’s equa-tion

r · U = r2� = 0 (6.44)

which can be solved using a wide variety of well established techniques. Once the veloc-ity potential � is known, the velocity field is generated by di↵erentiation (6.29) and thepressure field is determined from the Bernoulli integral (6.43).

6.6 The vorticity equation

Earlier we cast the momentum equation in terms of the vorticity. Now let’s derive aconservation equation for the vorticity itself. If we take the curl of the momentum equationthe result is


r⇥

0

BB@

@U

@t+�⌦⇥ U

�+r

✓h+

U · U2

+

◆� Trs�

1

⇢

✓4

3µ+ µv

◆r�r · U

�+

✓µ

⇢

◆r⇥ ⌦

1

CCA = 0 (6.45)

or

@⌦

@t+r⇥

�⌦⇥ U

��r⇥ (Trs)�

✓4

3µ+ µv

◆r⇥

r�r · U

�

⇢

!+ µr⇥

✓r⇥ ⌦⇢

◆= 0

(6.46)

Using vector identities, (6.46) can be rearranged to read

@⌦

@t+ U ·r⌦�

�⌦ ·rU

�+ ⌦r · U �rT ⇥rs�

✓4

3µ+ µv

◆r⇥

r�r · U

�

⇢

!+ µr⇥

✓r⇥ ⌦⇢

◆= 0

(6.47)

For inviscid, homentropic flow, rs = 0, µ = 0, µv = 0 , the vorticity equation reducesto

D⌦

Dt=@⌦

@t+ U ·r⌦ =

�⌦ ·rU

�� ⌦r · U (6.48)

Equation (6.48) is interpreted to mean that, in an inviscid fluid, if the flow is initiallyirrotational it will always remain irrotational. Conversely, if the flow has vorticity to beginwith, then that vorticity can be convected or amplified through stretching or volume changebut cannot disappear.

If the flow is incompressible and viscous, the vorticity equation is

D⌦

Dt=@⌦

@t+ U ·r⌦ =

�⌦ ·rU

�+

✓µ

⇢

◆r2⌦ (6.49)

For planar, two-dimensional flow,�⌦ ·r

�U = 0 and there is only one non-zero component

of the vorticity, ⌦z, which satisfies the di↵usion equation

D⌦z

Dt=

✓µ

⇢

◆✓@2⌦z

@x2+@2⌦z

@y2

◆(6.50)


This equation is of the same form as the incompressible equation for the transport ofthe temperature. See the energy equation in (6.2) with the viscous dissipation term ne-glected.

DT

Dt=

✓

⇢

◆r2T (6.51)

In two dimensions vorticity di↵uses like a scalar such as temperature or concentration.But note that vorticity can be positive or negative and where mixing between oppositesigns occurs, vorticity can disappear through di↵usion. In three- dimensional flow, thevorticity can be amplified or reduced due to vortex stretching or compression arising fromthe nonlinear term

�⌦ ·r

�U .

6.7 Fluid flow in three dimensions, the dual stream func-tion

The equations for particle paths in a three-dimensional, steady fluid flow are

dx

dt= U (x, y, z)

dy

dt= V (x, y, z)

dz

dt= W (x, y, z)

(6.52)

The particle paths are determined by integrating (6.52)

x = f (x, y, z, t)

y = g (x, y, z, t)

z = h (x, y, z, t)

(6.53)

where (x, y, z) is the vector coordinate of a particle at t = 0. Elimination of t among thesethree relations leads to two infinite families of integral surfaces; the dual stream-functionsurfaces


1

= 1

(x, y, z)

2

= 2

(x, y, z)(6.54)

The total di↵erential of the stream function is

d =@

@xdx+

@

@ydy +

@

@zdz (6.55)

Use (6.52) to replace the di↵erentials in (6.55) and note that on a stream surface d = 0,the stream-functions (6.54) are each integrals of the first order PDE

U ·r j = U@ j

@x+ V

@ j

@y+W

@ j

@zj = 1, 2 (6.56)

A given initial point, (x, y, z), defines two stream-surfaces. The velocity vector through thepoint lies along the intersection of the surfaces as implied by (6.56) and shown in figure6.1.

Figure 6.1: Dual stream-function surfaces.

Given the dual stream-functions, the velocity field can be reconstructed from

U = r 1

⇥r 2

(6.57)


6.8 The vector potential

The velocity field of an incompressible flow can be represented as the curl of a vectorpotential, A also called the vector stream function.

U = r⇥ A (6.58)

Use of the vector potential in (6.58) guarantees that the incompressibility condition r·U =0 is satisfied. In two-dimensional flow only the out-of-plane component of A is nonzeroand this corresponds to the stream function discussed in Chapter 1. Take the curl of(6.58).

r⇥ U = r⇥�r⇥ A

�(6.59)

Now use the vector identity

r�r · A

��r2A = r⇥

�r⇥ A

�(6.60)

Equation (6.59) becomes

⌦ = r�r · A

��r2A (6.61)

6.8.1 Selection of a Coulomb gauge

There is a certain arbitrariness to the vector potential and we can take advantage of this tosimplify (6.61). Suppose we add the gradient of a scalar function to the vector potential.Let

A0 = A+rf (6.62)

Since r ⇥ rf ⌘ 0 the velocity field generated by A0 or A is identically the same.

r⇥ A0 = r⇥ A (6.63)

Imagine that we are given A0. We can always find a function f such that

r · A0 = r2f (6.64)


which implies that

r · A = 0 (6.65)

This is called choosing a Coloumb gauge for A. In e↵ect we are free to impose the condi-tion (6.65) on A without a↵ecting the velocity field generated from A when we take thecurl.

Using (6.65) in (6.61) generates the Poisson equation for the vector potential.

r2A = �⌦ (6.66)

parIn order for (6.66) to be useful one has to be given the vorticity field ⌦. The vectorpotential is then determined using standard techniques for solving equations of Poissontype. The velocity field is then generated by taking the curl.

The vector potential is directly related to the dual stream-functions discussed in the pre-vious section.

A = 1

r 2

= � 2

r 1

(6.67)

6.9 Incompressible flow with mass and vorticity sources

Consider an incompressible flow constructed from a distribution of mass sources, Q(x, t)and distributed sources of vorticity, ⌦(x, t). The velocity field is constructed from a linearsuperposition of two fields.

U = Usources + Uvortices (6.68)

The velocity field generated by the mass sources is irrotational and that generated by thevorticity sources is divergence free. The continuity equation for such a flow has a sourceterm

r · U = r · Usources = Q (x, t) (6.69)

and the curl of the velocity is

r⇥ U = r⇥ Uvortices = ⌦ (x, t) (6.70)


In terms of the potentials the velocity field is

U = r�+r⇥ A (6.71)

The potentials satisfy the system of Poisson equations

r2� = Q (x, t)

r2A = �⌦ (x, t)(6.72)

This approach allows one to construct fairly complex flow fields that can be rotational whileretaining the simplicity of working in terms of potentials governed by linear equations (6.72)and the associated law of superposition (6.68). Notice that the theory of potential flowis exactly analogous to the theory of potentials in electricity and magnetism. The masssources coincide with the distribution of electric charges and the vorticity coincides withthe electric currents.

6.10 Turbulent flow

Turbulent flows are characterized by unsteady eddying motions with a wide range of scalessuperimposed on a slowly changing or steady mean flow. The eddies mix fluid rapidly andare responsible for much higher rates of mass, momentum and heat transport than wouldoccur if the flow was laminar. The primary di�culty with analyzing or numerically com-puting turbulent flows stems from the fact that, at high Reynolds number, the the rangeof scales involved can be very large with velocity gradients at the finest scale, where mostkinetic energy dissipation occurs, of order Re larger than the gradients at the largest scalethat drive the flow. Computations designed to fully simulate the flow must be resolvedon such a fine grid that often only a few simulations can be carried out even at moderateReynolds number. Very high Reynolds numbers are still beyond todays largest comput-ers. Experiments designed to measure a turbulent flow are faced with a similar challengerequiring extremely small measurement volumes as well as the ability to measure over alarge field comparable to the overall size of the flow.

Recall the equations of motion, repeated here for convenience


@⇢

@t+

@

@xi(⇢Ui) = 0

@⇢Ui

@t+

@

@xj

✓⇢UiUj + P �ij � 2µSij +

✓2

3µ� µv

◆�ijSkk

◆� ⇢Gi = 0

@⇢ (e+ k)

@t+

@

@xi

✓⇢Uiht � (@T/@xi)� 2µUjSij +

✓2

3µ� µv

◆�ijUjSkk

◆� ⇢UiGi = 0

(6.73)

In this section we will discuss an alternative to fully resolving the unsteady flow introducedby Osborne Reynolds in 1895 in which each flow variable is decomposed into a mean andfluctuating part. When the equations of motion are suitably averaged, the result is anew set of equations that relate the mean flow to the correlated part of the fluctuatingmotion.

Imagine an ensemble of realizations of the time dependent flow that all satisfy the same setof initial and boundary conditions, perhaps in the form of a series of laboratory experimentsrepeated over and over. Decompose each flow variable into an ensemble mean and afluctuation

F (x, y, z, t) = F (x, y, z, t) + F 0 (x, y, z, t) (6.74)

where F is a generic variable. The mean is defined by

F =1

N

NX

n=1

Fn (6.75)

where the number of realizations of the flow is N and the index refers to the nth memberof the ensemble. All members of the ensemble are aligned to have the same time origin andevolve with the same clock which is reset each time a new realization is initiated.

Apply the decomposition and averaging to a linear term in the equations of motion suchas

\@@xj

(P �ij) =\@

@xj

⇣⇣P + P 0

⌘�ij⌘=

@

@xj

⇣P �ij

⌘+

\@@xj

(P 0�ij) =@

@xj

⇣P �ij

⌘(6.76)

Any term that is linear in the fluctuation is zero. Nonlinear terms are more complex. Thedecomposition of the density, pressure and velocity is


⇢ = ⇢+ ⇢0

P = P + P 0

U = U + U 0

V = V + V 0

W = W +W 0

(6.77)

Consider averaging the momentum flux term ⇢UV .

[⇢UV =\

(⇢+ ⇢0)⇣U + U 0

⌘⇣V + V 0

⌘=

[⇢U V +\⇢0U V +\⇢U 0V + \⇢0U 0V +\⇢UV 0 + \⇢0UV 0 + \⇢U 0V 0 + \⇢0U 0V 0 =

⇢U V + d⇢0U 0V + d⇢0V 0U + ⇢[U 0V 0 + \⇢0U 0V 0

(6.78)

The transformation of the kinetic energy is

[⇢UV = ⇢U V +⇣d⇢0U 0V + d⇢0V 0U + ⇢[U 0V 0 + \⇢0U 0V 0

⌘= ⇢U V + ⌧xy|turbulent (6.79)

The terms in parentheses can be thought of as e↵ective stresses due to the back- groundfluctuations. Similar terms arise when generating the turbulent normal stresses in com-pressible flow

[⇢UU = ⇢U U +⇣2d⇢0U 0U + ⇢[U 0U 0 + \⇢0U 0U 0

⌘= ⇢U U + ⌧xx|turbulent

[⇢V V = ⇢V V +⇣2d⇢0V 0V + ⇢[V 0V 0 + \⇢0V 0V 0

⌘= ⇢V V + ⌧yy|turbulent

(6.80)

Time dependent terms are ensemble averaged in the same way.

c@⇢@t

=1

N

NX

n=1

@⇢

@t

��n

=@⇢

@t+d@⇢0@t

=@⇢

@t

d@⇢Ui

@t=

1

N

NX

n=1

@⇢Ui

@t

��n

=@⇢ bUi

@t+

\@⇢0 bUi

@t+

\@⇢Ui0

@t+

\@⇢0Ui0

@t=@⇢ bUi

@t+@[⇢0Ui

0

@t

(6.81)


Since fluid properties such as µ, muv and are temperature dependent they must also beincluded in the decomposition and averaging process.

6.10.1 Turbulent, incompressible, isothermal flow

If the flow is incompressible, the density is constant. If the temperature is also constantthe equations of motion simplify to

@Uj

@xj= 0

@Ui

@t+@UiUj

@xj+

1

⇢

@P

@xi�✓µ

⇢

◆@2Ui

@xj@xj= 0

(6.82)

Substitute the Reynolds decomposition (6.77) with density constant into (6.82) and en-semble average the equations.

@Uj

@xj= 0

@Ui

@t+@UiUj

@xj+@\Ui

0Uj0

@xj+

1

⇢

@P

@xi�✓µ

⇢

◆@2Ui

@xj@xj= 0

(6.83)

In incompressible flow the only nonlinear term is the convective term and the only terminvolving the fluctuations is the last term on the right which can be viewed as an e↵ectiveturbulent stress term, the Reynolds stress.

⌧ij |turbulent = �⇢\Ui0Uj

0 (6.84)

The advantage of this approach is that the entire unsteady motion does not need to beresolved, only the correlations (6.84) are needed. The fundamental di�culty is that theReynolds stresses constitute an additional set of unknowns and the equations (6.83) areunclosed. The field of turbulence modeling is basically all about searching for modelequations that can be used to relate the Reynolds stresses to the mean flow so the system(6.83) can be closed.

In free shear flows the laminar stress term in (6.83), µ@2Ui/@xj@xj is generally much smallerthan the Reynolds stress term and some simplification can be achieved by neglecting theviscous term in the momentum equation altogether. In wall bounded flows this cannot be


done. The Reynolds stresses tend to dominate the flow away from the wall but near thewall the velocity fluctuations are damped to zero and the viscous stress term is responsiblefor the friction on the wall. We will have more to say about the flow near a wall in Chapter8.

6.11 Problems

Problem 1 - Derive equation (6.6) beginning with the Navier-Stokes equations. Do thesame for equation (6.47).

Problem 2 - Show that for homentropic flow of an ideal gas rP = a2r⇢ where a is thelocal speed of sound.

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