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Volume

Dynamics

1

values

of the

universal

constant R

=

8.3143

R = 8314.3 R

=

1545.3

R = 1.9859 R

=

1.9859

R = 0.082056

gas R

J/mol-K

J/kmol-K ft-lbf/lbmol-R

Btu/lbmol-R kilocal/kmol-K

m3-atm/kmol-K

\357\277\274

JKfe

Dynamics

.

of

board

A. H-S.

Civil

Ang

Engineering\342\200\224Systems

Donald S. Berry University

Gere

StanfordUniversity

/. Stuart

Hunter

Princeton

and Applied Mechanics

Civil Engineering

Engineering

Statistics

University

T. William Lambe

Civil

Engineering\342\200\224Soil

V. Whitman

Institute of

Massachusetts

Technology Perry L. McCarty Standford

Dale

Environmental

Engineering

University

Industrial Engineering

Don Phillips

Texas

A

&

M

University

Chemical

Rudd

University of Wisconsin

RobertSteidel University

of

Mechanical

Engineering Engineering

California\342\200\224

Berkeley

R.

N. Cornell

Probability

TransportationEngineering

Northwestern

R.

and

of Illinois

University

James

engineering

advisors,

Civil

White University

Engineering\342\200\224Structures

Mechanics

I

Volume

Dynamicstr

Zucrow Zucrow

Maurice). Maurice).

Joe D.Hoffman of

School

New

Mechanical

York

Engineering

JOHN WILEY Santa Barbara

&

SONS,

London

Purdve

University

INC.

Sydney Toronto

rights

No

part of the

book

this

may

written

Sons, Inc.

simultaneously

be reproduced

nor translated

transmitted, without

Published

reserved.

All

Wiley &

by John

1976, \302\251

Copyright

permission

any

by

a machine

into

in

of the publisher.

I.

QC168.Z81976 0-471-98440-X the

Hoffman,

II. Title.

joint author.

in

references and

bibliographical

1. Gas dynamics.

Printed

Data

:

dynamics.

Includes

ISBN

means,

language

in Publication Library of CongressCataloging Zucrow, MauriceJoseph,1899-1975. Gas

Canada.

United

10 987654321

533'.2

76-6855

(v. 1) States

of America

index.

Joe D.,

1934-

nor

preface science of gas

to the

introduction

an

is

book

This

or graduate level. It

at the senior and scientist. course

is also

dynamics,designed

a

for

the

for

written

first

engineer

practicing

The presentation of the subjectmatteris sufficiently so that the student clear can understand the material withoutthe assistance ofan instructor. a review Therefore, of fundamental is the derivations of a re in included, principles equations presented and illustrative detail, many examples are worked out completely.Exerciseproblems arepresented for at the end of each chapter. Tables of the functionsrequired solving the illustrative and the exercise problems are includedin the appendixes. examples A in fluid mechanics and thermodynamics is desirable forthe studyof background courses in those disciplines are sufficient. gasdynamics. Introductory undergraduate The materialin this bookis too lengthy to cover in a one-semester completely we discuss in detail the basic concepts presented in course; at Purdue University, of each chapter,andthe application those to the flow of a perfect gas. As concepts timepermits,realgas effects are discussed, and applications such as nozzles, diffusers,

and the like,are

presented.

self-contained, beginning with a review 1), followed by a detailed derivationof thegoverning

is completely

book

The

(Chapter

principles

fundamental

of

equations

of 2), and an extensivetreatmentof theclassical steady one-dimensional (Chapter 3 to 9). The generalfeaturesof the gas dynamics flow of an inviscid fluid are discussed(Chapter10). multidimensional adiabatic of linearized flow (Chapter 11),the method introduction to the concept and two-dimensional flow (Chapter 12), and unsteady steady supersonic are An one-dimensional (Chapter 13) presented. appendix gives a brief to the concepts of numerical analysis that are employedin the book. tables are included to aid in working problems.

of fluid

flow (Chapter

topics

steady

An

of

characteristics

flow

introduction

Extensive

The materialin this book closely

Zucrow's

for

and

Flow

of

the

that

believe

We

examples,

out

worked

the

application

of

the

to Propulsion Engines, and Flow. The major Fluid

Compressible is the

emphasis

on the

solving real gas dynamic problems. learns by doing. Consequently, numerous illustrative student in complete detail, are presented in each chapter

methods

numerical

Aircraft

books

existing

application

and

Application of

Thermodynamics

several books already Missile Propulsion,

the

in

that

resembles

written about gas dynamics, example, Volume 1, Thermodynamicsof Fluid Shapiro's Dynamics and difference between this bookand the for

to illustrate

theoretical

analysis

to real problems.

Becauseof the inevitable conversion the international SI system), the illustrativeexamplesand exercise to

system employ

problems

(i.e., the

of units that

system

of

units.

The tables that

are includedtoaid

in

conventional gas dynamic

flow

working

the

functions,

atmosphere, and the presented for the generationof each tables if he desires. thermodynamic

of

properties the

tables,

properties

physical of

so the

tabulations

include

problems

air.

Computer

user can

of the

of the standard are programs

construct additional

vi

PREFACE

The

all

of

are introduced briefly. The materialincludes methods in the book: approximation and employed of linear algebraic equations, the solutionof nonlinear systems and the solution of ordinary differential equations. Numerical

of numerical

concepts numerical

the

of

solution

interpolation, the

integration,

functions,

examplesare presentedto 1 of

Volume

Missile

and

Aircraft Applicationto

Propulsion

Purdue

at

book

this

analysis

is based on a

Propulsion, and Engines over

University

Volume I, class notes

the past

and several of the illustrative

The commentsof the and the classnotes

at

most

University

in

Purdue

helpful

deserve

combinationof MauriceZucrow'sbook Thermodynamics of FluidFlowand course employed in a gas dynamics

special

taught

12 years

examples

students

were

at Purdue

methods.

these

illustrate

this

Hoffman.

Joe

by

who

University

in the

Many

figures

from Zucrow's book. have used Zucrow's book

taken

are

book

of the

of this book. Twocolleagues generous assistance.

preparation

for their

acknowledgments

Professor H. DoyleThompsoncontributed and criticisms ideas, during the writing of the book. ProfessorPeter invaluable assistance in provided is obtaining thermodynamic property data. Their generousassistance The excellent work performed our Miss by typist, CynthiaHoffman,deserves comments,

Liley

greatly

appreciated.

special

note.

the joint effort of a teacherand oneof his students. a labor of love for hisprofession, after Zucrow, performed from Purdue University. For Joe Hoffman,it was an opportunity book

This

was

it was

teacher, counselor, and friend,Doc achieve. Doc only a few people

Zucrow,

ever

of the manuscript. He

book,on

he

which

contribution

to

his

worked

be

will

missed

diligently

illustrious

Zucrow

in

a

warm

passed

personal

away

during

Professor

For his

retirement

with his that relationship to work

the final review

students, and by his colleagues, even on the last day of his life,

friends. This

isa

fitting

final

career.

Joe D. Hoffman West

Lafayette,

Indiana,

1976

contents

1

1 Review of FundamentalPrinciples

2

for

Equations

Governing

3 General

Compressible

Features of the Steady

Fluid

Flow

70 102

Fluid

Compressible

4 Steady One-Dimensional IsentropicFlow with Area 5 Steady One-Dimensional Flow with Friction 6 Steady One-Dimensional Flow with Heat Transfer 7

160

Change

243

299 327

ShockWave

422

Waves

8 Expansion

9 Mass Addition, Combustion

and

Waves,

of

to

Introduction

Flow

12 Introduction to

13 The

One-

Steady

461

10 General Features the Steady an Inviscid Compressible Fluid

Steady

Generalized

Flow

Dimensional

11

a

of

Flow

One-Dimensional

the

with

Two-Dimensional

Homentropic

Flow of

511 Characteristics

of

555 with

Application

Irrotational Supersonic Flow

Characteristics

of

Method

Adiabatic

Perturbations

Small

Method

Multidimensional

Applied

to Unsteady

to

580

One-Dimensional

622

Flow

Appendix

A

Numerical

Appendix

B

The Method of Characteristics Appendix

669

Analysis

in

Two

Independent

Variables

695

C

Tables

698

Index

761

VII

Volume

Dynamics

1

1

of

review

fundamental

principles 1-1 1-2

2

1

CHAPTER

FOR

NOTATION

PRINCIPAL

4

INTRODUCTION

1-3 DIMENSIONS (a)

4

UNITS

AND

4

Dimensions

5

of units

(b) Systems

(c) Principleof dimensionalhomogeneity 1-4

THE

(f)

1-5

The

continuum

The

Knudsen

8

POSTULATE

CONTINUUM

(a) Classification of the at a point in (b) Density (c) Velocity at a point in (d) Stress at a point

(e)

8

forces actingon a a

body

of

fluid

9

continuum

a

body

postulate

10

of fluid

11

applied

to a

12

gas

13

number

15

PERFECT SUBSTANCES

16

(a) The perfectsolid (b)

The

fluid

perfect

16

17

(c) The perfect liquid

17

(d) The perfectgas

1-6

1-7

18

FLUIDS

REAL

comments

on perfect and

(a)

General

(b)

Viscosity

(c)

Viscosity

data for gases

THE

SIMPLE

THERMODYNAMIC

Change

real fluids

18 20

(a) Properties (b)

8

of

22 23

SYSTEM

of the simple thermodynamicsystem state for a simple thermodynamic system

23 26

1-8

REVERSIBLE

1-9

WORK

28

1-10

HEAT

30

1-11

THE

FIRST

AND

IRREVERSIBLE

LAW OF

(b)

energy

30

THERMODYNAMICS

(a) Mathematical statement of the first Stored

27

PROCESSES

law

of

thermodynamics

30

30

2

REVIEW

PRINCIPLES

OF FUNDAMENTAL

(c) Internal (d)

(e)

Enthalpy

32

1-12

THE

1-13

NEWTON'S (b)

(c)

32 OF THERMODYNAMICS

LAW

SECOND

(a)

1-14

31

heats

Specific

(f)

31

energy

work

Flow

LAWS OF

Mathematical Momentum

of a

Momentum

of

of

kinetic

the

GAS

37

theory

37 38

intensity

39

40 40

(g) Thermal

(h)

41

conductivity

41

Diffusion

43

44

change

45

(e) Entropy change Process

(f)

46

equations of thermally

Mixtures

gases

perfect

AND

GASES

IMPERFECT

46 48

diagram

(g) Entropy-enthalpy (h)

53

GAS TABLES

(a) Specificheat data Gas

(b)

54

57

tables

1-17 THE ACOUSTICSPEEDAND

THE

MACH

NUMBER

(a) The a.cousticspeed The

(b)

OF

PROPERTIES

a cp

specific =

~cp

Ct

e

CHAPTER

1

acceleration.

area.

~cv

FOR

65

acoustic speed.

A

cv

o4

ATMOSPHERE

THE

heat at constant

fncp, molar

specific =

heat

pressure.

specific heat at constant at constant volume.

pressure.

specific heat at constant volume. mass fraction of speciesi. rrii/m,

mcv, molar

=

specific

stored

energy.

^ CA

Mach number

1-1 PRINCIPALNOTATION a

41

43

state

heat relationships

Enthalpy

(d)

of

GAS

41

equation of state

(b) Caloricequation (c) Specific

OF THE PERFECT

PROPERTIES

THERMODYNAMIC (a) Thermal

1-18

39

constant

Boltzmann's

Viscosity

(f)

36

36 37

THE PERFECT

AND

(d) Avogadro's number and (e) Specific heats

1-16

motion

Temperature

(c)

1-15

laws

of particles

(a) Basic considerationsof (b) Pressure

of Newton's

particle

a system

33 35

MOTION

expressions

THEORY

KINETIC

ENTROPY

AND

PRINCIPAL

1-1

E

force.

F

the

g

local acceleration due

gc

factor

g0

standard

dimension

force.

to

gravity.

due

acceleration

Newton's second law. (at sea level,45\302\260latitude,

defined by to gravity

of proportionality

9.80665 m/s2 (32.1740 ft/sec2). h

specific

k K

Boltzmann's = \342\200\224 v(dp/dv),

Kn

= LI I,

enthalpy.

1.38054

constant,

\342\200\242

10\"23

J/K.

modulus.

bulk

Knudsen number.

L

characteristiclength,or

m

mass.

m

molecular

M M

= Mach momentum.

n

unit

N

moles.

NA

Avogadro's

the

dimension

length,

weight. or the

number,

V/a,

dimension

mass.

vector.

normal

6.02252

number,

\342\200\242

1023

molecules/mol.

of species i.

JV;

moles

p

static pressure.

Q

heat.

R

=

R

universal gas constant,8314.3 J/kmol-K. =

for

constant

gas

Rim,

Reynolds

LVp/n,

s

specific entropy.

S

entropy.

t

a specific

or

time.

vector.

unit

T

the dimension time,

u

specific

U

internal

tangential internal

energy.

energy. volume.

v

specific

V

volume.

V

velocity

magnitude.

V

velocity.

W

work.

=

mole

fraction

specific

heat

Nt/N,

of species i.

GreekLetters y

=

cp/cv,

ratio.

0

the dimension temperature.

k

thermal

X

mean

H

absolute

conductivity. free

path or

v

= fi/p,

p

density.

t

shear

a

gas.

number.

absolutetemperature,

t

Xt

3

stored energy.

F

Re

NOTATION FOR CHAPTER 1

dynamic

molecules. viscosity,

kinematic viscosity.

stress.

stress.

C

of gas

dt

in

a vacuum)

=

4

REVIEW OF

PRINCIPLES

FUNDAMENTAL

Subscripts

denotes

i

i.

species

Other

denotes molar basis.

1-2

INTRODUCTION

Gas

the causesand from the arising of compressible fluids, particularly gases,and isa branch the more general of fluid dynamics. Gas dynamics brings togetherconcepts principles several branches of science, including mechanics, thermodynamics, and chemical and magnetic effects generally kinetics. Nuclear, electrical,

motion

science concerned with

is the

dynamics

effects

of

science

and

from aerodynamics,

are not considered part ofthe

of

science

This

chapter

for

basis

theoretical

gas

dynamics.

the fundamental concepts and principlesunderlying the analytical methods of gas dynamics.As in the case

reviews

is problem in fluid dynamics, the analysisofa gas dynamic problem interrelations between the followingfour fundamental physicallaws.

1. The

2. Newton's second law

3. The

of

law

first

of mass.

conservation

the

of

law

of motion.

thermodynamics.

second law of thermodynamics.

4. The

These particular

the properties of the flowing fluid or the under The law of the conservation of mass consideration. process of The the other three laws, as concepts underlying are reviewed mass, briefly in this chapter. laws to a flowing fluid, the properties of above fundamental

are

laws flow

fixed

of

system

of

independent

either

is

appliedto a

self-explanatory.

In applyingthe of

brief reviews Consequently, of perfect substances are presentedin thermodynamic properties of its major because to gas dynamics, the Furthermore, importance in some detail. discussed are perfect gas and

elastic

the

the of the continuumconceptand

considered.

be

must

fluid

based

the

of any on the

this chapter. propertiesofthe 1-3 DIMENSIONSAND

names

the

are

Dimensions

UNITS14

physicalquantities of

for characterizing physical quantities.The are force F, mass M, length L, time T, dynamics

employed in

interest

gas

0. Units are the namesgiven

and temperature

to

certain

magnitudes

dimension length may be but, once arbitrary of a

chosen as a standard of measurement. For the dimension example, of measurement are measured in units of meters.Thebasicunits The is manner. present section chosen, they must be employedin a consistent with characteristics of dimensions and units. concerned several of the inherent

1-3(a) Dimensions Two

of

kinds

dimensional

enter

quantities

quantities

and

quite

dimensionless

interms one or magnitudeexpressed a length a velocity meters, many of

of

acceleration of

quantity

so

so

meters many per has no dimensional

into engineering measurements: A dimensional quantity has its quantities. more basic units of measurement; for example, of so many meters per second,or an

generally

second per second, etc. On the otherhand,a dimensionless category whatsoever and is, therefore,a purenumber.

1-3

coefficient,suchas ratio of two similardimensional quantities,

A dimensionless

may be a

quantity

orifice plate, the several dimensional quantities an

the

discharge

or

arranged to

give

result.

dimensionless

a

5

UNITS

AND

DIMENSIONS

for

coefficient the

The

of

product

numerical

of a dimensionlessquantity is independent of the size of the fundamental set of units is employed. units employed for evaluating it, providedthata consistent The choice of either the principaldimensions or the size of the basic units for measurement is arbitrary. It is basedentirely the of a physical expressing magnitude In general, the principal dimensions may be any mutually on convenience. independent set that is convenient to use. Experience has demonstrated, however, that in the field of fluid four dimensions suffice. They are length L, mechanics, principal M or force F, and temperature0. The magnitude ofany physical time mass T, either in the field of fluid mechanics is expressible measurement in termsof units having the foregoing principal dimensions. Mass and forceare relatedby Newton's second law of motion; that is, force oc x law of motion is mass acceleration. In dimensional form, Newton's second then the dimensions F oc ML/T2. If mass M is taken as the primary of dimension, as the primary dimension, then the dimensions force F are ML/T2.IfforceF is chosen in of dimensions, of massM are FT2/L.Thus, either mass M or force F any system the other be chosen as the or quantity is then a secondary may primary dimension; magnitude

dimension.

derived

1-3(b)

of Units

Systems

Like the selectionof the primary

basic units of

of the

selection

the

dimensions,

brevity, is arbitrary, beingbasedsolely is the large number of this of different units employedin on convenience. Evidence and physicsfor the four dimensions M, L, T, and F. Thereis, however, engineering second law of motion, termedthe condition a restriction for imposed by Newton's of that mustbe a basic units. satisfied consistency, by system Consistency requires the 1 unit force = that the basicunitsmustsatisfy numerically relationship: following

1

x

mass

unit

Prior

1 unit

basic

the

termed

hereafter

measurement,

units for

acceleration.

systems of units

to 1971, several

were

the

in

used

widely

States.

United

(EA) EnglishEngineering(EE)system,(2) the In 1971 Metric (3) the English Gravitational (EG) system,and system. system, thatthe UnitedStates and the U.S. Bureau of Standardsrecommended to a modified Metric system called the Standard Internationalsystem(the SI

They were (1) the

English

(4)

Absolute

the

adopt

convert

and the

Table

1.1

1.1

Table

system).

Basic

the basic

presents

units of the

aforementioned

five

recommended symbol for each unit.

of Units

System

English

English

Engineering

Absolute

(EE)

Quantity

English

Gravitational

System

System

(EG)

(Metric)

(SI)

(EA)

foot

Mass

pound

(lbm)

pound

Time

second

(sec)

second (sec)

second

Force

pound (lbf)

poundal

pound (lbf)

dyne

Temperature

rankine (R)

rankine

rankine

kelvin

foot

(ft)

(lbm) (pdl)

(R)

International

Metric

Length

(ft)

of units,

Systems of Units

in Several

Units

systems

centimeter

foot

(ft)

slug

(slug)

(sec) (R)

(cm)

meter

gram (g)

kilogram

(s)

second

second

(dyne)

(K)

(m)

(kg)

(s)

newton (N)

kelvin (K)

6

OF FUNDAMENTAL

REVIEW

PRINCIPLES

of measurement in systems, including the new SI system, is given standard for each unit of measurement in the SI system was adopted the General Conference of Weights and Measuresin 1960. The standard for the by unit is the kilogram (kg), a cylinder of platinum-iridium of mass alloy kept at the International Bureau of Weights and Measuresin Paris. Of all the standard units, the mass unit is the only basic unit still defined by an artifact. The standardfor the is defined as 1,650,763.73 wavelengths,in vacuum, unit of length is the meter (m), which of the radiation between the levels 2p10 and 5d5 of corresponding to the transition the krypton-86 atom. The standard for the unit of time, the second (s), is defined as the time interval required for 9,192,631,770cyclesof radiation to the corresponding levels of the ground state of the cesium-133 transition betweenthe two hyperfine atom. The degree kelvin the standard for the unit of temperature, is defined as 1/273.16 (K), of the of the triple point of water. The newton temperature (N) is thermodynamic the standard for the unit of force in the SI system. Its magnitude is defined as the force required to acceleratea mass of 1 kg at the rate of 1 m/s2.Themole(mol) is the for the amount standard of substance of a system. The mole contains as many of carbon-12. When the mole entities as there are atoms in 0.012 elementary kilogram must be specified and may is used,the elementary entities be atoms, molecules, ions, of such particles. The standard for the electrons, other particles,or specified groups unit of measure of a plane angle is the radian is defined as the angle with (rad), which its vertex at the center of a circleand subtended to the radius. by an arc equal in length During the period that the SI system is being adopted in the United States, there will be at least five different consistent systems of basic units being employed. The relationships between them (and other systems of units) are readily determinable by the consistency condition to them. applying brief

A

history

4. The

Reference

1. The EE system. 1 pound

2. TheEA

force

(I Ibf)

= 1 pound

system.

1 poundal

3. The EG system.

(1 pdl)

force

mass

= 1 pound

(I Ibm)

mass

x 32.1740 ft /sec2

(1 Ibm)

x 1 ft/sec2

x 1 ft/sec2 force (1 Ibf) = 1 slug mass (1 slug) Metric system. 1 gram 1 dyne force (1 dyne) \342\200\224 mass (1 g) x 1 cm/s2 1 pound

4. The

5.

The

SI system. 1 newton force

(1 N) = 1 kilogram

Table C.I* presents of interest

in

gas

quantities

in

the

and

Gravitational (EG),

Table pertinent

(1 kg)

x 1 m/s2

formulas for several of the physical quantities included in Table Also C.I are the units of the physical dynamics. (EE), English Absolute (EA), English Engineering English Standard International (SI) systems of units. the

dimensional

C.2 presents the values to fluid mechanics.

In Section

mass

l-3(a),

it

is pointed

of someof out that

the

universal

physical

Newton's second law

of

constants motion

that

states

are

that

Focma

To

the

convert

denoted

by

into an equation, a factor of proportionality introduced into the above relationship.Thus

above

l/gc, is

=

F

* Tables labeled

C.I, C.2,etc.,are in

Appendix

C.

\342\200\224 ma

proportionality,

(1.1)

DIMENSIONS

1-3 From

and F. Table1.2

for m, a,

employed

= ma/F.The magnitude and unitsofgc

1.1, gc

equation

presents

of gc

values

the

in gasdynamics. units that are commonlyemployed

Table1.2

of

Factor

The

gc for

Proportionality

units

the

on

depend

for several

7

UNITS

AND

systems of

Several Systems of Units

System

of

Mass

Time

Length

Units

Force

of Proportionality

Factor

32.1740lbm-ft/lbf-sec2

EE

ft

sec

lbm

lbf

EA

ft

sec

lbm

pdl

EG

ft

slug

lbf

1.0

SI

m

kg

N

1.0

Metric

cm

g

dyne

1.0g-cm/dyne-s2

sec s s

gc

1.0lbm-ft/pdl-sec2 slug-ft/lbf-sec2

kg-m/N-s2

factor a conversion similar to factor gc is not a physicalquantitybut = = 1m 100 cm, 1 N-m 1 J, 1 kg 1000 g, and so forth. For a consistent of = 1 and from equation the of be omitted numerical value units, therefore, 1.1, gc

The

merely

=

set

may,

to yield

F = gc is

Since

values

' employed. appropriate

A

body

force acting

external

force, and

(d)

when

(1.2)

be omitted

making

the

from

numerical

units of many

of this

equations

the

calculations,

be determinedand

of proportionality must

factors

for the

are presented

1.1.

Example

other

factors

dynamics

gas

the

gc and

of

conversion

The

to

a physical quantity, it will

not

Remember, however, that

book.

ma

of the physicalparameters

pertinent

on the inside coversof thisbook.

of mass 5

kg is given an

accelerationof 10

Calculate

m/s2.

on the body in theseunits:(a)

(b) dynes,

newtons,

(c) pounds

poundals.

Solution

(a)

F =

ma

tb,

,-

\342\200\224(SW

F

(d)

In

=

ma

this

= (5

kg) the

example,

(^)(^)

(iog)

(^j

(log) units

appear explicitly,andit is

such as

gc

making

numerical

factors, if

original

is

any,

physical

-0\302\253

dyne

= 1.124

lbf

(o.2248O9 ^

(\302\243\302\243)

of each

its numerical value.

with

together

(l05)

ma = (5 kg)

F =

(c)

=

immediately

By

(32.174 g)

^)

(o.2248O9

physical quantity are so

the

doing, obvious

units

whether

written

of the or not

= 36.165pdl in

the

equation

resulting quantity a conversion factor

the units in the equationswhen is highly recommended. The required conversion computations themselves in terms of the units of the then suggest automatically units of the calculated physicalquantity. and the desired quantities

required.

The

technique

of including

8

however,

were,

examples

All

in the answers.

obtained

were

units

desired

the

illustrative

in the

followed

not

is

that

recommendation,that practice,becauseofspace of the examples presented in this book. originally worked in the recommendedmanner to assure

the foregoing strong

Despite limitations,

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

1-3(c) Principle of DimensionalHomogeneity An

relationships between states that all of the

mathematical

the

to

pertaining

principle

important

which dimensional physicalquantitiesis homogeneity, terms of an equationexpressing between physical relationship physical variables must have the same dimensions. Thus,the units evaluating as the dimensions each term in an equation may be quite of each long the term are the same. However, terms of a dimensionally homogeneous term in must be the same are to be units. combined, equation expressed that the conversion factors are introduced into the numerical time, computations. of

that

actual

an

in

employed

as

different

when

each

1-4

of discrete viewpoint a material substanceis composed particles; or it or atoms. In dealing with is assumed fluids, gaseous, liquid

a microscopic

From

is, molecules

that

6

POSTULATE5

CONTINUUM

THE

At

of the quite generally that the detailedmolecularstructure with a fluid by a continuum that makesit possibleto deal

fluid

may

be replaced

a macroscopic

on

scale. of fluid

differential element of a body The continuum postulate assumesthat contains a tremendousnumberofmolecules that the average statistical properties of the molecules containedin an elementaryvolume representthe of the fluid in the region of that elementary volume.Consequently, the properties model is a satisfactory one only for those situations the dimensions of the body of fluid under consideration, or ofa material within the fluid, are very large when compared the molecular distance average the between the molecules comprising the fluid.In other continuum postulate is satisfied if an infinitesimal change in the volume of the fluid influences an every

and

macroscopic

continuum

where

characteristic

body

with

words,

of molecules.

number

large

exceedingly

some of the propertiesofa fluid

Before defining

helpful to review briefly

it is

it. If

strain

or

deform

static forces; otherwisethey

forces

external

The

substantially

termed

are

act on a

that

A

force actingperpendicular to

parallelto of

a

A

surface.

body

and electrostatic

acting

force

is

normal

to

the

and

into a normal or shearing, force acting

be decomposed

a tangential,

and

two

groups: with any orientation

have

may

may

into

segregated

that is distributed over the entirevolume

is one

forces caused

attraction,

gravitational

by

fields, and the like.

The forceper unit area

the

force

for example, the

of material;

body

magnetic

the

be

may

force

surface

surface

the

are termed

they

time,

Fluid

of

fluid

of

with

invariant

Body

body of fluid

and (2)bodyforces. respect to the surface of the body the

postulate,

forces.

dynamic

(1) surface forces,

continuum

the

a body of fluid.Suchforces

act on

may

of the Forces Acting on a

Classification

1-4(a)

that

forces

the

the forces are

on

based

area,

a body

on

the stress is

(dimensions

called a

F/L2) is termed stress.

normal

If

a stress.If

a normal

stress

in closer contact acts in the direction for bringingthe particlescomprising stress. acts in the direction that tends with each other, it is termed a compressive as a tensile stress. to cause separationofthe particles, known the

If

it

is

it

body

1-4 1-4(b)

at a Point in a

Density

The smallest containsa

Continuum volume

elementary

is

postulate,

Let

1.1

P{x,y,z)

surrounded

dm.

a fluid particle; let Si^' illustrates schematically

termed

Figure

denote an arbitrary an

by

Let

comprising of molecules

d\"T,

number

large

sufficiently

p denote

denote that

of

function

seen

is

It

the

exhibits

bif

1. bmjbif approaches an

that

and

plot

following

\342\200\224>-Domain

of

molecules

are

noted

continuum

line

Asymptotic \342\200\224thatdefines

the

Figure 1.2

Determination

density

of

the

p

density

bif.

the

ratio

shrinks, the curve characteristics.

continuum.

-Domain where effects of

at a point

point

P is

elementary mass of

as bif

asymptoticvalue;that is,

in a

volume

Elementary

the

fluid

Then

(1.3)

homogeneous.

Figure 1.1

that

Assume

if.

volume

dm

=

Now let 5V shrink about the pointP(x,y,z),

qualitativelyin Fig. 1.2.

a

by

density of the mass enclosed by

the average

continuum

of fluid enclosed

that contains the

bif

mass of fluid that

volume.

limiting

point insideof if.

volume

elementary

an elementary for satisfying the

a large mass

p

as a

9

POSTULATE

CONTINUUM

THE

in

a continuum.

the

material

bm/bif,

as is done

presenting bmjbif becomes

more

10

FUNDAMENTALPRINCIPLES

REVIEW OF

2.

becomes

bY~

After few

molecules,

or

enter

either

minimum value where it contains only a the value of bmfb'f fluctuates widely as one or moremolecules leave the magnitude of bmjb'V becomes 5V. Consequently, some

than

smaller

indeterminate.

the

Introducing denoted

is given

p,

by

volume

limiting

bY~', then the

density of the fluidat the point P,

by

p = Limit

1-4(c)

Let

the

denote

V

in the limiting

Body of Fluid

in a

a Point

at

Velocity

value

average

b\"V'

volume

contained of the velocity of the fluid instantaneously an arbitrary point P(x,y,z). In general, surrounding V=

to

Refer

1.3. Let

Fig.

(1.4)

(1.5)

V(x,y,z,t)

r denote the radiusvectorto thepoint

P

at any

instant

of time t.

Then V

=

(1.6)

V(r,t)

V(x.y,*,t)

Figure 1.3

is called

flow

The

t. In

with

invariant

at

Velocity

a point

in a

fluid.

flowing

unsteady if V varies with the latter case, V = V(r)

=

time t, and it is

termed

steady

if V

is

(1.7)

V(x,y,z)

the of that the velocity at the pointP(x,y,z)isindependent of the molecule instantaneous of the fluid particle closestto pointP.V is the velocity velocity of the center of mass of the fluid particle,enclosed at the instant that it by, by

It should

coincides

with

be noted

the

point

P. By

V V =

i=!

definition, l

v

l

total

momentum

contained

total mass inside

in

(1.8)

1-4 where

the ith particle, 1 -4(d)

the number of particlesinsideb'V',and

N denotes

m; denotesits mass,

and

Figure1.4a the actionof

a

illustrates

the

composing

body;

of

velocity

momentum.

between

a continuum, in equilibrium under F3, F4, F5, and F6. Becauseoftheactions to be

assumed

body,

F2,

Ft,

forces,

internal forces, the expression

forces

individual

solid

external

several

external

those

its

the

denotes

Point

Stress at a

of

denotes

m^\\\\

V;

11

POSTULATE

CONTINUUM

THE

forces are transmitted through the material to internal force as employedhere doesnotrefer

molecules, but to

the

their combined

effect.

(a)

\342\226\240 a =

t

a,

Figure 1.4 Stress at (b) Stress at a point.

a

Assume that the materialof the by

an

imaginary

of the Cartesian equilibrium

forces

the

under distributed

of Part II

ab that

plane

over

body

passes

coordinate axesx, combined

Forces acting

continuum, (a)

in a

point

y,

is

divided

on

into

a body.

two

through point 0, andlet

z.

Part

I of

the body

labeled I and

point

(force

per

unit

area)

be

the

F6 and the

actions of the external forces F5 and the area A of the plane ab, causedby the action the magnitude It is convenient to express

stress

0

II,

origin

is assumed to bein

on that of Part I.

forcein termsof the

parts,

it produces

of the

internal

material

of an internal

on the area

over

which

12

OF FUNDAMENTAL

REVIEW

Let the stress be denotedby

it acts.

the

over

uniformly

the area

over

acting

PRINCIPLES

area

A of

the

In

a.

case the

general

stress is not distributed

the plane ab (see Fig. 1.4a).LetF denote

A. Then the stressa at thearbitrary

0 in

point

force

resultant

plane

ab is denned

by

a

=

Limit

The

area comparable in size to the volume in Fig. 1.4b, is a vectorhaving a, illustrated

is an

8A'

where

(1.9)

^OA

SA^SA'

stress

same

the

as the

direction

force 5F, and it is ordinarily inclined respect to the unit normal vector n, drawn the outwardly which indicatesthe directionofthe SA. By convention, The stress a can be resolved normal points in the positivedirection, and versa. a normal stress parallel to n, and a into two perpendicular components, is in the direction of the stress that it is, tangential or shearing perpendicular n; t. Hence, the stress a is denned vector the vector following equation. tangential resultant

with

surface

vice

mutually

to

unit

by

a = nan + ta,

1-4(e) The ContinuumPostulate In the case where a body of gas is be so altitudes), the gas may Applied

at

density

Gas

a

to

a

low

may

for

briefly may

The each

of

motion

number process

air at high

atmospheric

criteria

(see be

gas molecules is molecule

also

Section

1-14). as

conceived number

Avogadro's

cules/mol).

(e.g.,

pressure

low that the applicability of the continuum It question. is, therefore, essential that analytical the to the application of the continuum limitations of the results obtained from the microscopic viewpoint

postulate open be available determining postulate.To do that some of the structureofa gasareconsidered According to the kinetic theory,a gas very large number of molecules (e.g., to

be

(1.10)

N,

=

composed 6.02252 \342\200\242 1023

of a

being

mole-

move with extremely rapid randommotions, but with the tremendous collisions greatly impeded by

the

of neighboring molecules and the walls ofthe containing vessel. The collision is assumed to be elastic and to take a very short it also conserves energy time;

and momentum. Becauseof the collisions, extreme are conditions possible. Some molecules may transfer their kineticenergies to molecules and attain velocities that are practically zero, other molecules attain may very high velocities. The remainderof the molecules attain velocities (kinetic energies) between the aforementioned two extremes.Functionsrepresenting for the molecules distribution Maxwell and have been derived velocity by the methods of statistical mechanics.7\"10 Boltzmann, by applying normal conditions of pressure and temperature a gas moleculemoves a short linear the another molecular free path, beforeit collides called distance, molecule. the time interval between consecutivecollisionsis Consequently, gas The average value of the free path foran assemblage molecules likewise short. very is termed the mean free path, denoted several of the transport A, and properties of a gas are relatedto for example, its viscosity, thermal and conductivity, coefficient. In general, for a given number of moleculesper unit volume, will be larger for small molecules than for large molecules;the probability ofa collision in the size (diameter) of the molecule. with a reduction of molecules For the collisions a Maxwellian velocity distribution,10 having two

other

while

will

the

equilibrium

by

Under

only

with

1

\"of

by

A;

diffusion

A

decreases

elastic

the

mean

free

path

A is

given by

1-4

13

POSTULATE

CONTINUUM

THE

d is the effective molecular and the number of moleculesper cubicmeter ofthegaseous diameter in meters.If m denotes the mass of a molecule and p the density

where

N is

then

continuum,

are of the quantities for calculatingthe mean free path readily from the molecular diameter d. The latteris generally calculated of the application of the kinetic theory to the calculationofthe transport properties the gas, such as its viscosity, heat conductivity,and self-diffusion. values of the molecular diameters for Table 1.3presents typical approximate

all general, obtainable except In

several gasesat standardtemperature pressure = 298.15 K. It is evident from Table 1.3that at and

X

STP

Path

X (at

f

Table 1.3 MolecularDiameterd and

Gas Argon

(He)

1

Several Gases

STP) for

'

10.9

Oxygen

22.9

Carbon dioxide

10.5

2.95

(O2)

(CO2)

8.39

3.30

Ammonia (NH3)

7.46

10.2

3.00

Number

postulate,the molecular lineardimension

For a gas to satisfy the continuum small compared to a significant characteristic flow field. By definition, the ratio X/L is termed the by

Kn.

l,m-r\302\253

d,m-10-\302\273

2.00

Knudsen

\342\200\2244(f)The

extremely

Gas

2.90

3.50

(N2)

Nitrogen

A,m-10\"8

d.m-10-\"

(A)

Helium

Free

Mean

is an

p = 1 atm and small distance.

is, at

that

(STP);

free

mean

L

Knudsen

path

must

be

to

the

pertinent

and is denoted

number,

Thus,11

= X/L

Kn

(1.13)

Re = LVp/fJ. and The Knudsen number Kn canbe related the Reynolds number M where a is the acoustic speed in the gas. Employing the the Mach V/a, = heatratio the gas, specific cp/cv, it can be shown that11 to

\342\200\224

number

for

y

Kn = \\.26Jy(M/Re)

For

be

ordinarily gas

or

values

small

assumed

the

to be

number

Reynolds

length L

Re, the characteristic

either a dimension

of the flow characteristic length

a dimension

significant

of

(1.14)

in the of a body immersed

passage. For largevaluesof is the thicknessS of the

Re

layer

viscous effectsare important; this and boundary layer (see Section 5-10). The variablesL, as indicated below.11 1 S/L = yJl/Re 1.15 is not applicable to hypersonicflows (M > Equation values of Re, Kn is, therefore, large by solid walls where

layer

Re,

Re

Re

(i.e.,

fluid

of of

1),

a more

to the

adjacent

fluid

d are

\302\273

is

related

\302\273

may

flowing

termed

the

generally, (1-15)

7,

approximately).

For

given

Kn

s M/VRe

The continuum postulateis applicable those 0.01. Knudsen number is lessthan to

approximately

(1.16)

flows Accordingly,

for

the appropriate a gas may be assumed

which

14

FUNDAMENTAL

OF

REVIEW

to be

a continuum

When Table

PRINCIPLES

if

Kn

^ M/Re

< 0.01

and

Re ^ 1

Kn

^

< 0.01

and

Re

M/^Re

Kn > 0.01, the gas should 1.4 presents a classification

Table 1.4 Flow Regimes Knudsen Number Kn

0.01 < 0.1 < 3.0 < Based

\302\273 1

be treated as an assemblage of the flow regimes of a gas based

(1.17b) discrete

of

number.

the Knudsen

(1.17a)

on

particles. the

value

of

on the

Based

=

A/L

Kn




40

30

20

10

500

250

Temperature, Figure 1.13 on

(based

Viscosities of

Reference

surrounded it is called

some

common

1500

1250

1000

750

gases

2250

2000

1750

K

as functions

of

temperature

12).

by the control surface is termedthe either the surroundings or the environment.

and

system,

In

general,

the

to

external

everything

of the

behavior

the environment are described in terms of readily system and its interactions observableor measurable characteristics either the properties or the coordinates of the system. property a quantity having a value that dependsonly is, in general, on the state a system. Since thermodynamics is concerned only with the behavior of a system when it passes from one equilibriumstate to the are those giving a suitable macroscopic descriptionof the selected properties in the characteristics of the system after it has passedfrom one produced with

called

A

of

another,

macroscopic

changes

equilibrium state

to

another.

state of a

described be the system by specifying its corresponding independentpropertiesor their Hence, is in the of a functional between system expressed relationship

The

equilibrium

may

equivalents.

form

dependent

one

system,

or

that

gravitational

independent simple system

and

properties

system denned

is free

the

pertinent

independent

properties.

magnitudes the

of of a

state

one of its

In the case of a simple

of the effects of relativity,nuclear,capillarity, electric, magnetic,

two shearing stresses, the state is fully specified any by In is a a assumed t o be properties. general, thermodynamicsystem (see Section 1-15). Thus, if z is a dependentpropertyof a simple by the independent properties x and y, then effects

and

z =

f(x,y)

(1.31)

1-7

25

SYSTEM

THERMODYNAMIC

SIMPLE

THE

the by equation its actual determined by experiment. conditionsconnecting In general,the

The functional relationship between properties of is termed an equation of state, and form An equation of state presents the equilibrium z and the independent property properties x and y. mass If

or

involved, the

then

the

x and y are properties z will change by property

dependent

d denotes an exact differential. V^n

ferentationof equation1.31.

intensive,

by the partial

is obtained

dz

for

expression

the

amounts dx and dy, dz, where the operator

amount

small

the

of

independent

the small

by

changed

dependent

thermodynamic

but they may

scalars,

of the

values

the

is

be either on the massinvolved. extensive, dependent

y are

and

x

properties

symbolized

type

dif-

Thus

=

dz

+ i^f]

) dx

(~

=

dy

+

Mdx

(1.32)

Ndy

If equation 1.32has a solution,it mustbe possible to derive the total differential dz variables x and (properties) y. by differentiating some function of the independent In those cases where the aforementioned is the dz is procedure possible, quantity if the M and N are selected called an exact or perfectdifferential.In general, functions arbitrarily, dz will not be an exact differential. Hereafter, when a differential,such is not the result asdz, is inexact, it will be written as dz, signifying that the differential

of differentiating an actual

of

function

the

functional relationship z =

f(x,y) exists = and its partial derivatives(M df/dx and = If a

a

for

N

the region of the xy dx

M

expression

under

plane

8M

It is

calculus

the

in

shown

f (M where

(a,b)

and

=

N dy)

+

dx

if equation

that,

the

for the

continuous

function

throughout

linear differential relationship

reciprocal

- dN

a 33)

1.33 is satisfied,then

{*'y (M dx

+

N

dy)

=

f(x,y)

- f(a,b)

(1.34)

and final points for the integrationalongcurve

the initial that for an

are

(x,y)

are

df/dy)

if the

and

system,

simple

then,

consideration,

to be exact, it must satisfy

+ N dy

variables.

independent

C.

exact differentialexpressionthe of the line Equation on the initial and final coordinates of the integral depends only dy) |c (M dx + C. Hence, the line integral is of an exact differentialexpression of the the end of the paths (curves) connecting points integration. differential the line of a integral expression has the aforementionedproperty, thenthe the differential from which is formed is called eithera expression stateora A corollary to the foregoing is that for a pointfunction the function. line the value for all curves (paths) connecting end has same same points. Consider now the special case where the initialand of the for the points path line integral of an exact differentialexpression identical coordinates. The path In that is then a closed curve,suchas a complete case, cycle. 1.34

shows

value

N

path

independent

If

function

point

the

integral

final

have

thermodynamic

jc(M

Hence,

the

(complete

sources

line

cycle)

or sinks,

is zero;

properties as follows.

exact

the only

differential

dN/dx

of the

are

= 0

(1.35)

expression taken over a

closed

curve

limitation is that thereareno singular points,

within the region boundedby

derivatives dM/dy and The

of an

integral

dx + Ndy)

the

closed

curve

and

that

such

the partial

continuous.

exact linear differential expressionmay

be

as

summarized

26

integral between two

1. Its line

line integral

of the curve (path)

is independent

points

around a closedcurve

vanishes.

cycle)

(complete

satisfy relationship dM/dy = dN/dx. linear differential integral expression taken over an arbitrarily the expression is an exact differential. selected closed curve vanishes, then If a linear differential of satisfies one the conditions 1,2, 3, or 4 expression any it satisfies all of them. above,

second partial

3. Its

4. If the 5.

of a

line

property

Change

The equilibrium

of a

state

Thermodynamic

System system is completely defined

simple thermodynamic

thermodynamic properties. Hence,eachproperty

by

be taken

may

of any two independent properties. Thus a thermodynamic property the requirements as discussed in Section l-7(a). for a point function, the selected for describing the state of a simplethermodynamic properties be its absolute pressure p, its specificvolume v, and its absolute temperature

a function

satisfies

Let system

a Simple

for

any two independent

to be

above requirements is saidto be a

system.

State

of

that satisfies the

a function

the

of

the reciprocal

derivatives

In thermodynamics,

1 -7(b)

end

them.

connecting

2. Its

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

t;

actually any properties may be employed for defining its equilibrium state. its of state, equation expressed in terms of p, v and f, is given by any Accordingly, one of the following functional relationships. Thus, three

p= If there is a change from

v =

p{v9t\\

one

t = t{p,v)

v(p,t\\

to another, then

state

equilibrium

(1.36)

*-(s).*+(D.*

fdt\\ \342\200\224,

=

dt

fdt\\ , \342\200\224 dv

+

dp

For the total (equation

1.33)

differentials

must

dp,

be satisfied,

dt to be exact, the reciprocal and the corresponding line integral

dv, and

differential expression taken around As an example, consider the equation v

In

that

case,

dv = 0,

where the

the indicated

subscripts changes.

=

and equation dv

v

attached

(1.37c)

\\SvJp

\\8pjv

a closedcurve

relationship for the

pertinent

vanish.

must

of state

v(p,t)

= constant

(1.38)

1.37(b)becomes

\342\200\224

to dp

Rearrangement

and

dt

denote

of equation

{dv/dp)t

that

v remains

1.39yields

constant

during

AND IRREVERSIBLE

REVERSIBLE

1-8

PROCESSES

27

or /dt\\

fr)v\\

fdn\\

=

The partial derivatives (dp/dt)v the

of

coefficients

-

=

P

I

are

(dv/8t)p

(1.41) to the thermometric

related

Thus,

system.

/?\342\200\236

and

-1

\342\200\224 I

=

change at constant

coefficient

of pressure

coefficient

of volume change

volume

(1.42)

\\StJv

and

1 v

In general,

fSv\\

the coefficients /?\342\200\236 and

in temperature are small, it

1-8

REVERSIBLE A reversible

any

influence

the

system,

difference

to

applied

in

change

direction

the

in

the

causing the

fiv

fip are

and

one that is in a state

infinitesimal proceed

of the temperature.

If the changes

constants.

PROCESSES

is defined as

process

will cause the processto

to a

fip are functions be assumed that

IRREVERSIBLE

AND

so that

in its path

may

coefficient)

(dilatation

pressure

\\dt/p

at constant

the

of

at

equilibrium

all

producing change. When heat

influence

of that

points

the change

is added

to change its state is the temperature if an arbitrarily small Consequently, a or compression, expansion depending values of the mechanical effects produced from or to the system are equal to each

system state.

equilibrium

temperaturedifferenceproduceseither on its sign,and the corresponding for the same quantitiesof heat is other,then theprocess

an

absolute

transferred

reversible.

process involving friction cannotbe reversible becausea of work amount equal to the work required for overcomingthe frictionmust be before the direction of the process can be reversed. supplied It is the assumed that work expended in overcoming friction is instantly friction invariably into heat. Since accompanies the flow of real fluids, real are irreversible. As a matter of fact, all natural or processes always are irreversible. tend to proceed in a direction They processes toward never reverse themselves is a t the and unless work done, equilibrium is

It

from

apparent

this that a

finite

always

generally

converted

flow

spontaneous

given

expense

of

some

other

system,

to reverse

them.

consider a closed Consider the followingexamples natural First, processes. in different volume containing an inert gas that has velocities flow parts of the volume. Because of the irreversible momentum the viscous flux, process over the entire volume. variations in the velocity are reduced to zero in a variations Second, consider a closed region containing having irreversible of these variations are eliminated the temperature; by process two different heat conduction. Third, considera closed containing species of inert gases nonuniformly distributedso that there variations in the variations are eliminated in time the irreversible of the these species; of mass diffusion. Irreversible processes, such as the three here, process are the flux of and thus a finite characterized some by involving property time the state of equilibrium. Reversibleprocesses interval to establish of fluxes and thus require infinite timefor their characterized the absence by that is, they are infinitely slow processes. of

different

of

eventually

material

finally

volume

are

concentrations

by

discussed

require

are

completion;

28

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

elementary thermodynamicsthat the definitionof same is based on the concept of thermal equilibrium. Thus, bodiesareat thermal with each Sincethe when are in other. equilibrium temperature they t is one of the simple system, the abovedefinition of the coordinates the thermodynamic states. restricts study of the system to its equilibrium temperature its take the system to change statemust the causing place an processes its the be rate so that in all of intermediate assumed slow infinitely system may do the above requirement, even not to be in equilibrium. satisfy processes in a finite time. In because complete themselves if they could be frictionless, is its the an the more rapid actual degree of irreversibility greater general, from a thermodynamic and the poorer is its standpoint. frictionless The reversible processes of thermodynamics are, therefore, that take infinite time to complete themselves.Suchprocesses are, processes It will

be recalled from

the

temperature

of

temperature

at

Moreover,

states

Actual

they

process,

efficiency

imaginary

however,

changes of state. Although are of great value because they

accomplishing they

processes,

they

depart

basis

a

furnish

actual processes.Furthermore, the results be which processes can dealt reversible mathematically, processes, based to real processes modifications made applicable introducing

for judging

the efficiency of

obtained

by

with

assuming

be

can

real

from

considerably

for

efficient

most

the

by

on

experience.

1-9

WORK

In

as energy

is defined

work

thermodynamics,

system wherethe sole

effect

the

to

external

system

in

across

transit

could have

boundaries

the

been the raisingor lowering

of a

a weight.

of

is based primarily

Classical

thermodynamics signifies that mass

on the concept

of

a

closed

system,

The latter is an can neither enter nor system. identifiable collectionofmatterthatcanchangeboth shape and volume. Figure 1.14 illustrates a simple closed system enclosedin an impermeable control A. Assume that there is no friction, and let p denotethe uniformhydrostatic pressure on Let it be further assumed the A. that p is largerthan the by system which

the

leave

its

surface

exerted

p0,

of

Direction

of control surface

'

displacement

~7

^

'

Direction

of

normal

\\

\\

\\

\\

Original

^v

configuration

of control surface Final

configuration

of control

Figure 1.14

Work

caused

by the

displacement

surface

of the boundaries of a

system.

1-9 exerted by the environment on A, by the pressure Because of the pressure differencedp, the controlsurface A distorted to some new configuration A'. The work performed

hydrostatic

whole in displacing from

no friction,the

is, the

work

A

the

of

displacement

by

displaced

the

system

system

reversible.

is

performed

be

dp. and as a

moving boundary work. Because there is is a reversible process; that boundaries

is called

A'

to

amount

differential will

29

WORK

the static displacement of the controlsurfaceA, assumed here, be as a function of its volume V, in themanner pressure p of the systemmay plotted illustratedin Fig.1.15. The result is the curve which relates p and V for the change 12, from the the state 1 to state 2. equilibrium equilibrium Let WR denote the work done by a system that undergoesa reversible changeof from 1 to state 2. Then state state,

the reversible

For

WR

The subscript R the

Hereafter,

jp

d\"V~

attached to to

only

for

system

on its

work on the

positive

from

work

as

plane,

p\"T

in

Fig. work.

negative

the

as positive

surroundings

of state is a reversibleprocess. and it should be understoodthat W =

the change

that

work,

work

boundary

moving

The general

environment.

its

against

(1.44)

pd-T

change of state.

a reversible

Equation 1.44is the expression

system

J}2

be omitted,

jR will

subscript

applies

denotes

W

=

done by a

convention is to regardwork The

versa.

vice

and

done

closed by

representation

1.15, leads to the followingrule for lies to the right If the area \\p d\"V of an

the

of

distinguishing observer

is positive,and viceversa. moving the In the where the such as that illustratedin Fig.1.16, net work is the difference between the positive work and the negativework. It is area the enclosed by the curves (paths) formingthe closed by (cycle). If the were an exact differential line its expression p expression, integral around the closed of 1.16 would vanish instead of Fig. being equal to the enclosed = Section the differential SW is inexact. The work Hence, p [see l-7(a)]. on the process or path, that is, on functional between depends relationship the work of the process, is a closed curve, path

direction

in

case

the

curve

given

d\"V

curve

area

d\"V

the

p and V.

Equation 1.44 is applicableto eithera system there is no friction.In the casewhere the friction

of a machine,but the only to the fluid.

Figure

1.15

plotted

on

Determination the

pV- plane.

of

work

working

for a

fluid

process

is

or

frictionless,

Figure 1.16

its is

working confined

equation

Work

for

substance, provided to the components 1.44 may be applied

a cyclic

process.

30 1-10

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

HEAT

Experience has shown that

the boundaries of a For example, when a

across

transferred

be

may

energy

in the system in the absenceof macroscopic system. is brought into contact with its surroundings at a system at a temperature flows differenttemperature, from the higher temperature region to the lower changes

given

energy

in the system. temperatureregion,unaccompanied any macroscopic changes This energy transfer processistermedheat. is defined as energy in transit across the boundariesofa by a temperature difference between the system and its surroundings. the heat transferred that to a system is definedas and convention, from a system is denned as A process heat transferred that permits a one that transfer is termed diabaticprocess, does not permit heat transfer is termedan adiabaticprocess. on the process or path. Consequently, heat is not a work, depends is written Its differential as SQ to indicate that the inexact. by

heat

Hence,

caused

system

By

heat,

positive

heat.

negative

while

like

Heat,

1-11

THE FIRST

is

on

based simplest

of

ratio

the

OF

LAW

THERMODYNAMICS

and its validity thermodynamics is based entirely on experience, the fact that no phenomenon has been uncovered that contradicts it. statement is that heat and work are mutually interconvertible and that is a fixed quantity termed the mechanicalequivalent of heat, conversion J (note that the symbol J is also used to denote the unit of energy the

law of

first

The

Its

denoted

by

joule). The first law

gives no information

be convertedinto when a system convertsheatinto to the amount vice (and produced to a system that can

the

work

of

-11 (a)

of heat

the transfer

amount

corresponding

the

denote

heat

transferred states that

of heat so

denoted

converted J.

by

Thermodynamics

performed

j(6Q

- SW) =

involving

processes

the

that

transferred

of work

of

series

series of

processes

the system and

to or from

during the cycle.Then,experience

that

shows

Equation

Stored

1-11(b)

of

work.

5Q

the

versa).

is a constant

versa)

Law

It merely

amount

of the

is subjected toa and the performanceof Assume

form a completecycle.Let SW

(or

ratio

system of mass m that

a closed

Consider

of the First

Statement

Mathematical

vice

of heat

amount

the

regarding

work

work,

1

is

differential

property.

1.45 is

defines

property

(1.45)

statement ofthefirst

of

law

thermodynamics.

Energy

It follows from expression

the basic mathematical

Q

in

Section 1-7(a) that because the

vanishes

1.45

equation

of a closed

a property

for a

line

of

integral

differential

the

closed curve, the quantity (5Q \342\200\224 SW)

system. Let E, termedthe

stored

energy,

the

denote

by equation 1.45. Hence,

denned

dE = 8Q comprises all of the formsof 5W

The

stored

energy

E

system. Both 5Q and

E

may

include

the

3W

following

are

not

kinds

(1.46)

energy

that

are energies

can

in the

be stored

in transit. In

general,

storable;

they

of stored

energy: thermal energy (internal

energy

U); kinetic electrostatic,

chemical,

encountered

in

of

forms

aforementioned

E = U

unit mass basis, the

On a

stored

internal

E -

+

(1.47)

mgz

is denoted V2 \342\200\224

= u +

unit

per

energy

m\342\200\224

E

m

where u is the

V2

+

energy

=

e

+

2

gz

mass basis, equation

a unit

mass.

On

de

-

SW

SQ

1.46

(1.48)

mass basis.

a unit

on

determined

also

are

SW

=

where

by e,

becomes where SQ and

(mgz);

let

Hence,

energy.

gravity

the first three of the

one need consider only

mechanics,

effect of

forms of energy.For mostofthesituations

and other

nuclear,

fluid

OF THERMODYNAMICS31

by the

caused

energy

potential

(mV2/2);

energy

FIRST LAW

THE

1-1

to both reversible and irreversible Equations 1.46and 1.48 processes. In an irreversibleprocessthe energySEfis in effects overcoming dissipative such as friction, turbulence, etc. Suchformsof are converted into completely the mechanical work the random molecular motions of the molecules. SW will be smaller than p d'f (seeequation when Hence, energy dissipating effects are involved, equations 1.46 and 1.48become - pdf - p = and de = (1.49) SQ SEf Sef apply

expended

energy

Moreover,

1.44).

dE

Internal

1-11(c)

dv

SQ

Energy

Energy can be storedin theatomsand

of a

molecules

because of their

system

motions.

is also called energy), and the kinetic energies of translation, vibration, and rotationof the comprises the electronic and cloud. energy caused by the state ofthe particleelectron a closed of mass m having no kinetic and potential energies. Let system heat transferred to the system and the denote the work corresponding 1.49, the resulting equation performed during a change of state. Then, change in the internal of the is given by system - pdr - p = = and dU = SQ(1.50) SQSQ SUf Suf

That

of

form

energy U (it

internal

is termed

energy

thermal

particles, Consider

SW

SQ

from

energy

du

SW

where

effects of

1-11(d) Flow

a system of fixed mass mass Sm. Assume that

illustrates

to

A

is

a

small

joins the system mass m. Let into the system. The work

comprising m and flow

work

by

Sm,

SWnow.

v =

the

where

the

by

Sm the

moving

the

dissipative

the

m enclosed the mass Sm

is

on the

A.

boundary across

forced

surroundings, denoted

surroundings

distance

fixed

a

by

by

p,

combined

A and force

Sm

system,

dx into the originalsystem,is called

given by =

drVjSm. Physically,

intoa region

of the

pressure

performed

It is

5Waoyi

where

by

Work

Adjacent

the

caused

energy

friction.

1.17

Figure

the internal

the increase in

denotes

SUf

dv

pressure

= {pdA)

Fdx

p drV is p.

is

the

dx = p dt\"

work

The value

required

=

pv

Sm

to compress

of p d'V is negativewhen

(1.51)

the volume if mass

is forced

32

OF

REVIEW

FUNDAMENTAL

PRINCIPLES System

n

+

mass m

fixed

of

Tangent

Fixed external A

boundary

Figure 1.17

work

Flow

into the system On a unit mass

because

of mass

crossing

the

when (see Fig. 1.17),and positive basis, equation 1.51becomes =

SWa

1-11(e)

of a system.

boundary

mass

is forced out

of the

system.

(1.52)

pv

Enthalpy

By

the enthaply

definition,

per

of

unit

h =

Like the

specific

mass.

unit

internal

Differentiating

energy u, equation =

dh

From

equation

1.50, a change in

Combining

equations 1.50and

reversibleprocess,Suf

=

pv are

and

h

(1.53) thermodynamic

+ (pdv

du the

=

+

v

(1.54)

dp)

energy is given by

internal

specific

\342\200\224 \342\200\224 dv

3Q

duf

5Q

0, and SQR

p

(1.50)

-

8uf

+

vdp

(1.55)

equation 1.55 reduces to = dh vdp

where the

R denotes that Q is transferred in a subscript is and adiabatic reversible, that is, isentropic, then process

dh

properties per

1.53 yields

1.54yields

dh

For a

is

u + pv

both

=

du

substance, denoted by h,

of a

mass

- vdp =

dh

--dp

reversible

(1.56) manner.

= 0

If the

(1.57)

P

1-11(f) Specific Heats In

general,

the

specific

heat of

a substance,denoted c =

*\302\243 at

by

c, is

defined by (1.58)

THE SECOND

1-12

an inexactdifferential, the is conducted. interest process Since SQ is corresponding

and

readily

the

are

particular

constant-volume

to

in the thermospecific heat implies a heating processto attainthe change state. The specificheats cv and cp defined by equations 1.59 and 1.60are

dynamic

partial dervativesof

reversibleprocessit is

Even in a

heatinto

is

work

cannot

a gas

expand; of the

concerned

by a

the

the First

the

Law,

the

of Law

Second

the

system

loses

of a

capability For

the

a closed

capability

It

work.

heat

every

equilibrium

or process moves for spontaneous

change

First

is convertedinto

of

As

that can

of heat

that can based is entirely

Law,

As

that matter. The

regarding

quantity

to a

it to

movements

the

mentioned,

already

be converted into work is

Law

is the

concerned

Second

only

Law that is

be converted into useful work. Like on experience. if left undisturbed, will change system, or rest, and the process is

the system toward the equilibrium state, associated with the The change. property

system for spontaneous change, is calledits entropy the system that undergoes a cyclic changewhere

dQR is receivedreversiblyat

cp,

the heat added

random

the

control.

quantity

natural Accordingto the Second spontaneously and approacha state irreversible.

and

energy

with

of the heat that

with

all of

transform

the thermalenergyassociated

information

the portion

with

cv

that the transformationof degradation of a portion of the

gas molecules is not subjectto complete

gives

specific

that is less useful. For example,thethermalenergy associated be converted entirely into mechanicalwork by permitting

this is because

no

for

heat

has demonstrated

accompanied

First Law is not concerned with and

to

possible

Experience

inevitably

a form

into

supplied

work.

useful

into

system

not

derivatives.

partial

AND ENTROPY

THERMODYNAMICS

OF

LAW

SECOND

such

The

place.

taking

process

to

thermo-

themselves

are

thus

and

properties,

thermodynamic

dynamic properties, which are independentof theactual term specific heat is thus a misnomerwhen applied Because general usuage has firmly established the name will be adhered to in this book. that terminology

with

cp,

respectively. It can be

heating,

name

The

THE

and

cv

that13

shown

1-12

heats

specific

constant-pressure

the heat transfer

how

on

c depends

of

value

Of

33

ENTROPY

AND

THERMODYNAMICS

OF

LAW

the absolutetemperature t,

we

may

s.

heat

per

unit mass

write13

(1.61)

Equation

1.61 is of the sameform

simplesystem.Hence, by

entropy

by

Clausius

(1850).

1.35,

equation

a thermodynamic

defines

SQR/t

the

as

If s

defines

which property

denotes the entropy

of the

a property of

a

system, termed

per unit massfora systemthen,

definition,

as

=

(1-62)

two states is the 1.61 shows that the changein entropybetween given Equation for an them.Hence, irreversible the same for all processesconnecting entropy change value of for can be measured the any arbitrarily selectedreversible by process J SQ/t

processesconnecting the

same

initial

and

final

states.

\342\226\240\342\200\242

34

PRINCIPLES

OF FUNDAMENTAL

REVIEW

system that experiencesa reversiblechange state to another, equation1.50reduces

For a closed equilibrium

Substituting for

=

from

SQR

for du from

Substituting

into

1.62

equation

one

+

du

(1.63)

(1.64)

pdv gives

=

tds

dh

- vdp =

P

processes, thermodynamic

only

they

(1.65)

dp

reversible

are

values

integrated

having

the

-

-

dh

for Although equations 1.64and 1.65 derived valid for irreversibleprocesses because involve that is, they involve only exact differentials of

1.63, we obtain

equation

equation 1.54intoequation1.64

*

they are also properties; that

are

process.

1.50, expressed on

equation

Combining

the following

equation for both

the entropy change

a unit mass basis,

with

reversibleand irreversible

equation

1.64

processes.

=

=

ds

Thus,

from

+ pdv

du

tds =

gives

0)

to

SQR

independent

=

(Suf

(1.66)

during a processchange

state

from

du

+ p

J2 1

1 to

state

2 is

given

by

dv

t

where the integrationbetweenstate1and state2 is performed along any arbitrarily selected reversible process or processesconnecting the states1 and 2. = 0 and It follows from equation 1.66 that for a reversibleprocess, SQ = t ds, Suf = whereas foran irreversibleprocess, > 0 and SQ < t ds. In the casewhere 0, SQ Suf = 0, the the process is said to be adiabatic.When is reversible. In the process Suf = = c ase where the is said to be isentropic, a reversible 0, SQ special process Suf = and adiabatic ds 0. Consequently, process,

ds ^ the

where irreversible

\342\200\224

(1.68)

to reversible processes

equality

applies

(natural)

processes.

and the inequalityapplies to

= 288.9 Air is compressed from the initial state where K and tl \342\200\242 = = = 566.7K 1.38 105 to a new state where and 4.14 105 N/m2 Pi t2 p2 N/m2. the change in the entropy of the gasfor the following Calculate cases: (a) a reversible adiabatic compressionfrom p^ to p2 followed by a reversible isobaric heating to the final temperaturet2,and (b) an irreversible between the same states where process

1.4.

Example

\342\200\242

the

work

caused by

Assume

y

expended friction

=

in

remains

is equivalent

friction

overcoming in the

gas. These

to 232,600 J/kg and the heat

two processes are illustratedin Fig.1.18.

1.40.

Solution

(a)

If the air is assumed to be

a

perfect

gas

(see Section

(seeFig.1.18), compression

1-15) then, for an isentropic 0-28571

4.14

s, Figure

1.4.

From equation 1.65,for an isobaricprocess(dp A = ds

Section

From

MOTION 35

s2

for Example

Sketch

1.18

LAWS OF

NEWTON'S

1-13

for a

l-15(d),

perfect gas, dh

=

0)

dk \342\200\224

=

(a)

cp dt,

where

= 1004 J/kg-K for air.

cp

Thus, c\"dt \342\226\240

For reversible

(b)

process. have the

1-13

361.4

395.4

The laws be referred will be specified. primary

average

inertialcoordinate

which

system,

no rotation

and

laws can be written Every

as

insofar as it may

appliedto To

the

irreversible

SQ/t.

The

irreversible process

every

be

change

the

action

body,

there

to

which

they

and customary form simplest which comprises inertial coordinate system, of fixed stars, or to a secondary position a set of axes that moves with a constant have

their

inertial

primary

to

referred

(see

system

either

inertial

Reference

system,

14).

Newton's

follows.

persists in its state

particle

2. The rate of

comprises

relative to the

of a systemof particlesis

If the motion

3.

as

three laws of motion,the reference frame

Newton's

when they are referredto eithera a set of rigid axesfixedrelativeto the

1.

by

given

MOTION

OF

presenting

velocity

Btu/lbm-R)

(0.0864

J/kg-K

same entropy change. LAWS

Before

566.7 =

the irreversible process the entropy changeds is not process in part a, however,connectsthesamestate points Since entropy is a property,the reversible and the process

NEWTON'S

are to

Sl = 1004 In

-

s2

uniform

motion

in

a straight

to change that state. (quantity of motion) is proportional the in direction place of the lineof

line except

by force

compelled of

momentum

and

it takes

is an

of rest or

equal and opposite reaction.

to the force

actionoftheforce.

36

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

the mass m is given

to Einstein,

According

by

mn

m0 is

where

Mathematical

1-13(a)

Let M =

equal to its

to be

assumed

mechanic

fluid

all

For

be

the body at rest,

the mass of

speedof light.

the

denote

rriV

velocity

in this

applications

rest mass

of a

momentum

body, and c is the the fluid mass will book, of the

m

m0.

Laws of Motion

of Newton's

Expressions

the

is

V

particle or a body.By

law

second

Newton's

of motion,

^at

^ at 1.70 shows that

Equation

the externalforce

is equal

equation

for example,

direction,

to the derivative with 1.70, it follows that the

the x direction, denoted

where ax denotesthe acceleration

the

mass

termed

the

of

Newton's third

FAB on

force

the

exerts

customarily

law,

dV*

dM*

v

m in the x direction. reaction principle, implies and

Momentum,

m,

=

FBA

three

coordinate the components

let

and

x, y,

and z be its

Cartesian coordinates.

of the external forceF actingon theparticle, X

=

Y

=

j(mu) at

=

=

y =

dx/dt,

with

time,

dy/dt, then

jt(mw)

=

and z = dz/dt.

Y

=

Z =

d2x/dt2,etc.

Y,

and

mv mw

= =

the Z

then

(1.73) (1.74)

jt(my)

X = mil =

x = u =

to

=

Z =

a

At

j(mi) at

jt(mv)

where

(1.72)

of V referred denote its velocity vector. The components If X, axes will be denoted by u, v, and w, respectively.

time

invariant

0

t let V

given

x

force

opposite

of a Particle

Consider a particleof mass

where

that if body

A. Hence,

on body

denote

mn

body B exertsthe equal

body B, then

F^B + 1-13(b)

(1.70)

is

X,

FBA

From

M.

vector

in

A

F

vector

respect to time ofthemomentum component ofF acting an arbitrary by

M

(1.75)

jt(mz)

If the mass of the particleis

constant,

mx

(1.76a)

my

(1.76b)

mi

(1.76c)

1-14

Let a denote the corresponding acceleration vector. Hence, 1.78

Equation

1-13(c)

a

of

Momentum

on the

problems the interestisina

it

may

acting

between

1.72,

F =

ma

+

+ k'z

yy

(1.77) (1.78)

= ma.

of Particles

System

1.70 is based

Equation

\\x

37

GAS

PERFECT

Then

if the mass m is constant,F

that

shows

a =

THE

AND

THEORY

KINETIC

a

of

dynamics

a large

such

comprising

system

In most

particle.

single

number

engineering of particles that

continuum. In that case, in additionto theexternal forces on the there are internal forces because of the actions and reactions particles, the particles. satisfies equation By Newton's Third Law, each pair ofparticles so that the net effect of the internalforcesin a body is zero. For an arbitrary to

assumed

be

be a

the masses

system of particles having

V1;V2,. .., denoted

force,

force

external

by

jth particle is F,-and itscorresponding Hence, for the system of particles,thenetexternal

of particlesis

on the assemblage

F, acting

velocities

the corresponding

with

on the

acting

is nij{dVj/dt).

momentum

in

change

the

Vn,

,mn

m2,...

mu

F= EF,= iUmjVj)

(1.79)

j=\\at

j=i

or

F

is contained

a

within

the momentum acting

on

that

mass

of fluid

mass of fluid.

large number

of particles

As

inertial

an

of reference.

frame

GAS

PERFECT

viewpoint of classical thermodynamics as appliedto gases, harmonized with the microscopic viewpoint of the kinetictheory of gases, gives of the that into some of the characteristics cannot perfect insight important gas obtained from thermodynamic considerations alone. The

8Q)

the instant of time t, the rate of changein is dM/dt, which is equal to the net external noted the velocity and acceleration earlier, in

THE

(1

of a

composed

terms in equation 1.80must be measured 1-14 KINETICTHEORY AND

jyj

(mV)

A at

surface

control

that

for

F

at

1.80 shows that ifa massoffluid

Equation

force

^at

when

macroscopic

1-14(a) BasicConsiderations the Kinetic Theory One of the aims of the kinetictheory of gasesis of state for the perfectgas(see equation is based on the molecular structure matter, a gas

an

be

of

the

1.99).

equation

of

composed

motions. centimeter

of

a very

For

large

of

interpretation

According

may be

to that

the

theory,

thermal

which

conceived as being

move random extremely rapid conditions the number of molecules a cubic at standard

number

of molecules that

example, of air is approximately

with

in

2.705

\342\200\242

1019.

Although

the

molecular

velocities

are

molecule under standard extremely large (e.g.,the average velocityof thehydrogen molecule is greatly is 1840 m/s approximately),the motionof each conditions of number molecules. with the tremendous collisions neighboring impeded by meanfreepathX is very small. in 1 the molecular as out Section -4(e), Consequently, pointed

38

REVIEW OF

PRINCIPLES

FUNDAMENTAL

to equation

According

1.11,

is

X

given

by

(L81)

i

where N is the numberofmolecules per

d is

and

meter

cubic

of

the effective diameter

the moleculein meters.

Because the molecule of

follows

of

movement

the

rather

slow

with the neighboringmolecules,eachgas an extremely torturous path. For that reason, the diffusionofa gas, gas molecules through neighboring gas molecules,is inherentlya of collisions

number

large

process.

different independent motions, calleddegrees offreedom, on its structure. In it have translational motions and depending may general, rotational and the atoms vibrate with to of a polyatomic moleculemay motions, respect a each other. A monatomic molecule has 3 degreesof translational freedom only; diatomic molecule has 3 of and 2 of 5 of translation rotation rigid degrees freedom, about two axes at right anglesto the longaxis of the molecule. If the atoms of a diatomic moleculecanvibrate to each other, the degrees of freedomare with respect to 7 increased because of the vibrational triatomic degrees of freedom. A nonlinear molecule will have at least 6 degreesof freedom, and 3 for rotation 3 for translation about three mutually if the atoms vibrate, 3 for vibration,or axes and, perpendicular A

9

gas

degrees

will have

molecule

of

8

in all.7'

freedom

gas it is assumedthat the

For the perfect

of

degrees

only

are

freedom

the 3

for

translation (same as for a monatomic gas). is further assumed that the molecules are perfectly elastic and that the only interactions the molecules are the collisions; that is, there are no attractiveforces. kinetic is based on applying the laws of Newtonianmechanics to theory but because of the large number of moleculesinvolvedit becomes a molecules, to apply that statistical methods to the system with the consequence necessity the the resultsobtained of the properties of the gas. values averaged The diameter of a molecule d exerts a predominating influence on the transportproperties the gas. Physically speaking, if two molecules approach each otherand the between than d, then the molecules their centers is smaller It

between

The

the

mathematical

are

effective

of

distance

must collide.Thecollision is in the original directions and changes molecules. molecules are deformable, then the effective speeds of the colliding it is diameter d will be smallerfor molecules than experiencing for those experiencing low-speed collisions.Consequently, the of a will be a function of the speed the collision. molecule by

accompanied

If

the

collisions

high-speed

diameter

effective

of

The

rms velocity

^JC1 is different from c

the

=

=

(j-\\

mean

are related

~c. They

velocity

(1.82)

0.92lVc2

In this section,results from the kinetic internal properties of a gas as temperature, addition, the results for the transportproperties diffusion of a perfect gas are discussed. derived

for such

theory

thermal

viscosity,

thermodynamic are

etc.,

energy,

pressure,

by

presented.

In and

conductivity,

1-14(b) Pressure

Intensity

Consider number

of

a mass molecules

enclosed in a containerhaving the volume V, and m is the

of gas in

the

volume

\"V. If

mass of each

molecule,

N is

the then

the

KINETICTHEORY AND

1-14

of the gas

density

p is

by

given

Accordingto the kinetic

pressure

of the

impacts

C2

square of the velocitiesof theparticles.

of the

mean

the

to

due

(1-83)

^

intensity p acting on the walls of the molecules on thosewalls. Let denote the gas

the

theory,

Nm

=

P

container is

It

=

Hence,in

of

view

1.85

Equation kinetic

(L84)

\\{KE)\" p is given by

intensity

pressure

that7

be shown

can

the pressure intensity p is directlyproportionalto the per unit volume of the gas.

that

of translation

energy

the

1.83,

equation

shows

39

GAS

PERFECT

THE

1-14(c) Temperature to

According

to

proportional

the

kinetic

the

translational

motionsofthe

gas

the absolute temperature t of a perfectgas is kinetic energy (KE)tr associated with the random

theory,

is

which

particles,

It follows

from

1.86

equation

- NmC1 oc t

=

(KE)tr

given mass of gas, the velocity

for a

that,

temperature of the gas. It should the internal energy of the gas.Hence, the temperature with the

increases

(1.86)

be

It

is shown

then the internal

l-15(a) that (see equation1.99)

in Section

=

equations

mRt

(1.99)

1.84 and 1.99yields ^

1-14(d)

Avogadro's

Accordingto Avogadro, equal t and pressurep, contain same

of

volumes

the

amountofa substance atomsin 0.012 carbon that

number. The NA are Boltzmann's

of

all

as many

That

when

gases,

of molecules.

number

contains

12.

(1.87)

Constant

and Boltzmann's

Number

kg

y/G*

constitutes

fixed.

p-T

Combining

(KE)tr

is fixed,

if

energy of the gasis

that

noted

term

elementary

denoted

number,

held at

the same temperature

The mole3 is definedas the entities as there are carbon

by

NA,

is

called

Avogadro's

= 6.02252 It has been shown experimentallythat 1023 particles/mol. defined in Section number universal constant and R, l-15(a), gas termed both universal constants. Their ratio is likewisea universalconstant, \342\200\242

NA

Avogadro's

which

constant,

k =

\342\200\224 =

is denoted

1.38054

by k.

\342\200\24223

10\"

J/K

Hence =

Boltzmann's

constant

(1.88)

NA

Boltzmann's

constant

k is

the gas constant

for a singlemolecule ofa gas. If

z

denotes

40

FUNDAMENTALPRINCIPLES

REVIEW OF

the moleculardensity,the numberof molecules per

unit

then

volume,

1.99

equation

becomes p = zkt

(1.89)

1-14(e) Specific

Heats

to

According

kinetic

the

the heat 3Q requiredto raisethe

theory,

of the kinetic amount dt is stored in the gasin theform molecules. For a monatomic gas the kinetic energy is obtainedfrom on a molar basis,

gas

by an

a

of

temperature

of its

energy

1.87;

equation

thus,

(1.90)

(KE)tr=^Rt

heat capacity of the gas at constantvolume.Sincecv is the unit in temperature, increase in the internal kinetic energy per moleofgasper change it follows from equations 1.58and 1.90that

Let

=

cv

mcv

=

the molar

=

rhSQ

cvdt = -Rdt

dt =

mcv

(1.91)

and =

cv

Equation1.92

indicates

of

gives values

t, and

cB

that

for

in

good

gases. Polyatomicgases

l (8314.3) =

=

?-R

agreement of

the

of

of

the

to Maxwell's equipartition

those

than

larger

~cv

neglect

rotational and vibrationalmotions According

ofp and gas the specific heat is independent with the measurements for monatomic

a perfect

values

have

The differenceis attributedto

(1.92)

12,471J/kmol-K

the

given

required

energy

by equation 1.92. for supporting the

molecule.

polyatomic

law,9'10the kinetic

of

energy

a gas

molecule

6/2R.

For

of since the monatomic divided among all of its degrees Hence, freedom/ in 3 directions), the energy molecule has 3 degrees of freedom(translation per mole = 5 (3 of gas is ~cvt = 3/2 Rt. Consequently, for a rigid diatomic moleculewith / = = translational and 2 rotationaldegreesof freedom), 20,785 J/kmol-K, a ~cv 5/2R valuein much betteragreement with measurements. For a rigid polyatomic molecule

is equally

and3 rotational of freedom), with/ = 6 (3 translational because rigid molecules, there are additionaldegreesof the atoms; the latter phenomenonbecomesmoreand more degrees

=

~cv

freedom

of

the

of

vibration as

important

the

gas

increases.

temperature

1-14(f)

non-

Viscosity

It is pointedoutin Section l-4(e)

that

the gas molecules to

heat

and

conductivity

The viscosity molecules.

It

can

be

such

molecular

diffusion

the

kinetic

properties

transport

the mean of the gas as

relates

theory

coefficient.

of a gas arisesfromthe transportof themomentum shown that for a perfect gas the viscosity \\i is given

If the gas density p decreases, themeanfree pX

tends

to

remain

constant.

of

the

gas

by9

(1.93)

li=i-pXc

product

free path of its viscosity,

path

X increases

The viscosity

in such

a manner that

the

p. is, therefore, proportionalto

1-15

THE

OF

PROPERTIES

THERMODYNAMIC

PERFECT

t. Consequently, c, which depends entirely on the gastemperature with t. /i depends only on t and increases

a

for

41

GAS

gas,

perfect

the viscosity

Thermal

1-14(g)

Conductivity

The thermal conductivity in a gas arisesfrom spatial The latter cause the moleculesto transport the mean

internal motions. Accordingto the

kinetic

denoted

is given

k,

by

that

parameter

number

Prandtl

(1.94)

\\icv

in the study of viscousheattransfer

is important

Pr, defined

-

- pckcv s

is

problems

=

(1.95)

cph/k

equations 1.93, 1.94,and 1.95gives

Combining

=

Pr

1.96

2

2c -^

= -y

5 c,

is

gas

(y

=

A better

5/3).

approximation

by9

given

Pr

(1.96)

5

to a monatomic

strictly

applies

is the Euckenrelationship, which

=

(1.97)

^L-s

Equation 1.97givesthe sameresultas 1.96

a

for

for

approximation

1-14(h)

the

by

Pr

Equation

thermal

by9

k = A

random and of a gas, conductivity of their

energy

the

theory,

its temperature.

in

differences

other

gases

at ordinary

monatomic

gas

is a

and

good

temperatures.

Diffusion

Diffusionin a gasmixturearises in gradients to species comprising the mixture.According self-diffusion of a gas, denoted D, is given by from

the

kinetic

the

of the

concentrations theory,

individual of

coefficient

the

by

D =

(1.98)

/SDcA

intermolecular model.

a parameter that depends on the be employed for the diffusion of gasesthat are nearly also gases, a much more complicated relationship applies. where

1-15

ftD is

OF

PROPERTIES

THERMODYNAMIC

The thermodynamic state of a properties; p, v, t, u, h, and s. 1-15(a)

PERFECT

THE

perfectgas is

identical.

may

nonsimilar

For

GAS in

described

1.98

Equation

terms

of

the

following

Thermal Equation of State demonstrated It has of a singlechemical been

species

volume

v,

and

temperature

of

gas

t are

for a

that

experimentally of

molecular

homogeneous system, composed m, the pressure p, specific

weight

related by13

R = Limit p-o

(\342\200\224 )

\\t J

=

the gas

constant

42

PRINCIPLES

OF FUNDAMENTAL

REVIEW

The gas constant

For real gases, R varies only on the kind of gas involved. and because of the volume temperature gas pressure occupied by the gas of intermolecular moleculesand the presence forces. A gas for which R is a constant, or very nearly so, is termed a thermally gas. For such a gas perfect R depends

the

with

=

pv

Thedimensionless by Z.

as the

1.99 is known

Equation

ratio

p =

Rt, thermal

pv/Rt

p1T =

pRt,

equation

is termed the

mRt

(1.99)

of state for a perfect gas. compressibility factor, which

is denoted

Hence,

_~

p

pv

Rt~

pRt

The compressibilityfactor Z is, in general, a function of two of independent properties a simple thermodynamic system.Usually Z is plotted as a function of p/pcr for different constant values of t/tcn where pcr denotes the critical pressure and tcr is the critical the curve of Z temperature* For values of t/tcr > 2 or p/pcr < 0.05 approximately, = Z versus a horizontal line the value I. is p/pcr having Consequently, for practically relative and with that are low to that are pressures pcr temperatures high compared \302\253 Z it be that that the behaves in accordance with assumed I; is, tcr9 gas equation may 1.99. Because a gas is thermally does not imply that its specific heats cp and cv perfect are constant [see Section l-15(b)]. as the mole3, denoted by mol, is defined as the amount of The mass unit known a substance that contains as many elementary entities as there are atoms in 0.012 kg entities is Avogadro's number NA, discussed of carbon-12. That number of elementary m is the measure of the mass of a mole in Section of a l-14(d). The molecularweight are for substance. Thus, the units of molecular (mass per mole), example, weight in a volume if is denoted kg/kmol, g/mol, etc. If the number of molesof a substance is then mass denoted and the m, N, corresponding by by m

been shown empirically that R and the molecular weight

It has constant

\342\200\224 fnN

the

constant the gas as

gas

m of

R is related

to a universal

gas

follows.

m

1.99 may

equation

Thus,

be transformedto read

(1.100) of R

values

The

inside cover

are presented Substituting

for different systems of units book. The thermodynamic

this

of

m =

fnN

into

equation

* tcr pcr

of the by

NRt

on the gases

(1.101)

number

thermal equation of state for a thermally perfect gas in of moles N in the volume ir. Equation 1.101 is generally

chemists

and physicists

in

preference

to equation

1.99.

no matter how large a pressure is temperature above which a gas cannot be liquefied minimum pressurefor liquefying a gas at the temperature tcr.

is the

is the

common

1.101 expresses the

Equation

employed

are presented several

1.100 yields

p-T = terms

measurement

properties for

C.3.

Table

in

of

applied;

Caloric

1-15(b)

OF THE

PROPERTIES

THERMODYNAMIC

1 -15

PERFECTGAS

43

of State

Equation

The following functional relationship,known

written for the internal

of

energy

gas.

Thus

u =

u(t,v)

any

caloric

the

as

of state,

equation

may be

Differentiating the above expressionyields

It canbe

Hence,it

shown

in the

that,13

a thermally perfect gas,

case of

above that the internal energy of equations on its a in with gas depends only temperature, conclusion of gases [see Section l-14(c)]. Accordingly, a thermally perfect

perfect

the

from

follows

a thermally

u

the

harmony

theory

gas,

for

u =

u{t)

Combining equations 1.59and 1.102, gives du = 1.103, we

equation

Integrating

a thermally

for

perfect

gas,

dt

cv

(1.103)

obtain

u-

=

uo

dt

cv(t)

(1.104)

\302\243

t0 is

where

an arbitrary reference temperature where

1.104becomes

u

If t0

and u0 are

-

book,

form

only

uo. licv

is constant, equation (1.105)

t0)

chosen as zero, equation1.105becomes

having a constant value of cv

1.106is one this

=

u

cv(t -

u0 =

u

A gas

kinetic

of

the

a gas

caloric

is

equation

that is both

=

(1.106)

cvt a calorically

termed

of state and

thermally

perfect gas, withoutqualification.

the

Otherwise

perfect

for such a will

gas.In the

remainder

of

termed a perfect called an imperfect gas. will be

calorically gas

gas, and equation

be

1-15(c) SpecificHeatRelationships For

a

thermally

perfect

gas,

SQ

Differentiation

Law yields for a

the First =

cvdt

+ pdv

of the thermal equationof state

pv

pdv

Combining equation1.107 and

equation

reversibleprocess,

+ vdp = 1.108

Rdt yields

(1.107) =

Rt

gives

(1.108)

44

the case

For

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

where 8Q is addedisobarically (dp = cp =

)

Let

=

y

that

1.110

and

= =

into

cp/cv

(1.110)

equation

(1.111)

1) -

yR/(y

cP

y

obtain

R

+

cv

R/(y -

cv =

Substituting

0), we

heat ratio for the gas; thenit followsfrom

the specific

denote

cjcv

=

(1.112)

1)

1.110 gives

equation

**2/+i

'

c

c

(,,13,

f

f

number of degreesof in the as discussed in Section molecules, = a rigid for diatomic a 3, l-14(e). For a monatomicgas,/ 5, and gas, / = 6. to the kinetic Thus, according rigid gas, / polyatomic / is the

where

freedom

=

for

theory,

1.

monatomic

For

7 =

gases

1.67.

5/3=

2. For diatomicgases 7/5 1.40. = = 3. For polyatomic gases 1.33. 8/6 The above results are accuratefor monatomic gases, and least accurate for polyatomic gases. =

=

y

y

1-15(d)

Section 1

enthalpy h is denned in

property

thermodynamic

\342\200\224

dh = cvdt

Equation 1.114,

when

+

=

d(pv)

perfect gas, since

(cv

=

h-ho

cp(t

h

as

the

of

enthalpy

If its

datum.

tB

and

h0 are

value at 400 K is h

to be =

of state

for a perfect gas.

a perfect gas,

100.83

kJ/kg-K,

is measuredabove

calculate

value of cp.

Solution ~ h2

h2

u

K

hx

- u

=

cp(t2

-

tt)

100.83

400 -

=

300

1.0083

chosen (1.117)

equation

assumed

air,

(1.116)

t0)

cpt

caloric

the

of

form

=

Consequently,

l-15(b)].

1.115 becomes

where h = h0. If

where t0 is an arbitrary reference temperature as zero, equation 1.116becomes

Example1.5. The

(1.114)

(1.115)

equation

cp

useful

= cp dt

Section

[see

cv

1.117 is a

R) dt

P cp(t)dt

For a caloricallyperfectgas, is constant from equation 1.110, is also constant, and

Equation

+

yields

integrated,

h-ho=

K

gases,

A differential

1 l(e).

1.54. For a thermally change in enthalpy dh is given by equation = cv + R, equation = du 1.54 transforms to cv dt and cp

300

diatomic

Change

Enthalpy

The

for

accurate

less

kJ/kg-K

(0.2408

Btu/lbm-R)

the average

THERMODYNAMICPROPERTIESOF THE

1-15

Change

Entropy

1-15(e)

45

GAS

PERFECT

The entropy changefor a reversible and 1.65. 1.62,1.64, processis equations by For a thermally perfectgas, = Rt, du = cv dt, and dh = cp dt. Substitutingfor and 1.64, we obtain p into equation given

pv

du

Substituting for

v in

and

dh

1.65 yields

equation

ds = cpd*r-Rd^ t

1.119 yields

of equation

Integration

Jdt For a perfect gas,cp

=

cp

so that

constant,

In

R

t

s

=

In

cD

^R^dt \\y-lj

a perfect

gas,

is

y

y

t

Hence Furthermore,

1.119,

\302\273dp

t

(1.123)

p

-

1 (dp

y

\\p

1 In

(

of equation 1.124 yields

integration

/v - 1\\

=

p + constant

(1.125)

tp-iy-Diy= constant for

a perfect

gas,

pv =

Rt, so that

(1.126)

equation 1.126

may

be

transformed

read pvy =

Hence,ifa perfectgasundergoes temperature are related equation an

by

equation

1.127.

To

state, the supercript that

and

(1.124)

that

so

constant,

In

by

1.112

equations

which

t

to

(1.122)

J

from

dt

For

... + constant

,

change of state,ds = 0 and, dt ' t

from

(1-121)

U>

*J

|_p

For an isentropic it follows that

constant

+

= c. In

constant

+

P)

(1.120)

1.120 integrates to

equation

s = cplnt be transformed to read

1.121 may

constant

+

p

Rlnp

Equation

(1.119)

p

change

of state.

be

will

Hence, for an

change

isentropic

1.126,

that a state

indicate

prime (')

pp~y = constant

hereafter

and

(1.127)

its pressure

is arrived at attached

then

of state,

by

and volume are an

to the

isentropic changeof state

isentropic

property

from

t'2

(P2y

i

()

its pressure and

state

related

change

attained 1 to

state

(L128)

of by

2'

46

1 \342\200\22415(f)

PRINCIPLES

OF FUNDAMENTAL

REVIEW

Process

Equations

a gas can pass from

by which

Processes

least,infinitein number.In general, the

to another

state

one

are of

processes

specific

following

abstract at

are, in the

the greatest

significance in engineering.

1. The isothermalchange

=

(dt

adiabatic

2. The

0).

change (8Q = 0).

3. Theisobaric

=

(dp

change

0).

isovolumic change (dv = The isentropic change (ds =

4. The 5.

It

be

may

=

pv\"

from

recalled

be

will

state

0). that thermodynamics cases of the general

elementary as

regarded

constant.

0).

special

Thus, n = =

0 corresponds

n

=

oo corresponds

n

=

y corresponds

Flow processesthat takeplace

the exhaust

as the flow of gases through the nozzles of a turbine, the

such

rapidly,

nozzleofa rocket

to dt = 0 to dp = 0 to dv = 0 to ds = 0

1 corresponds

n

all of the abovechangesof so-called polytropicchange

bladepassages turbojet engine, are substantially an axial or turbine, air inlet ducts, and diffusers, compressor = of the extremely short time interval availablefor the adiabatic 0) because heat. Such are never isentropic (ds = 0)because however, processes, or

motor

of

(SQ

of

transfer

are

they

since

irreversible,

invariably

and eddies, all

wall friction,

of

they are accompanied cause

which

by

internal

that

losses

energy

turbulence,

friction,

the entropy a nozzle, through

increase

in discussing flow of gases of the fluid stream.Consequently, it is assumed asa approximation or turbine, tacitly compressor, of state are reversibleadiabatics,that is,isentropics, the resulting then modified to take into accountthe of irreversibility. the

the changes

that

first

equations

and

are

effects

Diagram

Entropy-Enthalpy

1-15(g)

Most gas dynamic processesmay

illustrated

be

the process on For a perfect gas,

by plotting

graphically

an entropy-enthalpydiagram, called a Mollier diagram. the of three however, h, and t are all directly proportional. Hence, may be The characteristics employed in constructing the diagram. general in the of the entropy-enthalpy diagram for a perfectgasare section. present state denoted Integrating equation 1.118 between a generalstateanda by generally

u

any

entropy-enthalpy

developed

reference

the

subscript

where

volume

o yields

constant v on the ts

s(v,t) =

of integration

the

+

(r/y

as

Equation1.130 areexponential shows

that

curves

on

the

of equation

= constant v'^1^\"

toivjvf-1^ curves

Lines

of

constant

1.129 by

specific

employing

Then

a parameter.

t =

(1.129)

(v/v0)

be zero.

means

by

v

R In

is chosen to

plane may be determined

constant values of

Integrating

In

cv

of v =

\\jp =

(1.130)

constant, termed v-linesor p-lines,

ts plane.

equation 1.119, yields

s(p,t)=

cp

In

(t/to)

-

R In (p/p0)

(1.131)

1-15

where the constant of integration is chosen

called eitherp-linesor

are

isobars,

t =

air

the

having

gas

perfect

from

= y

GAS

47

by

given

the abcissa

t and

temperature

PERFECT

of constant pressure,

Curves

zero.

be

to

calculated

curves,

is

thus

THE

to(p/poyy-i)lyeslc\" = constant

Figure 1.19presents

assumed to be a

OF

PROPERTIES

THERMODYNAMIC

(1.132)

1.130 and 1.132, for air, R = 287.04J/kg-K.The ordinate

equations

1.40 and

is the dimensionlessentropys/R.Thefigure

1500

$

2

1000

8.

500

I

Dimensionless

Figure

(p =

1.19

Characteristics of the lines

constant)

presents

in the

entropy,

I

17

18

and the isobaric

(also hs and us planesfor

a perfect

three curves of constant volumeratio v/v0

pressure ratio p/p0.It

I

16

s/R

(v = constant)

isovolumic

ts plane

15

14

13

12

11

10

=

pjp,

gas).

and

four curves

of constant

and 1.132 that, as stated earlier, of constant are than those exponential. v/vo steeper for constant p/p0; this is becausecp > cv, so that exp {s/cv) > exp (s/cp).At constant values of s, the values of p increase with the values of v = 1/p decrease. t, while = 294.4 K and The reference conditions for the curvesof are p0 = 1 atm. t0 for is arbitrary, it was chosen as sJR = 12 Since value s/R

both setsof

curves

from

follows

the reference convenient rangeof

values

1.130

equations

The

are

curves

Fig.1.19

is

obtained

for

s/R.

so that a

48

Perfect Gases

of Thermally

Mixtures

1-15(h)

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

do not interact chemically, of that Considera mixturecomposed gases air. that the constituent as for exampleatmospheric gases, hereafter termed each of the the species, are in thermal equilibrium other. Accordingly, of such a gas is at the t.The mixture temperature species mixture are determined in this section. for Daltori's law to be The conditionsspecified the requirements were if so that all the others absent, applicable; that is, eachspecies several

perfect

Assume

each

with

properties

thermodynamic

above

satisfy

as

behaves

the_

to the sum of the partial gas mixture, denoted by p, is equal _static jpressure pressurespt of each of the species. The partial pressure pt is defined asjhejrjressure -T by itself at that each species would exert if it occupied the entiremixturevolume that the mixture temperature t. For n constituents, it follows of

the

P=

l

of the

gas mixture, denoted by ~V, is equal species. The partial volume yt

would occupy

the

at

itself

by

to the

r Let m denote the

=

sum of the the

as

defined

p and

pressure

law;

Amagafs

is

of the

each

satisfies

perfect gases also

of thermally

A mixture

(1-133)

Pi

t ;=

temperature

that

is^the

partial

volumes

volume Y'i

of

each species

volume

that

t of the

mixture.Thus

rt

t

mass of the gasmixture.

Then

n

m

=

Yj

mi =

'''

ml + m2 +

+

mn

i=l

Let

the specific

v denote

volume of the gasmixture,andletp v

By

the partial

definition,

given

=

1

r

p

m

denote

its

density.

Then

\342\200\224 = \342\200\224

specific volume v(

the

and

partial

density

pt of

species ;', are

by

1

_

~

Pi For

rth

the

p; is

pressure

species

mt

of the mixture, having

obtained from equation

f

the molecular

weight

mh

the

partial

1.99.Thus,

(S) By

mass

the

definition,

fraction

Q is given by C^

Substituting for

w;

from

equation

1.135

TL =

PL

m

p

into

equation

(1.135) 1.134 gives

\302\273-*(%)-\"(\302\245)-~\342\204\242

where

R{

=

The

obtained

=

P

where p is the density of

the

gas

thethermalequationof state

for

the

is

R

is given

is

pressure,

\"

= E CtR,

pt\302\243

;=i

i=i

mixture.

Equation

a

gas.

perfect

'

(1.137)

CtRt

be put in the form of

1.137 may

Thus,

p = pRt

for the gas mixture and, in view

gas constant

effective

49

Thus,

\"

n? '

mixture

the

termed

brevity

1.136into equation1.133.

by substituting equation mt

where

for

the gas mixture,

p of

pressure

OF THE PERFECTGAS

PROPERTIES

THERMODYNAMIC

1-15

of

1.137,

equation

by

R=t

=

CiR> i

? m

= 1

,

of weight oil the gas mixture.Hence,a mixture as a the effective\" thermally thermally perfect gas having perfect gases_behaves 1.138. m defined jnolecular weight by equation For a mixtureof chemically the definitions and gases, reacting thermally perfect R in that case, results presented above are stillvalid.The effective constant gas

m

where

is

the

molecular

effective

inert

of the mass fractions however, is not a constant,since The tHermaTequation\"of state for a perfectgas, equation 1.101, which is repeatedbelow.

Ct changes.

each

=

p-T

Let for

NRt

by

(1.101)

becomes

1.101

i, equation

species

is given

basis,

of the ;'th species presentin the mixture.Then,

of moles

number

the

denote

iV;

molar

a

on

Pt-r = NtRt

Combining the aboveequationandequation1.101

gives

-=~

p

where

XA

mole

isjhe

of the

fraction

= *;

N

ith species, and, N

=

t

i= 1

In terms of the partial

volume

Vh

above

the

fraction volume

The

the partial

relates

1.139

Xh and equation V to the

mole fractions

Xt

written as

NtRt

V-

N-

^




Thus,

p

-

=

Kin\342\200\224

obtain

we

(1.149)

Rlnp

has p p N/m2 At

\342\200\242

1300

105

\342\226\240

105

N/m2.

of increasing

the

= increasing the pressure from the values of cp. is

N/m2

1500 K, influence on

little

105

689.5

temperatures,

= 1034

effect

the

above

the

= 0 to p

from p

\342\200\242

68.95

more

of the temperature

a function

as

air

for

cp

ranging

pressures

that, except at

show

curves

The

is important,

in many practical engineering applications. It accurate data on its thermodynamic properties. Measurements effect of temperature on cp for air is appreciable, but the influence fluid

important

than

less

1 percent at t

0.31

0.30 10,000(689.5-106)

0.29

1200

5000

0.28

(137.9- 105)

-2000

-1000 (68.95-106)

f

0.27

E

Z

(344.7-105)

-500(34.47-105)

noo 0.26

I

-p

=

0

0.25

0.24

1000

0.23

0.22

3000

2000

1000

900

Temperature t, I

I

I

I

1500

1000

500

1.20 Specific heat cp for air as a function as a parameter (basedon Reference 16).

of

Figure

1.21 presents

the specific heat

pressure p as a parameter.The

pressureon

y

is

significant

Because of the

assumed that

the

relatively

instantaneous

(p = 0). Moreover,that

only small

curves

at low

as

air

for

a function

expect, that the

would

one

of t, with the

effect of

temperatures.

influence

values

as

show,

y

K

with pressure

temperature

ratio

2500

2000

Temperature t, Figure

4000

R

of

of cp are

conclusionis

reasonably

defined

for air, it

on cp

pressure by

correct

the

curve for

most

is generally

for zero

pressure

gases.

Conse-

1-16 IMPERFECTGASESAND

GAS

55

TABLES

1.60

for

1.50

= 4000

p

Ibf/in.2 (275.8

\342\200\242

106

N/m2)

(137.9- 105)

2000

1000 (68.95\342\200\242 105)

1.40

1.30 3000

2000

1000

Temperature t,

4000 R

I

I

Figure pressure

Temperaturet, Specific heat ratio y for air parameter (based on Reference

1.21 as a

2000

1500

1000

500

2500

K

as a function

of temperature

with

16).

the experimental data on cp and y for gases of the gas temperature alone.18\"20 assume quite generally that cp is a function for correlating Several equationshave been the empirical cp data for proposed t. different as functions of the gas temperature polynomial, gases Equations having and also exponential, in forms are available. The correlatingequationspresented this are based on polynomials of the followingform: book quently,

the

equations

for correlating

= cp

{a +

bt +

ct2 + dt3 +

etA)R

(1.158)

values of the coefficients a,b,... cp has the same units as R. Table 1.6presents 21 were for several pure gases.21The specific heatdatain Reference obtained by a least squares curve fit of the data in Reference 20. The curve fits were obtainedfor two the low temperature interval extending from intervals, temperature adjoining to 5000 K. 300 to 1000 1000 K, and the high temperature interval extendingfrom The data are constrained to be equalat 1000 K. Curve fits for 483 substances are in Table 21.Also presented 1.6 are values of the properties presentedin Reference

where

h0

and

o

-1.20198 3.61510

-0.00693536-14.2452

0.103646

-0.578997

R

-1.061162.35804

-0.00652236

-2.03197

1.48914

of

-1.04752

2.15560

0.0998074

units

V10\"3

0.0362016-0.00289456

-1.61910 3.69236

the

cp has

e-1012

-6.76351

1.51549-0.572353

2.98407

t is

2.32401 -0.632176 -0.225773

-1.20815

2.89632

+ et*)R,

d- 109

3.62560 -1.87822 7.05545 3.62195 0.736183 -0.196522

3.71009 CO

-

b-103

3.67483 N2

Coefficientsin the Equationforthe Specific Heatat Constant

Gases

Several

for

cp

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

300-1000

1000-5000

300-1000

-0.322700

1000-5000

6.63057

0.866901

10.7071

300-1000

1000-5000

300-1000 2.69879 1000-5000

14.6218

to be

that tabulated

below.

Chemical

Molecular

Composition,

Species

Formula

Nitrogen

N2

28.016

78.11

Oxygen

O2

32.000

20.96

Argon

A

39.94

Weight

Percent by

Volume

0.93

the specific heat data presentedin Table1.6and determine at constant pressure of air as a functionof temperature. Employ

Solution

For

air,

equation

1.144 on 3

a molar basis, gives

the

specific

heat

AND GAS

GASES

IMPERFECT

1-16

TABLES 57

Introducing the values presentedin Table1.6

yields

Temperature

103

Gas

N2

XiCi

-0.943686

2.87041

o2 A

0.759926

Air

3.65359

\342\226\240

10\"

106

&

-0.493793

1.81528

-0.393675

1.47882

-1.33736

3.29421

\342\226\240

1012

K

Range,

300-1000

-0.176351

-1.41763

0.451813

-1.91142

0.275462

300-1000

300-1000

0.023250

300-1000

N2

2.26232

1.18375

-0.447065

0.0779596

-0.00509462

1000-5000

O2

0.759161

0.154304

-0.041191

0.00758786

-0.00060670

1000-5000

A

0.023250

Air

3.04473

From

above

the

1000-5000

table,

= ( 3.65359

cp

in the range

1.91142c3

0.275462c4

109

1012

106

103

(b)

K

5000

0.0855475c3

0.488256c2

+

+

3.04473

1000-5000

K

3.2942k2

1.33805c (

-0.00570132

0.0855475

300 to 1000

1.33736*

-

and in the range 1000to cp =

-0.488256

1.33805

where, in both equations, is in K. The units unit mass basis, cp = cp/m.Values of R in cover of this book.

of R determine

c

_

the units of

On

109

106

103

0.00570132c4\\

units

different

are

cp.

on the

presented

(c) a

inside

1-16(b)GasTables pertinent to high-temperature gas dynamic the for heats of gases presented in Table 1.6are too processes,the equations specific use. cumbersome for Considerable labor is saved by employing tabulated general values of the pertinent thermodynamic properties of the gases, asis done for air in For

the

making

Table

C.4.

It is shown in cv

dt

analyses

thermodynamic

and

=

dh

Sections cp dt.

1\342\200\22415(b)

and

Consequently,

1\342\200\22415(d)

the values of

tabulated as functionsof temperature the employedin calculating and h. Substituting if

u

h

where Section

h =

h0R at

14-3(a)],

are

=

C

h0R =

t0.

+

f' (a

Values

presented

= [h

at

for

that

u

a thermally and h for a values

instantaneous

1.158 into equation

equation

+ bt + ct2 +

perfect gas du = real gas may be of cv and cp are

dt3 +

dt

et*)R

1.115gives (1.159)

including the energy of formation[see in Table 1.6. Integrating equation 1.159yields of ho,

bf_

2

d?_

et~

R

(1.160)

58

OF

REVIEW

From

PRINCIPLES

FUNDAMENTAL

1.53 and 1.99, we obtain

equations

h-pv

= h-Rt

presented

in Table

=

u

When the values of

1.160

b, etc.,

a,

h0,

(1.161) 1.6 are substituted into equations

Table C.4 presents the valuesof u and h, and accounting cv cp, for air at one atmosphere pressure as a function Tables of for several other gases are presented in Reference22. temperature. The entropychangeds for a perfect gas is given by equation 1.119 and is repeated and

determinable.

u are

h and

1.161,

for variable

here

convenience.

for

Thus,

ds =

Equation 1.162

be

may

both

its

temperature values

entropy

from equation and its pressure.

is apparent

It

temperature.

If

t0

s

where

s-= s0 =

p0 are

and

=

(1.162)

p

gas by expressing cp as

a functionof

entropy of a gas depends Consequently, the compilationof a table 1.162 that the

on

of

them at different temperatures for severalvalues the base temperature and pressure, respectively,

calculating

requires

of constantpressure. forthe entropytable,

where

a real

to

applied

cp%-Rd-^ t

0, then

0 and the parameter(f>

is

-

0

Pc,\342\200\224-Kin\342\200\224= J'\302\260 t

Kin

\342\200\224

defined

(1.163)

po

p0

by

dt

(1.164)

and is a

function only of the gas temperature.

equation

1.164 gives

=

fa

s-t

4>0R

+

- + b +

ct + dt2 +

\\

J

1.165 yields

equation

4>

where

cj>

=

values of

4>

oR at for

=

t =

Uo

+ a

In

t

+

t0. Values of 0

bt

are

+

^-

^

+ ~\\

in Table

presented

is calculated from s2 - Sl =

an isentropic

(p2

R

(1.166)

1.6. Table

_

p

the subscript o

isentropic process,

the tabulated

(pi

to denotethe

cp

R base

of

values

(f>

- R In (\342\200\224 )

process {ds = 0), equation dp

Employing

+

C.4 presents

air.

in entropy change 1.163. Thus, equation A

For

(1.165)

et3)Rdt

J'o\\t

Integrating

into

1.158

equation

Substituting

by

means

of

(1.167)

1.162reducesto

dt

(1.168)

t condition

for

the

table,

then, for an

1-16 The pressure ratio pr = p/p0is termedthe valued functionof the Solving

relative

temperature

and

pressure

t.

equation

to be

is seen

for pr

1.169

Pi

= =

tt and

temperatures

t2, the pressure

= Pr2 =

JPllPo)

( (

Prl

(Pl/Po)

Pi

1.170

(1.170)

by

given

Thus,

the

between

state

of

change

a single-

gives

pr = e*1*

For an isentropic ratio p2/Piis

59

TABLES

GAS

AND

GASES

IMPERFECT

the relative pressuresare of significance.

the ratios of may be written as only

equation

Hence,

e4.36600E+00\302\273

9.69515E+00,

a molar

Calculate

SI(1,2)==

4.30528E+00,

**.36600e

-1.23926E-U6,

2. 27413E-10,

HI(4,2)= -4.896l**E

= 2.35804E+00,

SI(3,1 =

2. 00219E-09,

HI(3,2)=-7.45375E+02,

+ 04,

-1K83775E

o# OOOOOE+00.

-6.60709E-06,

HI(1,2) =-9.05862E+o2, HI(2,2)=-l,20198E+03,

=-7,45375E+02,

SI(2,1 =

modified

3.09817E-03,

0 0

EE

units,

for is

of cp, h, and THERMO.

W(l)=28.016,

X(2 )=0.209*T**4/4.0+Alb,J.K)*T**5/5.0>*R*GT

SF=X*(SI-089 = 0.8

r\\c

=

138.861 kJ/kg

kJ/kg

(47.760

0.8

(59.700

Btu/lbm)

Hence,

h2

=

ht +

Entering TableC.4at

h2

Ahc

=

=

440.190

301.329

kJ/kg,

+

138.861 gives t2

= 440.190 kJ/kg

= 437.43 K.

Btu/lbm)

=

62

From

(c)

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

1.167,

equation

-

=

As

- R In

0!

'2 =

and temperature Calculate

=

(3.0)

K

=

410.12

K,

Hence,

kJ/kg-K.

2

1.9.

Example

7.0844

C.4, 02 = 7.0189

-

=

an

for

(f>

kJ/kg-K.

Table

and from As

of

value

the

be

also

may

being

In

0.28704

=

7.0189 = 0.0655kJ/kg-K(0.01564

Btu/lbm-R)

is compressed isentropically from an initial 300 K, respectively, to a final pressure of and the compression work (a) assuming that the and (b) accounting for the variation are constant,

pressure 50atm.

temperature.

Solution

Table

From

(a)

C.3, m = 44.010,

845.73

=

l'2

/V

Ah'c

From

=

fl

(t

)

- ti) = 845.73(719.5 = 300 Table D.l.B, Volume 2, for K, Ji1 =

h2

=

-K

cp(t'2

From

kJ/kmol-K.

300)

1.167,

equation =

\342\200\224

for s2

sx

In

+ 8.3143

213.927

\342\200\224

j

Table

D.l.B,

2, for

Volume t'2 =

A*

=

4>2

1.10.

Example

A

15-m3

246.453

and

643.7 K

^A

=

(y j

=

354.8

kJ/kg

and 0t =

= 69 kJ/kmol, 0, we obtain

= 246.453

kJ/kmol-K

kJ/kmol-K, W2

=

15>029~ 69 =

kJ/kmol

15,029

339 9

44.010

m

air at

contains

tank

105 pt = 5.0 \342\226\240 N/m2

dischargedinto the atmosphere contained in the tank is reducedto of its the process is adiabaticand frictionless, air remaining in the tank. Take into account The air is

through

half

one

calculate

the

with

From

1.288.

K

=719-5

30\302\260

tx

213.927

air

=

y

/cn\\(l.288-1.0)/l.288

\\ly-Vly

UJ

From

and

J/kg-K,

1.128,

equation

(b)

= cp

a

until

nozzle

original

and

value.

pressure

and

variation

of

the

=

500 K.

mass

of air

fx

that the of the temperature the specific heat of Assuming

temperature.

Solution

to be isentropic.FromTableC.4,for tx = 500 K, vrl = 23,124 and prl = 6.2062. Sincethe final mass of the gas in the tank is onehalfoftheoriginal the final volume mass, specific v2 = 2v1. Hence,

The

process

is assumed

Vr2

=

2vrl

=

2(23,124)

= 46,248

1-17 From Table C.4,for final

the

pressure

\342\200\224

\\PrlJ

=

5.0

)

\342\200\242

105

a compressible and speed,

acoustic

\342\200\242 ) = 1.9058 105 N/m2(27.64

9^

(

THE ACOUSTIC SPEED AND For

Acoustic

flow

the

Those

process.

acoustic

physics)that the acousticspeed,

denoted

called the

where the isentropic Hence, dp/p = equation bulk

is given

a,

by

modulus

(see any

text on elementary

by

It is pointed

in

out

will

is focused

on

depend

in

the manner

on the main

gives the

=

Ks

a

=

perfect

y = cp/cv.

following expression for the

at which a

sound

depends on its length. that impartsdirectional wave

to

energy

or

wave

The

in the form

=

small

to the

external

if the

Introducing Ks =

(ypv)112

pressure fluid

the

fluid

acousticspeedin a

=

(11)

and

gas,

W

The speed

the

fluid,

a

l/v, and

if a fluid

that,

body of compressible

yp where

p =

density

If

layer, and the fluid is assumedto be

isentropic,then

the

v(dp/dv)s,

is compressible,the magnitude which the compressionis executed. the attention

1 -5(d)

Section

\342\200\224

be written

may

the

When

medium.

fluid

a

a sound

which

with

speed

(U79)

=

Ks

1.179

\342\200\224dv/v.

K

are

properties

7)

of

the

called

Speed

speed, also called the sonic speed,is the or a small in is propagated wave, pressuredisturbance, fluid may be consideredtobea continuum, it can be shown The

lbf/in.2

of smalldisturbances, flow, the speed of propagation the ratio of the flow to the acoustic velocity speed, of

The

Hence,

NUMBER

MACH

THE

Mach number,are importantproperties defined and discussedin thissection.

1-17(a)

and t2 = 381.05K (685.9R).

= 2.3655

pr2

46,248,

63

p2 is:

Pi = Pil

1-17

=

vr2

MACH NUMBER

AND THE

SPEED

ACOUSTIC

THE

boundary

compression process is yp

perfect

into

equation

gas.

Thus

(1.181)

(yRt)1'2

pressure

disturbance

travels in a fluid

creates a disturbance in in the immediate vicinity particles wave

1.179

the fluid of

the

the contiguous particles, the energizedparticles to them and, by that process, the disturbance is propagated energy impartdirectional of fluid. the collisions between the fluid particles the body Briefly, throughout constitute the mechanism which the by pressure wave propagates itselfand givesdirection to the motion of the fluid particles. Since the speed of propagationofa smallpressure translational is of the same order of magnitudeas themean disturbance wave) (sound is disturbance of the fluid particles, the transmission of a smallpressure speed influenced density of the fluid. by the molecular disturbance.

By

colliding

with

64

PRINCIPLES

FUNDAMENTAL

OF

REVIEW

is of the molecules density is large, so that the mean path in the medium with small a small disturbance i s very small, pressure in air The of a disturbance loss. foregoing would applyto energy propagation sea-level air contains at standard sea level, example, where 1 m3 of standard molecularmean and the corresponding value for the 2.705 1024 molecules path 10~8 m. At increasing the molecular of is 7.37 altitudes, density atmosphericair decreases as for example in the ionosphere, it is and, extreme altitudes,

molecular

If the

free

propagated

the

for

\342\200\242

free

\342\200\242

at

vanishingly

small.

where the molecular density is molecular mean the small, be that it so is or than the wave of the large may equal longer length If that is the case, a large number ofair molecules sound canmove,ina distance 1 wavelength, from the high-pressure region of the sound to to its lowwithout with other B ecause t he and pressure region colliding particles. in the high- and low-pressure regions ofa are the transport different, pressures of molecules from a high-pressureregionofthe to a low-pressure region without collisions tends to equalizethe pressuresand temperatures the wave. The net effect is a dissipation of the it to become damped. Since of the wave, causing the length of the free I governs the propagation of sound waves, it is path the criterion distinguishes the realm of fluid mechanics (continuum) from that At

the

altitudes

high

very

free path

to

wave.

wave

equal

temperatures

wave

wave

in

energy

mean

that

of freeparticle

flow.

The propulsion,

the

when

acoustic speed Section

of fluids, jet speed in the flow compressible and arises from the fact that the supersonic flight phenomenaoccurring relative between a fluid and a body is largecanbe related to the velocity Section discusses the a ssociated with the 3-8(b) speed. briefly phenomena of small pressure disturbances in a compressiblefluid. propagation a rigorous derivation of equation 1.180. presents

of the

importance

of 3-8(a)

acoustic

1-17(b) The Mach

Number

fluid large relative speed between a body and the compressible with of the fluid, the variationofitsdensity it, the compressibility speed, surrounding influences the properties of the flow field. The ratioof the local speed of the fluid V, = called M to its acoustic the local Mach number speed a, V/a, is a dimensionless

Where

there

is a

criterion of the

flow

M

a perfect

For

phenomena. V

V

a

( yRt)112

gas a =

(yRt)112,

so

that

(1.182)

or V2 M2

V2

directed

j random

kinetic energy T-\342\200\2247

kinetic

(1.183)

energy

The physical significance ofthe Machnumbercanbe readily grasped by considering 1.183. The velocity V measures the directed motion of the gas particles, equation and V2 measures the kinetic a for energy of the directed flow. The acousticvelocity a given to y/7, which is proportional to the random velocity gas is proportional of the gasparticles a2 is a measure of the kineticenergy Hence, (see equation 1.86). associated with the random motions of the gas molecules.Consequently, M2 = V2/a2, set of conditions, may be regardedasa measure for a given of the ratio of the kinetic to the kinetic energy of directed fluid flow energy of random molecular motion.

1-18 PROPERTIESOF THE

flows through a

A gas

1.11.

Example

static temperature is 1800 its R = 322.8J/kg-K.Calculate

passage

sonic

local

the

ratio

heat

specific

K,

a

with

y

800 m/s.

Its local

constant the Mach number. the gas

and

1.25,

and

velocity

65

of

speed

=

ATMOSPHERE

Solution

From equation1.181,

a=

=

(yRt)1'2

=

[1.25(322.8)(1800)]1/2

From equation1.182,

800

=

852.2 1-18

Atmosphericair is one

the

of

dynamics.It is

of

value,

of the atmospheric air, the

Since

to

therefore, p

a

is

the

while

depends

viscosity

and

pressure

and

at a

atmosphere

of of

properties

physical

on the atmospheric

only

of atmospheric air are, however, it has become the common Earth, air as functions of the atmospheric

temperature

the

above

the

gas

function

surface functionsof the altitude practice to expressthe altitude rather than of its pressure of

0.9387

mixtures pertinent to the scienceof some of its properties. of both the pressure and temperature

gas discuss

important

The atmosphericdensity

temperature

(2796.0 ft/sec)

OF THE ATMOSPHERE

PROPERTIES

temperature.

852.2 m/s

temperature.

given altitude change

Since

the

and

pressure

somewhat

the

with

latitude,

the interests of in day, and weather,it has becomenecessary to adopt the so-called International Standard Atmosphere standardization and convenience Based on that as a standardof reference. which has been adopted by standard, the National Aeronautics and Space Administration all major countries, practically has a series of empirical equations for definingthe temperature, (NASA) developed air as functions of the altitude.23 Table 1.9 and of atmospheric pressure, density the basic of the standard atmosphere, and Table 1.10presents characteristics presents asa its chemical Table C.5 presents the properties of the atmosphere composition. It should be noted that the reliabilityofthetabulated function of geometric altitude. the for altitudes exceeding 20,000 m has notbeencompletely of properties atmosphere

season,

time of

established.

Basic

1.9

Table

Characteristics

of the Sea

Level U.S. StandardAtmosphere2

SI Units Temperature

Pressure

Density

Gas

constant

Specific heat

ratio

to

288.16K

Po

1.01325105

Po

1.225

R

287.04 J/kg-K

53.35

ao

Viscosity

P-o

Mean

viscosity

Vo

59 F

\342\200\242

7 =

Acousticspeed Kinematic

cp/cv

N/m2

kg/m3

Molecular

weight

2116.22

lbf/ft2

0.076474

lbm/ft3

ft-lbf/lbm-R

1.40

1.40

340.29m/s

1116.4 ft/sec

1.4607-

1.572310\"4ft2/sec

1.7894-10\"5kg/m-s 1.2024-10~5lbm/ft-sec

=

P-o/Po

10\"5m2/s

free path

of air molecule

EE Units

K

6.6328-

m0

28.9644

10\" 8m

2.1761-

10\"7

28.9644

ft

66

REVIEW OF

FUNDAMENTALPRINCIPLES

Table 1.10 Compositionof

Clean

Near

Air

Atmospheric

Dry

Sea Level23

Constituent

Percent

Nitrogen

by

Volume

78.084

20.9476

Oxygen

0.934

Argon

0.0314

dioxide

Carbon

Hydrogen

0.00005

Neon

0.001818

Krypton

0.000114

Xenon

O.OOOOO87

Helium Methane Nitrousoxide

For

0.0002

0.00005

one layer

separating

four

into

divided

be

may

layers having markedly different characteristics.Eachlayer shell of different thickness, but there is no sharply dividing

spherical

surface

surrounding Earth

the atmosphere

convenience

gaseous a

0.000524

approximates

bounding

from the next. Instead,the characteristics of one

layer

next layer.The characteristics each of the four merge gradually distances from the surface layers are, however, distinctly different.In theorder their of Earth, the layers are calledthe troposphere, the stratosphere, ionosphere, and All of the layers have the commoncharacteristic the exosphere. that,as moves farther and farther from the surface of Earth, the density the gas in each layer continues to diminish and, at extremealtitudes,it approaches The characteristics of atmospheric air are subject to wide variations level sea to extreme altitudes. At the lower altitudes, the of molecular collisions is very large, and the air behavesas a continuum As the altitude Section 1-4). is increased, the mean and the frequency of molecular collisions increases, path at theextreme air is a continuum. no diminishes; and, altitudes, longer the lower regions of the atmospherecollides a body through moving a of air molecules, some of which rebound and collide other number large to and some of which are deflected into the path of the molecules, body the At altitudes above it that be assumed the 150,000 m, repeat process. free is that in such regions there can be no practically space, signifying into those of the

of

of

the

one

of

zero.

from

frequency (see

free

the

with

Consequently,

with

moving

may

atmosphere

resistance

aerodynamic

to

nor

flight,

can there

be any lift.

where the atmospheric that, Analysesofthe problemsof high-speed and surface heating largeat density is substantial, the drag becomes high speeds, the arises owing to frictional are removed boundary layer. These limitations if the conducted several at altitudes kilometers above the surface of Earth. indicate

flight

very

effects

flight

in

is

The propulsion

engines

for

such

accomplishing

consuming engines.Currently,rocket for conducting such flights. jet

propulsion

is

however,

cannot,

flights the

only

be air-

conceivable

means

REFERENCES

1.

Dimensional Analysis, Yale University Press, and Formsof \"Model Buckingham, Experiments Empirical the American Society of MechanicalEngineers, Vol. 37, 1915. P. W.

2. E.

Bridgman,

New

Haven,

Equations,\"

1922. Transactions

of

67

PROBLEMS

3. E. A.

National Units,\" NASA SP-7012,Second Edition, and Space Administration, 1973. of Measurement Systems,\" NBS SP 304A, National Bureauof Standards,

Aeronautics

4.

5.

\"Brief

History

October1972. H.

Mechanics

Advanced

Rouse,

New

The Nature of a Gas,

7. L.

B. Loeb,

8. R.

L. Sproull,Modern

10. J. F. Clarke

M.

and

pp.687-692,

12. J. T. R.

New

The

1931.

York,

1956.

York,

Dynamics of

Her

Edinburgh,

Fundamentals

2,

Chap.

Wiley,

Real Gases,Chapter2, Butterworths,

H.

J.

edited by

H. Emmons,

N.J., 1958.

Office, 1972.

Stationary

Majesty's

Dynamics,

Units,\" National EngineeringLaboratory,

in Metric

of Gases

Gas

of

Princeton,

Press,

University

\"Viscosity

Watson,

13. G. N. Hatsopoulosand

of General

Principles

Keenan,

Thermodynamics,Wiley,

New

1964.

York,

Page, Introduction to TheoreticalPhysics,

15. K. K. Kelly,

of Mines,

Bulletins 476,

Experiment

Gay,

Vasserman, Y. Z. Kazavchinskii, Air

and

\"The

Components,\"

1949;477,1950;and Specific

N.J., 1959.

Princeton,

Nostrand,

1960.

584,

of Certain

Heats

Gases over Wide

Bulletin 30, Cornell University

Engineering

1942.

October

Station,

Van

D.

and Temperatures,\"

Pressures

of

Ranges

17. A. A.

Bureau

U.S.

Ellenwood, N. Kulik,and N. R.

F. O.

Air

Shapiro, McGraw-Hill,

Kruger, Introduction to Physical Gas Dynamics,

Schaaf and P. L. Chambre,

S. A.

V. A.

and

\"Thermophysical

Robinovich,

National Technical

TT70-50095,

Properties of

Information Service,Springfield,

1970.

Va.,

18. T.

New

Wiley,

McChesney,

Princeton

16.

York,

A. H.

1964.

London,

14. L.

1959.

New

2 by

1965.

New York,

11.

Section

Mechanics,

Wiley,

Physics,

C. H.

and

Vincenti

G.

Fluid

1961.

York,

W.

Chap. 2, Wiley,

of Fluids,

L. Streeter, Handbook of

6. V.

9.

System of

International

\"The

Mechtly,

J. L. Duran, R.

P. Thinh,

Predictions,\"

Fifth

W.

and

pp. 98-103,

by R.

\"Equations Improve C*

S. Kaliaguine,

January 1971.

Chemical Data,\"ChemicalEngineers Handbook, H. Perry and C. H.Chilton, Section Hill, New York, 3, McGraw

R. Gambill,

edited

Edition,

and

Ramalho,

Processing,

Hydrocarbon

19. P. E. Liley

S.

\"Physical and

1973.

20. D.

R.

et

Stull,

21.S.

Reference B. J.

and

Gordon

Equilibrium

J. Kaye, Gas Tables,Wiley, \"U.S. Standard Atmosphere, 1962,\" Superintendent December 1962. Office, Washington, D.C., J. H.

Reflected and

Shocks,

and

Space

1971.

Administration,

23.

Rocket Performance,Incidentand Detonations,\" NASA SP-273, National Aeronautics

Compositions,

Chapman-Jouguet 22.

Thermochemical Tables, Second Edition, NSRDS-NBS37, Data Series, National Bureau of Standards,June 1971. of Complex Chemical McBride, \"Computer Program for Calculation

JANAF

al.,

Standard

National

Keenan and

New

York, of

1945. U.S.

Documents,

Government

Printing

PROBLEMS

1.

The

mass

gravity

of a body at sea level, at at sea level is given by g0

where

6 is

(b) at the

=

the equator,is 10,000 kg.

9.8066

because

acceleration

The

of

- 0.0259cos 26m/s2

the latitude in degrees. Calculatethe weight of the body North Pole.

at

sea

level (a)

at

45\302\260 latitude,

2. Derivean equationrelatingthe weight of a body at any altitude z to its weight Wo at sea level. and go is its sea-levelvalue, Plot a curve oig/go, whereg is the local acceleration gravitational as a function of the altitudez. At the equator the diameter of Earth (2R) is 12,756km. in Table 1.3 (a) at 10atm and 298K, 3. Calculate the mean free path X for the gases considered and at 0.1 atm and 298 and at 1atm 596 K. K, (c) (b)

68

REVIEW OF

FUNDAMENTALPRINCIPLES

4. Most solid a additive,

powdered aluminumas which fuel results in A12O3 appearing in the combustion products.At the in solid in the liquid phase exists temperatures occurring typical propellant rocket motors, A12O3 as very small spherical Those particles are accelerated by the drag force exerted on particles. them by the acceleratinggas.Determine or not rarefied whether gas-flow effects should be the in Assume that the mean the nozzle. expectedduring expansion process propulsive = 1 micron the at a point near the particlediameter (1CT6 m), and that gas properties Dp = 980 K, y = 1.30, exitof the propulsive nozzle are V = 3000 m/s, p = 0.35\342\226\240 105 t N/m2, \342\200\242 = = = R 430 J/kg-K, and p. 3.0 10\"5 The particle velocity Vp 2000 m/s. kg-m/s. \342\200\242 A cubic meter of water originally at 293K and a static pressure of 1.013 105 N/m2 is subjected to a pressure of 4826\342\200\242 105 walls of that the the containerarenot deformed N/m2. Assuming is and that there no change in the temperatureofthe water, of the calculate the final volume water. for

propellants

5.

6. The equation

A

variable

of air as a N-s/m2, the viscosity n of

in

Calculate

1.30.

in m2/s,

7.

p.,

viscosity

motor

rocket

y

moving

of F

circulation

the

about

air,

the results.

Plot

= 2x, in

m. The x and y

move the particle from of F at the point where of

the

the x

lines x

where

point =

\342\200\224 \302\261 2,

x = 1m

to that

2 m.

along a given

velocity

tangential

by the

- 2y

10x2

=

y

1 for \302\261

path. Find the

a two-dimensional

components =

u

v are

Y =

and

2xy

enclosed

rectangle

flow having the velocity

u and

of

are given by

Calculate (a) the work required to = the where x 2 m, (b) magnitude 8. The circulation T is the line integral

\342\200\224

x

a circular

v =

and

y

the components, in m/s, in

9. Calculatethe circulation around

x and

the

x2

y

respectively, in m.

y directions,

of radius

path

\342\200\224

R about the centerof a

free

vortex

= constant).

(VR Air

km.

30,000

X = x2 +

10.

function of the temperature t in K is given by v air, in N-s/m2, and the kinematicviscosity

as a function of the altitudez forthe range z = 0 to force the curve F, in N, acts on a particle along

components

where

contain finely

applications

is allowed

to expand

to state

B (wherepB

x,

coordinate

from an intial

state A

pA

(where

=

2.068

\342\200\242

105

N/m2

and

tA

=

333 K)

N/m2 K). Calculate the changein the specific of the air, and show that the changein entropy is the same for (a) an isobaric process entropy A to some from intermediate state C followed by an isovolumic from C to B, and change from A to some intermediate state D followedby an isentropic (b)an isothermal change change from D to B. 11. The specificheat at constant of a body is determinedby heating it internally with pressure an electric coil connectedto a 12V source of electricity. When the current flowing through the coilis2 A, the temperature of the mass of the body coil increases plus by 30 K in 2 min. If the massofthe system is 0.25 kg (includes mass of the coil) and the specificheat of the coil and the body are identical, calculate the mean specific heat of the body. 12. The specific heat of air at constantpressure, in 300 to 1000 K, is given by equation the range in 1.7. How are to heat 5 kg of air at constant pressure (b) Example many joules required from 300 K to 350 K? (Check the answer by means of Table C.4). 13. During a process,9000J of heat leaves the system and enters the surroundings, which are at a temperature of300K.(a) If the change in the entropy of the system is \342\200\224 is the 30.0 J/kg-K, If the in the entropy of the system process reversible, irreversible, or impossible? (b) change is \342\200\22420 is the process reversible, irreversible, or impossible?(c) If the change in the J/kg-K is \342\200\22440.0 is the process reversible, irreversible, or impossible? entropy of the system J/kg-K, 14.A body with a mass of 500 kg movesso that its component in the directions of the velocities y,

z

=

axes

1.034

are

\342\200\242

Vx

components of its momentumin in

kg-m/s.

and

10s

=

150 the

m/s,

directions

tB = 305

= Vy

200

of the

m/s, and

coordinate

Vz

=

250

m/s.

Calculate

the

axes and its total momentum,

69

PROBLEMS

a mass of 50 kg has a velocity of 20 m/s in the positive direction of the y axis. a body having a mass of 75 kg and moving with a velocity of 30 m/s in the negative directionof the y axis. What are the velocities of the bodies after impact,

15. An object with It collides with a

assuming

16.

collision?

elastic

perfectly

in a

of 1.0 \342\200\242 105 N/m2 centrifugal compressor from a pressure to a pressure of 6.0 105 N/m2. The initial temperature is 290.0 K. UsingTableC.4, calculate in temperature, (b) the changein internal (a) the change (c) the work imparted to the energy, the and the value of the specificheat cp for the air, neglecting velocity change, (d) average Air

is compressed

isentropically \342\226\240

process.

compression

17.

is

Air

an insulated

in

expanded

temperature of the airis 1400K.The calculate

the

(a)

cylinder equipped

initial

The

piston.

the final volume. UsingTableC.4, the gas, and (c) the pressure

(b) the work removedfrom

in temperature,

change

is 1/10

volume

original

frictionless

a

with

ratio.

18. The isentropic

is of a turbojet engine is 0.86.The pressureratio at altitude, the work for Calculate 10,000-m operated (a) per kg the compressing the air (neglect velocity change), (b) What is the exit temperature? of the turbine of a turbojet engine 19.The isentropic is 0.90. The pressure ratio of the efficiency turbineis 3.0.The inlet temperature to the turbine is 1000 K, and the unit is operated at Will this drive the compressor and auxiliariesof the turbojetof turbine 10,000-m altitude, (a) Problem 18? the velocity (Neglect change.) (b) What is the temperatureof the gas leaving

compressor

is being

the unit

and

the

of

efficiency

5.2

turbine?

20.

from cycle consists of an isentropiccompression

A gas-turbine

5 atm pressure, an isobaricheating expansion

isentropic

the

from

work, in J/kg, (b) the heat addedin isobaric \342\200\224 (d) the net work (turbine work compressor the answer of part e added), (f) Compare

cp = 1.0043J/kg-Kand

21. Calculatethe

22. Calculate

y

=

temperature 1.40.

the enthalpy change for of

efficiency

Table

and

J/kg-K

for a

0.88,

using cp and

of cp and y from Table at 23. An airplane is flying

an

polynomial

versus cp

equations

corresponding

y

at

compressor of 265 K

employing

interpolating the

results

assuming

an

with

(a) by

using

of 0.83,

efficiency

isentropic

C.4, and (b)

Table

a pressure

by using cp

=

turbine

to

the

illustrates

N2 in the for (a) N2 from

t for

polynomials

obtained

at a velocity

of

270

km/hr.

Calculate

the Mach

air. determination

the

of a

second-order interpolating

temperature range 1000to 5000K. to

300

1000 to 5000 K. (c) O2 from 25. In illustrative Example1.7,two fourth-order for the temperature ranges 300to 1000K and second-order

calculated

on heated air with an isentropic operating and an inlet 2.3, temperatureof 1090K (a) by using 1090 K from Table C.4, and (c)by using an average value a

of 10,000 m

altitude

A.I in Appendix A for

efficiency

C.4.

number of the airplanerelative 24. Example

thermal

ratio of

a pressure

C.4, (b) by

the

compressor

(c) the turbine work, in J/kg, the thermal efficiency (net work/heat

the cycle.

1.4 throughout

change inlet

an

and

1.0036

y

enthalpy

ratio of 5.2,

=

(e)

work),

(a) the

calculate

in J/kg,

heating,

with

to 1000K, and an

conditions

outlet

compressor

pressure. Use Table C.4,and

to atmospheric

pressure to

1 atm

K and

295

interpolating

24.

K, (b)

O2

from

polynomials

300

to

1000

for cp versus

K, and

t

for

to 5000 K are determined. Determine for the aforementionedtemperature ranges,

1000

for air

in Problem

1000

the

Determine

air two

2

governing

for

equations

compressible fluid flow PRINCIPAL

2-2

INTRODUCTION

2-3

FOR CHAPTER

NOTATION

2-1

2

70 72

DESCRIPTION OF A CONTINUUM control volume, and control surface

MATHEMATICAL (a)

System,

(b)

Extensive

intensive

and

73

73 74

properties

(c) The property field or Lagrangian approach (d) The system

(e)

The

control

(f)

The

substantial

or Eulerian derivative

volume

2-4 RELATIONSHIPBETWEEN

THE

75

75

SYSTEM

77 AND THE

APPROACHES

2-5

84

MASS

2-6 NEWTON'SSECONDLAW OF MOTION The net external force acting on a body of fluid (a) equation for a control volume (b) The momentum form of the momentum equation (c) Differential

2-7

THE

OF THERMODYNAMICS a control volume control volume a by for a control volume equation form of the energy equation

LAW

FIRST

07

\302\260

88 89

89 92

(a) Heattransfer to

94

(b)

Work

94

(d)

Differential

done

(c) The energy 2-8

THE

2-9

SUMMARY

2-1

PRINCIPAL

SECOND

OF THERMODYNAMICS

LAW

95

96 98 99

NOTATION FOR

B

vector

body

dA

vector

differential

e

specific stored energy.

E 70

CONTROL VOLUME

79 OF

CONSERVATION

76

approach

total

stored

force

energy

CHAPTER2

per unit area. of a

mass.

system.

2-1 PRINCIPALNOTATION F

vector force.

g

acceleration caused

h

specific

ij,k m

unit vectors in x, y, mass of a system.

m

mass

M

vector momentum of a

n

general

2

71

enthalpy.

intensive

p

Q

heat.

system.

property.

extensive

absolute

respectively.

vector in i direction.

unit normal general

z directions,

and

of flow.

rate

N

s

CHAPTER

gravity.

by

n;

FOR

property.

static pressure.

specificentropy.

S

total

u

specific

internal

energy, or x-component of velocity.

v

specific

volume,

or

=

V V

velocity

vector.

\342\226\240f

control

volume.

w

z-component

W

work.

of velocity.

y-component

w2)1/2, magnitude of velocity.

+ v2 +

(w2

a system.

of

entropy

of velocity.

letters

Greek

a

angle

between V and dA.

density. Other

p

5

inexact

V

vector

differential.

\342\200\224

Vector

+

\\AX

\342\200\242

B

=

+

}Ay

AXBX +

x

=

B

grad

=

ld~x

+

'

i

+M\302\253

k =

j

hk(A

-

tBy

AyBx)

*dfz

dA,v

dAz

dz

z

dA\\

t w

x^ 0

1

I

y

I

1

'

d{)

.

dz

dA

; 1

Jl

dz)

c

k= = j j = k k = =

\\

dy

\\dy 18

of this

remainder

the

A CONTINUUM

OF

DESCRIPTION

MATHEMATICAL

deriving the governing equationsfor the flow

Before

73

A CONTINUUM

derived in Chapter 2 are fundamentalto the analyticaltechniques

The equations in

OF

DESCRIPTION

MATHEMATICAL

of

a fluid,

it is

useful to formalize

in mathematical terms conceptssuchas control control volume, surface, andEuleriandescriptions extensive and intensive properties, propertyfield,Lagrangian and the substantial some of those flow, derivative; concepts are discussedto a system,

of

limited

in

extent

2-3(a)

1.

Chapter

ControlSurface

Control Volume, and

System,

Thebasic governing ofmatterthat is

flow

laws

a

termed

system

_externaljp_tfie

illustratesa

eitEerTKe^MrroMndings

or the environment. Figure 2.1

space.

through

moving

system

assemblage

neither

Mass

system.

is called

are related to a fixedidentifiable enters nor leaves a system.Everything

processes

System at t

time

at

System

time

Figure

A system

2.1

moving

through

U

space.

volume is denned as an flow. In general,the control may - A

control

spaceTThthis book,

only

however,

.control respect

inertial,

volume

to

an

control

is

inertial

it is

moving,

frame

volumes

are

fixed

imaginary

volume

rigid

assumed

may

control

considered.

volumes

a fluid whkh and its position in

through

its shape

change

to move at

of reference, see

volume

are considered, and

if the

a constant velocity(i.e.,-with

Section 1-13); no accelerat.ing,or non-

.,.

surface a control /^control surface is the imaginary completely enclosing permeable volume. Figure 2.2a illustratesthe conceptsof system, control and control volume, the control volume is fluid. isolatedin a body of flowing Mass can surface when flow across the control surface A except where A is parallel to the fluid velocity 2.2b illustrates the case where the control surface is in contact sector V,_ Figure one solid boundaries; with or more no fluid can be transportedacrossthe latter A that Fluid of boundaries. course, across portions of the controlsurface may flow, are either not in contact with solid boundariesor parallel to the fluid vector V. velocity

74

GOVERNING EQUATIONSFOR

FLUID

COMPRESSIBLE

FLOW

y

System at

time

t1

Control

surface

A

Control volume y

(a)

volume

Control

Control

surface

A

(b)

The relationship between a system, control surface, and control Figure volume, control surface, and control volume in a flowing fluid with (a) System, no solid boundaries, (b) System, and control control volume in the surface, of solid boundaries. presence 2.2

Extensive and Intensive

2-3(b)

extensive

An For

example,

property the

is one volume

Properties that depends of a system,

on the quantity ofmassunderconsideration. the mass of a system,and the momentum

as is possible,extensive properties denoted by An exception etc. letters; for example, U for internal energy,S capital entropy, it is mass, which is denoted m even is an extensive general though property. extensive is the N. denoted property by symbol An intensive is one having a value that is independent of the property mass under consideration. There are two intensive _pf types of propertiesrFirsT, and those pressure which are not temperature, explicitly dependent on the amount mass but have magnitudes state that are representative of the involved, of the of Second, there are intensive properties that are the specific system. a

of

system.

In this

book, as far

are

for

A

by

amount

_

like

of

overall

values

extensive

mass),

properties;

for example,

specific entropy s

specific internal energy u

per

energy

(internal

(entropy per unit mass),and specific

enthalpy

h

(enthalpy

unit

OF

DESCRIPTION

MATHEMATICAL

2-3

The second type of intensivepropertyis generally denoted A is case letters. n. general intensive property denotedby the symbol For a material substance that satisfiesthe continuumpostulate (seeSection unit

per

mass).

defined

n is

,.lim Am-o

lower

1 -4),

the value of

\342\200\224

Am

dm

(2.1)

the generalextensiveproperty

is defined

by p

n

a

for

N

J

density

dN

AN \342\200\224=

N = f Since

by

by n=

Hence,

75

CONTINUUM

A

is given

system

by

dm

(2.2)

System

= dtnldV [see Section1-4(b)],equation

transforms

2.2

to

N

where the integrationis

a flowing

with

associated

properties

mass

=

V occupied by the system. below for the following presented fluid; mass, momentum, stored energy, and entropy.

m=

\\

pd-T

momentum = M = stored

entropy

The the

2-3(c)

= E

energy

= S

the

=

d'V

Vp

T J

t\302\273e(t)f

SECOND

NEWTON'S

2-6

as

written

(g)

where M(t) is the

total mass

of

air,

water,

volume V2. The secondtermin equation (f)

f

Substituting

f

PV-dA=

P(-jVe)

(g) and

equations

(i)

may

be

integrated

(-j

in control

hardware

simply

= pVeAe

dAe)

= 2pAe(V0 -

kt)

(h)

(h) into (f) yields

dM/dt + Equation

\342\226\240

is

rocket

solid

and

2pAe(V0

kt)

= 0

(i)

to yield

dM =

-

2pAe(V0 JQ'

-

kt) dt

(j)

92

FLUID FLOW

COMPRESSIBLE

FOR

EQUATIONS

GOVERNING

from which

(~\\

momentum

rocket may now be determined by 2.56, to control volume V2- Thus,

on the

force

reaction

The

equation

equation,

Mg,

The body force caused by gravity acting on A2 is (-R). of the water inside the control volumeis (\342\200\224 Hence, }VC). velocity

the

and

\342\200\242

v(Pv

jA2

the

applying

force

surface

only

\342\200\224

+

dr

(P\\)t

The

(k)

is

(1) becomes

equation

- Mg =

-R

The thirdtermin equation

(i) and

equation

\342\200\224

(-

R = -(g

+

\342\200\224

into

kt)

=

-

(n) yields

equation

+ 2pAe{V0 -

kM

1~J

I

(o)

kt)2

and

(m)

-

2pAe [Vot

(n)

at

(o) into equation

\\MO

k)

.

-V\342\204\242

at

(Vo

VCM)

(k) and

equations

Substituting

=

Vc

(m)

as

written

|(-KM)=-Mf ot Substituting

- pAeV2

VCM)

(-

be

may

(m)

\342\200\224

for

solving

+

-

2pAe(Vo

The above example is a good illustration of the application an It demonstrates equation to unsteady flow problem. vividly

solution, careful attentionmust be

the

to

given

of

and

assumptions

R yields

(p)

kt)2 momentum

the

to obtain a

that

approximations

required.

2-7 THE FIRST It

is shown

THERMODYNAMICS

OF

LAW

in Section

l(b)

be written

may

thermodynamics

for

that

1\342\200\2241

mass of a

a unit

in the form

SW

SQ

limited to

a

and

kinetic,

thermal,

per unit massof

fluid

the

fluid,

flowing

(2.64) of

forms

stored

energy

associated

with

mass will be assumedto be energies. Consequently, the stored energy

stored

per unit

energy

potential

is

e =

For

of

law

(see equation1.48)

de = comprises the changein all of the

where de the fluid. For

continuum, the first

a system having

V2

u +

the total mass

m,

equation2.64becomes

dE

=

SQ

\342\200\224

+

the

(2.65)

gz

total

stored

energy

is E

= me, and

- SW

(2.66)

where SQand to the entire apply system. To be mathematically unambiguous, it might be preferableto other in for Q and 2.64 to symbols equation denote that they are intensiveproperties.Nevertheless, the convention here is to employ in the for representing the heat and Q and equation energy SW

employ

W

adopted

W

work

of terms, respectively,regardless

mass

the

of

unit

system.

the

time),

Q and

When

Q and

symbols

whether

the lack of precisionin the not

W

may

unit mass or

Any

employed. the

per

disadvantages W is more

Q and

symbols

the total

rate basis(i.e., arising from

than offset

by

to be

have

clearly defined

W is

Q and

on a

different symbols for the many which different conditions under determined. In a specificapplication,thebasisfor the terms

to define

having

Q and

be of

meaning

based

are

they

THERMODYNAMICS93

to be determined on a

W are will

W

LAW OF

FIRST

THE

2-7

the

by

for

symbol

stored

the

energy term. To

illustrate,

W refer to a unit mass basis;when E is employed, Q and of the to and Ware rates system; when de/dt is employed, apply Q and is unit when and for mass W are rates the entire mass; per dE/dt employed, Q e

when

is

Q and

employed,

mass

entire

the

W

ofthe system.

Figure2.12 physical the first developing expression

The

indicates

figure

Figure 2.12 a flowing

of mass

Control

shaft

schematically

bQ\342\200\224' '\342\200\224System volume

at

volume employed for for a control volume. shear work, heat transfer, and the flux the control

of thermodynamics

work,

t

time for

law

the

for

for

model

the

illustrates

deriving

the first law

of thermodynamics for

fluid.

across the boundaries

fluid particle,

of the controlvolume.Equation

2.66

but it may be transformedintoa rateequation

by dt. Thus,

ilF \342\200\224 =

dt

thetimerate Equation 2.67expresses

of

therefore,a substantial

derivative.

80

for

a system

is

valid

of E

(2.67) following

a fluid particle

and is,

Hence,

DE

~Dt

=

SQ

-

SW

a

by dividing

- SW

change

for

(2.68)

94

the

In

+

=

^

j^

into

+

T\"

lj\\_p{u lj\\_p

pertinent

properties

e=

+

u

+

V2/2

yields

2.34,

of the first law

of

be

the

for

thermodynamics

must

terms

work

system

in terms of

expressed

the

volume.

control

the

to

&V and n =

+ gz)p

V2/2

gz)\\dr

volume, the heat and

to a control

+

(u

equation

the transformation

complete

flow

= E =

for N and n

gz. Substituting

To

case, N

present

FLUIDFLOW

FOR COMPRESSIBLE

EQUATIONS

GOVERNING

Heat Transfer to a ControlVolume

2-7(a)

Theamount

of

volume, denoted

control

to a

transferred

heat

of heat transferred to the mass transferred to a system is considered

instantaneously positive.

caused

Q = conduction

as is donein References 10

14

and

for heat

Heat

transfer

may

radiation

+

properties

volume, the to each other,

control

be related

must

and convection

and in Reference15

heat transferprocesses effect the of heat transfer is, however, retained in the general the general influence of heat transfer is includedin the doing, in solving one-dimensional problems. and may be considered

transfer

arenot

the details of the

book

this

In

radiation.

by

form

considered;

so

By

SQ.

governing

equations

2-7(b)

Done

Work

by

a Control

The work done by is of

and

done

two main a

by

be

(2.70) the

of

properties

conduction

for

amount

Thus,

+ convection

To express SQ explicitly in terms of the flow different heat transfer processesandtheflow

heat

the

general,

radiation.

and/or

convection,

conduction,

by

In

the

volume.

control

the

occupying

is

5Q,

by

Volume on the

volume

control

a

types. First, there is shaft shaft

rotating

denoted

work,

of the system. Examplesof shaft

the boundaries

crossing

is denned as positivework by Wshaft; that is, the work

surroundings

othermachines.

done to operate compressors, hoists, and Second, there crosses the boundaries is the work done by the surface forces where a fluid of a control volume, or where the volume is moving; the former of a control boundary is work Section 1-1 The surface can forces, as shown earlier, flow [see l(d)]. be resolved into normal forces due to pressure forces due to shear for the work done by shear stresses will not be developed stresses. Expressions further herein; for a detailed of shear see Reference 10. Instead, the work, will be denoted and in that general work done by shear stresses retained by WiheaT are

work

the

works

called

stressesand tangential

discussion

form in the

The mass surroundings

work

appropriateequations. done

out of the push

by

the

pressure

stresses may be

control volume(positive

or

work),

mass

into

the

either the work

the

done

work

received

in

pushing

when the

volume (negative work).

control

Figure 2.13 illustratesthe determination stresses. The normal forceactingon of

the

infinitesimal

the

d\302\245n

=

flow

work

vector

p dA

area

because dA is

of normal

given by (2.71)

The differential force dFn given by equation 2.71 is the force exerted by the fluid a term. inside the control volume on the surroundingsand is, therefore, positive The rate of doing work is obtained by the of dFn in the component multiplying direction of the velocityV by the magnitude of the latter. Thus, 3Wn

3

Wn

= =

dFnV pV

cos a \342\200\242=

dA

=

\342\226\240 d\302\245n

V

(2.72)

\342\226\240

pv(p\\

dA)

(2.73)

THE FIRST

2-7

95

THERMODYNAMICS

OF

LAW

System

Figure

Flow work done by

2.13

a control

mass crosses

when

forces

normal

A

surface

:ontrol

surface.

for mass leaving the control volume and negative forces is likewise for mass work SWn by pressure flowing entering, positive into which is in accordance out of the control volume and negative for massflowing it, is positive, and vice versa.In done with the conventionthat the work by the system

Since (p\\

\342\226\240

dA)

for mass

is positive

the

the

summary,

across

done

The total

does

system

control

the

surface, done

work

by

entire control surface A.

when it pushes mass work and vice versa. forces is obtained the normal

=

determined The

rate

positive

(2-74)

dA.

that

fluid

is instantaneously

occupying

by

given

(2.75)

for a ControlVolume

The Energy Equation

2.68,2.69, Equations

2.70,

the first

law

and

2.75, for

thermodynamics

of

is

\342\226\240

dA)

2-7(c)

the

over

and negative work. The signof the work

of the

the control volumeis

SWn

integrating

by

\342\200\242dA)

Lpv{pV

LdW\"=

scalar product V at which work is done by the

that

by total

both

includes

2.74

volume,

Thus,

w\302\253

Equation

out of the control

when

combined, yield a control volume; it is

the equation representing

called the

equation.

energy

Thus,

Wshaf, +

-

Wshear

'

\"

\342\226\240 \342\226\240

V2

Q

h

+

= 0

\342\200\224

+

(2.76)

gzj(pV-dA)

Note that the flow work caused the specific internal energy u

expression (u

+

enthalpy

h. Since

combination

in

the

pv).

By

pressure

the

stored

definition

the internal surface

in

by

integral,

stresses,

convenient

work

the latter is

pv

always

with

is combined

in equation 2.69 to

[see Section 1-1 l(e)] flow

dA),

pv(pV

term

energy

energy u and the it is

\342\226\240 \\A

the

occur

yieldthe specific

in

to employ the specificenthalpyh

of

96

FLUIDFLOW

FOR COMPRESSIBLE

EQUATIONS

GOVERNING

as in equation 2.76. It should be recalled,however, the stored term. Since the latter specifies unsteady energy the

work term, the internal

energy

Differential Form of the

2-7(d)

The

surface

converting

the

control

volume

+

+

^fi

integral.

2.76 the resultto a

^

+

U\302\253

by

result is

size. The

of differential

\"

appears

not contain a flow

does

is obtained from equation energy equation into volume integrals and then applying

integrals

p. Derive by

flow

pt.

A second

rate m,

PROBLEMS

Figure

2.4.

for Problem

Sketch

2.14

101

time

of change of the density of the liquid in the tank, and in the tank to reachthe value pf < pt. density (b) required fluid flows steadily through a convergingconicalflow 5. An incompressible inviscid passage from a distance x is measured and wall where the an inlet diameter a, Do semiangle having the density is p. Assume that body forcesare negligible, The mass flow rate is m and the inlet. for the acceleration Derive an and that the flow is uniform at each cross section. expression as a function of x. of the fluid particles tank from which water flows through a well-rounded 6. Figure 2.15illustrates a cylindrical circularorifice at the bottom of the tank. Determine an expressionfor the height h of the water column in termsof Ax, A2, ho, g, and the time t. Assumethat AY \302\273A2. is moved toward the jet at a 7. A flat plate oriented perpendicularly to a horizontaljet of water = 6.0 from surface of the V water strikesthe and flows vertically jet plate velocity m/s. The the plate.The massflow rate of the jet from a stationary nozzleis 50kg/s, and the velocity of it at a the force that must be applied to the plate to maintain the jet is 15.0m/s.Determine

an expression for (a) the the

constant

rate

for the

time

velocity.

8. Figure 2.16illustratesa reducing the elbow

with

discharge

is rotated

a

uniform

velocity

90 deg

S

from

V1 that

4

at the

*v

I* h

0

\\\\

Figure 2.15

\342\200\242

pressure x>\\ = 14.80 10s N/m2 enters = 6.0 of the m/s normal to the inlet area.The direction = m and that of 0.30 of the inlet. The inlet diameter D^ \342\200\242 = and 12.00 105 the water at the exit section p2 N/m2, \342\200\242 flow is uniform at the exit 105 Assume that the N/m2. Water

of D2 = 0.15m.Thepressure = 1.0135 the atmospheric pressure p0 section. Calculate (a) the forceexertedby the pipe on the water, (b) the forceexertedby the water on the pipe, and (c) the forcerequiredto holdthe stationary. pipe flow rate of propellants m = on a thrust stand. The mass A rocket motor is fired statically = exit plane Ve at the nozzle 2500 m/s, and the static pressure 9.0 kg/s, the averagejet velocity Determine the to the of the exit is on surrounding atmosphere. equal pressure plane acting to the thrust stand by the rocket motor. the force transmitted

the exit

9.

elbow.

Sketchfor

Problem

2.6.

Figure

2.16

Sketch for Problem 2

3

general

the

of

features

steady one-dimensional

of a

flow

compressible fluid PRINCIPAL

3-2

INTRODUCTION

3-3

THE ONE-DIMENSIONAL

3-4

CONSERVATION

3-5

103

CHAPTER 3

FOR

NOTATION

3-1

104

OF STEADY DYNAMICS flow Frictionless (a)

ONE-DIMENSIONAL FLOW

FOR STEADY

MASS

OF

105

FLOW CONCEPT

FLOW ONE-DIMENSIONAL

(e) (f)

3-6

with

110

110

fluid (b) Frictionless flow of an incompressible Adiabatic frictionless flow of a compressible (c)

(d) Flow

112

112

fluid

113

friction

flow

Incompressible

flow

Compressible

with

friction

of a

perfect

114

gas

in

the

presence

of friction

fluid (g) External forces acting on a flowing THERMODYNAMICS OF STEADY ONE-DIMENSIONAL FLOW with work Flow and heat transfer (a) (b)

Adiabatic

flow

with no

120

120 122

external work

(c) Isentropic (d) (e)

3-7

THE

Isentropic SECOND

DIMENSIONAL

122

3-8

102

flow of a discharge LAW

114 115

flow

Adiabatic

109

123 124

perfect gas speed

for a

perfect gas

OF THERMODYNAMICS

FOR STEADY ONE125

FLOW

SOME GENERAL EFFECTS (a) Speed of propagation of

OF COMPRESSIBILITY ON small

disturbances

FLUID

FLOW

125 125

3-1 (b) Pressure disturbances in (c) Compressibility factor STAGNATION

3-9

(OR TOTAL)

NOTATION

PRINCIPAL

FOR

131

132

CONDITIONS

(a) Stagnation enthalpy (b)

134

134

temperature

Stagnation

136

(c) Stagnation pressure (d)

3-10

Maximum

Dimensionless

139 140

corresponding

3-12

3-13

3-14 3-1

145 mass

isentropic

THE IMPULSE

AND

rate

of flow

FUNCTION

THE PRESSURE

AND

147

rate

COEFFICIENT

150 151

152

coefficient

153

NOTATION FOR

PRINCIPAL

CHAPTER3

a

speed of

a0

stagnation

a*

critical (i.e.,sonic)speedof

A

flow

cp

specific heat at

cv

specific

D

drag force.

sound.

speed

hydraulic

friction

F

vector

9

acceleration

G h

static specific

=

sound.

area.

constant

pressure.

heat at constant

f

=

of sound.

cross-sectional

2

M

148

151

pressure

pressure

144

144

SUMMARY

H m m m

GAS

gas

on the mass flow

compressibility for maximum

Dynamic

(b) The

A PERFECT

FOR

a perfect

area

PRESSURE

DYNAMIC

(a)

141 142

for

equation

THRUST

STREAM

141

a*

characteristic speeds

3-11 THE CONTINUITY RELATIONSHIPS

(c) Effect of (d) Condition

to

M*

velocity

(a) The continuity (b) Critical flow

139

speed

isentropic

(b) Critical speed of sound (c) Thermodynamic properties between the (d) Relationships (e)

138

properties

DYNAMICS

GAS

OF

SPEEDS

CHARACTERISTIC

(a)

of stagnation

in terms

change

Entropy

137

acoustic speed

(e) Stagnation (f)

137

density

Stagnation

103 128

fluid

a compressible

CHAPTER 3

volume.

diameter.

in the

factor

Fanning equation

force.

pA + mf

A,

stream

rhV,

mass

enthalpy.

specific

stagnation

mass.

mass

flow

molecular

thrust,

to gravity. flux.

due

rate.

weight.

Mach number.

enthalpy.

and impulse

function

104

GENERAL FEATURESOF M*

=

p

absolute static pressure.

P

FLOW

ONE-DIMENSIONAL

STEADY

V/a*, dimensionless

velocity.

stagnation

pressure.

=

q

pressure.

dynamic

pV2/2,

heat.

Q

gas constant.

R

universalgasconstant.

R

s

specific

t

absolute

T

stagnation temperature.

u

static

V

velocity

V

vector velocity.

entropy.

static

internal

specific

^max

maximum

f7\"

volume

temperature. energy.

magnitude. speed.

isentropic of a control

volume.

W

work.

x

displacement

z

position in a gravitationalfield.

flow path.

along

Letters

Greek

=

a

=

y

Mach

sin~1(l/M),

heat

specific

cp/cv,

p

static density.

p0

stagnationdensity.

angle. ratio.

Superscripts

3-2

by an isentropic process.

is reached

state

the

that

denotes

*

criticalcondition,

M

where

=

1.

INTRODUCTION

In Chapter 2, the governing equations for

integraland

equations

a

are

their

and

nonlinear

highly

are

solutions

general

for

form

differential

the

of a

flow

volume

a control

fluid are

developed in both

(see Tables 2.1

and 2.2).Those

equations in three space variablesand time, nonexistent. Consequently, they must be practically set of

simplifiedbefore may be employed for obtaining useful results. it is that Thus, as a first approximation, (1) the. flow is steady^(2) the fluid properties are each location where mass crosses the control surface, and (3)the are body negligible. forces they

assumed

at

uniform

of

effects

Assumption

of the fluid

flow

particles

accompany

(ysignifies; that, at each, move

most

invariant

are

properties

along

of the

with

cross

of a

section

time. In

flow passage,

the magnitudes

reality, steady flow exists only-if-

practical

the

Because of the turbulenceand eddies cases of fluid flow, a practical is not steady

streamlines.

that

flow

are detectable of flow with the though no variationsin the most refined measuringinstruments.Ifthe of a fluid is constant, rate of flow it may be imagined that streamlines be drawn in the fluid in such a manner can that motion are to the average direction of the flow. The actual tangent be conceived To be composed of a steady motion alongthe may thereby streamlines with random disturbances of velocity superimposed on irregular

in a

strict sense, even

rate

measured

fixed

they

aforementioned

fluid

3-3 the net effect of

them,

the disturbances on

rate

mass

the

of flow

105

CONCEPT

FLOW

ONE-DIMENSIONAL

THE

of the main fluid

being zero.

area of the cross section of a flow passage is Assumption ,(2)'signifies eitherconstant changes so gradually that at each cross section thefluidproperties in the flow direction under consideration. In all other only change the

that

or,

appreciably3

directions the

Such

ignored.

are

fLows

one-dimensional

called

is justified for

A^sumptionj3)

take place so slowly

flow properties

the

in

changes

the

that

may

they

be

flows. of

flow

forces

of body

effects

the

gases\"\"where

are usually insignificant.

The resultsobtained

the

under

for

accurate

internal

many

(i.e.,s flows

flows for

areHiefuTqualitatively

three

aforementioned

inside of solid boundaries),and the results external

understanding

are quite

assumptions

flows.

subject the governing equationsare developed, in a form of to the flow fluid.1-6 In general, applicable assumptions, the solution or numerical of those equations must be obtained either graphical A fourth assumption is introduced that the working fluid behavesthertechniques. with both the thermal and caloric equations of state in accordance a perfect the 1-15). The results obtained governing equations gas (see Section aforementioned if derived the four be modified, by applying assumptions from the simplifying assumptions may be takeninto so that deviations any Tn ThiiTchapter, the

to

three

aforementioned

any

by

modynamically for

from

may

necessary, account.

The

of this

developments

chapter

show that the localstaticproperties of a

flowing

and t, when expressed as ratios of theircorresponding local stagnation M. for values P, p0, and T, depend only on the local Machnumber Consequently, a compressible flow variable be considered be the each to a function of fluid, may if the Mach number M is either known, or calculable Mach numberM.Consequently, at somepointin a flow field, then the ratios p/P, p/p0, and t/Tat that pointareknown,

fluid

and

3-3

p, p,

versa.

vice

THE ONE-DIMENSIONALFLOW The

equations

governing

three

space

CONCEPT

variables:

for analyzing fluid flow involvefour independent time. Furthermore, because of the mathematical

and

dimensions

of those equations, simplifying assumptionsare introduced,moreoften than not, to obtain a physical model that is moreamenable mathematical Themost and most commonly to analysis. important is that the flow is that one-dimensional; is, that all of the employed approximation over cross -. fluid properties are uniform section of the flow passage. every three different for a fluid Figure 3.1aillustratesdiagrammatically velocity profiles = D 2R. They correspond to a one-dimensional flowing in a circularductof diameter which has a uniform a turbulent the flow, developed velocity profile, fully flow?(with number Re = DVp/p.,ranging from 104 to and a fully developed 105), Reynolds difficulties

encountered

in

deriving

general

solutions

the

laminar

velocity

profile

laminar flow.

The equation solutionofthe

for

Navier-Stokes

equations

The velocityprofileforthe developed From the of experimental amount

(see

equation

turbulent

fully

vast

data

in Fig. 3.1a 5.172), and is

illustrated

flow

is based

that has been

is an exact

on experiment.

accumulated on the

106

GENERAL

OF

FEATURES

FLOW

ONE-DIMENSIONAL

STEADY

y

laminar

developed

Fully

viscous

flow

turbulent

developed

Fully

viscous

flow

(a)

Figure 3.1. Velocity profiles of one-dimensional, laminar, (b) Flow in a converging-diverging region of a pipe.

in several and

typical

turbulent

nozzle,

flow

profiles

(c) Developing

velocity profiles for the turbulent equation has beenderived.7'

flow

(a) A

passages,

velocity

of

flow

fluids

a

in

in

in

comparison round the

pipe, entrance

pipes,

the following

empirical

8

(3.2)

^Equation

the value

3.2

is known

as the power law equation(seeSection5-10).In

of n is determined

by

of the ReynoldsnumberRe.

experiment.

The

following

table

presents

that

n as

equation

a function

3-3 n for

Exponent

\342\200\224-

=

Re

n

Power Law VelocityProfiles 4-103

2.3

60

6.6

\342\200\242

1.1

104

of

divergence

being

the

Fromthe possiblefor the of the actual

foregoing velocity

flow profile

velocity

flow

\342\200\242

3.2

106

106

10

10

in the throat and profiles of the flow is inviscid,the viscous

velocity

3.1a).

Fig.

(see

In a

\342\200\242

layer near the wall. Figure3.1cillustrates velocity profile in the. entrance region ofa pipe. it is apparent that different configurations are few examples, medium. Whether or not the assumption of a flowing profile case depends on how significantly is justified in a particular the uniform differs from velocity profile of one-dimensional

the

one-dimensional

2

106

thin boundary

the

to

of

development

\342\226\240

8.8

The bulk

nozzle.

converging-diverging confined

1.1

105

the

schematically

effects

\342\200\242

7.0

Figure3.1billustrates a

107

CONCEPT

FLOW

ONE-DIMENSIONAL

THE

flow, because of its uniform

one-dimensional

are calculatedso that actual overall values of the mass rate, the flow properties

as

represent,

they

flow

and

momentum,

(see Fig. 3.1a), the as possible,

profile

velocity

closely

kinetic

energy

deviates to flow. If the velocity profile of the actualflow corresponding from the uniform profile of the corresponding one-dimensionalflow, appreciably velocity the calculated values of the flow properties will be in error. For mostpractical the error is ordinarily within the desiredaccuracytolerance. turbulent flows, however, in a pipe, having a velocityprofilesimilarto Fora fully developed turbulent flow in Fig. that illustrated 3.1a, the one-dimensional approximation is quite accurate. in the throats of converging and convergingthe flows It isalsoquite accurate for The xif the latter nozzles (see Fig. 3.1fc). and in the divergence dtverging\"nozzles, in diffusers the flow where the bulk of the is alsosatisfactory for approximation

actual

the

of the

outside

remains

flow

assumption

significant and

shapes,

complicated

having

to

lead

may

boundary

illustratedin Fig.3.1c. approximation

flow at

flow

one-dimensional

The

as

only

equations cross

each

2.1) can

Table

layer adjacent to the walls.Theone-dimensional errors in laminar flows, flows inside passages with developing flows, such as that ducts in as employed

concept

in gas dynamics is an

far as the flow model is concerned,but notinsofar are concerned. Once the approximationof uniform section is made, the integral forms of the governing

be applied to the simplifiedflow

The

model.

exact

as

the

governing-

flow

properties

(see

equations

differential

forms

from the integral of the governing equations(seeTable2.2)are derived equations the flow the on model. restrictions Consequently, being placed any invalid for the differential are aforementioned subject one-dimensional (i.e., a equations model. A new set of differential equations are, therefore,derivedin uniform flow) flow model by employing the integral the thissection for one-dimensional steady

without

formsofthe

equations

governing

2.1.

in Table

presented

an The one-dimensional flow approximation is exactfor the flow through tube. infinitesimal stream Thus, many of the general featuresthat characterizelarge-scale, the streamlines of a multidimensional flows one-dimensional are also along present

flow. In general,the one-dimensional change g of the so-calledfl driving g potential

is

approximation

flow

following

flow

friction, heat

transfer,

flow

should

passage

potentials

driving

and

be

mass

large,

and the

in the direction in this book: area

is small

are-considered

addition.

Furthermore,

if the

reasonable

rate

of

of flow. change, wall

the radius of curvature ofthe

profiles of the flow

properties

should

remain

\\

108

ONE-DIMENSIONAL FLOW

OF STEADY

FEATURES

GENERAL

I

(

/'

will

model

flow

dimensional

, ,,^0'(-

Accordingly, it is to be expectedthat the oneand in poor for flow passages such as pipej^lbows,

section.

cross

flow

each

at

similar

'\"

be

wherein a transition from laminarto turbulentflow occurs.Jtjihould be_noted or bulk flow model considers changesonly in the average that the one-dimensional the it disregards values of the flow properties in the directionof flow; completely in the direction normal to the streamlines. variations in the flow properties a duct

Consider the incompressibleflow of a fluid in a pipe for three different the same mass flow rate m: (1) one-dimensionalflow,(2) fully cases,allhaving to 3.1a. The and (3) fully developed turbulent flow. Refer laminar flow, Fig. developed 3.1. Thus, laminar velocity profile is given by equation 3.1.

Example

(a)

where R is

the inside radius

the

of

For Reynoldsnumbersin the neighborhood 105, by

3.2.

equation

= \302\253

Assume that all \"max

V,

for

and

turbulent

the

of

is well approximated

the centerline velocity.

umax denotes

and

pipe,

velocity

profile

Thus,

y\\1/7

1

\"max!

--j

three velocityprofiles

flow rate

mass

same

the

yield

m. Determine

the laminar and turbulent flows in terms of theone-dimensional both and momentum the ratios of the actual to the one-dimensional calculate energy for each velocity profile. velocity

flow

kinetic

Solution

For the

rhi-D =

MX_D

(a)

For

the

\342\200\242 \\

dA

pV

=

Momentum

=

Kinetic

&!.\342\200\236

=

f =

equations

=

pu(2%y dy)

=

\342\200\242

dA)

=

dA)

(pV

2np

dy

$*^

'

(c)

npR2V =

(d)

npR2V2

y dy =

|pR2V3

(e)

(subscript /): ( 1

wmax

\\ \342\200\242^u

(c) and

(f)

7T

\\

V

-

-2 )y K. J

\\

dy

=

/

pR2um^

(f)

yields

tW

=

2V

(g)

and kineticenergy The calculation of the momentum 2 \\

E, =

P JO u2y

=

=

dy

uy

\\

2np

\342\200\242

\342\200\224

PR

Inp

= 2np

\302\243

flow case

laminar

l-D):

(subscript

f V(pV JA

energy

rhi

Combining

case

flow

one-dimensional

yields

2

2np

Combining equations (d) and

(h)

yields

M

3

UJ

CONSERVATION OF

3-4

equations (e) and (i) yields

Combining

IT\"

=

2

W

turbulent flow case (subscriptt):

For the

(b)

109

FLOW

ONE-DIMENSIONAL

STEADY

FOR

MASS

=

m,

J\302\260

-

\342\200\224

=

dy

y

\342\200\224

oU

KJ

\\

(1)

pK2wmax

(c) and (1) yields

equations

Combining

1

umax

2;tp

/60\\ V

=

\"max

(m)

[^J

For the momentum

cr 2np JQ

(n) the ratio

(d) and

equations

f -

v

mLx (1

j

\\2/7

turbulent flow:

4971

=

ydy

)

case of a

in the

fluid

the

of

energy

=

Mt

From

and the kinetic

\342\200\224

(n)

PR2u2max

is

MJMl_D

=

102\302\260

(p)

w~

Y~ (p) and

Equations for

appropriate

3-4

this

2.1, the

2.86.

equation

(q)

the one-dimensional

flow

turbulent

pipe flow,

model

may

but equations (j) and

developed

(k)

be indicate

flow.

laminar

FLOW

ONE-DIMENSIONAL

for governing the conservationof massis resultant equation is calledthe equation. the of the of of mass is form conservation integral given

the equation

section, Table

From

1-058

considerable error for a fully

one-dimensional

steady

=

OF MASS FOR STEADY

CONSERVATION In

(q) indicate that

developed

fully

lead to

it may

that

EtjE^.Dis

(o), the ratio

(e) and

equations

Combining

a

derived

flow. The

continuity

law

by

Thus,

^ p,

d-T +

\342\226\240=

0

dA

pV

jA

(2.86)

flows, the first term in equation2.86iszero. to Fig. 3.2, which illustrates the flow through a streamtube.LetAx and A2 and the inlet and exit areas, Vt velocities over the areas Al V2 the mean and values of the and the corresponding px p2 densityofthefluid. Applying 2.86 to Ax and A2 yields

For steady Refer

denote

and

A2,

equation

f p\\

Integrating equation3.3

gives

the

stream

dA

+

following

f

= 0

pV-dA equation

for the

(3.3) mass rate of flow

through

tube. m =

Equation

the

\342\226\240

3.4 applies

p-^A^Vy

=

p2A2V2

=

pAV'\342\200\224

flow to the steady one-dimensional

constant of

any

(3.4) fluid.

110

3.2.

Figure

3-5

OF STEADY

GENERAL FEATURES

Steady

DYNAMICS OF

flow

FLOW

ONE-DIMENSIONAL

STEADY

momentum

The

tube.

a stream

through

FLOW

ONE-DIMENSIONAL

one-dimensional

for a steady

equation

the extendedto determiningthe external

flows are considered first. The resultsare then friction. Finally, a method is presentedfor

3-5(a)

of

effects

for

account

forces

acting

fluid.

a flowing

on

Inviscid

below.

derived

is

flow

Frictionless Flow

The integral

of

form

is repeated

-

dr

fluid

which

-dA

pn,

+

Fshear =

nr

\302\243

direction (i.e., equation 2.55)

+

dV Jy (put\\

\342\226\240

flows in

the x direction. For a steadyfrictionless flow,

through

passage

=

(put)t

(3.5)

dA)

Mj(pV

\302\243

flow a portion of a frictionless one-dimensional

3.3 illustrates

Figure

for the xt

equation

below.

renumbered

and

BiP

momentum

the

Fshear

= 0,

and equation 3.5 reduces to

Btp dr -

enters,

dA

=

\342\200\224

so that

attraction,

The

i dA.

\342\200\224

direction, of

component

the

body

which, z axis

the

where

k,

in the

force

axis is not orthogonalto

the

thevariablesp,p, and Fis

Fig.3.3,

n;

x uniform

=

and

dA,

that caused

convention,

by

x direction is For

axis.

over

ft

\342\200\224ipAg

any

x axis.

the

with

the assumed

gravitational

by

to act

is assumed

makes the angle

that

note

dz;

in the negative z the in Fig. 3.3 the z Thus,

flow, each of

one-dimensional

area.

cross-sectional

arbitrary

inviscid flow.

V, and V = iV. On the on the face where mass

=

\302\261i, ut

an

termed

be

will

force considered is

body

only

B = g,

(3.6)

dA)

Ui(p\\

\302\243

volume, dA = i

control

the

leaves

mass

where

\342\226\240

dX

forces

For the flow model illustratedin

face

\342\200\242=

are no shear

in which there

a flow

Hereafter,

pn, \302\243

For the

element of differential volumeof lengthdx, the surfaces which 3.6 equation the must be integrated are the inlet area exit area A + dA, and the stream tube over

A,

boundaryarea

a),

(dA/sin

respect to the A differential

across

few

a is

x axis. on

comments size

where

are

the

in order.

the angle

made

the

by

passage

employed in analyzing By convention, all of the flow properties

conventions

which mass enters will

be assignedthe nominal

values

p,

boundary

with

control volumesof on p,

the

V, V2/2,

surfaces

pA, etc.,

and it will be assumedthat positivechanges in these occur in the direction properties of flow. Thus, at the exit area, the properties are + dp, p + dp, V + dV, V2/2 + p etc. In are addition, d(V2/2), pA + d(pA), positive changes in the driving potentials

3-5

111

FLOW

ONE-DIMENSIONAL

STEADY

OF

DYNAMICS

(p + dp/2) dA

=

FB

(pAdx)

=

FB

- -(pAdx)

FBX

=

FB

B = g=

g

-k(pAdx)g

g cos ,

-pAgdz

-kg

= X COS P

Z

dz = dx

cos

/?

assumed.

Hence,

situation,

any

a

results

be

will

obtained

throughoutthisbook. to

Returning

+

pA

3.6, assign

the properties

dV) to the exit area A

m{V +

the average static pressureintensityis by (p

A

+

dA.

In a

specific flow

+

p

+

dp/2,

and

pA

dA.

On

which

wFto

area A, and

the inlet

the stream acts on the

tube boundary

area

acting on the boundarysurfaceof the

a.

dA/sin

is

desired.

stream

tube

is

+ dp/2) dA. Hence, equation3.6becomes

\342\200\224

+

dz

pAg

pA

+

dp-

p

2

Equation 3.7 is simplified terms. The resultis m

=

pAV,

from

dz

equation

-

- d(pA) =

pA

like

-pAg Substituting

-

dA

canceling

by

terms

A dp

and

3.4, into equation

dp + pVdV pgdz = is the well-knownBernoulliequation.

3.9

Theintegrated

of

form

r dp 1-

Jo

V2 \342\200\224 V

2

equation

gz

=

3.9

constant

m(V

+

neglecting

dV)

- mV

(3.7)

the higher-order

= mdV

+

Equation

is

area

flow.

of that force acting in the directionof flow

component total force

the

Consequently, given

the

only

However,

exit

a frictionless

If, however, property changes may turn out to be negative. is employed in developing the governingequations,the The convention discussed above will be followed consistent.

equation

and

d(pA)

the

and

A

in

of the

all

or

is

tube

of a stream

element

the inlet area

convention

consistent

a differential

acting on

Forces

3.3.

Figure

(3.8)

3.8 gives

0

(3.9)

is

(i.e., the

Bernoulli constant)

(3.10)

112

ONE-DIMENSIONALFLOW

OF STEADY

FEATURES

GENERAL

Equation 3.10 is valid

one-dimensional for steady frictionless flow, and holds only Bernoulli constant will in general, different streamline. The have, along streamlines in a multidimensional flow. In the constant the different along case where the flow field is uniform, as in the assumed one-dimensional flow, the Bernoulli constant has the same value over the entire flow field. To complete the a given values

between the density of the fluid integration of equation 3.10,the relationship p and its static pressure p must be known. 3.9 and 3.10 are forms of the momentum Equations equation for the steady onedimensionalflow of an inviscid fluid. In Section 3-6, the energy equation is developed in the absence of heat and work.One form for a steady one-dimensional flow of that is identical to equation 3.9. Hence, the energy equation is equation 3.40,which and momentum energy equations for the steady one-dimensionaladiabatic flow of an inviscid fluid are equivalent. That fact the two equations are not does not mean that It merely the same result, and that they independent. signifies that both equations yield are satisfied by the same expression.

3-5(b) Frictionless

Flow

If the steady

an

of an incompressible liquid is considered, or gasesunder are negligible the changes in density (e.g., low-speed subsonic gas because 3.10 yields p = constant, the integration of equation

inviscid flow

then,

-p +

V2 \342\200\224-

Equation 3.11is the integrated flow of an incompressible fluid.

form of

Flow of a

Frictionless

Adiabatic

+

2

P

3-5(c)

Fluid

Incompressible

where

conditions flows)

of

gz

= constant

(3.11)

Bernoulli's equation for

Compressible

the

steady

frictionless

Fluid

it is desirable, the flowing medium is a compressiblefluid, as stated to express the properties of the flow in terms of its Mach number 3.9, we obtain equation Rewriting

When

in

3-2,

M

+

ydV

d\302\261dA+

dp p

For the

special

case

of an

from

\342\200\224

V/a,

= 0

gdz

adiabatic frictionless flow,

(see Section 1-12).Consequently,

Section

the

(a) entropy

remains

constant

1.180,

equation

dp

For most g dz

% 0.

flow

processes

gases,

involving

Combining equations (a)

^Ov

P

3.12 is an alternate Equation dimensional adiabatic frictionless

form

(i.e.,

the body forces are negligible, so that introducing g dz = 0, we obtain

(b) and

and

of

Bernoulli's

isentropic)

(3.12) for the steady equation flow of a compressiblefluid. all of the frictional effectsare

one-

that according to Prandtl, the is large. confined to the where Hence, the equations velocity gradient boundary layer and derived in Sections to the of a flow field 3-5(c) apply 3-5(a), 3-5(b), region in that the fluid flow that is located outside of the boundary layer; particular region flow. be assumed to be an inviscid may It

may

be assumed,

the steady one-dimensionalflow

Consider

ONE-DIMENSIONAL FLOW

113

Friction

with

Flow

3-5(d)

OF STEADY

DYNAMICS

3-5

a fluid

of

through

a passage

having solid

boundaries such as that illustrated Fig. 3.4. Let SF^ denote the x component of the on the fluid element, and let 3Ddenotethex component acting friction force the internal caused drag forces by obstructions, such as screensand struts, the The forces SFf and 6D oppose the motion of the in fluid. and thus in

wall

of

fluid

submerged

on an element of Figure 3.4. Additional forces acting a stream tube due to friction and internal obstructions.

act in the

the flow

same directionas the pressure

A

force

with

directionis

illustrated

friction given

from

Fig.

dz

+

dp

Section

[see

3.4, the net external

Hence, for

3-5(a)].

force acting in the x

by

-{pgA

Substituting

in

into

3.13

equation

(3.13)

+ 5F< + 5D)

A dp

3.8 and introducing m

equation

=

we

pAV,

obtain

+ pgA dz

+ pAVdV

A dp

The wall friction force SF^

be

may

+

SFf

3D

+

in terms

expressed

=

0

(3.14)

of the hydraulic

and an

is defined experimental friction coefficientf, which the average value of the wetted perimeter later by equation 3.16. Let (WP) denote The of the fluid element is dx, so that the surface wetted for the flow passage. length the If radius for fluid element is dx. m denotes the flow (WP) hydraulic passage by the the diameter and 3 denotes then, by definition, corresponding hydraulic

characteristics

of

the

flow

passage

m

\342\200\224 =

having

a circular

is

cross

is

defined

by

equal

to the diameter

the

tangential force \\pV2

where t is

(WP)

for a

flow

passage

section.

The friction coefficientf

f-T

perimeter

it is

that

so

chosen

(3.15)

;

wetted

4

The definition of 3

flow area

3

=

area)

(wetted

the shear stressat the

wall.

Fanning

equation. Thus,

6F