t4 leyes de conservacion fluid mechanics

8/13/2019 T4 Leyes de Conservacion Fluid Mechanics

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C H A P T E R

4

Conservation Laws

O U T L I N E

4.1. Introduction 96

4.2. Conservation of Mass 96

4.3. Stream Functions 99

4.4. Conservation of Momentum 101

4.5. Constitutive Equation for

a Newtonian Fluid 111

4.6. Navier-Stokes Momentum

Equation 114

4.7. Noninertial Frame of Reference 116

4.8. Conservation of Energy 121

4.9. Special Forms of the Equations 125

4.10. Boundary Conditions 137

4.11. Dimensionless Forms of the

Equations and Dynamic

Similarity 143

Exercises 151

Literature Cited 168

Supplemental Reading 168

C H A P T E R O B J E C T I V E S

To present a derivation of the governing

equations for moving fluids starting from the

principles of mass, momentum, and energy

conservation for a material volume.

To illustrate the application of the integral

forms of the mass and momentum

conservation equations to stationary,

steadily moving, and accelerating control

volumes.

To develop the constitutive equation for

a Newtonian fluid and provide the Navier-

Stokes differential momentum equation.

To show how the differential momentum

equation is modified in noninertial frames of

reference.

To develop the differential energy equation

and highlight its internal coupling between

mechanical and thermal energies.

95Fluid Mechanics, Fifth Edition DOI: 10.1016/B978-0-12-382100-3.10004-6 2012 Elsevier Inc. All rights reserved.


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To present several common extensions and

simplified forms of the equations of

motion.

To derive and describe the dimensionless

numbers that appear naturally when the

equations of motion are put in dimensionless

form.

4.1. INTRODUCTION

The governing principles in fluid mechanics are the conservation laws for massmomentum, and energy. These laws are presented in this order in this chapter and can bestated in integral form, applicable to an extended region, or in differential form, applicablat a point or to a fluid particle. Both forms are equally valid and may be derived fromeach other. The integral forms of the equations of motion are stated in terms of the evolutionof a control volume and the fluxes of mass, momentum, and energy that cross its controsurface. The integral forms are typically useful when the spatial extent of potentially compli

cated flow details are small enough for them to be neglected and an average or integral flowproperty, such as a mass flux, a surface pressure force, or an overall velocity or acceleration, isought. The integral forms are commonly taught in first courses on fluid mechanics wherthey are specialized to a variety of different control volume conditions (stationary, steadilymoving, accelerating, deforming, etc.). Nevertheless, the integral forms of the equationare developed here for completeness and to unify the various control volume concepts.

The differential forms of the equations of motion are coupled nonlinear partial differentiaequations for the dependent flow-field variables of density, velocity, pressure, temperatureetc. Thus, the differential forms are often more appropriate for detailed analysis when fieldinformation is needed instead of average or integrated quantities. However, both approachecan be used for either scenario when appropriately refined for the task at hand. In the devel

opment of the differential equations of fluid motion, attention is given to determining whena solvable system of equations has been found by comparing the number of equations withthe number of unknown dependent field variables. At the outset of this monitoring effort, thfluids thermodynamic characteristics are assumed to provide as many as two equations, thethermal and caloric equations of state (1.12).

The development of the integral and differential equations of fluid motion presented inthis chapter is not unique, and alternatives are readily found in other references. The versionpresented here is primarily based on that in Thompson (1972).

4.2. CONSERVATION OF MASS

Setting aside nuclear reactions and relativistic effects, mass is neither created nodestroyed. Thus, individual mass elementsdmolecules, grains, fluid particles, etc.dmay

be tracked within a flow field because they will not disappear and new elements will nospontaneously appear. The equations representing conservation of mass in a flowing fluidare based on the principle that the mass of a specific collection of neighboring fluid particle

4. CONSERVATION LAWS96


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is constant. The volume occupied by a specific collection of fluid particles is called a materialvolume V(t). Such a volume moves and deforms within a fluid flow so that it always containsthe same mass elements; none enter the volume and none leave it. This implies that a materialvolumes surfaceA(t), a material surface, must move at the local fluid velocityuso that fluidparticles inside V(t) remain inside and fluid particles outside V(t) remain outside. Thus,a statement of conservation of mass for a material volume in a flowing fluid is:

ddt

ZVt

rx, tdV 0: (4.1)

wherer is the fluid density. Figure 3.18 depicts a material volume when the control surfacevelocity b is equal to u. The primary concept here is equivalent to an infinitely flexible,perfectly sealed thin-walled balloon containing fluid. The balloons contents play the roleof the material volume V(t) with the balloon itself defining the material surface A(t). And,

because the balloon is sealed, the total mass of fluid inside the balloon remains constant asthe balloon moves, expands, contracts, or deforms.

Based on (4.1), the principle of mass conservation clearly constrains the fluid density. Theimplications of (4.1) for the fluid velocity field may be better displayed by using Reynoldstransport theorem (3.35) withFr and bu to expand the time derivative in (4.1):Z

Vt

vrx, tvt dV

ZAt

rx, tux, t,ndA 0: (4.2)

This is a mass-balance statement between integrated density changes withinV(t) and inte-grated motion of its surfaceA(t). Although general and correct, (4.2) may be hard to utilizein practice because the motion and evolution ofV(t) and A(t) are determined by the flow,which may be unknown.

To develop the integral equation that represents mass conservation for an arbitrarilymovingcontrol volumeV*(t) with surfaceA*(t), (4.2) must be modified to involve integrations

overV*(t) andA*(t). This modification is motivated by the frequent need to conserve masswithin a volume that is not a material volume, for example a stationary control volume.The first step in this modification is to set Fr in (3.35) to obtain

d

dt

ZVt

rx, tdVZVt

vrx, tvt dV

ZAt

rx, tb,ndA 0: (4.3)

The second step is to choose the arbitrary control volume V*(t) to be instantaneously coinci-dent with material volumeV(t) so thatat the moment of interest V(t)V*(t) andA(t)A*(t). Atthis coincidence moment, the (d/dt)!rdV-terms in (4.1) and (4.3) are not equal; however, thevolume integration ofvr/vtin (4.2) is equal to that in (4.3) and the surface integral ofru,noverA(t) is equal to that overA*(t):ZVt

vrx, tvt dV

ZVt

vrx, tvt dV

ZAt

rx, tux, t,ndA Z

Atrx, tux, t,ndA

(4.4)

4.2. CONSERVATION OF MASS 97


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where the middle equality follows from (4.2). The two ends of (4.4) allow the central volumeintegral term in (4.3) to be replaced by a surface integral to find:

d

dt

ZVt

rx, tdVZ

Atrx, tux, t b,ndA 0, (4.5

where u andb must both be observed in the same frame of reference; they are not otherwise restricted. This is the general integral statement of conservation of mass for an arbitrarily moving control volume. It can be specialized to stationary, steadily movingaccelerating, or deforming control volumes by appropriate choice of b. In particularwhen b u, the arbitrary control volume becomes a material volume and (4.5) reduceto (4.1).

The differential equation that represents mass conservation is obtained by applyingGauss divergence theorem (2.30) to the surface integration in (4.2):

ZVt

vrx, tvt dV Z

At

rx, tux, t,ndA ZVt

vrx, tvt

V,rx, tux, tdV 0: (4.6The final equality can only be possible if the integrand vanishes at every point in space. If thintegrand did not vanish at every point in space, then integrating (4.6) in a small volumaround a point where the integrand is nonzero would produce a nonzero integral. Thus(4.6) requires:

vrx, tvt

V,rx, tux, t 0, or; in index notation vrvt

vvxi

rui 0: (4.7

This relationship is called thecontinuity equation. It expresses the principle of conservation omass in differential form, but is insufficient for fully determining flow fields because it is

a single equation that involves two field quantities, r and u, and u is a vector with threcomponents.

The second term in (4.7) is the divergence of the mass-density flux ru. Suchflux divergencterms frequently arise in conservation statements and can be interpreted as the net loss aa point due to divergence of a flux. For example, the local r will increase with time iV,(ru) is negative. Flux divergence terms are also calledtransportterms because they transfequantities from one region to another without making a net contribution over the entire fieldWhen integrated over the entire domain of interest, their contribution vanishes if there are nsources at the boundaries.

The continuity equation may alternatively be written using (3.5) the definition ofD/Dandvrui=vxi uivr=vxirvui=vxi [see (B3.6)]:

1

rx, tD

Dtrx, t V,ux, t 0: (4.8

The derivativeDr/Dtis the time rate of change of fluid density following a fluid particle. Iwill be zero forconstant density flow where rconstant throughout the flow field, and fo



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incompressibleflow where the density of fluid particles does not change but different fluidparticles may have different density:

Dr

Dth

vr

vt u,Vr 0: (4.9)

Taken together, (4.8) and (4.9) imply:V,u 0 (4.10)

for incompressible flows. Constant density flows are a subset of incompressible flows;rconstant is a solution of (4.9) but it is not a general solution. A fluid is usually calledincompressible if its density does not change with pressure. Liquids are almost incompress-ible. Gases are compressible, but for flow speeds less than ~100 m/s (that is, for Machnumbers< 0.3) the fractional change of absolute pressure in an air flow is small. In thisand several other situations, density changes in the flow are also small and (4.9) and(4.10) are valid.

The general form of the continuity equation (4.7) is typically required when the derivative

Dr/Dtis nonzero because of changes in the pressure, temperature, or molecular compositionof fluid particles.

4.3. STREAM FUNCTIONS

Consider the steady form of the continuity equation (4.7),

V,ru 0: (4.11)The divergence of the curl of any vector field is identically zero (see Exercise 2.19), so ruwillsatisfy (4.11) when written as the curl of a vector potential J,

ru VJ, (4.12)

which can be specified in terms of two scalar functions: J cVj. Putting thisspecification for J into (4.12) produces ru Vc Vj, because the curl of any gradientis identically zero (see Exercise 2.20). Furthermore, Vc is perpendicular to surfaces ofconstant c, and Vj is perpendicular to surfaces of constant j, so the mass fluxru Vc Vj will be parallel to surfaces of constant c and constant j. Therefore,three-dimensional streamlines are the intersections of the two stream surfaces, orstream functions in a three-dimensional flow.

The situation is illustrated in Figure 4.1. Consider two members of each of the families of

the two stream functionsca,cb,jc,jd. The intersections shown as darkened linesin Figure 4.1 are the streamlines. The mass flux _mthrough the surfaceAbounded by the fourstream functions (shown in gray in Figure 4.1) is calculated with area element dAhavingnasshown and Stokes theorem.

Defining the mass flux _mthroughA,and using Stokes theorem produces

4.3. STREAM FUNCTIONS 99


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4.4. CONSERVATION OF MOMENTUM

In this section, the momentum-conservation equivalent of (4.5) is developed fromNewtons second law, the fundamental principle governing fluid momentum. Whenapplied to a material volumeV(t) with surface areaA(t), Newtons second law can be stateddirectly as:

d

dt

ZVt

rx, tux, tdVZVt

rx, tgdVZAt

fn, x, tdA, (4.13)

whereruis the momentum per unit volume of the flowing fluid, gis the body force per unitmass acting on the fluid withinV(t),fis the surface force per unit area acting onA(t), andnisthe outward normal onA(t). The implications of (4.13) are better displayed when the timederivative is expanded using Reynolds transport theorem (3.35) with Fruand bu:Z

Vt

v

vtrx, tux, tdV

ZAt

rx, tux, tux, t,ndA

ZVt

rx, tgdVZAt

fn,x, tdA:(4.14)

This is a momentum-balance statement between integrated momentum changes withinV(t),integrated momentum contributions from the motion ofA(t), and integrated volume andsurface forces. It is the momentum conservation equivalent of (4.2).

To develop an integral equation that represents momentum conservation for an arbitrarilymoving control volumeV*(t) with surfaceA*(t), (4.14) must be modified to involve integra-tions over V*(t) and A*(t). The steps in this process are entirely analogous to those taken

between (4.2) and (4.5) for conservation of mass. First setF ru in (3.35) and rearrange itto obtain:Z

Vt

v

vtrx, tux, tdV d

dt

ZVt

rx, tux, tdVZ

Atrx, tux, tb,ndA 0: (4.15)

Then chooseV*(t) to be instantaneously coincident withV(t) so that at the moment of interest:ZVt

v

vtrx, tux, tdV

ZVt

v

vtrx, tux, tdV,

ZA

t

rx, tux, tux, t,ndAZ

A

t

rx, tux, tux, t,ndA,

ZVt

rx, tgdVZVt

rx, tgdV, andZAt

fn,x, tdAZ

Atfn,x, tdA:

(4.16a, 4.16b, 4.16c, 4.16d)

4.4. CONSERVATION OF MOMENTUM 101


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Now substitute (4.16a) into (4.15) and use this result plus (4.16b, 4.16c, 4.16d) to conver(4.14) to:

d

dt

ZVt

rx, tux, tdVZ

Atrx, tux, tux, t b,ndA

ZVt

rx, tgdV ZAt

fn, x, tdA: (4.17

This is the general integral statement of momentum conservation for an arbitrarily movingcontrol volume. Just like (4.5), it can be specialized to stationary, steadily moving, accelerating, or deforming control volumes by appropriate choice ofb. For example, whenbuthe arbitrary control volume becomes a material volume and (4.17) reduces to (4.13).

At this point, the forces in (4.13), (4.14), and (4.17) merit some additional description thafacilitates the derivation of the differential equation representing momentum conservationand allows its simplification under certain circumstances.

The body force, rgdV, acting on the fluid element dVdoes so without physical contact

Body forces commonly arise from gravitational, magnetic, electrostatic, or electromagnetiforce fields. In addition, in accelerating or rotating frames of reference, fictitious body forcearise from the frames noninertial motion (see Section 4.7). By definition body forces ardistributed through the fluid and are proportional to mass (or electric charge, electricurrent, etc.). In this book, body forces are specified per unit mass and carry the units oacceleration.

Body forces may be conservative or nonconservative.Conservative body forcesare those thacan be expressed as the gradient of a potential function:

g VF or gj vF=vxj, (4.18where F is called the force potential; it has units of energy per unit mass. When the z-axipoints vertically upward, the force potential for gravity is F

gz, wheregis the acceleration

of gravity, and (4.18) produces ge

gez. Forces satisfying (4.18) are called conservativbecause the work done by conservative forces is independent of the path, and the sum ofluid-particle kinetic and potential energies is conserved when friction is absent.

Surface forces, f, act on fluid elements through direct contact with the surface of theelement. They are proportional to the contact area and carry units of stress (force per uniarea). Surface forces are commonly resolved into components normal and tangential tothe contact area. Consider an arbitrarily oriented element of areadA in a fluid (Figure 2.5)Ifn is the surface normal with components ni, then from (2.15) the components fj of thsurface force per unit area f(n,x,t) on this element arefjnisij, where sijis the stress tensorThus, the normal component off is n,fnifi, while the tangential component is the vectof e(n,f)nwhich has componentsfke (nifi)nk.

Other forces that influence fluid motion are surface- and interfacial-tension forces that acon lines or curves embedded within interfaces between liquids and gases or between immiscible liquids (see Figure 1.4). Although these forces are commonly important in flows withsuch interfaces, they do not appear directly in the equations of motion, entering insteadthrough the boundary conditions.



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Before proceeding to the differential equation representing momentum conservation, theuse of (4.5) and (4.17) for stationary, moving, and accelerating control volumes havinga variety of sizes and shapes is illustrated through a few examples. In all four examples,equations representing mass and momentum conservation must be solved simultaneously.

EXAMPLE 4.1

A long bar with constant cross section is held perpendicular to a uniform horizontal flow of

speedUN, as shown in Figure 4.2. The flowing fluid has density r and viscosity m (both constant).

The bars cross section has characteristic transverse dimensiond, and the span of the bar is l with

l[ d. The average horizontal velocity profile measured downstream of the bar is U(y), which is less

thanUN due to the presence of the bar. Determine the required force per unit span, eFD/l, applied

to the ends of the bar to hold it in place. Assume the flow is steady and two dimensional in the plane

shown. Ignore body forces.

Solution

Before beginning, it is important to explain the sign convention for fluid dynamic drag forces.

The drag force on an inanimate object isthe force applied to the object by the fluid. Thus, for stationary

objects, drag forces are positive in the downstream direction, the direction the object would

accelerate if released. However, the control volume laws are written forforces applied to the contents of

the volume. Thus, from Newtons third law, a positive drag force on an object implies a negative force

on the fluid. Therefore, the FD appearing in Figure 4.2 is a positive number and this will be borne out

by the final results. Here we also note that since the horizontal velocity downstream of the bar, the

wake velocityU(y), is less than UN, the fluid has been decelerated inside the control volume and

this is consistent with a force from the body opposing the motion of the fluid as shown.

The basic strategy is to select a stationary control volume, and then use (4.5) and (4.17) to

determine the forceFDthat the body exerts on the fluid per unit span. The first quantitative step in

the solution is to select a rectangular control volume with flat control surfaces aligned with the

coordinate directions. The inlet, outlet, and top and bottom sides of such a control volume areshown in Figure 4.2. The vertical sides parallel to the x-y plane are not shown. However, the flow

does not vary in the third direction and is everywhere parallel to these surfaces so these merely

need be selected a comfortable distance l apart. The inlet control surface should be far enough

upstream of the bar so that the inlet fluid velocity isUNex, the pressure ispN, and both are uniform.

H

d x

yU

U(y)

U

FDinlet o

utlet

top

bottom

FIGURE 4.2 Momentum and mass balance for flow past long bar of constant cross section placedperpendicular to the flow. The intersection of the recommended stationary control volume with thex-yplane isshown with dashed lines. The force eFD holds the bar in place and slows the fluid that enters the controlvolume.



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The top and bottom control surfaces should be separated by a distanceHthat is large enough so that

these boundaries are free from shear stresses, and the horizontal velocity and pressure are so close

to UN and pN that any difference can be ignored. And finally, the outlet surface should be far

enough downstream so that streamlines are nearly horizontal there and the pressure can again be

treated as equal topN.

For steady flow and the chosen stationary volume, the control surface velocity is b0 and thetime derivative terms in (4.5) and (4.17) are both zero. In addition, the surface force integralcontributes eFDex where the beam crosses the control volumes vertical sides parallel to the x-y

plane. The remainder of the surface force integral contains only pressure terms since the shear stress

is zero on the control surface boundaries. After setting the pressure to pNon all control surfaces,

(4.5) and (4.17) simplify to:ZAt

rux,ndA 0, andZA

ruxux,ndA ZA

pNndAFDex:

In this case the pressure integral may be evaluated immediately by using Gauss divergence

theorem:

ZA

pNndA ZV

VpNdV 0,

with the final value (zero) occurring becausepN is a constant. After this simplification, denote the

fluid velocity components by (u,v) u, and evaluate the mass and x-momentum conservationequations:

Zinlet

rUNldyZtop

rvldxZ

bottom

rvldxZ

outlet

rUyldy 0, and

Zinlet

rU2Nldy

Ztop

rUNvldxZ

bottom

rUNvldxZoutlet

rU2yldy FD

where u,ndAis: eUNldyon the inlet surface,vldxon the top surface, evldxon the bottom surface,andU(y)ldy on the outlet surface where l is the span of the flow into the page. Dividing bothequations by rl, and combining like integrals produces:

Ztop

vdxZ

bottom

vdxZH=2

H=2UN Uydy, and

UN

0B@

Ztop

vdxZ

bottom

vdx

1CA

ZH=2

H=2

U2

yU2

N

dy FD=rl:

Eliminating the top and bottom control surface integrals between these two equations leads to:

FD=l rZH=2

H=2UyUN Uydy,



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which produces a positive value ofFDwhenU(y) is less thanUN. An essential feature of this anal-

ysis is that there are nonzero mass fluxes through the top and bottom control surfaces. The final

formula here is genuinely useful in experimental fluid mechanics since it allows FD/lto be deter-

mined from single-component velocity measurements made in the wake of an object.

EXAMPLE 4.2

Using a stream-tube control volume of differential length ds, derive the Bernoulli equation,

()rU2 gzp/rconstant along a streamline, for steady, inviscid, constant density flow whereUis the local flow speed.

Solution

The basic strategy is to use a stationary stream-tube-element control volume, (4.5), and (4.17) to

determine a simple differential relationship that can be integrated along a streamline. The geometry

is shown in Figure 4.3. For steady inviscid flow and a stationary control volume, the control surface

velocityb0, the surface friction forces are zero, and the time derivative terms in (4.5) and (4.17)are both zero. Thus, these two equations simplify to:Z

Atrux,ndA 0 and

ZAt

ruxux,ndAZVt

rgdVZ

AtpndA:

The geometry of the volume plays an important role here. The nearly conical curved surface is

tangent to the velocity while the inlet and outlet areas are perpendicular to it. Thus, u,ndA is: eUdA

on the inlet surface, zero on the nearly conical curved surface, and[U(vU/vs)ds]dA on theoutlet surface. Therefore, conservation of mass with constant density leads to

ds

p

A

U

x

y

z

stream tube

g

p + s( )dsA + s( )ds

U+ s( )dsextrapressure

force

p

A

U

FIGURE 4.3 Momentum and mass balance for a short segment of a stream tube in steady inviscid constantdensity flow. Here, the inlet and outlet areas are perpendicular to the flow direction, and they are small enoughso that only first-order corrections in the stream direction need to be considered. The alignment of gravity andstream tube leads to a vertical change of sin qds dzbetween its two ends. The area difference between the twoends of the stream tube leads to an extra pressure force.



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erUArUvU

vsds

AvA

vsds

0,

where first-order variations in U and A in the stream-wise direction are accounted for. Now

consider the stream-wise component of the momentum equation recalling that u Ueuand settingg egez. For inviscid flow, the only surface force is pressure, so the simplified version of (4.17)becomes

rU2ArUvU

vsds

2AvA

vsds

rgsin qAvA

vs

ds

2

dspA

pvp

vs

ds

2

vA

vsds

pvp

vsds

AvA

vsds

:

Here, the middle pressure term comes from the extra pressure force on the nearly conical surface of

the stream tube.

To reach the final equation, use the conservation of mass result to simplify the flux terms on the

left side of the stream-wise momentum equation. Then, simplify the pressure contributions by

canceling common terms, and note that sin q dsdz to find

rU2ArUUvU

vsdsA rUAvU

vsds

rgA vA

vs

ds

2

dz vp

vs

vA

vs

ds22

Avpvsdsvp

vs

vA

vsds2:

Continue by dropping the second-order terms that containds2 ordsdz, and divide byrAto reach:

UvU

vsds gdz1

r

vp

vsds, or

dU2=2 gdz 1=rdp 0along a streamline:

Integrate the final differential expression along the streamline to find:

1

2U2

gzp=r a constant along a streamline: (4.19)

EXAMPLE 4.3

Consider a small solitary wave that moves from right to left on the surface of a water channel of

undisturbed depthh(Figure 4.4). Denote the acceleration of gravity by g. Assuming a small change

in the surface elevation across the wave, derive an expression for its propagation speed, U, when

the channel bed is flat and frictionless.

Solution

Before starting the control volume part of this problem, a little dimensional analysis goes a long

way toward determining the final solution. The statement of the problem has only three parameters,



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U, g, and h, and there are two independent units (length and time). Thus, there is only one

dimensionless group,U2=gh, so it must be a constant. Therefore, the final answer must be in the

form:U const, ffiffiffiffiffighp , so the value of the following control volume analysis lies merely in deter-mining the constant.

Choose the control volume shown and assume it is moving at speed b Uex. Here we assumethat the upper and lower control surfaces coincide with the water surface and the channels fric-

tionless bed. They are shown close to these boundaries in Figure 4.4 for clarity. Apply the integral

conservation laws for mass and momentum.

d

dt

ZVt

rdVZ

Atrub,ndA 0, d

dt

ZVt

rudVZ

Atruub,ndA

ZV

t

rgdVZ

A

t

fdA:

With this choice of a moving control volume, its contents are constant so both the d/dtterms are

zero; thus,ZA

ruUex,ndA 0, andZ

AtruuUex,ndA

ZVt

rgdVZ

AtfdA:

Here, all velocities are referred to a stationary coordinate frame, so that u 0on the inlet side ofthe control volume in the undisturbed fluid layer. In addition, label the inlet (left) and outlet

(right) water depths as hin and hout, respectively, and save consideration of the simplifications

that occur whenhouthin houthin=2for the end of the analysis. Let Uoutbe the horizontalflow speed on the outlet side of the control volume and assume its profile is uniform. Therefore

uUex,ndAis eUldyon the inlet surface, and(UoutU)ldyon the outlet surface, where l is(again) the width of the flow into the page. With these replacements, the conservation of mass

equation becomes:

rUhinlrUoutUhoutl 0, or Uhin UoutUhout,

U

g

h

x

y

FIGURE 4.4 Momentum and mass balance for a small amplitude water wave moving into quiescent waterof depthh. The recommended moving control volume is shown with dashed lines. The wave is driven by theimbalance of static pressure forces on the vertical inlet (left) and outlet (right) control surfaces.



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and the horizontal momentum equation becomes:

r00UhinlrUoutUoutUhoutl Zinlet

pn,exdAZoutlet

pn,exdAZtop

pn,exdA:

Here, no friction terms are included, and the body force term does not appear because it has no hori-

zontal component. First consider the pressure integral on the top of the control volume, and let

yh(x) define the shape of the water surface:

poZ n,exdA po

Z dh=dx, 1ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1 dh=dx2

q ,1, 0l ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi1dh=dx2q dx

poZ

dhdx

dx po

Zhouthin

dh pohouthin

where the various square-root factors arise from the surface geometry; po is the (constant) atmo-

spheric pressure on the water surface. The pressure on the inlet and outlet sides of the control

volume is hydrostatic. Using the coordinate system shown, integrating (1.8), and evaluating the

constant on the water surface produces p po rg(he

y). Thus, the integrated inlet and outletpressure forces are:Z

inlet

pdAZ

outlet

pdAZtop

pon,exdA

Zhin0

porg

hiny

ldy

Zhout0

porg

houty

ldypohouthinl

Zhin0

rghinyldyZhout0

rghoutyldy rgh2in2h

2out

2

!l

where the signs of the inlet and outlet integrals have been determined by evaluating the dot prod-

ucts and we again note that the constant reference pressure podoes not contribute to the net pres-

sure force. Substituting this pressure force result into the horizontal momentum equation produces:

r

0

0U

hinlrUout

UoutU

houtl rg

2

h2inh2out

l:

Dividing by the common factors ofr and l,

UoutUoutU

hout g

2

h2inh2out

,

and eliminatingUoutvia the conservation of mass relationship,Uout

hin

hout

U=hout, leads to:

Uhinhout

hout

U

hinhouthout

Uhout g

2

h2inh2out

:

Dividing by the common factor of (hin e hout) and simplifying the left side of the equation

produces:



15/75

U2hinhout

g2hinhout, or U

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffighout

2hinhinhout

s z

ffiffiffiffiffigh

p ,

where the final approximate equality holds when the inlet and outlet heights differ by only a small

amount with both nearly equal toh.

EXAMPLE 4.4

Derive the differential equation for the vertical motion for a simple rocket having nozzle area Aethat points downward, exhaust discharge speedVe, and exhaust densityre,without considering the

internal flow within the rocket (Figure 4.5). Denote the mass of the rocket by M(t) and assume the

discharge flow is uniform.

Solution

Select a control volume (not shown) that contains the rocket and travels with it. This will be an

accelerating control volume and its velocity b b(t)ez will be the rockets vertical velocity. Inaddition, the discharge velocity is specified with respect to the rocket, so in a stationary frame of

reference, the absolute velocity of the rockets exhaust is uuzez(eVeb)ez.The conservation of mass and vertical-momentum equations are:

d

dt

ZVt

rdVZ

Atrub,ndA 0, d

dt

ZVt

ruzdVZ

Atruzub,ndA

gZVt

rdVZ

AtfzdA:

Here we recognize the first term in each equation as the time derivative of the rockets massM, andthe rockets vertical momentumMb, respectively. (The second of these identifications is altered when

the rockets internal flows are considered; see Thompson, 1972, pp. 43e47.) For ordinary rocketry, the

Ae

e, Ve

b(t)

g

z

Ve

M(t)

FIGURE 4.5 Geometry and parameters for a simple rocket having massM(t) that is moving vertically atspeedb(t). The rockets exhaust area, density, and velocity (or specific impulse) areAe,re, andVe, respectively.



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rocket exhaust exit will be the only place that mass and momentum cross the control volume

boundary and heren eez; thusub,ndA Veez,ezdA VedAover the nozzle exit. Inaddition, we will denote the integral of vertical surface stresses by FS, a force that includes the

aerodynamic drag on the rocket and the pressure thrust produced when the rocket nozzles outlet

pressure exceeds the local ambient pressure. With these replacements, the above equations become:

dM

dtreVeAe 0, d

dtMb reVebVeAe MgFS:EliminatingreVeAebetween the two equations produces:

d

dtMb Veb

dMdt

MgFS,

which reduces to:

Md2zRdt2

VedMdt

MgFS,

wherezRis the rockets vertical location and dzR/dtb. From this equation it is clear that negativedM/dt (mass loss) may produce upward acceleration of the rocket when its exhaust discharge

velocity Ve is high enough. In fact, Ve is the crucial figure of merit in rocket propulsion and iscommonly referred to as the specific impulse, the thrust produced per unit rate of mass discharged.

Returning now to the development of the equations of motion, the differential equationthat represents momentum conservation is obtained from (4.14) after collecting all four terminto the same volume integration. The first step is to convert the two surface integrals in (4.14to volume integrals using Gauss theorem (2.30):ZA

t

rx, tux, tux, t,ndAZV

t

V,rx, tux, tux, tdVZV

t

v

vxiruiuj

dV, and

ZAt

fn, x, tdAZAt

nisijdAZVt

v

vxi

sij

dV,

(4.20a, 4.20b

where the explicit listing of the independent variables has been dropped upon moving toindex notation. Substituting (4.20a, 4.20b) into (4.14) and collecting all the terms on onside of the equation into the same volume integration produces:

ZVt

v

vtruj

vvxi

ruiujrgj v

vxisij

dV 0: (4.21

Similarly to (4.6), the integral in (4.21) can only be zero for any material volume if the integrand vanishes at every point in space; thus (4.21) requires:

v

vt

ruj

vvxi

ruiuj

rgj

v

vxi

sij

: (4.22



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This equation can be put into a more standard form by expanding the leading two terms,

v

vt

ruj

vvxi

ruiuj

rvuj

vt uj

vr

vt v

vxirui

rui

vujvxi

rDujDt

, (4.23)

recognizing that the contents of the [,]-brackets are zero because of (4.7), and using the defi-nition ofD/Dtfrom (3.5). The final result is:

rDuj

Dt rgj v

vxi

sij

, (4.24)

which is sometimes calledCauchys equation of motion. It relates fluid-particle acceleration tothe net body (rgi) and surface force (vsij/vxj) on the particle. It is true in any continuum, solidor fluid, no matter how the stress tensor sij is related to the velocity field. However, (4.24) doesnot provide a complete description of fluid dynamics, even when combined with (4.7)

because the number of dependent field variables is greater than the number of equations.Taken together, (4.7), (4.24), and two thermodynamic equations provide at most 1 326 scalar equations but (4.7) and (4.24) contain r,uj, and sijfor a total of 13913unknowns. Thus, the number of unknowns must be decreased to produce a solvable system.The fluids stress-strain rate relationship(s) or constitutive equation provides much of therequisite reduction.

4.5. CONSTITUTIVE EQUATION FOR A NEWTONIAN FLUID

As previously described in Section 2.4, the stress at a point can be completely specified bythe nine components of the stress tensor s; these components are illustrated in Figures 2.4 and2.5, which show the directions ofpositive stresses on the various faces of small cubical andtetrahedral fluid elements. The first index of sijindicates the direction of the normal to thesurface on which the stress is considered, and the second index indicates the direction inwhich the stress acts. The diagonal elements s11, s22, and s33 of the stress matrix are the

normal stresses, and the off-diagonal elements are the tangential or shear stresses. Althoughfinite size elements are shown in these figures, the stresses apply on the various planes whenthe elements shrink to a point and the elements have vanishingly small mass. Denoting thecubical volume in Figure 2.4 bydVdx1dx2dx3and considering the torque produced on it bythe various stresses components, it can be shown that the stress tensor is symmetric,

sij sji, (4.25)by considering the elements rotational dynamics in the limit dV/0 (see Exercise 4.30).Therefore, the stress tensor has only six independent components. However, this symmetryis violated if there are body-force couples proportional to the mass of the fluid element, suchas those exerted by an electric field on polarized fluid molecules. Antisymmetric stresses

must be included in such circumstances.The relationship between the stress and deformation in a continuum is called a constitutive

equation, and a linear constitutive equation between stress sijand vui/vxjis examined here. Afluid that followsthe simplest possible linearconstitutiveequation is known as a Newtonian fluid.

In a fluid at rest, there are only normal components of stress on a surface, and the stressdoes not depend on the orientation of the surface; the stress is isotropic. The only second-order

4.5. CONSTITUTIVE EQUATION FOR A NEWTONIAN FLUID 111


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isotropic tensor is the Kronecker delta,dij,from (2.16). Therefore, the stress in a static fluidmust be of the form

sij pdij; (4.26wherepis thethermodynamic pressurerelated to r andTby an equation of state such as that foa perfect gas p rRT(1.22). The negative sign in (4.26) occurs because the normal components of s are regarded as positive if they indicate tension rather than compression (seeFigure 2.4).

A moving fluid develops additional stress components, sij, because of viscosity, and thesstress components appear as both diagonal and off-diagonal components within s. A simplextension of (4.26) that captures this phenomenon and reduces to (4.26) when fluid motionceases is:

sij pdijsij: (4.27This decomposition of the stress into fluid-static (p) and fluid-dynamic (sij) contributions iapproximate, becausepis only well defined for equilibrium conditions. However, moleculadensities, speeds, and collision rates are typically high enough, so that fluid particles (a

defined in Section 1.8) reach local thermodynamic equilibrium conditions in nearly all fluidflows so thatp in (4.27) is still the thermodynamic pressure.The fluid-dynamic contribution,sij, to the stress tensor is called the deviatoric stress tensor. Fo

it to be invariant under Galilean transformations, it cannot depend on the absolute fluidvelocity so it must depend on the velocity gradient tensor vui/vxj. However, by definitionstresses only develop in fluid elements that change shape. Therefore, only the symmetripart ofvui/vxj,Sijfrom (3.12), should be considered in the fluid constitutive equation becausthe antisymmetric part ofvui/vxj, Rij from (3.13), corresponds to pure rotation of fluid elementsThe most general linear relationship between sijandSijthat producessij0 whenSij0 is

sij KijmnSmn; (4.28

where Kijmnis a fourth-order tensor having 81 components that may depend on the local thermodynamic state of the fluid. Equation (4.28) allowseachof the nine components ofsijto blinearly related toallnine components ofSij. However, this level of generality is unnecessarywhen the stress tensor is symmetric, and the fluid is isotropic.

In an isotropic fluid medium, the stressestrain rate relationship is independent of theorientation of the coordinate system. This is only possible ifKijmnis an isotropic tensor. Alfourth-order isotropic tensors must be of the form:

Kijmn ldijdmnmdimdjngdindjm (4.29(see Aris, 1962, pp. 30e33), wherel,m, andg are scalars that depend on the local thermodynamic state. In addition, sij is symmetric in i and j, so (4.28) requires that Kijmn also bsymmetric ini and j, too. This requirement is consistent with (4.29) only if

g m: (4.30Therefore, only two constants,m and l, of the original 81, remain after the imposition of material-isotropy and stress-symmetry restrictions. Substitution of (4.29) into the constitutivequation (4.28) yields



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sij 2mSijlSmmdij;whereSmm V,uis the volumetric strain rate (see Section 3.6). The complete stress tensor(4.27) then becomes

sij pdij2mSijlSmmdij, (4.31)

and this is the appropriate multi-dimensional extension of (1.3).The two scalar constants m and l can be further related as follows. Settingij, summing

over the repeated index, and noting that dii3, we obtainsii 3p 2m3lSmm,

from which the pressure is found to be

p 13sii

2

3ml

V,u: (4.32)

The diagonal terms ofSijin a flow may be unequal. In such a case the stress tensor sij can haveunequal diagonal terms because of the presence of the term proportional tom in (4.31). We cantherefore take the average of the diagonal terms ofsand define amean pressure(as opposed tothermodynamic pressurep) as

ph13sii: (4.33)

Substitution into (4.32) gives

p p

2

3m l

V,u: (4.34)

For a completely incompressible fluid we can only define a mechanical or mean pressure,because there is no equation of state to determine a thermodynamic pressure. (In fact, the

absolute pressure in an incompressible fluid is indeterminate, and only itsgradientscan be deter-mined from the equations of motion.) The l-term in the constitutive equation (4.31) dropsout whenSmm V,u0, and no consideration of (4.34) is necessary. So, for incompressible

fluids, the constitutive equation (4.31) takes the simple form:

sij pdij2mSij incompressible, (4.35)wherep can only be interpreted as the mean pressure experienced by a fluid particle. Fora compressible fluid, on the other hand, a thermodynamic pressure can be defined, and itseems that p and p can be different. In fact, equation (4.34) relates this difference to therate of expansion through the proportionality constant my l2m/3, which is called thecoef-

ficient of bulk viscosity. It has an appreciable effect on sound absorption and shock-wave struc-

ture. It is generally found to be nonzero in polyatomic gases because of relaxation effectsassociated with molecular rotation. However, theStokes assumption,

l23m 0; (4.36)

4.5. CONSTITUTIVE EQUATION FOR A NEWTONIAN FLUID 113


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is found to be accurate in many situations because either the fluids myor the flows dilatation rate is small. Interesting historical aspects of the Stokes assumption can be found inTruesdell (1952).

Without using (4.36), the stress tensor (4.31) is:

sij

pdij

2mSij

1

3Smmdij

mySmmdij: (4.37

This linear relation between sand S is consistent with Newtons definition of the viscositycoefficient m in a simple parallel flow u(y), for which (4.37) gives a shear stress osm(du/dy). Consequently, a fluid obeying equation (4.37) is called a Newtonian fluidwhermandmymay only depend on the local thermodynamic state. The off-diagonal terms of (4.37are of the type

s12 mvu1vx2

vu2vx1

,

and directly relate the shear stress to shear strain rate via the viscosity m. The diagonal termof (4.37) combine pressure and viscous effects. For example, the first diagonal component o

(4.37) is

s11 p2mvu1vx1

my

2

3m

vumvxm

,

which means that the normal viscous stress on a plane normal to thex1-axis is proportional tothe extension rate in thex1direction and the average expansion rate at the point.

The linear Newtonian friction law (4.37) might only be expected to hold for small strainrates since it is essentially a first-order expansion of the stress in terms of Sij aroundsij 0. However, the linear relationship is surprisingly accurate for many common fluidsuch as air, water, gasoline, and oils. Yet, other liquids display non-Newtonian behavior amoderate rates of strain. These include solutions containing long-chain polymer molecules

concentrated soaps, melted plastics, emulsions and slurries containing suspended particlesand many liquids of biological origin. These liquids may violate Newtonian behavior inseveral ways. For example, shear stress may be a nonlinearfunction of the local strain ratewhich is the case for many liquid plastics that are shear thinning; their viscosity drops withincreasing strain rate. Alternatively, the stress on a non-Newtonian fluid particle may dependon the local strain rate and on its history. Such memory effects give the fluid some elastic properties that may allow it to mimic solid behavior over short periods of time. In facthere is a whole class ofviscoelastic substances that are neither fully fluid nor fully solidNon-Newtonian fluid mechanics is beyond the scope of this text but its fundamentals arewell covered elsewhere (see Bird et al., 1987).

4.6. NAVIER-STOKES MOMENTUM EQUATION

The momentum conservation equation for a Newtonian fluid is obtained by substituting(4.37) into Cauchys equation (4.24) to obtain:



21/75

r

vuj

vtui

vuj

vxi

vp

vxjrgj

v

vxi

"m

vuj

vxivuivxj

!my

2

3m

vumvxm

dij

#, (4.38)

where we have used (vp/vxi)dijvp/vxj, (3.5) withFuj, and (3.12). This is theNavier-Stokesmomentum equation. The viscosities, m and my, in this equation can depend on the thermody-namic state and indeed m,for most fluids, displays a rather strong dependence on tempera-ture, decreasing withTfor liquids and increasing with Tfor gases. Together, (4.7) and (4.38)provide 134 scalar equations, and they contain r,p, andujfor 1135 dependentvariables. Therefore, when combined with suitable boundary conditions, (4.7) and (4.38)provide a complete description of fluid dynamics when r is constant or when a single(known) relationship exists between p and r. In the later case, the fluid or the flow is saidto be barotropic. When the relationship between p and r also includes the temperature T,the internal (or thermal) energyeof the fluid must also be considered. These additions allowa caloric equation of state to be added to the equation listing, but introduces two more depen-dent variables,Tande. Thus, in general, a third field equation representing conservation ofenergy is needed to fully describe fluid dynamics.

When temperature differences are small within the flow,mandmycan be taken outside the

spatial derivative operating on the contents of the [,]-brackets in (4.38), which then reduces to

rDuj

Dt vp

vxj rgjm

v2uj

vx2imy

1

3m

v

vxj

vumvxm

compressible: (4.39a)

For incompressible fluids V$uvum/vxm0, so (4.39a) in vector notation reduces to:

rDu

Dt VprgmV2uincompressible: (4.39b)

Interestingly, the net viscous force per unit volume in incompressible flow, the last term onthe right in this equation, can be obtained from the divergence of the strain rate tensor orfrom the curl of the vorticity (see Exercise 4.38):

mV2u

j mv

2uj

vx2i 2mvSij

vxi m v

vxi

vuj

vxi vuivxj

! m3jik

vuk

vxi mV uj: (4.40)

This result would seem to pose a paradox since it shows that the net viscous force depends onthe vorticity even though rotation of fluid elements was explicitly excluded from entering(4.37), the precursor of (4.40). This paradox is resolved by realizing that the net viscous forceis given by either a spatialderivativeof the vorticity or a spatial derivativeof the deformationrate. The net viscous force vanishes when u is uniform in space (as in solid-body rotation), inwhich case the incompressibility condition requires that the deformation rate is zero every-where as well.

If viscous effects are negligible, which is commonly true away from the boundaries of theflow field, (4.39) further simplifies to theEuler equation

rDu

Dt Vprg: (4.41)

4.6. NAVIER-STOKES MOMENTUM EQUATION 115


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4.7. NONINERTIAL FRAME OF REFERENCE

The equations of fluid motion in a noninertial frame of reference are developed in thisection. The equations of motion given in Sections 4.4 through 4.6 are valid in an inertiaframe of reference, one that is stationary or that is moving at a constant speed with respecto a stationary frame of reference. Although a stationary frame of reference cannot be defined

precisely, a frame of reference that is stationary with respect to distant stars is adequate foour purposes. Thus, noninertial-frame effects may be found in other frames of referenceknown to undergo nonuniform translation and rotation. For example, the fluid mechanicsof rotating machinery is often best analyzed in a rotating frame of reference, and modernlife commonly places us in the noninertial frame of reference of a moving and maneuveringvehicle. Fortunately, in many laboratory situations, the relevant distances and time scales areshort enough so that a frame of reference attached to the earth (sometimes referred to as thelaboratory frame of reference) is a suitable inertial frame of reference. However, in atmospheric, oceanic, or geophysical studies where time and length scales are much larger, theearths rotation may play an important role, so an earth-fixed frame of reference must often

be treated as a noninertial frame of reference.

In a noninertial frame of reference, the continuity equation (4.7) is unchanged but themomentum equation (4.38) must be modified. Consider a frame of reference O0102030 thatranslates at velocity dX(t)/dt U(t) and rotates at angular velocity U(t) with respect toa stationary frame of reference O123(see Figure 4.6). The vectors U and U may be resolvedin either frame. The same clock is used in both frames so t t0. A fluid particle P can blocated in the rotating frame x0 x01,x02,x03or in the stationary frame x x1, x2, x3, andthese distances are simply related via vector addition:xXx0. The velocityuof the fluidparticle is obtained by time differentiation:

u dxdt

dXdt

dx0

dt U d

dtx01e01x02e02x03e03

Udx01dte01

dx02dte02

dx03dte03x01

de01dt x02

de02dtx03

de03dt Uu0U x0, (4.42

1

2

3

O

1

2

3

O

x x

u u= U + + x

X(t)

(t)

P

FIGURE 4.6 Geometry showing the relationship between a stationary coordinate system O123and a noninertiacoordinate system O0102030 that is moving, accelerating, and rotating with respect to O 123. In particular, the vectoconnecting O and O0 is X(t) and the rotational velocity of O0102030 is U(t). The vector velocityu at point P in O123ishown. The vector velocity u 0 at point P in O0102030 differs from u because of the motion of O 0102030.



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where the final equality is based on the geometric construction of the cross product shown inFigure 4.7 for e0

1

, one of the unit vectors in the rotating frame. In a small timedt, the rotation ofO0102030causes e01to trace a small portion of a cone with radius sina as shown. The magnitudeof the change ine01isje01j sinajUjdt, sodje01j=dt sinajUj, which is equal to the magni-tude ofUe01. The direction of the rate of change ofe01is perpendicular to U ande01, which isthe direction ofUe01. Thus, by geometric construction, de01=dt Ue01, and by directextension to the other unit vectors,de0i=dt Ue0i (in mixed notation).

To find the accelerationaof a fluid particle at P, take the time derivative of the final versionof (4.42) to find:

a dudt

ddt

U u0Ux0 dUdt

a02Uu0dUdt

x0U Ux0: (4.43)

(see Exercise 4.42) where dU/dt is the acceleration of O0 with respect to O, a0 is the fluidparticle acceleration viewed in the noninertial frame, 2Uu0 is the Coriolis acceleration,(dU/dt)x0 is the acceleration caused by angular acceleration of the noninertial frame, andthe final term is the centripetalacceleration.

In fluid mechanics, the acceleration a of fluid particles is denoted Du/Dt, so (4.43) isrewritten:

Du

Dt

O123

D0u0

Dt

O0102030

dUdt

2Uu0dUdt

x0U Ux0: (4.44)

This equation states that fluid particle acceleration in an inertial frame is equal to the sum of:the particles acceleration in the noninertial frame, the acceleration of the noninertial frame,the Coriolis acceleration, the particles apparent acceleration from the noninertial frames

angular acceleration, and the particles centripetal acceleration. Substituting (4.44) into(4.39), produces:

r

D0u0

Dt

O0102030

V0prgdU

dt 2Uu0dU

dtx0U U x0

mV02u0 (4.45)

| |dt

e1 (t)

e1 (t+ dt)

d e1

FIGURE 4.7 Geometry showing the relationship between U, the rotational velocity vector of O0102030 , and thefirst coordinate unit vector e01 in O

0102030. Here, the increment de01 is perpendicular to U and e01.

4.7. NONINERTIAL FRAME OF REFERENCE 117


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as the incompressible-flow momentum conservation equation in a noninertial frame of reference where the primes denote differentiation, velocity, and position in the noninertial frameThermodynamic variables and the net viscous stress are independent of the frame of reference. Equation (4.45) makes it clear that the primary effect of a noninertial frame is the addition of extra body force terms that arise from the motion of the noninertial frame. The termin [,]-brackets reduce togalone when O0102030 is an inertial frame (Uconstant and U 0)

The four new terms in (4.45) may each be significant. The first new term dU/dtaccountfor the acceleration of O0 relative to O. It provides the apparent force that pushes occupantsback into their seats or makes them tighten their grip on a handrail when a vehicle accelerates. An aircraft that is flown on a parabolic trajectory produces weightlessness in its interiowhen its accelerationdU/dtequalsg.

The second new term, the Coriolis term, depends on the fluid particles velocity, not on itposition. Thus, even at the earths rotation rate of one cycle per day, it has important consequences for the accuracy of artillery and for navigation during air and sea travel. The earthangular velocity vector U points out of the ground in the northern hemisphere. The Coriolisacceleration2U utherefore tends to deflect a particle to the right of its direction of travein the northern hemisphere and to the left in the southern hemisphere. Imagine a low-drag

projectile shot horizontally from the north pole with speedu(Figure 4.8). The Coriolis acceleration 2Uu constantly acts perpendicular to its path and therefore does not change the speed u othe projectile. The forward distance traveled in time t is ut, and the deflection is Uut2. Thangular deflection isUut2/utUt, which is the earths rotation in timet. This demonstratethat theprojectile in fact travels in a straight line if observed from outer space (aninertialframe)its apparent deflection is merely due to the rotation of the earth underneath it. Observers onearth need an imaginary force to account for this deflection. A clear physical explanation othe Coriolis force, with applications to mechanics, is given by Stommel and Moore (1989).

FIGURE 4.8 Particle trajectory deflection caused by the Coriolis acceleration when observed in a rotating framof reference. If observed from a stationary frame of reference, the particle trajectory would be straight.



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In the atmosphere, the Coriolis acceleration is responsible for wind circulation patternsaround centers of high and low pressure in the earths atmosphere. In an inertial frame,a nonzero pressure gradient accelerates fluid from regions of higher pressure to regions oflower pressure, as the first term on the right of (4.38) and (4.45) indicates. Imagine a cylin-drical polar coordinate system (Figure 3.3c), with the z-axis normal to the earths surfaceand the origin at the center of a high- or low-pressure region in the atmosphere. If it is

a high pressure zone,uRwould be outward (positive) away from the z-axis in the absenceof rotation since fluid will leave a center of high pressure. In this situation when there is rota-tion, the Coriolis acceleration2U u 2UzuRe4is in the4direction (in the Northernhemisphere), or clockwise as viewed from above. On the other hand, if the flow is inwardtoward the center of a low-pressure zone, which reverses the direction ofuR, the Coriolisacceleration is counterclockwise. In the southern hemisphere, the direction ofUzis reversedso that the circulation patterns described above are reversed. Although the effects ofa rotating frame will be commented on occasionally in this and subsequent chapters, mostof the discussions involving Coriolis forces are given in Chapter 13, which covers geophys-ical fluid dynamics.

The third new acceleration term in [,]-brackets in (4.45) is caused by changes in the rotation

rate of the frame of reference so it is of little importance for geophysical flows or for flows inmachinery that rotate at a constant rate about a fixed axis. However, it does play a role whenrotation speed or the direction of rotation vary with time.

The final new acceleration term in (4.45), the centrifugal acceleration, depends stronglyon the rotation rate and the distance of the fluid particle from the axis of rotation. If therotation rate is steady and the axis of rotation coincides with thez-axis of a cylindrical polar

FIGURE 4.9 The earths rotation causes it to budge near the equator and this leads to a mild distortion ofequipotential surfaces from perfect spherical symmetry. The total gravitational acceleration is a sum of a centrallydirected accelerationgn(the Newtonian gravitation) and a rotational correctionU

2R that points away from the axisof rotation.

4.7. NONINERTIAL FRAME OF REFERENCE 119


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coordinate system so that U (0, 0, U) andx0(R,4,z), then eU(U x0) U2ReR. Thiadditional apparent acceleration can be added to the gravitational acceleration g to definaneffective gravity ge g U2ReR (Figure 4.9). Interestingly, a body-force potential for thenew term can be found, but its impact might only be felt for relatively large atmospheric- ooceanic-scale flows (Exercise 4.43). The effective gravity is not precisely directed at thcenter of the earth and its acceleration value varies slightly over the surface of the earth

The equipotential surfaces (shown by the dashed lines in Figure 4.9) are perpendicular tothe effective gravity, and the average sea level is one of these equipotential surfacesThus, at least locally on the earths surface, we can write Fe gz, where z is measuredperpendicular to an equipotential surface, andgis the local acceleration caused by the effective gravity. Use of the locally correct acceleration and direction for the earths gravitationa

body force in the equations of fluid motion accounts for the centrifugal acceleration and thfact that the earth is really an ellipsoid with equatorial diameter 42 km larger than the polardiameter.

EXAMPLE 4.5

Find the radial, angular, and axial fluid momentum equations for viscous flow in the gaps

between plates of a von Karman viscous impeller pump (see Figure 4.10) that rotates at a constant

angular speedUz. Assume steady constant-density constant-viscosity flow, neglect the body force

for simplicity, and use cylindrical coordinates (Figure 3.3c).

Solution

First a little background: A von Karman viscous impeller pump uses rotating plates to pump

viscous fluids via a combination of viscous and centrifugal forces. Although such pumps may be

inefficient, they are wear-tolerant and may be used to pump abrasive fluids that would damage the

z

z

R

axial

inflow

radial

outflow

FIGURE 4.10 Schematic drawing of the impeller of a von Karman pump (Example 4.5).



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vanes or blades of other pumps. Plus, their pumping action is entirely steady so they are excep-

tionally quiet, a feature occasionally exploited for air-moving applications in interior spaces

occupied by human beings.

For steady, constant-density, constant-viscosity flow without a body force in a steadily rotating

frame of reference, the momentum equation is a simplified version of (4.45):

r

u0,V0

u0

V0p

r

2U

u0

U

U

x0

mV02u0:

Here we are not concerned with the axial inflow or the flow beyond the outer edges of the disks.

Now choose the z-axis of the coordinate system to be coincident with the axis of rotation. For this

choice, the flow between the disks should be axisymmetric, so we can presume that u0R, u04, u

0z,

and p only depend on R and z. To further simplify the momentum equation, drop the primes,

evaluate the cross products,

Uu UzezuReRu4e4

UzuRe4Uzu4eR, and U Ux0 U2zReR,and separate the radial, angular, and axial components to find:

ruR

vuR

vRuzvuR

vzu24

R!

vp

vRr2Uzu4U2zR m1

R

v

vRR

vuR

vR

v2uR

vz2 uR

R2!

r

uR

vu4vR

uzvu4vz

uRu4R

r2UzuR m

1

R

v

vR

Rvu4vR

v

2u4vz2

u4R2

!

r

uR

vuzvR

uzvuzvz

vp

vzm

1

R

v

vR

RvuzvR

v

2uzvz2

!:

Here we have used the results found in the Appendix B for cylindrical coordinates. In the first

two momentum equations, the terms in [,]-brackets result from rotation of the coordinate

system.

4.8. CONSERVATION OF ENERGY

In this section, the integral energy-conservation equivalent of (4.5) and (4.17) is developedfrom a mathematical statement of conservation of energy for a fluid particle in an inertialframe of reference. The subsequent steps that lead to a differential energy-conservationequivalent of (4.7) and (4.24) follow the pattern set in Sections 4.2 and 4.5. For clarity andconciseness, the explicit listing of independent variables is dropped from the equations inthis section.

When applied to a material volume V(t) with surface area A(t), conservation of internal

energy per unit mass e and the kinetic energy per unit mass ()juj2 can be stated:d

dt

ZVt

r

e1

2juj2

dV

ZVt

rg,udVZAt

f,udAZAt

q,ndA, (4.46)

4.8. CONSERVATION OF ENERGY 121


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where the terms on the right are: work done on the fluid in V(t) by body forces, workdone on the fluid in V(t) by surface forces, and heat transferred out ofV(t). Here, q ithe heat flux vector and in general includes thermal conduction and radiation. The finaterm in (4.46) has a negative sign because the energy in V(t) decreases when heat leaveV(t) and this occurs when q,n is positive. Again, the implications of (4.46) are bettedisplayed when the time derivative is expanded using Reynolds transport theorem

(3.35), ZVt

v

vt

rer

2juj2

dV

ZAt

rer

2juj2

u,ndA

ZVt

rg,udVZAt

f,udAZAt

q,ndA:

(4.47

Similar to the prior developments for mass and momentum conservation, this result can bgeneralized to an arbitrarily moving control volume V*(t) with surfaceA*(t):

d

dtZVt

re

1

2juj2dV Z

At

re

r

2juj2u b,ndA

ZVt

rg,udVZ

Atf ,udA

ZAt

q ,ndA, (4.48

whenV*(t) is instantaneously coincident withV(t). And, just like (4.5) and (4.17), (4.48) can bspecialized to stationary, steadily moving, accelerating, or deforming control volumes byappropriate choice of the control surface velocity b.

The differential equation that represents energy conservation is obtained from (4.47after collecting all four terms under the same volume integration. The first step is to

convert the three surface integrals in (4.47) to volume integrals using Gauss theorem(2.30): ZAt

rer

2juj2

u,ndA

ZVt

V,reur

2juj2u

dV

ZVt

v

vxi

r

e1

2u2j

ui

dV,

(4.49

ZAt

f,udAZAt

nisijujdAZVt

v

vxisijujdV, (4.50

and ZAt

q,ndAZAt

qinidAZVt

V,qdAZVt

v

vxiqidA, (4.51



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where in (4.49) u2j u21u22u23 because the summation index j is implicitly repeated.Substituting (4.49) through (4.51) into (4.47) and putting all the terms together into thesame volume integration produces:Z

Vt

v

vt

r

e1

2u2j

vvxi

r

e1

2u2j

ui

rgiui

v

vxisijuj

vqivxi

dV 0: (4.52)

Similar to (4.6) and (4.21), the integral in (4.52) can only be zero for any material volume if itsintegrand vanishes at every point in space; thus (4.52) requires:

v

vt

r

e1

2u2j

vvxi

r

e1

2u2j

ui

rgiui

v

vxi

sijuj

vqivxi

: (4.53)

This differential equation is a statement of conservation of energy containing terms forfluid particle internal energy, fluid particle kinetic energy, work, energy exchange, andheat transfer. It is commonly revised and simplified so that its terms are more readilyinterpreted. The second term on the right side of (4.53) represents the total rate of workdone on a fluid particle by surface stresses. By performing the differentiation, and then

using (4.27) to separate out pressure and viscous surface-stress terms, it can be decom-posed as follows:

v

vxi

sijuj

sij

vujvxi

ujvsij

vxi

pvuj

vxj sij

vujvxi

!uj

vp

vxj uj

vsij

vxi

!: (4.54)

In the final equality, the terms in the first set of (,)-parentheses are the pressure and viscous-stress work terms that lead to the deformation of fluid particles while the terms in the secondset of (,)-parentheses are the product of the local fluid velocity with the net pressure force andthe net viscous force that lead to either an increase or decrease in the fluid particles kineticenergy. (Recall from (4.24) that vsij/vxj represents the net surface force.) Substituting (4.54)

into (4.53), expanding the differentiations on the left in (4.53), and using the continuity equa-tion (4.7) to drop terms produces:

rD

Dt

e1

2u2j

rgiui

pvuj

vxjsij

vuj

vxi

!

ujvp

vxjuj

vsij

vxi

!vqivxi

(4.55)

(see Exercise 4.45). This equation contains both mechanical and thermal energy terms.A separate equation for the mechanical energy can be constructed by multiplying (4.22) byujand summing over j. After some manipulation, the result is:

rD

Dt

1

2u2j

rgjujuj

vp

vxjuj

v

vxi

sij

(4.56)

(see Exercise 4.46), where (4.27) has been used for sij. Subtracting (4.56) from (4.55), dividingby r1/y, and using (4.8) produces:

De

Dt pDy

Dt1

rsijSij

1

r

vqivxi

, (4.57)

4.8. CONSERVATION OF ENERGY 123


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where the fact that sijis symmetric has been exploited so sij(vuj/vxi) sij(SjiRji)sijSwithSijgiven by (3.12). Equation (4.57) is entirely equivalent to the first law of thermodynamics (1.10)dthe change in energy of a system equals the work put into the system minuthe heat lost by the system. The difference is that in (4.57), all the terms have units of powerper unit mass instead of energy. The first two terms on the right in (4.57) are the pressure andviscous work done on a fluid particle while the final term represents heat transfer from the

fluid particle. The pressure work and heat transfer terms may have either sign.The viscous work term in (4.57) is the kinetic energy dissipation rate per unit mass, and i

is commonly denoted by 3 1=rsijSij. It is the product of the viscous stress acting on a fluidelement and the deformation rate of a fluid element, and represents the viscous work put intofluid element deformation. This work is irreversible because deformed fluid elements do noreturn to their prior shape when a viscous stress is relieved. Thus, 3 represents the irreversiblconversion of mechanical energy to thermal energy through the action of viscosity. It ialways positive and can be written in terms of the viscosities and squares of velocity fieldderivatives (see Exercise 4.47):

3h1

rsijSij

1

r2mSijmy2

3m

vumvxm

dij Sij 2nSij1

3

vumvxm

dij2

myr

vumvxm

2

, (4.58

wheren h m=ris the kinematic viscosity, (1.4), and

sij mvuivxj

vujvxi

!my

2

3m

vumvxm

dij (4.59

for a Newtonian fluid. Here we note that only shear deformations contribute to 3 whenmyor when the flow is in incompressible. As described in Chapter 12, 3 plays an important rolin the physics and description of turbulent flow. It is proportional to m (and my) and the squarof velocity gradients, so it is more important in regions of high shear. The internal energyincrease resulting from high 3 could appear as a hot lubricant in a bearing, or as burningof the surface of a spacecraft on reentry into the atmosphere.

The final energy-equation manipulation is to expressq i in terms of the other dependenfield variables. For nearly all the circumstances considered in this text, heat transfer is caused

by thermal conduction alone, so using (4.58) and Fouriers law of heat conduction (1.2), (4.57can be rewritten:

rDe

Dt pvum

vxm2m

Sij

1

3

vumvxm

dij

2 my

vumvxm

2 vvxi

kvT

vxi

, (4.60

wherekis the fluids thermal conductivity. It is presumed to only depend on thermodynamiconditions, as is the case for m and my.

At this point the development of the differential equations of fluid motion is complete. Thfield equations (4.7), (4.38), and (4.60) are general for a Newtonian fluid that follows Fourier

law of heat conduction. These field equations and two thermodynamic equations provide13127 scalar equations. The dependent variables in these equations are r, e,pT, anduj, a total of 111137 unknowns. The number of equations is equal tothe number of unknown field variables; therefore, solutions are in principle possible for suitable boundary conditions.



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Interestingly, the evolution of the entropy s in fluid flows can be deduced from (4.57) byusing Gibbs property relation (1.18) for the internal energy de Tds e pd(1/r). Whenmade specific to time variations following a fluid particle, it becomes:

De

Dt TDs

Dt pD1=r

Dt : (4.61)

Combining (4.57), (4.58), and (4.61) produces:Ds

Dt 1

rT

vqivxi

3T 1

r

v

vxi

qiT

qirT2

vT

vxi

3T

, (4.62)

and using Fouriers law of heat conduction, this becomes:

Ds

Dt 1

r

v

vxi

k

T

vT

vxi

krT2

vT

vxi

2 3T: (4.63)

The first term on the right side is the entropy gain or loss from heat conduction. The last twoterms, which are proportional to the square of temperature and velocity gradients (see (4.58)),represent theentropy productioncaused by heat conduction and viscous generation of heat.

The second law of thermodynamics requires that the entropy production due to irreversiblephenomena should be positive, so that m, k, k> 0:Thus, explicit appeal to the second law ofthermodynamics is not required in most analyses of fluid flows because it has already beensatisfied by taking positive values for the viscosities and the thermal conductivity. In addition(4.63) requires that fluid particle entropy be preserved along particle trajectories when theflow is inviscid and non-heat-conducting, i.e., whenDs/Dt0.

4.9. SPECIAL FORMS OF THE EQUATIONS

The general equations of motion for a fluid may be put into a variety of special forms whencertain symmetries or approximations are valid. Several special forms are presented in this

section. The first applies to the integral form of the momentum equation and correspondsto the classical mechanics principle of conservation of angular momentum. The secondthrough fifth special forms arise from manipulations of the differential equations to generateBernoulli equations. The sixth special form applies when the flow has constant density andthe gravitational body force and hydrostatic pressure cancel. The final special form for theequations of motion presented here, known as the Boussinesq approximation, is for low-speed incompressible flows with constant transport coefficients and small changes in density.

Angular Momentum Principle for a Stationary Control Volume

In the mechanics of solids bodies it is shown that

dH=dt M, (4.64)whereMis the torque of all external forces on the body about any chosen axis, and dH/dtisthe rate of change of angular momentum of the body about the same axis. For the fluid ina material control volume, the angular momentum is

4.9. SPECIAL FORMS OF THE EQUATIONS 125


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H ZVt rrudV,

where r is the position vector from the chosen axis (Figure 4.11). Inserting this in (4.64produces:

d

dt

ZVt

rrudVZVt

rrgdVZAt

rfdA,

where the two terms on the right are the torque produced by body forces and surface stressesrespectively. As before, the left-hand term can be expanded via Reynolds transport theoremto find:

ddt

ZVt

rrudV ZVt

v

vtrrudV

ZAt

rruu,ndA

ZVo

v

vtrrudV

ZAo

rruu,ndA

ddt

ZVo

rrudVZAo

rruu,ndA;

whereVoand Aoare the volume and surface of a stationary control volume that is instantaneously coincident with the material volume, and the final equality holds because Vo does novary with time. Thus, the stationary volume angular momentum principle is:

d

dt

ZVo

rrudVZAo

rruu,ndAZVo

rrgdVZAo

rfdA: (4.65

FIGURE 4.11 Definitionsketch for the angular momen-tum theorem where dm rdV.Here the chosen axis points outof the page, and elementalcontributions to the angularmomentum about this axis arer

rudV:



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The angular momentum principle (4.65) is analogous to the linear momentum principle(4.17) whenb0, and is very useful in investigating rotating fluid systems such as turboma-chines, fluid couplings, dishwashing-machine spray rotors, and even lawn sprinklers.

EXAMPLE 4.6

Consider a lawn sprinkler as shown in Figure 4.12. The area of each nozzle exit is A, and the jet

velocity isU. Find the torque required to hold the rotor stationary.

Solution

Select a stationary volumeVowith areaAoas shown by the dashed lines. Pressure everywhere on

the control surface is atmospheric, and there is no net moment due to the pressure forces. The

control surface cuts through the vertical support and the torqueM exerted by the support on the

sprinkler arm is the only torque acting on Vo. Apply the angular momentum balance

ZAo

rruu,ndA ZAo

rfdA M,

where the time derivative term must be zero for a stationary rotor. Evaluating the surface flux terms

produces: ZAo

rruu,ndA arUcos aUA arUcos aUA 2arAU2 cosa:

Therefore, the torque required to hold the rotor stationary is M 2arAU2 cosa. When thesprinkler is rotating at a steady state, this torque is balanced by both air resistance and mechanical

friction.

FIGURE 4.12 Lawn sprinkler.



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Bernoulli Equations

Various conservation laws for mass, momentum, energy, and entropy were presented inthe preceding sections. Bernoulli equations are not separate laws, but are instead derivedfrom the Navier-Stokes momentum equation (4.38) and the energy equation (4.60) undevarious sets of conditions.

First consider inviscid flow (m

my

0) where gravity is the only body force so that (4.38reduces to the Euler equation (4.41):

vujvt

uivujvxi

1r

vp

vxj vvxj

F, (4.66

where F gz is the body force potential, g is the acceleration of gravity, and the z-axis ivertical. If the flow is also barotropic, thenrr(p), and

1

r

vp

vxj v

vxj

Zppo

dp0

rp0, (4.67

wheredp/ris a perfect differential, pois a reference pressure, and p0 is the integration vari

able. In this case the integral depends only on its endpoints, and not on the path of integration. Constant density, isothermal, and isentropic flows are barotropic. In addition, thadvective acceleration in (4.66) may be rewritten in terms of the velocity-vorticity crosproduct, and the gradient of the kinetic energy per unit mass:

uivuj

vxi u uj

v

vxj

1

2u2i

(4.68

(see Exercise 4.50). Substituting (4.67) and (4.68) into (4.66) produces:

vujvt

vvxj

264

1

2u2i Z

p

po

dp0

r

p0

gz

375 uuj, (4.69

where all the gradient terms have been collected together to form the Bernoulli functionBthe contents of the [,]-brackets.

Equation (4.69) can be used to deduce the evolution of the Bernoulli function in inviscidbarotropic flow. First consider steady flow (vuj/vt0) so that (4.69) reduces to

VB u u: (4.70The left-hand side is a vector normal to the surfaceBconstant whereas the right-hand side ia vector perpendicular to bothuandu (Figure 4.13). It follows that surfaces of constant Bmuscontain the streamlines and vortex lines. Thus, an inviscid, steady, barotropic flow satisfies

12u2i

Zppo

dp0rp0gz constant along streamlines and vortex lines: (4.71

This is the first of several possibleBernoulli equations. If, in addition, the flow is irrotationa(u 0), then (4.70) implies that



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1

2u2

i Zp

po

dp0

rp0gz constant everywhere: (4.72)

It may be shown that a sufficient condition for the existence of the surfaces containingstreamlines and vortex lines is that the flow be barotropic. Incidentally, these are calledLamb surfacesin honor of the distinguished English applied mathematician and hydrodynam-icist, Horace Lamb. In a general nonbarotropic flow, a path composed of streamline and vortexline segments can be drawn between any two points in a flow field. Then (4.71) is valid with theproviso that the integral be evaluated on the specific path chosen. As written, (4.71) requiresthat the flow be steady, inviscid, and have only gravity (or other conservative) body forcesacting upon it. Irrotational flow is presented in Chapter 6. We shall note only the importantpoint here that, in a nonrotating frame of reference, barotropic irrotational flows remain irro-tational if viscous effects are negligible. Consider the flow around a solid object, say an airfoil(Figure 4.14). The flow is irrotational at all points outside the thin viscous layer close to thesurface of the body. This is because a particle P on a streamline outside the viscous layer startedfrom some point S, where the flow is uniform and consequently irrotational. The Bernoulliequation (4.72) is therefore satisfied everywhere outside the viscous layer in this example.

FIGURE 4.13 A surface defined by streamlines and vortex lines. Within this surface the Bernoulli functiondefined as the contents of the [,]-brackets in (4.69) is constant in steady flow. Note that the streamlines and vortexlines can be at an arbitrary angle.

FIGURE 4.14 Flow over asolid object. Viscous shearstresses are usually confined toa thin layer near the body calleda boundary layer. Flow outsidethe boundary layer is irrota-tional, so if a fluid particle at Sis initially irrotational it willremain irrotational at P becausethe streamline it travels on doesnot enter the boundary layer.



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An unsteady form of Bernoullis equation can be derived only if the flow is irrotational. Inthis case, the velocity vector can be written as the gradient of a scalar potential f (called thvelocity potential):

uhVf: (4.73

Putting (4.73) into (4.69) with u 0 produces:

V

264vfvt

12jVfj2

Zppo

dp0

rp0gz375 0, or vf

vt1

2jVfj2

Zppo

dp0

rp0gz Bt, (4.74

where the integration function B(t) is independent of location. Here f can be redefined toincludeB,

f/fZtto

Bt0dt0,

without changing its use in (4.73); then the second part of (4.74) provides a second Bernoull

equation for unsteady, inviscid, irrotational, barotropic flow:

vf

vt1

2jVfj2

Zppo

dp0

rp0gz constant: (4.75

This form of the Bernoulli equation will be used in studying irrotational wave motions inChapter 7.

A third Bernoulli equation can be obtained for steady flow (v/vt 0) from the energyequation (4.55) in the absence of viscous stresses and heat transfer (sijqi0):

ruiv

vxie1

2u2j ruigi

v

vxjrujp=r: (4.76When the body force is conservative with potential gz, and the steady continuity equationv(rui)/vxi0, is used to simplify (4.76), it becomes:

ruiv

vxi

ep

r1

2u2j gz

0: (4.77

From (1.13)hep/r, so (4.77) states that gradients of the sum h juj2/2gz must bnormal to the local streamline directionui. Therefore, a third Bernoulli equation is:

h12juj2 gz constant on streamlines: (4.78

Equation (4.63) requires that inviscid, non-heat-conducting flows are isentropic (s does no

change along particle paths), and (1.18) implies dp/r dh when s constant. Thus thpath integral !dp/r becomes a function h of the endpoints only if both heat conductionand viscous stresses may be neglected in the momentum Bernoulli equations (4.71), (4.72)and (4.75). Equation (4.78) is very useful for high-speed gas flows where there is significaninterplay between kinetic and thermal energies along a streamline. It is nearly the same as



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(4.71), but does not include the other barotropic and vortex-line-evaluation possibilitiesallowed by (4.71).

Interestingly, there is also a Bernoulli equation for constant-viscosity constant-density irro-tational flow. It can be obtained by starting from (4.39), using (4.68) for the advective accel-eration, and noting from (4.40) that V2u Vu in incompressible flow:

r

Du

Dt rvu

vtrV12juj2ruu VprgmV2u VprgmV u: (4.79)When r constant, g Vgz, and u 0, the second and final parts of this extendedequality require:

rvu

vt V

1

2rjuj2 rgzp

0: (4.80)

Now, form the dot product of this equation with the arc-length element eudsds directedalong a chosen streamline, integrate from location 1 to location 2 along this streamline,and recognize that eu,V v=vsto find:

rZ21

vuvt,ds Z

2

1

eu,V

12rjuj2 rgz pds rZ

2

1

vuvt,ds Z

2

1

vvs

12rjuj2 rgzpds 0:

(4.81)

The integration in the second term is elementary, so a fourth Bernoulli equation for constant-viscosity constant-density irrotational flow is:

Z21

vu

vt,ds

1

2juj2 gzp

r

2

1

2juj2 gzp

r

1

, (4.82)

where 1 and 2 deno

t4 leyes de conservacion fluid mechanics

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