Third Edition
Engineering Fundarnerntals An Introduction to Engineering Saeed Moaveni M i n nesota State U n iversity,
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Tt{rotvtsoNEngineering Fundamentals: An Introdudion to Engineering, Third Edition by Saeed Moaveni
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Preface
CHANGES IN THE THIRD EDITION The third edition, consisting of 20 chapters, indudes a number of new additions that were incorporated in response to sugestions and requests made by professors and students using the second edition of the book The major changes include:
. A new section on Engineering Technology . Additional Ethics Case Studies . Additiond secdons on M,ffifAB " " " " "
Additional Professional Profiles Additiond Impromptu Designs Additional Engineering Marvels Case Studies Additional problems A new website to offer additional information for instructors and students including PowerPoint slides for each chapter
ORGANIZATIOl'I This book is organized into six parts andZ} chaprcrs; Each chapter begrns by stating its objectives and concludes by summarizing what the reader should have gained from studying that chapter. I have included enough m*erial for two semeterlong courses. The reason for this approach is to give tfre instructor sufficient materials and the fexibiliry to choose specific topics to meet his or her nee&. Relevant, weryday examples with which students can associate easily are provided in each chapter. Many of the problems at the conclusion of each chapter are handson, requiring the student to gather and analyze information. Moreover, information collection and proper utilization of that information are encourâ‚¬ed in this book by asking studene to do a number of assignments that require information gathering by using the Internet as well as employing uaditional methods. Many of the problems at the end of each chapter require studena to make briefrepoms so that they learn that successfirl engineers need to have good written and oral communication skills. To emphasize t}re importance of teamwork in engineering and to encourage group panicipation, many of the assignment problems require group work; some require the panicipation of the entire class. The main parts. of the book are:
Pr.rrlcs
Part 0ne: EngineeringAn Exciting Profession In Pan One, consisting of Chapters I through
5, we
inuoduce the students to the engineering
profession, how to prepare for an exciting engineering career, the design process, engineering communication, and ethics. Chapter 1 provides a comprehensive inuoduction to the engineering profession and its branches. It introduces the students to what the engineering profession is and explains some of the common uaia of good engineers. Various engineering disciplines and engineering organizations are discussed. In Chapter 1, we also emphasize the fact that engineers are problem solvers. They have a good grasp of fundamental physical and chemical lan's and mathematics, and apply these fundamenel lars and principles to design, develop, test, and supervise the manufacnrre of millions of products and services. Through the use of examples, we also show that there are many satisfying and challenging jobs for engineers. W'e pointed out that atthough the activities of engineers can be quite varied, rhere are some personaliry uaits and work habia that
typifr most of today's successfi.rl engineers:
. Engineers are problem solvers. " Good engineers hane a firm grasp of the fundamental principles that can be used to solve many different problems. . Good engineers are analytical, detailed oriented, and creative. . Good engineers have a desire to be lifeJong learners. For example, they uke continuing edu. . " . . . "
cation classes, seminars, and workshops to stay abreast of new innovations and technologres. Good engineers have written and oral communication skills that equip them to work well with their colleagues and to convey their expenise to a wide range of clients. Good engineers have time management skills that enable tlem to work productively and efficiendy. Good engineers hane good "people skills" that allow them to interact and communicate effectively with various people in their organization. Engineers are required to write reports. These reports might be l""gthy, detailed, and technical, containing graphs, charts, and engineering dt*i"gs. Or the may take the form of a brief memorandum or an executive sutnmary. Engineers are adept at using computers in many &fferent ways to model and analyze various practical problems. Good engineers aaively panicipate in local and national disciplinespecific organizations by attending seminars, worlchops, and meetings. Many even make presentations at professional meetings. Engineers generally work in a team environment where they consult each other to solve complo< problems. Good inteqpersonal and communication skills have become increasingly im
portant now because of the global market. 1, we also enplain the difference berween an Engineer and an engineering technologist, and the difference in their career options. In Chaprcr 2, the transition from high school to college is explained in terms of the need to form good study habits and suggestions are provided on how to budget time effectively. In Chapter 3, an introduction to engineering desbn, 'We show that engineers, regardless of their teamwork, and standards and codes is provided. background, follow cerain steps when designing the products and services we use in our everyday lives. In Chapter 4, we explain that presentations are an integral pan of any engineering pro1ecr. Depending on the size of the project, presentations might be brief, lengthy, freguent,
In Chapter
Pnsrecr and may follow a certain format requiring calculations, graphs, charts, and engineering draltrings. In Chapter 4, various forms of engineering communication, including homework presentation, brief technical memos, progless reports, detailed technicd reports, and research papers are explained. A brief introduction to PowerPoint is also provided. In Chapter 5, engineering ethics is emphasized by noting that engineers design many products and provide many services that affect our quality of life and safety. Therefore, engineers must perform under a standard ofprofessional behavior that requires adherence to the highest principles ofethical conduct. A large number ofengineering ethics related case studies are also presented in this chapter.
Part Two: Engineering FundamentalsConcepts Every Engineer Should Know In Part Two, consisting of Chapters 6 duough 13, we focus on engineering fundamenals and introduce students to the basic principles and ph1'sical larvs that th.y *ill see over and over in some form or other during the next four years. Successfirl engneers hane a good grasp ofFundamentals, which they can use to understand and solve many different problems. These are concepts that wery engineer, regardless ofhis or her area ofspecialization, should know In these chapteis, we emphasize that, from 6ur obseryation of our surioundings, we hare learned that we need only a few physical quandties to flrlly describe events and our surroundings. These are length, time, mass, force, temperature, mole, and electric current. We also sr
plain that we need not only physical dimensions to describe our surroundings, but also some way to scale or divide these physical dimensions. For example, time is considered a physical dimension, but it can be divided into both small and large portions, such as seconds, minutes, hours, days, years, decades, centuries, and millennia'W'e discuss comrnon s;rctems of unin and emphasize that engineers must know how to conveft from one qystem of units to another and alwErs show the appropriare units that go with their calculations. \V'e also explain that the phlnical lars and formulas that engineers use are based on observations of our surroundings. W'e show that we use mathematics and basic physical quantities to express our observations, In these chapters, we also explain that ttrere are many engineering design variables that arg related to the fundamental dimensions (quantities). To become a successfi.rl engineer a student must first firlly understand these fundamental and related variables and the penaining governing larvs and formulas. Then it is imporant for the student to know how these variables are measured, approximated, calculated, or used in pracdce. Chapter 6 explains the role and imponance of fundamental dimension and units in analof engineering problems. Basic steps in the analysis of any engineering problem are disysis cussed in detail. Chapter 7 inroduces length and lengthrelated variables and explains their imponance in engineering work For example, the role of area in heat transfer, aerodynamics, load distribution, and stress analysis is discussed. Measurement of length, area, and volume, dong with numerical estimation (such as trapezoidal nrle) of these values, are presented. Chapter 8 conslders time arid timerelated engineering paiameters. Periods, ftequencies, linear and angular velocities and acceleradons, volumetric flow rates and fow of traffic are also discussed in Chapter 8.
Pnuecs Mass and massrelated parameters such density, specific weight, mass florv rate, and mass moment of inenia and their role in engineering analpis, are presented in Chapter 9. Chapter 10 covers the imporance of force and forcerelated parameters in engineering. 'W'hat is meant by force, pressure, modulus of elasticity, impulsive force (force acting over time), work (force acting over a distance) and moment (force acting at a distance) are discussed
in detail. Temperarure and temperaturerelated parameters are presented in Chapter 11. Concepts such as temperafure difference and heat transfer, specific heat, and thermal conductivity also are covered in Chapter 11. Chapter 12 considers topics such as direct and alternating current, elecuicity, basic circuits components, pov/er sources, and the tremendous role of electric motors in our wery day life. Chapter 13 presents energy and power and explains the distinction berween these two topics. The importance ofunderstandingwhat is meantbywork, energy, power'watm, horsepo$/er,
and efficiency is emphasized in Chapter 13.
Part Three: Computational Engineering ToolsUsing Available Software to Solve Engineering Problems 15, we inuoduce Microsoft Excel and MAI1AB computational tools that are used commonly by engineers to solve engineering probtwo lems. These compuationd tools are used to record, organize, analyze data using formule", and present the resulm ofan analysis in chart forms. MAILAB is also versatile enough that students cln use it to write their own programs to solve complor problems.
In Part Three, consisting of Chapters 14 and
Part Four: Engineering Graphical eommunication: Conveying Information to 0ther Engineers, Machinists, Technicians, and Managers In Part Four, consisting of Chapter 16, we inuoduce students to the principles and nrles of engineering graphical communication and engineering sfmbols. A good grasp of these principles '!fe explain that engiwill enable students ro convey and undentand information effectively. neers use technical drawings to convey useful informadon to others in a standard manner. An engineering drawing provides information, such as the shape of a product, its dimensions, materials from which ro fabricate the product, and the assembly steps. Some engineering drawings are specific to a particular discipline. For enample, civil engineers deal with land or boundary' topographic, consrrucdon, and route survey drarings. Electrical and electronic engineers, on the other hand, could deal with printed circuit board assembly drawings, printed circuit board drill plans, and wiring diagrams.'W'e also show that engineers use special symbols and signs to convey their ideas, analyses, and solutions to problems.
Part Five: Engineering Material SelectionAn lmportant Design Decisioll fu
engineers, whether you are designing a machine pa$, a toy, a frame of a car, or a structruer the selection of materids is an important design decision. In Pan Five, Chapter 17, we look
Pnrrecn
vtl
more closely at materials such as metals and their allop, plastics, glass, wood, composites, and concrete rhat commonly are used in various engineering applications.'We also drscuss some of the basic characteristics ofthe materials that are considered in design.
fart
Six: Mathematics, Statistics, and Engineering EconomicsWhy Are They lmportant? In Part Six, consisting of Chapters 18 through 20,we introduce students to important mathematical, statistical, and economical concepts. We orplain that engineering problems are mathematical models of physical situations. Some engineering problems lead to linear models, whereas
others result in nonlinear models. Some engineering problems are formulated in the form of differential equations and some in the form of integrals. Therefore, a good undersanding of mathematical concepts is essential in dre formulation and solution of many engineering problems. Moreover, statistical models are becoming common tools in the hands of practicing engineers to solve quahty control and reliability issues, and to perform failure andyses. Civil engineers use statistical models to study the reliabfity of construction materials and stnrctures, and to design for flood control, for example. Elecuical engineers use statistical models for signal processing and for developing voicerecognition software. Manufacturing engineers use statistics for qualiry control assurance of the products they produce. Mechanical engineers use statistics to study the failure of materials and machine parts. Economic factors also play important roles in engineering desigrr decision making. Ifyou design a product that is too expensive to manufacnue, then it qrn not be sold at a price that consumers can afford and still be profitable to your company.
CASE STUDIESENGINEERING MARVELS To emphasize that engineers are problem solvers and that engineers apply physical and chemical laws and principles, along with mathematics, to design products and services that we use in our everyday lives, seven case studies are placed throughout the book These projects are ruly engineering marvels. Following Chapter 7, the design of New York City W'ater Tunnel No. 3 is discussed. The design of the Cateqpillar 797 Mining Tiuck, the largest mining mrck in the world, follows Chaprcr 10. FollowingChapter 13, relwantinformationaboutthe design ofthe Hoover Dam is discussed. The design of the Boeing 777 :spresented.in a case sudyfollowing Chapter 16. Finally, the Pratt and Whimey Jet Engine is discussed at the end of Chapter 17. There are assigned problems at the end of those case studies. The solutions to these problems incorporate the engineering concepts and laws that are discussed in the preceding chapters. There are also a number of engineering ethics case studies, from the National Society of ProFessional Engineers, in Chapter 5, to promote the discussion on engineering ethics.
IMPROMPTU DESIGNS I have included seven inexpensive impromptu designs that could be done during class times. The basic idea behind some of the Imprompru Designs have come from theASME.
viii
Pnsrecn
REFERENCES In writing this book, sweral engineering books, Veb pages, and other marerials were consulted. Rather than giving you a list that contains hundreds of resources, I will cite some of the sources that I believe to be usefirl to you. I think all freshman engineering students should ov"n a handbook in their chosen field. Currendy, there are many engineering handbooks anailable in print or electronic format, including chemical engineering handbools, civil engineering handbools, electrical and electronic engineering handbooks, and mechanical engineering handboolis. I also beliwe all engineering students should own chemistry phpics, and mathematics handbooks. These texts c:rn serve as supplementary resources in all your classes. Many engineers may also
findusefirl theASHRAEhandbook, the,FzndammalVolume,byrheAmerican SocieryofHeating, Refrigeratin& and Ah Conditioning Engineers. In rhis book, some data and diagrams were adapted with permission from the following sources:
Baumeisrer, T., et al., Marh's Handbooh,8th ed., McGravr Hill, 1978. Electrical Viing, 2nd ed., AA VIM, 1981. Electric Motors,5th ed., AAVIM, 1982. Gere, J. M, Mechanics ofMateriak,6th ed., Thomson, 2004. Hibbler, C., Mechanics ofMateriak,6th ed., Pearson Prentice Hall'
k
SnndardAtrnoEhne, Vashington D.C., U.S. Government Printing Office, 1962, 'Weston, K C,, Energt Conuersion, Vest Publishing, 1992. U.S.
AEKNOWLEDGMENTS my sincere gratitude to the editing and production tenm at Thomson Engineering, especially Hilda Gown. I am also gratefirl to Rose Kernan of RPK Editorial Services, Inc., and NewgenAustin I would also like to thank Dr. Karen Chou of Minnesota State University, who reviewed the second edition carefirlly and made valuable suggestions. I am also thankful to the follouring reviewers who offered general and specific comments: N.lly M.Abboud, Universiry of Connecticut; Barbara Engerer, Valparaiso Universiqt Alex J. Fowler, Universiry of Massachusetrs, Darrmouth; Peter Golding, University of Texas at El Paso; Fadh Oncul, Fairleigh Dickinson University. I would also like to thank the following individuals for graciously providing their insighu for our Student and Professional Profiles sections: Katie McCullough, Celeste Baine, NahidAfsari, Dominique L. Green, Susan Thomas, and Ming Dong. Thank you for considering this book and I hope you enjoy the third edition.
I would like to
enpress
Saeed
Moaueni
Gontents
Preface nl
PART 01{E:
 1.1 1.2 1.3 1.4 t5
ENGTNEERTNGAN ExcrTrNG
pRoFEssroN z
flntroduction to the Engineering Profession 4 Englneerlnq Work lg All Around
You
5
Englneerlng as a Professlon and Conmon Tnlts of Good Enqlneers 8 Conmon Traltg of Good Englneers 9 Englneerlng Dlsclpllnes 12
Accredltotlon Board for Englneerlng ond Technolory OBET) Professlonal
ltl
Proflle 23
Summary 23
Prcblem
23
lmBromptu Deslgn
2 2J 2.2 23 2,4 2,5 2.6
I
26
Freparing for an Engineering
eareer
Maklng the Transltlon from Hlgh School to
Colleqe 28
BudgetlngYourllne 28 Dally Studylng and Prepantlon
31
Get lnvolyed wlth an Englneerlng
0rganlzetlon 36
Your Graduatlon
Plan
?7
37
Gher Conslderatlons 3? Student
Proflle 38
ProfesslonalProflle 39
Sumnary 40
Problens 40
3 3.1 3.2 3.3 3.4 3.5 3.6
fintroduction to Engineering Design 4l EnglnecrlngDeslgnPfocess 42 Englneerlng Economlss 4i8
Materlal Selectlon 49
Teamwork 5l Common Tralts of Good
Teams 52
Confllct Resolutlon 53
lx
X
Coxrrrrs
3.7 3.8 3.9 3.10 3.ll 3.12 3.13 3.14 3.15
Prolect Schedullng and Task
Chart
53
Alternetlves 55 Patent, Trademark, and Copyrlght 56 Englneerlng Standards and Godes 5l Evaluatlng
Eramples of Standards snd Codes 0rganlzatlons ln the Unfted Exomples
ol lnternatlonal Standards
and
Drlnklng lTater Standards In the Unlted
States 60
Codes 62
States 67 States 68
0utdoor Alr Quallty Standards In the Unlted lndoor Alr Quallty Standards In the Unlted
states
70
Sunmary 72 Problems ?3 lmpromptu Deslgn
4
75
Engineering Communication 81
41 4,2 4.3 4.4 4.5 4.6 41 4.8
5
ll
Communlcatlon Skllls end Presentetlon of Englneerlng Baslc Steps lnvolved In the Solutlon of Englneerlng
Work 82
Problems 82
Homework Presentatlon 85 Progress Report, Erecutlve Sunmary end Short DetalledTeshnlcal
Report
Menos
0ral Gommunlcatlon and Presentatlon 90 PowerPolnt Presentatlon 91 Englneerlng Graphlcal Communlcatlon 100
Summary
101
Problens
102
Engineering
5.1 52 5.3 5.4
Ethics
Englneerlng
Ethlcs
105
106
The Gode of Ethlcs of the l{otlonal Socle$ of Prolerslonal .07 Gode ol Ethlcs for Englneers
Greed
Englneer's
111
Summary 1l?
Problem
117
Englneerlng Ethlcs: A Case Study from NSPE& 120
PART TWO:
ENGINEERING FUNDAMENTALS_
CONCEPTS EVERY ENGINEER SHOULD
5
87
87
Fundamental Dimensions and
6.1 6.2 6.3 6.4 65
KNOW 124
Units
126
Englneerlng Problemsand Fundamental Dlmenslons 127
Systens of
Unlts
128
Unlt Converslon 134 Dlmenslonal Homogenelty 136 Numerlcal versus Symbollc
Solutlons
138
Englneers l0?
CoNrprrs 6.6 6:l 6.8
Slgnlflcant Dlglts(Flgures) 139
Systems l4l
Englneerlng Components 8nd
Phyrlcal Laws of 0bservatlons In Englneerlng
ltlil
Summary t46 Problems
7
147
Length and LengthRelated Parametcrs tr52
7.1 7.2 73 7.4 7.5 7.6 7:I 7.8
Length as a Fundamental Dlmenslon 15il
length
Measurement of
157
llomlnal Slzes versus Actual
Slzes
160
lko lengths l6il StralnasRatloofTlrolengths 163 Radlans as Ratlo of
Area
16:!
Volume
172
Second Moment of
Summary
Areas 17
182
Problems 18:l lmpromptu Design
lll
188
An Englneerlng Marvel: The Nerv York Clty Water Tunnel No.
8 Time and TimeRelated Parameters 195 8.1 8.2 8.3 8.4 8.5 8.6
Tlme os e Fundanentol Dlmenslon 196 .98 Measurement ot nme
PerlodsandFrequencles Flow of
Trofflc
Parameten Inyolvlng length and
AngularMotlon Summary
215
Problems
215
Mass as a Fundamental
Measurement of
Parameters 218
Dlmenslon
Mass
219
222
Denslty, Specltlc Volume, snd Speclfls Mass
FlowRate 224
lnertla 2A
Mass Moment of
Momentum
Conservatlon of
205
214
Mass and Mass'Related
9.1 9,2 93 9.4 95 9.6 9.1
llme
212
ProfesslonalProflle
9
201
203
224
Mass 229
Sumnary 232 Problems 232 IMPROMPTU DESIGN
IV
235
onvlty
222
3*
189
xi
xii
Covrenrs
l0
Parameters 236
Force and ForceRelated
10.1 102 10.3 10.4
What We Mean by
Force
237
Newtonl Laws In Mechanlcs
241
Pressure snd StressForce Actlng Over an
Area
244
Modulus of Elastlclty, Modulus ot Rlgldlty, and Bulk Modulug of Gompresslblllty 256
105 10.6 l0:l
Dlstence 264
Moment, TorqueForces Actlng at a
WorkForce Actlnq 0ver a Dlstance 268 Llnear lmpulseForces Actlng 0ver
Summary
271
Problemr
271
IMPROMPTU OESIGII
V
Tlme
269
2r/
An Englneerlng Marvel: Caterplllar 797 Mlning
lf
278
Temperature and TemperatureRelated Parameters 282
It.l 1L2 ll.3 11.4 tls 11.6
Tempenture os a Fundrmentol Dlmemlon 283 Measurement of Temperature and lts
Unlts
286
Tenperature Dlfference and HeotTransfer 293 lhermal Comfort, Metobollc Ratg and Clothlng Insuletlon 306 Some Tenpemture'Related Materlals
Heatlng Values of
Fuels
ProlesslonolProflle
n2
Truck*
Summary
315
Problems
315
Propertles 309
312
314
Electric Currcnt and Related Parameters 319
121 12.2 123 l2A 12.5
Electrlc Gunent as a Fundenental Dlmenslon 320
Voltage
321
Dlrect Cunent and Alternatln0
Cunent 323
Electrlc Clrcults and Components 326 Electdc
Motors 334
ProfesslonalProflle 3il7 Summary 337 Problems 338
13 Energy and Power 341 l3.l 13.2 13.3 13.4 135
Work, Mechanlcal Energy,Thermal
Energy 342
Gonservatlon of EnergyFlrst Lrw of Thermodynonlcs 348 Understandlng What tfe Mean by
lfottsand Horsepower
Efflclency 355
351
Power 350
CoNrsNrs Student
Proflle 362
ProfesslonalProf,le 363 Summary 364 Problems 364 IMPROMPTU DESIGN
V[
366
An Englneerlng Marvel: Hoover
PART THREE:
Dam
367
coMpuTATroNAL ENGTNEERTNG
T00Ls
USING AVAILABLE SOFTWARE TO SOLVE ENGINEERING
14 Efectronic 14,1 14.2 l/t3 14.4 t4J 14.6 14:l 14.8
PROBLEMS 370
Spreadsheets 372
Mlcrosoft ErcelBaslc Cells and
ldeas 37il
lhelr Addresses 374
Crestlng Formulas In
Ercel
375
Functlons 383 Uslng Ercel Loglcal Functlons 387 Uslng Excel
Ploftlng wlth
Ercel
389
Matrlx Computatlon wlth
Ercel
400
CurveFlttlngwlthErcel 407 Sumnary 4ll Problems
15
412
MATLAE 419
15.1 152 15.3 15.4 155 15.6 15.7
MAT[ABBaslc
ldeas 420
Uslng MATLAB Bulltln
Functlons 429
Plottlng wlth MATUIB tl38 lmportlng Excel and Other Data Flles Into MATLAB 445
MatrhGomputatlongwlthMATLAB 447 Curve Flttlng wlth MATLAB 450 Symbollc Mathematlcs wlth MATLAB 451
ProfesslonalProflle 454 Summary 455 Problems 455
PART F0UR:
ENcTNEERTNG
cRApHrcAL coMMUNrcATroN
CONVEYING INFORMATION TO OTHER ENGINEERS, MACHINISTS,
TECHNICANS,ANDMANAGERS 460
16
Engineering Drawings and
16.1 16.2 16.3 16.4
lmportance ot Englneerlng
Symbols 462
Dnwlng 46il
0rthognphlc Vlews 4&l Dlmenslonlng and Toleranclng 467
lsometrlsVlew 469
xiii
xiv
CoxrrNrs
16.5 16.6 16:I 16.8 16,9
Sectlonal
Vlews
472
Clvll, Electrlcal, and Electronlc Solld
Dnwlngs
476
trlodellng 476
Why Do We l'leed Englneerlng
Symbols? tl83
Examples of Common Symbols used In Clvll, Electrlcal,
and Mechanlcal Englneerlng tl85
summary rts? Problems 487 An Engineerlng Marvel: Boelng 777* Commerclal
PART FIVE:
ENGTNEERTNG MATERIAL
Alrplane 493
sELEcTtoN
AN IMPORTANT DESIGN DECISION 498
l7
Engineering
17.1 172 t73 17.4
Materials 500
MaterlalSelectlon
501
Electrlcal, Mechanlcal, and lhermophyslcal Propertles of Materlals 503
Materlals 509
Some Common Solld Englneerlng
ilaterlals
Some Common Fluld
ProfesslonalProflle
519
521
Summary 522 Problems 522
VII
IMPROMPTU DESIGI'I
525
An Englneerlnq Marvel: The Jet
PART SIX: WHY ARE THEY
fB
Englne* 526
NnrnrMATrcs, sTATlsTtcs, AND IMPORTANT? 530
ENGTNEERING
Mathematics in Engineerinq 532
18.1 182 18.3 18.4 18.5 t8.5 18.7
trlathematlcal Symbols and Greek
LlnearModels
Alphabet 533
5:15
Nonllnearilodels
541
Erponentlal and logarlthmlc Models 546
MatrlxAlgebn
Glculus
551
562
Dlfferentlal Eguatlons 570
Summery 572 Problems 57!l
EcoNoMtcs
Corrslrrs
19
Probability and Statistics in Engineering 577
19.1 19.2 193 19.4 t95
ProbabllltyBaslc ldeas 510
StatlstlcsB$lc ldeos 59 FrequencyDlstrlbutlons 580 Heasuras ol Gntral Tendency Yarlotlonlllean, Medlon, and Standard
t{ormilDbtrllutlon
587
Sumn8rt 594 hoblems 594
20 Engineering 20.1 mA 20.3 2OA 20J 20.6 ZOJ 20.8 20.9
Carh Flow
Economics 597 Dlagnm 598
SlmpleandCompoundlntenit 599 FutureUlorthofaPnsentAmount 600 Efrectlve lnteregt
nde
60ll
PresentWorthofFutureAmount Present Worth of Serles
605
hynent or
hnutty
605
FuturEl{orthofSerlshynnnt 606 SummaryofEnglnesrlngEconomlcaAndysb 609 Chooslng the Best AlternatlvesDeclslon
Summory
617
PruUem
617
AppEndlr 620
Credlts 625 Index 626
l{aklng
613
Devlatlon 582
XV
Part
ENCTNEERING Al. Excmttc PnoFEssron
1
[email protected]@ltryq'@qt
i
In Part One of this book, we will introduce you to the engineering profession. Engineers
.aregroblem solvers. They have a good grasp of fundamental physical and chemical laws and mathematics and apply these laws and principles to design, develop, test, and
supervise the manufacture of millions of products and services. Engineers, regardless of
their background, follow certain steps when designing the products and services we use in our everyday lives. Successful engineers possess good communication skills and are
team players. Ethics plays a very important role in engineering. As eloguently stated by
the National Society of Professional Engineers (NSPE) code of ethics, "Engineering is an important and learned profession. As members of this profession, engineers are
expected to exhibit the highest standards of honesty and integrity. Engineering has a
direct and vital impact on the quality of life for all people. Accordingly, the services provided by engineers require honesty, impartiality, fairness and equity, and must be dedicated to the protection of the public health, safety and welfare. Engineers must
perform under a standard of professional behavior which requires adherence to the highest principles of ethical conduct." In the next five chapters, we will introduce you to the engineering profession, how to prepare for an exciting engineering career, the design process, engineering communication, and ethics.
Chapter I
Introductlon to the Englneering Professlon
Chapter 2
Preparlng for an Engineerlng Career
Chapter 3
Introductlon to Englneering Design
Chapter 4
Englneering Communlcatlon
Chapter 5
Englneering Ethlcs
CHAPTE,R.
1
INrnoDUcrroN To THE ENcTNEERTNG PnoFEssroN [f
ngineers are problem solvers.
1
Successful engineers possess good
L
communication skills and are team
players. They have a good grasp of
fundamental physical laws and mathematics. Engineers apply physical and chemical laws and mathematics to design, develop, test, and supervise the
manufacture of millions of products and services. They consider important
factors such as efficiency, cost, reliability, and safety when designing products. Engineers are dedicated to lifelong
learning and service to others.
1.1
ENcTNEERTNc
Wom Is Ar.r. Anomlo You
ofloa are not yet ceTtain loa uant to study mginening during the nextfour years in college and may haue questions simikr to thefollowing
Possibfu sorne
I
really uant to study mgineering? is mgineeringand what do mgineers do? 'Vhat are some ofthe areas ofspecialization in mgineering? How many differmt mgineering disciplines are tbere? Do I want to become a rnechanical enginen, or shoull.I pursue ciuil mgineering? Or would I be happier becoming an electrical mginen? How will I hnow rhat I haue piched the bestfeldfor rue? Wll the dernand far rn! area of speciakm.tian be hi.gh whm I gradaate, and'
Do
Vhat
bryond that? The main objectiues of this chapter Are to prouidz sorne answers to these and other questi.ons loa may haue, and to i.nnodace you to tlte engineeri.ng profasion
and its aaious brancltes.
1.1 Engineering Work ls All Around
You
Engineers make products and provide services that make our lives beter (see Figue 1.1). To see how engineers contribute to the comfon and the betterment of our weryday lives, tomor
row morning when you get up, just look around you more carefirlly. During the night, your bedroom was kept at the right temperature thanks to some mechanical engineers who designed the heating airconditioning, and ventilating q/stems in your home. When you get up in the morning and turn on the lights, be assured that thousands of mechanical and elecuical engineers and technicians at po$'er plants and power stations around the country are making certain the fow of electricity remains uninterrupted so that you have enough pov/er to turn the lights on or flun on your TV to watch the morning news and weather repon for the dty. Th" TVyou are using to get your morning news was designed by electrical and elecronic engrneers. There are, ofcourse, engineers from other disciplines involved in creating the final product; for example, manu6curing and industrial engineers. When you are gening ready to take your morning shower, the clean water you are about to use is coming to yor[ home thanks rc civil
f,
Figurel.l
Eramples of products and serviees designed by engineers.
CnerrsR
I
IxrnopucrroNTo rnrExcwnsRrNc Pnorsssrox and mechanical engineers. Even if you live out in the country on a farm, the pump you use to bring water from the well to your home was designed by mechanical and civil engineers. The water could be heated by natural gas that is brought to your home thanls to the work and effon of chemical, mechanical, civil, and peuoleum engineers. After your moming shower, when you ger ready to dry yourself with a towel, think about what types of engineer worked behind the scenes to produce the towels. Yes, the cotton towel was made with the help of agriculturd, industrial, manufacnuing chemical, petroleum, civil, and mechanical engineers. Think about the machines that were used to pick the cotton, transport the cotton m a factory clean it, and dye it to a pretty color that is pleasing to your eyes. Then other machines were used to weave the fabric and send it to sewing machines that were designed by mechanical engineers. The same is uue of the clothing you are about to wear. Your clothing may contain some polyester, which was made possible with the aid of petroleum and chemical engineers. "V'ell," you may say, "I can at least sit down and eat my breakfast and not wonder whether some engineers made this possible as well." But the food you are about to eat rras made with the help and collaboration ofvarious engineering disciplines, from agricultural to mechanicd. Lett sayyou are about to have some cereal. The milk was kept fresh in your refrigerator thanls to the efforts and work of mechanical engineers who designed the refrigerator components and chemical engineers who investigated alternative refrigerant fluids with appropriate thermal properties and other environmentally friendly propenies that can be used in your refrigerator. Furthermore, electrical engineers designed the control and the electrical power units. Now you are ready to get into your car or take the bus to go to school. The car you are about to drive was made possible with the help and collaboration of automotive, mechanical, electricd, electronic, material, chemical, and peuoleum engineers. So, you see there is not much that you do in your daily life that has not involved the work of engineers. Be proud of the decision you have made to become an engineer. Soon you will become one ofthose whose behindthescenes efforts will be taken for granted by billions of people around the world. But you will accept that fact gladly, knowing that what you do will make peoplet lives bener.
Engineers Deal with an Increasing World Population 'V'e
as people, regardless ofwhere we live, need the following things: food, dothing, shelter, and warer for drinking or cleaning purposes. In addition, we need various modes of transportation to get ro different places, because we may live and work in different cities or wish to visit friends and relatives who may live elsewhere. Ve also like to have some sense of security, to be able to 'We need to be liked and apprecixedby our friends and family, as well. relax and be entertained. At the turn of the 20th century, there were approximately six billion of us inhabiting the eafth. As a means of comparison, it is imporent to note that dre world population 100 years ago, at the turn of the 19th century, was one billion. Think about it. It took us since the beginning of human enistence to reach a populadon of one billion. It only took 100 years to increase the population by fivefold. Some of us have a good standard of living, but some of us living in developing countries do not. You will probably agree that our world would be a bemer place if wery one of us had enough to eat, a comfortable and safe place to live, meaningfirl work to do, and some time for relaxation. According to'the Iatest estimates and projections of the U.S. Census Bureau, the world
population will reach 9.3 billion people by the year 2050. Not only will the number of people inhabiting the eanh continue to rise but the age strucrure of the wodd population will also
1.1
EncnrBBnnlc W'onr Is Ar.r. A,nouuu You
1
0 1950
1960
lwo
1980 1990 2W
20LO
Year (a)
Over 1 00 population (thousands)
* estimate (b)
E figure t.2 (d ilre
latest pmjection of world population growth. (b) Ihe latest estimate of U.S. elderly population growth.
Smrce: Dacr hom U.S. Census Bureau.
people at least 65 yars of agewill more than 1.2). Figure double in the next 25 yers How is this information relevant? Well, now that you harre decided to study to become an engineer, you need to realize that what you do in a few years after your graduation is very importanr ro all ofus. You will design producs and provide services especially suited to the needs and demands of an increasing elderly population as well as increased numbers of people of all ages. So prepare well to become a good engrneer and be proud that you have chosen the engineering profession in order to conuibute to raising the living standard for weryone. Todry't world economy is very dynamic. Coqporations continually employ new technologies to maximize efficiency and piofits. Because ofthis ongoing change and emerging technologies, new jobs change. The world's eldedy
populationthe
(see
are created and orhers are eliminated. Computers and smart
elecronic devices are continuously
reshaping our way of life. Such devices infuence the way we do things and help us provide the
Cger"rsn
I
lNrnopucrroN To nrs
ENcrNnnRrNG PRoFEssroN
necessities of our livesclean water, food, and shelrcr. You need to become a lifelong leamer so that you can make informed decisions and anticipate as well as react to the global changes
caused by technological innovations as well as population and environmental changes. According to the Bureau of labor Statistics, U.S. Department of Labor, among the fastestgrowing occupations are engineers, computer specialists, and qystems analysts.
1.2
Engineering as a Profession and Common Traits of Good Engineers In this section, we will first discuss engineering in a broad sense, and then we will focus on se'W'e lected aspecs of engineering. will also look at the traits and characteristics common to many engineers. Next we will discuss some specific engineering disciplines. As we said earlier in this chapter, perhaps some ofyou have not yet decided what you want to study during yolu college years and consequendy may hane many questions, including: Wh.at is engineering and what do engineers do? !7hat are some of the areas of specialization in engineering? Do I really want to study engineering? How will I know that I have picked the best field for me? \Xrril the demand for my area of specialization be high when I graduate, and beyond that? The following sections are intended to help you make a decision that you will be happy with; and dorit worry about finding answers to all these questions right now You have some time to ponder them becluse most of the coursework during the fust year of engineering is similar for all engineering students, regardless oftheir specific discipline. So you have at least a year to consider various possibilities. This is true at most educational institutions. Even so, you should mlk to your advisor early to determine how soon you must choose an area of speciallzation. And dont be concerned about your chosen profession changrng in a way that makes your education obsolete. Most companies assist their engineers in acquiring further training and education to keep up with technologies. A good engineering education will en"h*grg able you to become a good problem solver throughout your life, regardless ofthe particular problem or situation. You maywonder during the ner3ffls
8,,
X
Problem 8.27 8.28.
A
car starts from rest and accelerates to a speed of 60 mph in 20 s. The acceleradon during this period is const4nt. For the next 20 minutes the car moves with
the constant speed of 60 mph. At this time the driver ofthe car applies the brake and the car decelerates to a firll stop in 10 s. The variation of the speed of the car with time is shovrn in the accompanying diagram. Determine the total distance traveled by dre car and the average speed ofthe car over this distance. Also plot the acceleration of the car as a function of time.
{)
v)
\3in.
Ttme (s) Prollen 8.28
CHAPTER
9
Mess AND MassRELATED PeneMETERS
kayaker maneuvers his kayak in a whitewater slalom event. Mass is
another important fundamental dimension that plays an important role in
engineering analysis and design. Mass provides a measure of resistance to
motion. Knowledge of mass is important in determining the momentum of moving objects. In engineering, to represent how
light or heavy materials are, we use properties such as density and specific gravity.
218
9.I
Mnss,Ls a FurquAnrsNrAl DrunNsroN
2t9
The objectiue of this cbaptn is to innoduce the concept of mass and massrelated qaantities encountered in mgi,ne*ing. We will begtn by disca*ing the building blocks of all matten Atnrns and molecales. We will thm introduce the concEt of rnass in terms ofa quantitatiue wre*sare ofthe amaunt of atoms possesed by a substance. Ve will tbm defne and discuss other rnassrelated engineering qaantities, such as density, specifc grauity, rnass rnornent of inertia wornentum, and mass flow rate. In this chapter, we will also consider conseraatiln ofmass and its application
in
9.1
engineering.
Mass as a Fundamental Dimension As we discussed in Chapter 6, from their daytoday observations humans noticed that some things were heavier than others and thus recognized the need for a phpical quantity to describe that observation. Early humans did not fi.rlly undersnnd the concept of gravity; consequently, the correct disdnction beween mass and weight was made later. Let us now look more carefirlly at mass as a physical variable. Consider the following. \Zhen you look around at your surroundings, you will find that matter exists in various forms and shapes. You will also notice that matter can change shape when its condition or its surroundings are changed. All objects and living things are made of matter, and matter itself is made of atoms, or chemical elements. There are 106 known chemical elements to date. Atoms of similar characteristics are grouped together in a table, which is called, the periodic able of chemical ebments. An example of the chemical periodic table is shown in Figure 9.1. Atoms are made up of even smaller particles we calI ebc*ons, protoru, and neunons.Inyow fust chemistry class you will study these ideas in more detail, ifyou have not yet done so. Some of you may decide to study chemical engineering, in which q$e you will spend much more time srudying chemistry. But for now, remember thar atoms are the basic building block of all matter. Atoms are combined naturally, or in a laboratory setting, to create molecules. For example, as you already know, water molecules are made of two atoms of hydrogen and one atom of orygen. A glass of water is made up of billions and billions of homogeneous water molecules. Molecules are the smallest pordon of a given matter that still possesses its characteristic propenies. Matter can exist in four states, depending on its own and the surrounding conditions: solid, liquid, gaseous, or plasma. Let us consider the water r:hat we drink wery day. As you already know, under certain conditions, water exists in a solid form that we call ice. At a standard atmospheric pressure, vrater exists in a solid form as long ia temperanue is kept under 0oC. Under standard atmospheric pressure, if you were to heat the ice and consequendy change its temperature, tfre ice would melt and change into a liquid form. Under sandard pressure at sea lwel, the water remains liquid up to a temperature of 100'C as you continue heating the water. If you were to c,ury out this experiment further by adding more heat to the liquid water, eventually the tiquid water changes its phase from a liquid into a gas. This phase ofwater we commonly refer to as stqrm. Ifyou had the means to heat dre water to wen higher temperatures, temperatures exceeding 2000"C, you would find that you can break up the water molecules into their atoms, and wentually rhe atoms break up into free electrons and nuclei that vre call plasma.
220
Cnerrsn
9
Mess ellp MessRnrrrso PARAMsTEns
VItrA
IA 1
,
H
He
IUA TVA VA YIA
IIA
r.0079
..
'!1 sâ‚¬.
Li
i*"
g:Ot:,
6.941
IITR TVB
,tm
K 39.098
al
,
n
,@:
21
:fr,
Sc
Ti
Fe
Co
Ni
44.956
47.88
50.942
51,W6
54.938
55,M5
58.933
58.69
s9
N
41
42
4{l
44
,+5
.s
Z
zg
27
IB w Cu
IIB 30
Zn 65,39
46
63546 47
48
Rb
si
Y
Zr
Nb
Mo
Tc
Ru
Rh
Pd
Ag
Cd
85.468
8?.62
88.906
9122/
92.W
95.94
98
t01.07
102.906
106.42
107.868
t12.4t1
72
73
74
76
T7
78
80
La
Hf
Th
w
Re
Os
Ir
ft
79
Au
Hg
138_906
178.49
180.948
183,M
186,2t7
19013
19222
195 nfl
196.967
200J9
89
1M
105
107
108
109
110
,1r1
Ac
Rf
Db
Sg
Bh
I{s
Mt
Uun
261
262
x3
262
zo
zffi
269
62
63
Sm
Bu
5036
t'r.9a
Cs 132q]5
87
r*r, ;],;S:r'
Fr
l;:Bs.
223
"lx*1. m.028
Lanthadide series
Actinide series
H
VB YIB VIIB 25 23 24 v Cr Mn
V,M
tigureg.l
5A
59
60
Ce
Pr
Nd
Pm
r40.115
140.90E
144.24
145
90
ot
g2
Pa
U
Th
,n
fia&
,31.M6 2qnmg
1
u
ffi
##ji
11
Na
j
i'41:i $q9s!j
1,;'9,:i
'se
10
Ne 16
Ar 39,948
Kr
i
ffi
49,1?3t
:FT
}i
112
4W
20.180
:'& j'!
Uuu Uub
& Gil
VIIA
83.8
& Xe t3r29 86
Ri 222
118
e6
66
67
e8
6S
m
71
Tb
DJ
Ho
Er
Tm
Yb
Lu
113.M
OU
158,925
1625
t6l.9j
96
97
98
9S
100
101
1m
t74.967 '103
Es
Fm
Md
No
Lr
157
Np
Pu
Axl
Cm
BK
Cf
21',1.lJ48
244
441
)a'7
247
,51
161
16R
,\e
t6t
lhechemicalelementstodate(2001).
Vell, what does all this have to do with mass? Mass provides a quantitative measure ofhow many molecules or atoms ate in a given object. The matter may change its phase, but its mass remains constant. Some of you will mke a class in dynamics where you will learn that on a macroscopic scale mass also serves as a measure of resistance to motion. You already know this from your daily observations. Vhich is harder to push, a motorcycle or a truck? As you know, it takes more effort to push a truck. \V'hen you want to rotate something the disuibution of the mass about the center of rotation also plays significant role. The further anray the mass is " located from the center of rotation, the harder it will be to rotate the mass about that axis. A measrue of how hard it is to rotate something wirJr respect to center of rotation k called mass 'We moruent of ineltia will discuss this in detail in Section 9.5. The other massrelated parameter that we will investigate in this chapter is momentum. Consider two objects with differing masses moving with the same velocity. \Dfhich object is harder to bring to a stop, the one with the small mass or the one with the larger mass? Again, you abeadyknow the answer to this quesdon. The object with the bigger mass is harder to stop. You see that in the game of football, too. If two players ofdiffering mass are running at the same speed, it is harder to bring down the bigger player. These observations lead to the concept of momentum, which we will explain in Section 9.6. Mass also plays an imponant role in storing thermal energ;r. The more massive something is, the more thermal energy you crrn store within it. Some materials are better than others at
9.1 M,lss es n FuNoervrsNTAL DuvrsNsroN
'
bePorc brayirrg this SW,
I
shoul/
221
hatv e coqirAeveA
storing thermal energy. For example, water is better at storing drermal energ;r than air is. In fact, the idea of storing thermal enerry within a massive medium ir firlly utilized in the design of passive solar houses. There are massive brick or concrete foors in the sunrooms. Some people even place big barrels ofwater in the sunrooms to absorb the available daily solar radiadon and store the thermal energy in the water to be used overnight. 'We will discuss heat capacities of materials in more detail in Chapter 11.
222
Cru,r"rnn
9
M.e,ss
9.2
axp MessRBr.erBp PanemsrnRs
fuleasurement of Mass The bihgram is the unit of mass in SI; it is equal to the mass of the international prototype of the kilogram. In practice, the mass of an object is measured indirecdy by using how much something weigtrs. The weight of an object on earth is the force that is exerted on the mass due to the gravitational pull of the earth. You are familiar with spring scales that measure the weight of goods at a supermarket or bathroom scales at home. Force due to gradry acting on the unknown mass will make the spring stretch or compress. By measuring the defection of the spring, one can determine the weight and consequendy the mass of the object that created that deflection. Some weight scales use force transducers consisting of a metallic member that behaves like a spring erccept the deflection of the metal member is measured electronically using suain gauges. You should understand the difference beween weight and mass, and be carefirl how you use tlrem in engineering analysis. Ve will discuss the concept of weight in more detail in Chapter 10.
9.3
Density, Speeifie Volume, and Speeific Gravity In engineering practice, to represent how light or how h exy materials are we often define propenies that are based on a unit volume; in other words, how massive something is per unit volume. Given I cubic foot ofwood and 1 cubic foot ofsteel, which one has more mass? The steel of course! The densi.ty of any substance is defined as the ratio of the mass to tlre volume that occupies, according to
densiw: T*t ' volume
it
(g.l)
Density provides a measure of how compact the material is for a given volume. Materials such *.t"ory or gold with relatively high values of density hare more mass per 1 ff volume or I m' "r volume than those with lower densiry values, such as water. It is imponant to note that the density ofmattsr changes with temperanue and could also change with pressure. The SI unit for densiry is kg/m3, and in BG and U.S. Ctrstomary qystem the density is expressed in slugs/ff, and
lb/ff
, respectively.
Specifu oolume, which is the inverse of densiry, is defined by specificvolume
:
volume
(e.2)
mass
Specific volume is commonly used in the srudy of thermodynamics. The SI unit for specific volume is m3/kg. Another cornmon way to represent the heaviness or lighmess of some material is by comparing its density to the density of water. This comparison is called rhe specifu graai.ty of amaterial and is formally defined by specific Savity
=
densiw of a material density of
[email protected]"C
(e.3)
9.3 Dwsrrr, Sprcrrrc VoluuE, AND Spscrrrc
Gnevrrr
223
It is imponant to note that specific graviqf is unitles becanse it is the ratio of the value of rwo densities. Therefore, it does not matter which qptem of units is used to compurc the specific gravity ofa subsance, as long as consistent unirc are used. Specfu weight is another way to measure how truly hearry or light a material is for a given volume. Specific weight is defined as the ratio of the weight of the material by the volume that it occupies, according to specific weight
=
weight
(e.4)
volume
Again, theweight of something on eanh is the force that is er<ened on its mass due to gravity. Ve will discuss the concept ofweight in more detail in Chapter 10. In this chapter, it is left as an exercise (see Problem 9.4) for yoa to show the relationship berween density and specific weight:
specificweight
:
(density)(acceleration due to gravity)
The density, specific gravity, and the specific weight of some materials are shown in Table 9.1.
iffi Material
Deneity (kg/u)
Specific
Gravity
SpecificWeight (N/mu)
Aluminum
2740
2.74
Asphalt
2110
2.11
26,990 20,700
Cement
1920
1c),
18,840
clry
1000
1.00
9810
Fireclay brick
[email protected]"c
t.79
Glass (soda lime) Glass (lead)
2470
2.47 4,28
17,560 24,230
4280 2230
Glass (Pyrex) Iron (cast)
72r0
'7
Iron (wroughr)
[email protected]'c
7.70
Paper
930 7830 7860 690 550
0.93
Steel
(mild)
Steel (sainless 304) \trood (ash)
Vood (mahogany) W'ood (oak) !(ood (pine)
750 430
2.23
)1
7.83 7.86 0.69 0.55 0.75 0.43
4l,gg0 21,890 70,730 75,540 9120 76,810 77,L10 6770 5400
7360 4220
Ftuids Standard air Gasoline
Glycerin Mercury SAE 10
Vater
Woil
1,.225
720
t260 L3,55A
920
[email protected]"c
0.0012 0.72 1.26 13.55 0.92 1.0
t2 7060 12,360 132,930 9030 9810
224
Cneprsn
9
Mess exp MessRsLATEp PlnetvrsrEns
9.4
Mass Flow Rate discussed the significance ofvolume fow rate. The mass flow rate is another dosely relared parameter that plays an imponant role in many engineering applications. There are many industrial processes that depend on the measurement offuid flow. Mass fow rate cells engineers how much material is being used or moved over a period of time so that they can replenish the supply of the material. Engineers use fl.owmeters to measure volume or mass forv rates of water, oil, gas, chemical fluids, and food products. The design of any fow measuring device is based on some known engineering principles. Mass and volume flow measurements are also necessary and common in our weryday life. For example, when you go to a gasoline service station, you need to know how many gallons of gas are pumped into the tank of your car. Another example is measuring the amount of domesdc water used or consumed. There are over 100 types ofcommerciallyavailable flowmeters to measure mass or volume flow rates. The selection of a proper flowmeter will depend on the application type and other variables, such as accuracy, cosr, range, ease ofuse, service life, and rype offuid: gas or liquid, dirry or slurry
In Chapter 8, we
or corrosive, for example.
Themassfuwrateissimply defined by the amount ofmass that flolrs through something per unit of time. mass
^ rate flow
mass : :
(e.5)
ume
Some of the more common units for mass fow rate include l1g/s, lg/min, lg/h or slugs/s or lb/s. How would you measure the mass fow rate of water coming out of a faucet or a drinking fountain? Place a cup under a drinking fountain and measure the time that it takes to fill
the cup. Also, measure the total mass of the cup and t}re water and then subtract the mass of the cup from the total to obain the mass ofdre water. Divide the mass ofthe water by the time interval it took to fill the cup. 'We can relate the volume flow rate of something to its mass flow rate provided that we know the density of rhe fowing fluid or fowing material. The relationship between the mass flow rate and the volume fow rate is given by mass
^ rate flow
mass : : .
(density)(volume)
tlme
=
umâ‚¬
/, ,/volume\ : \oensrry;l :\umel I
(9'6]
(densiry)(volume fow rate)
The mass flow rate calculation is also important in excavation or tunneldigging projects in determining how much soil can be removed in one day or one week, trlring into consideration the parameter of the digging and transport machines.
9.5
lVlass Moment
of Inertia
As menrioned eadier in this chapter, when it comes to the rotation of objecrs, the disuibudon of mass about the center of rotation plrys significant role. The funher arvay the mass is Iocated from the center of rotarion, the harder it is to rotate the mass about the given center of
.
9.5
I
I
rigurc9.Z
Ihe mass moment of inedia of a loint mass.
Mess MoMsur or INrnrra
225
rigureg.l
ltass moment of inertia of a system consisting of three point masses.
rotation. A measure of how hard it is to rotate something with respect to center of roadon is calledmass monent ofinertia. All ofyou will take a class in physics and some ofyou may even rake a dynamics class where you will learn in more depth about the formal definition and formulation of mass moment of inenia. But for now, let us consider the following simple siruations, as shown in Figarc9.Z. For a single mass particle m"located.at a distance rfrom the axis of rofation z z the mass moment of inerda is defined bv
f": lrn
(e.1)
Now let us expand this problem to include a system of mass particles, as.shown in Figure 9.3. The mass moment of inenia for the system of masses shown about the zz axis is now
fu=
/21m1
* 4*r+ 4*u
0.8)
Similady, $/e crrn obtain tlre mass moment of inenia for a body, such as a wheel or a shaft, by summing the mass moment ofinertia of each mass panicle rhat makes up the body. fu you uke calculus classes you will learn that you c:rn use integrals instead of summations to evaluare rhe mass moment of inenia of continuous objects. After all, the integral sign / is nothing but a big "So sign, indicating summation.
I":
t Jr2
drn
(9.9)
The mass moment of inenia of objects with various shapes can be determined from Equa
rion (9.9). You will be able to perform this integation in another semester or two. Examples ofmass moment ofinenia formulas for some typical bodies such as a cylinder, disk, sphere, and a thin rectangular plate are given on the following pâ‚¬e.
Cneprsn 9
Mess
lNp MessRrrersp
PARAMeTnRs
0isk
I
I,^: 1mRz
(er0)
2
tircular tylinder
I,:l*R2
(ell) z I I
,R
I
z
Splrcre
I:
a
mR" 5
t9.12)
9.5 Mnss Morurr.rr or Ixrnrre ilrte
Ihin fledangalar
L":
221
f,*w
(eJ3)
Determine the mass moment inenia of a steel shaft that is 2 m long and has a diameter (/) of 10 cm. The density of steel is 786A1 ( n)
\_,v
I
t'
weight of the book is pushing down on the desk, simuhaneously the desk also pushes up on the book Otherwise, the desk won't be supporting the book. Newton's third lar states that for every acdon there exists a reaction, and the forces of action and reaction harre the same magnitude and act along the same line, but they have opposite directions.
Newton's taw sf Gravitation The weight of an object is the force that is exened on the mass of the object by the eartht gravity. Newton discovered that any two masses, m1 and til2t atttta'ct each other with a force that is equal in magnitude and acts in the opposite direction, as shown in Figure 10.7, according to the following relationship:
Figurcl0.7
The gravitational attraction
two nasses.
of
r'E 
Grn,m'
:=
J
(r0.6)
10.2 Nsv'rox's Levs
rN MBcneNrcs
243
where
F: G: rz1 :
attractive force (N)
7n2:
rtrass
r:
universal gravitational constant (6.673
x
l011 m3lkg's2)
rmss of panicle 1 (kg, see Figure 10.7)
ofparticle 2 (1g,
see
Figure 10.7)
distance benlreen the center of each particle (m)
Using Equation (10.6), we c:rn determine the weight of an object having a mass m, ont*'e eanh, by substinningmfor the mass of particle 1 and substituting for the mass ofpanicle 2 the mass of Earth (Mr,*:5.97 x 7024 kg), *d using the radius of Earth as the distance between the center of each panicle. Note that the radius of Earth is much larger than the physical dimension of any object on Eanh, and, thus, for an object resting on or near t}re surface of Eanh, r could be replaced with Rsrr6 : 6378 X 103m as a good approximation. Thus, the weight (W) of an object on the surface of Eanh having amass rn is given by GMp^tyrn
ff:
Ril.h
And after letting
GMwa
g:
rZ, r
v=
We Cirn
Wtlte
\Earth
*g
fl0.7)
Equation (10.7) shows the relationship amongtheweight ofan object, its mass, and the locd acceleration due to gravity. Make sure you firlly understand this simple relationship. Because the eanh is not truly spherical in shape, the value ofg varies with latitude and longitude. However, for most engineering applications, g: 9.8 m/s2 or g: 32.2 ft/s2 is used. Also note that the value ofg decreases as you get funher away from the surface of Earth. This fact is evident from examiningEquation (10.6) and, the relationship forgleading to
Equation (I0.7). Determine the weight of an orploration vehicle whose mass is 250 kg on Eanh. !7h.at is the mass of the vehicle on the Moon (3Moo, : 1.6 mA2) and the planet Mars (gr* : 3.7 ml*)? 'Vhat is the weight of the vehicle on rhe Moon and Mars? The mass of the vehicle is 250 kg on the Moon and on the planet Mars. The weight of the
fromW'=
vehicle is determined
:
on Eanh: Yr I
On
I
L.
:; t;il;t:
rl:i.,t..,r]::l:':.:
r.s)(l.a
S)
:
2450
N
Moon:
'!7'
: (25oks)(tt:) :400N
Mars:
'!7'
= (2io ks)(r
:::l',:'
:.:
On :::i:
(250
mg:

: 
r$) :925N .::::l,I:=,rr
244
Cnlpren
l0
Foncn eNo FoncsREr,Arso PenervrnrsRs The space shutde orbits the Eanh at altitudes from as low as 250 km (155 miles) to as high as 965 km (600 miles), depending on its missions. Determine the value of gfor an asffonaur in a space shutde. If an astronaut has a mass of 70 kg on the surface of Eanh, what is her weight when in orbit around Earth? When the space shuttle is orbiting at an altitude of 250 km above Eanh's surface:
/
g:_ :f
:
GMr^* _leezz
ff
(70
ks)('rS)
x l0.
^'
#)o.,
x
*)O., KE.
x
1024ks)
=
9.o7 mr's:
:635N
At an altitude of 965 km:
GMa,*
o:
(u.urux ro' \
R2
ff=
tzo
rgl(z.aa
l(6378
x
ro3
+ 9(J x
ro'z4ks)
103)ml2
738 mlt2
*) = 517N
Note that at these altitudes an astronaut still has a significant weight. The near weighdess conditions that you see on TV are created by the orbital speed of the shutde. For o"
The total resistance to heat flow is given by
)R: &+&+&+ &+Rs+& : q:
0.I7 + 0.81 + I.32 + 11.0 + 0.45 + 0.68 (z^ua"
 4"*a")l _ (70  2ox15o) 14.43 )n
:
14.43
:520 Btu h
The equivalent thermd resistance circuit for this problem is shown in Figure
filn Insulation batt
resistance
To=20"F
ii:li.,tirfi
Figure
12.
Inside film
Outside resistance
I
1 1.
4
ll.l2
i:tir
The equivalent
themal resistance for
.   ,l:d,,i 
Example
.,Ll'tir! .    ::iri
= 70"F
ll.ll.
r
.:..1!i1.

Windchill Faetor As all of you know, the rate of heat transfer from your bodyto the surrounding air increases on a cold, windy day. Simply stated, you lose more body heat on the cold, windy day than you do on a calm, cold day. By now you understand the difference between natural and forced convection heat transfer. The windchill inden is intended to account for the combined effect of wind speed and the air temperature. Thus, it is supposed to account for the additional body heat loss that occurs on a cold, windy day. Howwer, keep in mind that the common windchill correlations are based on a series of experimens performed in a very cold climate, where the time that it took to freeze water in a plastic container under various surrounding air temperatures and wind conditions was studied. A common correlation used to determine the windchill index is
\?cr:
(10.45

v+ n\/v)(33

where
: I/: f :
!V'CI
windchill index (kcal/m2.h) wind speed (m/s) ambient air temperature (oC)
T,)
(fl.2r)
11.3
TSMpSRATURE
DrrrnnsNcsANo HEAI TnaNsrsn
".r
Ambient Temperafirf,e ("C)
Wind
305
Speed
(lm/h) 20
3.6
30 40
1.0
50 60 70 80
4.7

1.9
1 7 3.2 3.4
*2.8 6.0 8.1 9.5 r0.4
5
10
15
20
9.2 *r2.9
15.5 19.9
'r',
28.4 33.8 37.4 39.8
34.8 40.7
 l),.*
22.0 26.8 30.0 32.2 33.7 34.6
18.2
11.0

11.3
19.3
18.9
'7
24.7 25.9 26.8 27.2
35.r
_25
4r.2 47.7 52.0 55.0 57.A 58.2 58.8
44./
4r.5
47,4 49.2
42.5 43.0
50.3 50.9
ro 47.6 54.6 59.4 62.6 64.7 66.1.
66.8
and the value 33 is the body surFace temperature in degrees Celsius. The more common equiv
alent windchill temperaflre 7*oi,rr*,
Qoi,,r*,
:0.045(5.27v0'5
+
fC) 10.45
is given

by
0.28v)(7"
33)
+
33
(11.U',t
Note that in Equation (1L,22),Zis expressed in kilometers per hour (km/h). Thble 11.6 shows the windchill temperafiues for the range of ambient air temperanrre 30oC < T^ < 10'C and wind speed of 20 km/h < V< 80 km/h.
Radiation All maner emits thermal radiation This nrle is true as long
as the body in question is at a nonzero absolute temperature. The higher the temperature of the surface of the object, the more thermal energT is emimed by the object. A good example of thermal radiation is the heat you can literally feel radiated by a fire in a fireplace. The amount of rafiant energ;,' emimed by a surface is given by the equation
q: soAT!
fl1.23)
where q represents the rate of thermal energy, per unit time, emitted by the surface; e is the emissivity of the surface, 0 a 1, and o is the StefanBolv,mann constaint (c = 5.67 X . 108 s7/m2 Ka); A represents the area of rhe surface in m2, and is the surface temperanre of the object enpressed in Kelvin. Emissivity, e, is a properry of the surface of the object, and its value indicates how well the object emits thermal radiation compared to a black body (an ideal perfect emirer). It is important to note here that unlike the conducrion and convection modes, heat transfer by radiation can occur in a vacuum. A d"ily orample of this is the radiation of the sun reaching the earth's atmosphere as it tranels through a vacuum in space. Because all objects emit thermal radiadon, it is the net energy exchange among the bodies that is of
(
(
{
interest to us. Becarse of this fact, thermal radiation calculations are generally complicated in nature and require an indepth understanding of the underlying concep$ and geometry of the problem.
306
Cner.ren
1l
TSMpTRATuRE eNp
TruprnarunrRrr,nrso
Pan.ervrsrgRs
day, the fat roof of a tall building reaches 50oC in temperature. The area of the roof is 400 m2. Estimate the heat radiated from this roof to the sky in the evening when the temperanre of the surrounding air or sky is at20"C. The temperature of the roof decreases as it cools doqrn. Estimate the rate of enerry radiated from the roo{, assuming roof temperatures of 50,4O,30, and 25oC. Assume e = 0.9 for the roof. 'W'e can determine the amount of thermal energy radiated by the surface from Equation (11.23). For roof temperature of 50oC, we get
On a hot suruner
q
= aaAT!
:
Ps1(s.ez
x
ro'(;fu) )am mz)(tzz K)4 : 222,ooov
The rest of the solution is shown in Thble 11.7.
rffilffi i..
'ri.l
Surfacâ‚¬
Temperature
Cc)
. liri..
7(K)
(XQ
= TCq +n3
50
323
4A
313 303 298
30 25
:.
Surface Temperature
il.lrl*lrr.'rt'?i:l,liTl.iiii.li..:

Energy Emitted by the Surfrcâ‚¬ (w)
' q ='esAT: (0.9)(5.67 (0.9)(5.67 (0.9)(5.67 (0.9)(5.67
x 108X400X32r4 x 10)(400)(313)a : x rO 1(400X303)a : x 108X400X298)a :
222,0a0 196,000
V w
* 161,OOO * 172,sOO
ijir]i'ii.qi
,li
Most of you will take a hear transfer or a transport phenomenon class during your third will learn in more detail about various modes ofheat transfer. You will also learn how to estimate heat transfer rates for various situations, including the cooling of electronic devices and the design of fins for transformers or motorcycle and larrn mor,\r'er engine heads and other heat errchangers, like the radiator in your c:u or the heat enchangers in firrnaces and boilers. The intent of this section was to briefy inuoduce you to the concept of heat uansfer and year where you
its various modes.
11.4 Thcrmal Comfort, Metabolic Rate, and Clothing Insulation Human thermal comfoft is of special importance to bioengineers and mechanical engineers. For example, mechanical engineers design tJre heating, ventilating, and airconditioning (HVAC) qrstems for homes, public buildings, hospitals, and manufacturing facilities. Vhen sizing the HVAC qrstems, the engineer must design not only for the building's heat losses or gains but also for an environment that occupants feel comfonable within. Vhat makss us thermally comforable in an environment? As you know, the femperaflrre of the environment and the humidity of the air are among the imponant factors that define thermal comfon. For example, most of us feel comfonable in a room that has a temperanre of70'F and a relative humidiry of40 to 50o/o. Of course, ifyou are enercising on a ueadmill and watchingTV then perhaps you feel more comfonable in a room with a temperanue lower than 70"F, say 50' to 60oF. Thus, the level of activity is also an important factor. In general, the amount of energy that a
LL.4
Tnrnrvrar, CouFonr, Msranor.,rc Rrts, AND Clorrrnqc
Genemtiorr
Heat
@tulf..ff) .
:
INsurarron
307
Heat Gi:neration
'' ,(mer)
,1
Resting Sleeprng
t3
0,7
Seated, quiet
1U
Standine, relaxed
22
1.0 1.2
Walkingon Lnel Surface
2 miles/h 3 miles/h 4 miles/h
2.A
J/ 48
2.6
xe
7A
OfueWorh
18
Reading, seated .W'riti"C
TyPc
'.20'
Filing, seated Filins, stan.liqg
... ,1.0 I,0"
,,.:
.ir18
r
,'
DriuinglFlying
:'
Car Aircraft, routine
,
Afu craft, insbrument
r.4
33 44
',.
House cleaniqs
W'restling,_
'
1.6 rc 2.0
37 to'63
2.0 to 3.4
,
'::,
''i ': '
44 to
'
competidve
3:2
to,37
29;
looEm8
Caliqihenics/errcrcise. Tennis, singles Basketball
2,4 ,"
'
MiscelldneoatiHousiwork'
1.0 to 2.0 ',1,2':,,
1.8
.
59'
M is e lknaus lrisurc Aaia ities
r,
.'
22'
landing
,
1.2
'18 ro 37
Aircraft, combat Hearyvehicle
Dancing *"1n1
, t;1
..',:1
,
22 26
.
8l
2,4w4.4
55 ro'74 66 ta74.
3.0 to 4.0 3.6 to 4.0
90 to,140 130 to'160
7.0 to8.7
5,Aw7.6 ,l.,.
Source:
i
Amsicat Society of Heatingi Reftigeradng and AirConditioning Engineen.
â‚¬e, gender, size, and acdvity level. A person's body temperature is controlled by (1) convective and radiarive heat transfer to the surroundings, (2) sweating, (3) respiration by breathing surrounding air and orhaling it at near the body's temperanue, (4) blood circulation near the surface of the skin, and (5) metabolic rate. Meabolic rate determines the rate of conversion ofchemical to therrnal energTwithin a person's body. The metabolicrate depends on the person's activitylevel. Aunit commonly used to â‚¬xpress the metabolic rate for an a/erage person under sedentary conditions, per unit surface area is called met; '!0'lm2 or, in U.S. Customary units, 1 met = 18.4 Btu/h. ff. For an met is equal rc 58.2 .verage person, a heat uansfer surface area of 1.82 m2 or 19.6 fi3 was assumed when defining dre unit of met. Thble I 1.8 shows the metabolic rate for various activities. As you expect, clothrng ako affects thermal comfort. A unit that is generally used to o<press the insulating value of person generares depends on the person's
308
Cner'rnn11
TieMpBRATURxeNoTnMpsRATURERrr.ersoPanerrnreRs
;,fi1,,,X1.1;T;'
r', . .
f l'. q,t..,uill..lirlLi*ui;,
InsulationValue (do)
i
I clo = 0.155 m2.'C/\tror I do = 0.S8"F.ff.h/Btu
Clothing
'
Valking shorts, shortsleeve shirt Trousers, shonsleeve shift Tiousers, longsleeve shin
0.36 0.57
Trousers, longsleeve shirt, plus suit jacker
0.96
Same as above, plus vest and
0.51
l.t4
Tshirt
Sweatpants, sweatshirt
I I
KneeJength skin, shortslewe shin, panty hose, sandals Kneelengh skirt, shortsleeve shirt, full slip, panty hose KneeJength skirt, longsleeve shin, halfslip, panty hose, longslewe sweirter
1.10
Same as above, replace sweater Soarce: Americat Society of
a.74 0.54 a.67
with suit jactet
t.04
Heating Refrigerating, and AirConditioning Engineers.
clothing is called clo. I clo is equal to 0.155 m2. oC/\% or, in U.S. Customary units, 1 clo 0.88"F . ff . h/Bru. Thble 11.9 shows the insulating values ofvarious clorhing.
:
Use Thble 11.8 to calculate the amount of enerry dissipated by an average adult person doing the following things: (a) drivine a qr for 3 h, (b) sleeping for 8 h, (c) walking at the speed of 3 mph on a level surface fot 2h, (d) dancing for 2 h.
(a) Using
average values, the amount of energT dissipated by an ayerage adult person
driving a
carfor3his
(ts t rz)(39)t,u \ z ./ \h.,, 2 (b) Using
average values, the
6fc)Qh):1617Bru
amount of energy dissipated by an average adult person sleeping
for8his
/ ntu \. t3(ffir(te.6rc)(8h)
: 2038 Bru
(c) ITalking at the speed of 3 mph on a lwel surface for 2hl
U
^t(#)tt',rfl(zh)
lil n
1882Btu
(d) Dancing for2h
(t+")(#)," 6re)eh):2450sn,
H tli ll
ji*F&:(..:,Liattf"ii.t
:
In Problem II.23,you :allll.
t
:

=f
.

are asked
to converc these results from Btu to calories.
:.lirrr].ali1ri.{l:x'1fif.1tiiltEi,!1]i
lf .5
trl.5
Sons TgMpsnerunnRnr,rrno MersRrAL PnopsRTlss
309
Some TemperatureRelated Material Properties
Thermal Expansion As we mentioned earlier in this chapter, accounting for thermal eFansion and contraction of materials due to temperature fuctuations is important in engineering problems, including the design of bridges, roads, piping systems (hot water or steam pipes), engine blocks, gas turbine blades, elecuonic devices and circuits, coolanrare, tires, and in many manufacnrring processes. In general, as the temperature of a material is increased, the material will expandincrease in
lengthand if the temperature of the material is decreased, it will contractdecrease in length. The magnitude of this elongation or contraction due to temperature rise or temperature drop depends on the composition of the material. The coefficient of linear thermal expansion provides a measure of the change in lengdr that occurs due to any temperanue flucnrations. This effect is depiced in Figure 11.13. The coefficient oflinear expansion, ap is defined as the change in the length, AZ, per original length Z per degree rise in temperature, AZ and is given by the following relationship:
o'=
AL
fiF
fl1.24)
Note that the coefficient of linear e4pansion is a properry of a material and has the units of 1/"F or l/oC. Because a 1" Fahrenheit temperature difference is equal to a 1" Rankine temperanue difference, and a 1o Celsius temperaftue difference is equal to a 1 Kelvin temperature fifference, the units of a1 can also be expressed using 1/oR and 1/K The values of the coefficient of linear expansion itself depend on temperanue; however, average values may be used for a specific temperature range. The values of the coefficient of thermal expansion for various solid materids for a temperature range of 0"C to 100"C are given in Thble 11.10. Equation (I1,24) could be expressed in such a way as to allow for direct calculation ofthe change in the length that occurs due to temperarure change in the following manner:
LL:
qtL
LT
fl1.25)
For liquids and gases, in place of the coefficient of linear expansion, it is customary to define the coefficient of volumetric thermal expansion, ay. The coefficient of volumetric er<pansion is defined as the change in the volume, LV, per original volume Tper degree rise in temperaflre, AZ, and is given by the following relationship:
AV or: vtf
01.26)
Note rhat the coefficient of volumetric expansion also has the units of lloF or 1/"C. Equation (1 1.26) could also be used for solids. Moreover, for homogeneous solid materials, the
il
fiqurell.l3
The expansion of a material
caused by an increase in its temperature.
310 Cner"rsRll
ThtvrpnnerunsaNpTielrpsRATuRERnranspPenanusrsRs
(lfC)
MeanYalue of ar
I nrick r Bronze r Cast iron I Concrete Glass (plate) Glass (Pyrex) r Glass (thermometer) Masonry Solder , Stainless steel (AISI 31O Steel (hardrolled) Steel (softrolle0 Wood (oak) normal to fiber 1 Vood (oak) parallel to fiber I Wood (pine) parallel to fiber Source:T. Banmeister,
a
2.5
x
5.3
x 10"'6
10.0
X 106
2.9 5.5 3.3
5.9 X 106 8.0 X 106 5.0 x 106 1.8 X 106 4,5 x l}G
L065.0
x 2.9 x
13.4
X
x X
106 106 106
4.4x rc6 2.8
X
106
x 106 2.5 x 106 t.4x 1062.8 x 1.0
x 106
7.4X
LO6
106
1.06
x 106 x 106 x 106 L.7 x LO6 1.5 x 106 1.7 x LO6
l}G
1.6
5.6 X 106 6.3 x 10e 3.0 X 106 2J X 106 3,0 X 106
3.1 3.5
aL, Marh's Handbooh.
relatioruhip between the coefficient of linear enpansion and the coemcient of volumetric expansion is given by
0'y:3a1
flr.27)
Calculate the change in length of a 1000ftlong stainless steel cable when its temperature changes
by 100"F.
IZe will use Equation (11,.25) and Thble 11.10 to solve this problem. From Thble 11.10 the coefficient of thermal expansion for stainless steel is q. : 2.9 X 106 lfF. Using Equation (11.25), we have
AL: arl 67: l!uru,=.,,,,,..
,,.
liiiiiaiiitrltlii,iilli''
:!.
(2.9
1
X
106 1l'F)(1000 ft)(100"F)
=
0.29
ft :
 l:..:,
3.48 in. a.r:'rr,r!. ,i.r,iiiL,:iir,::i:
Speeifie Heat Have you noticed that some materials get hotter than others when exposed to the same amount of thermal enerry? For example, if we were to expose 1 kg of water and 1 kg of concrete to a heat source that puts out 1 00 J every second, you would see that the concrete would experience a higher temperature rise. The reason for this material behavior is that when compared to water, concrete has a lower heat capacity. More explanation regarding our observation will be
given in Example 11.16.
lL.5
Sorar ThupBnerunpRsr^arED MrrsRrAL Pnoprnrrss
311
Specfu beatprovides aquantiativewayto showhowmuch thermal energyis required rc raise the temperanue of a I kg mass of a materid by 1' Celsius. Or, using U.S. Customary units, the specific heat is defined as the amount of thermal enerry reguired to raise the tempeftrnue of a lJb mass of a material by 1' Fahrenheit. The values of specific heat for various materials at consant pressure are given in Thble 11.11. For solids and liquids in the absence of
any phase change, the relationship among dre required thermal energ;r (Ea.a), mass of the given material (rn),its specific heat (e), and the temperature rise (Tn a  d"i6j that will occur is grven
by
F,h.,r: mc(766 7kd
fl128)
where
4r*a : ln: c: T6ra T6,*j:
thermal energ)r (J or Btu) mass
(kgorlb)
specific heat (Jlkg. K or
temPerafiue rise ("C or
J/kg.'C or Bn/lb. .'R or Btr:/lb. . "F)
K
oF
or "R)
Next we will look at nvo examples that demonsuate the use of Equation (11.28).
ii
i,
l:_.::...:.
i
Spt$fic lleat (Y[S. IO
Maretial
1
eir(atatrngsptrgricpressure) Alqgrinuq (pure) Alumiirum'alloy'2024.T6 (4.5% copper,l.5 7o Mg; 0.6% Mn) ' r AsPhali ',, 
,
,
i
i
903
)
875
l
,
j
l
I
Bronze (907o
spper,
1
0olo.
Brass (707o coppetl 30o/o
aluminutn)
tnc) ,
,.
380
:
Brick (fue clay)
Concrercl
l"
l.
:
roPpâ‚¬r (P$e)
Glass
,
lprre)
.Sqinless steels
Lead
I l.
I
:
i
Pap"l Pletinurn (pue)
Sand
.
(AISI 302, 304, 316, 347)
960
,
,
880
l
750 '. L29
:
Gold Iron
,1 , 4f0,
\
I
r
:.r
t,
O
r29 t340 r33,
r
Silicon
7r2
1
Silver
i!
Zilrrc
235 389
i:Vater(iquid)
:"
arso
i 1
447
48b,477,46t,n
,
, r
r
"
312
CnarrsR
1l
Truprnarunp ANo ThtvrppnerunsRr,r.ereo
PenervrsrsRs
An aluminum circular disk with a diameter, d, of 15 cm and a thickness of 4 mm is er<posed to a heat source that pua out 200 J every second. The density of the aluminum ts 2700 k/'. Assuming no heat loss to the surrounding, estimate the temperature rise of the disk after 15 s. 'WewillmakeuseofEquation (11.28) andThble ll.ll tosolvethisproblem, butfirstwe need to calculate the mass of the disk using the information given. mass
=m=
(density) (volume)
: V: ja, (U.U;rro.; : 110.5 m)2 (o.oo4 m) : 4 ' 4', m: (7.06858 x l05 m3)(zZoo kg/*') :0.191kg
Volume
F,h*,r:
mc(766

X t05
m3
Z.a)
:
(0.191
ksx875llks, KX[",r
(Ta*t
7fu6"1)
:
200J
7.06g58
1.2K

7ir.a)
(werysecond)
And after 15 s the temperature rise will be 18'C or 18
K
,1,u,
'We
hane enposed 1 kg ofwater, 1 kg of brick, and 1 kg of concrete, each to a heat source that puts out 100 J every second. Assuming that all of the supplied energy goes to each material and they were all initially at the same temperature, which one of these materials will have a greater temperarure rise after 10 s? 'We can answer this question using Equation (i 1 ,28) and Thble 1 1.1 I . \7e will first look up the values of the specific heat for water, brick, and concrete, which are dot* = 4180 J/kg. K db,i.I : 960 Jlkg. K and f,oo* : 880 J/kg. K. Now appl)4ng Equation (11.28), Eh*; : ruc(76o1 4."J to each situation, it should be clear *rat although each material has the same amount of mass and is exposed to the same amount of thermal energF, the concrete will e4perience a higher temperatrue rise because it has the lowest heat c:rpacity value among the three given materials.
1
  l
n1.6 Fleating Values of Fuels As engineers you need to know what the heating value of a fuel means. !Zhy? \7here does the energy that drives your c:r come from? Vhere does the energy that makes your home warm and cozy during the cold winter months come from? How is electricity generated in a conventional power plant that supplies pov/er to manufacuring companies, homes, and offices? The answer
rc all of these questions is that the initial energ;r comes from fuels. Most conventional fuels drat we use today to generate power come from coal, natural gas, oil, or gasoline. All these fuels consist ofcarbon and hydrogen. When a fuel is burned, whether it is gas, oil, etc., thermal energy is released. The heating value of a fuel quantifies the amount of enerry that is released when a unit mass (kilogram
ll.6
Density (lb/gal)
Heating Value (Btu/gal)
I
6.950 to 6.675
$7,040 to 132,900
2 4
7.296ro 6.960 7.787 to7.396
141,800 148,100 150,000 152,000 155,900 108,500
.940 to 7 .686 8.080 to 7.890 8.448 to 8.053 to 6.2
7
5H 6
6.0
Gasoline Source: Ameicen Society of
, , i
to 137,000
to 143,100 to to to to
146,800
149,400 151,300 117,000
Higher Heating Value
Musselshell, Montana Emrcy, Utah Pi[c, Kenrucky Cambria, Pennsylvania
HeatingValue
(Btu/lbJ
HeatingValue (Btu/CI
12,075 13,560 r5,040 15,595 13,770
Williamson, Illinois McDowell, I7est Virginia
, i . I ,
15,600
Source:Babcock and \Filcox Company, Stearn:
Ia
313
Fuer.s
Heating Reftigerating and AirConditioning Engineets.
@tu/lbJ
Source ofGas
and
Vlrurs or
Grade No.
5L
Coal &om C.onnf andState of
HsATrNc
Generation
23,170 22,9A4 22,477 21,824 20,160
Pennqylvania
Southern California
Ohio Iouisiana Oklahoma
Soarce: Babcnik and
Wil*r
op*y,
Steam:
and 30
@
60'F
in. Hg
Ir29 1116
964
t0a2 974
Ia Gmaation ar) Ilsr.
Use.
or pound) or a unit volume (cubic meter or cubic foot) of a fuel is burned. Different fuels have different heating values. Moreover, based on the phase of water in the combustion product, whether it is in the liquid form or in vapor form, nvo different heating values are repomed. The higher heating value of a fuel, as the name implies, is the higher end of energy released by the fuel when the combustion byproducts include water in liquid form. The lower heating value refers to the amount of energy that is released during combustion when the combustion products include water vapor. The typicd heating values of liquid fuels, coal, and natural gas are given in Thbles 1 1 .12 through 1 1.14.
How much thermal energ;r is released when 5 lb. of a coal sample from Emroy, Utah is burned? The total amount of thermal energy, Er*a, released when some fuel is burned is determined by muldplying the mass of the fael, m" by the headng value of fuel, .F1y. Ea,,na 1
= mHv:
(5
rb)13,560(tr)
: 67,800 Btu
314
CneprsR
ll
Thuprnerunr ervo TnMpsRATURERrranro Panaurrsns Calculate the total amount of thermal energy released when 60 is burned inside a gas fumace.
Eh*"r Ii .*liiGi*!i?ir,Ir::i iili'
:
(60
ff) tmr(#) :
60,120 Btu
ff
of natural gas from Louisiana
Pnonr,sMs
315
SUil4TilARY Now that you have reached this point in
r:he
text
"
You should have a good undersanding of what temperafi.ue rneans, how it is measured, and how heat transfer occurs.
'
You should know the difference among temperanre scales and understand how they are related.
r(.c) :
I
;lT
(.F) 
r(.F):]lr{"r)l*r,
321
7(K):r('C)+273.15 r(K)
' ' ' ' ' '
:
a
r('R)
t.(.*)
fl.3.
a. \7hat b. \fhat
areas.
is a normal body temperature? is ttre temperature range that clinically is re
ferred to as fever?
fl.4.
body? Is this value consunt?
d. \V'hat is a comforable room temperature
range?
is its significance in terms of human thermal comfon? \Vh.at is the role of humidity?
e. 'What is the operating temperame range of the
11.2.
condenser in a household refrigerator? The condenser secdon of a household refrigerator is the black tubing in the back ofthe refrigerator where heat is being rejected to the surrounding air. Using Excel, or a spreadsheet ofyour choice, create a degrees Fahrenheit to degrees Celsius conversion table for the following temperanue range: from 50oF to 130oF in increments of 5oF.
Alcohol thermometers range
of
scale
will

czrn measure
temperatures in the
100oF to 200oF. Determine the
read the same number as a thermometer
with
a Celsius scale.
c. What is a normal surface temperanrre of your '!7hat
?.,o,
rcmperanre at which an alcohol thermometer with a Fahrenheit
items. !?'rite a brief report discussing how tfrese values
in their respective
:
You should understand what absolute zero temperaflrre meurns. You should know vrhat we mean by the term heat, know its common units, and be familiar with different modes of heat transfer. You should know what the Rvalue or Rfactor for insulation means. You should be able to perform some simple heat transfer cdculations. You should be familiar with some thermophpical properties of materials such as thermal conductivity, specific heat, and thermal expansion. You should undersnnd what the heating value of a fuel means.
flJ. Investigate the value of temperature for the following are used
7(R):rfF)+45e.67
1r.5.
Obain information about K, E,
and" Rtype
thermo
couple wires. Vrite a brief repon discussing their accuracy, temperanue range of application, and in what application they are commonly employed. A manufacnrer of loosefill cellulose insulating material provides a table showing the relationship between _
Rvalue
(units?)
R40 R32
R24
Rr9 R.T3
Thickness
ll 9 6.5 5.25 3.5
:__ _
(in.)
i
',
316 Cner"rnnll
ThupnnrrtrnnnNpTieMpnRATunrRnretnoPenelrsrERs
the thickness of the material and its Rvalue. The man
the accompanying table. Assume an inside room tem
ufacturer's data is shown in the accompanying able. Calculate the thermal conductivity of the insulating materid. Also, determine how thick the insuladon should be to provide Rvalues of
peranre of 68'F and an outside air temperature of 10'R with an exposed area of l5O tC. Calculate the
a. &30
b. il.6.
R20 Calculate the Rvalue for the following materials:
a. 4in.thick brick
b. lOcmthick brick c. l2in.thick concrete slab d. 20cmthick concrete slab e. lcmthick human fatlayer
llt.
il.8.
Calculate the thermal resisfance due ro convection for the following situations: a. warm water with / : 200 !7'lm2. K b. warm air with h : I0 \7/m2. K c. warm moving air (windy situation) I : 30 V/m2. K A typical exterior masonqF wall of a house, shown in the accompan)4ng figure, consists of the items in
heat loss through the wall.
n.9. In order to increase the thermal
resistance of a typical exterior frame wall, such as the one shown in Example 11.11, it is customary to use 2 X 6 studs instead of 2 X 4 studs to allow for placement of more insulation within the wall cavity. A typical exterior Q x 6) frame wall of a house consists of the materials shown in the accompanyingfigure. Assume an inside room temperature of 68oF and an outside air temperature of 20oF with an exposed area of I50 fe. Determine the heat loss drough this wall.
Problen
ll.9
Resistance
(h. fif . "F/Btu) 2. Proilem
ll.8
Resistance
,
Items
, l' Outside film resistance r (winter, 15 mphwind) ' 2. Facebrick (4 in.) I 3. Cement mortar (l/2 in.) I 4. Cinder block (8 in.) , 5. Airspace (314'n.) i 6. Gypsum wallboard (1/2 in.) I 7. Inside film resistance (winter)
(h.tr.T/Btu)
I I I i i I r
1. Ouaide film
(*int"r,
15
resistance
mphwind)
2. Siding wood (ll2 X 8 lapped) 3. Sheathing (ll2in. regular) 4. Insuladon ban (5 1/2 in.) 5. Gypsum wallboand (1/2 in.) 6. Inside film
resistance
(wi"t"r)
a.r7 0.81
r,32 19.0
0.45 0.68
0.r7 a.44 0.1
r.72 7.28 a.45 0.68
ll.l0. A typical ceiling of a house consists of items shown in the accompanying able. Assume an inside room temperature of70'F and an actic air remperaflrre of 15"F, with an exposed area of 1000 ff. Calculate the heat loss through the ceiling.
PnoslErvrs
317
to the surroundings and perfect thermal contact at the common interFace of the sample and the copper.
Problen
ll.l0
Recista[ce
(h.
Items aaic 6lm resistance 2. Insulation batt (6 in.) 3. Gypsum wallboard (1 /2 in.) 4. Inside 6lm resistance (winter) 1. Inside
ff
. 'F/Btu) 0.68 19.0 a.45
Cold water in Warm water out Problem
ll.l4
0.68
Thernocouple Temperature
Iocation fl.lt. Estimate the change in the length of a power trans
mission line
in your sate when the temperanre
changes by 50"F.'Write a brief memo to your instructor discussing your findingp. fl12. Calculate the change in 5mJong copper wire when its
CC) 120
",
100 85 ,71
temperature changes by 130'F.
Determine the temperature rise that would occur lll5. Use the basic idea of Problem II.l4 o design an apwhen 2 kg of the following materials are exposed to paratus that could be constructed to measure the thera heating element putting our 500 J. Discuss your mal conductivity of solid samples in the range of 50 to assumPuons. 300 \f/m.K Write a brief repon discussing in detail a. The material is copper. ttre overall size of the apparatus; components of the b. The material is aluminum. appararus, including the heating source, the cooling c. The material is concrere. source, insulation materialq and the measurement elellJ4. The thermal conductivity of a solid material can be dements. In the repon indude drawings and a sample extermined using a setup similar to the one shown in the periment. Estimare the cost of such a setup and briefy accompanying figur". The thermocouples are placed discuss it in your reporL Also, write a brief experiat 2.5cm intervals in the known material (copper) and mental procedure that could be used by someone who the unknown sample, as shown. The known material is unfamiliar with the apparatus. is heated on the top by a heating elemenr, and the bot11.16. A copper plate, with drmensions of 3 cm X 3 cm X tom surFace of the sample is cooled by running water 5 cm (length, width, and thickness, respectively), is or,h"ough the heat sink shown. Determine the thermal posed to a thermal energy source that puts out 150 J conductivity of the unknown sample for the set of data every second, as shovm in the accompanying 6gure. given in the accompanying table. Assume no hst loss The density of copper is 8900 kg/*'. Assuming no
lt.t3.
318
Cnrprsn
11
Tlluprnerunr eNo TetvrprRATunrRrrerro Paneurrnns
heat loss to the surrounding block, determine the tem
peranrre rise in the plate after
l0
1r.19.
seconds.
How would you use the principle given in Problem 11.16 to measure the heat output of something? For example, use the basic idea behind Problem 11.16 to desrgn a setup that can be used to measure the heat
150 J fl.20.
Problem
11.16
output ofan ironing press, Calorimeters are dwices that are commonly used to measure the heatingvalue of fuels. For example, aJunkem flow calorimeter is used to measure the headng vdue of gaseous fuels. A bomb calorimeter, on the other hand, is used to measure the heating value of liquid or solid fuels, such as kerosene, heating oil, or coal.
11.t7.
An aluminum
plate,
with dimensions of 3 cm X
3 cm X 5 cm Qength, width, and thickness,
thermal energ/ source that puts out 150 J *"ry second as shown in the accompanying figure. The density of aluminum is 2700 kg/*'. fusuming no heat loss to the surrounding block, determine the temperature rise in the plate after
tively), is enposed to
two types of calorimeters.
a brief repoft discussing the principles behind the operation of these
a
10 seconds. fl18.
Perform a search to obtain information about these 'Write
respec
Use the basic idea of Problem 11.16 to design an apparafts that could be constructed to measure the heat capacity of solid samples in the range of 500 to 'Write a brief report discussing in detail 800 J/kg . K the overall size ofthe apparatus; components ofthe apparanrs, including the heating source, the supponing block, insulation materials; and the measurement elements. In the repon include dranrings and a sample experiment. Estimate the cost of such a setup, and briefy discuss it in your report. AIso, write a brief experimental procedure that could be used bysomeonewho is unfamiliar with the apparatus.
calorimeters. n.fl.
Refer to Thbles 11.12 and 11.13 to answer this question. \7hat is the maximum amount of energy released when a 10Jb sample of coal from McDowell, 'West Virginia is burned? Also calculate dre amount of energT released when 15 fii of natural gas from Okla
homa is burned. n.n. Contact the natural gas provider in your city and find out hor/ much you are being charged for each fii of naftual gas. Also, contact your electric company and determine how much they are charging on averâ‚¬e per kVh usage of electricity, If a hotair gas firrnace has an efficiency of 94o/o and an electric heater has an effi' ciency of 1000/0, what is the more economical way of heating your home: a gas furnace or an electric heater? 11.23. Convert the results of Example 11.13 from Btu to calories.
CFIAPTER
ErncrRrc CunRENT AND RnrerED PeneMETERs ir
nOineers understand the importance
lf r
of electricity and electrical power and the role they play in our
everyday lives. As future engineers you
should know what is meant by voltage, electric currenf, and know the difference between direct and alternatinq cunent. You should also know the various sources
of electricity and understand how
elechicity is generated.
L2
320
Cner'reR
12
Errc'rmc CunnrNreNo Rruarnp Pl,nnrvrsrsRs
nee* t0 und.erstand the fundarnetutali of electricity and rnagnetsee all the deahes, appliances, and rnachines that are driaen by electrical power. The objectiue of this chapter is to innoduce the fundarnental,s of electicity. Ve will briefu discuss what is rneant by elecnic charge, electric current (both abemating cunent, ac, and direct current, dr), electrical resistance, and uobage. We will ako dzfne what is rnea.nt by an elec*ical circuit and its comltonmts. We will then looh at the role of elecnic rnotors in our eueryday liues Euery mgineer
isrn. Look around" and you will
and identify thefactors that mgineers consider when selecting a motorfor a spectfc application.
12.1 Electric Current as a Fundamental Dimension As explained in Chapter 6, based on our understanding of our physical world today, we need snm fundammtal or base dimmsions to correcdy express the physical lavrs that govern our world. They are length, rnass, ti.me, tetnperatare, electric cunent, awount of sabstance, a\d Iuminous intmsity. Moreover, with the help of these base dimensions we can derive all other necessary physical quantities that describe how nature wodrs.
In the prwious
chapters, we discussed the role of the fundamental dimension length and lengthrelated parameters such as area and volume, the fundamental dimension mass and massrelated parameters, the fundamental dimension time and timerelated parameters, and the fundamental dimension temperature and temperaturerelated parameters in engineering analysis and design. 'We now turn our attention to the fundamental dimension cunent As explained in Chapter 6, it was not until 1946 that the proposal fot amperc as abar,e unit for elecric curent was approved by the General Conference on Weights and Measures (CGPM). ln 1954, CGPM included arnpere among the base unia. The ampere is defined formally as rhat constant current which, if maintained in two straight parallel conductors of infinite length, of negligible circular cross secdon, and placed I meter apan in a vacuurn, would produce between these conductors a force equal to 2 X 107 newton per meter of length. To better understand what the ampere represents, we need to uke a closer look at the behavior of material at the subatomic level. In Chapter 9 we explained what is meant by atoms and molecules. An atom has three major subatomic pardcles, narnely, electrons, protons, and neutrons. Neutrons and protons form the nucleus of an atom. How a material conducts elecuicity is influenced by the number and the arrangement of elecuons. Elecrons have negative charge, whereas protons have a positive charge, and neutrons have no charge. Simply stated, the basic law of electric charges states that unlibe charges atnact each othn uhile lihe charges repel"In SI units, the unit of charge is the coulomb (C). One coulomb is defined as the amount of charge that passes a point in a wire in 1 second when a current of 1 ampere is fowing through the wire. In Chapter 10, we explained the universal lavr of gravitational atuaction between two masses. Similarly, there exists a larp that describes the attractive electric force between two oppositecharge particles. The electric force exerted by one point charge on anor:her is proponional to the magnitude of each charge and is inversely proponional ro rhe square of the distance between the point charges. Moreover, the electric force is attractive
12.2 Vor,recr if the
charges have opposite signs, and
it
is repulsive
if the
321
charges have the same sign. The elec
tric force between two point charges is given by Coulomb's lar: hq,q,
Frc: # r'
(r2.r)
wherch:8.99X10eN.nfl*,hand4z(C)arethepointcharges,andristhedistance(m) berween them. Another important fact that one must keep in mind is that the electric charge is conserved, meaning the elecuic charge is not created nor destroyed, it can only be transferred from one object to another. You may already know that in order for water to flow through a pipe, a pressure difference must exist. Moreover, the water flows from the highpressure region to the lowerpressure re
gion. In Chapter
1 1, we also explained that whenever there is a temperature difference in a medium, or between bodies, thermal energy fows from the hightemperature region to the low. temperature region. In a similar way, whenwer t"here exists a di:fference in electric potential between two bodies, electric charge will flow from the higher elecric potential to the lower potential region. The fow of charge will occur when the two bodies are connected by an electrical conductor such as a copper wire. The fow of elecuic charge is called eloctric carrent ot simply, cunent The elecuic current, or the flow of charge, is measured in amperes. One ampere or "amp" (A) is defined as the fow of 1 unit of charge per second. For example, a toaster that draq/s 6 amps has 6 units of charge flowing through the heating element each second. The amount of current that flows through an electrical element depends on the electrical potential, or voltage, available across the element and the resistance the element offers to the flow of charge.
12.2 Voltage Wbage rcpraena the amount ofworh required ro moye charge benveen two points, and the amount of charge that is moving between the nryo points per unit time is called caffent Elzc*omotioeforce (emfl represents the elecric potential di.fference between an area with an excess offree electrons (negative charge) and an areawith an elecnon deficit (positive charge). The voltage, or the electromotive force, induces qurent to flow in a circuit. The most common sources of elecrricity are chemical reaction, light, and magnetism.
Batteries All ofyou have used batteries for different purposes at one time or another. In all bameries, elecuicity is produced by the chemical reaction that takes place within the battery. \7hen a device that uses batteries is on, its circuits create paths for the electrons to flow through. Vhen the device is turned oF, there is no path for the eledrons to flow, thus the chemical reaction stops. A baftery cell consists of chemical compounds, internal condudors, positive and negative connections, and dre casing. Examples of cells include sizes N, AA, AAA, C, and D. A cell thar c.uurot be recharged is called aprimary cellAn alkaline bamery is an example of a primary cell. On the other hand, a secondary cell is a cell that can be recharged. The recharging is accomplished by rwersing the current fow from the positive to the negative areas. Lead acid cells in yoru cru battery and nickelcadmium (NiCd), and nickelmetal hydride (NiMH) cells are examples of secondary cells. The NiCd bameries are some of the most cornmon rechargeable
322
Cnrprsn
12
Erecrnrc CunnrNr eNo RsLAreo Pannurrnns
6 volts
H
fa
Figurel2.l
Batteries connected in a series arrangement.
1.5 volts
m
1.5 volts
Figurcl2.2
Batteries connected in a parallel arrangement.
batteries used in cordless phones, toys, and some cellular phones. The NiMH batteries, which are smaller, are used in many smaller cellular phones because of their size and capacity. To increase the voltage output, batteries are often placed in a series arrangement. If we connect bateries in a series arrangement, the baaeries will produce a net voltage, which will be the sum of the individual batteries placed in series. For example, if we were to connect four l.5volt batteries in series, the resulting potential will be 6 volts, as shown in Figure I2.LBatteries connected in a parallel anangement, as shown in Figure 12,2, produce the same voltage,
but more current.
fhotoemission Photoemisssion is another principle used to generate electricity. \7hen light strikes a surface that has cerain properties, electrons can be freed; thus elecuic power is generated. You may have seen examples of photovoltaic devices such as light meters used in photography, photo' voltaic cells in handheld calculators, and solar cells used in remote areas to generate elecuicity. Photovoltaic devices are becoming increasingly common in many applications because they do not pollute the environment. In general, there are two ways in which the sun's radiation is convened to electricity: photothermal and photovoltaic. In photothermal plants, solar radiation (the sun's radiation) is used to make steam by heating water fowing through a pipe. The pipe runs duough a parabolic solar colledor, and then the steam is used to drive a generator. In aphotovohaic cell, the light is converted direcdy to electricity. Photovoltaic cells, which are made ofgallium arsenide, have conversion efficiencies of approximately 2lo/o.The cells are
12.3 Drnscr CunnrrvreNo
323
Ar.rsRNarrNc CunnsNT
Transmission
line
E
IA Disposal to
River
I
figuretZ.l
Source: Cnrrtesy
Aschematicofasteampowerplant.
ofXcel Energy.
grouped together in many satellites to create an afiay that is used to generate electric pov/er. Photovoltaic solar farms are also becoming common. A solar farm consists of a vast area where a great number of solar arrays are put togetfrer to convert the suris radiation into electricity.
Power Flants Elecricity that is consumed at homes, schools, malls, and byvarious industries is generated in a power plant. \?'ater is used in all steam powergenerating plants to produce elecuicity. A simple schematic of a power plant is shown in Figure 12.3. Fuel is burned in a boiler to generate heat, which in turn is added to liquid water rc change its phase to steam; sterrm passes duough rurbine blades, turning the blades, which in effect runs the generator connected to the turbine, creating electricity. The elecricity is generated by turning a coil ofwire inside a magnetic field. A conductor placed in a changing magnetic field will have a current induced in it Magnetism is the most common method for generating elecnicity. The lowpressure steam leaving the turbine liquefies in a condenser and is pumped through the boiler again, dosing a cycle, as shown in Figure 12.3. U.S. elecuicity generation by fuel type for the year 2005 is shown in Figare 12.3a.
n2.3 Dircct Current and AlternatinE Direa ctorent (/r)
Current
is the flow of electric charge that occurs in one direction, as shown in Figure 12.4(a). Direct surrent is typically produced by batteries and direct current generators. In the late 19th century, given the limited understanding of fundamentals and technology and for economic reasons, direct current could not be transmitted over long distances. Therefore, it was succeeded by alternating current (ac). Direct curent r'eas not economically feasible
324
CHeprsn
12
Er,rcrnrc Cunnnnr aNo RBrlrBp Penervmruns
2W ?1
i
1500
B
T
1000
s00
0
Coal
Petroleum
Natural gas
Nuclear
Renewable/Other
Fuel type
E
(a) Direct
I
Figure
12.4
fiqure
t2.3a
U.S.
elechici$ generation by fuel type (2005).
curent
(b) Alternating current
The dhect and altemating currents.
to transform
because of the high volages needed for longdistance transmission. However, dwelopments in the 1960s have led to techniques that now allow the uansmission of direct crurent over long drstances.
Alternating anrent (ac) is the flow of electric charge that periodically reverses. As shown in Figure 12.4 (b), the magnitude of the current stans from zero, increases to a maximum value, and then decreases to zero; the fow of elecuic charge rwerses direction, reaches a ma(imum value, and returns to zero again. This fow pattern is repeated in a ryclic manner. The time interval between the peakvalue of the current on two successive cycles is calledrheperiod, and the number of cycles per second is called thefreqamcy. The peak (maximum) value of the alternating current in either direction is called rhe amplitudz. Alternating current is created by generators at power plana. The current drarn by various electrical devices at your home is alternating current. The alternating current in domestic and commercial power use is 60 cycles per second (herz) in the United States.
Kirchhoff 's Current Law Kircbboffl current la.u
is one of the basic laws in electricity t}rat allows for the analysis of currents in electrical circuits. The larr states that at any given time, the sum of the currents enter
ing a node must be equal to the sum of the current leaving the node. This statement
is
12,3 DrnBcr CunnnrviawpAr,rsRNATnc Cunnrrr
I
325
Figurel2.5
Ihe sum of curents entering a rode
[u$
squal the sum of
the cunent leaving the node:
i2
h+
h.
^+
demonstrated in Figure 12.5. As explained earlier, physical lavrc are based on observations, and the Kirchhoff's current laey is no er>.Amount = 1000:250:3000; >>
IntereEt_Rate = 0.06;0. 01:0.08; Interest_Earaed. = (.Anount' ) * (IatereEt_Rate), fprlnrf (' \n\n\r\L\r\r\r\r\r rnrereEr Ratet ) ;fprLntf (' \a\t .Anount\t\t' ) I . .. fprLatf ( ' \t\t e"g[' , InterestRate) ; fprlntf ( ' \n' ; ; dtsp ( lihount ! ,IDterest Earned] ) >> >> >>
Interest Rate
Dollar Amount
0.06
o.a7
1000
50
t250
/)
70 87.5
80 100
1500
90
105
t20
t750
105
t22.5
140
2000 ?250 2500
120 135
t40
160
r57.5
180
150
175
?:750
r65
r92.5
3000
180
2t0
204 220 240
15.2
Usu.rc MATLAB Bur.rnr Frnrcrrorvs
429
,i,
[l
Figurc
15.5
The commands and result
for tmmple
14.2 (Revisited).
t:t
On the last command line, note that the three periods . . . (an ellipsis) represents a continuation marker in MAILAB. The ellipsis means there is more to follow on this command line. Note the use of fprinLf and disp commands. The final result for Example 14.2 (Revisited) is shown in Figure 15.5. ,l
15.2 [fsing MATLAB Built'in MATLAB offers
a large selection
Functions of builtin functions that you qln use to analyze data. fu we
discussed in the prwious &apter, by builtin funcdons we mean standard functions such as the sine or cosine of an angle, as well as formulas that calculate the totd value, the average value,
or the sandard deviation of a set of data points. The MlfILAB functions are available in various categories, including mathematical, trigonometric, statistical, and logical functions. In this chapter, we will discuss some of the common firnctions. MAI,AB offers a Help menu that you can use to obain information on various commands and functions. The Help button is marked by a question mark ? located to the left of the current directory. You can also rype help followed by a command name to leam how to use the command. Some examples of commonly used MATLAB functions, alongwith their proper use and descriptions, are shown in Thble 15.7. Refs to Example 15.2 when studying Tirble 15.7.
The following set of values will be used to introduce some of MAlfABt builtin functions. Mass: U02 lts 99 rc6 rc3 95 97 102 98 96l.WrenstudyingTableL5.T,theresults of the executed functions are shown under the "Result of the Example" column. More examples of MlflLAB's Functions are shown in Thble 15.8.
430
Cnu'rrn 15 MAILAB
'#;;
i meqn "'q i j i
;**;oor
;t
*i;i:;
It
sums the values
It
calculates the average value
dre data in
It daermines
max
*n"
a
in
a
,;
*orr1* ogthu n="pl" ,,,i til;lilrdfn
gilen array.
1013:
of
101.3
given array.
the largest value
in
the given array; ,It determines ttre smalle* value irt the given array, It calculates the standard deviadon fot the values in the giveqrena)r It sorts the values in the give$' ,,, , array in ascending order. It returns the value of zr,
i
r min ,l
sEq
:
joir; I,:
3.14151.926535897.. i
'
,.,'i;
.
It retums tangent value of the
[email protected] argument. The argument must be
in radians,
cos
It
renrrns cosinevalue ofthe argument. The argument must be
in
It
returns sine value of the argument.
in radians, s
,',,,,,
frg*su**l**rh,h:ifT:,
;l
sqrt (x) f actorial (x)
Retums the square root ofvalue.t
ky"^ *:.:*e of factorial of*. For example, factorial(5) wlll return: (5)(4)(3)(2)(r) : 120.
Trigonornzt ix Functions (x) This is the inverse cosine function ofx. It is used to determine the value of an angle when its cosine value is kno\trn. asin (x ) This is the inverse sine function of*. It is used to determine the value of an angle when its sine value is known. atan (x) This is the inverse tangent function of r. It is used to determine the value :. of an angle rylieg its glgent value is knor,rn. '',1 ,,
acos
,
',
'
kponcntial and Ingarithmic Functiot s exp
(x)
los (x) 1og10 (x) 1og2 (x)
Retums the value of a". Retums rhe value of the natural logarithm of *. Note that
* must
greater than 0.
Returns the value ofthe common (base 10) logarithm of.r. Returns the value of the base 2 loSanthm of x.
be
15,2 Usnc MAIIAB Bur.rnv FuNcrroxs
431
Usiag MAil.AB, compure the aver4ge (ari hmecic mean) and the sandard deviation of the density of water data given in Thble 15.9. Refer to Chapter 19, Section t9.4, to learn about what the vdue of the standard deviation for a set of data points represents.
1,,
GroupAFindingq
GroupB Flndings
1.,
ffi* The final resuks for Example 14.4 (Revisite$ are shown in Figure 15.6. The commands leading to these results are:
MAILAB
>>DensJ.tyA = [1020 1015 990 1060 1030 950 975 1020 980 950] t >>DeasLtyB = [950 9AO 890 1080 1120 900 ].040 1150 910 10201; >> DensLty_A_Avela![e = nean (Denstty_A) DensityA_Avâ‚¬rdgâ‚¬ = 1000.00 >> DenEl.ty_B_Aver?![â‚¬ = mean (DensLty_B) DensiLyBAvâ‚¬rdgâ‚¬ = 1000.00 > > Standard._Xlevlat lon_For_croup_l\ = std ( Dengity_A) StandardDeviat i on_For_Group_A = 34.56 > > Standard_DevLat Lon_For_Group_B = std ( Deagity_B ) SLandard_Deviat. ion_For_Group_B = 93 .22
432
CHer"rsR
15
MAILAB
e
Edt uew w& wln*w
He
rdi&'
[email protected]*l&jt
I
figumls.O
Command l{indow
iATIAB'S
for trample
14.4
(Revhited).
Li.!l:l:r.':.t:
I::
:l:t:,:r
tlt: ;l:a:a:lt: _t:
::ri.,::::::
The Loop eontrol
.
 forand while commands
When writing a computer program, often it becomes necessarl to execute a line or a block ofyour computer code many times. MAIf,AS providesprand ubilocnmmands for such situations.
forloop
Using theprloop, you can execute number of times. The sfnax of aforloop is
The
a
line or a block of code a specified (defined)
for index = sLartvalue : incremenL : endva1ue a line or a block of your computer code end For example, suppose you want to evaluate the function ! : x2 * 10 for r values of 22.00 22.50,23.00,23.50, and24.00. This operation will result in correspondingTvalues of 494.00, 516.25,539.00,562.25,and 586.00. The MlftIAB code forthis enample then couldhave the following form:
x = 22.0; Fnr
i

1.1.R
Y=xn2+IO i
disp(lx',y'l) x = x + 0.5;
end
I5.2
UsrNc MAIIA,B BunrrN FuNc"rrors
433
Note that in the preceding example, the index is the integer i and its startvalue is 1, it is incremented by a value of 1, and ia endvalue is 5.
rfile Loop Using the whilzloop,you c:ur execute condition is met. The qyntax of a while loop is
The
whi
a
line or a block of code until
a
specified
Le control I ing  expres s ion
a 1ine or a block of vour comouLer
code
end 'With the while command,
as
long
as
the controllingexpression is true, the line or a block of code using rhe whib com
will be enecuted. For the preceding example, the MIIILAB
code
mand becomes:
x = 22.0; while x >pressur6=[20 tS 22 26 L9 1,9 2L L27i >>fprlotf('\t Line Pressure (pgt) \t VaLve FosLtlon\n\n');
for L=1:8 if pressure(t) >=20 fPrLntf (' \t e"9 \t\t\ts\t\t else
fpriatf ( '\t
end end
I
rigurel5.8
The solution
ol Ennple
14.5 (Revhited).
?"9
\t\t\t\t\t
OFEN\n'rBreEsure(L) CITOSED\n'
)
,pregsure (L) )
436
Cneprsn
15
MAILAB
The Mfile As explained previously, for simple operations you c:m use M,{'l,AB's Command lTindow to enter variables and issue commands. However, when you write a program that is more than a few lines long, you use an Mfrle. It is called an Mfile because of its .zz extension. You can cre
Mfile using any text editor or using MAIT,ABI Editor/Debugger. To crqrte an MfiIe, open the Mfile Editor and MlfIl,AB opens a new window in which you can type your program. fu you type your program, you will notice that MAILAB assigns line numbers in the left column of the window. The line numbers are quite useftrl for debugging your program. To save the file, simply click File * Save and type in the filename. The name of your file must begin with a letter and may include other characters such as underscore and digits. Be carefirl not to name your file the same as a MAILAB command. To see if a filename is used by a MATIAB command, type exist ('filenamd) in the M,$LAB's Command'S7indow. To run your program, click on Debug + Run (or use the function key F5). Don't be discouraged to find mistakes in your program the first time you attempt to run it. This is quite normal! You can use the Debugger to find your mistakes. To learn more about debugging opuons, type help ate an
dtbug
It
n
the
MATU.B Command !7indow.
has been said that when Pascal was 7 years old, he qrme up with the formula
rdn f l\ ttt
determine the sum of 1,2,3, . . . , through n. The story suggests that one day he was asked by his teacher to add up numbers I through 100, and Pascal came up with the answer in few minutes. It is believed that Pascal solved the problem in the following manner: First, on one line he wrote the numbers I *uough 100, similar to
1 2 3 4.. ,........99
100
Then on the second line he wrote the numbers backward
100 99 98
97
...........2
1
Then he added up the numbers in the nvo lines, resulting in one hundred identical values
of
101
101 101 101
101
...........
101
101
Pascal also realized that the result should be divided by
1 through 100 twiceleading to the answer:
.
approach and came up with the formula
100(10 I )
;:
n(n+ r)
 ,
2since
he wrote down the numbers
5050. l,ater, he generalized his
15.2
Flle Edt
Tqt
Bdffitr ,
Cell
[email protected] Dehrg De$ktop Wrdow
&EB
UsrNc MAILAB Burr,trN Funcrrous
437
Ftalp
"leitmTl aelc*eareeli:
F{iF
rtil 4.. ":,i :
g Ask the usar lo input the uFper vaLue ,, upper_value=input('Please input thâ‚¬ upper vlaue of the nunLters.'), . 4;'t ,:
,1i
:
:l
6:, i,
q iFt
1: for
'8,.:
a! 1,:
1.0
i i
;:1 Figure
fhF
crrn onil:l
r^
zero
surn=o,
k=1:1:upper_value sum=sum+kt
end
g Print the results
fprintf('\n
?he eum of numbors from 1 to $g is eqaul to: tg\nr,upper_value'sum)
15.9 lhe lrfile for Enmple
15.3.
Dcktop Wftew Heb
rFilei
rgjeFi3e d.H? shortcuts E Howto Add m
Figure
15.10
uJffis tgfd
The results 0f Example 15.3.
Nexr, we will wdte a computer program using an Mfile that asls a user to input a value for a and computes the sum of 1 through n. To make the program interesting, we will not make use of Pascal's formula; instead, we will use a for loop to solve the problem. \(e hane used MATI"A.B'S Editor to create the program and have named it ForIoopF,xample"m, as shown in Figure 15.9. In r:he shown program, the 7o symbol denotes comments, and any text following tle %o symbol will be ffeated as comments by MAII,AB. Also, note that you can find the Line (Ln) and Column (Col) numbers corresponding to a specific location in your program by moving the cursor. The line and the column numbers are shown in the right side, bocom corner of the Editor window. As you will see, the knowledge of line and column numbers are use'We frrl for debugging your program. run the program by clicking on Debug * Run, and the result is shown in Figure 15.10. , ..,:ri:i:1rtrj.I i:
. .i
438
Cneprun
15
MAIL.AB
tr5.3 Plotting with MATLAB M.{flAB
offers many choices when it comes to creating chats. For example, you c:m crezrte columl charts (or histograms), contour, or surface plorc. As we mentioned in Chapter 14, as an engineering student, and later as a practicing engineer, most of the charts that you will create will be tclt We charts. Therefore, we will explain in deeil how to ueztte an
xy
charts,
x!
chart.
Starting with a 10 cm X 10 cm sheet of paper, what is the largest volume you c:rn create by cuning out r cm X r cm from each corner ofthe sheet and then folding up the sides? See Figure 15.11. Use Ml[f,AB to obtain the solution. The volume created by cutting out r cm X r cm from each corner of the 10 cm X 10 cm sheet of paper is given by V: (10  2r)(10  2.a)*. Moreover, we know that, for r : 0 and x : 5, the volume will be zero. Therefore, we need to create a range of x values from 0 to 5 using some small increment, such as 0.1. We then plot the volume versus r and look for the maximum value of volume. The MAIIAB commands that lead to the solution are:
I The
>>x = 0:0.1:51 >>volune = (102*x) .*(102*x) .*xi >> Blot (x,vohme) >>tLtle ('VoJ.ume aE a functlon of x') >> xLabel ( 'x (cn) ' )
rigurct5.ll l0 cm
x
l0 cn
sheet in
Erample 15.4.
>> >>
I
y1abeL ( 'VoLu.me
grid. mLnor
figure
15.12
(cm^3 ) ' )
The IIAIIAB Command lindow
for txample 15.4.
The MAILAB CommandWindow for ExampleL5.4 is shown in Figure 15.12. The plot of volume versus r is shown in Figure 15.13.
ld
Figurc
fi1ii.lftrxT,ff{idi.iRBiEt:
15.13
nIF,Il
The plot ot volune versus
r
for Erample 15.4.
?j
Let us now discuss the MAIT.AB commands that commonly are used when plotting data. command plots yvalues versus xvalues. You can use various line types, plot symbols, or colors with the command pLoE(x,x,E), where s is a characer suing that defines a particular line type, plot symbol, or line color. The E can take on one of the properties shown in Thble 15.12.
The pLot (x, y)
DataSvmbol
lb
lo ,o
ic im iv ik
Blue Green
o
Point Circle
Red
x
rmark
+
Plus Star
cy* Magenta
ltllow Black
Solid
Doned Dashdot Dashed
Square
d
Diamond Triangle (down)
Triangle (up) Tiiangle (left) Triangle (right)
439
440
I
Cnaprsn
15 MAILAB
Figurcl5.l4
l'lATLABl Line Property Editor.
For example, ifyou issue the cornmand pL
ot, ( x, y t t k*   ),
MAILAB will plot the curve using a black solid line with an * marker shown at each dara point. If you do not spec' ify a line color, MAlf.AB automatically asslgns a color to the plot. Using the tttle ( 'text' ) commandr /ou crrn add text on top of the plot. The
xlabel
(
t
text' )
command creates the tide for the xaris. The text that you enclose will be shown below the xaxis. Similady, the ylabeL ( 'text' ) command creates the tide for the yaxis. To turn on the grid lines, type the command grLd on (or just grtd). The command grLd off removes the grid lines. To flrn on the minor grid lines, as shown in Figure 15.13, type the command grLd between single quotation marks
mlnor.
L5.3 Plorrrrcvrrn MAILAB
I
441
tigurelS.tS
The plot of
tranple
15.3 with
nodified properties.
Generally, it is easier to use the Graph Property Editor. For enample, to make the curve line thicker, change the line color, and to add markers to the data points (with the rnouse
pointer on the curve) doubleclick the left mouse button. Make sure you are in the picking mode first. You may need to click on the arrow next to the print icon to activate the picking mode. After doubleclicking on the line, you should see the line and the Marker Editor window. As shovrn in Figure t5.T4,weincreased dre line thickness from 0.5 to 2, c,hanged the Iine color to black, and set the daapoint marker sryle to Diamond. These new seaings are reflected in Figure I 5.15. Next, we will add an arrow pointing to the maximum value of the volume by selecting the TixtArrow under *re Insert option (see Figure l5.LG), and add the text'Maximnm volume occurs at x: 1.7 cm." These additions are reflected in Figure 15.17. V'e can also change the font size and style and make the title or the axes labels boldface. To do so, we pick the object that we want to modify and then, from the Menu bar, select Edit and then Current Object Properties . . . . Then, using the Propercy Editor shown in Figure 15.18, we c:rn modi& the propenies of the selected object.'We have changed the font size and the font weight of dre tide and the labels for F,xample 15.3 and shown the changes in Figure 15.19.
\firh MAILAB, you c:ln generate other types of plots, including contour and surface plos. You can also control the r and yaxis scales. For enample, the MAIT,AB's
Loglog(x,y)
usesthebasel0logarithmicscalesforrandyaxes.Note.randyarethevariplot. The command LogLog (x, y) is identical to the plot (x, y) ,
ables that you want to
K
figurets.t6
Using the lnsert Tert Arow
options, you can add anows or text to the plot.
I
Figurel5.lT
The solution of
M2
tnmple
15.4.
15.3 Pr.orrrNcwrnr MAILAB
t
443
Figurel5.l0
MAIIAB5
Ie*
Property Editor.
it
uses logarithmic axes. The command semlLogx(x,y) or semilogy(xry) plot with basel0 logarithmic scales for either only dre raxis oryaxis. Finally, it is wonh noting that you can use the hold command to plot more than one set of daa on the
except
creates a
same chan.
A reminder, when creating an engineering chart, whether you are using MAIT,AB, Excel, other drawing software, or a freehand dranring; an engineering chan must contain proper labels with proper units for each axis. The chart must also contain a figure number with a tide exptaining what the chart rcpresents. If more than one set of data is plomed on the same chart, the chan must also contain a legend or list showing symbols used for different data sets,
43
Cueprnn
I
15 MAIAB
FigurelS.lg
The result of Erample 15.4.
Using the resula of F:rample 14.1, create a graph showing the value of the air density as a funcdon of temperature. The Command'l7indow and the plot of the density of air as a function of temperature are shown in Figures 15.20 and 15.21, respectivelp
I
Figure
15.20
Command $lindow
for Erample t4.6 (Revhited).
15.4 Iuponrrxc E:rcrr,aNo Orrrrn Dere
L
Figure
15.?l
Frr.ss rNro
MAil,AB
445
Plot of density of air for Erample 14.6 (Revhited).
i] t1
IttA,.i,
r:ff1&1i.lilr,,t,t 
.i,r"lllta?nji:11lftlfiil!:
15.4
ii,!liijitii!.'i iiirxlitiijliirrrtii

ii,rXitilrl!ii111ta lr,iilril'liTrlTi1.ir'rrira
l:
Jli:1ii,1ir,irlrirl.. .'lTlLr.li4lltl
Nmporting Exeel and Other Data Filcs into MATLAB At times, it might
be
convenient to impon data files tlat were generated by other programs, such
MAilAB for additional analpis. To demonsrate how we go about imponing a daa file into MATLAB, consider the Fxcel file shown in Figure I5,22.The Excel file was created for Example 15.4 with two columru: the r values and the corresponding volume. To impon this file into MAIIAB, from the Menu bar, we select File *1sn Irnport Data . . . , t}ren go to the appropriate directory, and open the file we want. The Impon'Wizard window, as
Excel, into
in Figure L5.23,r ,ilappear next. Select the "Create vectors from each column using column nameso and the \Pizard will iport the data and will sare them as x and voXu:ne
as shown
variables.
Now lett the
MIIIIAB
Figure 15.25.
'We
that we want to plot the volume as a function of r. then simply type commands that are shoiyn in.Figtire 15,24.Tlte resulting plot is shown in
say
*r; ; .';I"***"
t11
:A
,
,:.9.i,
:'l
x
:.t:,
2 0.1
0.0 9.6
A:
na
18.4
sl
0.3
20.5
6
9.4
l,1.9 40.5
8l
0 0.
116.â‚¬
0.â‚¬
58.4
r:,$
10
I
[:]
D:I
rrolumo it:,;
iril!,
51.8
lr;
m,{ 6{.(
iii
fa\=rX.i;, r"#?fizr;; ri
I
Figurel5.Z
The Ercel data
L
;rr:
file used in tranple 15.4.
Figurul5.Z3
iAIIABI lmport ltizad.
@:..:
n aiI*, a
fo';
4'g.i'1i;;**l.rryffii!
Ar Sat rtst d, el,el ..qdfl!8 Ealtn trs l&r E ls &m $Frg 61ffid o!e*d @l,ablo, ln rhe
[email protected] @tt!Dâ‚¬R. >
[email protected] ta.wlwel > td.tla
[email protected] e a
[email protected] ot r'l I ilibâ‚¬t (,a
[email protected]'l > Itd!â‚¬r, (
[email protected] le.3lrl
il
t *ld Figurel5.Z4
The commands leading to the plot
shown in Figure 15.25.
46
>l
"
"
.
t.
::
""ll.f
:';: "'
15.5 Mernor CouputetroNs wrrg MAILAB
L
Figure
15.25
Plot of uolume venus
r
using data imported frorn an Ercel file.
15.5 lVatrix Computations with MATLAB fu
explained earlier, MATI/,B offers many tools for matrix operations and manipuladons. Thble 15.4 shows er) A=[0 5 0;8 3 7;9 2
91
A
0 50 8 37 929 >> B=tA 6 2;7 2 3;1 3 El
4 0 z
72 3 1 3 4 >> C=[1; 2; 5l 12 5
>> A+B
ans
=
4It2 15 5 10 1
10 5
>> AB alfD
_A

_1
')
r14 tt
f
LJ
>>3*A
0 15 0 24 9 21zt
o
zt
>>A*B
35 10 15 60 75 35 31 77 50 >>A*C AIfD

10 33
>>det (.4)
ans
I
_4tr' Figurel5.26
Ihe solution to Example 15.5.
448
=
41
15.5 Mernnr CoupurerroNs wrrH MAILAB The formulation of many engineering problems leads to
a
449
qrstem of dgebraic equations. As you
will learn later in your math and engineering classes, there are a number ofwaln to solve a set of linear equations. Solve the following set of equations using the Gauss elimination, by invening dte [r4] matrix (the coefficients of unknowns), and multiplying it by i6] matrix (the values on the right hand side of equations. The Gauss elimination method is discussed in detail in Section 18.5. Here, our intent is to show how to use MAITAB to solve a set of linear equations.
2qI x2* 4:13 3x1*2x2*4xu=i2 5qx2t34:17 For this problem, the coefficient matrix [r4]
"nd
the righthand side matrix
Ibl
are
",:l', ii].,,u,,, ={1i\
I(e will first use the Mltil,AB matrix left division operator \ to
solve this problem. The
\
operator solves the problem using the Gauss elimination. W'e then solve the problem using the inv command.
To get started, select "MATLAB Help" from the Help menu.
>>A
=
12 7.
Li3 2 4;5 1 3l
11 IJ
24 l_ 3
J
>>b = lL3t32iL71
1\ 
l_
t
JZ L7 >>
X=
x = .4,\b
2.0000 5.0000 4.0000 >>x = invla'1 *5 X= 2 .0000 5.0000 4.0000
450
CHaprsn
15
MAILAB
:
:
Note that if you substitute the solution xr: 2, x2 5, and, x3 4 into each equation, you find that they satisfy them. That 2(2) + 5 + 4= 13,3Q) + 2(5) + 4(4) 32, and
k
:
5Q)5+3(4)=17. llr.:    :il r.r:i
15.6
"..:.l
Curve Fitting with MATLAB In Section 14.8, we discussed the concept of curve fitti"g. MAIIA.B offers a variety of curve' fitting options. V'e will use Example L4.ll to show how you can also use MAlfAB to obtain an equation that closely fits a set of data points. For Example 14.11 (Revisited), we will use the command POLYFIT (x, y, n) , which determines the coefEcients (ca, c1, c21 . . . t e) of a polynomial of order a that best fits the data accordrng to:
!= cox'*
c1x"l
*
c2xo2
*
car"3
* "' *
co
Find the equation that best fits the following set of data poins in Thble 15.13. In Section 14.8, ploa of data points revealed that the relationship betrveeny and x is quadradc (second order polynomial). To obtain the coefficients of the second order polynomial that best fie the given daa, we will rype the following sequence of commands. The MAIIA,B Command Window for Example 14.11 (Revisited) is shown inFigte 15.27:
>>format compact
))x=0:0.5:3 ))! = t2 O.75 0 0.25 0 0.75 2J >> COefficients = poLtrfit (x,y,2)
0.00
2.00
0.50
4.75
1.00 1.50
0.00
2.00 2.50 3.00
0.25 0.00 Q,75
2.04
I
Figuru
15.27
T[e Comnand Window for Example
l4.ll (Rwhited).
15.7 Upon execution of the d0 :
iji 1i
,l
1, c1 : 3t
polyfit
and c2
=
Snvmor.rc M.mrBuerrcs wrrH
command,
MAILAB
Mdfl{B will return the following
2, which leads to the equationy
:
x2

451
coefficients,
3x t 2.
.''...'.''
fl5.7 Symbolic Mathematies with
MATLAB
use MAI.AB to solve engineering problems with numerical values. In this section, we briefy explain the syrnbolic capabilities of MATLAB. In rymbolic mathematics, as the name implies, the problem and the solution are presented using symbols, such as x instead of numerical values. \Wb will demonstrate MALAB's symbolic capabilities using F.xamples 15.7 and 15.8.
In the previous sections, we discussed how to
\7e will use the fo[owing functions to perform MAnA.Bt qymbolic operations,
as
Thble 15.14.
f(*):*5x+6 rtQ): * t
: (x+ 5)2 fsk):5*tr2xY
fi(r)
I
Function
Description of the Function
ExamFle
It
Flx = qfm('x^25\+6') F2x = sfm('x3')
creates a
sfmbolic
function.
'When
F3x F4x possible,
it facorizes
the function into simpler terms. simplifies the funcdon.
simplify
It
expand co11ect,
It expands the function. It simplifies a s)'mbolic expression
solve
It
aryn]
It
vavlve
nt \ f!,
min,
max)
by collecting like coefficients. solves tfie orpression for its roots. plots the function f in the range of min and max.
= sfm('(x+5)^2') = sym('51y+2\y')
Flx = x^2*5*x*6
F2x: x3 = (x*5)^2
p31
F4x=
5\y+2\y
faao(Fxl)
(x2).(x3)
simplifr(Flx/F2x) expand(F3x)
x2 x^2*10\*25
collea(F4x)
7*x2*y
solve(Flx)
x=2andx=3
ezplot(F1x,0,2)
See
Figure 15.27
shown in
452
Cru.prnn
15
MAilAB
t
Figuru
15.28 lhe ezplot
for Erample l5J;
see lhe last row in
labh
1514.
::::l.'I
Solutions of Simuftaneous Linear Equations In this secdon, we will show how you to a set oflinear equations.
c:rn use
MAIIABI symbolic
Consider the following three linear equations with three unknowns:
2x*y*z=t3 3xt 2y * 4z:32 5x !t3z=t7 In MAILAB, the solve equations. The basic form
solvers to obtain solutions
I
y, arrdz
to obein solutions to symbolic algebraic of the solve command is solve('eqal r, teqn2t,
command is used
. , 'eqnr ) . fu shown below, we define each equation first and then use the goLve command to obtain the solution.
>>equatioa 1 = '2*x+y+z=13'i >>equatJ.on 2 = I 3*2i1!*y+4*z=32' i >>equat,Lon3 = '5*xy+3*z=L7' i >>[x,Yrz] = golve(equatlon l,equatlon 2,equation 3)
15.7 The soludon is given
by, :
2,
!:
Svmnor,rc
5, and a
:
Menrpuencs wrrn MAILAB
453
4. The MAIL"EB Command Vindow for
Example 15.8 is shown in Figure 15.29.
H '::
Figure
15.29 I[e
::1r: :
solution of the set of linear equations discussed in Eranple 15.8.
ll
As we said at the beginning of this chapter, there are manygood textbools that discuss the capabilities of MAIfAB to solve a firll range of problems. Here, our intent was to introduce only some basic id6as so that you can perform some essential operadons,'fu you continue your engineering education in other classes, you will lsarn more about how to use MAILAB effectively to solve a wide range of engineering problems.
454
CrrnPrsR
15 MAILAB
PnonreMs
455
SUMMARY Now that you hane reached this point in the text you should:
.
MAILAB is a tool that can be used to solve engineering problems. Moreover, you MAILAB to present the results of an analysis in chan form. You can input your own formulas or use the buihin funcdons provided by MAII,AB. know that can use
o know how to edit the content of a MIILAB file.
.
be familiarwith MAILABI builtin functions. o know how to import data files into MAILAB.
"
. . .
r5J.
Using the
know how to creirte a proper engineering chart using MAILAB. know how to perform mauix computations with MAIIAS. be
familiarwith MAIIA.8I curvefiting capabilities.
be familiar with
MAILABI symbolic mathematics capabilities.
MAfLAB Help menu, discuss howthe fo[
lowing functions are used. Create a simple example, and demonsuate the proper use of the function. a. ABS (X)
b. TIC, TOC
c. SIZE (x) d. FD( (x) e. FLOOR
(x)
f.
CEIL (x) g. CALENDAR r5.2.
Create a table that shovss the water pressure in lb/in'z in a pipe located at the base of the water tower as you
varythe heightofthewater in incremen$ of 10 ft.Also, plot the water pressrue 0b/in1 versus the height of water in feet. V'hat should the water level in the water tower be to create 80 psi of water pressrue in a pipe at the base of the water tower? 15.3. As we explained in Chapter 10, viscosit,' is a measure of how easily a fuid fows. The viscosity of water can be determined from the following correlation.
In
Chapcer 10, we discussed fluid pressure and the role of water towers in small towns. IJse MAILAB to create a able that shows the relationship berween the height of water above ground in the water tower and the water pressure in a pipeline located at the base of the water tower. The relationship is given by
p: :
p
:
g: h:
/:\ : ,r1g\T"/
where ;r,
=
viscosiry (Nls. rn'z)
T = temperature (IQ ct=2.414x 105NA.m2
pgh
where
P
11
c2= 247.8K water pressure at the base of the water tower in pounds per square foot (lbiff)
density ofwater in slugs per cubic foot (p 1.94 slugs/ff) acceleration due to gravity (S
= 32.2 ftl*)
height ofwater above ground in feet (ft)
:
cz= l40K Using MAITA3, create a able that shows the viscosas a function of temperature in the range of 0'CQ73.15 K) to 100"C(373.15 K) inincrements of 5'C. Also, create a graph showing the value of viscosity as a function of temperature.
ity of water
456 15.4.
CHePrsR
15 MAILAB
Using MAIIAB, create a table that shows the relationship between the unia of temperature in degrees Celsius and Fahrenheit in the range of 50oC to
The frontal area./4 represents the frontal projecdon area and could be approximated simply by multiplying 0.85 times the width and the height of a rectangle that oudines the ftont of the car. This is the area that you see when you view the car from a direction normal to the front grill. The 0.85 factor is used to adjust for rounded corners, open space below the bumper, and so on. To give you some idea, rypi"d drag coefficient values for sports c:us are between 0.27 to 0.38 and for sedans are between 0.34 to 0.5. The power requirement to overcome air resistance is computed by
of the cart
150"C. Use increments of 10"C. 15.5.
t5.6.
15.7.
t5.8.
t5.9.
1510.
Using M,{ILAB, create a table that shows the relationship among units of the height of people in centimeters, inches, and feet in the range of 150 cm to 2 m. Use increments of 5 cm. Using MAILAB, create a table that shows the relationship among the unim of mass to describe people's mass in kilograms, slugs, and pound mass in the range of 20I
opamp
logic symbol (or)
v
gt,,^ 90"
natural gas line
elbow
45'etbow
+D
all,,
confiolihput,,;
Output if all eontrol input
siguals are on
Si!'nals are
Outputrif
,. confiolinput
,
i,
NAI{D
NOT
off
I Ftuinblng and Ptptng Symbols
ilt'll
1
: sil*ouo
i^
r=:l
il
Oval batltub
X Iloor sink
Showersall
fr&iirlpool bath
t,,,
iH
':,, Tows
iry
r
'45"
Elbow
Chillpddrinking rilâ‚¬tsr t:
f
90o
supply
Elbow
drinking waterretqr,ni
Chillâ‚¬d
r
,i::
tF Connecting
Source: @ f 999 American
Tirhnical Publishen, Ltd.
t
_
*T" +
l*o*T
pip joint
86
tl
It*ijl
Hot wate.r
,lEl: Expan5ionjoint
,,,
Straightsize
'.i
gfoss
rD.'
64072A 56064A
6.37.r
i.
5"56.3
na"E a Gte, Mechania ofMa*rizh Reprinted with permission J. Nelson, a division of Thomson l,ea.ming: www.thomsonrights.com
Material
I
Aluminumallrys
., Brass
Btorze
I
Cast iron (tension) Cast iron (compression) Concrete (compression) Copper alloys
Yield Strength (MPa) Ultimate Stren$h (MPa)
35500 70554 8269A
na290 55760
Gtass Plate glass Glass fibers
Magnesium allqrs
Nickel
s02sp 100620
Plastics
Polyethylene Rock (compression)
l l
Granite, marble, quaru Limestone, sandstone Rubber
100550 2AA520
l
11
,
l
200830
69480 340L400 lO70 230830 301000 70: 700020,000 1404/+A 310760

I
l
r
I
r
i
l
t.
Nylon
'
of
4080 728
l
i
50280 2o2AA 720
l
'
l
Continued
Matetial Steel
Yteld Strength
@a)
Illtimate Strength (MPa)
,
Highsrength
340'1000
550tzAO
Machine
550860
Spti"e
34A7oA 4001600
Sainless
280700
7001900 4001000
520
900
Tool Steel wire
2801000
Smrctural steel
?00700
Titanium alloys
7601000
5501400 340*830 9001200 1400*4000
Tungorcn Wood (bending) Douglas
fu
3050
Oak Southern pine \Food (compresion parallel to gain) Douglas
fir
Oak
5080 50100
4A60
50=100
,3050
4A70 3050 4070
30*40 3050
Southern plne Source: Adapted fromJ.
4060
M.
Gere, Mechani.cs of Materia.b. Reprinted with permission of Nelson, a division
Thomson l,earning: www.thomsonrights.com
',..___ Coefficient ofThermal
Material
Erpansion
Aluminum allop
(IfC) x ld
23
Bronze Cast iion Concrete Copper alloys Glass
Magnesium
allqn
Nickel
f.ryansion
(ffF) x
106
13
t9.t21.2
Brass
:,
Coefficient ofThernal
i0.611.8
T8_21
9.911.6
9.9r2
5.55.6
7r4
48
16.6r7.6
9.29.8
5 11 26.L28.8
L4.516.A
36
l3
aa
Plastics
Nylon Polyethylene
Rock:
70L4A
4080
140290
80150
59
35 70l 10
,
130*200
Rubber
l0t8
5.59.9
T4
8.0
Strucnual
t2
9.6 6.5
I ltailum aroys Tungsten
4.3
Steel
r,
Highstrengti Stainless
*Note
thr you nust muhiply the coefficients of thermal expansion.
8.
co"ma"""
lr I
giri i" .hi"
4.56.A 2.4
"bi"
bt to:d io obeio th"
actual
""lues
Source:AdaptcdfromJ.M.Gere,Mechan;csofMaterlak ReprintedwithpermissionofNelson,a division of Thomson learning: www.thomsonrights,com
of
of
17.3
17.3
Sotur CouuoN Sor.ro ENcrrsesRrNc
Some Gommon Solid Engineerinq Materials In this section, we will briefly examine the chemical composition and common of some solid materials. \[e will discuss light metals, copper and its alloys, iron and steel concrete, wood, plastics, silicon, glass, and composite materials. Most of you will take a materials class during your sophomore or junior year and will learn more in depth t the atomic stnrcture of various materials. Here our intent is to introduce you to common materials and their applications.
[ightweight lvletals Aluminum, titanium, and magnesium,
because of their small densities (relative to lightweight meak. Because of their relatively high ratios, ligfrnveight metals are used in many structural and aerospace applications. Alrrminum and its allop have densities that are approximately onethird the of steel. Pure aluminum is very soft, thus it is generally used in electronics applicati and in making refecors and foils. Because pure aluminum is soft and has a reladvely tensile suength, it is alloyed with other metds to make it stronger, easier to weld, and to resistance to corosive environmen$. Aluminum is commonly alloyed with copper zlnc (Zn), magnesium (Mg), manganese (Mn), silicon (Si), and lithium (Li). The National Sandards Institute (ANSD assigns designation numbers to specifr alloys. Generally speaking aluminum and its alloys resist corosion; they are easy to ill *[ cut and can be brazed or welded. Aluminum parts can also be joined using adhesives. are good conductors of electricity and heat and thus have relatively high thermal and low electrical resistance values. Aluminum is fabricated in sheets, plates, foil, rods, and #fre and is extruded to make window frames or automodve pams. You are already familiar withl everydry enamples of common aluminum products, induding bwerage cms, household alumifrum foil, nonrust staples in tea bags, building insulation, and so on. Tiani'lm has an excellent strengthtoweightratio. Titanium is used in where relativelyhigh temperatures, exceeding400" up to 600oC, are expeced. Ti are used in the fan blades and the compressor blades of the gas turbine engines of and military aiqplanes. In fact, without the use of titanium alloys the engines on aiqplanes would not hrve been possible. Like aluminum, titanium is alloyed with metals
commonly referred to
Lightwelght and durable, aluminum alloys are used
to produce a wide range of productsfrom high' performance engines to soda cans.
as
510
CnrprsR
17
Enqcrxrrmxc
Memnns
Because of their excellent strengthtoweight ratio and resistance to corrosion, titanium alloys are used to produce a range of productsfrom bike frames to frames for glasses
to improve its properties. Titanium alloys show excellent resistance to corrosion. Titanium is quite expensive compared to aluminum and it is heavier than aluminum, hardng a density which is roughly onehalf that of steel. Because of theit relatively high strengthtoweight ratios, titanium alloys are used in both commercial and military aiqplane airframes (fuselage and wings) and landing gear components. Titanium alloys are becoming a metal of choice in manyproducts; you c:rn find them in golf clubs, bicycle frames, tennis racquets, and spectacle frames. Because of their excellent corrosion resistance, danium allop have been used in the tubing in desalination plants as well. Replacement hips and other joints are examples of other applications where tianium is currendy being wed. Vith ia silverywhite appearance, magnesium is another lightweight metal that looks like aluminum, but it is lighter, haring a density of approximately 1700 kg/m3. Pure magnesium does not provide good strength for strucnrral applications and because of this fast, it is alloyed with other elemenc such as aluminum, manganese, and zinc to improve its mechanical characteristics. Magnesium and its allop are used in nuclear applicadons, in drycell batteries, and in aerospace applications and some automobile parts as sacrificial anodes to protect other
metals from corrosion. The mechanical propenies of the lighnveight metals are shown in Thbles 17.1 through 17.4.
Copper and
lts Alloys
Copper is a good conductor of electriciry and because of this properry is commonly used in many electrical applications, including home wiring. Copper and many of its alloys are also good conductors of heat, and this thermal propeny makes copper a good choice for heat
17.3
SoMe CoMrvroN Sor.rp ENcrNnsRrNc
5rl
l:'...:: .. . .t: ,,..
r..
.i
.
:l
r.;i
.''
Examples of products made of copper alloys.
exchanger applications in air conditioning and refrigeration systems. Copper alloys also used as tubes, pipes, and fitdngs in plumbing and heating applications. Copper is alloyed ith zinc, tin, aluminum, nickel, and other elements to modifr is properties. When copper is with
zinc it is commonly called brass. The mechanical propenies of brass depend on the position ofpercent copper and percent zinc . Bronzc is an alloy ofcopper and tin. alloyed with aluminum and is referred to as alurninurn bronze. Copper and its used in water tubes, heat exchangers, hydraulic brake lines, pumps, and screws.
comis also
are also
lron and Steel Steel is a common material that is used in the framework of buildings, bridges, the pliances such as refrigerators, ovens, dishwashers, washers and dryers, and cooki Steel is alloy of iron with approximately 2o/o or less carbon. Pure iron is soft
"n
good for structuial applications, but the bddition of even a small amount of hardens it and gives steel bener mechanical propenies, such as grearcr strength. The of steel can be modified by adding other elements, such as chromium, nickel, silicon, and tungpten. For example, chrornium is used to increase the resistance of rosion. In general, steel can be classified into three broad groups: (1) the carbon ing approximately 0.015 to 2o/o carbon (2) lowalloy steels having a maximum of elements, and (3) highalloy steels containing more than 8%o of allolng elements. constitute most of the world's steel coruumption, thus you will commonly find body ofappliances and cars. The lowalloy steels have good strength and are machine or tool pars and as strucural members. The highalloy steels, such as stai could contain approximately l0 to 30o/o chromium and could contain up to 35o/o
ofaputensils.
thus not
to lfon
to corcontaln
,llolnng steels
in the used as steels,
The
512
Cueprnn
17
ErcrNnrnrNc MrreRrALs
Alloy steels, because of mechanical properties such as superior strength, are used in a wide range of
applicationsfrom structural components to machine parts.
l8/8 stainless steels, which contain 18% chromium and 8olo nickel, are commonly used for tableware and kitchenware products. Finally, cast iron is also an alloy of iron that has 2 to 4o/o carbon. Note that the addition of extra carbon to the iron changes ia propenies completely. In fact, cast iron is a britde material, whereas most iron alloys containing less than 2o/o carbon are ductile.
Conerete in construction of roads, bridges, buildings, tunnels, and dams. !7hat is normally called concrete consists of three main ingredienrc: aggregate, cement, and water. Aggregate refers to materials such as gravel and sand, and cement refers to the bonding material that holds rhe aggregate toger:her. The types and size (fine to coarse) of aggregate used in mfing concrete varies depending on the application. The amount of water used in making concrete (watertocement ratio) could also influence its strength. Of course, the mixture must have enough water so that the concrete can be poured and hane a consistent cement Today, concrete is commonly used
to aggregate in making concrete also affecs the sffengh and durability of concrete. Another facor that could infuence the cured strength of concrete is the temperature of its surroundings when it is poured. Calcium chloride is added to cementwhen the concrete is poured in cold climates. The paste that completelywraps around all aggregates. The ratio of amount of cement used
17.3
Solrs CoMMou Sor.rp ENcrNssRrNc
Examples of concrete used in construction.
addition ofcalcium chloride will accelerate the curing process to counreract the of the low temperature of the surroundings. You may have also noticed as you walk by poured concrete for a driveway or sidewalk that water is sprayed onto the concrete for some time after it is poured. This is to control the rate ofcontraction ofthe concrete as it sets. Concrete is a britde material that can suppoft compressive loads much better than it does tensile loads. Because of this fact, concrete is commonly reinforcedvithsteel bars or steel mesh that consists of thin metal rods to increase its loadbearing capacity, especially in the sections where tensile sffess.is expected. Concrete is poured into forms that contain the metal mesh or steel bars. Reinforced concrete is used in foundations, foors, walls, and columns. Anotfrer common construction practice is the vse of 1>recast conc?ete. Precast concrete slabs, blocks, and structural members are fabricated in less time with less cost in facory settings where surrounding conditions are controlled. The precast concrete parts are then moved to the construction site where they are erected. This practice sanes time and money. As we mentioned, concrete has a higher compressive strength than tensile srength. Because of this fact, concrete is alsoprestessed in the following manner. Before concrete is poured into forms that have the steel rods or wires,
514
Cner"rsn
17
ENcrr.rEERINc M,lreRrALs
the steel rods or wires are suetched; after the concrete has been poured and after enough time in the rods or wires is released. This process, in turn, compresses the concrete. The prestressed concrete then acts as a compressed spring which will become uncompressed under the action of tensile loading. Therefore, the prestressed concrete section will not experience any tensile stress undl the secrion has been completely uncompressed. It is important to note once again the reason for this practice is t*rat concrete is weak under tension. has elapsed, the tension
Wood Throughout history wood, because of ia abundance in many parts of the wodd, has been a material of choice for many applications.'Wood is a renewable source, and because of is ease of workability and its strength, it has been used to make many products. Wood also has been used as fuel in stoves and fireplaces. Tod"y, wood is used in a variety of producs ranging from telephone poles to toothpicla. Common examples of wood products include hardwood fooring roof trusses, furniture frames, wall suppors, doors, decorative items, window &ames, trimming in luxury c:us, tongue depressors, dothespins, baseball bats, bowling pins, fishing rods, and wine barrels (see Figure 17.1). I(ood is also the main ingredient that is used to make various paper producs.'$Thereas a steel structural member is susceptible to rust, wood, on the other hand, is prone to fire, termites, and rotting.'Wood is anisotropic material, meaning that its properties are directiondependent. For example, as you may already know, under axial loading (when pulled), wood is stronger in a direction parallel to a grain than it is in a direction across the grain. However, wood is stronger in a direction normd to the grain when it is bent. The properties ofwood also depend on its moistue contenq the lower the moisture content, the stronger the wood is. Density ofwood is generally a good indicadon of its strength. As a rule of thumb, the higher the density ofwood, the higher its strength. Moreover, any de' fects, such as knots, would affect the loadcarrying capacity ofwood. Of course, the location of the knot and the e*ent of the defect will direcdy affect its strength. Timber is commonly classified as sofwood md hardwood" Softwood timber is made from rees that have cones (coniferous), such as pine, spruce, and Douglas fir. On the other hand, hardwood timber is made &om ffees that have broad leaves or hane fowers. Examples of hardwoods include walnut, maple, oak, and beech. This classification of wood into softwood and
I
tigurel7.t
Eramples of wood producls.
17.3
Sotvrs CoMMox Sor.ro ENcrxsBRrNc Merenrer.s
sr5
hardwood should be used with caution, because there are some hardwood timbers that are softer than softwoods.
Plastie s In the latter pan of the 20th century, plastics increasingly
became the material of choice for many applications. They are very lighnveight, strong, inexpensive, and easily made inro various shapes. Over 100 million metric tons of plastic are produced annually worldwide. Of couirse, this number increases as the demand for inexpensive, durable, disposable material grows. Most of you are familiar with examples of plasdc products, induding grocâ‚¬ry "lt "dy and trash bags, plastic soft drink containers, home deaning containers, vinyl siding polyvinyl chloride (PVC) piping, valves, and fittings that are readilyavailable in home improvement centers. Styrofoamru plates and cups, plasdc forks, knives, spoons, and sandwich bags are other examples ofplastic products that are consumed every day. Pollrners are the backbone of what we call plastics. They are chemical compounds that have very large, molecular, chainlike strucnrres. Plastics are often classified into nvo categories: 'When thermoplastics and tbennosas. heated to certain temperatures, the thermoplastics can be molded and remolded. For example, when you recyde Styrofoam dishes, they can be heated and reshaped into cups or bowls or other shapes. By contrast, thermosets can not be remolded into other shapes by headng. The application of heat to thermosets does not soften the material for remolding instead, the material will simply break dovrn. There are many other ways of classifying plastics; for instance, they may be classified on the basis of their chemical composition, or molecular strucrure, or the way molecules are arranged, or their densities, For example, based
Because of their low cost of production, polymerbased products are commonly used in our everyday lives.
516
Cnl,;prnn
lT
Er.rcnvsrRrxc MarsRrALs
on their chemical composition, polyethylene, polypropylene, polyvinyl chloride, and polystyrene are the most commonly produced plastics. A grocery bag is an example of a product made from highdensity polyethylene (HDPE). Honrwer, note that in a broader sense polyethylene and polystyrene, for example, are thermoplastics. In general, the way molecules of a plastic are arranged will influence its mechanical and thermal properties. Plasdcs have relatively small thermal and electrical conductivity values. Some plastic materials such as Styrofoam cups are designed to have air trapped in them to reduce the heat conduction even more. Plastics are easily colored by using various metal oxides. For example, titanium oxide and zinc oxide are used to give aplasdc sheet its white color. Carbon is used to give plastic sheets their black color, as is the case in black uash bags. Depending on the application, other additives are also added to the polymers to obtain specific characteristics such as rigiditp flexibility, enhanced strength, or a longer life span that excludes any change in the appearance or mechanical properties of the plastic over time. As vrith other materials, research rs beingperformed wery day to make plastics stronger and more durable and to control the aging process, to make plastics less suscepdble to sun damage, and to control water and gas diffirsion ,ttough them. The lamer is especially important when the goal is to add shelflife to food that is wrapped in plasdcs. Those of you who are planning ro study chemical engineering will take semesterlong classes that will explore polymers in much more detail.
Silieon Silicon is anonmetdlic chemical element that is used quite extensivelyin the manufacruring of ransistors and various electronic and computer chips. Pure silicon is not found in nature; it is found in the form of silicon dioxide in sands and rocls or found combined with other elements such as aluminum or calcium or sodium ormagnesium in theform thatis commonlyreferred to x silicates. Silicon, because of its atomic structure, is an excellent semiconductor, a material whose electrical conductivity propenies can be changed to act either as a conductor ofelectricity or as an insulator (prwentor of elecricity flow). Silicon is also used as an alloying element with other elements such as iron and copper to give steel and brass cerain desired characteristics. Be sure not to confuse silicon lwth siliconed which are qirnthetic compounds consisting of silicon, oxygen, carbon, and hydrogen. You find silicones in lubricana, varnishes, and waterproofing products.
A computer chip.
17.3
SovB Coruuon Sorrp ExcrxnsRrxc M.e,rrnrars
517
Glass commonly used in products such as windows, light bulbs, TV CRI tubes, housewares drinking glasses, chemical containers, bwerage and beer conainers, and decorative items (see Figure 17.2). The composition of the glass depends on its application. The most widely used form of glass is sodalimesilica glass. The materials used in making sodalimesilica Glass is
such
as
glass include sand (silicon dioxide), limestone (calcium carbonate), and soda ash (sodium car
bonate). Other materials are added to create desired characeeristics for specific applications. For example, bode glass contains approximately 2o/o aluminum oxide, and glass sheea contain abortt4o/o magnesium oxide. Metallic oxides are also added to give glass various colors. For example, silver oxide gives glass a yellowish stain, and copper oxide gives g[ass its blueish, greenish color, the degee depending on the amount added to the composition of the glass. Optical glasses have very specific chemical compositions and are quite expensive. The composition of optical glass will influence its refractive index and its lightdispersion properties. Glass drat is made completely from silica (silicon dioxide) has propenies that are sought after by many indusuies, such as fiber optics, but it is quite expensive to manufacture because the sand has to be heated to temperatures exceeding 1700"C. Silica glCIs has a low coefficient of thermal expansion, high electrical resistivity, and high ffansparency to ultradolet light. Because sfica glass has a low coef,ficient of thermal enpansion, it can be used in hightemperature applications. Ordinary gla.ss has a relatively high coefficient of thermal enpansion; therefore, when its temperaue is changed suddenly, it could break easily due to thermal sffesses dweloped by the temperafure rise. Cookrvare glass contains boric oxide and aluminum oxide to reduce its coefificient of thermal expansion. Fiber Glass Silica glass fibers are commonly used today in fiber optics, thm branch of science that deds with transmining data, voice, and i*â‚¬o thto"gh thin glass or plastic fibers. Every day, copper wires are being replaced by transparent glass fibers in telecommunication to connect comPuters together in nenrcorls. The glass fibers rypically have an outer diameter of 0.0L25 mm (12 micron) with an inner transmitting core diameter of 0.01 mm (10 micron). Infrared light signals in the wavelength ranges of 0.8 to 0.9 m or 1.3 to 1.6 m are generated by lightemitting diodes or semiconducor lasers and travel ,hto"gh the inner core of glass fiber.
I
FigurelT.Z
Eramples of glass products.
518
Cnar'rsn
17
ENctNenRrNc Memnrers
Examples of fiber optic cables made by Corning.
The optical signals generated in this manner can travel to distances as far as 100 km without any need to amplify them again. Plastic fibers made of polymethylmethacrylate, polystyrene, or polycarbonare are also used in fiber optics. These plastic fibers are, in general, cheaper and more flexible than glass fibers. But when compared to glass fibers, plastic fibers require more amplification of signals due to their greater optical losses. They are generally used in networking computers in a building.
ComBosites lightweight and good suength, composite materials are becoming inoeasingly the materials of choice for a number of producr and aerospace applications. Today you will find composite materials in military planes, helicopters, satellites, commercial planes, fastfood restaurant tables and chairs, and many sponing goods. They are also commonly used to repair bodies ofautomobiles. In comparison to conventional materials, such as metals, composite materials can be lighter and stronger. For this reason, composite materials are used extensively in Because of their
aerospace applications.
Composites are created by combining two or more solid materials to make a new material that has properties that are superior to those of the individual components. Composite materials consist of two main ingredients: matrix material and fibers. Fibers are embedded in matrix materials, such as aluminum or other metals, plastics, or ceramics. Glass, graphite, and silicon carbide fibers are examples of fibers used in the construction of composite materials, The strength of the fibers is increased when embedded in the matrix material, and the composite material created in this manner is lighter and stronger. Moreover, once a crack starts in a single material, due ro either excessive loading or imperfections in the material, the crackwill propagate to the point of failure. In a composite material, on the other hand, if one or a few fibers fail, it does not necessarily lead to failure of other fibers or the material as a whole. Furthermore, the fibers in a composite material can be oriented either in a certain direction or many directions to offer more strength in the direction of expected loads. Therefore, composite materials are designed for specific load applications. For insance, if the expected load is uni
L7.4
Soun CouuoN Fluro
Merrnnrs
519
axial, meaning that it is applied in a single direction, then all the fibers are aligned in dre direcdon of the expected load. For applications expecting multidirection loads, the fibers are aligned in dlfferent direcdons to make the material equally strong in various directions. Depending upon what type of host matrix material is used in creating the composite material, the composites may be classified into three classes: (1) polymermatrix composites, (2) metalmatrix composites, and (3) ceramicmatrix composites. We discussed the characteristics of matrix materials earlier when we covered metals and plastics.
11.4
Some Common Fluid Materials Fluid rcfets to both liquids and
gases. Air and water are among the most abundant fluids on eanh. They are imponant in sustaining life and are used in many engineering applications. W'e
will briefly
discuss them next.
Air We all need air and water to sustain [ife. Because air is readily available to us, it is also used in engineering as a cooling and heating medium in food processing, in controlling thermal comfon in buildingr as x conuolling medium to nun equipment on and off, and to &ive power tools. Compressed air in the tires of a car provides a cushioned medium to transfer the weight of the car to r:he road. Understanding the propenies of air and how it behanes is imponant in many engineering applications, including understanding the lift and drag forces. Better undersanding of honr air behaves under cenain conditions leads to the design of bemer planes and automobiles. The earth's atmosphere, which we refer to as air, is a mixrure of approximately 787o nitrogen , 2lo/o oxygen, and less than lo/o argon. Small amounts of other gases are presenr in eanh's atmosphere, as shown inTable 17.5.
Yolume byPercent
Niuogen (N2) Oxygen (O2)
.freon(tu)
."
78.084 20.946
a.994
Small amounts of othn gases are presmt in a*nosphere iwkding
Neon (Ne) Helium (He) Methane (CHa) Krypton (IG) Hydrogen (H) Nitrous oxide (NzO) Xenon (Xe)
0.0018 0.000524 0.0002 0.000114 0.00005 0.00005
0.0000087
520
Cneprsn
17
ENcrxsrRrNc M^rrnnrar.s There are other gases present in the atmosphere, including carbon dioxide, sulfirr dioxide, and nirogen oxide. The amosphere also conains water vapor. The concenuation level of these gases depends on the altitude and geographicd location. At higher altitudes (10 to 50 km), the eanh's atmosphere also contafurs ozone. Even though these gases make up a small percentage of
eartht atmosphere, they play a significant role in maintaining a thermally comfortable environment for us and other living species. For example, the ozone absorbs most of r:he ultraviolet radiation arriving from the sun that could harm us. The carbon dioxide plqrs an important role in sustaining plant life; howwer, if the atmosphere contains too much carbon dioxide, it will not allow the earth to cool down effecdvely by radiation, 'Water vapor in the atmosphere in the form of clouds allows for ffansport of water from the ocean to land in the form of rain and snow.
Humidity There are two common ways of expressing the amount of water vapor in air: absolute humidity or humidiry rado, and relative humidiry. The absolute humidity is defined as
the ratio of mass ofwater vapor in a unit mass of dry air, according to absolute
humidiw
:
mass of water vapor mass
of dry
"i.
(kg) fl7r)
(k)
For humans, the level of a comfonable environment is better expressed by relative humidity, which is defined as the ratio of the amount ofwater vapor or moisture in the air to the maximum amount of moisture that the air can hold at a given temperature. Therefore, relative hu
midity is defined
as
relative humidity
:
the amount of moisture in the air (kg) the maximum amount of moisture that air can hold
(fu)
(r7"2)
Most people feel comforable when the relative humidity is around 30 to 50o/o. The higher the temperatrue ofair, the more water vapor the air can hold before it is firlly saturated. Because of its abundance, air is commonly used in food processing, especially in food drying processes to make dried fruits, spaghetti, cereals, and soup mixes. Hot air is transponed over the food to absorb water vapors and thus remove them from the source. Undersunding how air behaves at given pressures and temperanrres is also imporant when desigrung c:rs to overcome air resistance or designing buildings to withstand wind loading.
Water lbu
already know that wery living thing needs warer to sustain life. In addition to drinking water, we also need water for washing, l..r"dty, grooming, cooking, and fire protection. You may also know that twothirds of the eartht surface is covered with watec but mosc of this water quulot be consumed direcdy; it conains salt and other minerals that must be removed firsr Radiation from the sun evaporates water; water vapors form into clouds and eventually, under farorable condirions, water vapors turn into liquid water or snow and fall back on the land and the ocean. On land, depending on the amount of precipitation, paft of the water infihrates the
soil, part of
it may be absorbed by vegeation, and part runs
as streams
or rivers and collects
into naturd reservoirs called lakes. Surface water refers to water in reservoirs, lakes, rivers, and strâ‚¬,!ms. Groundwater, on the other hand, refers to the water that has inflrated the ground; surface and groundwaters eventudly reflrrn to the ocean, and the water cycle is completed. As
L7.4
Sorvrs CoMnroN FLUID
MersRrALs
521
we said eadier, everyone knows that we need water to sustain life, but what you may not redize is that water could be thought of as a common engineering material! Water is used in all steam polvergenerating plana to produce electricity. fu we explained in Chapter 13, fuel is
burned in a boiler to generate heat, which in turn is added to liquid water to change its phase to steam; steam passes through turbine blades, turning the blades, which in effect runs the generator connected to the rurbine, creating electricity. The lowpressure steam liquefies in a condenser and is pumped ,hto"gh the boiler again, closing a cycle. Uquid water stored behind dams is also guided thto"gh water turbines located in hydroelectric pov/er plane to generate elecuicity. Mechanical engineers need to understand the thermophysical propenies of liquid water and steam when designing power plants. 'W'e also need water to grow fruits, vegetables, nuts, cofton, trees, and so on. Irrigation channels are designed by civil engineers to provide water to farms and agricultural fields. Vater is also used as a cutting tool. Highpressure water containing abrasive partides is used to cut marble or metals. Vater is commonly used as a cooling or cleaning agent in a number of food processing plants and industrial applications. Thus, water is not only transpofted to our homes for our domestic use but it is also used in many engineering applications. So you see, understanding the propenies of vrater and how it can be used to transport thermal energy, or what it akes to uanspoft water from one location to the next, is important to mechanical engrneers, civil engineers, manufacnrring engineers, agricultural engineers, and so on.'W'e fiscussed the Environmental Protection Agency (EPA) standards for &inking water in Chapter 3.
522
CHepreR
17
ENcnmsRrNcMerrnrer,s
SUMMARY Now that you hare reached this point in the text
.
You should understand that engineers select materials for an application based on characteristics of materials, such as strength, densrty, corrosion resistance, durability, toughness, the ease
. .
of machining, and manufacrruability. Moreover, you should understand that materid
cost is also an imporant selection criterion. You should be familiar with common applications of basic materials, such as light metals and their alloys, steel and im allgrs, composite materials, and building materials such as @ncrete, wood, and plastics. You should be familiar with the application of fuids, Euch as air and rsater, in engineering. You should also be familiar with the composition of air and what the term hurnidity means.
l7J.
Identi& and list
lt.L
used in your car. Name at least five different materials that are used in a
a. b. c. d. e.
at least ten different materids that are
refrigerator. r7.3.
11.4"
t7.5.
17.6.
111.
Identify and list at leasr five different marerials that are used in your TV set or computer. List at least ten different materials that are used in a building envelope. Ust at least five different materids used to fabricate window and door frames. List the materials used in the fabrication of incandescent light bulbs. Idendfy at least ten producs around your home that
11.9.
a briefreport discuss the advanages and disadvanages of using Styrofoam, paper, glass, stainless steel, and ceramic materials for coffee or tea cups. Every day you use a wide tange of paper products at home or school. These paper producs are made from different papergrades.'Vood pulp is the main in
it with
some chemicals. Investigate the composition, processing methods, and the annual consumption rate
grades of paper products in the United States, and write a brief report discussing your findings. The paper products to investigate should
of the following
include:
paper towels As you already know, roofing materids keep the water
from penetrating into the roof srucnue. There is a wide range of roofing produca available on the market today. For example, asphalt shingles, which are made by impregnating a dry felt with hot asphah, are used in some houses. Other houses use, for example, wood shingles, such as redcedar or redwood shingles. A targe number of houses in California use interlocking clay tila as roofing materials. Investigate the propenies and the characteristics of various roo'Write a brief repon discussing your fing materials.
In
gredient used in making a paper product. It is common p'. Lctice to grind the wood first and cook
sanitary papers glassine and waxing papers bagpaper boxboard
f,
r7J0.
make use ofplasdcs. t7.8.
pnnungpaPers
firdi"gt. t711.
Adhesives are used extensively to bond together parts made of the same or dlfferent materials. Discuss the characteristics that should be considered when select
ing an adhesive for an application. For example, when selecdng an adhesive engineers consider the adhesivet ability to bond dissimilat materials, cure time, service temperatrue, suength, and so on. Investigate the advantages and disadvanages ofvarious adhesives, such as natural adhesives and thermosetting, ttrermoplastic, and sFnthedc elastomers,
Pnosr.EMs 1112.
Visit a home improvement center (hardware/lumber smre) in your town, and ffy to gather information
about various types of insulating materids that can be used in a house. W'rite a brief report discussing advanuges, disadvanages, and the characteristics ofvarious insulating materials, including their thermal characteristics in terms ofRvalue. t7J3. Investigate the characteristics oftitanium alloys used in sponing equipment, such as bicycle frames, tennis racquets, and golf shafa. Vrite a brief summary report discussing your findings. lt.t4. Investigate the characteristics of titanium allop used in medical implants for hips and other joint replacemen6. 'lV'rite a brief summary report discussing your findings. Iil5. Cobaltchromium alloys, stainless steel, and titanium allop are three common biomaterials that have been used as surgical implants. Investigate the use of these biomarerials, and write a brief report discussing the advantages and disadvantages of each. ltJ6. According to the Aluminum Association, in 1998,
102 billion aluminum qrns were produced and of these, 62 billion cans were recycled. Measure the mass
17"17.
of ten aluminum cans, and use an average mass for an aluminum can to estimate the total mass of aluminum cans tjrat was recycled. As we discussed in this chapter, when selecting materials for mechanical applications, the value of modulus of resilience for a materid shows hour good the material is at absorbing mechanical energ;r without sus'
airirg
any permanent damage. Another imponant characteristic of a material is its ability to handle overloading before it &acnres. The value of modulus of toughness provides such information. Ipokup thevalues ofmodulus ofresilience and modulus oftoughness for the following materials:
a. titanium b. steel t7.t8.
17J9.
Investigate and discuss some of the characeristics of the materials r:hat are used in bridge construction. As we discussed in this chapter, the strengthtoweight
ratio of material is an imponant criterion when selecting material for aerospace applications. Calculate the average strengtfrtoweight ratio for the following materials: aluminum alloy, dtanium alloy, and steel. Use Tables 17
.2 and 17.3 to look up appropriate values.
17.20.
523
average, will an aluminumalloy tennis racquet be if it is made ftom titanium alloy' Obtain a tennis rac{luet, and take appropriate mea
How much heavier, on
sluements to perform your analysis. v.n. Tensile test machines are used to measure rhe mechanical propenies of materials, such as modulus of elasticiry and tensile suength. Visit the Veb site of the MTS Systems Coqporation to obtain information on test machines used to rc$ the strength of materids.'Write a brief repon discussing your fitdi"gs. fi"n. Endoscopyrefers to medical examinadon ofthe inside of a human body by means of insening a lighted optical instrument thtough a body opening. Fiberscopes op' erate in the visible warelengths and consist of nryo major components. One component consists of a bundle of fibers that illuminates the examined area, and the other component ffansmi$ the images of the exam
ined area to the eye of the physician or to some display device. Investigate the design of fiberscopes or t*re fiberoptic endoscope, and discuss your findings in a
briefreport. 17.21.
\7"24.
Ct;nol
tableware glass that sparkles is sought after by many people as a sign of affiuence. This crystd commonly contains lead monoxide. Investigate the propenies of crysal glass in detail, and write a brief report discussing your fitrdi"f . You all have seen grocer)'bags that have labels and printed information on them. Investigate how information is printed on plasdc bags. For example,
a
cornmon practice indudes using
a
wetinking
process; another process makes use of lasers and heat transfer decals. Discuss your findings in a brief fePort. 17.25. Teflon and Nylon are trade names of plastics that are used in many products. Iook up the adual chemical name of these products, and give at least five examples ofwhere they are used. 17.26. Investigate how the following basic wood products are made: plywood, pardcle board, veneer, and fiberboard. Discuss your findings in a briefreport. Also investigate common methods of wood preservation, and discuss 'What your findings in your report. is the environmenal impact of both the production and use of ueated wood producs in this question? t7rI. Investig'ate the common uses of co$on and its typical properties. Discuss your findings in a brief repon.
524 17.28.
Csepren
17
ENcrNnnnrNc MereRrALs
As most of you know, commercial uansport planes cruise at an altirude of approximately 10,000 m (33,000 ft). The power required to maintain
fight depends on air drag, or resistance, at that altitude, which may be estimated by the following level
reladonship:
Power
17.30.
I : o*coulu
where p,;" is alrrri.y of
the given akitude, Cp rep"i, ", resenr the drag coefficient of the plane, r4 is the planform area, and U represens the cruising speed of the plane. Assume that a plane is moving at constant speed
and, Cp remaining constant, determine the ratio of power that would be required
if the plane is cruising
at 8000 m and when the plane may be cruising at 11,000m. 17.29.
Investigate the average daily water consumption per capia in t}re United States. Discuss the personal and public needs in a brief report. Also discuss facors such as geographical location, time of the year, time of &" d"y, and cost of consumption pafterns. For example, more water is consumed during the early morning hours. Civil engineers need to consider all
these factors when designing water systerns for cities. Assuming that the life expectancy of people has increased by five years over the past decades, how much additional water is needed to sustain the lives of 50 million people? Vhen a ring gets stuck on a finger, most people reson to water and soap as a lubricant to get the ring off. In earlier times, animal fat was a common lubricant used in wheel axles. Moving pans in machinery the piston inside your car's engine, and bearingr are examples of mechanical componen$ that require lubrication. A
lubricant is a substance that is introduced between the parts that have relative motion to reduce wear and friction. The lubricants must have characteristics that
are suitable
for a given siruation. For instance, for
liquid lubricants, viscosity is one of the important propenies. The flash and fire and the cloud and pour are examples of orher characteristics that are examined when selecting lubricants. Investigate the use of petroleumbased lubricants in reducing wear and friction in today's mechanical componens. Vrite a brief repon discussing the application and characteristics
poins
of liquidpetroleumbased and solid lubricanm that are commonly used, such as SAE 10\f40 oil and graphite.
525
The
Jet Engine
*
Introduction You think you've got problems? How would you like to be a molecule of air minding your own business at 30,000 feet when all of a sudden you get mugged by five tons of Pratt & !7'himey
jet engine? Over the course of the next 40 thousandths of a second you, Mr. or Ms. Molecule, will be beaten duough 18 stages ofcompression, singed in a firrnace heated o neady 3,000 degrees Fahrenheit, enpanded through a turbine and pushed out the back with a wicked headache and the bitter knowledge as the aircraft screams on that your ugly ordeal has been merely for the sake of geaing Aunt Marge to Phoenix for a much needed rest. That's basically what happens wery day in the skies above as thousands of Pratt engines power aircraft from place to place while the passengers and crew inside those aircraft experience only the steady and reassuring thrum that resuls from Prant precision manufacturing.
The princife of jet propulsion has been demonstrated by anyone who has blown up a balloon for a child and accidenally let go before tpng it closed. The air stored up inside the *Materiah were adapted with permission 6om Pratt Matthew Broder.
526
&
Vhimey, a Unired lirhonologies Company. Tort by
TnrJrr
ENcnw
527
balloon accelerates as it rushes to esc{rpe through the narrowest part of the balloons neck. This acceleration, or change in the speed of air, combines with the weight of the air itself to produce thrust. It is the thrust that sends the renegade balloon zipping madly about dre room. The original turbojet engine, which debuted in scheduled commercial service in 1952, rccomplished this same uick on a breathmking scale and with pinpoint control. The progress of jet engine technology in the past four decades has been to increase the amount ofair going into the engine, and change the speed of that air with ever greater efEciency. AII the examples in this anicle are drarnrn from the mightiest family of engines ever to fy, the Pratt 6c'Whitney 4000s.
The
Jet Engine
Fan The propulsion process begins with the huge, 9footdiameter fan at the front of the engine, spinning 2,800 times a minute at takeoff speed. That fan sucks in air at the rate of 2,600 pounds per second, or enough to vacunm out r:he air from a 4bedroom house in less t}ran half a second.
Compression As the air leaves the fan it is now separated into two sffeams. The smaller sueam, about 15 percent of the total volume of air, is called primary or core air and enters the fust of two compressors that are spinning in the same direction as the fan itself. As the primary air passes through each stage of the two compressors, both its temperature and pressure rise.
Gombustion '$7hen compression is complete, the air, now 30 times higher in pressure and 1,100 degrees hoter, is forced duough a fumace or combustor. In the combustion chamber, fuel is added and burnt. The air's temperature soars even higher, and the air is finally ready to do tJre nvo jobs for which it has been so hastily prepared.
Turbine The fint job is to blast tht"gh the blades of two turbines, sending ttrem whirling just like the wind spinning the arms of a windmill. The whirling turbines turn the shafis that drive both compressors and the fan at the front of the engine. This process, in which the engine extracts energJr from the air it has just captured, is what allows modern jets to operate with such high fuel efficiency.
Exhaust The second job is to push the aiqplane. After passing duo"gh fr" rurbines, the hot air is forced through the exhaust opening at the back of the engine. The narrowing walls of the exhau$ force the air to accelerate and, just as with the balloon, the weight of the air combined with its acceleration drives the engrne, and the airplane aaached to it, forward.
528
CHeprsR
17
EnrcrNsERrNc
Fan
Mamnrers
Air or Bypass Air
The larger air stream exiting the fan, representing 85 percent of the total, is called fan air or blpass air, because
it
bypasses this entire process.
The engine itself is shrouded in a metd casing cdled the nacelle, shaped roughly like a sidewqrc ice cream cone with the bottom cut off. Bypass air is forced through the ever narrower space benreen the nacelle s'all and the engine, picking up speed along the way. Because of its huge volume, bypass air needs only to accelerate a small amount to produce an enormous kick ofthrust. In the P\f4084 engine, bypass air accounts for 90 percent ofthe thrust, and has the added benefits ofkeeping the engine cooler, quieter and more fuel efficient.
lrrTiili:illii=:.iililil,:::lli:ii.rilialill
.]
Part
MerHErvrATrcs, Srarrsrrcs, AND ENcTNEERTNG EcoNoMrcs WHv Anr
Ttlrv lmpontnnt?
Engineering problems are mathematical models of physical situations with different
forms that rely on a wide range of mathematical concepts. Therefore, a good understanding of mathematicalconcepts is essential in the formulation and solution of many engineering problems. Moreovel statistical models are becoming common tools in
the hands of practicing engineers to solve quality control and reliability issues, and to perform failure analyses. Civil engineers use statistical models to study the reliability of construction materials and structures, and to design for flood control. Electrical engineers use statistical models for signal processing and for developing voice'
recognition software. Manufacturing engineers use statistics for quality control assurance of the products they produce. Mechanical engineers use statistics to study
the failure of materials and machine parts. Economic factors also play important roles in engineeringdesign decision making. lf you design a product that is too expensive to manufacture, then it cannot be sold at a price that consumers can afford and still be
profitable to your company. In the last part of this book, we will introduce you to important mathematical, statistical, and economical concepts.
/
ctrapter tB
Mathematics In Englneering
Chapter 19
Probablllty and Statistics in Engineering
Ghapter 20
Engineerlng Economics
1B
CF{APTER
MaTHEMATICS IN ENcINEERING fiE
n general, engineering problems are
I mathematical models of physical I
situations. For example, a model
,:t:
Stoppitg
Spod Speed ('nph) (ft/')
Sight (ft)
This simple model estimates the distance
0
0.0
0
5
'74
2l
L+,/ 22.4
47
29.3
rr4
10 T5
a driver needs in order
to stop his car
traveling at a certain speed after detecting an hazard. The model proposed by the American Association of State
20 25 30 35
40
,

2A1
5r.3
)?5 o
=tn
trs E
E10
v5
fl
tigurelg.?
The cumulative.f requency
histogram for Enmple 19.2.
5059
5069
s079 s089 Scores
5099
t7 23
26
582
Cneprnn
19
PnosABrlrry ^lNp Srerrsrrcs rN ENcTNEERTNc 30 A
b2s o
?..20
.,
,(2
o15
E10
o5
L
a Figurelg.3
Ihe cumulative'frequency
0r50
polygon for ftalnple 19.2.
80
70
90
100
Scores
convey the same information as contained in Thble 19.2. Howwer, it might be easier for some people to absorb the information when it is presented graphically. Engineers use graphical communication when it is the clearer, easier and more convenient way to convey information.
19"4 Measures of Central Tendency and VariationMean, Median, and Standard Dcviation secdon, we will discuss some simple ways to examine the central tendenry and variations within a given daa set. Every engineer should have some undersanding of the basic fundamentals of statistics and probabiliry for analyzing experimental data and experimental errors. There are alwaln inaccuracies associated with all experimenml observadons, If several variables are measured to compute a final result, then we need to know how the inaccuracies associated wirh these intermediate measurements will influence the accuracy of the final result. There are basically two types of observation errors: systematic errors and random errors. Suppose you were to measure the boiling temperanrre of pure water at sea level and standard pressure with a thermometer that reads 104'C. If readings from this thermometei are used in an experiment, it will result in systematic errors. Therefore, cystematic srrors, sometimes calledfr*ed errors, are errors associated with using an inaccurate instrument. These errors can be detected and aroided by properly calibrating instruments. On the other hand, random *rors are generuted by a number of unpredictable variadons in a given measruement situation. Mechanical vibra' tions of instruments or variadons in line volage friction or humidity could lead to fuctuations in experimental observations. These are o
, ,,
,
*f: !?q
it' i*,"
Frequency Midpoint Rarye
fn
5059 6A69
3
7A79
8089 9A99
i
tc
)4.>
74.9
64.5
74.9
9
//+.>
6
84.5
74.9 74,9
3
94.5
74.9
tc
(*
 7)'f
*20.4 1248.5 540.8 10.A *0.4 1.44 552.96 9.6 19.6 1152.5 i ' E(rV)2f=34g6
Using Equation (I9.7), the mean of the scores is
>("/)= re47 = 74.9 i: T 26 Similarly, using Equation (19.8), we cdculate the sandard deviation, as shown in Table 19.8.
n: 2f: nt:25
26
It4%
\25
. l,,11'rarit..I,
Normal disuibution is discussed next.
11.8
.l .: .:f.rli::,:lli.',i.I
 
rtii
:,


_:l
19.5 NonrvrarDrsrnrsurroN
587
19.5 NormalDistribution In Section 19.1, we explained what we mean by a statistical e"Feriment and outcome. Recall that the result ofan experiment is called an outcome. In an engineering situation, we often perform enperiments that could have many outcomes. To organize the outcomes of an experiment, it is customary to make use of probability distributions. A probability distribution shows the probability values for the occurrence of the outcomes of an ."T'eriment. To better undersand the concept of probability disuibution, let's turn our affenuon to Example 19.2. Ifwe were to consider the chemistryrcst as an experimentwith outcomes represented bystudent scores, then we can calculate a probability value for each range of scores by divifing each frequency by 26 (the total number of scores). The probability disuibution for F:rample L9.2 is given in Tirble 19.9. From examining Tirble I9.9, you should note ttrat the sum of probabilities is 1, which is rue for any probability disuibution. The plot of t}re probability distribution for Example 19.2 is shown in Figure 19.4. Moreover, if this was a grpical chemistry test with rypical
Range
5059
6069 7079 8089
9099
Probabilitv
Frequency 3
0.1 15
26
0.192
26
1
9
0.346
26 5
0.231
26 3
0.115
26
}p=l
h
0.250 0.200
P
0.150 0.100 0.050
D
Figurelg.4
Plot of probability distribution
for Elample
19.2.
0.0m
i
:
1
I
588
CH^IITSR
19
PnosABrlrryeNo Sransrrcs rN Er.IcrNEsRrNc students, ttren we might be able to use the probability distribudon for this class to predict how srudenrs might do on a similar test next year. Often, it is difficult to define what we mean by a typical class or a tfpical test. However, ifwe had a lot more students take this test and incorporate their scores into our analysis, we might be able to use the results of this eirperiment to
predict rhe ourcomes of a similar tesr to be given later. As the number of students taking the tesr increases (leading to more scores), the line connecting the midpoint of scores shown in Figure 19.4 becomes smoother and approaches a bellshaped curve.'We use the next example to funher explain this concept.
In order to improve the production time, the supervisor of assembly lines in
a manufacturing sening of computers has studied rhe time that it takes to assemble ceftain parts of a computer at various stations. She measures the time that it takes to assemble a specific part by 100 people at different shifts and on different dqrs. The record of her study is organized and shown in Thble 19.10.
TABLE
I
1
19.19
Dgta Pgrtaininq to Example 19.4
TimeThatlt t"ker aPenson
itoAssemblethe
.
Part (minutes) Frequency
Probability 0.05
5
8
0.08
I
7 8
II
0.I
t5
0.15
9
17
0.17
10
t4
o.r4
1l
t1
0.13
l2
8
0.08
t3
6
0.06
r4
3
0.03
Based on data provided, we have calculated the probabilities corresponding to the time intervals that people took to assemble the parts. The probability distribution for Fxample 19.4 is shown in Thble 19.10 and Figure 19.5. Again, note that the sum of probabilities is equal to 1. Also note that if rve were to connect dre midpoints of time results (as shown in Figure 19.5), we would have a curve that approximates a bell shape. As the number of data points increases and the intervals decrease, the probabilitydistribution curve becomes smoother. A probability distribution that has a bellshaped cun'e is called a nornaal distribution. The probability distribution for many engineering experiments is approximated by a normal
distribution.
19.5 NonruarDrsrnrsurroN
589
0.18 0.16 0.1"4
o.L2 0.10 0.08 0.06 0.04
I
0.02
tigurelg.S
0.00
8910rL Time (minutes) 
...tdrnn
The detailed shape of a normaldistribution curve is determined by its mean and standard deviation values. For example, as shown in Figwe 19.6, an enperiment with a small standard deviation will produce a tall, narrow curve; whereas a large sandard deviarion will result in a shon, wide curve. However, it is important to note that since the normal probabiliry distribution represents all possible outcomes of an experiment (with the total of probabilities equal to 1), the area under any given normal distribution should always be equal to 1. Also, note normal disaibution is qfmmeuical about the mean. In statistics, it is custornary and easier to normalize the mean and the sundard deviation values of an enperiment and work with what is called the standard norrnal dis*ibutioa which has a mean value of zero (x : 0) and a standard deviation value of 1 (s = l). To do this, we define what commonly is referred to as a z score acrr,rding to
z: xV
fi9.e)
J
In Equation (19.9), arepresenc the number of sandard deviations from the mean. The mathematical function that describes a normaldistribution curve or a standard normal curve is ratfier complicated and may be beyond the level of your qurent understanding. Most of you will learn about it later in your statistics or engineering classes. For now, using Excel, we have generated a table that shows the areas under ponions of the standard normaldisaibudon curve, shown in Thble 19.1 1 . At this stage ofyour education, it is imponant for you to know how to use the table and solve some problems. A more detailed explanation will be provided in your future classes.
f
Ihe shape of a nonnal dishibntion curve
[y its
will next demonsffaJe how to
use
Thble 19. 1 I , using
A
Figurelg.6
determined
'We
as
mean and
standard deviation.
/\
/*:""\ Small standard
large standard
deviation
deviation
a
number of example problems.
iiii
ffi
Note that the standard normd curve is qmrmetrical about the mean.
[email protected] Mean:0
z =3.N
[email protected] t:i o
e t
0.0000 i O,AL 0.0040 i, a.oz 0.00s0 1 ,0.03 0.0120 i ,'A.M 0.0160 1 i0.05 0.0199 10.06 a.0239 iit':a.97 0.0279 I ]O.OS 0.0319 1
i
Z
A
0.51 0,1950 r.o2 a.3461
O.52 0.1985 r.03 0.3485 0.53 0.2019 La4 a350s 0.54 0.2054 1.05 0.3531
Z 1.53
r.54 t.55
t,56
0.2088 r.06 0.3554 1.57 0.2123 r.07 a3577 1.58 0.2157 r.08 0.3599 t.59 0.2190 1.09 0.3621 1.6 0.2224 1.1 0.3643 1.6r i ioos 0.0is9 0.6 0.2252 r.rl 0.i665 r.62 i lO.r 0.0398 0.61 0.2291 r.r2 0.3686 1.63 i)t ,0.11 0.04i8 a.62 0.2324 t,r3 0.3708 l.g* ,o.tz 0.a478 0.63 0.2357 L.r4 0.3729 r.65 r 0.13 0.0517 O.g+ 0,2i89 r.15 0.3749 r.65 t o.r4 0.0557 a,65 0.2422 1.16 0.3770 r.67 i o.t5 0.0596 0.66 0.2454 r.r7 0.3790 1.6s i .,o.ta '0.06i6 0.67 0.2486 t.tg 0.3810 1.69 ',,,,a.L7 '0.0675 0.68 0.2517 1.r9 0.3t30 t,7 i.r0.r8 ,0.A714 0.69 0.2549 rA 0.3849 r.7r i ',0.19 0.0753 oJ 0.258a L.zr 0.3869 lJz ', 0a' :,0,029s o,7l 0.26il 1.22 0.3s88 L:73
0.55 0.56 o,57 o.58 0.59
A
0.4370 0.4382 0.4394 0.4406 0.4415 0.4429 0.4441 0.4452 0,4463 0.4474 0.44s4 0.4495 0.4505 0.4515 0.4525 0.4535 0.4545 0.4554 0.4564 0.4573 0.4582
z
Ai
A
z
2.O4 0.4793 '::2,55 0.4946 ,.a6 2.O5 0.4798 2.56 0.4948 ?,O7 2.06 0.48A3 '2;57 0.4949 3,A8 2,O7 0.4805 ,2,58 0.4951 3.09 2.A8 0:4812 2.59 0.4952 3.L 209 0.4817 2.6 0.4953 3.r1 2.t 0.4821 2.61 0.4955 3.r2 2.rr 0..4826, 2.62 0.4956 3.r3 2,t2 0;483A " 2161' 0.4957 !.L4 2.13 0;4814' \& 0.4959 3.15 2,14 0.4s35 2,65 0.4960 3.16 2.L5 0:4842 ,',2.66 0.4961 5.L7 2.16 0,4s46 2,67 ' 0.4962, 3.18 2,17 0,;4850' '2.68, 0.4963,' 3.r9 2.18 0,4854 2.69 0.4964 3.2 2.t9 0,4857 :2:7 0.4965 3.21 22 0,4861 2.7t ' 0.4966 3.22 2,2t 0,4864 2.72 0.4967 3.23 2;22 0.4s65' 2.73 0.4968 324 223 0,4871 ',2,74 0.4969 3,25 2.24 0.4875 2J:5 0.4970, ,.26
0.4989 0.4959 0.4990 0.4990 0.4990 A.4991 0,4991
0.4991 0.4992 0.4992 0.4992 0:4992 0.4993 014993
0.4993 0.4993 0.4994 0.4994 0.4994 0,4994 0.4994 Continucd
590
19.5 Nonruar DrsrnrsurroaT
o.2t 0.0832 0.72 022 0.0871 0,73 o.23 0.091A O,74 0.24 0.0948 0.75 0.25 0.0987 0.76 0.26 0.1026 0,77 a.z7 0.1064 0.7& 0.28 0.1103 0.79 a.29 0.1141 0.8 0.3 0.1179 0.81 0.31 0.1217 0.82 o.32 0.1255 0.83 0.33 0.1293 0.84 a.3.4 0,1i31
0.85
o.35 0.1368 0.86 0.36 0J4A6 0.87 a$7 0.1443 0.88 0.38 0.1480 0.E9 439 4.15t7 0.9 o.4 0.1554 0.91 o.At 0.1591 0.92 0.42 0.1628 0.93 o.43 0.1664 0.94 o.u 0,1700 a.95 a,45 0.1736 0.96 0.46 0.1772 0.97 o,47 0.1808 0.98 0.48 0.1844 0.99 o.49 0.1879 I 0.5 0.1915 1.0r
a.2642 0.2673 0.27A4 0.2734 0.2764 0.2794 0.2823 0.2852 0.2881 0.2910
L,23
t.24
L25 L.26
r.27 1.28
L29 L.3
t,'r 1,32
0,29i9
L.33
0.2967 0.2995 a.3023 0.3051 0.3078
t.Yt
0.ir06
r,35 t.36 r.37 r.38 1.39
0.3133 L,4 0.3159 r.4l 0.3186 1.42
0.j212
r.43
0.3238 t.M 0.3264 r.45 0.3259 t,46 0.3315 L.47 0,3340 1.48 0.3365 t.49 0.3389 t.5 0,3413 1.51 0.3438 t.52
0.3907 t,74 0.3925 r.75 0.3944 rJ6 0.3962 r,77 0.3980 1.78 0.3997 L:79 0.4015 1.8 0.4032 1.8r 0.4049 t.82 0.4066 1.83 0.4082 t.84 0.4099 1.85 0.4115 1.86 0.4131 r.87 0.4147 1.88 0.4162 1.89 0.4177 r.9 0.4192 1.91 0.42A7 L92 0.4222 19t 0,4236 t.94 0.4251 t.95 0.4265 r.96 a.4279 r.97 0.4292 1.98 0.4306 r.99 0.4319 2 0.4332 2.Or 0.4345 2.02
a.457
2.O5
0.4591 0.4599 0,4608 0.4616 0.4625
2.29
0.463i
2.3
0.4641 0.4649 0.4656 0.4664 0.4671 0.4678 0.4686 0.4693 0.4699 0.4706 0.4713 0.4719 0.4726 0.4732
0.47i8
2.25 2,26
2,27 2.28
2.3t 2.32 2.33
2.34 2.35
2.36 2.37
238 2.39 2.4
2.4t 2.42 2.43
2.4* 2.45
0.4744 2.46 0.4750 2.47 0.4756 2.4 0.4761 2.49 0.4767 2.5 0.4772 2.5r 0.4778 2.52 0.4783 2.53 0.4788 2.54
591
2J6 0.4971 ?.27 0.4995 2,77 0.4972 3,28 0.4995 2.78 0,4973 3.29 0.4995 2.79 0.4974 33 0.4995 2,8 0.4974 3.?t 0.4995 2.8r 0.4975 3.32 A.4995 2.82 0.4976 3.33 0.4996 2.83 0.4977 3.34 0.4996 2.84 0.4977 3.35 0.4996 2.85 0.4978 3.36 0.4996 2.86 0.4979 3.37 0.4996 2.87 0.4979 3.38 0.4996 0.49r t 2.88 0.4980 3.39 0.4997 0.4913 2.89 0.4981 3.4 0.4997 0.4916 2.9 0.4981 3.4r 0.4997 0.4918 2.9r 0.4952 3,42 0.4997 0.4920 2.92 0.4982 3.43 0.4997 0.4922 2.93 0.4983 3.4* 0.4997 0.4925 2.94 0.4984 0.4997 '.45 0.4997 0.4927 2.95 0.4984 1.46 0.4929 2.96 0.4955 3,47 0.4997 0.4878 0.4881 0.4884 0.4887 0.4890 0.4893 0.4896 0.4898 0.4901 0.4904 0.4906 0.4909
0.4931 2.97 0.4985 5,48 0.49i2 2.98 0.4956 3.49 0.4934 2.99 0.4986 3.5 0.4936 0.4987 3,5r ' 0.4938 3.0r 0.4987 3.52 0.4940 0.4987 3.53 '.O2 0.4988 0.4941 3.O3 0.4943 3.04 0.4958 0.4%5 3.05 0.4989 3.9
0.4997 0.4995 0.4998 0.4998 0.4998 0.4998
0.5000
Using Table 19. 1 1, show tiat for a standard normal distribution of a data set, approximately 68%o of the data will fall in the interval of s to s, about 95olo of the data falls between 2s to 2s, and approximately all of the data points lie benreen 3s to 3r. In Thble t9.ll, z: 1 represents one standard deviation above the mean and 34.13o/o of the total area under a sandard normal curve. On the other hand, s, : I represen$ one sran; dard deviation below the mean and 34.73o/o of the total area, as shown in Figure 19.7. There, fore, for a standard normal distribution, 6870 of the data fall in the interval of a :  I to g = 1 (r to s). Similarly, z: 2 and z = 2 (nvo standard dwiations below and above the
s92
Cnepren
19
PnosABrrrrv.oxo Srarrsrrcs nr ENcnvssRrNc mean) each represent 0.4772o/o of the toal area under the normal curve. Then, as shown in Figure 19.2 95o/o of the data fall in the interval of 2s to 2s. In the same way, we can show
rhat99,7o/o(forz= 3then.r4 =0.4987and.2:3thenA:0.4987) ordmostallofthe
data points lie between
3s to 3s. A  0.4772 + 0.4772*0.95
A=03413+0.3413=0.68
z:1.00 e:1.00 e=1.00 e=1.00
;
tigure
19.7
The area under a norlnal curve
for
z=2.00
z
=2.ffi
Example 19.5.
For Example 19.4, calculate the mean and standard deviation, and determine the probability that it will take a person between 7 and, ll minutes to assemble the computer parts. Refer to Table 19.12 when following solution steps.
r Time r (minuts) ;x
Frequency
*f
tc fr
5
25 48
4.22
89.04 82.95
77 r20
2,.22
t.22
54.2t' 22.33
x
6 7
 v)'f
f
1l
8
(*
4 )1
9
1'7
r53
Q.22
a.82
10
r4
140
1l t2
L3
r43
0.78 1.78
4r.r9
B
96
14
6
t4
3
78 42
2.78 47R 4.78
2*f=gzz
x: 2*f n :
K:9.22
minutes
': ,ffi:'{#:2'28minutes
8.52
6r.83
.
s5.73 168.55
2(xx)2f=515.16
19.5 NonuelDrstnnrrrron
593
The value 7 is below the mean valae (9.22) and the z value corresponding to 7 is deter
minedfrom
xV 79.22 ,: ;: rE:
o.97
From Table l9.ll, A = A334O. Similarly, tlre vdue 1 1 is above the mean value and the sscore corresponding to 11 is computed from
It  9.22 :0,78 z: xV t 2.28 : From Thble l9.ll, A= 0.2823, Therefore,
the probability that it will take a penon be+ 0.2823 0.6163 as shoqrn
tween 7 and I L minutes to assemble the compffer paft is 0.3340
:
in Figure 19.8.
A
z0.97
I
:
03340 + 0.2823 = 0.6t63
z=0.78
figuretg.g
Area under tfie probability dishibuthn curtte for Emtnple 19.6.
For Example 19.4, deterrnine the probability that it will take a person longsr than 10 minutes to assemble the computer parts. For this problem, the .6score is
xV ro9.22 Zff:0.34 =;=: ": From Thble 19.11, A = 0.1331. Since we wish to determine the probabiliry that it tak"s longer than 10 minutes to assemble the part, we need to calculate the area 0.5  0.1331 : 0.3669, as shov"n in Figure 19.9. The probability t6o il lyill take a person longer than 10 minutes to assemble tfre compurer pan is approximately 0.37.
594
Cneprsn
19
Pnonasrr,rrverp Sterrsrrcs
rN ENcrrvnsR$Ic
A:0.L331"
z = 0.34
I
Figurelg.g
Areas under the probability distribution curve for Erample 19.7. ril.
In closing, keep in mind that the pqpose of this chapter was to make you arvare of the importance ofprobability and statistics in engineering, not to provide a detailed coverage ofstatistics. As you ake setistics classes and advanced classes in engineering, you will learn much more about stadsticd concepts and models.
SUMMARY Now that you have reached this point in the text n You should undersand the imporant role of statistics in various engineering disciplines. . You should be familiar with basic probability and statistics terminologies. . You should have a good underssnding of frequency distribution and cumulative frequenry disnibution and what kind of information they provide. . You should have a good grasp of statistical measures of cenual tendenry and variation. . You should know how to compute basic statistical information such as mean, variance, and sandard deviation for a set of daa points. . You should know what is meant by normal disuibution and standard normal distribution. . You should know how to use Thble 19.1 1.
of a test for an
engineering class of 30 students is shown here. Organize the data in a manner similar to Thble 19.1 and use Excel to creare a
l9J. The
scores
19.2.
t9.3.
histogram.
For Problem 19.1, calculate the cumulative frequency and plot a cumulativefrequency polygon. ForProblem 19.1, usingEquations (19.1) and (L9.6), calculate the mean and sandard deviation of the class scores.
Scores: 57, 94, 8I, 77, 66, 97, 62, 86, 7 5, 87, 9I, 7 8, 61, 82, 74,72,70, 88, 66,75, 55, 66, 58, 73, 79,
51,63,77,52,94
t9.4.
For Problem 19.1, using Equations (19.7) and (19.8), calculate the mean and standard deviation of the class scores.
PnosLEnIs 595 r9.5.
For Problem 19.1, calculate the probability distribu
19.9.
tion and plot the probabilitydistribution curve. r9.6.
In order to improve the production time, the supervisor of assembly lines in a manufacturing setting of cellular phones has studied the time that it akes to assemble cerain parts of a phone at various stations. She
 i
2 x 4 Lumber Steel Spherical Balle (cm) Width (ir.)
The record of her study is organized and shown
3.50
3.40
1.00 0.95 1.05 1.10 1.00 0.90 0.85
20
3.65 3.35
0.95
28
3.60
0.90
in the accompanying able.
3.55
Time that it takes a person to assemble the part (minutes)
Frequency
4
r5
5
6
are
]':"""1
measures the time that it akes to assemble a specific pan by 165 people at different shifts and on different days.
Determine the average, variance, and standard devia
tion for the following parts. The measured values given in the accompanying table.
3.45 3.60 3.55 3.40
',
I
1.05
34
191.
t9.8.
8
28
9
24
10
L6
Plot the data and calculate the mean and sandard deviation. For Problem 19.6, calculate the probabiliry disuibution and plot the probabilitydisuibution curve. Determine the average, variance, and standard dwi
ation for the following parr. The measured values are given in the accompanying able.
Screw length
(cm)
19.X0.
ance, and standard deviation for the cereal boxes. Does
the manufacturer's information noted on the box fall
within your meruurement? l9.ll. 19J2.
Pipe Diameter (in )
t.25
2.55 2.45 2,55 2.35
1.18 1.22 1.15
2.60
r.l7
2.40 2.30
L.r9
2.40 2.50
1.18 1 la
2.50
t.25
The next time you make a trip to a supermarket ask the manager if you can measure the mass of at least 10 cereal boxes ofyour choice. Choose the same brand and the same size boxes. Tell the manager this is an assignment for a class. Repon the anerage mass, vari
Repeat Problem 19.10 using tluee other products, such as cans ofsoup, tuna, or peanu6. Obtain the height, age, and mass ofplayers for your favorite professional basketball team. Determine the average, variance, and sandard deviation for the height, age, and mass. Discuss your findings. Ifyou do not like basketball, perform the experiment using daa from a soccer team, football team, or a sports team of your choice.
r9J3.
For Example 19.4, determine the probability that it will take a person benrreen 5 and 10 minutes to assemble the computer parts.
1914.
For Example 19.4, determine the probabiliry that
will take a person longer than 7 minutes to
1.22
it
assemble
the computer parts. 1915.
For Problem 19.6 (assuming normal distribution), determine the probability that it will take a person between 5 to 8 minutes to assemble the phone.
596 Class
Csaprcn
PnosABrlrrr.al.tn Sranrsrrcs rx ErvcrxnrnrNc
are
pafonned in
class.
Your instructor will pass along an unopened bag of Hershey's kisses. You are to estimate the number of kisses in the bag and write it down on a piece of paper. Your instructor will then collect the daa and share the results with the class. Your assignment is to organize the data per your insructor's suggestion and calculate the mean and standard dwiation. Compute the prob
ability distribution. Does your data disuibution ap
19.17.
proximate a normal distribution? Answer any additional questions that your instructor might ask. Your instructor will ask for a volunteer in class. You are to estimate his or her height in inches (or in cm) and write it down on a piece of paper. Your instructor will then collect the daa and share the results with the class. Your assignment is to organize the data per your
instrucort suggestion and calculate the mean and sandard deviation of the data. Compute the probability
1918.
a normal disuibution? Answer any additional ques
Experi.mentsProblems I 9. I 6 througb I 9.20 are acpn
immx that 19.16.
19
distribution. Does your data disuibution approximate a normal disuibution? Answer any additional quesrions that your instructor might ask Your instrucor will ask for a volunteer from class. You are to estimate his or her mass in lb. (or in kg) and write it down on a piece of paper. Your instructor will then collect the data and share the results with the class. Your assignment is to organize the data per your insmrctor's suggestion and calculate the mean and sandard dwiation of the data. Compute the probability disuibution. Does your data disuibution approximate
19.t9.
$"20,.
t9.2r.
tions that your instructor might aslc You are to write down on a piece of paper the number ofcredits you are taking this semester. Your irutructor will then collect the data and share the resuhs with the class. Calculate the mean and standard deviation of the data. Assuming a normal distribution, determine the probability that a student is taking benreen 12 to 15 credits this semester. !?hat is the probability that a student is taking less than 12 credits? You are to write down on apiece of paper how much (to the nearest penny) money you have on you. Your instnrctor will then collect the data and share the results with the class. Your assignment is to organize the daa per your instructort suggestion and cdculate the mean and scandard deviation of the data. Assuming a normal distribution, determine the probability that a student has between $5 to $tO.'What is the probabiliry that a student has less than $10? You are to ffiite down your waist size on a piece of pa
per.
If you don't know your waist size, ask your in
structor for
a
measuring tape. Your instructor will then
collect the dam and share the results with the class. Your assignment is to organize the data per your instruccort suggestion and calculate the mean and standard deviation of the data. Assuming a normal disuibution, determine the probability that a student will '\V'hat have a waist size that is less than 34 inches. is the probability that a student will have a waist size that is between 30 in. to 36'n.?
CHAPTER
ENcTNEERTNG
r lr r
conomic considerations play
20
EcoNoMrcs
a
vitat role in product and service development and in engineering
design decision making pfocess.
597
598
Cnepren20 ExcrxnrnrNcEconourcs
in Chapter 3, economicfactors alwals play important roles i.n mgineering daign decision making. Ifyou drsigrc a product that is too erpmsiae to rtantr.faaure, then it cAn not be sold at a price that consurners can afford and still be
As we explained
prof.table to your colnpanJ/. Thefact is that companies design producx andprouide smtices not only to rnahe our liaes bener but ako to make rnone/ In this section, we will discuss the basics of engineering econornics. The information proaidcd here not only applies to mgineering projecx bwt can ako be applied to financing A car or A house or borcowingfrorn or inaesting mons! i.n bank* Some of you rnd! want to apply the knowledge gained here to dzterrnine lour srudent ban payrnmts or lour credit card paltments. Therefofâ‚¬, t!)e aduise you to deuelop a good und.erstanding of mgineering economics; the information presmted here could help you manage yoar rnone! more wisely.
20.1 Gash Flow Diagrams Cash flow diagrams are visual aids drat show the flow of costs and revenues over a period of time. Cash flow diagrams show whm the cashfuw occurs, tlte cashfuw magnitudc, and' whaher the cash fnw is out ofyur pocha (cost) or into your pocha (reumae). It is an imponant visual tool that shows the timing, the magnitude, and the direction of cash flow. To shed more light on the concept of the cash fow diagram, imagine rhatyouare interested in purchasing a new car. Being a firstyear engineering student, you mqf not hane too much money in your sarings account at this time; for the sake of this example, let us say that you have $1200 to yotu name in a sarings account. The car that you are interested in buying costs $ 15,500; let us further assume that including the sales tax and other fees, the total cost of the car would be $16,880. Assumingyou can afford to put down $1000 as a down payment for your new shiny car, you askyour bank for a loan. The bank decides to lend you the remainder, which is $15,880 at 87o interest. You will sign a contract that requires you to pay $315.91 every month for the next five years. You will soon learn how to calculate these monthly payments, but for now let us focus on how to dravr the cash fow di"Sr. The cash fow diagram for this activity is shown in Figure 20.1 . Note in Figure 20.1 the direction of the arrows representing the money given to you by the bank and rhe payments that you must make to the bank over the next five years (60 months).
$1s,880
I
Figure2o.t
A cash flow diagram
for honowed
money and the monthly payments.
$315.91
20.2
Srtvrpr,s AND
CoupouNp [NrsREsr
599
cash fow diagram for an investment that includes purchasing a machine that costs $50,000 with a maintenance and operating cost of $1000 per year. It is expected that the machine will generate revenues of$15,000 per year for five years. The expected salvage value of the machine at the end of five years is $8000. The cash flow diagram for the investment is shown in Figure 20.2. Agaun, note the directions of anows in the cash flow diagram. Ve hare represented the initial cost of $50,000 and tfi.e maintenance cost by arrows pointing down, while the revenue and the sdvage value of the machine are shown by arrows poindng up.
Drav the
$15,000
$8000
I
0
2
3
4 $1000
t
Figure?0.Z
T[e cash flow diagram
lor Erample 20J.
$50,000
i,
it  ii.1rlfi:lii:ilrllltirriinllilf,illi'riitl
ili.i.liri1ii,r,.l:ir',r.::i
.,:iirlrr,i,rililllii'i.li
20.2 Simple and Compound Interest Interest is the exua money in addirion to the borrowed amount thar one must pa)r for the purofhadng access to the borrowed money. Simph interastis rhe interest that would be paid only on the initial borrowed or deposited amount. For simple interest, the interest accumulated on the principle each year will not collect interest itself. Only the initial principal will collect interest. For example, if you deposit $100.00 in a bank at 6% simple interest, after six years you will have $136 in your account. In general, ifyou deposit the amount P ar a rate of io/o for a period of n years, then the total future valu e F of the P at the end of the ath year is grven by pose
F:P+(PXrXtx):P(r+ni\
(20r)
Compute the future value of a $ 1500 deposit, after eight years, in an account that pqrs a simple interest rate of 7o/o. How much interest will be paid to this account? You can determine the future value ofthe deposited amount using Equation (20.1), which results
in
F:
P(r
*
ni) = r5oo[1 + 8(0.07)]
:
$2340
And the total interest to be paid to this account is
intere*: (p)(n)(i) l:. :::: .
:
(1500x8x0.07) = 634s
600
Cno,prsn20 Excnwrnruc Ecoxourcs
Interestfor BalanceattheEnd Bqfnring of the Year (dollars and cents)
t
tfieYear 60lo
at
(dollars
and
oftheYear, Includirg the Interest (dollare
cents)
and cents)
106.00
,
t12.36
'
3
r12.36
6.00 6.36 6.74
4
I 19.10
7.r4
rv,6.24
,
5
126.24
r33.8r
'
6
133.81
7.57 8.02
100.00 106.00
I 19.10
t4 t.65
Simple interests are very rare these days! Almost all interest charged to borrow accounm or interest earned on money deposited in a bank is computed using conqpound interest The concept of compound interest is discussed next.
Compound Interest Under the compounding interest scheme, the interest paid on the initial principal will also collect interest. To better understand how the compound interest earned or paid on a principal works, consider the following example. Imagine that you put $100.00 in a bank that pq/s you 67o interest compounding annually. At the end of the first year (or the beginning of the second year) you will have $106.00 in your bank account. You have earned inrerest in the amount of $6.00 during the first year. Howwer, the interest earned during the second year is determined by ($106.00X0.06) : $6.36. That is because the $6.00 interest of the first year also collecs 6%o interest, which is 36 cena itself. Thus, the total interest earned during the second year is $6.36, and the total amount available in your account at the end ofthe second year is $1 12.36. Computing the interest and the total amount for the thfud, fouffh, fifth and the sixth year in a similar fashion will lead ro $141.83 in your account at the end of the sixth year. Refer to Thble 20.1 for detailed calculations. Note the difference between $100.00 invested at 6%o simple interest and 60/o interest compounding annually for a duration of six years. For the simple interest case, the total interest eamed, after six years, is $36.00, whereas the total interest accumulated under the annual compounding case is $41.83 for the same duration.
20.3
Future Worth of a Present Amount Now we will dwelop a general formula that you can use to compute the future value Fof any present amount (principd) 4 after nyears collecting zolo interest compounding annually. The cash
fow
di"S"
for this situation is shorrn in Figure 20.3.In order to demonsffate, stepby
step, the compounding effect of the interest each year, Table 20.2 has been developed. As shown
in Thble 20.2, sardng with the principal P, at the end of the first year we will hare P * Pi or P(l + i). During the second year, dre P(l * i) colleca interest in an amount ofP(l + i)i' au;.d. by adding the interest to the P(l * i) amount that we stamed with in the second year, we will have a toal amount of P(L + i) + P(l * i)i. Fae'oling out the P(1 * /) term, we will hare
2O.3 Ftmrnr Wonmr or e PnEseNT Arvrorlxr
I
601
Figure20.3
The cash flow diagram for future
wort[ of a deposit made in the bank today.
Balance at the
Year
Begimiryof
Interest for
theYear
the Year
i1
P
t3 :4 i
P(l + r) P(t + ilz PU + i)3
ll ,A
5
Balance at the Eqd
P+(P)(i)=P(r+i) P0+i) +P(l + DQ)=P(L+i)z
(P)(t P(t + t)(i) P(r
+
P0 + il2 + P(l + i)2(t) : P(r + il3 P0 + il3 + I{1 + i)3(i): P(L + il4 P$+ il4 +P(l + i)4{i\: ?(1 +rt
ilz?)
P(l+
t)4
p0 + i)'Q) PIL + il4(il
P(r +
i),l
P(r +
ilt(i)
ofthe Year,
Inclu,f ing the Interest
P(l + i)L + P(l + i)"11;1: P(l + i)
P0 + il2 doilars at the end of the second year. Now by following Tirbl e 2A.2 youcan see how the interest earned and the total amounr are computed for the third, fourth, fifth, . . . , and the nth yan. Consequently, you c:rn see that the relationship beween the present worth P and the future value.Fof an amount collecting iolo interest compounding annually after a years is given by
F: P(r + i)"
(20.21
Compute the future value of a $1500 deposit made today, after eight years, in an account that of 7o/o that compounds annually. How much interest will be paid to this
pays an interest rate
account?
P,
i,
The future value of the $1500 deposit is computed by subsdruting in Equation Q0.2) for and zl, which results in the amount that follows:
F= p(L + i)n:1500(t + .07)8 = $2577.27 The total interest earned during the eightyear life of dris account is determined by calculating the difference beween the future value and the presenr deposit value.
L ..,._
inwest:
$2577.27

$1500
:
$1O77.27 ::.:,:
,:
:
:':::
r::I
602
CHrprsR20 ENcrNsrRrNcEcoxourcs Many financial instirutions pay interest that compounds more than once a year. For example, a bank may pay you an interest rate that compounds semiannudly (mice a year), or quanerly (four times ayqi, or monthly (12 periods ayar).If the principal Pis deposited for a duration of n yars and the interest given is compounded n periods (or zz dmes) Per year, then the future value.Fof the prin"tpal P is determined from
r=n(t.*)
(20.3)
Compute the future value of a $1500 deposit, after eight years, in an account that paln an interesr rate of 7o/o that compounds monthly. How much interest will be paid to this account?
To determine t'he future value of the $1500 deposit, we substitute in Equation (20.3) for P, i, rn and z. The substitution results in the future value shown next.
/ q{Z) = / s.s7\(8)(tz) I':1500[1+ =1500t1+ J 12/ r2l \ 
\
s262r.73
And the total interest is n
interest: $262I.73
iL,,r*,,r.lr.,l,..,,.

$1500
=
$11.21.73 _i ..
:r
__.i_1
The results of ExamplCI 20,2, 20,3, and 20.4 are compared and summarized in Table 20.3. Note the effecs of simple interest, interest compounding annually, and interest compounding monthly on the total future value of the $1500 deposit.
Exanple Numbet
D qration
Principal (dollars)
Example 20.2 Example 20.3
1500 1500
Example 20.4
1500
lnterest Rate
(yâ‚¬ars)
Futrrre Yalue (dollarc and centr)
Interest Earned (dollars and cents)
7olo simple
8
2340,00
840.00
compounding annually 7%o compounding
8
2577.27
1077.27
2621.73
112r.73
7olo
monthly
20.4 Errucrrw
20.4
603
INrsREsr RArs
Effeetive Interest Rate If you deposit $1OO.OO in a savings account, at 60/o compounding monthly, then, using Equation (20.3), at the end ofone year you will have $106.16 in your account. The $6.16 earned during the first year is higher than the stated 60lo interest, which could be understood as $6.00 for a $100.00 deposit over a period of one year. In order to aroid confusion, the stared or the quoted interest rate is called the noninal interest rate, and" the actual earned interest rate is called the effeaioe i,ntercet rate. The relationship benveen the nominal nte, i, and the effective rate, ia,, is given by
t&:
('*t)'
(20.4)
where rn represents the number of compounding periods per year. To bener understand the compounding effect of interest, let us see what happens if we deposit $100.00 in an account for a year based on one of the following quoted interests: 6%o compounding annually, 67o semiannually, 60/o qwterly,6olo monthly, and 6o/o daily. Thble 20.4 shows the difference among these compounding periods, the total amount of money at the end of one year, the interest sarned, and the effective interest rates for each case. \[hen comparing t]re five different interest compounding frequencies, the difference in the interests earned on a $100.00 investment, over a period of ayear, may not seem much to you, but as the principal and the time of deposit are increased this value becomes significant. To better demonstrate the effect of principal and time of deposit, consider the following orample.
Totel Number of Conpounding Compounding
Period
Periods
TotalAmount after I Year (dollars and cents)
6.00
60/o
10011+  l:106.09 2/ \
\'
6.09
6.090/o
/ I+=l:106.13 o.otr\4
6.13
6.130/o
6.16
6.160/o
6.18
6.180/o
Annually
I
100(1+0.0O:106.00
Semiannually
2
Monthly
D*ily
/
o.o6
1001
Quanerly
4/
\
12
365
lnterest Effective (dollars Interest andcents) Rate
1oo( r
*
Y)": 106.16 roo(r *#)"' = 106.18
604
CneprsR
20
ENcrNErRrNcEcoNourcs Determine the interest earned on $5000 deposited in a savings account, for 10 years, based on one ofdre following quoted interest rates: 670 compounding annually, semiannually, quarterly, mondrly, and daily. The solution to this problem is presented in Thble 20.5.
Period
Periods
Annually
5000(1
10
8954.23
/ *i), o.oe \m =9030.55 / o.oe \& : eo7o.oe 5o0o(1 .; ) 1 0.06\'20 5ooo(1 *i),,:eoe6e8
Qu".t"rly
40
Motrthlv
120
/ *d) o.o6 \ trto , = e11o'14
i
I
+ 0.00,lo:
and cents)
5000(1
Semiannually
ii'
(dollan
UsrngEq. (16,8) (dollars and cents)
Compounding Compounding
Dnily
5ooo[1
3650
':_'
,.
: :
::.:.::L:.,.:.:,
3954.23
'4030.55
4o7o.oe
4ae6.e8
4710'14
t'
:,.
Determine the effective interest rates corresponding to the nominal rates: (a) 7olo compounding monthly, (b) 16.50/o compounding monthly, (c) 6Vo compounding semiannually, (d) 9o/o
compounding quanerly. We can compute the (a)
i.r=
(b)i.n
i"6.
for each
case by
substituting for
/ * ; \ 1  / *t/o.oz \'2  I (t \, ;) 
i and m n
E4uation
(20 ,4) .
0.0722or7.22o/o
( o.t6j\'2  0.1780 orr7.80o/o = . *. i \, )  = (t  ,:o.o6oe or6.oeo/o
" ry)'
(d) z"s'
: (t .
?)n  ,:o.oe3o ore.3oo/o iirrlfifilii.lii.:liirrif.EJ
i
20.6 PnrsmmVoRrrr
20.5 Prescnt
oF Senrss PeuvrsNT
onArvNurrr
605
Worth of a Future Amount
Let us now consider the following situation. You would like to hare $2000 available to you for down payment on a car when you graduate from college in, say, five years. How much money do you need to put in a cenificate of deposit (CD) with an interesr rate o f 6.50/o (compounding annually) today? The relationship between the future and present value was dweloped earlier and is given by Equation (20.2). Rearanging Equation Q\.Z),wehxe a
' !(r+;)" p=
(20.5)
and substituting in Equation (20,5) for the future value we hane
2N9p ' = (1 + 0.065)t
:
4
the interest rate
i,
and the period
re,
$1459.76
This may be
a relatively large sum to put aside all at once, especially for a fustyear engineering student. A more realistic option would be to put aside some money each year. Then rhe question becomes, how much money do you need to put aside every year for the next 6ve years ar the given interest rate to hare that $2000 available ro you at the end of the fifth year? To answer this question, we need to develop the formula that deals with a series of payments or series of deposits. This situation is discussed next.
20.6
Present Worth of Series Payment or Annuity In this section, we will firsr formulate the relationship between a present lump sum, 4 and future uniform series payments,r4, and then from ttrat relationship we will develop the formula that relates the uniform series payments,4 to a future lump sum F. This approach is much easier to follow as you will see. To derive these reladonships, let us first consider a situation where we hane borrowed some money, denoted by P, at an annual interest rate i from a bank, and we are planning to pay the loan yearly, in equal amouns l, in n years, as shown in Figure 20.4.
I
Figure20.n
The casiflory diagram for a
borrowed sum of money and its equivalent seiles payntents.
605
Cnqprsn20
ElvcrrvsERnrc EcoNourcs To obtain the relationship between P andA, we will treat each future payment separately and relate each payment to its present equivalent value using Equation Q0.5); we then add all the resulting terms together. This approach leads to the following relationship:
AAAAA P+_ (r+4 (1 +;y*G;t+"'+OiAa*0*ny
Q0''61
As you can see, Equation (20.O is not very userfriendly, so we need to simplify it somehow. 'What ifwe were to multiply borh sides ofEquation Q0.6) by the term (1 + z)? This operation results in the following relationship:
P(t +
A,=+, A,r+"'+ . n..+;4* + i)=A+, "' (r !, (t + i)'z (t + i)"^' + i)' (r + ;)z' (t + 4r
QL.tt
Now ifwe subtract Equation (20.6) from Equation (20.7), we have
P(l
+i)P=A+L+r,A.+A * + i) rY1t*
\1
_
1r* o,
+"'+
A L(l + 4' (1 + i)',', (r + ;72'
I a
A
l_l!rI
A
A
O;;l;;* 1t* o'''A A I + (r + i)',t (l 4"1 Il
(20.8)
Simplifying the righthand side of Equadon (20.8) leads to the following relationship:
P(r+i)PA
i
â‚¬0.8b)
(r + i)"
And after simplifying the lefthand side of Equation (20.8), we have
P(i) =
A((L+i)"r)
(20.8c)
Q+i)"
Now if we divide both sides of Equation (20.8c) by i, wehave
tl p=nl\!il'. ' '^L i(r + i)'
ao.,
I
Equation (20.9) esablishes the relationship bewveen the present value of a lump sum P and its equivalent uniform series payments ,4. We can also rearrange Equadon (20.9) , to represent r4 in terms of Pdirecdy, as given by the following formula:
.l r)' ! 0 + i)'  t: 'L0 + i)"  r)
. P(i)(r + i)" :,l A:
.(t)(r
o.r,
?0.7 Future Worth of Series Payment To dwelop a formula for computing dre future worth of a series of uniform payments, we begin with the relationship betrveen the present worth and the furure worth, Equation (20.2), and then we substitute for P in Equation Q0.2) in terms of
.r4,
using Equation (20.9). This
20.7
Ftrnrns'WonrH or Srnrss PeuvrrNr
607
procedure is demonsuated stepby+tep, next. The relation berween a presenr value and a future value is given by Equation (20.2):
F
= P(r
* i)"
(20.2)
And the relationship between the present wonh and a uniform series
is
given by Equation (20.9):
[(t+z')'tl P=Al n * I L i(l+i)"
CO.tt
Substituting into Equation (20.2) for P in terms ofr4 using Equation (20.9),we have
P F
= P(r
*
i)" =
lfl + t)'
^L 4r;
1.l
n
l(1
+
,)'
(20.10
Simplifying ftuation (20.11) rCIults in the direct relationship between the future wonh Fand the uniform payments or depositsr4, which follows:
,=^lsTl
G0.ta
And by rearranging Equation Q0.12), we c:rn obain a formula for A in terms of future worth F:
.s=Fl I L(l + 4' ll
aomt
Now that we have all the necessary tools, we fllrn our attention to the question we asked eadier about how much money you need to put aside wery year for the next five years to have $2000 for the down payment of your car when you graduate. Recall that the interest rate is 6,50/o ampounding annually. The annual deposits are calculated from Equation (20.13), which leads to the following amounr:

s'oe\
n = ,oool L(l + 0.065),

I
:
$351.25
Puttingaside $351.26 in abankweryyear for the nortfiveyears maybe more manageable than depositing a lump sum of $1459.76 today, especially if you don't currendy have access ro that large a sum!
It is importanr to note thar Equations (20.9), (20.10), (20.12), and (20.13) apply to a situation wherein the uniform series of paymen$ or revenues occur annua$t, Vell, the next question is, how do we handle situations where the payments are made mondrly? For example, qu or a house loan paymens occur monthly. Let us now modi& our findings by considering the relationship benareen present value ? and uniform series paymenrs or revenuer4 that occur more than once a year at the same frequency as the frequency of compounding interest per year. a
608
CllePrsn
20
ENcrNsERrNc EcoNourcs For this situation, Equation Q0.9) is modified to incorporate the frequency of compounding interest per year, m, inthe following maruler:
l(r*.)_]'l tv *)" *l'*  )
n: nl \tT!/
(20.14)
Note that in order to obtain Equation Q\.I ),we simply substituted in Equation (20.9) for i, ilm. andfor n" nm.Equation (20.14) can be rearranged to solve for.r4 in terms ofPaccording to
(*)('.
,="[
*)
(' * t)* ,
Similarly, Equations Q0,12) and (20.13) can be modified for situations wherer4 occurs more than once a
[email protected]_at the same frequency as the compounding interestleading to the following relationship:
a= rL
('**)''""''
I (20.16)
t 7n
ltl
ml .l n:,17
t.7m '_l
(2017)
L\'*;)
Finally, when the frequency of uniform series is different from the frequency of compounding interest, i"6 must first be calculated to match the frequency of the uniform series. Ler us retum to the question we asked earlier about how much money you need to put aside for the next five ycrfs to have $2000 for the down payment on your car when you graduate. Now consider the situation where you make your deposis every month, and the interest rate is 6.50/o compounfing monthly. The deposits are calculated from Equation (20.17), which leads to the following:
*  lt :^"1 ^:'15;7ll
0.065 __..v
('*
T2
o.o6t
\('
n)
=
2)(5)
$28.29
1
Putting aside $28.29 in the bank every month for the next five years is even more manageable than depositing $l5t.Ze in a bank every year for the next five years, and it is cenainly more manageable than depositing a lump sum of $1459.76inthe bank todal L,,,u,,r,,,,.,,.,.,.,,,,
::,'',, .l 
l 
:t,r:al:l
2A.8 Suvuany or
ENcrNpsRrNc EcoNourcs ANer,ysrs
609
Determine the mon$Iy paymeng for a fiveyear, $10,000 loan ar an interest rare of 87o com
pounding monthly. To calculate the monthlypayments, we use Equation
^:,lV)l:?).1 L\'.;) rJ 20.8 Summary
:,o,ooo[(;,
LT
(2O.Ir.
x* )1 ):$20276
of Engineerinq Economies Analysis
The engineering economics formulas that we have dweloped so far are summarized in Thbles 20.6 and20.7. The definitions of the terms in the formulas are given here:
P: F:
costlump sum ($) future wofth, or fuguss ssstlump sum ($) I : uniform series payment, or uniform series revenue ($) i : nominal interest rare z"n'= effecrive interest rate a : number of years rn = number of interest compounding periods pet yeet present worth, or present
The interesttime factors shown in the fourth column of Thble 2A.6 arc used as shoncuts to artoid writing long formulas when waluating equivalent values ofvarious cash fow occurrences.
F
P
P
F
F= P(l * i)
(FIP, i, n)
o F '(1+i), f(t+i)"ll P:AI' i(r+i)' , , L l
I
(rxr + i)" I
F
:6+S
ffl+;\t1'1 (PtA,i.n):Ld;1
 A=Pl:l
(Atp,,,o)=l#++]
': ^Li)
(FtA,i,n):lga#]
LQ+irrJ f(l + r)' 1l
A
(PIF,i,n)
= (l + i)
.f
(4
I
^ "l1r * ;1,  t1
(AtTi,n)=
ltr#_,]
Cnerrsn20
ExcrNsERrNcEcoNomrcs
ur"
ffi, ro"J"
r*,= (,r + *))
i
t
r= i"(r * *)*
,:G# I
(,. *)* ,:^l 17*t\ry
,1
;\1+;)
r /;\/ A= ?l
P
L
,=
^f L
I ,="L
(;/(1+;)

J ;\ou1
776
\'*;)

'J

1 ;\@)t") I ' \'*;)
=l
I
ln
l:
nl
t:
rl,
('**)'""'I1
For example, when evaluating the series payment equivalence of apresent p.itt"ip"l, instead
of
writing,
O0 + i)" A pl 'L(l +i)"r)f wewtiteA: P(A/4 (Atp, L
t, n),where, of course,
J,il" .f o' :1,!'),0 L(r +i)"rJ
In this example,the (AIP, a z) term is cdledt\einteresttimefaaor,andit reads.d gSven P at io/o inreresr rate, for a duradon of n years,It is used to findA, when the present principal value P is given, by multiplying P by the value of the interesttime fa*or (AIP, A n). As an otample' the numerical values of interesttime factors for i = 8o/o are calculated and shown in Table 20.8.
I
1.08000000 1.16640000 7,219V1200 4 ,. 1.3664s396 r.4.6932808 t 6 r.58687432 r.71382427 7 1.85093021 2 5
t,
I
lo
,.
Ti; t2
0.92592593 0.85733882
4,92592593
p.79r53224
2.5770if)699
g:/3t02985
3.31212684 3,99271004 4,62287966
0.65058320 0.63016963 0.58349040 0.54026888 r.99900463 0.50024897 2.t5592500 0.46319349 2.33163900 0.42888286
251817012 0.397rr376 2.71962373 036769792
T3
2.93719362 0.34046104 3.L72169rr A3r524r7o 3.42594264 0.29189047
L4 15
l6
t7',, 3]0001805 0.27026895
l8
3.99601950 0.25024903
' 43\57On6,:
A.23171206
4.66095714 zt.: 5.03383372 ',?2 ;, 5.43654041 2t 5.87146365 6.34118074 2/t 25 6.84847520 26 7.39635321 7.98806147 2l 8.62710639 29 9.31727490 30 10.06265689 31 , 10.86766944
0.2145482r
19
I
2A:,,
',
n
:
1,7832&7'
5,VA697A06
5:/4663894 6.2468:8791
6.71008140 7,13896426 7.536fr7802 7.94377594
8.2U23698 8sr:94786e 8.85136916
91216381t
937188714 e.6fi35?92a
0.r0r8522r
0.19865575
9.818r4V41 r0,01680316
0.1539405r
rc20A7$66
o.og8o32w
0.09642217 ra.12875525 0.09497796 0J4601790 'r0.674n619 0.09367878
a.17045&5 a.j.36nr559 o.t:tz,arz^o a.094a1476
a,a80wg7r 0$6902949 0.060a7&4 ,0n52695.02
0.04652t8r ,0.04129685 0,a3682954 ,a,03297697
0,a2962943
0.a267\2n '0.a24r276' a,a2$522t:' O.OtgglZZS"
0'9199??07
10.37105895
60.89329557 66.76475922
0,ar642217 a:frt497796
73.10593995
0,01367878
0.13520176 0.12518682 0.11591372 0.10732752 0.09937733 0.09201605
10.80997795
0.0925W13 79.95U15r5 a.ar2507l'3 0.09144810 87.35076836 0.01144810 0.09048891 95.33882983 b.olor))rO logarithmic models, 54955 I
[email protected] air quality slandaids, 6870 Safe Drinking VaterAct, 65 Sufae Varer Tiernent Rule
(svTR),68 Enos, typs of in *pedmtal obsemtion,582 E*iq, L05I22. See afu Code ofethio
dsignprm,45 
32, 54L 546,
normal distdbution, 587 594
372
role of, 52 418, 445  447.
prcbab{rty,5V579
mlc K% oL J/JJla
@J1s.374376,380382 we fiaing 407411 FilIomlffi4 380382
sK6, )//)/y stopping sight
disuna,
532, 541543
temperatue distribution aaos
a
plain
wall,536538 time, rcle of in, 197198 Engineing profesi on, 2 l 22 Accrediation Bord for Enginedng and Technologr
(ABE$,
An
262
@.S.),
4748 Fibs (qptic) gl,s' 517518 Fill ommd,380382 Filet, dwings,468469
Fihn rcistane (o#cimt), 302303 Finishs, tm rcle o(,52 Fim lm of themodynmic,348350
emn, 582 nte,210211,224 volme,210211
Fixed
Fls
M,224
Fluid friction, re Vrmsity Fluid matqials, 5 1 9521 Foot (ft), U, S. Cutomaryuit oflength,
l)o
MAILAB,432
pmetm, 236
z8l
ra,
mingover
a,2442i6
bu& modulu of omprcibility, 263264 Caterpillar 797 miaiag t ,r"ls *Ctneerng Wg, z/dl6r omprshe (ultimate) strength, 261
d6^d,237 disvne, aarng at a, 2& 267 distane, acting ovo a 268269 facmr of ufety @,S,), 262
fnaton,240241 gaitanoml 237,242244 See
ako
imFul*, 269271 modulu of elasticity, 256 259, 260 modulu of rigidity, 259263 momertof a,?14267 Nswton! lM, 241244
fomulo,376380 tunaons3763W,387389 imponane in ngindhg, 372373 ioponing fila into MAfl{B, 445447
insningells, olmm, ad rom,376
eeprepmtion for,2740
logiel functiom, 387389
ommon uaits of 812
mari: compuradom,
400 
Hnke\lm,238,258259
lia6
Cells: Ercel fuoaiom uithmetic operatioro, 376
slution o[,8284
of*fety
&fuenheit fF), U.S. Cretomarymitof tmpemtue, 291293 Fmd (F), boic uit for qpacitors, 334 Faible rclution region, dcign prre,
xtage racrion,2.69271.
112
dsign altematim, 5556
ErqL
5
122
Evalvdot,45,5556
nonlina
models,
LzO
112
tm
Factor
Fora ad forerelated
enginert med, 111l12 enginsin& 106109 Ndonal Sciery of Profesional Enginen (NSPE), 107, 120 122 plagiaisn, 112 prcfmional mpomibility, I 12
Evalutoc
385
MATLAB,43O Extmion linc,468
for omd,
for euginm aademic dishoncty, 112 @e studis fo!, ll21I7, ode of, 107111
mthematia h,532 576
)4tJttu
Excâ‚¬I, 1
lryel onaminmt goal
@ntram
Pste,383
Exponential functiorc, 385, 430
lwel ontamimt (MCL)
onfliaofinterct,
logiel,387389
nigonomeaic 385
EPA, ree Environmqtal Protection Agqcy (EPA)
546549,549
togmmc, 16)
Xxmtiresrmic,87
(MsG),67
57 6,
.h ffowdiagm,598599
onomio m4 597619
r95217 See
Enginering organiatiom, involvement with,36
ooling
timerdated
mim
r97 198, 530 53r, 577596,597619
pamet6,
t52
tmpemtm od
lroimm
Enginering papu, 8586 Enginering prcblem, 8284, 127 128,
236281
innoduction to, 12J length md lengthrelatd
stâ‚¬I,511512 water,520527
pmetes,
specialiatioro of, ll, I3L4 technologr progm, 2l 22 U.S. Buru of Iabor Satistia for I world PoPrl'iion' altrâ‚¬ts of 68
387,387389 rc of, 3M
[email protected] arab{is qponential,3S5
relational opemton, 388 todsy ( ), 384
56
Air Act, 65, 6869 drinking water mdads, 5768 function of,65 indor air qualiry GACD, 70 hdmr air quality sandarls, /0l2
rcod,514515
l5l
o6,
Clm
503508
tianim,509510
128,3t9340 enerry md power, 34136! fore md forcrelared llmetes,
497.
o4,
wie
Enviromental enginering prcfesion o4, 1819 Enviromental Protation Agacy (EPA), 65,6772
soli4 509519
of labor Statisia for,
dimmiom md unis,
ad
l0
ofmatter, 502503
prcpenic
\Feb sitc for, 1213 Enginering findmenels, 124 369
time
wel,526
silion,5l
spaialiaion md, 1314
128,
mgineing
plasic,5l5516
nucla, 1920 petloleu, 19
Buru
(water vapor), 520
lightweight neals, 5095
phm
of Enginming
15
l5
mgn$im,510
19
materials,2021 minir& f,$
U.S.
F.sm,
ircn,511 Jet enghe,
14
ud Pmaie
Principla
)r/>rd
hmidity
wg6,5/)5/O Ercel functiom, 376
Fmdanentals of Enginering Eem (FE), 15 inuoduction m, 23, 426 Ntional Sociery of Profesioual Engins (NSPE),3
@nqetg'5L2514 opper md its allqre, 5 105 1 I dcign pr>2, >>5))+ mr:ftipliccion of, 553554 time, a a phpiel prcperty of, 206 Srundmommtof m,177182
bem, mmpls
of in, 180182
akoBn
(EPA)
Amaim
National Standtrds Insdtute
terials
(ASIM,6l62 FmFis & nomalistim
(ATNOR),63
599
British Stadrds Instituts (BSt), 63 Ce standarcls, 63 Chim Sate Bua of cllatry *4 Tirhniel Supwisioa (CSBTS), 63 @nmirms, methods of nmaging
nrle, 159, 621 Singu&r marix, 559 Sir, nomiasl w actnal, 160162
Deutschc Initut fiu Nomung @IN), 63
139
l4I
Silion,51 Simple
intact,
Sine
Slidc, PwuPoint pwntarioro, 9397, 9798, IOO mimation of, 9798 imting ftom otha PmrPoint 6lc,
mcta,94 of, 131
Solid linc, onhogra.phic dwings, 465 Solid marerials, 5095 19 Solid modeling, 476 482,493497 Bcing 777 omerciat airylane, engi
nuingmd493497 Bmlru
openriom for, 479, 481 bonomup, 478479
onputer nmaielly onrolled (CNC) mchines 47 imporane of in enginedng
42
84. 138138, 197198. See aho Engineing Problem; Linear
muatiom malFis of,83 defining the prcblm, 8283
mple ofpmtation
fur,84
ronoiel,
138139 pametric fom' 83
simpli$ing the prcblm, 83 197198
of, 8283 symbolic, 138139 time rcle of in, 197198 steps
(EPL),65,6712 of U.S., 6567
mmpls
(NFPA),62
[email protected],5860 orymiations for, 60 62, 064 outdmr air quality, 6870 Star