derivation of equations 2013

8/13/2019 Derivation of Equations 2013

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Contents

4 The Equations of Fluid Motion 3

4.1 One-dimensional mass conservation and momentum balance equa-

tions: a first look at the continuum equations for fluid motion 34.2 Alternative viewpoint: Following a fluid element The one-

dimensional Reynolds Transport Theorem . . . . . . . . . . . 6

4.2.1 The Leibniz formula . . . . . . . . . . . . . . . . . . . 6

4.2.2 One-dimensional mass conservation . . . . . . . . . . . 7

4.2.3 The transport theorem in one dimension . . . . . . . . 8

4.3 Acoustics: The study of small disturbances . . . . . . . . . . . 8

4.4 A one-dimensional approximation to a jet . . . . . . . . . . . 9

4.5 First remarks about kinematics . . . . . . . . . . . . . . . . . 10

4.5.1 Material derivative: the skydiver example . . . . . . . 10

4.5.2 The Lagrangian versus Eulerian representations of thevelocity field . . . . . . . . . . . . . . . . . . . . . . . . 12

4.6 A useful physical interpretation of the material derivative . . . 16

4.7 Conservation of mass: The continuity equation . . . . . . . . . 17

4.7.1 The meaning of the velocity vectoru . . . . . . . . . . 18

4.8 Reynolds Transport Theorem . . . . . . . . . . . . . . . . . . 19

4.8.1 Conservation of mass (again) . . . . . . . . . . . . . . 20

4.8.2 Special form of the Reynolds Transport Theorem . . . 20

4.9 Linear momentum balance . . . . . . . . . . . . . . . . . . . . 22

4.9.1 body forces . . . . . . . . . . . . . . . . . . . . . . . . 22

4.9.2 Surface forces . . . . . . . . . . . . . . . . . . . . . . . 23

4.9.3 Back to the linear momentum statement . . . . . . . . 24

4.9.4 The stress tensor and the state of stress . . . . . . . . 25

4.9.5 Cauchy stress equation of motion . . . . . . . . . . . . 27

4.10 Balance of Angular Momentum . . . . . . . . . . . . . . . . . 284.11 More Kinematics . . . . . . . . . . . . . . . . . . . . . . . . . 30

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4.11.1 Relative motion of nearby fluid elements . . . . . . . . 30

4.12 Constitutive equations: a Newtonian fluid . . . . . . . . . . . 34

4.12.1 Preliminary: Fluid statics . . . . . . . . . . . . . . . . 34

4.12.2 The stress tensor for fluids in motion (the main event) 35

4.12.3 Mean normal stress in a fluid (after Batchelor 1968) . . 39

4.12.4 On to the Navier-Stokes equation . . . . . . . . . . . . 40

4.13 Energy conservation in continua . . . . . . . . . . . . . . . . . 42

4.13.1 Mechanical Energy . . . . . . . . . . . . . . . . . . . . 42

4.13.2 Thermal Energy equation . . . . . . . . . . . . . . . . 43

4.13.3 Consequences of equilibrium thermodynamics . . . . . 44

4.14 Consequences of the second law: an equation for entropy vari-ations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46

4.15 Conditions for incompressibility . . . . . . . . . . . . . . . . . 48

4.16 The structure of conservation laws(something of a review) . . . 49

4.17 Appendix: Kinematics of material surfaces . . . . . . . . . . . 52

4.18 Appendix: Governing equations in component form . . . . . . 56

4.18.1 CONTINUITY EQUATION . . . . . . . . . . . . 56

4.18.2 NAVIER-STOKES EQUATIONS . . . . . . . . . 56

4.19 Appendix: The Navier-Stokes equations a little bit of history 61

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4 The Equations of Fluid Motion

4.1 One-dimensional mass conservation and momentumbalance equations: a first look at the continuumequations for fluid motion

Consider a gas of densitythat flows along a circular tube of constant radiusR.We shall denote the cross-sectional area A = (R2) and the axial direction asz. In the simple one-dimensional analysis given here we neglect viscous effectsin the gas, and so assume that the velocity is uniform across the tube. Weask how the fluid changes its velocity u owing to a pressure gradient p/zthat acts along the tube. The case A = constant is treated first and thengeneralized to flow in a circular tube of (slowly) varying radius R(z, t) (or areaA(z, t)).

The ideas introduced now will later be generalized to three dimensions,and expressed using vector notation. The influence of fluid friction (viscosity)will also be introduced and a general discussion of viscous effects leads nat-urally to the idea of stress expressed in terms of a second-order tensor. Thephysical arguments and mathematical steps are similar to those encounteredin this one-dimensional example.

To begin the continuum analysis we introduce three unknowns (z, t),u(z, t) and p(z, t), which enter the governing equations as

Physical statement unknowns in the equationconservation of mass , u

linear momentum balance ,u,pequation of state , p

Table 1: An example of the conservation laws and corresponding unknownsfor the one-dimensional example described in the text.

The final results will consistent of three equations for three unknowns,, u and p. To make the analysis more general we should consider the tem-perature variation (the temperature is then another unknown), which then

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requires introduction of an energy equation.

Now consider a fixed (control) volume in the tube as sketched in Figure1. The control volume has width z, and front (z) and back (z+ z) facesof areaA.

(z,t)FLOW

FLOW

control

volume

u(z,t)

p(z,t) p(z+ z,t)

Figure 1: The control volume selected to perform the one-dimensional massand linear momentum balances.

Conservation of mass is first written in words and then expressed mathe-matically as

time rate of changeof the mass in V =

rate of mass flowintoV

rate of mass flowout ofV (1a)

t(.A.z)

mass

= (.u.A) |z mass/time

(.u.A) |z+z (1b)

Since we are assuming that the cross-sectional area is constant then dividingthrough by zand taking the limit z 0 gives1

t

+(u)

z

= 0, (3)

which is referred to as the one-dimensional continuity equation.

Next consider the change of linear momentum of the mass of fluidin the fixed control volume V. This statement is Newtons second law for afluid flowing through a fixed control volume; the linear momentum can changedue to advection (flow) and due to forces that act. We focus on changes ofmomentum in the z-direction. The word, then mathematical, statements of

1Recall the definition of the derivative

df

dz

= limz0

f(z+ z) f(z)

z

. (2)

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4.2 Alternative viewpoint: Following a fluid element The one-dimensional Reynolds Transport Theorem

There is a conceptually useful alternative to think about conservation laws,which is to consider a fixed mass of material and follow it along its motion.

We then keep track of the time-rate-of-change of the quantity of interest (e.g.mass, linear and angular momentum, energy, etc.). We refer to this approachas thematerialrate of change. In the usual approach we form an integral of afinite amount of some quantity and then calculate the time-rate-of-change theintegral. The three-dimensional version of this idea is called the ReynoldsTransport Theorem and is introduced below. First, we consider the one-dimensional version, which is conceptually simpler and is analogous to theLeibniz rule you likely saw in a calculus course.

4.2.1 The Leibniz formula

The one-dimensional illustration of this idea is the Leibniz formula from cal-culus: time differentiation of an integral whose limits depend on time is givenby2

d

dt

b(t)a(t)

f(x, t) dx= b(t)a(t)

f

t dx+f(b(t), t)

db

dt f(a(t), t)

da

dt boundary terms

. (9)

The two terms on the right-hand side are contributions from the functionevaluated at the end points and further depend on the rate of change of the end

2The derivation of this result can be found in texts on advanced calculus. The basic ideacan be established by using the definition of the derivative as

d

dt

b(t)a(t)

f(x, t)dx = limt0

b(t+t)a(t+t)

f(x, t+ t)dx b(t)a(t)

f(x, t)dx

t (8a)

= limt0

b(t+t)a(t+t)

(f(x, t+ t) f(x, t)) dx

t + lim

t0

b(t+t)a(t+t)

f(x, t)dx b(t)a(t)

f(x, t) dx

t

(8b)

=

b(t)a(t)

f

tdx + lim

t0

b(t+t)b(t)

f(x, t)dx

t lim

t0

a(t+t)a(t)

f(x, t)dx

t (8c)

= b(t)

a(t)

f

t dx+f(b(t), t)

db

dt f(a(t), t)

da

dt , (8d)

where the simplification to the final result follows by the use of a Taylor series.

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points. Ifx denotes position, each of these terms have the form (velocity f).Theseboundary contributionsare important to recognize as is emphasized nextwhen considering variations of properties following a fluid element.

4.2.2 One-dimensional mass conservation

Consider a flow in one dimension as shown in Figure 1. We consider a fluidelement with a label , which has instantaneous position z(t), and speedu(t) =

dzdt

. Suppose that we follow a fixed mass of fluid that moves with thefluid. If the end points of the domain, from left to right in Figure 1, are twoarbitrarily chosen points, z = a(t) and z = b(t), then the mass is obtainedby integrating the density between points a and b. Since this mass does notchange (the velocity of the end points are those of the fluid), then dmdt = 0, orwe can write the integral

dm

dt = 0 =

d

dt b(t)

a(t) dz. (10)

Using the Leibinz rule (9) we can write

0 = d

dt

b(t)a(t)

dz= b(t)a(t)

t dz+(b(t), t)

db

dt (a(t), t)

da

dt. (11)

We can write the last two boundary terms as The integralun-does thederivative. Also,the velocity ata(t), is u = da

dtand similarly atthe rightend-point.

(b(t), t)db

dt (a(t), t)

da

dt =

b(t)a(t)

z(u) dz. (12)

Combining these results we have

0 = ddt

b(t)a(t)

dx = b(t)a(t)

t

dz+ b(t)a(t)

z

(u) dz (13a)

= b(t)a(t)

t + z(u)

=0

dz. (13b)

Since the limits of integration, a(t) and b(t), are arbitrary, we can concludethat for the integral to be zero, the integrand must be zero. Hence, we have theone-dimensional form of the continuity equation

t +

z(u) = 0. (14)This equation is the same as (3), which we derived above using a fixed controlvolume.

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4.2.3 The transport theorem in one dimension

The transport theorem relates to the rates of change of transportable quan-tities such as mass, energy, linear momentum, entropy, etc. We consideredthe case of mass above. We now treat this more general idea in one dimen-

sion. In general, we consider f(z, t) to denote a quantity/mass. Then, thetotal amount of the quantity between x = a(t) and x = b(t) is obtained byintegratingfbetween the two arbitrary points a and b. Hence, the time rateof change of this quantity following the fluid is

d

dt(some quantity) =

d

dt

b(t)a(t)

fdz (15)

We use the Leibniz and the step shown above for converting boundary termsto an integral involving the z-derivative of a function:

d

dt

(some quantity) = d

dt b(t)

a(t)

fdz (16a)

= b(t)a(t)

t(f) dz+

b(t)a(t)

z(fu) dz. (16b)

Again, this quantity can be written as a single integral, so we obtain theone-dimensional transport theorem:

d

dt(some quantity) =

b(t)a(t)

t(f) +

z(fu)

dz. (17)

If you wish, you can use the continuity equation (14) to simplify theintegrand of equation (17) to obtain the one-dimensional transport theorem

in the formd

dt(some quantity) =

b(t)a(t)

f

t +u

f

z

dz. (18)

Exercise: If you were interested in linear momentum, then in one dimensionwe keep track ofu or f = u in the equations above. In this case, equations(17) and (18) produce the left-hand side of one-dimensional linear momentumequations discussed above.

4.3 Acoustics: The study of small disturbances

Sound propagation can be analyzed by considering small perturbations fromequilibrium. For example, consider a liquid or gas with density 0and pressure

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p0. Small disturbance from this equilibrium state can be written

(z, t) =0+ (z, t); p(z, t) =p0+ p(z, t); u(z, t) = u(z, t), (19)

where the indicates a disturbance quantity of small magnitude: || 0and | p| p0. Substituting into the continuity and momentum equations, lin-earizing about the equilibrium state (this amounts to neglecting all quantitiesinvolving products of small quantities) and using an equation of state p()leads to the wave equation(you should show all of the steps)

2f

t2 =

dp

dc2

2f

z2 where f= ,p, u. (20)

The speed of sound c characterizes the propagation of disturbances. For adi-abatic motions of an ideal gas p = constant, where is the ratio of the

heat capacity at constant pressure to that at constant volume, c = p0/0330 m/sec at standard pressure and temperature.The above approach can also be used to consider sound propagation in

a uniform flow of speedU. The steps involved in linearization of the completeequations is a common approach to analyzing complicated problems beginningwith the full equations.

Figure 2: A jet of water issuing from a nozzle in a kitchen sink.

4.4 A one-dimensional approximation to a jet

As a nice every day example of the one-dimensional equations we consider ajet issuing vertically downward from a nozzle, e.g. your kitchen sink as in

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Figure 2. Let the radius of the nozzle be a0 and the volumetric flow rate fromthe nozzle be Q0. Also, we assume that the density of the liquid is constant.The radius a(z) of the jet and the speed of the jet u(z) both change alongthe length of the jet. So, this example is a case where the cross-sectional areaof the flow, A(z) = a(z)2, changes along the flow direction. We analyze the

steady motion using the continuity and momentum equations, and includingthe gravitational body force that acts on the jet:

d

dz(Au) = 0 and u

du

dz =

dp

dz+g (21)

The continuity equation can be integrated to yield a2u= a20u0 = Q0.

We now solve the linear momentum equation using the boundary con-dition that the initial velocity is u(0) = u0 =

Q0a20

and the condition that the

surface pressure is p = p0, the atmospheric pressure, for all z (we neglectsurface tension). In this case, the pressure gradient dp/dz = 0. Hence, the

momentum equation is simply (after a small rearrangement) Thinking back toyour basicphysics, thisequationexpresses theinterconversion ofkinetic andpotential energy,which given thechoice of signs, is12u2 gz =

constant.

du2

dz = 2g. (22)

To simplify some of the algebra, it is convenient to rescale variables as

U= u

u0, Z=

z

u20/(2g)

dU2

dZ = 1, withU(0) = 1 (23)

The solution of the ODE is

U(Z) = (1 +Z)1/2 . (24)

The shape of the jet follows from a

2

=Q0/u(z) ora(z)

a0= (1 +Z)1/4 . (25)

We conclude that the radius has decreased by a factor of 2 on a length scale

Z= 15 or z = 15u20

2g . If we had a liquid jet issuing at a speed of 1/2 m/sec,

then this length scale is about 0.2 m or 20 cm.

4.5 First remarks about kinematics

4.5.1 Material derivative: the skydiver example

An important idea that is utilized frequently in fluid dynamics is the materialderivative, which provides the variation with time of any quantity as measured

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Figure 3: The skydiver example: A skydiver measures the atmospheric temper-ature and we are interested in reporting this measurements relative to ground-fixed coordinates.

by an observer moving with the velocity u of the fluid. The basic idea behindthis use of measurement is illustrated by the skydiver example.

To begin with, we know that the atmosphere is thermally stratified. So,if a skydiver jumps out of an airplane and during descent measures the at-mospheric temperature distribution, then although the measured temperaturevaries with time, this information can be translated into a combination of thespatial and temporal variations by recording the divers vertical position as afunction of time. Let the temperature distribution measured by the skydiverbe Td(t). An observer on the ground keeps track of the temperature distribu-tionT(z, t) and, via communication with a diver whose instantaneous positionis zd(t), determinesT(zd(t), t). Clearly,T(zd(t), t) =Td(t).

Let us denote the (Lagrangian) time derivative observed by the diveras dTd/dt; we refer to d/dt as the total time derivative. Next, express theresult in terms of the temperature dependence recorded by the ground-based(Eulerian) observer. Thus, using the chain-rule

dTddt

=dT(zd(t), t)

dt =

T

t +

dzd(t)

dt

T

z , (26)

or introducing the vertical velocity withuz = dzd(t)dt we have

dTddt

total time derivative

=dT(zd(t), t)

dt = T

tlocal time derivative

+ uzT

z

convective contribution

.

(27)

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The right-hand side of (27) is the Eulerianrepresentation of the time variationof temperature measured by the skydiver moving with velocityuz (which maydepend on time).

For the fully three-dimensional problem where the velocity field is u =(ux, uy, uz) we are interested inT(x, y, z, t) and it should be apparent that (27)generalizes to

dTddt

=T

t +ux

T

x+uy

T

y +uz

T

z =

T

t + u T . (28)

The form of the right-hand side of (28) is very common and is called thematerial time derivative, typically denoted D/Dt (or here d/dt), of a quantitymoving with velocity u(x, t):

material time derivative: D

Dt

t+ u . (29)

Acceleration: The material derivative of the velocity (discussed below) is

Du

Dt =

u

t + u u. (30)

Exercise: Provide a physical interpretation of DuDt

by describing the meaningof u

t and u u.

4.5.2 The Lagrangian versus Eulerian representations of the veloc-ity field

Let us now inquire about ways to keep track of the fluid velocity. Two ideasare common, the Eulerianand the Lagrangianrepresentations.3 The formerproves to be more useful for performing numerical calculations, though thelatter is conceptually very helpful. In the Eulerian description all variables arereported as a function of position x and time t, where x is measured relativeto some appropriately chosen origin; this is the usual case for fields (e.g. thetemperature or electric field) as commonly measured in the laboratory. Forexample, the velocity is denoted u(x, t). Next we consider the conceptuallyuseful Lagrangian description.

3Both descriptions were originally given by Euler. For a short commentary, see thehistorical article by Truesdell included at the end of these notes.

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Figure 4: The Lagrangian labelling of fluid particles for following fluid motion.

By analogy with particle mechanics (e.g. cannon balls) we consider themotion of small pieces of fluid. We imagine labeling with dye, or a thin mem-brane that moves with the local velocity, a small region of the fluid and follow-ing the motion of the center-of-mass. From such a trajectory we may calculatethe velocity of the particle and then the acceleration.

From now on we shall refer to a vanishingly small (from a continuumviewpoint) volume of fluid, containing the same mass of fluid, as a fluid particleor a fluid element. In the continuum mechanics description we cannot attacha specific length scale to the size of the fluid particle but we can proceedwith confidence that the theoretical construct of a fluid particle is in accordwith our everyday notions of mass, velocity and acceleration as discussed inparticle mechanics (note that deformationof the fluid element is importantwhen resistance to fluid motion is discussed).

Consider the motion of a fluid element. We denote the position asx= X(t) orX(t; ) where the subscript labels the different particles (e.g.conceptually it is convenient to imagine labeling particles by their initial po-sitionX(0)). The Lagrangian velocity, denoted uL, then corresponds to thetime-derivative of position for this fluid particle,

uL(t; ) =X(t; )

t (particle label held fixed) . (31)

Likewise, the Lagrangian acceleration is

aL(t; ) =

uL(t; )

t =

2X(t; )

t2 , (32)which we shall see is important when applying Newtons second law to themotion of a fluid. The alert reader will observe a difficulty with the practical

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implementation of this idea for reporting the velocity field in most flows. Inorder to make use of equations (31) and (32), the position as a function oftime must be obtained foral lof the fluid particles in the fluid. In general, thisis not practical.

Let us express theseLagrangianideas in terms ofEulerianvariables, suchas the velocity field u(x, t). In particular, the Lagrangian positionX(t; ) issuch that

X(t; )

t =uL(t; ) =u(X(t; ), t) (33)

as the fluid particle occupies (Eulerian) position x = X(t; ) at time t. Wemay then note that the corresponding acceleration (following a particle) isobtained by using the chain rule:

aL(t; ) = uL(t; )

t =

u(X(t; ), t)

t (34a)

= ut x

+

X(t; )

t u=ut

x

+ u u, (34b)

where we have used (33) and further we have taken some care to emphasizewhich variables are held fixed. Although the particle acceleration seems arather rudimentary concept in basic mechanics, its representation via the Eu-lerian variables is more involved than one might think at first glance, and mostimportantly involves the velocity field in a nonlinearmanner (u u).4 Wewill next see how to apply these ideas to the applications of Newtons secondlaw (F= ma) to fluid parcels.

4The time derivative measured by an observer moving with the local fluid velocity u(x, t)can be expressed with elementary calculus by considering the change of position xin timetand taking the limit t 0:

aL = limt0

u(x + x, t+ t) u(x, t)

t (35a)

= limt0

u(x + x, t+ t) u(x + x, t)

t +

xjtuj

u(x + x, t) u(x, t)

xj

u

xj

(35b)

= u

t + u u . (35c)

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Figure 5: The meaning of the convective contribution u to the materialtime derivative. The sketch indicates how flow through a spatial gradientdI/dxgives rises to a time rate of change of a quantity (here I) measured bya moving observer.

Summary: The material derivative

D

Dt=

t+ u (36)

expresses the rate of change measured by an observer translating with veloc-

ity u. Consequently, if we consider some property FL(t; ) associated withfluid element , then the time-rate-of-change of that property as the materialelement is followed, expressed in terms of the Eulerian fieldF(x, t), is

DFL(t; )

Dt =

DF(x= X(t; ), t)

Dt =

F

t + u F (37)

Most importantly, the accelerationa of the fluid element when expressed inEulerian variables is

a=u

t + u u . (38)

The termuuis nonlinear in velocity, and this is the principal difficulty whenconsidering analytical solutions to the momentum equations (to be derivedshortly) that describe the motion of a fluid.

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4.6 A useful physical interpretation of the material deriva-tive

It is important to reiterate the physical interpretation of the two contributionsto the total time derivative following a moving observer, equation (36). The

first term on the right-hand side is the obvious term that arises from timevariations at a fixed position in space. The second term is a consequence oftranslation through a gradient, and this term contributes even if the field istime independent, i.e. () /t = 0. To interpret how translationu = 0 andand a nonzero spatial variation of some quantityI, i.e. I, contribute to thetime rate of change, refer to the physical picture in Figure 5.

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Figure 6: (a) Flow through a fixed control volume V, with bounding surfaceS and unit outward normal n. (b) A material control volume (fixed mass)translating and deforming through space as a result of motion of the fluid.

4.7 Conservation of mass: The continuity equationWe first derive a differential equation describing the variation of density in thefluid. This equation couples (x, t) andu(x, t).

The standard derivation takes a fixed volume V in space. Letndenotethe unit outward normal to the surface S. The rate at which mass enters asurface element dS is u n dS (note that n u > 0 when material leavesand < 0 when material enters; the dimensions ofu ndSare mass/time); itis +u n dSwhen material leaves the volume. Hence, conservation of massapplied to the fixed volume gives

ddt

V

dV time rate of

change of mass inV

= S

n udS net flow of mass

(acrossS) into V

. (39)

Since the region Vis fixed, differentiation and integration can be exchanged.Using the Divergence Theorem we arrive at

V

t + (u)

dV = 0, (40)

and, as the fixed volume Vis otherwise arbitrary, then we conclude that the

integrand must vanish everywhere in space:

t+ (u) = 0. continuity equation (41)

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Remark: Rearrange (41) to identify the material derivative (D/Dt) of. Wesee that

t+ u + u= 0

1

D

Dt = u . (42)

The left-hand side is the relative time rate of change of fluid density following

fluid motion. This statement may be easier to visualize if we introduce thespecific volume (volume per unit mass) V =1 and write (42) as

1

V

DV

Dt = u , (fractional rate of change of volume of a material element)

(43)which provides a physical interpretation of u. A non-zero divergence of thevelocity means that the density (or specific volume) of the material elementat that position is changing. How would you explain equations (42) and (43)to a friend?

4.7.1 The meaning of the velocity vector u

It is useful to consider the general form of a conservation law for a scalar f,which can represent some quantity per unit volume:

f

t + j= 0, (44)

wherej is the flux vector for the quantityf. As can be shown using the deriva-tion given in the previous section the flux of the quantityf(amount/area/time)isn jwheren is the unit outward normal to a surface element. If we consider

the density field , then the flux, mass/area/time, may be used to define thevelocity field: j= u. We can imagine an experiment where we measure j andso this identification defines the velocity vector u. Because of the definitionof linear momentum as the product of mass and velocity, then we may alsoview this definition as u as prescribing the linear momentum/mass of fluid.Alternatively, the mass flux density vector j = u can be interpreted as thelinear momentum per unit volume of fluid.

Next, letr denote the position vector. Note that multiplying the conti-nuity equation byr yields

(r)t

+ r j= 0. (45)

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With the identity (show this)rj= (jr)j, and integrating the continuityequation over some volume V, then (Landau & Lifshitz, section 50)

d

dt

V

rdV =V

jdV, (46)

where we have assumed that on all bounding surface n jrdecays sufficientlyfast for any boundary contributions to vanish. Equation (46) has a naturalphysical interpretation consistent with Newtons second law since the left-hand side is the time rate of change of the center-of-mass position while theright-hand side and the right-hand side is the total momentum in V.

4.8 Reynolds Transport Theorem

In order to think further about conservation laws, it is useful to considervolume integrals of linear and angular momentum, energy, etc., and to thinkabout the material rate of change of such integrals (i.e. the time rate of changefollowing a fixed mass that moves with the fluid). The volume integral is overa finite region, and the domain can change shape as the velocity field may havegradients. Hence, we consider time rates of change of integrals whose limitsdepend on time.

The transport theorem in three dimensions: The generalization of equation(9) to three-dimensions is needed in continuum mechanics, where it is knownas the Reynolds Transport Theorem. Therefore, we consider the time rate ofchange following a small material volume element Vm(t) whose surface pointsmove with the local fluid velocity u(x, t). The transport theorem states

d

dt

Vm(t)

f(x, t) dV =Vm(t)

f

t dV +

Sm(t)

n uf dS (47a)

=Vm(t)

f

t + (uf)

dV (47b)

Written in the order shown above this formula holds forany function f, be ita scalar, vector or tensor quantity. Equation (47a) is the three-dimensionalversion of (9).5 A geometric proof of this result is given in the handout fromWhitaker.

5

A proof of (47) follows by viewing each of the dV elements as small material volumes,and differentiating (following material points, so d/dt D/Dt) the left-hand side of (47a)

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4.8.1 Conservation of mass (again)

As a preliminary, reconsider the time rate of change of mass by following themotion of an arbitrarily chosen material volume Vm(t). The volume contains(instantaneously) a mass equal toVm(t)(x, t) dV, which is conserved by def-inition ofVm. Applying the Reynolds Transport Theorem,

d

dt

Vm(t)

dV = 0 Vm(t)

t+ (u)

dV = 0 . (50)

Since the volume element is arbitrary, then we arrive at the continuity equa-tion:

t+ (u) = 0. (51)

The continuity equation relates two unknown quantities, (x, t) and u(x, t),and so we must proceed onwards to develop more equations.

4.8.2 Special form of the Reynolds Transport Theorem

It is common to keep track of material properties by reporting them on a permass basis. Since the volume integrals above involve f dV, then the propertyf has dimensions property per volume. If we let f = F, then F hasdimensions property per mass. If we substitute f = F in equation (47)then you should be able to show

d

dt Vm(t)F(x, t) dV = Vm(t) Vm(t)

F

t

+ (uF) dV = Vm(t) DF

Dt

dV,

(52)

using the chain rule:

d

dt

Vm(t)

f(x, t) dV =

Vm(t)

Df

Dt dV +f

D(dV)

Dt

. (48)

We need D(dV)Dt

. Since we may view the volume dVas a set of points with boundary S,

the time rate of change of volume follows D(dV)Dt

=Su n dS=

dV

udV dV u,where the last identity follows in the limit that the volume tends to zero. Thus, the specific

volume, here equivalent to dV, evolves as D(dV)Dt

= dV u, and so (48) can be simplifiedto

ddtVm(t)

f(x, t) dV = Vm(t)

DfDt

+f u dV , (49)which can be rearranged to give equation (47b).

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which is referred to as the special form of the Reynolds Transport Theorem.We shall find it useful on several occasions below since most transport relationsinvolve a property per mass.

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4.9 Linear momentum balance

We now follow a material volume element, which by definition contains fixedmass and apply other conservation, or balance, laws. Newtons second lawrelates the change of linear momentum (mass times velocity) of the material

to the forces acting on the moving domain Vm(t). In words we examine

time-rate-of-change oflinear momentum in the volume

=

sum of all the forces (e.g. due to surface+ body forces) acting on the material

.

(53)The linear momentum of this element is simply

Vm(t)udV. As for the forces

that act, in continuum mechanics we recognize two kinds, surface forces andvolumeor body forces.

4.9.1 body forces

Body forces are long range forces which act (perhaps unequally) on all thematerial points that constitute the body. These forces arise from the materialbeing placed in a force field and so include gravitational (the most familiar),as well as magnetic and electric forces.

Letf(x, t) denote the body force/mass. Then, the total body force actingon the domain is

Vm(t)f dV . Examples of the body force include

(i) gravity: f=g, where g is the gravitational acceleration.

(ii) ferrofluids: example of a magnetic body force see theScientific Amer-ican article by Rosensweig attached at the end of this section.

(iii) electrohydrodynamics: When an ionic fluid (e.g. salt water) is placedadjacent to a surface, invariably the surface adsorbs some charge andthere remain mobile ions of opposite charge (counterions) in the fluidadjacent to the surface. These counterions move in the presence of anapplied electric field tangential to the surface. Ife(x, t) is the electriccharge density/volume, andE(x, t) is the local value of the electric field,then there is a body force/volume eEon the fluid.

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This effect is responsible for the electrophoretic motion of particles inliquid (colloidal science), and is the basis for electro-osmotically drivenflows.

(iv) magnetohydrodynamics: when a material carries an electric current (j=

current density) in a magnetic field B, there is a (Lorentz) body force/volumegiven byj B.

4.9.2 Surface forces

These are short range force, which represent two types of molecular scaleinteractions:

(i) intermolecular molecular forces between nearby fluid molecules. It isknown that there is short range molecular ordering in liquids and conse-

quently a force must be applied to slide one layer of molecules relativeto another, i.e. relative macroscopic motion requires a force to be ex-erted. Because continuum mechanics cannot resolve such molecular scalephenomena, we model this as a force/area exerted across a surface, andthis effect is one of the essential contributions to friction (i.e. viscosity)in fluids.

(ii) linear momentum exchange due to random thermal motions (fluctuationsabout the mean). A good example of this is a billiard ball gas consistingof hot (fast) and cold (slow) atoms or molecules. When fast and slowparticles exchange places across a surface marked in the material, then

the linear momentum on each side is changed. By Newtons second lawthe change of linear momentum is equivalent to the action of a force fromone side of the surface on the other side.

Notation for surface forces: If we now consider a small surface Sseparatingtwo regions of fluid, we write the force/area, or stress vector, exerted by fluidon one side of the surface on the other side as

limS0

F(n)S

=t(n) stress vector . (54)

The subscript n reminds us that in general we should expect the magnitude

and direction of this force to depend on the orientation of the surface chosenfor consideration. The stress vector,t(n) (with dimensions force/area), is animportant quantity to think about in continuum mechanics and we will have

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Figure 7: Notation: (a) the force F on a surface element S with unitnormaln; (b) A volume element in the shape of a pill box with thickness .

Use a force balance, and let 0 to demonstrate t(n)= t(n).

to undertake a few steps to relate it to details of the velocity field. For nowwe just assume that there is such an internal stress field in a material

4.9.3 Back to the linear momentum statement

We return to a linear momentum balance and use the word statement of New-tons second law (53) applied to a material volume element, which leads to

d

dt

Vm(t)

u dV =Vm(t)

f dV +Sm(t)

t(n) dS . (55)

The left-hand side can be rewritten using an identity for the Reynolds Trans-port Theorem (see exercises): we use the identity

ddt

Vm(t)u dV =

Vm(t)DuDt dVVm(t)

DuDt dV mass acceleration

= Vm(t)

f dV

body force

+ Sm(t)

t(n) dS surface force

, (56)

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whereDu/Dtis the acceleration following the motion of a small fluid element.This formula, though nice conceptually, is useless until we say more about theinterpretation oft(n).

4.9.4 The stress tensor and the state of stress

We shall now introduce some general features of the state of stressin a contin-uum. The arguments illustrate that the stress vector t(n) must have a specialform: t(n)= n T, where the stress tensorT is a second-order tensor.

6 Cauchyis responsible for introducing the conceptual framework of the stress tensor.It is a basic feature of all of continuum mechanics.7

6A special case useful when viscous effects may be neglected(as we shall learn, this cor-responds to high-Reynolds-number flows away from boundaries): The mathematical stepsare illustrated nicely. Suppose that the only forces transmitted across a surface are normalforces associated with pressure. Then,t(n)= np(x, t), and the integral linear momentumequation is

Vm(t)

Du

Dt dV =

Vm(t)

f dV

Sm(t)

np dS

Vm(t)

Du

Dt f+ p

dV =0, (57)

where the second equality follows from use of the Divergence Theorem. Since the materialvolume Vm(t) is arbitrary, the kernel must vanish everywhere:

Du

Dt =f p Eulers equation. (58)

7For a historical comment on this point, it is useful to quote Truesdell: (p. 236, 238 ofThe Creation and Unfolding of the Concept of Stress, Essays in the History of Mechanics,

Berlin: Springer-Verlag, 1968, p. 184-238, by C. Truesdell)From the above account, it is clear that every conceptual element in Cauchys theory

was to be found in one or another of the special theories constructed in the previous century.Moreover, in researches of Fresnel done in 18211822, with which Cauchy must certainlyhave been familiar, many of Cauchys results are more or less implied, although in Fresnelswork the concepts of stress and strain are always connected through a presumed linearlyelastic response.

Thus it might seem that Cauchys achievement in formulating and developing the generaltheory of stress was an easy one. It was not. Cauchys concept has the simplicity of genius.Its deep and thorough originality is fully outlined only against the background of the centuryof achievement by the brilliant geometers who preceded, treating the special kinds and casesof deformable bodies by complicated and sometimes incorrect ways without ever hittingupon this basic idea, which immediately became and has remained the foundation of the

mechanics of gross bodies.Nothing is harder to surmount than a corpus of true but too special knowledge; to reforge

the tradition of his forebears is the greatest originality a man can have.

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Figure 8: The state of stress as exhibited by showing the components on the6 faces of a cube about the point of interest. Here the stress is denoted with instead ofT.

Stress equilibrium: If the volume of a small material element is O(3) and we

let 0, then we see that volume integrals O(3

) vanish more quickly thansurface integrals O(2); see equation (56). Therefore,

lim0

1

2

Sm(t)

t(n) dS 0. (59)

Apply to a pill-box shaped element (see figure): t(n)= t(n).

Cauchy stress tetrahedron: t(n) = n T.

The components of the stress tensor, Tij, have very specific physicalinterpretations: Tij denotes the force per unit area in the j -direction on

a face with normal in the i-direction.

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4.9.5 Cauchy stress equation of motion

If we substitute t(n)= n Tinto (56) then we have

Vm(t) Du

Dt

dV = Vm(t) fdV+Sm(t) nT dS Vm(t) Du

Dt

f T dV =0 ,(60)

where the second equality follows by using the Divergence Theorem on theright-hand side. Since Vm(t) is otherwise arbitrary, then we conclude

Du

Dt =f+ T Cauchy stress equation of motion , (61)

which is known as the Cauchy stress equation of motion. This equation hasan enormous range of applicability since it applies to all continua. It remainsto relate the stress tensor Tto the gradients of the velocity field.

Applicability of the Cauchy Stress Equation: It is worth noting that equation(61) applies equally well to any material that can be treated within the con-tinuum approximation. Essentially, this means that the scale of interest formathematical and physical descriptions is largecompared to the typical scaleof constituents that make up the material. Hence, although we have intro-duced the continuum ideas thinking about single-phase fluids, the ideas applyequally well to materials such as sand, fluid-saturated soil and other poroussystems, etc. Of course, where the distinct character of the material enters isin theconstitutive equationthat relates the stress (T) to the velocity or strainand their gradients.

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Figure 9: The notation used to keep track of quantities for a moving controlvolume.

4.10 Balance of Angular Momentum

The angular momentum balance applied to a given mass (the material controlvolume) is

time-rate-of-change of angularmomentum in the material volume

=

sum of all the torques,(e.g. due to surface + body forces)

.

(62)Recall that the angular momentum of a particle is defined as the moment ofthe particles linear momentum. By analogy with the moment of a force, theangular momentum is sometimes referred to as the moment of momentum.

Relative to an origin from which we measure position x of a point in thevolume, we write the integral balance for the moment of linear momentum ofa material volume as

d

dt

Vm(t)

x udV =Vm(t)

x fdV+Sm(t)

x t(n)dS+body

couples +surfacecouples ,

(63)where the last two terms, motivated by analogy with the corresponding quan-tities in the linear momentum balance, indicate possible sources of angularmomentum. The magnitudes of these two terms are not significant for ordi-

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nary liquids (we comment on these two sources later, likely on homework # 3;see also the attached reprint from Nature, Angular momentum of continuaby Dahler and Scriven). Therefore, in the absence of such exotic contributions,we can use the special form of the Reynolds Transport Theorem (equation 52)to simplify the left-hand side of (63), and substitute t(n)= n T. We obtain:

Vm(t)

D

Dt(x u) dV =

Vm(t)

x f dV +Sm(t)

x (n T) dS . (64)

Since x here indicates material points, DxDt = u, but then u u = 0. Also,applying the Divergence Theorem to the surface integral on the right-handside of (64) yields

Vm(t)

x Du

Dt dV =

Vm(t)

x f dV Vm(t)

(T x) dV . (65)

Next note the vector identity (T x) = x ( T) + T: , where is the third-order permutation tensor; also T : = Tijjikek. After minorrearrangement we have

Vm(t)

x

Du

Dt f T

dV =

Vm(t)

T: dV , (66)

The kernel on the left-hand side is zero by the Cauchy stress equation ofmotion, and since the material volume Vm(t) is chosen arbitrarily, then T : = 0. We conclude from this that Tij = Tji or that T = T

T, i.e. T isa symmetric second-order tensor and has only six independent components.

Notice that this is a very general conclusion about anycontinua, independentof the molecular constitution, and holds so long as local couples (torques) arenot present.

Therefore, the linear and angular momentum balances have produced

Du

Dt =f+ T and T= TT . (67)

There are six independent components of the stress tensor T. It remainsto relate these components to the velocity field and in particular the gradientsof the velocity field.

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1961 Nature Publishing Group


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1961 Nature Publishing Group


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Figure 10: Relative motion of two nearby fluid elements.

4.11 More Kinematics

The stress in a fluid describes forces exerted by one portion of the fluid onadjacent portions. This stress is expected to depend on the rateat which thematerial is being deformed. As a preliminary we discuss the deformation ofa fluid. The mathematical description of relative motion and displacement ofmaterial points is called kinematics.

4.11.1 Relative motion of nearby fluid elements

Begin by considering the relative motion of two nearby fluid elements. Infigure 10 we denote by r the small vector element connecting two nearbyfluid elements, P and Q. The position of fluid element P is denoted by x.Expanding the velocity at Q (positionx + r) in a Taylor series about P thenwe have8

u(x + r) =u(x) + r u(x) +1

2rr: u(x) + . . . (Taylor Series) (70)

8For a function of a single variable, the Taylor series is

f(x+x) = f(x) +xdf

dx|x+ (68)

Likewise for a function of two variables,

f(x+x, y+y) = f(x, y) +x

f

x |x,y+y

f

y |x,y+ (69)

By identifying r = (x,y), we see that the first two terms of (69) are shorthand for thefirst two terms on the right-hand side of (70).

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The relative velocity u, correct to O(r2), is then

u(r) =u(x + r) u(x) =r u(x) , (71)

i.e., the relative motion is given by the velocity gradients, which is expectedsince, if there were no changes of velocity with respect to any direction, then

the fluid is in uniform motion.

The velocity gradient tensor ucontains information describing relativemotion of nearby fluid elements. Now decompose u into symmetric andanti-symmetric parts:

u = 1

2

u + uT

symmetric

+1

2

u uT

anti-symmetric

(72a)

= E + . (72b)

E is known as the rate-of-strain tensor and as the vorticity tensor. Thisdecomposition of the velocity gradient tensor into its symmetric and anti-symmetric parts, E and , respectively, is an important and useful idea. Wenext proceed togeometricinterpretations ofE and .

The vorticity tensor, : Since is, by construction, an anti-symmetric tensor We introduce thethird-order tensor= ijkeiejek

it has only three independent components. Hence can be represented interms of a vector, say, as

=1

2 . (73)

Write down the components of this expression: Note that indexnotation = ijkkeiej

ij =12(uj

xi uixj

) =12ijkk (74)

For k = 1 we have

1=u3x2

u2x3

, (75)

which is the 1component of the vector u. The same is true for the othercomponents, so we have = u. We summarize the above manipulationsby writing the relationships between the vorticity tensor , thevorticity vector, and the velocity vectoruas

=1

2 , = : , = u. (76)

We will now show that the vorticity vector has a natural interpretation interms of the local angular velocityof the fluid.

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Figure 11: Vorticity as local rotation.

Figure 12: Two flows with circular streamlines. (a) A flow that lacks vorticity(r = 0). (b) A rigid-body rotation, which is a flow with vorticity.

Exercise: Confirm the identities in (76).

Returning to the Taylor series (71), the relative velocity contribution due

to isr =

1

2r ( ) =

1

2rikijkej =

1

2 r, (77)

which is a rigid body rotation, at Q and taken about P, with angular velocity12

: the vorticity vector equals half of the localangular velocity of the fluid,as sketched in Figure 11.

It is easy to confuse the true meaning of this interpretation of vorticity.An important point to remember is that it describes localvariations and, inparticular, local rotations. Vorticity does not simply represent (circular) mo-tion along a closed path. Rather, vorticity corresponds to changing orientationof a fluid element in space. This distinction is made clear in figure 12.

The rate-of-strain tensor, E: To begin with, from (71) we have

u= r E + r . (78)

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Taking the inner product with r we have

r Dr

Dt =r E r (79)

where r r = 0 since is anti-symmetric. Also, we have written the

relative velocity of the two points as the (Lagrangian) time-rate-of-change ofposition moving with point P. Equation (79) may be written in terms of therate-of-separationr r= r2 of two nearby fluid elements: The time-rate-of-

change ofseparationdistance is givenin terms of therate-of-straintensor E.

1

2

D(r r)

Dt =r E r, (80)

which is completely determined by E.

Remark1: SinceE is symmetric, this tensor may be diagonalized. The eigenvectorsare called the principle directions of strain and a fluid element that begins

aligned along one of the eigenvectors remains aligned for all time; hence,the interpretation ofEas representing a pure straining motion.

Remark2: Symmetric and traceless version for E Show that trE = Eii = u.So,

E=

E 1

3( u) I

+

1

3( u) I (81)

Remark3: Simple shear flow, u = Gxey, where G is the shear rate interpret asequal parts rate of strain and vorticity.

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Figure 13: Statics: The normal stresses (pressure) on a surface.

4.12 Constitutive equations: a Newtonian fluid

4.12.1 Preliminary: Fluid statics

Consider the stresses acting in a static fluid, i.e. u = 0. Assuming that thebody force is gravitational, f=g, the stress equations of motion reduce to

T +g= 0 . (82)

The standard definition of a fluid is a material that cannot support tan-gential (shear) stresses without motion occurring. However, tensile/compressiveornormalforces can be supported. It then follows that the only stresses actingin a fluid at rest must lie normalto any surface (see Figure 13) and we call

this stress the static (equilibrium) fluid pressure,pe. So,

t(n)= npe , (83)

where the minus sign is introduced since we consider pressure as compressive,so acting opposite the outward normaln. It follows that the stress tensor hasthe form

t(n) = n T= npe T= peI. (84)

Substituting (84) into (82) gives

Fluid statics: pe+g= 0, [can integrate if(pe) is known]

(85)which is the governing equation for fluid statics. This equation may be used tocalculate forces on stationary objects like dams, to determine the equilibrium

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Figure 14: Violation of the state of rest.

of submerged bodies, to calculate the shape of static fluid interfaces such aspendant bubbles, sessile drops, etc.9

Remark: The pressure in equations (83-85) is the same quantity as discussedin equilibrium thermodynamics. Hence, for a single phase fluid, pe is relatedto the density and temperature of the fluid via an equation of state, e.g.

pe= pe(, ).

Before proceeding notice that the pressure in (84) clearly satisfies The trace of asecond-ordertensor is denotedtrT.pe =

1

3

trT. (recall trT= Tii= T11+T22+T33) (86)

Violation of the state of rest: Equation (85) allows determination of a sufficientcondition for convection u = 0 to occur. Taking the curl of (85) we have( p= 0)

g= 0 (g= constant). (87)

Hence, for there to be a state of rest so that equation (85) is satisfied, werequire g. This means that any density variations must be such that thedirection of variation is collinear with g. Furthermore, everyday experiencetells us that the situation is only stable provided the density increases in thedirection ofg. A case that illustrates the violation of this condition is shown

in figure 14, where a heated wall creates a temperature gradient, and hence adensity gradient, in the direction perpendicular tog; fluid motion must occurhere.

4.12.2 The stress tensor for fluids in motion (the main event)

We consider the stresses acting in a fluid in motion. In this case, the simpleconcept of an equilibrium pressure acting equally in all direction is lost.

It is useful to recall the introduction to viscosity given in elementarytextbooks. There an experiment is described where a flat plate is dragged at a

9Atmospheric pressure: A standard application is the vertical pressure distribution inthe atmosphere, which follows from dp

dz = g, wherez is measured vertically upwards and

the pressure is linked to the density by an equation of state (p).

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yx

U

H y Ux= U y H

Figure 15: A simple shear flow.

velocity U above a stationary plate separated by a distanceH(see figure 15).The force/area exerted on the plate (y-plane) in thex-direction is denotedyx

and is assumed to vary linearly with U. The relationship is written

yx = U

H =

duxdy

, (88)

where the material coefficient is the (shear) viscosity. We now extend thisdescription to more general flow situations, though retain the basic linearitywith the velocity gradient as in (88).

Begin by recalling the simple form of the stress tensor for a static fluid.The constitutive equation we propose should reduce to the static case T =peIwhen there is no motion. So, we write

T= peI + ( is called the deviatoric stress) , (89)

where =0whenu =0. Clearly,should be symmetric sinceTis symmetric.

We are now ready to relate the nonequilibrium stress to the velocityfield and, in particular, togradientsof the velocity. First, begin with some rea-sonable assumptions that are representative of a wide class of simple (typicallylow molecular weight) fluids.

(i) We do not expect the state of stress to depend on the translation of thereference frame, therefore we do not expect to depend explicitly onu. In other words, two observers translating relative to one another ata uniform velocity must measure identical stresses. Hence, we expect to be a function of the velocity gradients. We write = (u), or

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equivalently think ofas dependent on the rate-of-strain tensor E andthe vorticity tensor .

(ii) We do not expect the stress response of a fluid to be affected by a localrigid body rotation. This implies that the stress tensor is independent10

of and depends solely on E, i.e. =(E).(iii) Assume that the relationship betweenandEislocal and instantaneous,

i.e. it depends only on the value ofE at the point of interest x at timet. Nevertheless, structured materials (e.g. polymers) typically have astress response that depends on the historyof the flow.

(iv) Linearity: Assume that, similar to simple constitutive equations you havestudied previously (e.g. Hookes law, Ohms law, Ficks law, Fourierslaw), the relationship between the second-order tensor and the second-order tensorEis linear.11

We then make a mathematical statement: the most general linear rela-tionship between a second-order tensor and a second-order tensorE isof the form

=A : E index notation

ij =AijklElk. Note: A characterizes the material.(90)

This equation is accounting for the stress to depend on all possible linearcombinations of the nine components of the rate of strain tensor. Since

10This assumption frequently raises questions! First, note that Batchelor (An Introduc-tion to Fluid Dynamics, p. 144) comments in a footnote (most appropriate since this is afootnote) that It is taken for granted, in most expositions of fluid dynamics, that a devia-toric stress cannot be generated by pure rotation, irrespective of the structure of the fluid,simply on the grounds that there is then no deformation of the fluid; however, rigorous

justification for this belief is elusive.Second, if we did assume a dependence on , so also a dependence on the vorticity vector

, then we should not expect a difference in the state of stress if the body rotated onedirection or another. Therefore, the constitutive equation would at most involve terms pro-portional to the square ofand such terms would not enter into a linear theory (assumption(iv) above).

11In heat conduction, or diffusive mass transfer, you begin with a conservation statementfor a scalar quantity (say, energy or mass) in the form

t = q, where q is the flux

vector. In a stationary fluid, you are familiar with the simple idea (Fouriers law for thermalconduction or Ficks law for mass diffusion) that q= , from which you arrive at the

heat equation,

t =

2

(with = constant). However, a more generallinear

constitutivestatement is q = , where is a second-order conductivity tensor. For isotropicmaterials,= I.

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and E are symmetric we lose no generality by assuming

Aijkl= Ajikl and Aijkl = Aijlk . (91)

(v) The final condition necessary to complete the constitutive relationship

is to assume that the material is isotropic, which means that there isno preferred direction in the material. Physically this means that twoobservers who are simply rotated relative to another will characterize thematerial deformation (i.e., the tensor A) in identical ways. Mathemati-cally, then we expect the tensor A to be an isotropicfourth-rank tensor.Although the components of vectors and tensors generally depend onthe choice of coordinate system, isotropic tensors have the property thattheir components are the same in all cartesian reference frames. Themost general fourth-order tensor is a linear combination formed from ij(the only second-order isotropic tensor), hence

Aijkl= 1ijkl+2ikjl +3iljk . (92)

It follows from (91) that 2 =3 so Aijkl =1ijkl +2[ikjl +iljk ].Then,

ij =AijklElk =1ijEll+ 22Eij (93)

or in vector notation (use the identity trE= Eii = u)

=1( u)I + 22E. (94)

We have thus arrived at an expression for the stress tensor (let 2= )

T= (pe 1 u) I + 2E. (95)

The coefficient is known as the shear viscosity. We now have a usefulconstitutive expression relating the stress tensor in the fluid to the localvelocity gradients.

Before proceeding further we note that (95) may be recast in terms ofE 1

3( u) I, which is the local rate of stretching above any local

dilatation or compression. Hence we write the constitutive relationship The state of stressfor a Newtonianfluid is linear inthe velocitygradient andinvolves twomaterialconstants, theshear viscosity and the bulkviscosity.

T= (pe u) I + 2E 1

3

( u) I where = 1+2

3

.

(96) is referred to as the bulk, volume or second coefficient of viscosity.Both (95) and (96) relate T to uand involvetwo material parameters,

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and . Both parameters can be shown to always be 0 by using anargument based on the second law of thermodynamics.

An aside: In the classical theory of elasticity there are also two param-eters, the Lame constants, that appear in the stress-strain relationshipfor an elastic material.

4.12.3 Mean normal stress in a fluid(after Batchelor 1968)

Before proceeding further it is useful to discuss the mean mechanicalpressure, which refers to the physical interpretation of the force/area ascharacterized by the intuitive notion that pressure represents the localsqueezing force. Since a fluid in motion is not a thermodynamic equi-librium system, this discussion provides a connection with what wouldbe recorded by an instrument capable of measuring normal forces. Forexample, consider the average of the normal stressn t(n)over the surface

of a small spherical volume surrounding the point x:

1

S

S

nn: T dS. (97)

If the spherical surface has radius , then we may expand Tin a Taylorseries aboutx, T(x +n) =T(x) +n T(x) +. . ., which leads to

1

S

S

nn: T dS= 1

42

S

nn dS :T(x) +O() . (98)

The integral ofnn over a spherical surface must be an isotropic second-

order tensor, hence equals cI. The constant c is found by taking thetrace so that Recall

trI= ii = 3

cI=S

nn dS trace

3c= 42 c=4

3 2 . (99)

Hence,1

S

S

nn: T dS=1

3I: T =

1

3trT, (100)

represents the average normal stress over a surface surrounding xand weuse this as a mechanical definition of the pressure, p, in a moving fluid.

Accounting for the usual notion of pressure as compressive we define Mechanical

definition ofpressure: theaverage normalstress, p =limS0

1S

Snn :

T dS

p= 1

3trT. (101)

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If you think physically about pressure in fluids in motion, you reallyare considering p, which is not, in general, equal to the thermodynamicpressure. Also, notice that this equation has the same form as equation(86), but now the stress tensor T accounts for fluid motion.

The relationship betweenpe and p: From (96) we see that

trT= 3pe+ 3 u (102)

orp pe= u . (103)

It is instructive to view (103) as a linear perturbation away from equi-librium where u is the only scalar invariant of the velocity gradienttensor that is linear in u (a property is termed invariant if it is in-dependent of the choice of cartesian coordinate system used; it turnsout that there are three invariants for a second-order tensor). Also, we

expect on physical grounds that in a pure expansion u> 0, so thatthe measured pressure would be lessthan the equilibrium pressure, andthis result is consistent with the form of (103).

Exercise: Evaluate the following integral over a spherical surface S:SnnnndS. Hint: The result should be an isotopic fourth order tensor.

4.12.4 On to the Navier-Stokes equation

We can now put all of the above results together to obtain

t

+ (u) = 0 , (104a)

Du

Dt = g+ T and T= TT (104b)

T =

pe ( 2

3 ) u

I + 2E . (104c)

If the flow is isothermal, then pe() only and the above correspondsto 10 equations for the 10 unknowns (, u, T). On the other hand, iftemperature variations occur in the fluid, then we have an additionalunknown, the temperature, and we must incorporate the energy equation

(discussion in the next section).Some standard simplifications:

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(a) If the density is taken to be constant, then u = 0, and thecontinuity and Navier-Stokes equations have the common form Most common

form of theNavier-Stokesequations. u = 0 (105a)

Du

Dt = u

t + u u = p+2u +g. (105b)It is also important to remark at this point that incompressibil-ity is an approximation so that we have simply written the pressurein (105) as p since we are now neglecting the variations of pressurewith density. The pressure in (105) is no longer the thermodynamicpressure, but is now a dependent variable to be determined by si-multaneously solving the equations in (105), i.e. these representfour equations for the four unknowns, (p, u).

(b) If we assume that the two coefficients of viscosity are constant, then

we combine (104b) and (104c) to obtain

Du

Dt = pe+

+

3

( u) +2u +g , (106)

which is one rather general form of the Navier-Stokes equations.

The combination of an equation of state for pe() with (106) andthe continuity equation (104a) corresponds to a (nonlinear) systemof 5 equations for the 5 unknowns, (pe, u).

(c) The bulk, or second, coefficient of viscosity: The physical processesgiving rise to dissipation are the same for the two coefficients of

viscosity so we should expect and to be of the same order ofmagnitude; nevertheless, most measurements (and these are diffi-cult and rare) show / = O(1 100), so there is some variabilitydepending on the fluid system, and > . Because multiplies uwe should expect processes with significant compressibility tobe those where the influence of appears. Propagation of soundwaves relies on compressibility of the fluid and so it is in the damp-ing of sound waves that most attention to is generally given.

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4.13 Energy conservation in continua

Here we show that ideas you have previously studied in courses covering equi-librium thermodynamics can be written as differential equations appropriatefor fluid flow. The energy equation is typically needed when flows involve tem-

perature variations. This section makes extensive use of results from classicalthermodynamics; if you have not seen some of these ideas before then you willeither have to accept on faith some of the physical and mathematical ideasdiscussed or you can study them in standard thermodynamics textbooks.

We begin with the conservation laws (body force/mass g):

continuity:

t + (u) = 0

D

Dt = u(107a)

momentum: Du

Dt = T +g with T = TT .(107b)

The stress tensor depends on the material in question. In general, we take

T= peI + , (108)

where for an isotropic Newtonian fluid we have (this defines the deviatoricstress tensor)

T= (pe u)I + 2

E 1

3( u) I

. (109)

Herepeis the equilibrium thermodynamic pressure and two material constants,the shear viscosity and bulk viscosity , characterize the response of the

material to rate of deformation. We will work in terms ofpe, though it is alsocommon to present the discussion below in terms of the fluid pressure p (or pabove), which differs from pe according to p = pe u.

4.13.1 Mechanical Energy

If we take the inner product ofu and equation (107b) then we have

2

Du2

Dt = u ( T) +g u (110a)

= (T u) T: E +g u , mechanical energy equation(110b)

where the symmetry ofT allows the simplication T : u= T : E. Note that Verify thatT: u= T : E

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there is a useful physical interpretation: for a fluid element, the time-rate-of-change of kinetic energy is balanced by changes in potential energy (the lastterm on the right-hand side), as well as work done by the stresses acting on thesurface of the fluid element and changes due to the conversion of mechanical tothermal energy these last two effects are embodied in theu ( T) term and

will be seen more clearly below. Because of this exchange of different forms ofenergy, equation (110) is referred to as the mechanical energy equation.

4.13.2 Thermal Energy equation

Letedenote the internal energy per unit mass. Also, since our experience withequilibrium thermodynamics reminds us that temperature gradients producea heat flux, we denote the heat flux vector as q (energy/area/time). The totalenergy of a given parcel of fluid involves the kinetic and internal energy of thematerial. The first law of thermodynamics for conservation of total energy is

time rate of change of totalenergy of a material volume =

rate at which energy is added by transportacross bounding surfaces or internal generation(111)

+rate at which work is doneby forces acting on surfaces .

If we apply the first law for total energy to a fluid element, we are led to

2

Du2

Dt kinetic energy

+ De

Dt

internal energy

= (T u) +g u q +Q (112)

where Q denotes the heat generated internally per unit mass, say because of

local chemical reactions, the adsorption of light, etc.Subtracting (110b) from (112) gives (here we again use T = peI + )

De

Dt = T: E q +Q (113a)

= pe u

reversible

work

+ :E q +Q (113b)

Equation (113) is a form of the thermal energy balance. It remains to relatethe internal energy e to the temperature and for this we turn to standard

thermodynamical considerations.

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4.13.3 Consequences of equilibrium thermodynamics

Rather than quoting results that can be found in standard treatments of equi-librium thermodynamics, we derive the necessary results here. If desired, skipthe details and go directly to equation (122), where a local variation in internal

energy is related to temperature T, pressure and density variations, and is usedin the energy equation (113). Of course, some familiarity with thermodynamicconcepts are needed to follow the manipulations presented next. It is impor-tant to recognize that we are studying flow problems that naturally representsystems out of equilibrium. Nevertheless, locallywe consider the system tobe in equilibrium, and so use relationships from equilibrium thermodynamicssimply because the length scales (and time scales) for the system to equilibrateon the molecular scale are small (short) relative to the macroscopic variations.

For a single phase material there is an equation of state f(pe, , T ) = 0.The entropy/masss is an intensive variable, which, expressed as a function of

T andpe, i.e. s(T, pe), gives the differential relation

ds=

s

T

pe

dT+

s

pe

T

dpe . (114)

A basic identity issT

pe

= cpT

, wherecpis the constant-pressure heat capacity

per unit mass. Also, a Maxwell relation gives spe

T

= 12T

pe

. So, we find

T ds= cpdT+ T

2

T

pe

dpe . (115)

The first law for a single phase fluid is de = T ds pedv, wherev is the specificvolume. Sincev = 1, then de = T ds+ pe

2d. Substituting this result into

(115) gives

de= cpdT+ T

2

T

pe

dpe+pe2

d (116)

Hence, if we follow a fluid particle (d D/Dt), then from (116), we have

De

Dt =cp

DT

Dt +

T

T

pe

DpeDt

+pe

D

Dt . (117)

Using the continuity equation gives

De

dt =cp

DT

Dt +

T

T

pe

DpeDt

pe u. (118)

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Substituting into the thermal energy equation (113) leads to

cpDT

Dt =T : E q +Q+pe u

T

T

pe

DpeDt

(119)

and if we introduce the constitutive definition T= peI + then we have

cpDT

Dt = :E q +Q

T

T

pe

DpeDt

. (120)

We can next evaluate :E; for simplicity we only consider a Newtonianfluid (see equations (108) and (109)). It is convenient to write E= 13( u) I+

E 13( u) I

since then we obtain the simple result (note that we use the

short-hand notation A2 =AijAji)

:E = ( u)2 + 2E 13( u) I2

, (121)

which is a sum of squares, so :E 0. Physically, these terms represent the(irreversible) conversion of mechanical energy to thermal energy and alwayslead to temperature increases in a fluid, as implied by the form of (120).

If we further introduce Fouriers law for the heat flux, q = kT, wherek is the thermal conductivity (here assumed constant), then (120) and (121)yield the general form of the thermal energy equation (for Newtonian fluids):

cpDT

Dt

=k2T+ ( u)2+2E 1

3

( u) I2

+Q T

TpeDpe

Dt

this term is usually neglected

since it is smaller than the others

.

(122)This equation is most often used in the simplified form where convection andconduction terms dominate:

DT

Dt =

T

t + u T =

k

cp2T . (123)

There are also some cases where viscous dissipation matters, and for incom-pressible flows it is then necessary to consider:

DT

Dt =

k

cp2T+

2

cpE2 . (124)

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For example, in shear flows E =0, so that viscous effects produce heating inthe fluid, which alters the temperature distribution according to (124).

Final remark: Using the second law one can demonstrate that the materialfunctions , and k, that appear in the above equations are necessarily 0(and equality is only possible for the ideal case of a fluid without irreveribleprocesses, i.e. no friction or viscosity and no thermal conduction).

4.14 Consequences of the second law: an equation forentropy variations

The second law of thermodynamics is concerned with entropy variations, and inparticular how the entropy of a closed system (one that exchanges neither heatnor mass with its surroundings) can only increase. In continuum mechanics weapply this idea to a fluid element. We shall see that the inequality expressed

by the second law places constraints on the material coefficients that appearin constitutive equations.

The usual thermodynamic statement of the second law is that ds dqT

wheredqis the heat added per unit mass at temperature T. If we now considerthe heat flux vector q then the equivalent statement for a material volumeVm(t) is (n q>0 when heat is added since n is the outward normal)

d

dt

Vm(t)

sdV = Sm(t)

n q

T dS+

Vm(t)

T1QdV . (125)

Using the Divergence Theorem and the usual steps associated with the factthat Vm(t) is arbitrary yields

Ds

Dt

T1q

+T1Q (126)

We now introduce some of the thermodynamic statements we introduced ear-lier. Sincede = T ds + pe

2dwe can eliminate s in place ofe and so first arrive

at

De

Dt

pe

D

Dt T

T1q

+Q , (127)

or substituting from equation (119) we find

T: E q +T T1q +pe u 0 . (128)This result can be simplified further to

:E T1q T 0. (129)

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Introducing Fouriers law for the heat flux, q = kT and the Newtonianconstitutive equation for gives

( u)2 + 2

E 1

3 uI

2+

k

T (T)2 0. (130)

For real processes, we expect the inequality in this equation to be strictly >0.Since expansion, shear and temperature gradients are in principle independentprocesses, then we see that the three materials functions , and k (notnecessarily constants) are positive.

Exercise: Beginning with equation (129) derive (130).

Exercise: EvaluateI:

E 13( u) I

.

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mass conservation; the continuity equation t

+ (u) = 0balance of linear momentum f= body force/volume Du

Dt = T +f

balance of angular momentum (no unusual torques) T= TT

mechanical energy balance 2Du2

Dt =u ( T) +f u

first law of thermodynamics 2Du2

Dt +De

Dt = (T u) +f u q +Q

thermal energy equation DeDt

=T : E q +Q

constitutive equation T= peI + ;Newtonian fluid = ( u) I + 2

(E 13( u) I

)Table 2: An incomplete summary of the basic equations

4.15 Conditions for incompressibility

We begin with the continuity equation t + (u) = 0 and examine underwhat conditions the variations in density can be neglected. In particular, with = constant, then u = 0. Let us assume that the flow is isothermal.Hence, we inquire into the variations of densitythat occur during fluid motion

as a result ofvariations of pressurethat accompany flow. The modification tothe argument below to treat temperature variations is straightforward.12

Write the continuity equation as

u1x1

+ u2x2

+ u3x3

+u

+

1

t = 0 . (131)

We expect the pressure to change during the flow and from (131) we mayneglect the density variations relative to a typical term u1

x1provided

1.

The density variations that accompany a pressure change are

pp. A convenient measure of material compressibility is provided by theisentropic speed of sound, c2 =

p

s

so

1

p

c2 1. (132)

In general, fluid motions may be simply characterized in the limits (i)inertially dominated flows, where p= O(U2) and (ii) viscously dominatedflows, where p = O(U/). Thus, the density variations should be smallprovided The ratioU/c is

known as theMach number,M =U/c.inertially dominated flows :

U

2

c2 1 (133a)

12Best to add a footnote

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viscously dominated flows :

U

U2

c2 1 (133b)

The speed of sound in air (STP) is approximately 330 m/sec and hence from(133) the variations of density are expected to be only O(104) even for flow

speeds 1 m/sec. Consequently, even flows of gases (for example, breathing)are commonly treated as incompressible, u= 0.

4.16 The structure of conservation laws(something of a review)

As we have seen in the previous pages, classical physics postulates the ideaof conservation laws, which we develop to obtain differential equations: forexample, time rate of change of linear momentum of a particle is equal tothe sum of the forces acting on it. Aslso, as the first law of thermodynamicsstates that the total energy of the universe is conserved, then we concludethat across any specified boundary, the time rate of change of the total energyinternal to the boundaries is equal to the net rate energy is transferred acrossthe boundaries. Both statements emphasize that the time rate of change isbalanced by quantities acting across and within boundaries.

To provide a mathematical structure, we denote by (x, t) a quantity(e.g. linear momentum, internal energy, concentration of chemical species i)per unit mass. can be a scalar, vector, or tensor function. It is simplestto consider a volume fixed in space to which we apply the conservation law.Then, the generic conservation law states that

d

dt V (x, t)(x, t) dV = S(t) n udS Sn q dS , (134)where q denotes the vectorial (or tensorial) molecular flux, or amount of theconserved quantity per unit time per unit area. As is standard the unit normaln is directed outward from the volume.

Applying the Divergence Theorem to the integral on the right, and in-terchanging differentiation and integration on left (permissible as the volumeis fixed in space) we have

V

()

t + (u) + q

dV = 0 . (135)

Since the domain of integration has not been specified, it is in fact completelyarbitrary, and hence it must be true that the integrand is identically zero (you

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should now be familiar with this argument). Therefore

()

t + (u) = q . (136)

We summarize a few results in the table.

1. Combining (136) with the continuity equation yields

t +u = q . (137)

2. Consider transport of a chemical speciesi in a solvent. Denote the massfraction of species i as i. Then we can choose = i and with Fickslaw in the form q= Dii we have

it +u i = (Dii) . (138)

In the most common case (a dilute solution and an incompressible flow)= constant, can we let i= i, so that

it

+ u i= (Dii) . (139)

If we divide by the molecular weight of the species, and denote the molarconcentration as ci then

cit + u ci= (Dici) . (140)

The same equation applies to the number density of molecules of speciesi.

3. Linear momentum: The linear momentum per unit mass of fluid is sim-ply the (mass average) velocity u. The momentum flux across surfacesis denoted the (negative) of the stress tensor T so we have (could incor-porate body forces)

u

t + u u= T. (141)4. Consider the case of energy.

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5. Consider the case of angular momentum.

Exercise: First let = 1. Then, subtract this result from (136) to arrive atthe conservation equation in the form

t

+u = q . (142)

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4.17 Appendix: Kinematics of material surfaces

This section is for those who are interested in a little more about kinematics.

Here we present a short discussion of the kinematics of material sur-face elements, motivated by an error discovered in a paper by H. Oguz &

A. Prosperetti, A generalization of the impulse and virial theorems with anapplication to bubble oscillation, Journal of Fluid Mechanics, 218, 143-162(1990). The error in the article was basically using one of the tensor theoremswith the dyadic product written in the incorrect order. The presentation be-low follows closely that given by Batchelor, p. 131-132. Some of this materialnecessarily repeats aspects of the kinematic discussion given in earlier sections.

In introductory courses on fluid dynamics it is common to discuss somebasic issues pertaining to the kinematics of line elements of the fluid, includ-ing a discussion of the rate-of-strain and vorticity tensors responsible for thestretching and rotation, respectively, of individual elements. Here we extend

this discussion to surface elements.

Some preliminaries: A material element of the fluid refers to a set of masspoints which move according some velocity field. Typically, we have at ourdisposal an Euleriandescription of the velocity field u(x, t) which gives thevelocity at a certain location x at timet. Different fluid elements may occupythe same positionx at different times. In a Lagrangiandescription we imaginelabeling each of the individual fluid elements and following their individualmotion. If we label such line, surface, or volume elements, we speak ofmateriallines, surfaces or volumes, respectively. It is worth keeping in mind that a fluidelement (the same set of mass points) can distort as it is advected by the flow,

though the total mass remains constant.The concept of a material derivative lies at the basis of all discussions

and applications of kinematics. The material derivative provides that rate-of-change of a quantity as an observer moves with the local fluid motion, u(x, t).For example, consider some scalar quantity (x, t) associated with each pointin the fluid. If a set of points is labeled and followed, then the rate of changeof, denotedd/dt,13 as measured by an observer moving with the local fluidvelocity u(x, t), is

D

Dt =

d

dt =

t + u (143)

(For a derivation, see any standard fluid dynamics text and the notes at this13It is common to indicate the material derivative as D/DT. I tend to use the two

notations interchangeably as d/dtmeans the total time derivative.

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beginning of this chapter). The first term accounts for variations in thatoccur at a fixed position x and the second term accounts for variations in since the position of the element changes andmay vary with spatial position.The material derivative d/dt is thus defined as

ddt

t

+ u . (144)

Kinematics of line and volume elements:

We will now discuss the kinematics of line elements, volumes, and sur-faces respectively. In each case we will use the notation to represent a smallquantity and an exact expression for behavior in the neighborhood of a pointin the fluid x is found by taking the limit 0.

First recall that a small line element may change both its orientationand length owing to a velocity gradient. Mathematically,

d

dt = u (145)

where u(x, t) is the velocity gradient measured at the current location of theline element.

l

u(x)

u(x+ l) = u(x) + l u(x)

x

Next, consider a small material volume V, consisting of a labeled set ofmass points. As this set of points is advected by the flow, the occupied volumemay change, though the total mass remains constant (we would naturally saythat the density changes). The volume change is calculated by taking a smallelement dSof the bounding surface and moving it the direction of the normalcomponent of velocity u n. So we see that the rate-of-change of volume iscalculated according to

d V

dt

= S(t) u n dS (146)whereS(t)denotes the (material) surface bounding V.

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u

n

S dS

V

By the Divergence Theorem we see

d V

dt =

V(t)

u dV . (147)

For small material volumes we may approximate the right-hand side by aTaylor series about the center x, being content with the first term, so that

d V

dt = u(x) V +o(V). (148)

Hence, the rate-of-change of volume of a material element is proportional tothe volume itself and this expression is valid in the limit that the volumeelement becomes vanishingly small. From now on t

derivation of equations 2013

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