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,ffif-AB " " " " "
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 semeter-long 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 hands-on, requiring the student to gather and analyze information. Moreover, information collection and proper utilization of that information are encour€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: Engineering-An 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 discipline-specific 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 length-related 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 time-related 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 mass-related 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 force-related 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 temperature-related 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$/er,
and efficiency is emphasized in Chapter 13.
Part Three: Computational Engineering ToolsUsing Available Software to Solve Engineering Problems 15, we inuoduce Microsoft Excel and MAI1AB computational tools that are used commonly by engineers to solve engineering prob-two lems. These compuationd tools are used to record, organize, analyze data using formule", and present the resulm ofan analysis in chart forms. MAILAB is also versatile enough that students cln use it to write their own programs to solve complor problems.
In Part Three, consisting of Chapters 14 and
Part Four: Engineering Graphical eommunication: Conveying Information to 0ther Engineers, Machinists, Technicians, and Managers In Part Four, consisting of Chapter 16, we inuoduce students to the principles and nrles of engineering graphical communication and engineering sfmbols. A good grasp of these principles '!fe explain that engiwill enable students ro convey and undentand information effectively. neers use technical drawings to convey useful informadon to others in a standard manner. An engineering drawing provides information, such as the shape of a product, its dimensions, materials from which ro fabricate the product, and the assembly steps. Some engineering drawings are specific to a particular discipline. For enample, civil engineers deal with land or boundary' topographic, consrrucdon, and route survey drarings. Electrical and electronic engineers, on the other hand, could deal with printed circuit board assembly drawings, printed circuit board drill plans, and wiring diagrams.'W'e also show that engineers use special symbols and signs to convey their ideas, analyses, and solutions to problems.
Part Five: Engineering Material SelectionAn lmportant Design Decisioll fu
engineers, whether you are designing a machine pa$, 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 voice-recognition 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 STUDIES-ENGINEERING 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 Newgen-Austin 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
ENGTNEERTNG-AN 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 Length-Related 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 Time-Related 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 Force-Related
10.1 102 10.3 10.4
What We Mean by
Force
237
Newtonl Laws In Mechanlcs
241
Pressure snd Stress-Force 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, Torque-Forces Actlng at a
Work-Force Actlnq 0ver a Dlstance 268 Llnear lmpulse-Forces Actlng 0ver
Summary
271
Problemr
271
IMPROMPTU OESIGII
V
Tlme
269
2r/
An Englneerlng Marvel: Caterplllar 797 Mlning
lf
278
Temperature and Temperature-Related 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 Energy-Flrst 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 Ercel-Baslc 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[AB-Baslc
ldeas 420
Uslng MATLAB Bullt-ln
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
Probablllty-Baslc ldeas 510
Statlstlcs-B$lc ldeos 5|9 FrequencyDlstrlbutlons 580 Heasuras ol Gntral Tendency Yarlotlon-lllean, 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 Alternatlves-Declslon
Summory
617
PruUem
617
AppEndlr 620
Credlts 625 Index 626
l{aklng
613
Devlatlon 582
XV
Part
ENCTNEERING Al. Excmttc PnoFEssron
1
j@@ltryq'@qt
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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 air-conditioning, 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 behindthe-scenes 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 age-will 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
population-the
(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 lives-clean 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 ner
What ls Engineering and What Do Engineers Do? Engineers applyphysical and chemical lars and principles and mathematics to design millions of producs and services that we use in our weryday lives. These products include c:rs, com-
puters, aircraft, clothing, toys, home appliances, surgical equipment, heating and cooling equipment, health care devices, tools and machines that makes various products, and so on (see Figure 1.3). Engineers consider impoftant fadors such as cost, efficiency, reliability, and safety when designing these products. Engineers perform tests to make cerain that the producs they design withsand various loa& and conditions. They are continuously searching for ways to improve already existing products as well. They also design and supervise the construction of buildings, dams, highwap, and mass transit systems and the consnuction ofpower plants that
1.3
I
CouuoN Tnerrs or Gooo
ENcrNBsRs
rigunt.3
As an engineer you
will apply
physical and chemical laws and
pilnciples and mathematics to design various products and services.
supply power to manufacnrring companies, homes, and offices. Engineers play a significant role in the design and maintenance of a nationt infrastrucnrre, including communication qrsterns' public utilities, and nansportation. Engineers continuously develop new, advanced materials to make producs lighter and stronger for different applications. They are also responsible for finding suitable ways ro extract peuoleum, natural gas, and rarrr materials ftom the eanh, and they are involved in coming up with ways of increasing crop, fruit, and vegetable yields along with improving the safety of our food products. The following represent some comrnon queers for engineers. In addition to desigr, some engineers work as sales representatives for products, while others provide technical suppon and uoubleshoodng for customers of their products. Many engineers decide to become involved in sales and customer support, because their engineering background enables them to explain and discuss technical information and to assist with installation, operation, and maintenance ofvarious products and machines. Not all engineers work for private industries; some work for federal, state, and local governmen$ in various capacities. Engineers work in departmena of agriculture, defense, energ)r, and transponation. Some engineers work for the National Aeronautics and Space Administration (NASA). As you can see, there are many satisfring and challenging jobs for engineers.
1.3
Common Traits of Good Engineers Atthough the activities of engineers are quite varied, there are some personality trais and work habits that rypifr most of today's successfirl engineers. o Engineers are problem solvers. u Good engineers have a firm grasp of the fundamenal principla of engineering, which they can use to solve many different problems.
"
Good engineers are analytical, detailed oriented, and creative.
10
Cner.rsn
I
INrnooucrroN To trts EwcrNesRrNc PnorrssroN o Good
engineers have a desire to be lifelong learners. For example, they take continuing education classes, seminars, and workshops to stay abreast ofinnovations and new technologies. This is particularly important in today's world because the rapid changes in technology will re' quire you as an engineer to keep pace with new technologies. Moreover, you will risk being laid offor denied promotion ifyou are not continuallyimprovingyour engineeringeducation. . Good engineers, regardless of their area of specialization, have a core knowledge that can be applied to many areas. Therefore, well-trained engineers are able to work outside their area of specialization in otler related fields. For example, a good mechanical engineerwith awellrounded knowledge base can work as an automodve engineer, an aerospace engineer, or as a chemical engineer. . Good engineers have wrimen and oral communication skills that equip them to r /ork well with their colleagues and to convey their expenise to a wide range of clients. . Good engineers have time-management skills that enable them to work productively and efficiendy. o Good engineers have good "people skills" that allow them to interact and communicate effectively with various people in their organization. For example, they are able to communicate equally well with the sales and marketing experts and their own colleagues. o Engineers are required to write repor$. These repons might be lengthy, detailed technical reports containing graphs, chars, and engineering drawings, or they may take the form of brief memoranda or executive summaries. . Engineers are adept at using computers in many different ways to model and analyze various practical problems. . Good engineers actively panicipate in local and national discipline-specific organizations by attending seminars, workshops, and meetings. Many even make presentations at professional
"
meetings. Engineers generally work in a team environment where they consult each otfier to solve complex problems. They divide up the task into smaller, manageable problems arnong themselves; consequendy, producdve engineers must be good team players. Good interpersonal
and communication skills are increasingly imponant now because of the global market. For example, various parts of a car could be made by different companies located in different countries. In order to ensure that all components fit and workwell together, cooperation and coordination are essential, which demands strong communication skills. Clearly, an interest in building things or tfing things apart or solving prtzz.les is not all that is required to become an engineer. In addition to having a dedication to learning and a desire to find solutions, an engineer needs to foster certain anitudes and personaliry traits. These are some other facts about engineering that are worrh noting.
.
For almost all entryJevel engineering jobs, a bachelor's degree in engineering is required.
According to the U.S. Bureau of Labor Statistics:
.
The starting salaries of engineers are significandy higher than those of bachelor's-degree graduates in other fields. The oudook for engineering is very good. Good employment oppornrnities are expeced for new engineering graduates during 2008-2075. o Most engineering degrees are granted in electrical, mechanical, and civil engineering, the
.
parents
ofall other engineering branches.
In the year 2004,
engpneers
held 1.4 million jobs
(see
Thble 1.1).
The distribution of employrnent by disciplines is shown in Thble 1.1.
1,3 Couuox
Tnerrs oF GooD Excrxwns
t,
i
Total' all engineen
i i i 1
i I j i i i i I i I j i
\A*9,OOO
LOOo/o
Civit
237,000
Mechanical
225,A0A
16.4 15.6
Industrial Eleffrical
r77,004
11 1
156,000 143,000
10.8
Electronics, except computer Computer hardware
Materials
Nuclear Peroleum
17,000 16,000
1.2
9,7Q4
0.7
6,800 5,204 3,400 172,400
0.5
Aerospace
Environmental Chemical HealthandsaGry, erceptminingsaGty
Biomedical Marine engineen and naval archiects Mining and geological, including mining safery
furicultural AII other engineers i
9.9 5.3
77,000 76,A00 49,000 31,000 27,400 21,000
.--
-
<.,
3.4 2.1 1.8
l.) 1.1
0.4 0.2 11.8
_
The anerage starting salary for engineers is shown in Thble 1.2. The data shown in Tifile 1.2 is the result of the 2005 survey conducted by the National Association of Colleges and Employen.
Discipltnes or $pecialties Aerospace /aeronautical /asuonautical
Bachelortc
$50,993
furicultural
46,172
Bioengineering and biomedical
48,503
Chemical
53,8r3
Civil
43,679
Computer ElcctriCIl /elecronics and communications Environmental /environmental health Indusuial / manu6cnuing
52,464 51,888
47,384 49,567
Materials Mechanical Mining and mineral
50,982
Nuclear
51,182 6r,;516
Petroleum
50,236 48,643
Ph.D.
Master's
$62,930
$72,529
53,022
59,657 57260 79,59r 48,050 59,625 60,354 69,625 64,416 80,206 56,561
85,000
59,880
68,299
58,814
68,000
12
Cneprnn
I
INTnooucrroN To rgs
ENcTNBBRTNc Pnorussror.l
According to the U.S. Bureau of labor Statistics, in the Federal Government sector, mean annual salaries for engineers ranged from $100,059 in ceramic engineering to $70,086 in agriculrural engineering in 2005.
1.4
Engineering Disciplines Now that you hane a general sense ofwhat engineers do, you may be wondering about the various branches or specialties in engineering. Good places to learn more about areas of specialization in engineering are the'Web sites of variotrs engineering organizations.'W'e will orplain in Chapter 2 that as you spend a litde time reading about these organizations, you will discover many share common interests and provide some overlapping services that could be used by en'Web sites that you mqy find usefirl grneers of various disciplines. Following is a list of a few when searching for information about various engineering disciplines. Amerlcan Academy of
Engineers
Envlronmental
Englneers
www.enviro-eng$.org Amerlcan Instltute of Aeronautlcs
Astronautics www.aiaaorg
Amerlcan Instltute of Chemlcat
lnstitute of Electrical and Electronics www.ieee.org
and
Instltute of Industrlal Englneers www.iienet.org National Academy of Engineering
Engineers www'nae'edu
www.aiche.org The Amerlcan Soclety of Agrlcultural Englneers www.asae.org Amerlcan Society of Clvll Engineers www.asce.org Amerlcan Nuctear Soclety www.ans.org
NatlonalScience Foundation
wwwnsf'gov Natlonal Society of Black Engineers www.nsbe.org National Society of Professional Englneers www.nsPe.org
Society of Automotlve Englneers www'sae'org
AmericanSoc|etyforEng|neer|ngsocletyofHispanlcProfess|ona|Englneers
Educaflon www.asee.org Amerlcan Soclety of Heatlng, Refrigerating, and Alr'Conditlonlng Englneers www.ashrae.org Amerlcan Soclety of Mechanical Engineers www'asme'org Biomedfcaf Englneering Soclety
mecca.org/BME/BMES/Society
www'shPe'org
Society of Manufacturlnq Engineering www'sme'org Society ofWomen Englneers www.swe.org Tau Beta pi
(All'Engineering Honor Soclety)
www.tbp.org NASA
centers
Ames Raearch Cmter
www.arc.nasa.gov
1.4 Dryd* Fligltt
Research
Cmter
www.dfrc.nasa.gov Goddard Space Fligbt www.gsfc.nasa.gov
Cmtn
ENcrNreRrNc
Marsball Space Fligbt Cmter www.msfc.nasa.gov
Gbnn Researcb Cmter
www.jpl.nasa.gov
www.gfc.nasa.gov
Jobrcon Space Cmter www. jsc.nasa.gov
www.invent.org
Space
Cmter
www.lisc.nasa.gov
13
I^angley fus earc h Center www.larc.nasa.gov
Jet Propukion Laboratory
Kmnedl
Drscrpr,rxss
lnventors Hall of Fame Patent and Trademark 0ffice www.uspro.gov U.S.
For an additional listing of engineering-related W'eb sites, please see this book's companion 'W'eb
site.
What Are Some Areas of Engineering Specialization? There are over 20 major disciplines or specialties that are recognized by professional engineering societies. Moreover, within each discipline there enist a number of branches. For example, the mechanical engineering program can be traditionally divided into rwo broad areas: (1) thermal /fluid systems and (2) strucrural lsolid systems. In most mechanical engineering programs, during your senior year you can take elecdve classes that allow you to prusue your interest and
litilr,,.t_B ti:r,t:.i,i,):il
',)),, i:.:;:
Engineers are adept at using computers in many different ways to model and analyze various practical problems.
t4
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IvrnopucrroN To rIrB EncrNrrnrxc PnorrssroN broaden your knowledge base in these areas. So, for example, ifyou are interested in learning more about how buildings are heated during the winter or cooled during the summer, you will take a heating, ventilating, and air-conditioning class. To give you additional ideas about the various branches within specific engineering disciplines, consider civil engineering. The main branches of a civil engineering program normally are environmental, geotechnicd, water resources, transportation, and strucnual. The branches of electrical engineering mqy include power generation and ransmission, communications, control, eledronics, and integrated circuits.
Not all engineering disciplines
are discussed here, but you are encouraged to visit the \?'eb
sites ofappropriate engineering societies to learn more about a particular engineering discipline.
1.5
Aeereditation Board for Engineering and TeehnoloEy Over 300 colleges and universities in the United States offer bachelor's-degree programs in engineering that are accredited by the Accrediation Board for Engineering and Technology (ABET). ABET examines the credendals of the engineering program's faculty, curricular content, facilities, and admissions standards before granting accreditation. It may be wise for you to find our the accreditation status of the engineering program you are planning to anend. ABET maintains a'Web site with a list of all accredited programs; visit www.abet.org for more information. According to ABET, accredited engineering programs must demonstrate that their graduates, by the time of graduations, have
' an ability to apply knowledge of mathematics, science, and engineering; " an ability to design and conduct experiments, as well as to analyze and interpret data; . an ability to design a system, component, or process to meet desired needs; . an ability to function on multidisciplinary teams; . an ability to identifr, formulate, and solve engineering problems; "
.
an understanding ofprofessional and ethical responsibility; an ability to communicate effectiveh; the broad education necessary to understand the impact ofengineering solutions in a global and societal context; a recognition ofthe need for and an ability to engage in lifelong learning
o
a
"
an ability to use the techniques, skills, and modern engineering tools necessary for engineer-
o
.
knowledge of contemporary issues; and
ing practice. Therefore, these are the educational outcomes that are expected of you when you graduate from your engineering program. Bachelor's-degree programs in engineering are typically designed to last four years; however, many srudents take five years to acquire their engineering degrees. In a typical engineering program, you will spend the first two years srudying mathematics, English, physics, chemistry introductory engineering computer science, humanities, and social sciences. These first two years are often referred to as pre-engineering. In the last two y@rs, most courses are in engineering, usually with a concentration in one branch. For example, in a typical mechanical engineering program, during the last two years of your studies, you will take courses such as thermodynamics, mechanics of materials, fluid mechanics, heat transfer, applied thermodynamics, and design. During the last two years of your civil engi-
1.5
AccnnprmrroN Bo.eno ron ENcrNsrRrNG AND
TscuNolocv
15
neering studies, you €n expect to take courses in fluid mechanics, uansportation, geotechnical engineering, hydraulics, hydrology, and steel or concrete design. Some programs offer a general engineering curriculum; students then specialize in graduate school or on the job. Many community colleges around the country offer the first nvo years of engineering programs, which are normally accepted by the engineering schools. Some engineering schools offer five-year master's-degree programs. Some engineering schools, in order to provide hands-on experience, have a cooperative plan whereby studen$ take classes during the fust three years and then may take a semester off from stud)4ng to work for an engineering company. Of course, after a semester or two, students reflrn to school to finish their education. Schools that offer cooperative programs generally offer firll complements of classes every semester so that students can graduate in four years if they desire.
Professional EnEineer All 50 states and the District of Columbia require registration for engineers whose work may affect the safety of the public. As a first step in becoming a registered professional engineer (PE), you must have a degree from an ABET-accredited engineering program. You also need to take your Fundamentals of Engineering Exam (FE) during your senior year. The exam lasr abour eight hours and is divided into a morning and an afternoon section. During the morning session, you will answer multiple-choice questions in chemistry physics, mathematics, mechanics, thermodynamics, electrical and elecronic circuits, and materials science. During the four-hour afternoon session, you will answer multiple-choice questions specific to your discipline, or you may choose to take a general engineering exam, After you pass your FE exam, you need to gain four years ofrelwant engineering work experience and pass another eight-hour exam (the Principles and Practice of Engineering Exam) given by the state. Candidates choose an exam ftom one of 16 engineering disciplines. Some engineers are registered in several states. Normally, civil, mechanical, and electrical engineers seek professional registradoru. As a recent engineering graduate, you should expect to work under the supervision ofa more experienced engineer. Based on your assigned duties, some companies may have you attend workshops (shon courses that could last for a week) or a day-long seminar to obtain additional uaining in communication skills, time management, or a specific engineering method. fu you gain more knowledge and experience, you will be given more fteedom to make engineering decisions. Once you have many years of e
16
Cneprrn
L
lNrnooucrroN To rHp ExcrNnsRrNc PnorsssroN
A civil engineer at work.
geotechnical. Civil engineers work as consultants, construfiion supervisors, city engineers, and public utility and uansponadon engineers. According to the Bureau of labor Statistics, the job oudook for graduates of civil engineering is good because as population grov/s, more civil engineers are needed to design and supervise the construction of new buildings, roads, and water supply and sewage qFstems. They are also needed to oversee the maintenance and renovation of existing public strucures, roads, bridges, and airports.
Electrical and Electronic Engineering Electrical and electronic engineering is the largest engineering discipline. Electrical engineers design, develop, test, and supervise the manufacturing of electrical equipment, induding lighting and wiring for buildings, cars, buses, trafuis, ships, and aircraft; porver generation and transmission equipment for udlity companies; elec-
tric motors found in various producs; control
devices; and radar equipment. The major branches of electrical engineering indude power generation, power transmission and disuibution, and controls. Elecuonic engineers design, develop, test, and supervise the production of elecuonic equipment, including compurer hardware; computer nennork hardware; communi-
cation devices such as cellular phones, television, and audio and video equipmenq as well as
L.5 Accnsorrarrox
An electrical engineer
Boano ron. ENcrNrsRrNG AND TrcnNor.ocv
17
at
work.
measuring instrumenrc. Growing branches of electronic engineering indude computer and communication electronics. The job oudook for electrical and electronic engineers is good because businesses and government need faster compurers and better communication systems. Of course, consumer electronic dwices *ill pLy a significant role in job growth for electrical and electronic engineers as well. Mechanical Engineering The mechanical engineering discipline, which has evolved over the
ye:rs as new technologies have emerged, is one of the broadest engineering disciplines. Mechanical engineers are involved in the design, dwelopment, tesdng, and manufacturing of machines, robots, tools, power generating equipment such as steam and gas turbines, heating coolinp and refrigerating equipment, and internal combusdon engines. The major branches of mechanical engineering indude thermal /fluid systems and strucnrral /solid qystems. The job oudook for mechanical engineers is also good, as more efficient machines and power generating equipment and alternative energ;r-producing devices are needed. You will find mechanical engineers working for the federal government, consulting firms, vatious manufacturing sectors,
the automotive indwtry and other transportation companies. The other common disciplines in engineering include aerospace engineering, biomedical, chemical engineering, environmental engineering, peuoleum engineering, nuclear engineering and materials engineering. Aerospace Engineering Aerospace engineers design, dwelop, test, and supervise the manufacture of commercial and military aircraft, helicopters, spacecraft, and missiles. They may work on projects dealing with research and development of guidance, narriga.tion, and control qrstems. Most aerospace engineers work for aircraft and missile manufacnrrers, the Department of Defense, and NASA If you decide to pursue an aerospace engineering crreer, you should 'Washington, Texas, or Florida, because these are the states with expect to live in CaliFornia, large aerospace manufacnrring companies. According to the Bureau of l,abor Sstistics, the job oudook for aerospace engineers is expected to grow not as fast through the year 2010. One rea-
t8
CrrepteR
I
llrrnopucrroN To
ttE
ENcrNEEnrxc Pnorssstor.I
job growth is the decline in Department of Defense expenditures. However, population growth and the need to meet the demand for more passenger air traffic, commercial aiqplane manufacnrrers are expected to do well. son for this slower because of
Biomedical Engineering Biomedical engineering is a new discipline that combines biology, chemistry medicine, and engineering to solve a wide range of medical and health-related problems. They apply the laurc and the princrples of chemistry, biologr, medicine, and engineering to design anificial limbs, organs, imaging systems, and devices used in medical procedures. They also perform research alongside of medical doctors, chemists, and biologists to better undersrand various aspects of biological qrstems and the human body. In addition to their aaining in biologT and chemistry, biomedical engineers have a suong background in either mechanical or electrical engineering. There are a number of specializations within biomedical engineering, including: biomechanics, biomaterials, tissue engineering, medical i-.gu& and rehabilitation. Computerassisted surgery and tissue engineering are among the fastest growing areas of research in biomedical engineering. According to the Bureau of Labor Statistics, the job oudook for graduates of biomedical engineering is very good, because of t}re focus on health issues and the aging population.
Engineering As the name implies, chemical engineers
use the principles of chemistry related of problems to the production of chemsolve engineering sciences to a variery and basic icals and their use in various industries, including the pharmaceutical, electronic, and photographic industries. Most chemical engineers are employed by chemical, petroleum refining, 6.1m, paper, plastic, paint, and other related indusmies. Chemical engineers also work in metallurgicd, food processing, biotechnoloW xnd fermentation industries. They usually specialize in certain areas such as polymers, oxidation, fenilizers, or pollution control. To meet the needs of the growing population, the job oudook for chemical engineers is also good, according to the Bureau of l,abor Statistics.
Chemical
Environmental Engineering Environmental engineering is anorher new discipline that has grovrn out ofour concern for the environment. As the name implies, environmenal engineering is concerned with solving problems related to the environmenl They apply dre larvs and the principles of chemistry, biolory, and engineering to address issues related to water and air pollution control, hazardous waste, waste disposal, and recycling. These issues, if not addressed propedy, will affect public health. Many environmenal engrneers get involved with the devel-
opment of local, nationd, and international environmental policies and regulations. They study the effecs of industrial emissions and the automobile emissions that lead to acid rain and ozone depletion. They also work on problems dealing with cleaning up enisting hazardous waste. Environrnenal engineers work as consultants or work for local, State, or Federal agencies. According to the Bureau of Labor Statistics, the job oudook for graduates of environmental engineering is very good, because environmental engineers will be needed in greater numbers to address and conuol the environmental issues discussed above. It is important to note that rhe job oudook for environmental engineers, more than engineers in other disciplines, is affected by politics. For example, looser environmental policies could lead to a fewer jobs, whereas stricter policies could lead to a greater number of jobs.
L5
AccnBprrarroN Boeno ron EucrNrpnrNc
AND
ThcnNor-ocv
19
Manufacturing Engineering Manufacuring engineers dwelop, coordinate, and supervise rhe of manufacturing all types of products. They are concerned with mfing products efficiently and at minimum cost. Manufacturing engineers are involved in all aspecs of pro-
process
duction, including scheduling and materials handling and the design, development, supervision, and control of assembly lines. Manufacturing engineers employ robom and machine-vision technologies for production purposes. To demonstrate concepts for new products, and to save time and money, manufacruring engineers create prototypes ofproducts before proceeding to manufacture acnral products. This approach is cdledpratotyping Maaufacturing engineers are employed by dl types of industries, including automotive, aerospace, and food processing and packaging. The job outlook for manufacnrring engineers is expected to be good. Petroleum Engineering Petroleum engineers specialize in the discovery and production ofoil and naturd gas. In collaboration with geologists, petroleum engineers search the wodd for underground oil or natural gas reservoirs. Geologists have a good understanding ofthe properties
of the rocks that make up tle eanh's crust. After geologisa waluate the propenies of the rock formations around oil and gas reservoirs, they work with petroleum engineers to determine the best drilling methods to use. Petroleum engineers are also involved in monitoring and supervising drilling and oil extraction operadons. In collaboration with other specialized engineers, peuoleum engineers design equipment and proc€sses to achiwe the maximum profitable recovery of oil and gas, They use computer models to simulate reservoir performance as they experiment with different recovery techniques. If you decide to prusue petroleum engineering, you are most likely to work for one of the major oil companies or one of rhe hundreds of smaller, independent companies involved in oil exploration, production, and service. Engineering consulting fums, government agencies, oil field services, and equipment suppliers also employ petroleum engineers. According to the U.S. Depaftment of labor, large numbers of petroleum engineers are employed in Texas, Oklahoma, Louisiana, Colorado, and California, including offshore sites. Many American petroleum engineers also work overseas in oilproducing regions of the world such as Russia, the Middle East, South America, or Aftica. The job oudook for petroleum engineers depends on oil and gas prices. In spite ofthis fact, if you do decide to study petroleum engineering employment oppornrnities for petroleum engineers should be favorable because the number ofdegrees granted in petroleum engineering has traditionally been low. Also, petroleum engineers work around the globe, and many employers seek U.S.-trained petroleum engineers for jobs in other countries. Nuclear Engineeting Only a few engineering colleges around the country offer a nuclear engineering program. Nuclear engineers design, dwelop, monitor, and operate nuclear power equipment that derives its power ftom nuclear energ'y. Nuclear engineers are involved in the design, development, and operation of nuclear power plants to generate elecricity or to Power
Navy ships and submarines. They may also work in such areas as the production and handling of nuclear fuel and the safe disposal of its waste products. Some nuclear engineers are involved in the design and development of industrial and diagnostic medical equipment. Nuclear engineers work for the U.S. Navy, nuclear power udlity companies, and the Nuclear Regulatory Commission of the Department of Energy. Because of the high cost and numerous safery concerns on the part of the public, there are only a few nuclear power plan* under construc-
20
Crr,rprsR
I
lNrnopucrroN To rnn ENcrNnrnrNc Pnornsstol,t tion. Even so, the job oudook for nuclear engineers is not too bad, because currendy there are not many graduates in this field. Other job opponunities exist for nuclear engineers
in the departments of Defense and Energy, nuclear medical technology, and nuclear
waste
management. Mining Engineering There are only
a few mining engineering schools around the country.
Mining engineers, in collaboration with geologists and meallurgical engineers, find, extract, and prepare coal for use by utility companies; they also look for metals and minerals to efiract from the eanh for use by various manufacnrring industries. Mining engineers design and supervise the construcdon of aboveground and underground mines. Mining engineers could also be involved in the development of new mining equipment for extraction and separation of minerals from other materials mixed in with the desired minerals. Most mining engineers work in the mining industry, some work for government agencies, and some work for manufacturing industries. The job oudook for mining engineers is not as good as for other disciplines. The mining industry is somewhat similar to the oil industry in that the job oppornrnities are closely tied to the price of metals and minerals. If the price of these producrc is low, then the mining companies will not want to invest in new mining equipment and new mines. Similar to petroleum engineers, U,S. mining engineers mqy find good oppornrnities outside the United States. Materials Engineering There are only a few engineering colleges that offer a formal program
in materials engineering, ceramic engineering, or metallurgical engineering. Materials engineers research, dwelop, and test new materials for various products and engineering applications. These new materials could be in the form of metal alloys, ceramics, plastics, or composites. Materials engineers study the nature, atomic strucnrre, and thermo-physical propenies of materials. They manipulate the atomic and molecular structure of materials in order to creare materials that are lighter, sffonger, and more durable. They create materials with specific
mechanical, electrical, magnedc, chemical, and heat-transfer propenies for use in specific applications; for example, graphite tennis racquets that are much lighter and stronger than the old wooden racquets; the composite materids used in stealth militaryplanes with specific electromagnetic properties; and the ceramic tiles on dre space shutde that protect the shuale during reentry into the atmosphere (ceramics are nonmetallic materials that can withstand high temperatures). Materials engineering may be funher divided into metallurgical, ceramics, plastics, and other specialties. You can find materials engineers working in aircraft manufacturing; various research and testing labs; and electrical, stone, and glass products manufacturers. Because ofthe low number of current graduates, the job oppornrnities are good for matedals engineers.
Engineering Technology In the preceding texr, we introduced you to the engineering profession and its various areas of specialization. Let us now say a few words about engineering technolory. For those ofyou who tend to be more hands-on and less interested in theory and mathematics, engineering technology might be the right choice for you. Engineering technolog;r programs typi."lly require the
1.5
AccnsprrATroN Boeno ron ENcrxsnRrNG AND TecgNolocv
2l
A nuclear engineer at work.
knowledge of basic mathematics up to integral and differential calculus level, and focus more on the application oftechnologies and processes. Although to a lesser degree than engineers, engineering technologists use the same principles of science, engineering, and mathematics to assist engineers in solving problems in manufacnrring, construcion, product development, inspection, matntenance, sales, and research. They may also assist engineers or scientists in setting up experiments, conducting tests, collecting data, and calculating some results. In general, the scope of an engineering technologistt work is more application-orienrcd and requires less understanding of mathematics, engineering theories, and scientific concepts that are used in complex designs. Engineering technology programs usually offer the same type of disciplines as engineering programs. For example, you may obtain your degree in Civil Engineering Technology, Mechanical Engineering Technology, Electronics Engineering Technology, or Industrial Engi-
22
Cn^nprun
I
hvrnooucrroN To rHE ENcrNsERrNc PnornssloN neering Technology. However, if you decide to pursue an engineering technology degree, note that graduate srudies in engineering technology are limited and registration as a professional engineer might be more difficult in some states. The engineering technology programs are also accredited by the Accreditation Board for Engineering and Technology (ABET). According o ABET, the Baccalaureate engineering technology programs must consist of a minimum of 124 semester hours or 186 quarter hours of credit. Associate degree programs (two-year) must have a minimum of 64 semester hours or 96 quatter hours of credit. Moreover, each engineering technologr program must have five components: communications, mathematics, physical and natural science, social sciences and humanities, and a technical content. The technical content of a particular engineering technolory prograrn is focused on the applied side ofscience and engineering, and is intended to dwelop the skills, knowledge, methods, procedures, and techniques associated wirh that particular technical discipline. According ro ABET, an accredited engineering technology program must demonstrate that their srudents at the time of graduations have:
.
an appropriate masrery of the knowledge, techniques, skills and modern tools of their
disciplines, an ability to apply current knowledge and adapt to emerging applicadons of mathematics, science, engineering and rcchnology, . an ability to conduct, analyze and inreqpret experiments and apply experimental results to improve processes, . an ability to apply creativity in the design of systems, components or Processes appropriate to program objectives, . an ability to function effectively on teams, . an ability to identity, analyze and solve technical problems, ' an ability to communicate effectively, . a recognirion ofthe need for, and an abiliry to engage in lifelong learning, ' an ability to understand professional, ethical and social responsibilities, ' a respect for diversiry and a knowledg" "f contemporary professional, societal and global issues, and n a commitment to quality, timeliness, and continuous improvement.
.
Therefore, these are the educational outcomes that are expeced of you when you graduate form an engineering technology program.
Pnorrxrvrs
u{fr
SUMMARY Now that you have reached this point in the text
.
.
.
You should have a good understanding of the significant role that engineers play in our everyday lives in providing water, food, shelter, and other essential needs. You should have a good idea of the common traits and activities of good engineers. Engineers are problem solvers. They possess good oral and written communication skills and hane a good grasp of fundamental physical laws and mathematics. They apply physical and chemical larvs and mathematics rc design, dwelop, test, and supervise the manufacurre of millions of products and services. They are good team players. They consider important factors such as efficiency, cost, reliabiliry, and safetywhen designing products. Engineers are dedicated to
lifelong leaming and service to others. You should be familiar with the dlfferences among various engineering disciplines. You now know tlrat civil, electrical, and mechanical engineers represent a large percenage of the total number ofengineers.
--I
lillrilgt:Jilt\ 1.1. Observe your o$/n sunoundings. !?'hat are some of the engineering achievemen* that you couldn't do without todarf
n.2. Using the Internet, find the appropriate organization for the following list of engineering disciplines. Depending on yolu personal interest, preparc a brief
24
Cnerrsn
1
lNrnopuc"rroN To rHE ExcrNsrRnTc PnornsstoN
two-page report about the goals and missions of the organization you have selected. Bioengineering Ceramic engineering Chemical engineering Civil engineering Computer engineering Electrical engineering Electronic engineering Environmental engineering Industrial engineering Manufacturing engineering Materials engineering Mechanical engineering
t.8.
.
t.3.
To increase public awareness about the importance of engineering and
1.4.
t.5.
t.6.
't.7.
1.9.
to promote engineering
education
and careers among the younger generadon, prepare aad give a l5-minute !7eb-based presentation at a mall in your town. You need to do some planning ahead of time and ask permission from the proper authorities. If your introducrion to engineering class has a term project, present your final work at a mdl at the date set by your instructor. If the project has a competitive component, hold the design competition at the mall. Prepare a 15-minute oral presentation about engineering and its various disciplines, and the nerc time you go home present it to the juniors in your high school. fuk your college engineering deparrment and engineering organizations on your campus to provide engineering-related brochures to take dong. This is a class project. Prepare a W'eb site for engineering and its various branches. Elect a group leader, then divide up the tasls among yourselves. As you work on the project, take note ofboth the pleasures and problems that arise from working in a team environment. 'W'rite a brief repon about your experiences regarding this project. \Vhat are your recommendations for others who may work on a similar project? This is a team project. Prepare a'!?'eb-based presentation of the history and future of engineering. Collect pictures, short videos, graphs, and so on. Provide linla to major engineering societies as well as to major research and development centers.
This is a dass project. Each of you is to ask his or her parents to rhink back to when they graduated from high school or college and to create a list ofproducts and services that are ariailable in their everyday lives now that were not arrailable to rhem then. Ask them if they ever imagined that these products and services would be arailable today. To get yoru parents stafted, here are few examples: cellular phones, AIM cards, personal computers, airbags in cars, price scanners at ttre supermarket, E-Z Passes for tolls, and so on. Ask yotu parenm to explain how these products have made their lives better (or worse). This is a class project. Each of you is to compile a list of products and services that are not available now that you thinkwill be readily available in the next 50 years. Compile a complete list and present it to the class. You can postyour findings on Problems 8 and 9 to my 'Web page, so that the rest of the country can look at
your results. Ll0.
ilt.
Perform a W'eb search to obtain information about the number of engineers employed by specific area and their mean salaries in recent years. Present your findings to your instrucor. If you are planning to study chemicd engineering, invesdgate what is meant by each of the following terms: polymers, plastics, thernoplastics, and thermnsetting.
Give at least ten examples of plastic products that are consumed every day, Write a brief repon explaining
yourfindings. r.12.
Ifyou
are planning
to study electrical engineering, in-
vestigate how electricity is generated and distributed. 'W'rite
brief report explaining your findings, Electric motors are found in many appliances and devices around your home. Identify at least ten producm at home that use electric motors. Idendfy at least 20 different materials that are used in a
il3.
il4.
various products at home. t.15.
Ifyou
planning to study civil engineering, investihad, i.rnpact had, wind load, and snow load in the desigrr of strucnrres. 'Vrite a brief memo to your instructor discussing your are
gate what is meant by dcad I,oad, liae
findings. t.t6.
This is a group prqect.As you c:rn see from our discussion of the engineering profession in *ris chapter, people rely quite heavily on engineers to provide them
PnoslErvrs with
safe and reliable goods and services. Moreover, you ralizr that there is no room for mistakes or dis-
honesry in engineering! Mistakes made by engineers could cost not only money but also more imporandy lives. Thinkabout the following: An incompetent and unethical surgeon could cause the death aof at most one pefson at one time, whereas an incompetent and unethical engineer could cause the deaths ofhundreds of people at one time. If in order to siwe mon€y an unethical engineer desrgns a bridge or a pan for an
25
aiqplane that does not meet the safety requirements, hundreds ofpeoples'Iives are at riskJ Visit the \9'eb site of the National Society of Professional Engineers and research engineering ethics. Discuss why engineering ethics is so important, and explain why engineers are orpected to practice engineering using the highest sandards of honesty and integntf. Give o
ri 0bjective: To design avehicle from the materials listed below and adhere to the following nrles.
"I find that the harder I work, the more luck I
26
Tbomas
Jfferson (1 743 - 1 826)
irj
seem
to havel'
CHAPTER
2
PnnpARrNG FoR AN ENcINEERING CAnEER aking the transition from high school to college requires
extra effort. ln order to have
a
rewarding education you should realize
that you must start studying and preparing from the first day of class, attend class regularly, get help right away, take good notes, select a good
study place, and form study groups. You should also consider the time
management ideas discussed in this
chapter to arrive at a reasonable weekly schedule. Your education is an expensive
investment. lnvest wisely.
27
28
Cnaprrn
2
PnrpenrNc FoRAN ENcrNBsRlrc Canren
In this
chapter, you
will
be introdwced to some aery important sugestions and idzas
that could, iffollowed, make your mgineering edacation more rewarding Read this section uery carefully, and think about how yoa can adapt the strategies to
get the optirnurn bmef.tfromyour collegeyears.
ffired
here
Ifyou encoanter difrcahy inyour
studies, reread this chapterfor ideas to help you rnaintai.n some selfdiscipline.
2.1
Making the Transition from High School to Colleqe You belong to an elite group of students now because you are studying to be an engineer. According to rhe Chronicb ofHigher Educatioa approimately only 5%a of students who gaduate with a B.S. degree ftom universities and colleges across the Unircd States are engineers. You will be taught to look at your surroundings differendy than other people do. You udll learn how to ask questions to find out how things are made, how things work, how to improve things, how to design something from scratch, and how to taLe an idea &om paper to reality
and acually build something. Some of you may be on Four own for the veryfirst time. Making the transition from high school to college may be a big step for you. Keep in mind that what you do for the next four or five years will affect you for the rest of your life. Remember that how successfirl and.happy you are will depend primarily on you. You must take the responsibiliry for learning; nobody can make you learn. Depending on which high school you a*ended, you may not have had to study much to get good gades. In high school, most of your learning took place in class. In contrast, in college, most ofyour education akes place outside the class. Thereforg yo& may need to develop some new habits and get rid of some ofyour old habits in order to thrive as an engineering student The rest of this chapter presents sugescions and ideas that will help you make your college experience successfirl. Consider these suggestions and try to adapt them to yoru own unique situation.
2.2
Budgeting Your Time Each of us has the same 24 hours in a day, and there is only so much that a person can do on an average daily basis to accomplish certain thirS. Many of us need approximately 8 hours of sleep every night. In addition, we all need to have some time for work, friends and family, studying relaxation and recreation, and just goofing around. Suppose you were given a million dollars when you reached your adulthood and were told that is all the money that you would have for the rest of your life for clothing, food, entenainment, leisure activities, and so on. How would you go about spending the money? Of course, you would make reasonable efforts to spend and invest it wisely. You would carefirlly budget for
various needs, trying to get the most for your money. You would look for good sales and plan to buy only what was necessary, and you would atempt not to waste any money. Think ofyour education in a similar manner. Don't just payyour tuition and plan to sit in class and daydream. Your education is an expensive investment, one that requires your responsible management.
2.2
BupcsrrNc Youn Trtvrs
29
A srudent ar a private university went to her instructor to drop a class because she was not getting the grade she wanted. The instructor asked her how much she had paid for the class. She said tlat she had spent approximately $2000 for the four-credit class. The instructor happened to have a laptop computer on his desk and asked her the following question: If you bought a laptop computer from a computer store, took it home to install some software on it, and had some difficulty making the computer work, would you throw it in the trash? The student looked at her instructor as ifhe had asked a stupid question. He explained to her that her dropping a class she had already paid tuition for is similar in many ways to throwing ail'ray a computer the first time she has trouble with some software. Try to learn from this example. Generally speaking, for most of us learning is a lot of work at the beginning and itt not much fun. But often, after even a short period of time, learning will become a joy, something you work at that raises your own self-esteem. Learning and understanding new things can be downright exciting. Let us examine what you can do to enhance your learning during the next few years to make the engineering education you are about to receive a firlfilling and rewarding o
'!?ith
24 hours in a given day, we have, for a one-week period, 168 hours arailable.
Let's allocate liberal time periods to some activities common to most students. Refer to Thble 2.1,
when following this example. Notice that the time periods allocated to various activities in this able are very generous and you dont have to deprive yourselfofsleep or relaxation or socializing with your friends. These numbers are meant only to give you a reasonable staning point to help you budget your time on a weekly basis. You may prefer to spend an hour a day relaxing during the week and use the additional social hours on weekends. Even with generous relaxation and social time, this sample allows 68 hours a week to dwote to your education. A typical engineering student takes 1 6 semester credits, which simply means about 16 hours a week are spent in the classroom. You still have 52 hours a week to s$dy. A good nrle of thumb is to spend at lea$t 2 to 3 hours of studlng for each hour of class time, which amounts to at least 32 hours and ar most 48 hours a week ofstudying. Of course, some classes are more demanding than others and will require more time for preparation and homework, projects, and lab work. You still have from 4 ro 20 hours a week in your budget to allocate at yoru own discretion. You may not be an l8-year-old freshman whose parents are psytngmost of your tuition. You may be an older student who is changing cueers. Or you may be married and have children, so you mu$ have at least a part-time job. In this case, obviously you will have to cut back
{rliiji,, i
Activity
x (8 hourslday) Cooking and aung: Q days/week) X (3 hours/day) Grocery shopping Personal grooming: (7 daysiweek) X (1 hour/day) Spending time with family, lgirl/boy) friends, relaxing, playing sports, exercising watchingTV: (7 daln/weeD X (2 hours/day) Sleeping: (7 days/week)
Totat
li'liliri ff_'' RequiredTime perVbek
i6
(hours/week)
: 2l (hoursiweek) = 2 (hours/weeD 7 (hours/week)
=
14 (houn/week)
=
100 (hours/week)
-
Hour
Monday
Tuedef
Wednesday
Thor.daf
F id"y
Satffday
Soodry
7-8
Showt/
Shower/ Drcss/
Shower/
Showerl
Shower/
Exaa sleep
Exaa sleep
Dress/
Dras/
Dras/
Dress/
Breakfaet
Breakfast
Breakfast
Breakfast
Breakfast
CALCT]LUS CLASS
CALCULUS
St"dy
CALCULUS
English
cLrss
Showerl Dress/
Sbown/
CLASS
Breabfast
Breahfast
ENGI-ISH ",
Grocery
CTASS
shopping
OPEN TIOUR
Srudy Calculus
Grocery shopping
HOI]R
8-9
9-10
10-ll
ENGLISH
Studylntro.
ENGLISH
CLASS
to F.ng,
.CLASS
Study Intro. to Eng.
Study
INTRO"
Calculus
TOENG.
Stody H/SS
INTRO. TOENG.
t2-l t-2 2-3
t-4 4-5
Study Chemistry
H/SS CLASS
Relax
crass
Study Chemistry
Exercise
crJlss Lunch
Lunch
Lunch
Lunc,h
Lunch
Lunch
Lunch
CHEM.
Stody
CHEI\{.
Relax
Calculus
clJtss
Study Intro. to Fng.
Relax
cLtss Study H/SS
CHEM. I.AB
Study H/SS
Study Intro. to Eng.
Study Calcufts
cHEwt. I-{TB
Study H/SS
Exercise
CHEM.
Exercise
Exercise
Dinner
H/SS
H/SS
CHEIT,I;,.,,
5-6
Dinner
Dinner
6-7
Study Chemistry
Study
Study
C*alculqs
Calculus
Study Chemistry
Study
Study
Calculus
C,alculus
Study Intro.
Study
StudyInto.
to Frg.
Chemistrf
to E rs.
Stody
Study Chemistry
8-9
9-r0
English
l0-lt
30
,
,elA$$ti1ffi Study
Study H/SS
Stody
Study
Study Calculus
Study Chemistry
Snrdy
Calcrrlus
Exercise
Study Chemistry
Stndy Chemistry
Dinner
Dinner
Dinner
Relax
Study Calctlus
Relax
Relax
Dinner
Study
$tudylntro.
IAB
7-8
OPEN
crdss
CLASS
tt-12
.:,
Dress/
Study Chemistry
Calculus
English
Cdculus Srudy Calcrrlus
Recreation
Recreation
Strdy
Study
Recreation
Recreation
English
English
Relax/Get
Relar/Get
RelaxlGet
Relax/Get
ready for bed
ready for bed
ready for bed
$tudf
, to Eng.
Stody Chemictry
ready for bed
H/SS
$tndy Engfish
Recreation
Recreation
Relax/Get ready for bed
2,3
Darr-v SruovrNc AND
PnBpenerron 3l
some areas. For example, you may want to consider not mking as many credits in a given follow a five-year plan instead of a four-year plan. Depending on how many hours a week you need to work, you crn rebudget your time. The puqpose of this time budget example is mainly to emphasize the fact that you need to learn to manage your time wisely ifyou want to be successfirl in life. Every individual, just like any good organizadon, monitors his or her resources. No one wants you to flun into a robot and time yourself rc the second. These examples are provided to give you an idea of how much time is available to you and to urge you ro consider how efficiendy and wiselyyou are allocating and using your time. The point is that
in
semester and
budgedng your time is very important. \fith the exception of a few courses, most classes that you will take are scheduled for 50-minute periods, with a lO-minute break between classes to allow students to attend several classes in a row. The other important reason for having a lO-minute brek is to allow dme to clear your head. Most of us have a limited attention span and c:rnnot concentrate on a certain topic for a long period of time without a break. Tirking a break is healthy; it keeps your mind and body working well. Typi"rtly, as a fust-term freshman in engineeringyou mayhave a course load similar to the one shown here:
Chemistry (3) Chemistry l^ab (1)
Introduction to Engineering (2) Calculus (4) English Composition (3) Humanities /Social Science electives (3) Table 2.2 is an example of a schedule for a freshman engineering student. You already know your suengths and weaknesses; you may have to make sweral afiempts to arrive at a good schedule that will fit your needs the best. You m47 also need to modifr the example schedule shown to allow for any variabiliry in the number of credits or other engineering program requirements at your particular school. Maintain a daily logbook to keep track of how closely you are following rhe schedule and where time is being used inefficiently, and modifr your schedule accordingly.
2.3
Daily Studying and Preparation You sart sndying and preparingfrorn the uery frst day of classlIt|n always a good idea to read the material that your professor is planning to cover in class ahead of time. This practice will improve both your understanding and retention of the lecnrre materials. It is also impoftant to go over the material that was discussed in class again later the same day after the lecure was given. \Vhen you are reading the material ahead of a lecnre, you are familiarizing yourself with the information that the instructor will present to you in class. Dorft worry if you dorit firlly understand wery*ring you are reading at that time. During the lecrure you c:ur focus on t*re material that you did not firlly understand and ask questions.'S?'hen you go over the material after the lecture, werything then should come together. Remember to read before the class and study the material afur the class on the same day!
3?
Cnaprpn2
PnrperuNc ronaNENcrNurnrNc Car,Esn
Read the material your professor is planning to cover in class ahead of time
Attend Your elasses Regularly Yes, even ifyour professor is a bore, you c:m still learn a great deal from arending the dass. Your professor may offer additional explanations and discussion of some material that may not be well presented in your textbook Moreover, you can ask questions in class. If you hane read the material before class and have made some notes about the conceprs thar you do not firlly understand, during the lecture you c:m ask questions to clarifr any misunderstanding. If you need more help, then go to your professor's office and ask for
additional assistance. Get Help Right
Away
$7'hen you need some help, don t wait
till
the last minute to ask! Your pro-
fessor should have his or her office hours posted on the office door or be available on the Web.
The office hours are generally stated in r.he course syllabus. If for some reason you cannot see your professor during the designated office hours, ask for an appointment. Almost all professors are glad to sit down with you and help you out if you make an appointment with them. After you hane made an appointment, be on time and have your questions written down so that you remember what to ask. Once again, remember that most professors do not want you to wait until the last minute to get help! Any professor can tell you some stories about experiences with students who procrastinate. Recendy, I had a student who sent an e-mail to me on Sunday night at 10 : 05 p.rra. asking for an extension on a homework assignment that was due on the following day. I asked the student the ner(t mornin& "\Vhen did you start to do the assignment?" He replied, "at 10 : 00 r'.rr'r. Sunday night." On another occasion, I had a student who came to my office and introduced himself to me for the fust time. He asked me to write him a recommendation lerer for a summer job he
2.3 Derrv
SruovrNc AND PnnpenerroN
33
was applying for. Like most professors, I do not write recommendation lemers for students, or anyone, whom I do not acrually know. Get to know your professors and visit them often!
Notes
Everyone knows that it is a good idea to take notes during lecflrre, but some realize students may not that they should also take notes when reading the textbook" Try to [isten carefirlly during r:he lectures so you c:m identify and record the important ideas and concepts. If you have read ahead of dme the text materials that your proGssor is planning to cover in class, then you are prepared to write down notes that complement what is already in your terbook. You dorft need to write down werything tlat your professor says, wdtes, or projects onto a screen. The point is to listen very carefrrlly and write down only notes regarding the impoftant concepts that you did not understand when you read the book. Use wirebound notebools for your notes. Don't use loose papers, because it is too easy to lose some of your notes that way. Keeping a notebook is a good habit to dwelop now. As an engineer you will need m keep records of meetings, calculations, measurements, etc. with time and date recorded so you c:rn refer back to them ifthe need arises. Thus it is best that you keep the notes in a wirebound notepad or a notebook with the p€es sown into the binding so you won't lose any pages. Study your notes for at least an hour or two the same day you take them. Make sure you understand all the concepts and ideas that were discussed in class before you atTake Good
tempt ro do your homework assignment. This approach will save you a great deal of time in the long run! Dort't be among those students who spend as litde time as possible on undersanding the underlying concepts and try to take a shoncut by finding an example problem in the book similar to the assigned homework problem. You may be able to do the homework problem but you worit develop an understanding of the material. lTitlout a firm grasp of the
34
Cnaprsn
2
PnrpenrNc FoR AN ENcrNeERrNc Cenrsn
Not a good way to study!
will not do well on the exams, and you will be at a disadvantage later in and in life when you practice engineering. Tirke good legible notes so that you can go back to rhem later if you need to refresh your memory before exam time. Most of today's engineering textbooks provide a blank margin on the left and right side of each sheet. Don't be afraid to write in these margins as you studyyour book. Keep all your engineering books; dont sell them back to the bookstore-some day you may need them. If you have a computer, you may want to type up a slrmmary sheet of all important concepts. later, you can use the Find, command to look up selected terms and concepts. You maywish to insen links between related concepts in your notes. Digital notes may take some extra time to type, but they can save you time in the long run when you search for information. basic concepts, you
your other
classes
Select a Good Study Place You may already know that you should study in a comfortable place with good lighting. You do not want any distractions while you are studying. For example, you do not want to study in front of a TV while watching your favorite situation comedy. A library is certainlyagoodplace forstudying, butyou can makeyour dormroom oryourapaftmentroom into agoodplaceforsrudfng. lllk toyourroommate(s) aboutyourstudyhabits andstudytime. Explain to them that you prefer to study in your own room and appreciate not being disturbed while studying. If possible, find a roommate with a declared engineering major, who is likely to be more understandrng ofyour study needs. Remember that a bad place for studying is in your
girlftiendt or boyfriend's lap or arms (or any other acceptable engineering configuration). Anodrer usefirl idea is to keep your desk clean and avoid having apicnrre ofyoursweethean in front of you. You dorit want to daydream as you are studying. There is plenry of time for that later. Form Study Groups Your professor will be the first person to tell you that the best way to learn something is to teach it. In order to teach something though, you have to fust understand the basic concepts. You need ro study on your own first and then get together with your classmates
2.3 Danv SrupyrNc
lNp PnrranenoN
35
. - &" -'': A good way to study!
to discuss and explain key ideas and concepts to each other. Everyone in your sftdy group should agree drat they need to come prepared to discuss appropriate materials and rhat they dl need to contribute to the discussions. It should be understood that the srudy groups serve a different purpose than that ofa tutoring session. However, ifanother student in your class asks for assistance, help ifyou can by explaining new concepts you have learned. Ifyou have difficulty explaining the material to someone else, that could be an indication you dont firlly undersand the concept yourselfand you need to study the material in more detail. So remember that a good way to learn something is to form discussion groups where you explain ideas and concepts in your own words to others in the group. Be an actiue bamn not apassiue leamn! Prepare for Examinations Ifyou study and prepare from the first day ofclass, then you should perform admirably on your exams. Keep reminding yourselft'hat there is absolutely no substitution for dailystudying. Dorftwait until *re night before the exam to study! That is not the best time for learning new concepa and ideas. The night before the exam is the time for rwiew only. Just before an exarn, spend a few hours rwiewing your notes and sample problems. Make sure that you get a good nightt rest so you can be ftesh and think clearly when you take the enam. It may be a good idea to ask your insrructor ahead of time what rype of exam it will be, how many questions there will be, or what susestions she or he has to help you prepare better for the e:
36
Cneprsn
2
PnspARn{c FoR AN ENcrrrnnrNc Cennsn
AIChE
ffiWffiffi-
,emolfcnrWoffofW@dwre ,.1ii.ii'i:1.,
GIE
ff,
Join an engineering organization!
2.4
Getting lnvolved with an Engineering 0rganization There are many good reasons to join an engineering organization. Networking, panicipating in plant tours, listening to technical guest speakers, participating in design competitions, attending social events, .eking advantage of learning oppornrnities *gough shoft courses, seminars, and conferences, and obtaining student loans and scholarships are a few of the common benefits of belonging to an engineering organization. Moreover, good places to learn more about areas of specialization in engineering are the'Web sites of various engineering organiz* tions. As you spend a litde time reading about these organizations, you will discover that many share common interests and provide some overlapping services that could be used by engineers of various disciplines. You will also note that the primary plrrpose of these professional engineering organizations is to offer the following benefits:
l.
They conduct conferences and meetings to share new ideas and findings in research and development.
2. They publish technical journals, books, reports, and
magazines rc help engineers in partic-
ular specialties keep up-to-date. courses on curent technical developments to keep practicing engineers abreast of the new developments in their respective fields. They advise the federal and state governments on technology-related public policies. Th.y create, maintain, and distribute codes and standards that deal rrith correct engineering dCIign practices to ensure public safety.
3. They offer short 4. 5.
6. They provide
a networking mechanism through which you get to know people from different companies and institutions. This is imponant for two rq$ons: (1) If there is a problem that you feel requires assistance from ou*ide your organization, you have a pool ofcolleagues whom you have met at the meedngs to help you solve the problem. (2) \Vhen you know people in other companies who are looking for good engineers to hire and you are thinking about something different to do, then you may be able to 6nd a good match.
Find out about the local student chapters of nadonal engineering organizations on your qrmpus. Anend the fust meetings. After collecting information, choose an organization, join, and become an active particrpant. As you will see for yourself, the benefits of being a member of an engineering organizadon are great!
2.6 OrnrnConsronnpmoNs
2,5
37
Your Graduation Plan schools, there are three levels of admissions. First, you get admined to the university. For that to happen, you must meet certain requirernents. For example, you need to rank in the top xo/o of your high school class, have an ACT or SAI score of rooc, and hatre so many years of english, mathematics, sciences, and social studies. After you complete your freshman year, you
At most
may need to apply to the college in which the engineering program of your interest resides. To be accepted ro, say, the college of engineering at yoru university, you need to meet additional requirement. Finally, at the end ofthe second year, upon successfirl completion ofmath, chemistry, phpics, and basic engineering classes, then you need to apply and gain admission to a specific engineering program, for example, civil, electrical, mechanical, and so on. Make sure you meet with your advisor so that you understand what the requirements are for admission to the college and the specific program, becau$e at many universities, admission to an engineering program is highly selective. It is also a good idea to sit down with your advisor and plan your graduation. List all of the classes that you need to take in order to obtain your degree in four or five years. You can alwap modify this plan later as your interests change. Make sure you understand rhe prerequisites for each class and in which semester a class t1'pically is offered; a program flowchart will be quite usefirl. In order to make you aware ofyour social responsibfities as an engineer, you also are required to take a certain number of classes in social sciences and humanities. Think about your current interests and plan your social science and humanity electives as well. Again, dort'tworry about your inceresa changing, you can always modiS your plan. For those ofyou who currendy are studying at a community college and planning to rransfer to a university later, you should contact the university, learn about their engineering course ffansfer policies and requirements, and prepare your graduadon plan accordingly.
2.6
0ther Considerations Work Ifyour study
schedule allows, volunteer for a few hours a week to help those in need in your community. The rewards are unbelievable! Not only will you feel good about yourself but you will gain a sense of satisfacdon and feel connected to your community. Doing Volunteer
'\Iolunteering
could also help dwelop communication, management, or supervisory skills that you may not dwelop by just attending school. Vote in Local and National Elections Most of you are 18 or older. Thke your civic duties seri ously. Exercise your right to vote, and q'to play an active role in your local, state, or federal govemment. Remember that freedom is not free. Be a good responsible citizen, Get to Know Your Classmates There are many good reasons for getting to knorn' one or two other students in your classes. You may want to study with someone from class, or if you are absent from class, you hane someone to contact to find out what the assignment is or find out what was covered in class. Record the following information on the course syllabi for all your classes: the name of a sudent sitting next to you, his or her telephone number, and his or her
e-mail address.
38
Crg.epran2 PnspARrNc roRAN Encnrnrnnrc CenBrn Get to Know an Upper'Division Engineering Student Becomingacquaintedwith junior andsenior engineering students can provide you with valuable information about their engineering education ercperience and campus social issues. Ask your instructor to inuoduce you to a junior or a senior engineering student, and record the following information on your inuoduction to engineering dass syllabus: the name of an upper-division engineering student, his or her telephone number, and his or her e-mail address.
,r*"",-"* gram,
I
"i
I discovered tJrat
b,
to stand ous aod to tionship with my pq
paledmrnerhemorr once said, "the job
of
withlife, comfo* nicheiq thefieldof the development o
mewith
my
gar-
eet as an employee of the
So why not become an engineer? Environmental engineering has been my field of erpertise for the past ventilarion problems for rural .orrrt ooiorr, drainage, and dients. Afur working for seven years, I decided to go back 20 years. It allows me to inregrate all of the sciences and to universiry to complete a Master's degree while keeping to comb up with solutions to problems concerning our
:LH"ffiitJ;[ffi*15'"'ffffi#.1#r "t*ffff;-#:k frfli+ffiilfi
:frffi;Sffi:'i'"rffi:f:'*':T:;
neering is all about tadng your imagination run wild. gineers are probrem
*r"*i;;'.t;;;;'^dil
En-
t*-eup*idrinnovariverechnorogiestharwu;*:jH
i::"_1T1!l_l,l-y_":::1*:::':*:ry:g
is that solutions must now integrate some social aspects.
are given
ro ser a reer ror environmentar
*.
urin
l:: ;:.t* *o
.1.":; I?,"
jou,o
wil
Llc
design
a'
anaerobic_{i_
Pru.,u'c.' uy urE I'uPua-
vour friends can collect' for one
Xffh,**.i#trj;ff*:J:f"'#ffi
*"ffia,-*eeachoneoryo*r,i*J?.rar," *"rt. into a container of
produced. Put this food
If
solution requires too many changes in someone's life or roo much worh it will not be adopted. For erample, I worked with a smdl communiry in the hean of Montreal cenpr, lFe to " *"d. thi, ."nter .apily visible and accessibG to all rnembers of the mmmuniry: th.y could drop off their food waste on the way to work three rnornings per week, and being visible, the center drew customirs in no time. The project worked where many had friled before, becaus€ y/e a disunce out or'lheir a
Y";T*'co
T"*r':"*P**F
turbtdity of the both tbe
:#:J::f.H:i,l#*:$k
39
Cser'rrn2
40
Pnnpmrnc FoRAN ENcrNsnRrNc CnneEn
SUMMARV Now that you hane reached this point in the text
. . .
21.
You should use the ideas discussed in this chapter to make your ffansidon from high school to college smoothly. You should also consider the time management ideas discussed in rhis
chapter to arrive at a reasonable weekly schedule. You should realize that you must start studying and preparing from the first day of class, acend classes regularly, get help right avay, uke good notes, select a good study place, and
form srudy groups. You should knorv the imporance of joining an engineering organization and choose an organization, join, and become an active participant.
Prepare a schedule for the cunent semester; also prepare two additional alternative schedules. Discuss the pros and cons ofeach schedule. Select what you think is the best schedule, and discuss it with your instructor or advisor. Consider his or her suggestions and modtfy .h. schedule if necessary; then present the final schedule ro your instructor Maintain a daily logbook
to keep track of how closely you are following the schedule and where time is being used inefficiendy. Hour
Monday
Tuesday
Vednesday
'Write a one-page summary discussing each week's activities that deviated &om the planned schedule, and come up with ways to improve or modifr its shoncomings. Turn in a biweekly summary report to yoru
instructor or advisor. Think of this exercise as an ongoing test similar to other tests that engineers perform regulady to understand and improve things. 2.2. Meet with your advisor and prepare your graduation plan, as discussed in Section 2.5.
Thursday
Ftidry
7-8 8-9
9-
0
10- I
1l-
2
11
L-2
2-3
3-4 zl-)
5-6 6-7 7-8 8-9 9-10
l0-1
1
"I
have never let my schooling interfere
-Marh
Twain
(1 835
- I 9 I 0)
with my education."
Saturday
Su"dty
CHAPTER
INrnoDUcTroN To ENcINEERING DBSIGN
ngineers, regardless of their backgrounds, follow certain steps when designing the products and services we use in our everyday lives. These steps are: (1) recognizing the need
for a product or service, (2) defining and understanding the problem (the need) completely, (3) doing the preliminary research and preparation, (4) conceptualizing ideas for possible
solutions, (5) synthesizing the results, (6) evaluating good ideas in more detail, (7) optimizing the result to arrive at the best possible solution, and (8) presenting
the solution.
42
Cneprrn3 lxrnooucrroN To ENcrNsBRrNc DnsrcN In this chapter, we will introduce )/ou to the mgineering daign process. ,4s ule discussed in Chapter 1, mgineerc appfu pfusical and chemical laws and principles and rnathernatics to design ntillions ofproduas and seruices that we use in our eueryday liues. Here we will looh rnore closely at what the term dcign means and leam more about how engineas go about designing these produas and seruices. We will discuss the basic steps that most engi.neCIs follow when dcsignEngineers are problern solaers.
ing sornething. We will abo innoduce you to the econornic consid.erations, material selection, team worh, project scheduling and engineering standards and
an integralpart ofthe
d.esign process
codes-all
andprodua and seruice dcuelopmmt.
3.1 Engineering Design Proecss Let us begin by emphasizing what we said in Chapter 1 about what engineers do. Engineers apply physical laws, chemical laws and principles, and mathematics to dcign mtllions of products and services thatwe use in our weryday lives. These produca include cars, computers, aircraft, clothing, toys, home appliances, surgical equipment, heating and cooling equipment, heal*r care dwices, tools and machines that make various products, and so on. Engineers consider important factors such as cost, efficiency, reliabiliry, and safetywhen designing the products, and they perform rests to make certain that dre products drey design withsand various loads and conditions. Engineers are continuously searching for ways to improve already exisdng products as well. Engineers also fusign and supervise the constnrction of buildings, dams, highways, and mass transit sysrcms. They also fusign and supervise the construction of power plants that supply power to manufacnrring companies, homes, and offices. Engineers play a significant role in the dcsign and maintenance of nations' infrastrucnrres, including communicadon s)rstems, udlities, and transportation. They continuously dwelop new advanced materials to make products lighter and sffonger for different applications. Engineers are also responsible for findtng suitable ways and designingthe necessary equipment to extract peuoleum, natural gas, and rarnr materials from the earth. Let us now look more closely at what constitutes rhe dcsign procsss. These are the basic steps that engineers, regardless of their badrground, follow to arrive at solutions to problems. The steps include: (1) recognizing the need for aproduct or a service, (2) defining and undersanding the problem (the need) completely, (3) doing preliminary research and preparation, (4) conceptuali"ing ideas for possible solutions, (5) synfiesizing the findings, (Q evaluating good ideas in more detail, (7) opdmizing soludons to arrive at the best possible solution, (8) and presenting the final soludon. Keep in mind that these steps, which we will drscuss soon, are not independent of one another and do not necessarily follow one another in the order in which they are presented here. In fact, engineers often need to return to steps 1 and 2 when clienn decide to change design parameters. Quite often, engineers are also required to give ord and written progress reports on a regular time basis. Therefore, be aware of the fact that wen though we listed presentation of the design process as step 8, it could well be an integral part of many other design steps. Let us now take a closer look at each step, sarting with the need for a product or a service.
3.1 ExerNsERu{c Step
l:
DssrcN
Pnocsss
43
Recognizing the Need for a Product or a Service
All you have ro do is look around to realize the large number of producs and servicesdesigned by engineers-rhatyou use every day. Most often, we take these produca and services for granted until, for some reason, there is an interruption in the services they provide, Some of these existing producrc are constandy being modified to take advantage ofnew technologies. For example, cars and home appliances are constantly being redesigned to incoqporate new technologies. In addition to rhe products and the services already in use, new producs are being dweloped every day for the pupose of making our lives more comfonable, more pleasurable, and less laborious. There is also that old saying that "wery time someone complains about a situation, or about a ask, or curses a product, right there, there is an oppornrnity for a product or a service.o As you can tell, the need for products and services exists; what one has to do is to identifr them. The need may be identified by you, the company that you may eventually work for, or by a third-parry client who needs a solution to a problem or a new product to make what
it
does easier and more efficient.
Step 2: Problem Definition and Understanding One of the 6rst things you need to do as a design engineer is to firlly understand the problem. This is the rnost important step in any fusign process. If you do not hane a good grasp of what the problem is or ofwhar the client wants, you will not come up with a solution that is relevant to the need of rhe client. The best way to firlly understand a problem is by asking many questions. You may ask the client questions such as: How much money are you willing to spend on this project? Are there restricdons on the size or r*re type of materials that can be used? When do you need the product or the sewice? How many of these products do you need? Questions
44
Crraprrn
3
lrrnooucrroN To ENcrNsERrxc DssrcN often lead to more questions that will better define the problem. Moreover, keep in mind that engineers generally work in a team environment where they consult each other to solve complex problems. They divide up the task into smaller, manageable problems among themselves; consequendy, productive engineers must be good team players. Good inteqpersonal and communication skills are increasingly important now because of rhe global market You need ro make sure you clearly understand your ponion of the problem and how it fia with the other problems. For er
SteB 3: Research and Preparation Once you firlly undersand the problem, ,$ a next step you need to collect usefirl information. Generally speaki ng a good place to staft is by searching to determine if a product dready exists that closely meets the need of your client. Perhaps a product, or components of a product, already has been dweloped by your company that you could modifr to meet the need. You do not want to "reinvent the wheel"!As mentioned earlier, depending on the scope, some projects require collaboration with other companies, so you need to find out what is available through these other companies as well. Try to colled as much inforrnation as you can. This is where you spend lots oftime not onlywith the client but also with other engineers and technicians. Interner search engines are becoming increasingly imporant tools to gather such information. Once you have collected all peninent information, you must then reviewit and organize it in a suitable manner.
Step 4; eoneeptuolizatism During this
phase of desigrr, you need to generate some ideas or concepts that could offer reasonable solutions to yoru problem. In other words, without performing any detailed analysis, you need to come up with some possible ways of solving the problem. You need to be creative and perhaps develop several alternative solutions. Ac this sage ofdesign, you do not need to rule out any reasonable working concept. If the problem consists of a complex system, you need to identify the components of the qrstem. You do not need to look at details of each possible solution yet, but you need to perform enough analSrsis to see whether the concepts that you are proposing hane merit. Simply stated, you need to ask yourself the fo[owing question: I?buld the concepts be likely to work if they were pursued funher? Throughout the design process, you must also learn to budget your time. Good engineers hane time-management skills that enable them to work productivd and efficiendy. You must learn to create a milestone
chan detailing your time plan for completing the project. You need to show the time periods and the corresponding ssls that are to be performed during these time periods.
Step 5: Synthesis Recall from our discussion in Chapter 1 that good engrneers have a firm grasp of the fundamental principles of engineering, which they can use to solve many different problems. Good engineers are analyticd, detailed oriented, and crearive. During this sage ofdesign, you begin
3.1 ENcrNenRrNe Desrcx Procsss
45
to consider deails. You need to perform calculations, nrn computer models, narrow down the type of materials to be used, size the componens of the qfstem, and answer questions about how the producr is going to be fabricated. You will consult pertinent codes and standards and make sure thatyour designwill be in compliancewith these codes andstandards. ITewill discuss engineering codes and standards in Section 3.10.
Step 6: Evaluation Analyze the problem in more desil. You may have to identify critical design parameters and consider their infuence in your find design. At this stage, you need to make sure that dl calculations are performed correcdy. If there are some uncertainties in your analpis, you must 'S7hen
possible, working models must be created and perform experimental invesdgation. tested. At this stage of the design procedure, the best soludon must be identified from alternatives. Details of how the product is to be fabricated must be worked out firlly.
Step 7:0ptimization Optimization means rninimization or maximization. There are two broad types of design: a functional design and an optimized design. A funcdonal design is one that meets all of the preestablished design requiremenrc but allows for improvement to be made in cenain areas. To better understand the concept of a functional design, we will consider an example. Let us as-
l0-foot-tdl
ladder to support a person who werghs 1335 newtons (300 pounds) with a cerain factor of safety. \7'e will come up with a daign tfiat consists of a steel ladder that is 10 feet tall and can safely support the load of 1335 N (300 lb) at each step.
sume rhar we are ro design a
The ladder would cost a certain amount of money. This design would satisfy all of the requiremen$, including those of strength and size, and thus constinrtes a functional doigo. Before we can consider improving our design, we need to ask ourselves what criterion we should use ro optimize the design. Design optimization is always based on some particular crircrion, such as cost, srength, size, weight, reliability, noise, or performance. If we use the weighr as an optimizarion criterion, then the problem becomes one of minimizing the weight of the ladder without jeopardizing its strengt}. For example, we mqy consider making the ladder from aluminum. !7'e would also perform suess analysis on the new ladder to see ifwe could remove material from certain sections of the ladder without compromising the loading and safety requirements. Another imporranr fact rc keep in mind is rhar optimizing individual componens of an engineering qystem does not necessarily lead to an optimized system. For example, consider a thermal-fuid system such as a reftigerator. Optimizing *re individual componenr independendy-such as the compressor, the evaporator, or the condenser-with respect to some criterion does not lead to an optimized overall system (refrigerator). Traditionally, improvements in a design come from the process of starting with an initial design, performing an analysis, looking at results, and deciding whether or not we can improve the initial design. This procedure is shown in Figure 3,1. In the past few decades, the
optimization process has grown into a discipline that ranges from linear to nonlinear programming techniques. As is the case with any discipline, the optimization field has its own terminology. There are advanced classes that you can take to learn more about the design opti. mization process.
46
Cnaprun
3
IxrnopucrroN To ENcrNrsRrNc DssrcN
Can the design be improved?
il
riguret.t
An optimization procedure.
Step 8: Presentation a final solution, you need to communicate your soludon ro rle clienr, who may be your boss, another group within your company, or an outside customer. You may have to prepare not only an oral presentadon but also a wriften repoft. As we said in Chapter 1, engineers are required to write reports. Depending on the size of the project, these repora might be lengthy, detailed technical repofts containing graphs, charts, and engineering drawings, or
Now that you have
they may take the form of a brief memorandum or executive summaries. A reminder again that although we have listed the presentation as step 8 of the design process, quite often engineers are required to give oral and wrinen progress reports on a regular time basis to various groups. Consequendy, presentation could well be an inregral part of many other design steps, Because of the imporance of communication, we have devoted an entite chapter to engineering communication (See Chapter 4). FinaIIy, recall &om our discussion in Chapter i rcgarding the attributes ofgood engineers, we said that good engineers have written and oral communication skills that equip them to work well with their colleagues and to convey their expertise to a wide range of clients. Moreover, engineers hare good "people skills" that allow them to interact and communicate effectively with various people in their organization. For example, they are able rc communicate equally well with the sales and marketing experts and with their own colleagues in engineering. In step 7 of the design process, we discussed optimization. Let us now use the next enample to introduce you to some of the fundamental concepc of optimization and its rerminology.
3.1
EucrNnpntxc Dssrcx Pnocrss
47
Assume that you hane been asked ro look into purchasing some storage tanla for your company, and for the purchase ofthese tanlcs, you are given a budget of$ I 680. After some research, you find two tank manufacnrrers that meet your requirements. From Manufacrurer Ar /ou arn purchase I6-f€-capacity tanks that co$ $120 each. Moreover, ttre type of tank requires a floor space of 7 .5 ff. Manufacnrrer B makes 24-ff -capacity tanla that cost $240 each and that reof floor quir. a floor space of 10 ff . The snks will be ptaced in a section of a lab that has 90 the budgetcapacity within space available for storage. You are looking for thg greatest storage
fl
ary and foor-space limitations. How many of each tank must you purchase? First, we need to define the objeaiue functioa which is the funaion that we will attempt to minimize or maximize. In this example, we want to maximize storage capacity. \7'e can represent this requirement mathemaucallv as
maximize
Z:
l6x1
+ 24a
(3J)
subject to the following constraints: L20x1
-t 24Ax2=
7.5\ t 70x2<
1680
(3.2)
90
(3.3)
420
(3.4)
x2>
(3.5)
0
In Equation (3.I), Z is the objective funcdon, while the variables q and x2 arc called design 'and represent the number of 16-ff-capacity tanks and the number of 24-f€uariabl,es, apacity tanlis, respecively. The limitations imposed by the inequalities in Equations (3.2)(3.5) *e referred ro as a ser of cowtraints, Although there are specific techniques that deal with solving linear programming problems (the objective function and constrarnts are linear), we will solve this problem graphically to illustate some additional concepts. Ler us fust rwiew how you would plot the regions given by the inequalities. For example, to plot rhe region as given by the linear inequality 120.:c1 t 24Ax2< 1680, we must first Plot the line 1204 -t 240x2: 1680 and then determine which side of the line represents the region. For example, after plotting the line I2Ox1 i 240x2 = 1680, we can test points 11 : 0 and xz : 0 to see if they fall inside the ineguality region; because substitution of these points into the inequality sadsfies the inequdity, that is, (120X0) + (240)(0) < 1680, the shaded region represen$ the given inequality (see Figure 3,2(a)). Note that if we were to substitute a set of points ouaide the region, such ?s xl : 75 and x2 : 0, into the inequality, we would find that the
inequalityisnorsatisfied.TheinequalitiesinEquations (3,2)-(3., areplottedinFigure3.2(b). The shaded region shown in Figure 3.2(b) is called afeasible solution region. Every point within this region sadsfies the constraints. However, our goal is to maximize the objective function given by Equation (3.1). Therefore, we need to move the objective function over the feasible region and determine where its value is maximized. It can be shown that the maximum value of rhe objective function will occur at one of the corner points of the feasible region. By evaluating the objective function at dre corner points of the feasible region, we see that the maximum value occrus at .:cr = 8 and rz : 3. This evaluation is shown in Thble 3.1. Thus, we should purchase eight of the 16-ff-capacity tanks from Manufacrurer A and three of the 24-fC-capacity tanks from Manufacturer B to maximize the storage capacitywithin the given constraints.
48
Crnr"rsR
3
Iurnoouc:rox ro Encnrnrnrrc DssrcN
(0,9) (0,7)
(0,0)
(a)
. -. -..
(12,0) (140) (b)
,,
lri$rer.2 (a) Ihe region as given by
tte linear inequality
120 11
+
24012<
1680. (b)
I[e feasible solulion hr tranple 3.t.
It is wonh noting here that rnost of you will uke specific desigr classes during the nent four years. In fact, most of you will work on a relatively comprehensive desrgn project during your senior year. Therdore, you will leam more in depth about desiggr process and ia application specific to your discipline. For norr our intent has been to intoduce you to the design process, but keep in mind that more design is coming your way.
3.2
Engineering Economics
.......i................... I I I
Economic Factors always play important roles in engineering desigrr decision making. Ifyou design a product that is too expensive to manu&cnre, then it can not be sold at a price that consumers can afford and still be profiable to your company. The fact is that companies design
3,3 MarsRrAL SsLEcrroN
49
producs and provide services not only to make our lives better but also to make monef In Chapter 20,we will discuss the basics of engineering economics. The information provided in Chaprer 20 not only applies to engineering pror."ts but can also be applied to financing a car or a house or borrowing from or investing money in banks.
3.3 fulaterialSelection As design engineers, whether you are designing a machine part, a toy, or a frame for a car or a strucrure, the selection of material is an important design decision. Ttere are a number of factors that engineers consider when selecdng a material for a specific application. For o
to make producs lighter and stronger
for diferent applications.
In Chapter 17, we will look more closely at materials that commonly are used in various engineering applications. Ve will also discuss some of the basic physical characteristics of ma'We will examine the application and properties of comterials that are considered in design. mon solid materials; such as metals and their alloys, plastics, glass, and wood and those that solidi& over time; such as concrete. \?'e will also investigate in more detail basic fuids; such as air and water. By now, it should be clear that material propenies and material cost are imponant design factors. In general, the properties of a material may be divided into three groups: electrical, mechanical, and thermal. In electrical and elecronic applicadons, for orample, the electrical resistivity of materials is important. How much resistance to flotnt of electricrty does the material offer? In many mechanical, civil, and aerospace engineering applications, the mechanical propenies of materials are imponant. These propenies include the modulus of elasticiry, modulus of rigidity, tensile sUength, compression suength, suengtlr-to-weight ratio, modulus of resilience, and modulus oftoughness. In applications dealingwith fluids fliguids and gases), thermophpical properdes such as density, thermal conductiviry, heat capacity, viscosity, vapor pressure, and compressibiliry are important properdes. Thermd expansion of a material, whether solid or fuid, is also an important design factor. Resistance to corrosion is another imporant facor that must be considered when selecting materials. Materid properties depend on many factors, including how the material was processed, its age, its exact chemical composition, and any nonhomogeniry or defect within the material. Material properties also change wir-h temperature and time as the material ages. Most comPanies that sell materials will provide, upon request, informadon on the important propenies of their manufactured materials. Keep in mind that when practicing as an engineer, you should use the manufacnrrers' material property values in your design calculations. The properry values given in this and other te:
Electricat Resistivity-The value of electrical resistivity is a mffNure of resistance of material to flo''p of elecuicity. For enample, plastics and ceramics rypi.ally hare high resistivity,
50
Cneprnn
3
InrnopucrroN To ENcrNssRrNc DBsrcN whereas meuls typically hane low resistivity, and among the besr conducors of electricity are silver and copper.
Density-Density is defined as mass per unit volume; it is a measure of how compact the material is for a given volume. For example, the anerage density of aluminum alloys is 2700kglm3;when compared to steel density of is approximately one
third of the density of
7850k/tt',
aluminum has a densitlthat
steel.
Modulus of Elasticity (Y"*S'" Modulus)-Modulus of elasticiry is a me:uure of how easily a material will stretch when pulled (subject to a tensile force) or how well the material will shorten when pushed (subject to a compressive force). The larger the value of the modulus of elasticity is, the larger the required force would be to stretch or shonen the material. For example, the modulus of elasticity of aluminum alloy is in the range of 70 to 79 GPa (giga Pascal, grg" : 10e), whereas steel has a modulus of elasticity in the range of 190 ro 210 GPa; therefore, steel is approximately three times stiffer than aluminum alloys. Modulus of Rigidity (Shear Modulus)-Modulus ofrigidity is a measure ofhow easily a material can be twisted or sheared. The value of the modulus of rigidity, also called shear modulus, shows the resistance of a given material to shear deformation. Engineers consider the value of shear modulus when selecdng materials for shafts, which are rods that are subjected to twisting torque$. For example, the modulus of rigidiry or shear modulus for aluminum alloys is in the range of 26 rc 36 GPa, v,'hereas the shear modulus for steel is in the range of 75 to 80 GPa- Therefore, steel is approximateh three times more rigid in shear than aluminum is. Tensile Strength-The tensile suength of a piece of material is determined by measuring the maximum tensile load a material specimen in the shape of a rectangular bar or cylinder can carry without failure. The tensile suength or ultimate strength of a material is expressed as the maximum tensile force per unit cross-sectional area of the specimen.'!7hen a material specimen is tested for its suength, the applied tensile load is increased slowly. In the very beginning of the test, the matedal will deform elastically, meaning that if the load is removed, the material will return to its original size and shape without any permanenr deformation. The point to which the material exhibits this elastic behavior is called the yield point. The yield stength represen$ dre maximum load that the material can carry without any Permanent deformation. In cerain engineering design applications, the yield strength is used as tJre tensile strength. Conprcsion Strength-Some materials are stronger in compression than they are in tension; concrete is a good eirample. The compression strength ofa piece ofmaterial is determined by measuring the maximum compressive load a material specimen in the shape of recangular bar, rylinder, or cube can carry without failure. The ultimate compressive strength of a material is expressed as the maximum compressive force per unit cross-sectional area of the specimen. Concrere has a compressive strength in the range of 10 to 70 MPa (mega Pascal, mega
:
106).
Modulus of Resilience-Modulus of resilience is a mechanical properry of a material that shows how effecdve the material is in absorbing mechanical energywithout going through any Permanent damage. Modulus of Toughnecs-Modulus of toughness is a mechanical prcperry of a materid that indicates the ability of *re material m handle overloading before it fracnues.
Strength-to-Weight Ratio-As the term implies, ir is the ratio of strength of the material to its specific weight (weight of the material per unit volume). Based on the application,
3.4
Trervrvom
5t
engineers use eitfier the yield or the ultimate strength of the material when determining
the scrength-rc-weight ratio of a material.
Thermal F,xpansisn-The coefficient of linear expansion can be used to determine the change in the length (per original length) of a material that would occur if the temPerature of the material were changed. This is an important material properry to consider when designing products and strucnrres that are expected to experience a reladvely large temperanre swing during their service lives. Thermal Conductivity-Thermal conductivity is a property of materials that shows how good the material is in transferring thermal energT (heat) from a high-temperature region to a low-temperanre region within the material. Heat Capacity-Some materials are better than otlers in storing thermal energy. The value of heat capacity represents the amount of thermal energy required to raise the temperaflrre one kilogram mass of a material by one degree Celsius, or, using U.S. Customary units, r*re amount of thermal energ;r required to raise one pound mass of a material by one degree Fahrenheit. Materials with large heat capacity values are good at storing thermal energy.
Viscosity, vapor pressue, and bulk modulus of compressibilrty are additional fluid properties that engineers consider in design. value of viscosity of a fluid represenrc a measure of how easily the given fuid can flow. The higher the viscosity value is, the more resistance the fuid offers to fow. For example, it would require less energ;r to ffanspoft water in a pipe than it does to transport
Viscosity-The
motor oil or glycerin.
Vapor Pressure-Under the same conditions, fuids with low vapor-pressure values will not evaporare as quickly as those with high values ofvapor pressure. For example, ifyou were ro learre a pan of water and a pan of glycerin side by side in a toom' the water will evaporate and leare the pan long before you would nodce any changes in the level of glycerin.
B.'lk Modulus of Compressibility-A fluid bulk modulus
represents how compressible
the fluid is. How easily can one reduce the volume of the fluid when the fluid pressure is increased? For example, as we will discuss in Chapter 7, it would take a pressure of 2.24 x 10e N/m2 ro reduce I m3 volume of water by lo/o, or said another way, to a final volume of 0.99 m3.
3.4
Teamworlt A d"ttg"t team may be defined as a group of individuals with complementary expenise, problem solving skills, and talent who are working together to solve a problem or achieve a corunon goal. The goal might be providing a service; designing developing, and manufacturing a product; or improving an existing service or product. A good team is one that gets the best out of each other. The individuals making up a good team know when to compromise for the good of the team and its common goal. Communication is an essential pan of successfrrl teamwork. The individuals making up the team need also to clearly understand the role of each team member and how each task fits together.
52
Cnc'prnn
3
hvrnopucrroN To ENcrNrERrNc DrsrcN
3.5
Common Traits of Good Teams More and more, employers are looking for individuals who not only have a good grasp of engineering fundamentals but who can also work well with others in a team environment. Successfirl teams have the following components:
l.
3. 4.
The project that is assigned to a tslm must have clear and realistic goals. These goals must be understood and accepted by all members of the team. The team should be made up of individuals with complemencary expertise, problem solving skills, background, and talent. The team must have a good leader. The team leadership and the environment in which discussions take place should promote
5.
openness, respect, and honesty. Team needs and goals should come before individual needs and goals.
2.
In addition to these characreristics that make up a good team, Dr. R Meredith Belbin, in his book Managernmt Tbams: Vlty The! Succeed or Fail, identifies additional roles for good team membets. A team with members who represent the following secondary roles rends to be very successfirl.
The organizer is someone who is experienced and confident. This person is trusted by members ofthe team and serves as a coordinator for rhe endre project The organizer does not have to be the smanest or most creative member of rhe team; however, he or she needs to be good at
clarfing
goals and advancing decision making.
The creator is someone who is good at coming up with new ideas, sharing them with other team members, and lening the team develop the ideas funher. The creator is also good at solving difficult problems, but may hane problems with following certain protocols.
The gatherer is someone who is enthtrsiastic and good at obtaining things, looking for possibilities, and dweloping conracts. The motivator is someone who is energetic, confident, and outgoing. The motivator is good at finding wap around obstacles. Because the motivator is logical and doesnt like vagueness, he or she is good at making objecrive decisions.
The evaluator is someone who is intelligent and capable of understanding the complete scope of the project. The evaluator is also good at judging ourcomes correcdy. The team worker is someone who uies to get everyone to come togetheq because he or she does not like fricdon or problems among team members. The solver is someonewho is reliable and decisive and can flun conceprc into practical solutions. The ffnisher is someonewho can be counted on to finish his or her assigned task on time. The finisher is detail orientated and may worry about the teamb progress roward finishing the assignment.
There are many other facors thar influence team performance:
n the way a company is organized howprojects are assigned what resources are available to a team to perform their tasla
. .
" the corporate culture: whether
openness, honesty, and respect are promoted
3.7 Pnopc"r ScnrnurrNc.rNo
Thsr
Cnenr
53
These factors are considered external to the team environment, meaning that the team members do not have much control over them. Howwer, there are some factors that are internal to the team envirorunent, factors which the team can conuol. Communication, the decision making process, and the level of collaboradon are examples of factors that are controllable at the team level.
3.6
ConflictResolution of people work together, conflicts sometimes arise. Conflics could be the result of miscommunication, personality differences, or the way events and actions are interpreted by a member of a team. Managing conflicts is an important pan of a team dynamic. When it comes to managing conflicts, a person's response may be categorized in one of the following ways. There are those in a team environment who try to avoid conflicts. Although this may seem like a good approach, it demonsuates low assertiveness and a low lwel of cooperation. Under these conditions, the person who is assenive will dominate, making Progress as whole difficult. Accommodadng ream members are highly cooperative, but their low assertiveness could result in poor team decisions. Because ideas ofthe most assertive person in the group may not necessarily refect the best solution. Compromising team members demonstrate a moderate level of asserriveness and cooperation. Compromised soludons should be considered as a last resort. Again, by compromising, the team may have sacrificed the best solution for the sake of group unity. A better approach is the collaborative "conflict resolutiorf' approach, which demonstrares a high level of assertiveness and cooperation by the team. Vith this approach, instead of pointing a finger at someone and blaming an individual for the problem, the conflict is ueated as a problem to be solved by the team. The team proposes solutions, means of evaluation, and perhaps combines solutions to reach an ideal soludon. However, in order to reach a resolution to a problem, a plan with clear steps must be laid out. Good communication is an integral part of any conflict resolution. One of the most im-
Vhen a group
poftant nrles in communication is to make sure that the message sent is the message receivedwithout misundersmnding. Team members must listen to each other. Good listeners do not interrupt; drey allow the speaker to feel at ease and do not get, angry or criticize. You may want to ask relevant questions to let the speaker know that you really are listening. Now you have some idea about teamwork and what makes successfirl teams; next we will discuss project scheduling.
3:l
Project Scheduling and Task ehart Project scheduling is a process that engineering managers use to ensure that a project is completed on time andwithin the allocated budget. Agood schedule will assign an adeguate amount of time for various project acrivities. It will also make use of personnel and the available resources for planning, orgaruzing,and controlling the compledon of the project. Awell planned schedule could also improve the efficiency of the operation and eliminate redundancy in task assignments. An example of a simple project schedule and task assignment is sholvn in Thble 3.2. This schedule was used for a small design project for an introduction to engineering class.
il
3.8
3.8
Ever,uerrNc ArrsRNATrvBs
55
EvaluatingAlternatives Once you hane narrowed down your design to a fewworkable concepts, it is customary to use an evaluation table similar ro the one shown in Tbble 3.3 to evaluate alternative concepr in more detail. You stan by assigning a lwel of importance (I) to each design criterion. For ex-
ample,youmayuseascaleof
l
to
5,withI: l indicatinglitdeimponance, andl = 5 sig;m-
fying erftremely important. Next you will rate (R) each workable concePt in terms of how well itmeetseachdesigncriterion.YoumayuseascaleofR = 3 forhigh, R = 2andR = I formedium and low, respectively. Note thar the design criteria that we have listed in Thble 3.3 areto serve as an enample, not as absolute design criteria. The design criteria vary depending on a prgect. For your class projecr, you should list the design criteria that you feel are important. Moreover, note that it is customary ro divide rhe design criteria into positive and negative criteria. After you assign the I and R values ro your design options, you add the R X I scores for each design and select the daign wirh the highest overall rating. An example demonstrating how to evaluate alternarives is shon'n in Table 3.4.
Desiga
Doigp Critedon Positive
Originality Practicability
Manufacmabfity Reltabiliry Performance
Durabiliry Appearance
Profitability Other
N"g"tiw Production cost Operating cost Maintenance cost Time rc cornplete the project
Environmental impact
Other Net score
R
RXI
R
tr
RXI
55
Cnrprsn
3
INrnooucrrox ro ExcnrsrRrNc DEsrcN
D€sign
I
Desip
RXI
Derip Criterion
II
RXI
Positine a
i8
3
T2
Pracdcability Manufacnuability Reliability
3
r5
1
l0
3 3
a
t0
r5
4
Performance
3
r)
a
2
,
,8 ,8
a
Appearance
4 4
Profitability
5
3
r5
Originaliry
Durabiliry
4
3
5
10 8
L2
t0
O&er 99
Negative Production cost Opemting cost Maintenance cost Time to complete the project
5
2
10
4 3
2
U
2
6
5
3
Environmental impact Other Net ccore
3.9
5
L5
3 3
9
8
r5
l0
75
49
62 25
50
Patent, Trademark, and Copyright In the early days, trade information and invention were kept in the family and passed on from one generadon to the netrt. For example, when a plow maker came up with a better design, he kept the details of the design to himself and shared the specifications of the new invention only with his familn induding son(s), brothers, and so on. The new designs and invenrions stayed in the family to prot€ct the business and to prevent others from duplicating tle inventor's design. Howwer, new designs and inventions need to be shared if they are to bring abour imProvemen$ in weryone's lives. At the same time the person(s) who comes up with a new idea should benefit from it. Tirade information and invention, if not protected, can be stolen. So you can see that, in order for a govemment to promote new ideas and inventions, it must also provide means for protecting others &om stealing someone's new ideas and inventions, which are considered inte llecaal properry, Patents, uademads, service marls, and copyrights are examples of means by which intelIectual propemy is protected by Unired Sares lavr.s.
3.10
ErrrcrussRrNc Sreupenos eNp
Coors
57
Itatent tffhen you come up with an invention, in order to prevent others from making, using, or selling your invention, you may file for a Parcnt with the U.S. Patent and Trademark Office. The right given by the patent in the language of the statute and of the grant iaelf includes, "the right to exclude others from making, using, offering for sale, or selling" the invention in the United States or "imponing" the invention into the United Sates. It is important to understand tlat the patent does not grant tle inventor the right to make, sell the invention, but it excludes others from making, using, or selling the invention. or use, For a new parenr, an invention is protected for a rerm of20 years from the date on which the application for the patent was filed. The U.S. Parent and Trademark Office recommends that all prospective applicana retain the services of a registered parent aftorney or patent agent to prepare and execute their applications. A design parent is good for 14years from the time it was granted. A utility patent lasts for eirher 17 years ftom the dme it was granted or 20 years &om the earliest filing date, whichever is longer. A utility patent is issued for dre way anitem worl.c and is used; whereas a design patent protects the way an item looks. Mark A rademark is a name, word, or sfmbol that a corirpany uses to distinguish its products from others. It is imfortant to note that the uademark right issued to a company excludes others from using rJre same or sirnilar mark, but it does not prevent other companies from making the same ot ri*il.r products. Coke@ and Pepsi@ are examples of similar produas with different trademarls. A service mark is a name, word, or qymbol that a company uses to distinguish its services from others. A service mark is the same as a uademark, except that it applies to a service rather than a product. Trademark and Service
eopyright Copyright
is a
form of protection provided by the laws of the United States to the
authors of "original works of authorship," The copyright laws cover literary, dramatic, musical, fiistic, and other types of intellectual works and is obtainable for bot}r published and unpublished work The copyright laws protect the form ofexpression used by the authors, not the conrenr or the subject matter of their work. For example, an author can wrire a book about the fundamentals of physics. The copyright lavr protects the author's work from others copyrng the
It
not prevent others &om writing another book about the same fundamentals of physics, nor does it prevent others from using the fundamental laws of physics. For aworkcreated afterJanuary I, 1978, the copyright larnrs protect it for a term that is equal to the authort life plus 70 years after the author's death. For a piece of work that has two or more authors, the term extends to the last surviving authort life plus 70 years.k is also imponant to note that currendy there are no internadonal copyright larrs that would protect an author's work throughout the wodd. exacr way things were explained or described.
does
3.10 Engineering Standards and Codes Today's existing standards and codes ensure that we hane safe strucnues, safe transporradon qtsrems, safe drinking water, safe indoot/outdoor air quality, safe products, and reliable services. Standards also ericourage uniformity in the size of pans and components that are made by various manufacturers around the world.
58
Crr.lr'rsn
3
lNrnonucrroN To ENcrnnsRrNc DssrcN
Why Do We Need Standards and Codes? Standards and codes have been developed over the years by various organizations to ensure product safety and reliability in services. The standardization organizations set the authoria-
tive standards for safe food supplies, safe strucflrres, safe warer qrstems, safe and reliable elecuical systems, safe and reliable ransponation qFstems, saG and reliable communication qystems, and so on. In addition, standards and codes ensure uniformity in the size of parts and components that are made by various manufacturers around the wodd. In rodayt globally driven economy where parts for a product are made in one place and assembled somewhere
Width of tire in millimeters
Ratio of sidewall height-to-tread width. Range: 35 to 80. Higher numbers mean a smoother ride, but sloppier handling. Lower numbers mean a harsher ride. but crisper handling.
Radial construction Final digits of manufacture/s code tell when tire was made:091 on this example means 9th week of '01. Rubber hardens with age; look for a recent date.
Wheel diameter in inches
Maximum-load rating index. Typical range: 75 to 100. Higher means the tire can carry more weight. The amount ol weight is noted in small print elsewhers on the sidewall.
How well the tire on wet roads in govemment tests. Best: A. Worst: C.
How long the tread should last. Example: Tread rated 220 should last twice as as tread rated 1 10. lndex doesn't equal the specific number of miles ol wear. Number of plies (layers) of material making up the tire
Code forthe tire's maximum safe speed when properly inflated and in good condition. The code: S - 112 mph T- 118 mph U - 124 mph H - 130 mph V - 149 mph Z- 1S0-plus mph, as specitied by manulacturer
Standards and codes have been developed over the years to ensure that we have safe structures, safe transportation systems, safe electrical systems, safe drinking water, and safe indoorhutdoor air quality. A tire is a good example of an engineered product that adheres to such standards. There are many national and international standardization organizations that set these authoritative standards. Standards and codes also ensure uniformity in the size of parts and components made by various manufacturers around the world. Source : Rubber
Manufacturert Association.
3.10
ENcrNnsRrNc SraNomos eNp Copss
59
Men's Shirts
Europe
u.s.
36 14
37
14+
38
39
15
L5i
4t 16
42 r6i
44
zt)
46
1l
l2
43 17
Men's Shoes
, i
Europe U.S.
38 55
39
4l
42 43 8910
for uniformity and consistency in para and components and in the way they are made. These standards ensure that parts manufactured in one place can easily be combined with pars made in other places on an assembly line. An automobile is a good example of this concept. It has literally thousands of parts that are manufactured by various companies in different pars of the Unircd States and the world, and all of these parts must fit together properly. To shed more light on whywe need standards and codes, let us consider products that we all are familiar with, for example, shoes or shirts. In the United States, you are familiar with shoe sizes of 9 or 10 or 11 and so on, as shown in Thble 3.5. In Europe, the standard shoe sizes are 43 or 44 or 45 and so on. Similarly, the sandard shin sizes in the United States are 15 or l5l or 16 and so on, whereas in Europe the standard shin sizes are 38,39, or 4l and so on. If a shirt manufacnrrer in Europe wants to sell shirts in the United States, it has to label them such that people undentand the sizes so that they can choose a shirt of the correct size. Converseh if a shoe manufacnrrer from the United Stares wants to sell shoes in Europe, it has to label them such that the shoe sizes are understood by European customers. \Zould it not be easier if wery shirt or shoe manufacnuer in the wodd used uniform size identifications to eliminate the need for cross else, there exisn an even greater need than ever before
referencing? These simple examples demonstrate the need for uniformity in the size and the way products are labeled. Now, think about all possible parts and components that are manufacnued every day by thousands of companies around the world: parts and components such as
bolts, screws, nuts, cables, tubes, pipes, beams, geffs' painB, adhesives, spring!, wires, tools, lumber, fasteners, and so on. You see that if every manufacturer built products using its own standards and specifications, this practice could lead to chaos and many mis6t parts! Fortunately, rhere are existing inrernational standards that are followed by many manufacturers around the
world.
A good example ofa product thar uses international srandards is your credit card or your bankcard. It works in alt the AIM machines or store credit card readers in the world. The size of the card and the format of information on the card conform to the Intemational Organizadon of Standards (ISO), thus allowing the card to be read byAIM machines everywhere. The 35-mm camera film speed (e.g,, 100, 200, 400) is ano*rer o
60
Cneprsn
3
Ixrnooucrrox to ENcrxrenruc DrsrcN
Examples of products conforming to the lS0.
There are many standardization organizations in the wodd, among them various engineering organizations. Recall from our discussion in Chapter 2 *rat most national /international engineering organizations create, mainain, and distribute codes and srandards that deal with uniformity in size of paru and correct engineering design practices so that public safety is ensured. In fact, the American Society of Mechanical Engineers (ASME) discussed at its first meeting in 1880 the need for standardized sizes for screws. Here we will focus on some of the larger standardization organizations in the United Sates, Canada Europe, and fuia'We will brie{ly dacribe the role of t'hese organizations and how they may interact. Among ttre more well-known and internationally recogoized organizations are the American National Standards Institute (ANSI), the American Society for Testing and Materials (ASTM), the Canadian Standards Association (CSA), the British Standards Instizute (BSI), the German Deutsches Insdnt firNormung (DIl.{), the FrenchAsaciation Frangaise dc normalisariaz (AFNOR), the Swedish Sandardiserigen I Saeri.ge (SIS), the China State Bureau of Quality and Technicd Supervision (CSBTS)' the Internadonal Organization for Sandardization (ISO), and the European Union Ce marking.'!7e will briefy describe these organizations in the following sections.
3Jn
Examples of Standards and Codes Organizations in the United States The American National Standards lnstitute The American National Standards Institute (ANSI) was founded in 1918 by five engineering societies and three government agencies to administer and coordinate standards in the United States. TheANSI is a not-for-profit organization, which is suppored byvarious public and private organizations. The institute itself does not develop the standards, but instead it assists qualified groups, such as various engineering organizations, with the development of the standards and sets the procedures to be followed. Today, the American National Sandards Instinrte rePresents the interests of well over a thousand companies and other members. According to the American National Sandards Institute, there are over 13,000 approved ANSI standards in use today, and more staadards are being developed.
3.ll
Enervrprrs
or Srerroanps aNo Cooss OncerrzlrroNs
rN THE UNrrsp
Srarps
61
The American Soeiety for Testing and Materials (ASTM) Founded in 1898, the American Society for Testing and Marcrials (ASTM) is another not-forprofit organization. It publishes standards and test procedures that are considered authoritative technical guidelines for product safety, reliabiliry, and uniformity. The testing is performed by its national and international member laboratories. TheASTM collects and publishes the work of over 100 standards-writing committees dealing with material test methods. For example, ASTM sets the sundard procedures for tests and practices to determine elastic propenies of marerials, impact testing, fatigue testing, shear and torsion properties, residual suess, bend and fenure testing, compression, ductility, and linear thermal expansion. The ASTM also sets the standards for medical devices and equipment, induding bone cements, screws, bolts, pins, prostheses, and plates, and specificadons for allcys used in surgical implants. Electrical insulation and electronics-related standards are also set by ASTM. Additional examples of ASTM work include the following:
. . . . . . . . "
Test guidelines for evaluating mechanical propenies of silicon or other procedures for testing semiconductors, such as germanium dioxide.
Testing methods for trace metallic impurities
in electronic-gade aluminum-copper and
aluminum-silicon. Standards related to the chemical andysis of paints or detecdon of lead in paint, along with rests ro meaEure the physical propenies of applied paint films, such as film thickness, physical suength, and resistance to environmental or chemical surroundings. Standard procedures for evaluating properties of motor, diesel, and aviation fuels, crude petroleum, hydraulic fuids, and electric insulating oils. Test procedures for measurements of insulation properties of materials. Standard procedures for soil testing such as density characteristics, soil texture, and mois-
true content. Building-construction-related tests and procedures, such as measuring the structural performance of sheet metal roofr. Tests for evaluating the propenies of textile frbers, including co$on and wool. Standards for steel piping, tubing, and fittings.
You can find specific standards dealing with any of dre materials in these examples in the Annual EooA of ASTM, which includes large volumes of standards and specifications in the following areasr Iron and steel products
Nonferous metal products Metals test methods and analydcd procedures
Construction Petroleum produds, lubricants, and fossil fuels Paints, related coatings, and aromatics
Textiles Plastics
Rubber Electric insulation and elecronics 'Water and environmental technolory Nuclear, solar, and geothermal energy
62
Cnnr'rsn
3
lvrnopucrroN To
Er.rcrNsERrNc DssrcN
Medical devices and services General methods and instrumentation General products, chemical specialities, and end-use products
ASTM Standards are becoming available on CD-ROM and on-line. ASTM also publishes
a
number ofjournals: Cernmt, Connete
y't Aggregates
Geotechnical Tbsting Journa I
Journal of Cornposins Tbchnolog and Research Joarnal of Forensic Scimces Journal of Tbsting an) Epaluation
National Fire Proteetion Assoeiation (NFPA) Losses from fires total billions of dollars per year. Fire, formally defined as a process during which rapid oxidization of a material occuts, gives off radiant energy that can not only be felt but also seen. Fires can be caused by malfunctioning electrical sjrsterns, hot surfaces, and overheated materials. The National Fire Protection Association (NFPA) is a not-for-profit or-
ganization that was established in 1896 to provide codes and sandards m reduce the burden of fue. The NFPA publishes the National Electrical Codc@, rhe Life Safq Codz@ , the Fire Preuention Codzw , the National Fuel Gas Codz@ , and, the National Fire Alartn Codn@ ,It also provides raining and education.
Underwriters Laboratories (UL) The Underwriters l",aboratories Inc. (UL) is a nonprofit organization that performs product
U ItlLrr
safety tesn and certifications. Founded
n
1894, today Underwriters Laboratories has laboratories in the United States, England, Denmarlq Hong Kong, India Italy Japan, Singapore, and Thiwan. Its certification mark is one of the most recognizable mad<s on products.
@
3.12
Examples of International Standards and eodes The lnternational Organization for Standardization (lS0) As the name implies, the International Organization for Standardization, established in 1947, consists of a federation of national standards from various countries. The International Orga-
nizadon for Standatdizaion promotes and dwelops standards that can be used by all countries in the world, with the objective of facilitating standards that allow for free, safe exdrange of goods, producs, and services among countries. It is recognized by ia abbreviation, or short form, ISO, which is derived from'isos] a Greek word meaning "equal.o As you take more engineering classes, you will see the prefix iso in many engineering terms; for example, zlabar, meaning equal pressure, or isotherm, meaning equal remperature. ISO was adopted instead of IOS (International Organization of Standards) so that there would not be any nonuniformity in the way the abbrwiation is presented in other languages.
3.12 Ce
F,:ravprxs or INtsRNATIoNaT SteNrenos eNo
Copss
63
Standards
Ce standards. Before the formation of the European Union and the utilization of Ce standards, manufacurers in Europe and those exporting to Europe had to comply with different standards based on the requirements that were dictated by a specific country. Ce provides a single set efsafety and environmental standards that is used throughout Europe. The Ce marking on a product ensures conformity to European sandards. The letters Ce come from the French words Conformitd Europeenn€.
All producs sold in Europe musr now complywith
Other Internationally Recognized Standardization 0rqanizations The Bridsh Standards Institute (BSI) is another internationally known organization that deals with standardization. In fact, BSI, founded in 1901, is one of the oldest sandardization bodies in the world. It is a nonprofit organization that organizes and distributes British, European, and International standards. Other internationally recognized sandardization organizations include rhe German Deuttches Institutfir Normwng (DIN), the Frenchlrraciation FranEaise de nortnalisation (AINOR), the Swedish Sandardiserigen I Saerige (SiS), and the China State Bureau of Quality and Gchnical Supervision (CSBTS). Visit the Web sites of these organizations to obtain more information about them. A comprehensive list of various standardization organizations and their \X/eb addresses (if available) that you may find usefirl when searching for information about organizations dealing with engineering standards and codes follows. American Boller Manufacturers
Assoclation
(ABMA)
www.abma.org
American Instltute of Aeronautics and
Astronautics wwwaiaa.org
Air Condltionlng Contractors of America
American Instltute of Chemical Engineers
(AccA)
www.aiche.org
www.accLorg
The American Soclety of Agricultural Engineers www.asae.org
Amerlcan Gas Assoclation (AGA)
www.aga.org Associatlon of Home Appllance Manufacturers GHAM) www.aham.org
Air Movement and Control Assoclation GMCA} www.amcir.ofg
Amerlcan Society of Civil Engineers ww'l r.asce.org American Nuclear Society
wwv/.ans.org Amerlcan Society of Heating, Refrigerating and Alr-Condltloning Engineers (ASHRAE) www.ashrae.org
American National Standards lnstitute
Amerlcan Society of Mechanical Englneers
(AN5r)
(ASME
www.ansi.org
ww\p'.asme.org
Alr Gondltioning and Refrlgeration Institute (ARl)
Amerlcan Society for Testing and
www.ari.org
Internatlonal)
Materials (ASTM) www.asffn.org
64
Cneprun
3
InrnopuctroN to ENcnvsrnrxc DssrcN Amerlcan Water lVorks Associatlon (AWWA)
www.firwa.org Building 0fficials and Code Admlnistrators
International (BOCA) Brltish Standards Instltute (BSl) www.bsi.org.uk
International Fire Code lnstitute (lFCl) www.ifci.org Internatlonal 0rganizatlon for Standardization (lS0) www.iso.ch Mechanical Contnactors Associatlon of
China State Bureau of Quality and Technlcal
Supervlsion (CSBTS) www.csbts. cn.net /english
Amerlca (MCAA) wwv/[email protected] Manufacturers Standardlzation Society of
Councll of Amerlcan Bulldlng 0fflcials (CABO) (see also International Code
the Valve and Flttings Industry (MSS)
Council)
North Amerlcan lnsulation Manufacturers Associatlon (NAIMA) www.naima.org
www.indcode.org Compressed Alr and Gas lnstitute (CAGI)
Natlonal Assoclation of PlumblngHeatlng-Goollng Contractors (NAPHCC) www.naphcc.org
Canadian Gas Assoclation (CGA) wwv/.cga.ca Canadlan Standards Assoclatlon (GSA) www. csa-international. org
Cooling Tower lnstitute (CTl) France-Association Frangalse de
normallsation (AFNOR)
www.afnor.fr Germany-Deutsches Institut (DIN)
fiir
Normung
www.din.de
National Conference of States on Building Codes and Standards (NCSBCS)
National Electrlcal Manufacturers Association 0EMA) www.nema.org National Fire Protectlon Association (NFPA)
www.nfra.org Natlonal Research Council of Canada
Heat Exchange Instltute (HEl)
(NRCC)
Hydraulic Institute (Hl) Hydronic Institute (HYDI)
Occupatlonal Safety and Health Act (0SHA) www.osha.gov
Institute of Electrical and Electronics
Sheet Metal and Air-Conditioning
Engineers
Gontnactors' National Assoclation GMACilA)
www.ieee.org
Institute of Industrial Engineers
Soclety of Automotive Engineers wwlf.sae.org
www.iienet.org International Association of Plumbing and Mechanlcal Officlals (IAPM0) www.iapmo.org
lllumlnating Engineering Soclety of North Amerlca 0ESNA) www.iesna-org
Society of Manufacturlng Englneering www.sme.ofg Sweden-Standardiserigen I Sverige (SlS) www.sis.se
Underwriters Laboratorles (UL) www.ul.com
3,12
Exeivrpr:es
or Inrrnxerroxar, Srar,roenps eNp
Coors
65
Specific Examples of Standards and Codes in Use in the United States Next we will look at some specific standards and codes that are used in various engineering products and pracrices in the United States. Thble 3.6 shows examples of these standards and codes. In the next three sections, we will describe the standards for drinkingwater and the air that we breathe. These are standards that affect all of US, every day! The Environmental Protection Ag"n.y (EPA) of the United Sates sets the standards as required by the U.S. Congress. The Clean Air Act was signed into law it 1970, and the Safe Drinking W'ater Act was enacted by Congress
n1974.
Subject/Title
Publisher
Rderence
Afu Conditioning Room air conditioners
CSA
C22.2 No. Ll7-1970
General requirements for helicopter air conditioning (1970)
Compresorr ANSI/AHAM SAE
Computers
SAEARP292B
Protection of electronic
UL
and power boiletselectric (1991) Gas-fued low pressure stsam and hot water boilers
ANSr/UL384-1991
Primary safety controls for gas- and oil-fued appliances (1985)
ANSI 221.13-1987;
Solid state conuol for
Z2r.r3a-r989
Building Codec ASTM BOCA
building codes National building code,
BOCA
BOCA
ASME
ANSI/ASME 4r7.1-1990
Chimnep, factory built,
NFPA
ANSI/NFPA 211-1988
UL
ANSI/UL2444-
AGA
ANSI 221.23-1989;
1987
22t.23-L99r
UL
ANSI/UL
coolers
Refrigerated vending machines (1989) DrinLing water coolers (1e87)
CSA
C22.2No.132-
UL
ANSI/UL 541-T988
uL
ANSr/UL 399-1986
UL
ANSI/UL474-r9S7
ANSI
ANSI C84.1-1989
NET,IA
NEMA250-1991
NEMA
NEMA\TD1.I983
Dehumdifiers Dehumidifiers (1987)
103-1983
residential type and building heating appliances (1989)
Electrical 'r'ohage ratings
for
elecuical powerq/$ems and equipment Enclosures for electric
CXean Rooms
Procedural standards for ccnified testing of cleanrooms
ANSI/UL 372.1985
L973
Chimneys fueplaces,
UL
Cooletg
Milk
1lth ed. (1990)
andvents, and solid fuel burning appliances
ANSI/NFPA75. 1989
thermostats
ASTM
Safety codes for elevators and escalators
NFPA
appliances Gas appliance
ASTM sundards used in
ANSI/ASME 819.1-1990
computer/data processing equipment
Controlc
Heating water supply,
ASME
comprcssor sy$ems
(RA-C1)
Boilss
Chiongn,
Safery standards for air
NEBB.I988
equipment General requirements for wiring devices
(R 1989) Continued
NEMA
National elecuic code (1e90)
UL
Fuseholders (1987)
ANSI/NFPA.
Incinerators
70-1990
Incinerators, waste linen handling systens and equipment Residendal
and
ANSI/UL 5r2-1986
;
,
ARI
ARI110-90
refrigeraling equipment
:
: ,'tise
FquiPment (1991)
,
WTel ,
steady-
.ASTMC23G89
ASTM
state and thermal performance of building assemblies
ANSI/UL 916.
UL
Nadonal commercial and industrial insulation
1987 a
MICA
MICA
ASME
A$T4EMFC-3M:1t89
1988
standards
Fana
Dua desigr for residential
ACCA
[danual D
Measurements
winter and summer air conditioning Fans
flow
ASME
ANSI/ASMEPTC
UL
ANSI/UL 507
LL-L984 (R-1990)
,' ..]
,q:r{r:&,*
Measurement offluid in pipes, usi.g orifice, nozle, and venturi Measurement indusuial sound
l
of
ASTM
ASTME 119-88
fuetctsofbuilding
.ANSI/ASMEPTC
ASME
ANSI/ASMEPTC 19.13-i961
speed
Measurement
, 'constructionand ,' meterials Uniform fire code I standards (1991)
ASME
36-1955
Measurementofrorary
Fbe Protection Sandard method for ,,
UL
Iosulation ASHRAE AN$I/ASHRAEI IES 100.2-199r
,
),
Tbst method for
residential
I '.Eneigy management .:
1990,
(r973)
.r nameplate voltages
Enersyconservation in - Eaisting buildings-high
AI-ISI{NFPA82t.
incinerators
EretEY
Air condirioning and
NFPA
uncertainty
ASME
,ANSIIASMEI}TC I
IFCI
]
IFCI
Pressuremeasurem"*,,
:'
Fire doors and windorss
NFPA
ANSIINFPAsO1990
Fire protection handbook NFPA National fire codes (annuat) NFPA
19.1-1985
(R 1990)
NFPA
lrl.
ASME
79.2;t987, ANSI
Temperature measurement
'ANStlnSUnftC .
$SI/AST4EPTC 19.3.1974'
NFPN.
,(R
llsel
Mobile Homes aad Rpcle4tioaal' Yehidm :1, Fre€zerr
',
Household refriggrators,
A}IAM
rr- colnfinstigg: I
n.
re$lgeretofs-treezers, household freezers
iand
l
a[o
MH
RecEdonal vehicles '1. . ,
Recreadonal
UL
ANSI/UL250.1991
UL
ANSr/UL471-1992
.:
vehlcl+
Series-M1992
cAN/CSA.Z240 ,
Lowvoltage lighti"s fixnl{es fbr use in ':, I'
RV Series:M92
:NFPA50lGl990
wz34
recreadonal velricles',
freezers (1992)
Motors and Generators
Humidiffers [,Iumidifiers
66
,carylcsR-zz+o
l,
HRF.l
Household refrigerators and freezers (1983) Commercial refrigsrators
r
Mobile honres
ANSI/A}IAM;
Steam generating units
UL
ANSI/UL 563-1985
:
,.:
ANSI/ASil4E PTC
111e64:1R1e?1)
3,13 Dnnnauc lflrpn
Subject/Tide Morcrs and
generators NEMA
Electric motors
(1989) tll.
}{EMAMG r-1987
ANSI/UL 1004
SraNoenos rN THE Urylrsn Starss
Subject/Title
Publisher
National smndard plumbingcode (NSPC) Standard plumbing code
NAPHCC NSPC
sBCCr
1988
Thermal protection motors (1980
for
:
UL
ANSI/UL 547.
Pumps
r99l
Cenuiftgal pumps Centriftgal pump test
Pipe' Tubings, and Fiainge Power
piping
ASME
,dNSI/ASME 831.1-19S9
Re&igeration
piping
ASME
ASMEIANSI B.3r.5-1987
Specification for
polyvinyl ASTM
Electric swimhi"gpool purnpsf filtgrs and
chlorinators (1980
ASTMM2.89
Motot
AS{'M 842-89
Sound Measurement Measurement of industrial
NSF-I4
sound Procedural standards for measuringsound and
cbloride (PVC) plastic
operated water
HI UL
sizes
componenc NSF
and related materials
1993
SBCCI (199r) (R 19e2) ASME PTC 8.2-
r990
HI ANSI/UL 1O8I1985 i
ut
ANSr/UL 343-1985
ASME
ASME/ANSI PTC
NEBB
36-1985 NEBB-1977
UL
ANSI/UL 1025.
i
vibradon Space Heaterc
Plumbirg
Uniformplumbingcode IAPMO
Electric air heaters (1980)
(r991) Source: Amelican Society of
Reference
PumPs
Pipe Specification for seamless ASTM copper pipe, standard Plasdc piping
srandard (1988)
ASME
67
1980
Heating Reftigerating and Air-Conditioning Engineers.
3.13 Drinking Water Standards in the United States The U.S Environmental ProtectionAgency (EPA) sets the standards for the maximum level of conraminants rhat can be in our drinking water and still be considered safe to drink Basically, the EPA sea rwo standards for the level ofwater contaminents: (1) tbe maximum contaminant lwel goal (MCLG) and (2) the ma:
ment RuIe (S\PTR).
67
68
Ihe
CnaprBn3 InrnooucrroNroENcrNnsRrNcDrsrcN
the standard for our drinking water.
EPA sets
MCL in
3.14 0utdoor Air Quality Standards in the United States The source of outdoor air polludon may be dassified into dree broad categories: the stationary sattrces, the rnobib sources, and the natural sources. Exampla ofstationary sources include power plants, factories, and dry cleaners. The mobile sources of air pollution consist of cars, buses, truclis, planes, and trains. As the name implies, the sources of natural air polludon could include windblown dust, volcanic eruptions, and forest fires. The Clean AirAct, srhich sets the standard for six major air pollutans, was signed into larr n 1970. The Environmental Protection Agenry is responsible for setting standards for these six major air pollutan*: carbon monox-
ide (CO), lead (Pb), nitrogen dioxide (NOt, ozone (O3), sulfiu dioxide (SO),
and
paniculate matter (PM). The EPA measures r:he concentration lwel of these polluants in many urb"n areas and colleca air qualiry information by acrual me:rsurement ofpolluuats from thow
ofmonitoringsiteslocated *uoughoutthe country. Accordingto astudyperformedbythe EPA (1997), benveen 1970 and 1997, rhe U.S. population increased by 3lo/o and the vehicle miles traveled increased by l27o/o. Drlr:ing this period, the total emission of air pollutants from
sands
stationary and mobile sources decreased by 3 1olo because ofimprovements made in the efficiency of cars and in indusuial practices, along with the enforcement of the Clean AirAct regulations. Howwer, there are still approximately 107 million people who live in areas with unhealthy air qualiry. The EPAis continuouslyworking to set standards and monitor the emission of pollutants that cause acid rain and damage to bodies of warer and fish (currendy there are over 2000 bodies ofwater in the United Setes that are under fish consumption advisories), damage to the stratospheric ozone layer, and damage to our buildings and our national parla. The
3.14 Ourooonfun QulrrrrSrero.nnos ';'
;'*11.
rN THE UNrrso
I,.r1r. ,-. ; . .i-*-.r:
Sretrs
69
r.-.l.;
Contaminant MCGL
MCL
Source of Contaminant bv Industries
Antimony
6 ppb
6 ppb
copper smelting refining, porcelain plumbing firures, peuoleum refining, plastics, resins,
Asbestos
7
7 M.L.
asbestos products,
plastic pipes copper smelting, c:r parts, inorganic pigments, gray ductile iron, steel works, furnaces, paper
storage barceries
M.L.
(million
fiben per liter)
chlorine, asphalt felts and coating a[to parts, petoleum refining,
Barium
2ppm
2pp^
Beryllium
4 ppb
4ppb
copper rolling and drawing, nonferrous metal smelting, aluminum foundries, blast furnaces,
Cadmium
5 ppb
5 ppb
Chromium
0.1 ppm
0.1ppm
ztnc ard lead smelting, copper smelting, inorganic pigments pulp mills, inorganic pigments, copFer
Copper
1.3 ppm
1.3 ppm
smelting, steel rrorks primary copper smelting, plastic materials,
mills
petroleum refining
Cyanide Lead
Mercury
:#n-
0.2 pprn 15 ppb
2ppb
Nickel
2 ppb 0.1 ppm
0.1 ppm
Nitrate
l0 ppm
l0
Nitrite
I PPm 0.05 ppm 0.5 ppb
I PPm 0.05 ppm
Selenium
Thallium
ppm
2 ppb
pouJtry slaughtering, prepared feeds metal heat ueating, plating and polishing lead smelting steel works and blast furnaces, storage batteries, china plumbing fixtures elecuic lamps, paper mills petroleum re6ning, gray iron foundries, primary copper, blast furnaces, steel nitrogenous fenilizer, fenilizing mixing paper mills, canned foods, phosphate fertilizers metal coatings, peuoleum refining primary copper smelting, petroleum refining, steel works, blast furnaces
unheakhy air has more pronounced adverse health effects on children and elderly people. The human heahh problems associated with poor air quality include various respiratory illnesses and heart or lung diseases. Congress passed amendmen$ to the Clean Air Act in 1990, which required the EPA to address the effect of many toxic air pollutants by setting new standards. Since 1997, the EPA has issued 27 air standards that are to be firlly implemented in the coming years. The EPA is curendy working with the individual sstes in this counry to reduce rhe amount ofsulfur in fuels and setting more stringent emission sandards for cars, buses, trucls, and power plants. 'We all need to understand that air pollution is a global concern that can affect not only our health, but also affect our climate. It may trigger the onset of global warming that could lead to unpleasant natural evenr. Because we all contribute to this problem, we need to be avare of the consequences of our lifestyles and find ways to reduce pollution. Ve could caqpool
70
Cnar"rnn
3
INrnopucrroN To ENcrNneRrNc DssrcN
0utdoor pollution!
or take public transportation when going to work or school. \?b should not leave our cars running idle for long periods of time, and we can remind others to consume less enerry. We should conserve enerry around home and school. For example, turn off the li ght in a room rhat is not in use. \fhen at home, in winter, set the thermostat at 65o F or slighdy lower and wear a swearer to feel warm. During summer, at home set the air-conditioning thermostat at 78o F or slighdy higher. By consuming less enerry and driving less we can h"lp our environment and reduce air pollution.
3.15 Indoor Air Suality Standards in the United States In the previous section, we
discussed outdoor air pollution and the related health effeca. Indoor air pollution can also create health'risks. According to EPA studies of human exposure to airpollutants, the indoor levels ofpollutants may be two to five times higher than outdoor levels. Indoor air quality is imponant in homes, schools, and workplaces. Because most of us spend approximately 9Ao/o of out time indoors, the indoor air qualiry is very imporant ro orrr shon-term and long-term health. Moreover, lack of good indoor qu"lity can reduce pro"it ductivity at the workplace or create an unfavorable learning environment at school, causing sickness or discomfort to building occupanrc. Failure to monitor indoor air qualrry (AQ) or failure to prevent indoor air pollution can also have adverse effects on equipment and the physical appearance of buildings. In recent years, liability issues related to people who suffer dizziness or headaches or other illness related to "sick buildings" are becoming a concern for
3.I5
I
lNpoonfun Qunr,rrr
SrervoARDs rN rnE UNrrso
Stlrss
Outside Sources
Building Equipment
Component/Furnishings
Other Indoor Sources
Polluted outdoor air
IIYAC equipment
Componen*
Science laboratories, copy and
Pollen, dust, fungal
Microbiological growth on
Nearbysources
Microbiological growth in drip pans; ducmork, cofu, and humidifiers, improper venting of combustion producs, and dust or debris in ductwork
Loading docks, odors from dumpsters, unsanitary debris or building
Non-IfVAC equipment Emissions &om ofrce
spores, indusuial
emisions andvehicle emissions
exhausts near outdoor
equipment, and emissions
air intakes
from shops, labs and cleaning processes
Underground sourcc
print
areas,
food preparation
areas, smoking lounges,
cleaning materials, emission from trash, pesricides, odors and voladle organic compounds from paint, chalk, and adhesives, occupants with comrnunicable diseases, dryerase markers and similar pens, insects and other pests, personal hygiene products
Furdshings Emissions from new furnishings and floorings and microbiological growth on or in soiled or watet-
Radon, pesticide, and leakage from underground storage
soiled or water-damaged materials, dry ffp, that allow the passage ofsewer gas, materials containing volatile organic compounds, inorganic compounds, damaged asbestos, and materials that produce particle"s (dust)
71
tanls
damaged furnishings Source: EPAFaesheets'
EPA-402-F-964M, Ocober 1996.
building managers. According to the EPA, some cornmon heahh symptoms caused by poor indoor air quality are
. .
headache, fatigue, and shormess of breath; sinus congestion, coughing, and sneezing;
o eye, nose, throat, and skin irritation; o dizziness and nausea. As you know, some of these qfmptoms may also be caused by other factors and are not ne@ssarily caused by poor air quality. Stress at school, work, or at home could also create health
problems with symptoms similar to the ones mentioned. Moreover, individuds read differendy to similar problems in their suroundings. The factors that influence air quality can be classified into several categories: the heating, ventilation, and air-conditioning (ltVAC) system; sources of indoor air pollutants; and occupants. [n recent ysrs, we hane been enposed to more indoor air pollutants for the following reasons. (1) In order to save energy we are building tight houses that have lower air infiluation or erdluation compared rc older stnrctures. In addition, the ventiladon rates have also been reduced to save additional energy. (2) I7e are using more synthaic building materials in newly built homes that could give off harmfrrl vapors. (3) \[e are using more chemical pollutants, such as pesticides and household cleaners. As shon'n in Thble 3.8, indoor pollurants can be created by sources within the building or they can be brought in from the outdoors. It is imporant to keep in mind that the level of contaminants within a building can vary with time. For example, in order to protect foor surfaces
72
Crreprnn
3
InrnooucrroN To
ErrrcrNsERrNc DssrcN
from wear and tear,
it
is customary fo wax them. During the period when waxing is raking
place, it is possible, based on the type of chemical used, that anyone near the area to be exposed to harmfirl vapors. Of couse, one simple remedy to this indoor air problem is to wax the foor late on Friday aftemoons to avoid exposing too manF occupants to harmfi.rl vapors. Moreover,
this approach will provide some time for the vapor to be exhausted out of the building by the ventilation qistem over the weekend when the building is not occupied. The primarypurpose ofawell-designedheating vendlation, and air-conditioning (HVAC) system is to provide thermal comfon to its occupants. Based on the buildingls heating or cooling load, the air that is circulated.htoqh the building is conditioned by heating, cooling, humidifying or dehumidifying. The other imporant role of awell-designed HVAC system is to filter out the contaminants or to provide adequate ventilation to dilute air-contaminant levels. The air fow patterns in and around a building will also affect the indoor air qudiry. The air fow pattern inside the building is normally created by the H'\A.C qystem. Howwer, the outside air flow around a building envelope that is dictated by wind patterns could also affect the air flow pamern within the building as well. !?'hen looking at air fow pafterns, the important concept to keep in mind is that airwill always move from a high-pressure region to a low-pressure region.
Methods to Manage Contaminants (l) source elimination or removal, (2) source substitution, (3) properventilation, (4) exposure control, and (5) air cleaning. A good example of source elimination is not allowing people to smoke inside the building or not allowing a car engine to run idle near a building's outdoor air inake. In odrer words, eliminate the source before it spreads out! It is imponant for engineers to keep that idea in mind when designing the HVAC systems for a building-avoiding placing the outdoor air intakes near loading docls or dumpsters, for example. A good example of source substitution is to use a gende cleaning produfi rather than a product that gives offharmfirl vapors when cleanThere are several wa)rs to control the lwel of conaminants:
ing bathrooms and kitchens. Local exhru"t control means removing the sources of pollutants before they can be spread through the air distribution system into other areas ofa building. Everyday examples include use of an exhaust fan in restrooms to force out harmfirl contami' nants. Fume hoods are another example of local exhaust removal in many laboratories. Clean outdoor air can also be mixed with the inside air to dilme the contaminated air. The American Society of Heating Refrigerating andAir Conditioning Engineers (ASHRAE) has established a set of codes and standards for how much fresh outside air must be inuoduced for various applications. Air cleaning means removing harmfirl particulate and gases from dre air as it passes through some cleaning system. There are various methods that deal with air contaminanr removal, including absorption, catalpis, and use of air filters. Finally, you can bring dre indoor air quality issues to the attention of friends, classmates, and family. We all need to be avrare and try to do our paft to create and maintain healthy indoor air qualiry.
SUMMARY Now that you hane reached this point in the text
"
You should know the basic design steps that all engineers follow, regardless oftheir backgoun4 to design producs and services. These steps are: (1) recognizing the need for a
Pnorlrus
73
producr or a service, (2) defining and undersanding the problem (the need) completely, (3) doing the preliminary research and preparation, (4) conceptudizing ideas for possible solutions, (5) qynthesizing the results, (@ evaluating good ideas in more detail, (7) optimizing the soludon to arrive at the best possible solution, (8) and presenting the solution. You should realize that economics ptayr an important role in engineering decision making. You should realize thar the selection of material is an imporant design decision. You should be familiar with the cornmon traits of good teams. You should undentand the importance of project management. You should be familiar with the concepts of patent, uademark and copynght. You should know why we need to have standards and codes in engineering. You should be familiar wirh the role and mission of some of the larger standardization organizations in the world, such as ANSI, ASTM, BSI, CSA, ISO, CSBTS, SIS, Ce, AFNO&
e
6
3.1.
3.2.
3.3.
3.4.
and DIN. You should be familiar with the role of the EPA and the smndards it sets for drinking wacer, outdoor air quality, and indoor air quality. You should be able to name some of the sources of indoor and outdoor air polluants, You should be able to name some of r:he sources ofwater pollutants.
List at least ten products that already o
3.5. You hare seen bottle and container caps in use all aroundyou. Investigate the design ofbotde caps usedin the following producs: Pepsi or Coke botdes, aspirin botdes, shampoo botdes, mouthwash bottles, liquid deaning containers, hand lotion containers, aftershave bonles, ketchup or mustard botdes. Discuss what you
think are imponant design pararneters. Discuss the adwith each desigg.
vantages and disadvantages associated
3.6. Mechanical dips
are used to close bags and keep things together. Investigare the design ofvarious paper clips, hair clips, and potato chip bag clips. Discuss what you think are important design parameters. Discuss the
3.7.
advanages and disadvantages associated with each of three designs. Discuss in detail at least two concepts (for example, activities, processes, or methods) that can be employed grccery during buqy hours to befter serve customers stores.
3.8.
3.9.
^t
Discuss in detail at least fi/o methods or procedures that can be employed by airline companies to pick up your luggage at your home and drop it offat your final destination. In the near future, NASA is planning to send a spaceship with humans to Mars. Discuss important concerns and issues that must be planned for on this trip. Investigate and discuss issues such as how long it
would take to go to Mars and when the spaceship should be launched, considering that the distance benneen the earth and Mars changes based on where they are in their respective orbits around the sun. \fhat type and how much food reserves are needed for this trip? What type of orercise equipment should be on board so muscles wont atrophy on this long trip? \ffhat should be done with the waste? What is the energy requirement for such a trip? What do you think
74
CrreprsR
3
lrrnooucrroN To
ENcTNEERTNc Dssrcr.r
are the important design parameters for such a trip? a repon discussing your findings.
size, schedule number, inside diameter, outside
'lfrite 3J0.
You have been using pens and mechanical pencils for many yeius now. Investigate the design of at least five different pens and mechanical pencils. Discuss what you think are imporant design parameters. Discuss the advantage and disadvantage associated wirh each design.
Vrite a brief repon explaining
3J8"
3J9.
your
firditgr. 3.11.
3.t2.
3.t3.
3.r4.
This is an in-class design project. Given a 12 n.-by12 in. aluminum sheet, design a boat that can hold as many pennies as possible. \V'hat are some of the important design parameters for this problem? Discuss them among yourselves. You may want to choose a day in advance on which to hold a compedtion to determine the good designs. This is an in-class design project. Given a bundle of drinking stravrs and paper clips, design a bridge between two chairs that are 18 in. apan. The bridge should be dCIigned to hold at least 3 lb. You maywant to choose a day in advance on which to hold a competition m determine the design that carries maximum load. Discuss some of the imporant design parameters for this problem. Identify and make a list of at least ten products around your home that are certified by the Underwriters l,aboratories. Create a table showing hat sizes in the United States and Europe.
Continental
us.
5t 52
A 4
3.20.
Collect information on the American Whe Gage (A!fG) standards. Create a able for annealed copper wires showing the gage number, diameter in mils, cross-sectional area, and resistance per 1000 ft.
'Write
a brief memo to your instructor explaining the role and function of the U.S. Deparrmenr of Trans-
poradon (DOT). Obtain information on the standard sign typefaces used for highway signs in the United States. The Federal Highway Administration publishes a set of sandards called Swndard Alphabas for HighwaJ, Sign. 'Write a brief memo to yoru instructor enplaining your
firdi"S. 3.4.
Obtain information about the Nuclear Regulatory
Commission (NRC), which sets the sandards for handling and other activities dealing with radioacdve ma'lZrite terials. a brief report to yorrr instrucor regarding your findings. 3.A. Obain information about the classification of fire extinguishers. S7rite a brief repon explaining what is meant by Class A, Class B, Class C, and Class D fires. 3.n. Vrite a brief report detailing the development of safery belts in cars. \Vhen was tle first safety belt designed? Which was the first car manufacnrrer to incorporate safety belts as sandard items in its cars? 3.U. Investigate the mission of each of the following st urdards organizations. For ach of the organizations listed, write a one-page memo to your instructor about its mission and role.
53
a" The European Comminee
6t
in metdc and
Obein information about what the colors on trical resistor mean. Create
a
eble showing the electri-
colors: Blach Brown, Red, Orange, Yellow, Green, Blue, Violet, Gray, White, Gold, and Silver, and a column showing the values. Imagine that you are making this able for others to use; therefore, include at leasr two examples of how to read the codes on electrical resistors at the bonom of your able. Collect information on U.S. standard steel pipes (Ll4 in. to 20 in.). Create a table showing nominal
Standards Institute
c. PanAmerican Standards Commission (COPANT) d. Bureau of Indian Standards (BID) e. Hong Kong Standards and Testing Centre Ltd.
an elec-
(HKSTC)
cal resistor codes. Your table shouldhave a columnwith
3.17.
Elecuotechnical
(ETSD
U.S. units. 3.r6.
for
Sandardization (CENELEC)
b. European Telecommunication 315. Create a table showing wrench sizes
diam-
eter, wall thickness, and cross-sectional area.
f. Korea
Academy
of Indusuial
Gchnolory
(KAITECH)
g. SingaporeAcademy oflndustrialTecfinolog7 h. Sandards New Zealand (SNZ)
'Vrite
(PSB)
a brief repon explaining what is meant by ISO 9000 and ISO 14000 cerdfication. 9.26. Ask your local city water supplier to give you a list of the chemicals that it tests for in vour water. Also ask
3.25.
Prosr,BMs how your city water is being treated. You may want to contact your state depanment of hedth/environment m get additional information. For help in locating state and local agencies, or for information on drinkingwater in general, you can call the EPA's Safe Drinking 'W'ater Hotline (800) 4264791.You can also visit the EPA'Web site atwww.epa.org. You can also obtain information about the uses and releases of chemicals in yoru state by contacting the Community Right+oKnorv Hotlins (800) 535-0202. 3.21. Collect information on the Surface V'ater teatment 'Sfrite a brief Rule (S\7TR) standards set by the EPA memo to your instructor enplaining your findings. 3.28. Obain the EPAI consumer fact shees (they are now available on the \Feb) on antimonp barium, beryllium, cadmium, cfanid.e, and mercury. After reading the fact sheets, prepare a brid report explaining what they are, how they are used, and what health effecs are associated with them. 3.29. In 1970 the U.S. Congress passed the Occupational Sdety and Heahh Act. The following is a duplicate of the act, which reads:
Visit the Departrnent of I-abor's Occupational HealthAct (OSHA) home
Safety and
page atwww.osha.gov,
andwrite
a brief repon describing the rype of safety and health standards that are covered bv OSFL{
Public Law 91'595 91st Congress, S. 219
December 29,1970 As Amended by Public Law 101-552.
Section 310.l, November 5,1990 As Amended by Public Law 105'198,
July 16,1998 As Amended by Public Law 105'241
September 29,1998 An Act
healthful working conditions for working men and womery by authorizing enforcement of the standards developed under the Act; by assisting and encouraging the States in their efforts to assure safe and healthful working conditions; by providing for research, information, education,
To assure safe and
and training in the field of occupational safety and health; and
for other purposes. Be it enacted by the Senate and House of Representatives of the United States of America in Congress assembled, that this Act may be cited as the "Occupational Safety and Health Administration Compliance Assistance Authorization Act of
1998."
Define the pupose of the brainstorming session and the ground nrles. Examples of ground nrles: no criticism of ideas as they are being presented, one person ulls at one time and for an agreed upon period. Choose a facilitator to keep uack
ofground nrles.
Record all ideas where erreryone can see then, . Dorft worry about looking foolish, record all your ideas, It is a good practice not to associate a person with an idea, Think ofeach idea as a
group idea.
After Brainstoming Identifr the promising ideas. Dorit evaluate the ideas in deail yet! Discuss wEZs to improve a promising idea or promising ideas. Choose and list the ideas for deail evaluation. Evaluate the most promising ideas!
75
76
Cn.eprsn
3
lxrnooucrroN To ENcrxssRrNc DssrcN radios uansmit at 4 watts). The base station is also transmiaing at low power, Low-power transmitters have rwo advantages: . The transmissions ofa base station and the phones within its cell do not make it very far outside that c€ll. ThereFore, in the figure above, both ofthe purple cells can reuse the same 56 frequencies. The same frequencies can be reused extensively across the city. r The power consumption of the cell phone, which is normally bat-
An Engineering il|arvel Millions of people in the United
States and around the wodd use cellular phones. They are such grear gadges*wifi a cell phone, you can talL to anyone on the planet from just about anywhere! But have you wer wondered horv a cell phone works? Vhat makes it different ftom a regular phone? S?'hat do all those confiuing terms like PCS, GSM, CDMA and TDMA mean? This anide disclsses the technolory behind cell phones so that you can see how amazing they really are.
tery-operated, is relatively low. Low power means small baneries, and this is what has made handhdd cellular phones possible. The cellular approach requires a large number ofbase stations in a city ofany sizr. A typical large ciry can haye hundreds oftowers. But because so many people are using cell phones, coss remain low per user. Each carrier
in
each city also runs one central ofrce called the Mobile Glephone Switching Office (MTSO), This office handles all of the phone connections to the normal land-based phone system, and conuols all ofthe base stations in the region,
The eell Approach One of the most interesting things about a cell phone is thar it is really a radio-an entremely sophisticated radio, but a radio nonetheless. The telephone was invented by Alexander Graham Bell in 1876, and wireless communication qln trace its toots to dre invention of rfie radio by Nikolai Tesla in the 1880s (formally presented in 1894 by a young Italian named Guglielmo Marconi). It was only natural that these two great technologies would eventually be combined! In the dark ages before cell phones, people who really needed mobile communications ability instdled radio telephones in their cars. In the radio telephone system, there was one central antennatower per city, and perhaps 25 channels available on that tower. This central antenna meant thar the phone in your car needed a powerfrrl transmifter-big enough to uansmit 40 or 50 miles. It also meant that not many people ould use radio telephones-there just were not enough channels. The genius of the cellular system is the division of a city into small cells. This allov/s entensive frequency reuse across a city, so that millions of people can use cell phond simulaneously. In a typical analog c"ll phone qrstem in the United Sutes, the cell phone carrier receives about 800 frequencies to use across the ciry. The carrier chops up the city into cells, Each cell is rypically sizcd at about l0 square miles (25 square kilometers). Cells are normally thought ofas hexagons on a big hexagonal grid. Each cell has a base station that consists
ofa tower and a small building
containing the radio equipment A single cell in an analog sycem uses one-seventh ofthe available du-
it on the hexagortal grid, are each using one-seventh of the available channet so that
Cell Phone Codes .
Electronic Serial Number (ESN)-a unique 32-bit number programmed into the phone when it is manufacnued, . Mobile ldentification Number (MIN)-a l0 digit number dedved 6om your phone's number. . Slatem Idendfcation Code (SID)-a unique 5 digit number that is assigned to each canier by &e FCC. \?hile the ESN is considered a permanent part ofthe phone, borl the MIN and SID codes are programmed into the phone when you puchase a service plan and haue the phone activued
From Cell All cell phones
have special codes associated with them. These codes are used to identi$ the phone, the phone's owner and the service provider. Let's say you have a ceII phone, you tumed it on, and sorneone ries to call you. Here is what happens to the call: . Vllen you firs pcmrer up 6e phone it lisers 6r an SID (se sid$ar) on rhe control drannelTtreontrolcharurelhaspdrt 6equencydmthephoneand
plex voice channels. Thar is, one cell, plus the six cells a-round
each cell has a unique set offrequencies and there are no collisions: A cell phone carrier typi
. .
in
Each cell phone uses two ftequencies per call-a duplex chalnelso there are typiolly 395 voice channels per carrier. (The other 42 frequencies are used for control channels). T'here?ote, each cell has 56 or so voice channels anailable. In other wonds, in any cell, 56 people can be talking on their cell phones at one time. With digital transmission methods, the number of available channels increases. For example, a TDMA-based digital rystem can carry three times as many calls as an analog system, so each cell would harre about 168 channels arailable. Cell phones have low-power transminers in them. Many cell phones have mro signal strengtbs: 0.6 watts and 3 watts (for comparison, most CB
*By Marshall Brain and JetrTyison. Prinrcd from the Yeizan.amlleaing Center. Copyright O 1998-2001. Ho*stufrvorks,com, Inc Al rights reserved. fhttp. //www.howstu6arorks.com]
76
.
a
clEy.
to Cell
. . . . .
hsestadonusetotalkooneanothsaboutthingstikecallset-upanddunndchaoging. Ifdre phooe cannot fnd any conrrol channels o lissr tq it knows itis out ofnng+ urddispl4ra "no savice'rcg.. \7hen it receives the SID, the phone compares it to the SID programmed into the phone. If the SIDs match, the phone Lnows thar the cell it is communicating with is part of its home system. Along with the SID, the phone also uansmis a registration request, and the MTSO keeps aack ofyour phonet location in a databasethis way, the MTSO knovgs which cell you are in when it wants to rhg your phone, The MTSO gets the call, and it tries to find you. k lools in its &tabase to see which cell vou are in. The MTSO picls a frequency pair that your phone will use in that cell to uke the call. The MTSO communicates with your ohone over the conuol channel to rcll it what frequencio to *", *d otr"" yo* phone and the tower switch on those ftequencies, the call is connected. You are talk' ingby wo-wry ndio ro ifriend! As you move toward the edge ofyour cell, your cell's base station will note that your signal strengtJr is diminishing. Meanwhile the base station in the cell you are moving toward (which is listening and measuring signal strength on all &equencies, not iust its orrn one-
ANENcrNsERrNcMenwrseventh) reill be able to see your phonet signal suength increasing. base stations coordinate ilremselves drough the MTSO' and at some point, your phone gets a signal on a control channel telling it to change ftequencies. This hand o$swirches your phone to the new c€ll.
The two
analog. AMPS and NAMPS only operate in the 800
offer many of
n
MHz band and do not
t[e featues common in digital ce[utar
service suctr as e-mail
andVeb Lrowsing.
Along eomes Digital Roaming Ifthe SID on the control
channel does not match the SID programmed into your phone, then the phone knows it is roaming. The MTSO of the cell that you are roa$ing in contacts the MTSO of your home system, whic.h then chedrs its daabase to confirm that the SID of the ohone you are using is valid. Your home system verifies your phone to' the l6cal MTSO, which then tracks your phone as you move through its c€ils. And the amazing dring is thar all of this happens within secondsl
Cell Phones and CBs A good way to understand the sophistication ofa cell phone is to compare it to a CB radio or a w'alkie-talkie. Simplexvs. Duple* Both walkie-talkies and CB radios are simplor devices. That is, two people communicating on a CB radio use the same freguency, so only one person can talk at a time. A cell phone is a duplex device. That means that you use one frequency for talLing and a second, separate frequency for listening. Both people on the call can talk at once. Channelsr Awalkie-qlkie rypically has one channel, and aCB radio has 40 channels. A q2ical cell phone can communicate on 1,664 channels or more!
Range A walkie-talkie can transmit about one mile using a 0.25 wtt transmifier. A CB radio, because it has much higher power, can transmit about five miles using a 5 watt transmitter. Cell phones operate within cells, and they can switch cel.ls as they move around. Cells give cell phones incredible range. Someone using a cell phone can drive hundreds of miles and maintain a conversation the entire time because of the cellular approactr.
AMPS In 1983, the analog cell phone sandard
called AMPS (Advanced Mobile Phone Sptem) was approved by the FCC and first used in Chicago. AMPS uses a range of frequencies between 824 MHz and 894 MHz for analog cell phones, In order to encourage competition and keep prices low, the U, S. govemment required the presence of two carriers in every market, known as A and B carrien. One of the carriers was nomrally the local exchange carrier (LEC), a fancy way of saying the local phone company. Caniers-A and B are each assiped 832 frequencies: 790 for voice and another 42 for data A pair offrequencies (one for transmit and one for receive) is used to create one channel. The frequencies used in analog voice channels are qryically 30 kflzwide. The reason thar 30 kHz was chosen as the sundard size is because it gives you voice quality comparable to a wired telephone. The transmit and receive frequencies of each voice channel are separated,by 45 MHz to keep them ftom inteGring with each other. Fach carrier has 395 voice like channels, as well as 2l d"ta channels ro use for housekeeping ""aiui1iss Adregisration, paging etc. A version of AMPS known as Narrowbaad vanced Mobile Phone Service (NAMPS) incoqporates some digiul technology to dlow the sy$em ro carry about tluee times as many calls as the original version. Even though it uses digial technolory, it is still consideted
use the same radio technologT as analog phones but in a different way. Analog sy$ems do not frrlly utilize the signal between the phone and the cellular ne$/ork Analos siqnals cannot be compressed and 'manipulated toronitrg applies rc as easily as a true digital si[ni. T1r" many cable companies that are going to digiul-so they can fit more channels within a given bandwidth. It is amazing how much more efrcient digital systems can be. Digiul phones conven your voice into binary information (ls and 0s) and then compress it This compression allows between drree and ten cell phone calls to ocrupy the space of r single aneJog cell phone voice call. Many digial cellular qrstems rely on Frequency Shift Keying (FSK) to send data back and forth over AMPS. FSK uses two ftequencies, one for 'l"s and the otler for "0"s, ahemating rapidly between the tq/o to send.lig'ul information between dre cell tower and dre phone. Clwer modulation and encoding schemes are required to convert the analog informadon to digiul, cornpress it and conven it back again while maintaining aa acceptable level of voice quality. All tlis mearx that digiul cell phones hane o contain a lot ofprocessing pov'ed
Digial cell phones
o-"
Cellular Access Technologies There are duee common technologies used by cell phone nerworks for transmining information: . Frequency Division MultipleAccess (FDMA) . Time Division Multiple Access C|DMA) . Code Division MultipleAccas (CDMA) Althougir these technologies sound very intimidating you can get a good sense of how they work just by breaking down the tide of each one. The first word tells you what the access method is and the second word, division, lets you know that it splis calls based on that access method. . FDMA pur each call on a sepante freguency. . TDMA assigns eacJr call a certain ponion of time on a designeted
.
tlequency.
a unique code to each call and spreads it over the available ftequencies. The last pin of each name is multiple access. This simply means thar more than one user (muhiple) can rse (access) each cell. FDMA separates
CDMAgives
the spectrum into distino voice channels by slining it into uniform chunls of bandwidth, To bener undersmnd FDM.A, rhink of radio stations, Each station sends its signal at a different frequency within the available band. FDMA is used mainly for analog transmission. While it is certainly capable of carrying digfual informarion, FDMA is not cnnsidered to
be an elficient method for digital transmission. TDMA is the access me&od used bv the Elecuonics IndustryAlliance and the Telecommunications IndustryAssociation for Interim Ssndard 54
05-54) and Interim Standard 136 (IS-13O. UsingTDMA, a narrowband that is 30 kHz wide and 6.7 milliseconds Iong is split time-wise into dree time slots. Narrow band means channels in the traditional sense. Each conversation ges the radio for one-third ofthe time. This is possible because voice &ta that has been converted to digital informarion is compressed so that it takes up significandy less transmission space. Therefore, TDMA has three times the caprcity of an analog sptem using the same number of channels, TDMA systems operate in either the 800 MHz (IS-54) or 1900
MHz 0S-13O frequency bands.
78
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3
lxrnooucrroN To ENcrNrsRrNc DssrcN
TDMAis also used as the access technoloryfor Global System for Mobile communications (GSM). However, GSM implements TDMA in a somewhar different and incompatible way ftom 15-136. Think of GSM
Eual Band vs. Dual Mode
and 15-136 as two different operating systems that work on the same processot, Iike Windows and Linux both worLing on an Intel Pentium III. GSM systeos use encryption to make phone colls more secure. GSM operates in the 900 MHz and 1800 MHz bands in Europe and Asia and in the 1900 MHz (someti-es referred to as 1.9 GHz) band in the United Stares. It is used in digial cellular and PCS-based gntems. GSM is also the basis for Integrared Digital Enhanced Network (IDEN), a popular qntem introduced by Motorola and used by Nextel. GSM is the intemarional sandard in Europe, Australia and much of Asia andA6ica. ln covered areas, cell-phone-users can buy one phone that will work anywhere else the sandald is supported" To connect to the spe-
Ifyou travel
cific sen'ice prwiden in these different countries, GSM-users simply
Dual Band/Dual Mo&: The best of both worlds allows you to switch between frequency bands and transmission modes as needed. Changi"g bands or modes is done automatically by phones that support these options. Usually the phone will have a default option set, such as 1900 MHz TDMA, and will try to conneo at that frequency with *rat technolory fint. If it supporu dual bands, it will switch to 800 MHz if it cannot coonect ar 1900 MHz And if the phone suppofts more than one mode, it will try the digital node(s) first, then switc.h to analog. Sometimes you can eyen find Tri Mode phones. This term c:rn be deceptive. It may mean that the phone supports wo digital technologies, such as CDMA and TDMA, as well as analog. But it can also mean that it suppors one digital technology in oro bands and also offers analog support A popular version of the TiriMode qpe of phone for people who do a lot of intemational traveling has GSM service in tl-re 900 MHz band fot Europe andAsia, and the 1900 MHz band for the U.S. in addition to the analog service.
switch subscriber identification module (SIM) cards. SIM cards are smail removable disks that slip in and out ofGSM cell phones. Thry store all the connection data and identification numben you need to access a particular wireless serviceprovider. Uaforaurately, the 1900 MHz GSM phones used in the United States are not compatible with the intemational qntem. If you live in the United Stares and need rc have a cell phone access when you're overseas, the easieet thing to do is buy a GSM 900MHz/l800MHz cell phone for traneling. CDMA ekes an entirely diferent approach from TDMA CDMA, after dtgtizingdara spreads it out over the endre barrdwidth it has arailable. Multiple call" are overlaid over eac,h other on the channel, with each .oig"d i unique sequence code. CDMA is a form of spread spectrum, which simply means that data is sent in small pieces over a number of the discrete frequencies available for use ar anv time in the specfied range. AII the users uansmit in the same wide-band c.hunk of spectrum. Each usert signal is spread over the entire bandwidth by a unique spreading code, At the receiver, that same unique code is used to tecover the sigDal. Because CDMA rystems need m put an accurate time stamp on each piece ofa signal, it references the GPS system for this information. Between eight and 10 separare calls can be carried in the same dannel space as one anaIogAMPS call. CDMA tecbnologr is the basis for Interim Standard 95 (IS-95) and operares in both rhe 800 MHz and 1900 MHz ftequency bands. Idealln TDMAand CDMArue transparent to each other. In pmctice, high power CDMA signals wil raise the noise floor for TDMA receivers, and high power TDMA signals can cause overloading and jamming
ofCDMA
receivers.
What's the Difference Between Cellular and PCS?
a lot, you will probably want to look for phones that o$er dual band, dual mode or both. Lets take a look at each ofthese options. Dual Band: A phone ttrat has dual band capability .irr fr"quencies. This means that it can operare in both the 800"orit*r and 1900 MHz bands. For example, a dual band TDMA phone could use TDMA seryices in either an 800 MHz or a 1900 MHz qatem. Dual Mode In cell phones, mode refers to the type of transmission technologr used. So, a phone thar supported AMPS and TDMA could switdr back and fonh as needed. An imoortant factor to look for is that one of lhe modes is AMPS. This gives you analog service ifyou are in an area that doesn't have digiul suppon
.
.
.
Inside a Cell Phone scale, cell phones are some of the most intricare devices people play with on a daily basis. Modern digital cell phones can process millions ofcalculations per second in order to compress and decompress the voice stream. Ifyou ever take a cell phone apan, you will find that it contains iust a Few individual oarts: . An arnzingcirclti boad containing thl brains of rhe phone
On a "complerity per cubic inch"
. . . . . .
Anantenna A liquid cry*al displsy (LCD) A keyboard not unlike the one we saw in
a
TV remote control
A microohone Aspeaker
Abatterv
The circuit board is the heam of the systern. Here is an examole of a
Personal Communications Services (PCS) is a wireless phone service veqy similar to cellular phone service with an emphasis on petsonal service and
extended mobiliry. The term "PCS" is often used in place of digial cellular, but true PCS means that other seryices like paging, caller ID and e-mail are bundled into the sewice. 'While cellular was originally created for use in cars, PCS was designed Som the gound up for grearer user mobility. PCS has smaller cells and therefore requires a larger number of antennas to cover a geographic area PCS phones use frequencies between 1.85 and 1.99 gigahere, (1850 MHz-1990 MHz). Ti:chnicalln cellular s)'stems in the United States operate in tlrc 824-894 megahere (MHz) ftequency baqds; PCS operates in the 1850-1990 MHz bands. And while it is based on TDMA, PCS has 200 kHz channel spacing and eight time slots instead of the typical 30 kHz channel spacing aad three time slos found in digital cellular. Just like digital cellular, there are several incomparible standards using PCS technolory. Two of the most popular are Cellular Digital Packet Data (CDPD) and GSM.
Ihe &ont ofthe circtit boerd"
Ihe back ofthe circuit boarrl
AN ENcnvrnnrNc MARvEL In the photos abovg you see sweral computer chips. Letb talk about what some of the individual chips do. The Analogto-Digital and Digitalto-Analog conversion chips translate the outgoing audio signal from analog to digital and the incoming signal from digital back to analog. The Digial Signal Processor (DSP) is a highly cusromized processor designed to perform signal manipulation calculations at high speed. The microprocessor handles all ofthe housekeeping chores for the keyboard and display, deals with command and control signalingwith the base station, and also coordinates the rest of the functions on the board. The ROM and flash memory chips provide storage for the phone's operaring gntem and customizable feanues, such as the phone directory. The RF and power section handles power management and recharging and dso deals with the hun&eds of FM channels. Finally, the RF (Radio Frequenry) amplifiers simals in and out of tle antenna. ' Thehandle dispby Eas grown considerably in size as the number offeatures in cell phones have increased. Most phones curendy aailable ofer built-in phone directories, calculators and even games, And many ofthe phones incorporate some mre of PDA, or I0'eb browser. Some phones store cenain information, such as the SID and MIN codes, in intemal flash memorywhile otlers use e$ernal cards that are similar to Smartmedia cards. Cell phones have such tiny speaken and microphones that it is incredible how well most ofthem reproduce sound. As you can see in the picnrre above, the speaker is about the size of a dime and the microphone is no larger than the watch banery beside it. Speaking ofthe watch banery, this is used by the cell-phoneis intemal clock. chip. Vhat is amazing is that all of that funcdonality-which on.ly 30 years ago would have filled the endre floor of an ofrce building-norr fits into a package that sits comfortably in the palm ofyour hand!
Cell Phone Towers A cell phone tower is typicallya steel pole or lattice strucrure that dses hundreds offeet into the air. This cell phone towe! along I-85 near Greenville, S,C. is typical in the U.S.:
79
What They Can Do Cell phones provide a way of staying in touch and having instant communication at your fingenips. \7ith a cell phone, you can: . Call your sipficant other to let them know that you are on yourway
. . . . . . . . . . . .
home-
Contact the police or hospital ifyou have an emergency. Let the bos know that vou are stuck in traffic and will be late for that big meeting. Provide a way for others to contact you if you are always on the go. Call horne or work to check your messages while on the road. Store contact infonnation (names and phone numben). Make task or todo lists (some models). Keep trackand remindyou ofappointments (date book, calendar). Use the built-in calculator for simole math. Send or teceive e-mail (some mod-e,ls). Get information (news, enterainment, stock quotes) from the Interset (some rnodels). Play simple games (some models). Integrate other devices such as PDAs, MP3 playen, and GPS receivers (some models).
Features Here is a list offeatures that should be considered when lookine for a cell phone: Service plan
. '. . . . .
Mode Banery grye
Displal Included functions Special Features Size
'Price
Service Plan Before you set your sights on a particular make or model ofcell phone, you should decide on tle service plan that interests you Otherwise, you could find that the phone you want is not supponed by the plan you need.
$vlode Are you looking for aaalog or digftal? Do you prefer PCS or cellular? TDMA or CDMA? [f you harre read the Hor.r They Vork section, tfien you know what each ofthese terms mean. Look for dual mode/dual band phones ifyou uorel a lot,
Battery type
This is a modern towerwith three different cell phone providers riding on the same strucrure. Ifyou look at the base of the tower, you will see that eachproviderhas ia own equipment, andyou can also see howlitdeequipment is involved today (older towers often have small buildings at the base).
Cell phones use nn'o main bartery technologies: NMH (Nickel Meal Hydride)-high capacity battery that provides extla Dower for €xtended use Lii-ion (Lithium lon)-has a lot of power in a lightweight package but usuallycosts more than NMH baneries Nbte both the talk time and standby time when comparing phones. Also, check to see how long the battery takes to reclarge and whether a rapid charger is arailable. Most cell phone batteries are removable, but some of the smaller models have a built-in banery instead
80
Crrepnrn
3 lrrrooucrroN
To ENcrNaERrNc
Drsrcn
Display
Frice
AII cell phones ho'e LCD dirpl"n, but the specific features ofthe display qrn varn Size-A Iarge multi-line display is typically more expensive but neces-
Ifyou
sary if you plan to use the phone for wireless Intemet. Colorvs. monochrome-Most cell phones hese monochrome displap (16 graj/s), but a few are beginning to appear that have color. Cell phones with color screens need more memory and tend to be more expensive. Rcflecdve or backlit-Almost all cell phones have backlit screens, which are good for low light conditions.
Included Functions Most premium phones oIler all of these features while more econornical phones may only hare a Gw:
'. . . . . . . . . . . . . . . .
Phone Directory
Clock C.alculator Games Personalized/custom sounds
Appointment Reminder/Calendar Incoming number storage Automatic redial Last number recall
Mute/hold button One touch dialinghpeed dialing Vibrate mode
lock/Alarm Call forwarding Multiparry cdls
E-mail/text messaging Minibrowser
Special Features Some cell phones hane special features su& as: Wireless Internet Hands-free Headset /speakeqphone External volume/ringer control Rapid charger/buih-in charger
. . . . . . . . . .
Caradaoter
Modemfunction PC synchronization
PDA MP3 player
GPSreceiver 'When looking at phones with a headset or speakerphone connection, check to see ifthe plug is proprietary. Ifit is, then check the cost and availabiliry ofthe headset or speakerphone. All phones come with a charger, but not all of them "o-e with a rapid charger.
Size Think about how you plan to use the phone. Vill it mosdy be a car phonei Or do you plan to carry it in a pocket all day long? This will help you determine if that cheaper phone with the same features that weigts 10 ounces that onlyweiglx 5 ounces but costs twentyper-
is a bemer deal than the one
cenr more.
ate like most ofus, price is alweys a consideration. \Veigh the options carefully and make sure that you dorit pay for features that you probably will never use. Ifyou are not ineresrcd in wireless Intemet, rhet yot mxy not want to pay the extra bucla for a Vz{P-enabled phone.
FAQ Vhardoes a "dual-modeo p&onemean? A'dual-mode" phone allows the phone to utilize either a.ligttal signal, or ifnecessary, an analog signal. This additional abiliry will keep you connected in some temote regions that do not have digiul cellular capability yer.
What does the lide card in the back of a digital cdl phone do? The litde cad, called a SIM (subscribet identity rnodule) card, is what al' lows your phone to access the global networlc It contains your saned phone numbers, yorr phone number, and your PIN number, which you must enter in order to use lhe phone. The card is transferable, so that you may switch phones, and simply switch the SIM card into the new phone, and harre your phone number and otfier settings stay the same.
Hov can you send ermail 61 access tfie Inte.rnet from a digital cdl phond Ifyou hare a VA?-enabled phone, tfien you are able to access the wireless Internet. This version of the lntemet, is specifically designed to be easily usable on cellular phones and other portable devices. Even ifyour phone is not\FAP-enable4 you may still be able to send e-mail. Any phone that has SMS (short messaging service) capability, is able to send short tort messages to Intemet e-mail addresses, provided you purchase this service from your access
providet.
Hov do you add personal dng tones to a cell phonel Certain models ofcell phones allo,r the addition ofpersonal ring tones to your phone. The easiest way to do this (if it is available on your model) is to directly "compose" a ring using your cell phone. Instructions are included with your phone if this feature is available. Otherwise, you have two options. You can 6uy a data cable specific to your phone, and use software on a computer rc compose a song and donnload it to your phone, or you may visit a'Web site and have them send a special message that includes a ring tone to your phone (usually for a small fee). C.ool Faccs
Most newer digital cellular phones hare some soft of enterainment programs on them, renging from simple dice-*uowing games to memory and lggig pilzles. - Airproximately 20 percent of American teens (more gfuls than bq/s) own a cellular phone. Cellular phons are more popular in European countries than they are in the United States-with neaiy 213 of Europeans owning a phone, compared to only about 1/4 ofAmericans. The GSM sandard for digital cal phones was established in Europe in the mid-I980t-long before digital cellular phones became commonplace inAmerican culrure. The technology is now possible to locate a person using a cellular phone down ro a range ofa few meters, anlurhere on the globe. 3G (thid generarion wireless) phones may look more like PDAs, with features such as video-conferencing, advaaced personal calendar 6.rnctions, and muld-player gamiag,
CFIAPTER
4
ENcTNEERTNG CorvrM uNrcATroN [[] resentations are an integral part of ll' rt engineering project. Depending
E
on the size of the project, the
presentations might be brief, lengthy, frequent, and may follow a certain format requiring calculations, graphs, charts, and engineering drawings.
81
82
Cnaprsn4 ExcrNsBRrNcCouuuNrclrton Engineers areproblern soluers. Once thry haue obnined a solution to aproblern, they need to cornrnunicate effectiuely th:eir solution to uarious ?et?le inside or outsidt
their organization. Presentations Are An integralpart of any mgineeringproject.
'4s
an engineering student, you would be asked to present Jtour solution to assigned homework problems, write a technical report, or giue an oral presmtation to lour class, student organization, or An audimce at a student confermce. Later, as an mgineer, you could be asked to giue presmtations t0 lour boss, colleagues in your drsign group, sales and marketingpeople, the public, or to an outside customer. Depending on the size ofthe?roject, tltepresentations rnight be brief, lengtfui,freqaent, and mayfollow a certainforrnat. You rntght be asked to giue dcailed technicalpresentatilns conuininggrapbs, charts, and ercgineeri.ngdrawing, or thq may take the forrn of briefproject updates. In this chapter, we will expkin some of the corwnon engineeringpresentationformats. We will abo innodace you to Mieosoft@ PowerPoint, a tool that is used commonly to giue an attractiae and ffictiae presentadzn,
4.1
Communication Skills and Presentation of Engineering Work As an engineering srudent, you need to dwelop good written and oral communication skills. During the ner$ four or five years, you will learn how to s(press your thoughts, present a concept for a product or a service and an engineering analysis ofa problem and its soludon, or show your findings from experimental work. Moreover, you will learn how to communicate design ideas by means of engineering drawings or computer-aided modeling techniques. Staning right now, it is important to understand that the ability to communicate your soludon to a problem h as important as the solution itself. You may spend months on a project, but ifyou cannot effectively communicate to othersn the results of all your efforts may not be understood and appreciated. Most engineers are required to write reports. These reports might be lengthy, detailed, technical reports containing chars, graphs, and engineering drawings, or they may take the form of a brief memorandum or executive summary. Some of the more common forms of engineering communication are explained bdefly next.
4.2
Basic Steps Involved in the Solution of Engineering Problems Before we discuss some of the common engineering presentation formats, let us say a few words about the basic steps involved in the solution of an engineering problem. There are four basic steps that must be followed when analyzing an engineering problem: (1) defining the problem, (2) simplifring the problem by assumptions and esdmations, (3) performing the solution or analysis,
$d
(4)
verifing
dre resuhs.
Step 1: Defining the Problem Before you can obtain an appropriate solution ro a problem, you must 6rst thoroughly understand the problem irelf. There are many questions that you need to ask before proceeding to determining a solution. What is it exacdy that you want to analyze? ttr7hat do you really hnow
4.2 Basrc Srsps Irvor-wo
IN rrrE Sor.urrorv
or ENcnqrsRrNc
PnosLEMs
83
about the problem, or what are some of the things known about the problem?'S?hat are you looking for?'lfhat exacdy are you tr)4ng to find a solution to? Tirking time to undersand the problem completely at the beginning will save lots of time later and help to anoid a great deal of frustration. Once you understand the problem, you should be able to divide any given problem into two basic questions:'What is known? and \fhat is to be found?
Step 2: Simplifying the Froblem Before you can proceed with the analysis of the problem, you may first need to simplify it. Assumptions and Estimations Once you have a good understanding of the problem, you should then askyourself this question: Can I simplifr the problem by making some reasonable
and logical assumptions and yet obtain an appropriate solution? Undersan&ng the phpical lar.rs and the fundamental concepts, as well as where and when to apply them and their limitations, will benefit you gready in making assumptions and solving the problem. It is very impoftanr as you take diferent engineering classes in the next few years that you develop a good grasp of the fundamental concepts in each class that you ake.
Step 3: Performing the Solution or Analysis Once you have carefirlly studied the problem, you crn proceed with obtaining an appropriate solution. You will begrn by appl)nng the physical lans and fundamental concepts that govern the behavior of engineering systems to solve r:he problem. Among the engineering tools in your toolbox you will find mathematical tools. It is always a good practice to set up the problem in parametric form, that is, in terms of the variables involved. You should wait until the very end to substitute for the given values. This approach will allow you to change the value of a given variable and see its infuence on the final result.
Step 4: Verifying the Results The final step ofany engineering analysis should be the verification ofresults. Various sources of error can contribute to vnong results. Misunderstanding a given problem, making incorrec assumptions to simplify the problem, applying a physical lavr that does not truly fit the given problem, and incorporating inappropriate phpical properties are common sources of error. Before you present your solution or the results to your instructor or, later in your career, to yoru nranager, you need to learn to think about the calculated resuh. You need to ask yourself the following question: Do the results make sense? A good engineer must always find ways to check results. 'Sfhat if I change one ofthe given parameters. Howwould Askyourself this additional question: that change the result? Then consider ifthe outcome seems reasonable. Ifyou formulate the problem such that the final result is left in parametric form, then you can experiment by substituting different values for various parameters and look at the final result. In some engineering work, actual physical experimenr must be carried out to veri& one's findings. Staning today, get into the habit of asking younelf ifyour solution to a problem makes sense. Asking your instructor if you have come up with the right answer or checking the back ofyour textbook to match answers are notgoodapproaches in the long run. You needtodwelop the rneans to checkyour resulabyasking yourself tle appropriace questions. Remember, once you start working for hire, there are no answer bools. You will not want to nrn to your boss to ask ifyou did the problem right!
I
84
Figure
4.t
An erample ofengineering problem presentation.
4.3 HolrsvoRKPREsEIinAmoN
85
4.3 HomeworkPresentation Engineering paper is specially formatted for use by engineers and engineering students. The paper has duee cells on the top that may be used to convey such information as course number, assignment due date, and your name. A given problem may be divided into a "Given" section, a "Find" section, and a "Solutiono section. It is a good practice to drar horizontal lines to separate t}re known infornation (Given section) from the information that is to be found (Find section) and the analysis (Solution section), as shown in Figure 4.1. Do not write anything on the back of the paper. The grid lines on the back provide scale and an oudine for freehand sketches, tables, or plotting data by hand. The gdd lines, which can be seen from the front of tte paper, are there to assist you in &arving things or presenting tables and graphical information on the front ofthe page neady. These grid lines also allow you to presenc a freehand engineering draving with its dimensions. Your engineering assignments will usually consist of many problems, rhus you will present your work on many sheer, which should be stapled together. Professors do not generally like loose papers, and some may even deduct points from your assignment's total score if the assignment sheets are not sapled together. The steps for presenting an engineering problem are demonstrated in Example A.l.If you are presenting solutions to simple problems and you think you can show the complete solution to more than one problem on one p€e, then separate the two problems by a relatively thick line or a double line, whichever is more convenient for you.
Derermine the mass of compressed air in a scuba diving tank, given the following information. The internal volume of the ank is 10 L and the absolute air pressure inside the tank is 20.8 MPa. The temperature of the air inside the tank is 20"C. Use the ideal gas lav to analyze this problem. The ideal gas law is given by
PV= mRT where
P: 7: rn:
absolute pressure ofthe gas, Pa
volume of the gas, m3 mass,l
R: g"r"otrt*t,f{ 7:
absolute temperature, Kelvin,
Th" g* constant R fo r ir 's 287
K
JlkC. K At this time, do not worry about understanding the ideal gas lan'. This lawwill be explained to you in deail in Chapter 11. The purpose of this example is to demonstrate how a solution to an engineering problem is presented. Make sure you undersand and follow the steps shown in Figare4.2.
86
Crteprsn'4
ENcrNEBRnyc
Couuumcetrox
Always double-underline answers and state units
L
figurc
l.Z
An eranple
lndex answer
ol engineering homework presentation tor Erample 4.1.
ra{,ililrllillrEilillll
4.5 Dnrlrlro
4.4
ThcnlcrcAr, RepoRT
87
Progress Report, Excutive Summary, and Short Memos Progress Report repors are means of communicating to others in an organization or to the sponsors of a project how much progress has been made and which of the main objectives of the project hane been achieved to date. Based on the total time period required for a project, progress reports may be written for a period of a week, a month, sweral months' or a year. The format of rhe progress report may be dicated by a manager in an organization or by the project's Progress
sPonsors.
Executive Summary Executive summaries are means of communicating to people in top management posiuons, such as avice president of a company, the findings of a deailed study or a proposal. The executive summary, as the name implies, must be brief and concise. It is generally no more than a few pages long. In the executive summafy, references may be made to more comprehensive reports so that readers can obtain additional information if they so desire.
Short lvlemos Short memos are yer another way of conveying information in a briefway to interested individuds. Generally, short memos are under rwo pages in length. A general format for a shofi memo follows. The header of the memo contains information such as the date, who the memo is from, ro whom it is being sent, and a subject [ine. This is followed by the main body of the memo.
Date
May 3,2001 Fromr Mr. John Doe
To: Re:
Members of Project BudgetRequest
X
4.5 Detailed Technical Report Detailed technical reports dealing with experimental investigations generally contain the following items:
Title
The tide of a report should be a brief informative description of the repon contents. A sample of an accepable tide (cover) sheet is shown on the next page. If the report is [ong, a table ofcontents should follow the tide page.
88
Cner.rsn4
ENcrNssRrNc CouuulucarroN
Abstract This is a very important pan of a reporc because readers often read this part first to decide ifthey should read the repon in detail. In the abstract, in complete but concise sentences you state the precise objective, emphasize significant findings, and present conclusions and/or recommendations.
0bjectives The puqpose ofthe objectives section is to state what is to be investigated through the performance of the experiment. Be sure to list your objectives explicidy (e.g., 1., 2,, . . , , etc.) Theory and Analysis There are sweral purposes of the theory and analysis secrion:
n To state pertinent principles, laws, and equations (equations should be numbered); o To present analydcal models that will be used in the experimenq " To define any unfamiliar terms or qgmbols; and o To list imporant assumptions associated with rhe experimental doig". Apparatus and Experimental Procedures There are rwo main puposes of rhe apparatus and
experimenral procedures secdon:
1. To present
a list of apparatus and instrumentadon to be used, including the instrument
ranges, least count, and
identification numbers.
2. To describe howyou performed
the experiment. The procedure should be itemized (step 1., 2., etc,) and a schematic or diagram of the instrument setup should be included.
Data and Results The purpose of this section is to present the resuls of the experiment, as described in the stated objective, in a tabular and/or graphical form. These tables and graphs show the results of all your effons. Include descriptive information such as titles, colarnn or row head.ings, units, axis labek, and daa poinrr (data poins should be marked by O, El, A, etc.). All figures must be numbered and have a descriptive tide. The figure number and the tide should be placed below the figure. All tables must be numbered and have a descriptive tide as well. Howwer, for tables, the table number and the tide should be placed above the sble. It is sometimes necessary to note in this section that you have included the original data sheers in the appendix to yoru repoft. Discussion of the Results The puqpose of the results section is to emphasize and explain ro the reader the imponant resula of the experiment and point out their significance. lZhen applicable, be sure to compare experimenal results wich theoretical calculadons.
eonclusions and Recommendations The conclusions and recommendations section compares your objectives with your experimental results. Support your conclusions with appropriate reference materials. Be sure to state recommendations based on the conclusions.
Appendix The appendix
.
serves several purposes:
To provide the reader with copies of all original data sheets, diagrams, and supplementary notes.
4.5
Dpran-no Trcnrvrclr.
Rnponr
89
All State University Departnent of Mechanical Engineerilry
Course Tide
FqrerimentNo.
Erperiment Tide
Date Experiment C-ompleted
Students'Narnes
.
To displey sample calculations used in processing the daa. The sample calculadons should contain the following parts:
A tide ofthe calculation A statement of mathematical equation Calculation using one sample of daa
90
Crraprun
4
Encrxrnnuvc Couruur.ncenor,t References A list of references rhat have been numbered in the text must be included in the report. Use dre following format examples: For Books: Aurhor, dde, publisher, place ofpublication, date (year), and page(s). ForJoumalArticles: Aurhor, tide of ardcle (endosed in quotation mar[<s), name of journal, volume number, issue number, year, and page(s). For Intemet Materials: Author, dde, date, and URL address.
4.6
Oral Communication and Presentation 'We
communicate orally to each other all the time. Informal communication is part of our everyday life. V'e may talk about sports, weatler, what is happening around r:he world, or a homework assignment. Some people are better at expressing themselves than orhers are. Sometimes we say things that are misunderstood and the consequences could be unpleasant. When
it comes to formal presentations, there are certain nrles and suategies that you need to follow. Your oral presenation may show the results of all your efforts regarding a pro.iect that you may hane spent montlx or a yeax to dwelop. If the listener cannot follow how a product was designed, or hovr the analysis was performed, then all your efforts become insignificanr. It is very important that all information be conveyed in a manner easily understood by the listener. The oral technical presentation in many wqrs is similar to a wrinen one. You need to be well organized and have an oudine of your presenadon ready, similar to the format for a written rePort. It may be a good idea to write down what you are planning to present. Remember it is harder to erase or correct what you say after you have said it than to write it dovm on a piece ofpaper and correct it before you say it. You want to make wery effort to ensure that what is said (or sent) is what is understood (or received) by the listener. Rehearse your presentation before you deliver it to a live audience. You may want to ask a friend to listen and provide h.lpfitl suggestions about your style ofpresentation, delivery content of rhe talk, and so on. Present the information in a wry rhat is easily understood by your audience. Avoid using terminolory or phrases that may be unfamiliar to listeners. You should plan so that you worft overexplain key concepts and ideas, because those who are really interested in a specific arca of your talk can always ask questions later. Try to keep your alk to about half an hour or less because the attention span of most people is about 20 to 30 minutes. If you have to give a longer talk, then you may want to mix four presentation with some humor or tell some interesting related story to keep your audience's attention. Maintain eye contact with all of your audience, not just one or two people. Dorit wer hane your back to the audience for too long! Use good visual aids. Use presentation software such as Power Point to prepare your alk. When possible, incoqporate charts, graphs, animated &arvings, short videos, or a model. \07'h.en available, co demonsffate concepts for new products, you can make use ofprototyping technology and have a prototype ofthe product on hand for your presentation. You may also \ilant to have copies of the outline, along with notes on the important concepts and findings, ready to hand out to interested audience members. In summary, be organized, be well prepared, get right to the point when giving an oral presentation, and consider the needs and expectation ofyour listeners.
4.7 PovsRPorNT Pnrsnrrerrox
4.7
9l
PowerPointPresentation a Microsoft PowerPoint presentation, a tool that is used commonly used to give an attractive and effecdve presentation. If you are already familiar with creating a PowerPoint presentadon, you c:rn skip this secdon without tlre loss of
In this secrion, we will introduce you [o continuity.
Microsoft PowerPoint: Basic ldeas PowerPoint is a presentation tool that you can use to generat€ and organize your slides showing text, charts, graphs, and video clips. Vith PowerPoint, yorr crln also create supplementary materials, such as handoua for your audience and preparatory notes for your presentation. A typical PowerPoint window is shown in Figure 4.3. The main components of the PowerPoint window, which are marked and numbered, as shown in Figure 4.3, aret
1. Tide bac Contains the name of tlre current
active presentation.
2. Menu bar: Contains tte commands used by PowerPoint to perform certain tasks. 3. Toolbar buttons: Contains push buttons (icons) that execute commands used by PowerPoint.
4. Current slide
Shows the current active slide.
5. View buttons: Allows the user to switch to different \fiew options.
6. Drawing toolbar buttons: Contains push buaons for drarving. 7. Status bar: Gives information about the qurent presentation, including the template style currendy in use.
I
tigure4.3
The main components of the
PowerPoint window.
the design style or
92
Cnegrnn4
ENcrNssRrNc Couuurqrca.rroN
View 0ptions As you create your presentation slides, you maywant to choose one of the following View op-
tions, as shown in Figore 4.4.
Nomal
You use Normal view when creating or modifying slides for your presenation. In this view, you will see slides displayed one at a time. Slide Sorter: In this view, the thumbnail (small) size ofeach slide is displayed in a single window. Slide Show: You use this view option when you are ready to start yotu presenration. In the default mode, your slides are presented one slide at a time each time you click the Ieft mouse button or strike the Enter kqf. To exit the Slide Show mode, press the Escape key. Notes page '!7hen preparing for your presentation, it may be helpfirl to hare notes that go with each slide so ttrat you can rehearse for your presentation and have notes also available during your presentation. Using the Notes page view option, you can type notes in a teJft box that goes with each slide. The text box is shown below tlle reduced view of a given slide. \7h.en typing your notes in the text box, yov, mty warrt to increase magnification using the drop-down zoom arrow (fiom the toolbar), qfiich wi[ make it easier to see what you are tyPrng.
Templates PowerPoint offers a number of aruacdrre remplates that you can use in your presenadon. Each template has a graphic formar and a color scheme. To crqre a slide using a template, you need only to add tent, graphics, or pictures to that templare. PowerPoint allows you to use a
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Uiew options of PowerPoint.
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4.7
PovsRPorNT PnrsrNterrox
93
different template for a different slide. To cre:rte a presentation with a template go to Format then Slide D"sb . . . , as shown in Figure 4.5. Then select one of the arailable templates by clicking on it. The design templates are shown on rJre right-hand side of the PowerPoint window shown in Figure 4.6.
Slide Layout The PowerPoint Auto layout option offers over 25 arrangemena that you c:rn use to Present rhe contenr of your slides. The content of a slide may include text, chafts, graphs, and video clips. Each layout provides placeholders that allow you to insert text, tables, charts, pictures, media clips, diagrams, or organizational charts. To select a specific layout, go to Format then Slide Leyout. . . , as shown in Figue4.7. To insen text and objects, click the left mouse button inside the place holder and type or pasre rext or objects from other Microsoft documents. You can then use a number of formatting
Dg|ArTefitplates
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PowerPoint.
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Backgroundcolorwindow.
f
figuruLg
Accessingslidenaster.
tools available to conveft te:ft style to bullet tert, change font size, change color, or adjust paragraph spacing. Once you are done typing or insening tent or objects, click the left mouse burton outside the placeholder. You can also change the background color ofyour slides by going to Format then Ba-kg"ound . . . . Use the pull-down arrow to select the color of your choice and Apply it to the acrive slide or to all slides by pressing Apply to All. The Baclqground color window is shown in Figure 4.8.
Slide lvlaster For formal presentations, you may want to include your universiry logo or your student organization logo on the botom of every slide in your presentation. To do this, you need to creare a slide master by going to Mery then Master and Slide Master, as shown in Figure 4.9. Add the logo or any other items you want to the slide master in the boxes shown in Figure 4.10. Now you have a slide master that you c:rn use to create the slides for your presentation.
Inserting Slides from Another PowerPoint File You can insert slides from another PowerPoint file by going to Insert then Slides from Files . . . , as shown in Figure 4.Ll. Next, from &e Slide Finder dialog box, use the Browse. . . option to locate the file drat has the slides that you want insened into your new presentation. Choose the slide that you want under the Select Slide: option (you can use either the Normal or the Outline view to choose, as shown in Figures 4.I2 and,4.13). Note that if you want to maintain the format of the slide, you should toggle on the Keep sourse formatting box located on the lower corner of the Slide Finder dialog box.
Click to edit Master tifie C[.k to €dit Mster te]il 6fyl€6 -Sscord lel€l . Tf[ft] lsr€!
- F&Ior&"d
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Creating a slide masteL
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slidu from other files.
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Selecting Slide Transition.
Slide Transition Transitions give your slide presentations sound and visual effects as one slide disappears and r}re next slide appea$. You can select from a number of transition oprions by going to Slide Show and then Slide Thansition . . . , as shown in FigurCI 4.I4 arr,d.4;75. For example, you crn select Blinds Horizonal, Checkerboard Acloss, and so on. Norc the options are listed in alphabetic order. Next, using the drop-down arrow under the Speed: option, set the speed rate at a desired level of slow, medium, or fast. You can also add sound effects to the slide transitions by using the drop-down arrow under the Sound: option and selecting the one you like.
4.7
PowrnPonw PnssENTATroN
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Selecting Animation Schemes.
Animation Animation is yer another way of adding visual and sound effects to your slides. For enample, you c:m show a bulleted list of items one at a time, or you c:rn have each item fade away as you move ro the next irem. To use the Animation option, go to Slide Show then Animation Schemee . . . and pick the scheme which you like, as shown in Figures 4.16 and 4.17. Crarc a slide and ry various options to see how a panicular animation scheme worls.
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SetectingActionbuttons.
Action Buttons Action buttons allow you to move to
a specific slide
in another PowerPoint file, 'W'ord docu-
ment, or Excel file, without haning to leane your current presentation to access the file. The Action button is linked to the 6le, and upon clicking it, you automaticallywill go to dre file containing the desired slide or document. To create an Acrion burton, go rc Slide Show then Action Buttons and seled a button, as shown in Figure 4.18. Then go to the locadon on yow slide where you want to put the bufton, and while holding down the left mouse bwton, drag
4.7
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CnatinganActionbutton.
the mouse to create a rectangular box for the bumon, as shown in Figure 4.I9.T\e dialog box for Action Settings will appear nexr. You can then hyperlink (Hypedink to:) the Action
buttontoNextSlide(usethedrop-downarrow),URL...,orOtherFile....Youcanuse rhe Browse. . . burton to find the file thatyouwould like to connect to theAction bumon. You can also choose the Run Program: option to link the bumon to an executable file. There is also the Play Sound: option, which will run a sound file when the acdon button is pressed. The ac-
don bumon settings are shown in Figure 4.20. You can also change some of the Action button aftributes, such as color (with the mouse poinrer over the button) by double-clicking the left mouse bufton. Next, set the attributes, as shown in Figare 4.2I.
Rehearse Timing After you har.e prepared yoru presenadon, you need to rehearse your presentation before you deliver it to a live audience. Another reason for rehearsing your presentation is that, tfpi..lly, engineering presentations must be made within an allotted time period. PowerPoint offers an option that allows you to time yoru presenation. Therefore, you want to take advantage of this tool. Once you are ready ro reheane, go to Slide Show then Rehearse fimings, as shown in
Figxe4.22.
100
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Figure4.Z0 Actionbuttonsettings.
Figure4.Zl
Changingbuttonattributes.
Printing Slides, llandouts, and Notes Pages Once you harte put together your PowerPoint presentation, you may want ro print your slides, handouts, note pages, or an oudine view ofslides. To do so, go to File then Print. . . . These steps lead you to the Print dialog box shown in Figarc 4,23t
1' Print range
Select the slides that you want printed. You can pick the
Al[, Current slide,
or Slides range.
2, Print what 3.
Use the drop-down arrow and select what you want printed: Slides, Handouts, Note p4ges, or Outline view of slides. Color/grayscale: Use the drop-dovrn arrow, select from Color, Gra;rscale, Pure Black and
'White.
4.8 Engineering Graphical Gommunication Engineers use special kind of drawinp, called engineering drarings,
to conyey their
ideas
and design information about products. These dravvings portray viul informarion, such as shape of the produc, its size, type of materid ured and assembly steps. Moreover, machinists use the information provided by engineers or drafts pe$ons, on the engineering dralrings, to make the parts. For complicated qfsterns made of various paffs, the å also serve as a
Suruu.mv
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tiqwe4.22
Selecting Rehearse
U fiqurc4.Z3 Printwindowof
Iinings.
r01
PorlerPoint.
how-to-assemble guide, showing how the various para fit together, In Chapter 16, we provide an introduction to engineering graphical-communication principles. Ve will discuss why engineering dravings are important, how they are drawn, and what nrles must be followed to cr€i[e such drawings. Engineering rymbols and signs also provide valuable information. These sFmbols are a "language" used by engineers to corwey their ideas, solutions to problems, or analyses of cerain situations. In Chapter 16, we will also discuss the need for conventiond engineering symbols and will show some common rymbols used in civil, electrical, and mechanical
engineering.
SUMMARY Now that you hane reached this point in the text
o You should know rhe basic steps involved in the solution of engineering problems: !?hat
is
the problem? !(hat is to be determined? Can the analpis be simplified by making appropriate assumprions? \(hat are rtre physical [on's and principles that govern and pre&ct the be-
harior of the given problem? You should realize that you must alwqrs find yoru own wa)n to venS your solutions to a problem. There are no answer bools outside the classroom. o You should reatize that it is very important for engineers to knorp horr to communicate well wirh others both orally and in written form. You should also be familiar with various wqrc of giving an engineering presenration, including a home work presenation, Progress report' executive surnmary, short memo, and detailed rcchnical rePort.
.
Crrlr"rnn4
102
4J. 42.
4.3.
ExcrxrBnnvc CouuuNrcerrox
'W'rite
Investigate the operadon of various turbines. a briefrepon explaining the operation ofsrcam turbines, hydraulic rurbines, gas urbines, and wind rurbines. In a brief repoft, discuss why we need various modes of uansporation. How did they evolye? Discuss the role of public nansportadon, water transportation, highway transportation, railroad transponation, and air uansponation. Identifr the major components of a compurer,
'Vhen
engineering. preparing your presentation, consider a four or 6ve year detailed plan ofstudy, inyolvement \./ith ercra curricular activities, an internship, volunteer acdvities, and so on. Share your plans with 41.
4.8.
and briefly orplain the function or the role of each componenr.
4,L
Electronic communication is becoming increasingly important. In your own words, identifr the various situations under which you should write a lefter, send an e-mail, make a telephone call, or talk to someone in person. F.nplain why one panicular form of communication is preferable to the others available. You mey have seen examples of emoticons (derived from emotion and icons)-simple printable characters used in e-mrils to convey human facial expressions. Here are some examples of emoticons: ;
) or:-)
Iaryhing
: ( or:-(
Sad
:,(
or:.(
;)
or;-)
The
Ctyr"S
Sqprise4 Shocked
\pink Angry/Yelling
>:-(
Angry/Grumpy
:-*
Kiss
: -xx
Returning a kiss Hean (e.g.I <3 U) Broken Heart
folhwing
4.5.
Prepare
4.6,
school. Prepare
4.fl.
assignmenx could be done
Ifyour introduction to
engineering class has a terrn project, prepare a Poq/erPoint presentarion ofa lengh specified byyour instrucor and deliver it to your class at a date specified by your instructor. Prepare a 2O-minute PowerPoint presentation about
the history and future of engineering. Incorporate picnrres, short video clips, graphs, and so on in your presentation. Prepare a l5-minute PowerPoint presenadon about an engineering topic, such as alternative energy or an environmental issue that interests you and that you would like to learn more abour. Present it to ydur class. Visit the V'eb site of the National Society of Professional Engineers and research engineering ethics. Prepare a PowerPoint presentation showing why engineering ethics is so important and explaining why honesty and inregrity in engineering are essendal. Give
)rO
<3
410.
Smiling
:-D
:-O
4.9.
your classmates. From the subjects presented in this book, choose a topic, prepare a l5-minute PowerPoint presentation, and deliver it to your class.
as
group projeas.
a l5-minurc PowerPoint presentation about engineering and its various disciplines, and the next time you go home present ir to the juniors in your high
4"12.
413.
examples of engineering codes of ethics. Present an ethics-related case and involve the class by discussing it during your presentation. Visit the'Web sire of an engineering organization, such as ASME, IEEE, or ASCE, and prepare a PowerPoint presenation about this year's student design competitions. Deliver your presenertiori at one of your engineering organizaao n meetingp. Prepare a l0-minute PorverPoint presentation about
engineering professional rqgistration. Fxplain why professional registradon is imponant and what the requirements are. Probbms
4.14-4.20 in the Dan and
Results part of Section 4.5, all '4s tablzs and graphs must haae dzsni.ptiae infortnation such as titles, column or row headings, anits, and axis labels, and daa points mast be cbarfu marhed.. All fgures rnutt be numbered
described
a l5-minute PowerPoinr presenration abour
your plans to receiye a rewarding education and the preparadon that it takes to hane a successfirl career in
Pnogr.eMs and haae a dzsniptiue title.
The
fgure number and the ti.ilz
Spod (nph)
should be placed bebw tbefg.re. All ablcs rnust be nurnbered and haue a dzseiptiue ti.tle as well. Howetter, for ublzs, tbe tablz number and the titb should be placed aboae the able.
4.14. Plot the following data- Use two different y-axes. Use a scale ofzero to 100"F for temperature, andz'erc to 12 mph for wind speed. Present your work using the ideas discussed in this chapter and engineering papers. '-
Time
(p.m)
75 80
4 6
4
82 82
5
78
z 3
6 8
201 252 309 370 436
50 55 60 65
508
584 666
/>
755 844
80
941
)
/)
4
70
4
68
2
range
of -50" to 50"C.
cussed in this chapter and using engineering paper. 4J6. Create a table that shows the relationship between the units of mass in kilograms and pound mass in tie
range of 50 kg to 120 kg. IJse increments of 10 kg. Present your work incorporating the ideas discussed in this chapter and using engineering paper. 4Jt. The given data show the result of a model known as [email protected] distanrr' used by civil engineers to design roadwErs. This simple model estimates the distance a driver needs in order to stop his car, traneling at a certain speed, after detecting ehazxd. Plot the daa using engineering paper and incoqporating the ideas discussed in this chapter.
iSp*a i (nrh) il0
r55
30 40
i
Use Incremenm of 10oC. Present your work incorporating the ideas dis-
lo
t(
1)
i
Creare a table that shows the relationship beween the unia oftemperature in degrees Celsius and Fahrenheit
in the
Digtance (ft)
35
--
Tbmperatrgrc MndSpeed (mph) fF)
SCIppingSigbt
7A
I
4.15"
.-ii.-'-
103
StopprngSlght Distance (ft) 0 11
4/
r)
78
20
rt4
4.t8.
The given data represent the velociry distribution for a fow of a fluid inside a circular pipe with a radius of 0.1 m. Plot the data using engineering paper and incoqporating the ideas discussed in this chapter.
Radial distance' r(m) r = 0 correspondr the center of
to pipe
U(r) Velocity (mls)
-0.1
0
-0.09 :0.08 -0.07 -0.06 -0.05 -0.04 -0.03 *0.02
0.095
-0.01
0.18 0.255 0.32
0.3v5 0.42 0.455 0.48
0.495
0
0.5
0.01
a.495
a.02 0.03 0.04
0,48
0.05
0.375
0.06 o.0v
0.255
0.08 0.09 0.1
0.455 0.42
432 0.18 0.095 0
104 4J9.
Cuepmn4
Enrcnvr,BnrxcCo*ruuxrc^mron
In an annealing process-a proc€ss wherein mar€rials such as glass andmeal areheated to high temperarures and then cooled slowly to toughen them-thin steel plates are heated ro t€mpefttttues of 900"C and then cooled in an envirorunent with temperarure of 35"C. The resuhs of an annealing process for a thin plare is shovm belon'. Plot the daa using engineering paper incoqporating de ideas discussed in rhis chapter.
'Ifho
paper and incorporating the
ideas
"hapt€r.
D{cdioa
SNrgf*o,
X(mm)
F(N)
0
0
)
l0
10
20
r5
30
20
40
ta$edwhat is bitter does not knowwhat is sm'eet." Proanb
has never
-Gertnan
4,n. The relationship betqrcen a spring force and is defection is given below. Plot the resuhs using engineering
in tlis
CFIAPTER
ENcTNEERTNG I f -
5
Ersrcs
ngineers design many products and
ntouiC. many services that affect our quality of life and safe$. They
supervise the construction of buildings, dams, highways, bridges, mass transit systems, and power plants. Engineers
must perform under a certain standard of professional behavior which requires adherence to the highest principles of
ethical conduct.
105
r06
Cneprsn
5
ENcrNsnRrNc Emrcs
eloqumtly stated in the National Society of Profasional Enginens (NSPE) codz of uhics, "Engineering is an important and leamed profession. ,4s members of thh profession, mginens are exltected to exhibit the highest standards of honesty and intugriA. Engineering has a direct and uital impac-t on the qaaltry of Wfo, allpeople. According$t, the seruices proaid,ed. by engineers require honesty, impartiality, fairness, and equity; and must be dedicated to the protection ofthe public health, safety, and wefare. Engineers mu.st perform under a standard of professi.onal behaaior which requires adherence to the highest principles of ethical condact." In this chapar ae will discuss the importance of mgineering ethics and will present the National Society of Professional Engineers code of ethics in dztail. We will ako prouidz two cdse studies that you may uant to discass in class. '4s
5.1
Engineering Ethics Ethics refers to the srudy ofmorality and the moral choices rhat we all haye to make in our lives. Professional societies, such as medical and engineering, hane long esablished guidelines, standards, and nrles that govern the conduct of tleir members. These nrles are also used by the members of the board of ethics of the professional organization to interpret ethical dilemmas that are submined by a complainant. As we discussed in chapter 1, engineers design many producs, including cars, compurers, aircraft, clothing, to1's, home appliances, surgical equipment, heating and cooling equipment, health care dwices, tools and machines that make various produc$, and so on. Engineers also design and supervise the construction of buildings, dams, highways, and mass transit systems. They also design and supervise the consrucdon of power plants that supply power to manufacruring companies, homes, and offices. Engineers play a significant role in rhe design and maintenance of nadons'infrastructures, including communication systems, utilities, and transportation. Engineers are involved in coming up with ways of increasing crop, fruit, and vegetable yields along with improving the safety of our food products. As you ca.n see, people rely quite heavily on engineers to provide them with safe and reliable goods and services. There is no room for mistakes or dishonesty in engineering! Mistakes made by engineers could cost not only money but also more importandy lives. Think about the following: An incompetent and unethical surgeon could carrse at most the death ofone person at one time (when a pregnanr woman dies on the operating table, two deaths may result), whereas an incompetent and unethical engineer could cause the deaths ofhundreds ofpeople at one dme. If an unethical engineer in order to save money designs a bridge or a part for an airplane that does not meet the safery requirements, hundreds of peoples' lives are at risk! You rcdize that there are jobs where a person's mistake could be tolerated, For example, if a waiter brings you Coke instead of the Pepsi that you ordered, or instead of french fries brings you onion rings, you can live with that mistake. These are misakes that usually can be corrected without any harm to anyone. But if an incompetent or unethical engineer incorrecdy
"A man who has committed a mistake and doesrit correct it is committing another mistake."
-Confaci.us
5.3 Coos or Ernrcs ron ENcnrsrRs
107
designs a bridge, or a building or a plane, he or she could be responsible for killing hundreds of people. Therefore, you must realize why it is so imponant that as future practicing engineers you are expected to hold to the highest standards ofhonesry and integrity. In dre section that follows, we will look at an example of a code of er:hics, namely, the Nadonal Society of Professional Engineers code. The American Society of Mechanical Engineers, dre American Society of Civil Engineers, and the Institute of Electrical and Electronics Engiof ethics. They are typically posted at rheir \feb sites.
neers also have codes
5.2
The Code of Ethics of the National Society of Professional Engineers The National Society of Professional Engineers (NSPE) ethics code is very detailed. The NSPE ethical code of conduct is used in making judgrnents about engineering ethic-related cases that are brought before the NSPEs Board of Ethics Rwiew- The NSPE ethical code ofconduct follows.
5.3
Code of Ethics
for Engineers *
Preamble Engineering is an impornnt and learned profession. As members of this profession, engineers are orpected to exhibit the highest sandards of honesty and integrity. Engineering has a direct and vial impact on the quality of life for all people. Accordingly, the services provided by engineers require honesty, impardaliry, fairness and equity, and must be dedicated to the protection of t}re public health, safery and welfare. Engineers must perform under a standard of professiond behavior which requires adherence to the highest principles ofethical conduct.
l. Fundamental Canons Engineers, in the fulfillment of their professional duties, shall: and welfare of the public. Perform servrces only in areas oftheir competence. Issue public statements only in an objective and uuthfirl manner. Act for each employer or client as ftithfirl €ents or trustees.
1. Hold paramount the safety, health
2. 3.
4.
5. Avoid
deceptive acts. themselves honorably, responsibly, ethically, and lan'fully so as honor, reputation and usefiilness ofthe profession.
6. Conduct
ll. 1.
*
to enhance the
Rules of Practice Engineers shall hold paramount the safety, health, and welGre of the public. a. If engineers' judgment is overnrled under circumstances that endanger liFe or properry, they shall notifr their employer or client and such other audrority as may be appropriate. b. Engineers shall approve only those engineering documents which are in conformitywith applicable standards.
Sozrea; The National Society ofProfessional Engineen (NSPE).
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5
ExcrxBrnrNc Ernrcs
c.
Engineers shall not reveal facts, data, or information without the prior consent of the client or employer except as audrorized or required by law or this Code.
d.
Engineers shall not permit the use of their name or associate in business ventures with any person or firm which they believe is engaged in fraudulent or dishonest enteqprise. Engineers haring knowledge of any alleged violation of this Code shall report thereon to
e.
appropriate professional bodies and, when relevant, also to public authorities, and cooperate with the proper authorities in furnishing such informadon or assistance as may be required.
2.
Engineers shall perform services only in t-he areas of their competence. a- Engineers shall undertake assignments only when qualified by education or experience in the specific technical fields involved.
b.
Engineers shall not affix their signatures to any plans or documents dealing with subject matter in which they lack competence, nor to any plan or document not prepared under their direction and control.
c.
3.
Engineers may acc€pt assignments and assume responsibiliry for coordination of an entire project and srgn and seal the engineering documents for the entire project, provided that each technical segment is signed and sealed only by the qualified engineers who prepared the segment. Engineers shall issue public saternents only in an objective and truthful manner.
a. Engineers shall be objective and truthful in professional reports, statements, or testimony. They shall include all relevant and peninent information in such reports, sratements, or testimony, which should bear the date indicating when it was current. Engineers may express publicly technical opinions rhat are founded upon knowledge of the facts and competence in the subject marter. c. Engineers shall issue no statements, criticisms, or arguments on technical maters which are inspired or paid for by interested panies, unless they have prefaced their comments by explicidy identifring the interested panies on whose behalf they are speaking and by revealing the existence of any interest the engineers may have in the matters. Engineers shall act for each employer or client as faithful agen$ or tn$tees. a. Engineers shall disclose all known or potential conflicts of interestwhich could influence or appear to influence their judgment or the quality of r:heir services. b. Engineers shall nq1 accePt comPensation, financial or otherwise, from more than one parry for services on the same project, or for services penaining to the same project, unless the circumstances are fi:lly disclosed and agreed to by all interested panies. c. Engineers shall not solicit or accept financial or other valuable consideration, direcdy
b.
4.
or indirecdy, from ou*ide
agents
in connection with the work for which they are
rcsponsible.
d. Engineers in public
5.
service as members, advisors, or employees of a governrnental or quasi-governmental body or department shall not participate in decisions with respect to services solicited or provided by them or their organizations in private or public engineering practice. e. Engineers shall not solicit or accept a contract from a governmental body on which a principal or officer of their organization seryes as a member. Engineers shall avoid deceptive acs. a. Engineers shall not falsify their qualifications or permit misrepresentation of their or their associates' qualifications. They shall not misrepresent or exaggerate their responsibility in or for the subject matter of prior assignments. Brochures or other presentations
5.3 Coor or Ernrcs
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ENcrNsnRs
109
incident to the solicitation of employment shall not misrepresent percinent facts concerning employen, employees, associates, jointventurers or past accomplishments. b. Engineers shall not offer, give, solicit or receive, either direcdy or indirecdy, any contribution to influence the award of a contract by public authority, or which may be reasonably construed by the public as hardng the effect of intent to influence the amarding of a contract. They shall not offer any gift or other valuable consideration in order to secure work They shall not pay a commission, percentage, or brokerage fee in order to secure work, er.cept to a bona fide employee or bona fide established commercial or marketing agencies retained by them.
lll.
Professional 0bligations
1. Engineers shall be guided in all their relations by the highest sandards of honesry and integrity.
a. b.
Engineers shall acknowledge their errors and shall not drstort or alter the facts. Engineers shall advise their diens or employers when they beliwe a project will not be successfirl.
c.
d.
Engineers shall not accept outside employment to the detriment of their regular work or interest. Before accepti ng any outside engineering employment they will notifr their employers. Engineers shall not amempt to attract an engineer from another employer by false or misleading pretenses.
e.
2.
3.
Engineers shdl not promote their orrn interest at the expense ofthe dignity and integrity ofthe profession. Engineers shall at dl times strive to serve the public interest. a. Engineers shall seek oppornrnities to participate in civic affairs; career guidance for youths; and work for the advancement of the safety, health, and well-being of their community. b. Engineers shall not complete, sign, or seal plans and/or specifications that are not in conformity with applicable engineering standards. If the client or employer insists on such unprofessional conducq they shall notifr the proper authorities and withdraw from further service on the project. c. Engineers shall endeavor to extend public knowledge and appreciation ofengineering and its achievements. Engineers shall avoid all conducr or practice that deceives the public. a- Engineers shall avoid the use of $atements conaining a material misrepresentation of
fact or omitting a material fact.
b. Consistent with the foregoing, engineers may advertise for recruitment of personnel. c. Consistent with the foregoing engineers may prepare anicles for the lay or technicd press,
4.
but such articles shall not imply credit to the author for work performed by others. Engineers shall not disclose, without @nsent, confidential information concerning the business affairs or technical processes of any present or former client or employer, or public body on which they serve. a- Engineers shall not, without the consent of all interested pardes, promote or arrange for new employment or practice in connection with a specific project for which the engineer has gained pardcular and specialized knowledge. b. Engineers shall not, *'ithout the consent of all interested parties, panicipate in or represent an adversary interest in connecrion widr a specific project or proceeding in which
fl0
CneprBn
5
Er.rcrrveERrNc
Ernrcs
the engineer has
gined panicular specialized knowledge on behalf of a former client or
employer.
5.
6.
7.
8.
9.
Engineers shall not be infuenced in their professional duties by conficting interests. a. Engineers shall not accept financial or other considerations, including 6ee engineering destgqs, from material or equipment suppliers for specifying their product. b. Engineers shall not accept commissions or allowances, direcdy or indirecdy, from conuactors or other panies dealing with clients or employers of the engineer in connection with work for which the engineer is responsible. Engineers shall not aftempt to obtain employment or advancement or professional engage-
ments by untruthfully criticizing other engineers, or by other improper or questionable methods. a- Engineers shall not request, propose, or accept a commission on a contingent basis under circumstances in which their judgment may be compromised. b. Engineers in salaried positions shall accept part-time engineeringwork only to the extent consistent with policies of the employer and in accordancn with ethical considerations. c. Engineers shall not, without consent, use equipment, supplies, laboratory or office facilities of an employer to carry on outside private practice. Engineers shall not attempt to injure, maliciously or falsely, direcdy or indirecdy, the professional reputation, prospects, practice, or employment of other engineers. Engineers who believe others are gu.ilty of unethical or illegal practice shall present such informadon to the proper authority for action. a. Engineers in private practic€ shall not review the work of another engineer for the same client, except with the knowledge ofsuch engineer, or unless the connection ofsuch engineer with the work has been terminated. b. Engineers in governmental, industrial, or educational employ arc entitled to review and eyaluate the work of other engineers when so required by their employment duties. c. Engineers in sales or industrid employ are entided to make engineering comparisons of represented products with products ofother suppliers. Engineers shall accept personal responsibility for their profssional activities, provided, however, that engineers may seek indemnification for services arising out of their practice for other than gross negligence, where the engrneer's interests cannot otherwise be protected. a. Engineers shall conform with state registration lars in the practice ofengineering. b. Engineers shall not use association with a nonengineer, a corporation, or partnership as a "cloak" for unethical acts. Engineers shall give credit for engineering work to those to whom credit is due, and will recognize the proprietary interests ofothers. a. Engineers shall, whenever possible, name the person or persons who may be individually responsible for designs, inventions, writings, or other accomplishments. b. Engineers using designs supplied by a client reco gnizn that the designs remain the properry of the client and may not be duplicated by the engineer for others without express permission. c. Engineers, before undenaking work for others in connection with which the engineer may make improvements, plans, designs, inventions, or other records that may jusdfy copyngh* or patenfs, should enter into a positive agreemenr regarding ownership. d. Engineers' designs, dat& records, and notes referring exclusively to an employer's work are the employer's property. Employer should indemnify the engineer for use of the information for any pupose other than the original purpose.
5.4
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111
tebruary 2001 "By order of the Unired States Disuict Coun for the Disuict of Columbia, forner Section 1 1(c) of the NSPE Code of Ethics prohibiting competitive bidding, and all poliry sratements, opinions, nrlings or other guidelines interpreting im scope, have been rescinded as unlarvfully interfering with the legal right of engineers, protected under tle antitrust laws, to provide price information to prospective clients; accordingly, nothing contained in the NSPE
As Revised
Code of Ethics, policy statements, opinions, nrlings or other guidelines prohibits the submission of price quotations or competitive bids for engineering services at any time or in any amount." Conmitlee In order ro comecr misunderstandings which have been indicated in some instances since the issuance of the Supreme Court decision and dre entry of the
Statement by ilSPE Execulive
Final Judgment, it is noted that in its decision of April 25, 1978, the Supreme Coun of the United States declared: "The Sherman Act does not require competitive biddingl It is futher noted that as made clear in the Supreme Court decision:
1.
2. 3. 4. 5.
6.
Engineers and firms may individually refi.rse to bid for engineering services. Clients are not required to seek bids for engineering services. Federal, state, and local larrs governing procedures to procure engineering services are not affected, and remain in frrll force and effect. Srate societies and local chapters are free to actively and aggressively seek legislation for professional selecdon and negotiation procedures by public agencies. State registration board nrles of professional conduct, induding nrles prohibiting competitive bidding for engineering services, are not affected and remain in fiilI force and effect. State registration boards with authority to adopt nrles of professional conduct may adopt nrles governing procedures to obtain engineering services. As norcd by the Supreme Court, "nothing in the judgment prevents NSPE and its members
from attempting to influence governmental acdon . . .o.
Note:In regard to the question ofapplication ofthe Code to coqporations vis-a-vis real persons, form or type should not negate nor influence conformance of individuals to the Code. The Code deals wit}r professional services, which services must be performed by real pe$ons, Real persons in rurn establish and implement policies within business strucnures. The Code is clearly written to apply rc the engineer and items incumbent on members ofNSPE to endeavor to live up to its provisions. This applies to all pertinent sections of the Code. business
5.4
Engineer's Creed The engineer's creed, which was adopted by NSPE in 1954, is a statement of belief;, similar to the Hippocratic oath taken by medical practitioners. It was developed to state rhe engineering philosophy of service in a bridway. The NSPE engineer's creed is:
' . . .
To give the utmost of performance; To participate in none but honest enteqprise; To live and work according to the larvs of man and rhe highat standards of professional conduct; and To place
. '
service before profit, the honor and standing ofthe profession before personal advantage, and " the public welfire above all other considerations. In humility and with need for Divine guidance, I make this pledge.
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The engineer's creed is rypically used in graduation ceremonies or licensure certificare presentadons. These are additional definitions that should be studied carefully.
Academic Dishonesty-Honesty
is
very imponant in all aspects of life. Academic dishonesry
refers to behavior that includes cheating on tests, homework assignments, lab reports; pla-
giarism; Iying about being sick and not taking a test because of i$ signing t-he attendance sheet for another student, or asking another srudent to srgn the sheet for you in your absence, Universities have different policies that deal rvith academic dishonesry including giving the dishonest student a failing grade for the course or requiring the student to drop the class or placing a student on probation. Plagiarism-Plagiarism refers to presenting someone else's work as your own, You may use or cite the work of others inclufing information from journal anicles, books, online sources, TV or radio, but make sure that you cite where you obtain the information from. In Chapter 4,we discussed in desil how you should give proper reference in your oral and written communications. Confi6l of Interest-A confict between the individual's personal interests and the individual's obligations because of the position he or she holds. Contrace-Conffact is an agreement among two or more pafties, which they entered into freely. A legal conffact is one that is legally binding, meaning if not fulfilled it could have Iegal consequences.
Professional Responsibility-It is the responsibility associated with the mastery of special kind of knowledge that a person posse$ses and the use of knowledge for well-being and benefit of the sociery. Read the gasss-frern the following list-assigned to you by your instructor before class
and be prepared to discuss them in class.
The 2002 NSPE Milton F. Lunch Ethics Contest Facts Engineer A is a graduating senior with excellent credentials from X University. Engineer A has had a series of job interviews with engineering companies from around the U.S. Following interviews with several industrial companies, Engineer A decides to accept an offer with ABC Incorporated located in his hometown ofTownville and plans to notifyABC the following week. In the interim period, Engineer A receives a call &om Engineer B, an executive with KYZ Incorporated, a potential employer with whom Engineer A inrcrviewed. On behalf of XYZ, Engineer B offers Engineer A (at XYZt expense) a chance to visit XYZ's headquaners in Mountainville, a city located near a resort axea following Engineer A's graduation. Engineer A eadier had decided he would not accept a position with XYZ if offered a position byABC, because Engineer A wanted to live near Townville to be close to family and friends, and also because ABC provided better long-term professional oppornrnities. Howwer, after receiving the call from XYZ, Engineer A decides to accept tlre invitation to visit XYZ's headquarters and combine rhe trip with a post-graduation vacation, believing that the visit to XYZ will broaden Engineer A's knowledge of the employment market, as well as future professional opportunities with XYZ. A week after the trip, Engineer A calls ABC and informs the company that he will accept the position with ABC. 0uestion Vas
it ethical for
Engineer A to accept the invitation to visit )ClZ headquarters?
5.4 The 2003 NSPE Milton
F"
ENcnvsrn's Cnsro
ll3
lunch Ethics Contest
Facts Individual A is involved in a vehicular accident during which another individual, Individual B, is killed. During the police investigation, the police determine and recommend to prosecutors that the matter should be converted from a civil matter to a criminal prosecudon, and Individual A is charged with homicide. At p"n of the prosecutor's investigation, the prosecutor (Prosecutor C) retains Engineer D (an engineering expen) to review the record and provide a supponing forensic repon and testimony in the prosecution of Individud A. Following Engineer Dt extensive review of the record, Engineer D 6nds technical evidence to demonstrate that dre cause of the accident was noncriminal and determines that he c:tnnot support the prosecutiont theory that Individual At actions suppon a criminal prosecution. Learning of Engineer D's determination, Prosecutor C continues the prosecution of IndividualA, and does not call Engineer D as a wimess ar trial. The trial ends in a hung jury, and the case is expected to be retried in the near future. 0uestion
Vhat
are Engineer D's erhical obligations under the facts and circumstances?
The following is ethics related
cases
that were brought before NSPE's Board of Ethical
Review. These cases were adapted with permission from the National Society of Professional Engineers.
eonfidentiality of Engineering Report: Case No. 82'2 Facts Engineer A offers a home owner inspection service, whereby he undenakes to perform an engineering inspection ofresidences by prospective purchasers. Following the inspecrion, Engineer A renders a written report to the prospective purchaser. Engineer A performed this service for a client (husband and wife) for a fee and prepared a one-p€e written report, concluding that the residence under consideration was in generally good condition requiring no major repairs, but noting several minor items needing attention. Engineer A submitted his reporr to the client showing that a carbon copywlls sent to the real estate firm handling the sale ofthe residence. The client objected that such action prejudiced their interests by lessening rheir bargaining position with the owners of the residence. They also complained that Engineer A acted unethically in submitting a copy of the repoft to any others who had not been a party to the agreement for the inspection services, 0uestion Did Engineer A acr unethically in submining a copy of the home inspection report to the real estate firm representing the owners?
Gift Sharing of Hotel Suite: Case N0.87'4 Fach Engineer B is direcror of engineering with
a large governmenral agenq that uses many engineering consultants. Engineer A is a principal in a large engineering firm that performs services for that agency. Both are members of an engineering society that is conducring a trnto-day seminar in a disant ciry. Both plan to attend the seminar, and they agree to share costs of a nvo-bedroom hotel suite in order to have bener accommodations.
0uestion
\U?'as
it ethical for EngineerA and B to share
the
hod
suite?
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Credit for Engineering Work-Design Competition: Case No" 92-1 Facts Engineer A is retained by a city to desrgn a bridge as pan of an elwated highway q/stem. Engineer A then retains the services of Engineer B, a structural engineer with expenise in horizontal geometry, super strucnrre dobr, and elwadons to perForm certan aspects of the design services. Engineer B designs the bridge's three curved welded plate girder spans, which were critical elements of the bridge doig.t. Several months following compledoir of the bridge, Engineer A enters the bridge design into a national organizationt bridge design competition. The bridge design wins a prize. However, the entry fails to credit Engineer B for his part ofthe design. 0uestion
.Was
it
ethical for Engineer A to fail to give credit to Engineer B for his pan in the
design?
Services-Samc Serviccs for Different Clients: Casc No. 00-3 Facts Engineer A, a professional engineer, performs a naffic study for Client X as part of the clientt permit application for traffic flow for the development of a store. Engineer A invoices Client X for a complete traffic study. Later, Client X learns that part of the uaffic study provided by Engineer A to Client X was erlier developed by Engineer A for a dweloper, Client Y, at a nearbylocation and that Engineer A invoiced Client Y for the complete traffic sudy. The second study on a new project for Client X utilized some of the same rarv data as was in the repon prepared for Client Y. The final conclusion of the engineering study was essentially the same in both studies. 0uestion \?as
it ethical for Engineer A to charge Client X for the complete traffic
study?
Use of Alleged Hazardous Material in a Proeessing Facility: Case No. 99'11 Facts Engineer A is a graduate engineer in a companyt manufacnrring facility that uses toxic chemicals in its processing operations. Engineer Ais job has nothing to do with the use and control of these materials. A chemical called "MegaX" is used at the site. Recent stories in the news have reported alleged immediate and long-term human genetic hazards from inhalation of or other contact with Mega)(. The news items are based on findings from laboratory experimens, which were done on mice, by a graduate student at a well-respected universityt physiology department. Other scientists hane neither confirmed nor refuted the experimenal findings. Federal and local governments have not made official pronouncements on the subject Several colleagries outside ofthe company hane approached Engineer A on the subject and ask Engineer A to "do something" to eliminate the use of Megd( at the processing facility. Engineer A mentions this concern to her manager who tells Engineer A, "Dorit worry, we have an Industrial Safety Specialist who handles that." Two months elapse and Megax is sdll used in the facrory. The conroverry in the press continues, but since ttrere is no funher scientific evidence pro or con in the matter, the issues remain unresolved. The use of the chemical in the processing facility has increased and now more workers are exposed daily to the substance than was the case two months ago.
5.4 (luestion Does Engineer circumstances?
A
have an obligation
Software Design Testing:
ENcrNpsn's
Cnsso
115
to take fumher action under the facts and
NSPEBER ease $lo.
96'4
Facts EngineerA is employed by a software company and is involved in the design of specialized software in connection with the operations of facilities atrecting the public health and q"dity conuol, water quality control). A" p"n of the design of a parsafety (i.e., nuclear, "ir ticular software system, Engineer A conducrs srtensive testing, and although the tests demonstrate that the software is safe to use under existing standards, EngineerA is arrare ofnew draft smndards that are abour to be released by a standard setring organization-sandards that the newly designed software may not meet. Testing is exuemely cosdy, and the company's clients are eager to begin to move forward. The softrvare company is eager to satisfr its clients, protect the software company's finances, and protect existing jobs; but at the same dme, tfre management of the software companywants to be sure that the software is safe to use. A series of tests proposed by Engineer A will likely result in a decision whether to move forward with the use of the software. The tests are costly and will delay the use of the software at least six months, which will put the company at a competitive disadvantage and cost the company a significant amount of money. Also, delalnng implemenation will mean the state public service commission udlity rates will rise significandy during this dme. The company requests Engineer A's recommendation concerning the need for additiond software testing. 0uestion Under the Code of Ethics, does Engineer A hare a professional obligation to inform his company of the reasons for needed additional testing and his recommendacions that it be underaken?
Whistleblowing: Case No. 82'5 Facts Engineer A is employed by a large industrial company that engages in substantial work on defense prqects. Engineer At assigned duties relate to the work of subcontracors, including rwiew of rJre adequacy and acceptability of the plans for material provided by subcontractors. In the course of this work, Engineer A advised his superion by memoranda of problems he found with cenain submissions of one of the subcontractors, and urged man€ement to reject such work and require the subcontractors to comect the deficiencies he outlined. Management rejected the comments of Engineer A, pardcularly his proposal that the work of a panicular subcontractor be redesigned because of Engineer At claim that the subcontractor's submission represented er
il6
CneprsR
5
EncrNsERrrqc Erurcs 0uestion Does Engineer A have an ethical obligation, or an ethical right, to continue his effor$ to secure change in the policy of his employer under these circumstances, or to report his concerns to proper authority?
Academic Qualifications: ease No.
79'5
Facts EngineerA received a Bachelor ofScience dqgree in 1940 from a recognized engineering curriculum and subsequentlywas registered as a professional engineer in two states. Iater, hewas arvarded an earned "Professional Degree" from the same institucion. In 1960 he received a Ph.D. degree from an organization that anrards degrees on the basis of correspondence without requiring any form ofpersonal attendance or sudy at the institution and is regarded by state authorities as a "diploma mill.' Engineer A has since listed his Ph.D. degree among his academic qualifications in brochures, correspondence, and otherwise, without indicating its
naffe. 'Was
0uestion Engineer A ethical in citing his Ph.D. degree as an academic qualification under these circumsances?
Advertising-Misstating Credentials: Case No. 92-2 Facts Engineer A is an EIT who is employed by a medium-sized consulting engineering firm in a small ciry. Engineer A has a degree in mechanical engineering and has performed seryices almost exclusively in the field of mechanical engineering. Engineer A learns that the firm has begun a marketing campaign and in its literature lists Engineer A as an electrical engineer. There are other elecuical engineers in the firm. Engineer A alercs the marketing director also an engineer, to the error in the promotional literature, and the marketing director indicates that the error will be corrected: However, after a period of six months, the ertor is not corrected. 0uestion Under the circumstances, what actions, if any, should Engineer A take?
Advertising-Statement of Project Success: Case No. 79'6 [acts EngineerA published
an advertisement in the dassified section of a daily newspaper under tle heading, "Business Services," which read in firll "Consulting Engineer for Industry. Can reduce present process heating fuel consumption by 3Ao/o to 70o/o while doubling capaciry in same foor space. For more information contact Engineer A, telephone 123456:7890: 0uestion Sfas Engineer
At advertisement ethical?
[Jsing Technical Proposal of Another Without Consent: Case N0.83-3 Facts Engineer B submitted a proposal to a county council following an interview concerning project The proposal induded technical information and data that the council requested as a basis for the selection. Smith, a staff member of the council, made Engineer B's proposal available to Engineer A. Engineer A used Engineer Bt proposal without Engineer B's consent in dweloping another proposal, which was subsequendy submitted to the council. The extent to which EngineerA used Engineer B's information and data is in dispute berween the pardes. a
Prosr.Elrs
n7
0uestion W'as it unethical for Engineer A to use Engineer B's proposal without Engineer B's consent in order for Engineer A to dwelop a proposal *rat Engineer A subsequendy submined to the council?
Use of CD-ROM
for llighway Design: Case hlo. 98-3
Facts Engineer A, a chemical engineer with no facilities design and construcdon experience, receives a solicitation in the mail with the following information: "Engineers to&y cannot afford to pass up a single job that comes by, including construction projects that may be new or unfamiliar. Now-thanks to a revolutionary new CD-RoM-specifying designing and costing out any construction prorect is as easy as pointing and clicking yotu mouse-no matter your design experience. For instance, nwer designed a highway before? No problem. Just point to the'Highways'window and click Simply sign and return this lener today and you'll be among the first engineers to see how this firll-featured interacdve library ofstandard design can help you work faster than ever and increase your firmt profits."
Engineer
A
orders the CD-ROM and begins
to offer facilities design and construction
services.
0uestion \tras it ethical for EngineerA to offer facilities design and construcdon services under the facts presented?
SUMMARV Now that you have reached this point in the tex
.
You should understand the importance of engineering ethics and why you should live by
.
these codes of ethics. You should understand the engineer's creed and reason-s why you should take the pl"dge.
"Keep company with good men, and you'll increase their number."
-Ialian
Proverb
5.1. The following
is a series of questions peraining to the NSPE Code of Ethics. Please indicate whether the statements are tnre or false. These questions ,ue pro-
vided by the NSPE. Note: This ethics test is intended solely to rc$ individual knowledge of the specific language conained in the NSPE Code of Ethics and is not intended to measlue individud knowledge of engineering ethics
or the ethics of individual
engineers
or engineering
students.
1. Engineers in the fulfillment of their professional
2.
duties must careftrlly consi&r the safety, heakh and welfare of the public. Engineers may perform services outside of their areas of competence as long as they inform their employer or dient of this fact.
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ENcrNssRrNc Emrcs
3. Engineers may issue subjecdve and pardal statements if such starements are in writing and consistent with the best interesr of their employer, client or the public. Engineers shall act for each employer or client as faithful agents or trustees. Engineers shall not be required to engage in truthfi.rl acts when required to protect the public health,
14. Engineers may accept assignments and assume responsibility for coordination of an entire projecr and shall sign and seal the engineering documents
for the entire project, including each technical segment of the plans and documents . 15. Engineers shall strive to be objective and truthfirl in professional repons, statemenB or testimony,
with primary consideration for the best
of the engineert client or employer. The engineert
Engineers may not be required to follow the provi-
repofts shall include all relevant and pertinent information in such reports, sfatemen$ or testi-
could endanger or compromise ttreir employer or their
sions ofstate or federal lawwhen such acdons
client's interests, If engineers' judgment is overruled under circumstances that endanger life or properry, they shall no-
tify their employer or client and such other authoriry
as may be appropriate. Engineers may review but shall not approve rhose engineering documents that are in conformitywith applicable standards.
9. Engineers shall not reveal facs, data or information without the prior consent of the client or employer except as authorized or required by lar or this Code. 10. Engineers shall not permit the use of their name or their associate's name in business ventures with any person or firm that they believe is engaged in fraudulent or dishonest enteqprise, unless such enterprise
or activity is deemed consistent with applicable state 11.
interests
safety, and welfare.
or federal law.
Engineers having Lnowledge of any alleged violation of this Code, following a period of thirry days during which the violation is not corrected, shall re-
poffthereon to appropriare professional bodies and, when relwant, also to public authorities, and cooperate with the proper authorities in furnishing such information or assistance as may be required. t2. Engineers shall undenake assignments only when qualified by education or experience in the specific technical fields involved. 13. Engineers shall not affix their signatures to plans or documents dealing with subject matter in which they lack competence, but may affix their signatures to plans or documents not prepared under their direction and control where the engineer has a good faith belief that such plans or documents were competendy prepared by another designated parry.
mony, which shall bear the date on which the engineer was retained by the client to prepare the repofts. 16. Engineers may express publid technical opinions that are founded upon knowledge ofthe facts and competence in the subjecr matfer. 17. Engineers shall issue no statemenm, criticisms, or arguments on technical ma$ers that are inspired or paid for by interested parries, unless they have prefaced their comments by explicitly identifying the interested pardes on whose behalf they are speaking, and by revealing the existence of any interest tlre engineers may have in the matters. 18. Engineers may not panicipate in any matter involving a cqnfliff ofinterest ifit could influence or appear to influence their judgment or the quality of
their services. 19. Engineers shall not accept compensation, financial or otherwise, from more than one parry for services on the same project, or for services penaining to the same project, unless the circumstances are
firlly dis-
closed and agreed to by all interested parties.
20.
Engineers shall not solicit but may accept financial
or other valuable consideration, direcdy or indirectly, ftom ouride agents in connecdon with the work for which they are responsible, if such compensation ir firlly disclosed.
21.
Engineers in public service as members, advisors or employees ofa governmental or quasi-governmental
body or department may panicipate in decisions with respect to services solicited or provided by them or their organizations in private or public engineering pracdce as long as such decisions do not involve technical engineering mamers for which drey do not posses professional competence.
Pnonr.Bu 119 22. Engineers shall not solicit or accept a conffact &om a govemmental body on which a prin"lpal or officer of their organizadon serves as a member. 23. Engheers shdl not intentionally falsifr their qualifications or actively perurit writen misrep-
resentation of their or their associate's qualifications. Engineers may acc€pt credit for previous work performed where the work was performed during t}re period the engineer was employed by the previous employer. Brochures or other presequtions incident to the solicitation of employment shall specifically indicate the work
performed and the dates dre engineer was employed by the firm.
24. Engineers shall not offer, give, solicit, or
receive,
either direcdy or indirecdy, any conuibution to influence the an'ard ofa conuact by a public authority, or which may be reasonably construed by the public as having the effect or intent of influencing the arntard of a contract unless such conuibution is made in accordance with applicable federal or state election campaigr finance larvs and regulations. 25. Engineers shall acknowledge their errors after consultingwith their employer or client
A Case Studtt
From ].|SPE
*
The following is an ethics-related
case
that was brought before NSPE's Board of ethics review.
Facts Engineer A is a licensed professional engineer and a principal in a large+ized engineering fum. Engineer B is a graduate engineer who works in industry and has also worked as a student in Engineer At firm during a srunmer. Although Engineer B was employed in Engineer At firm, EngineerA did not have direct knowledge of Engineer Bt work Engineer B is applying for licensure as a professional engineer and requests that Engineer A provide him with a letter of refsence testi&ing as to Engineer B's engineering experience and that rhe engineer (Engineer A) was in direct charge of Engineer B. Engineer B was under the assumption that Engineer A had personal knowledge of Engineer B's work. Engineer A inquired about Engineer B's experience &om someone who had direct knowledge of Engineer B's experience. Based on the inquiry Engineer A provides the letter ofreference explaining the professional reladonship berween EngineerA and Engineer B.
Question it ethical for Engineer A to provide the lemer
'W'as
of reference for Engineer B anesting
to Engineer Bt engineering experience even though EngineerA did not hare direct control
as
of
Engineer B's engineering work?
References Section
IL3.-
Code ofEthics: Engineers shall issue public suternmts
onlt in an objectiue and. tru.th-
fulmanner. Section
II.3.a.-Code of Ethics: Engineas shall
stateTnents
be objectiae
and truthfal in professional re?orts,
or testimony. They shall i.ncludc all rebuant and pminmt information in
re?ortg statentents or testimonl, which should bear tbe dan indicating whm
it
*Materials were adapted with permission ftom the National Society of Professional Engineers (NSPE).
na
wa.s
sacb
carrent.
Drscussrox
121
II.5.a.-Code of Ethics: Engineas shall notfahify their qualifcations or pennit misrepresentarion of their, or their associates'qualifcatior.s. They shall not misrQ,resent or acaggerate their reEonsibiliry in orfor the nbjea maxer ofprior assignrnmts. Brochures or other presmatiorc incidznt to the soliciwtion of ernplayment shall not rnisr4tresent pminmt facts con-
Section
ceming ernpbyert, ernployees, associates, Section
aenttr.rers or
patt a.ccowpl^hments.
III.I.-Code
dard^s
Seai.on
joint
of Ethics: Engineers shall be pided in all tlteir rektions hy the highest stAnofhonesry and integrity.
III.8.a.-Code of Ethics:
ti.ce
Engineers shall conforrn
uith
state registration laws i.n tbe
prac-
ofmgineering.
Diseussion The Board has, on prior occasions, considered cases involving the misstatement of credentials of an engineer employed in a firm. In BER Case No. 92-1, Engineer A was an EIT who was employed by a medium-sized consulting engineering firm in a small ciry. Engineer A had a degree in mechanical engineering and had performed services almost exclusively in the field of mechanical engineering. Engineer A learned that the fum had begun a marketing campaign and in its literature listed Engineer A as an electrical engineer. There were other elecuical engineers in the firm. EngineerAalened dre marketing direcor, also an engineer, to the error in the promotional literature and the marketing director indicated that the error would be corrected. However, after a period of six months, the error is not corrected. In nrling that the firm should take actions to correct lhe error, the Board noted that the firm's marketing direcor has been informed by the engineer in question that the firm's markedng brochure contains inaccurate information that could mislead and deceive a dient or potential client. Under eadier BER Case No. 90-4, the marketing direcorhad an ethical obligation to take expeditious acdon to correct the error. The Board noted that the marketing director, a professional engineer, had an ethical obligation both to the dients and potential clients, as well as to Engineer A, to expeditiously correct the misimpression which mayhwe been created. The Board of Ethical Review can certainly understand in the present case the desire of Engineer A to assist another engineer (Engineer B) in enhancing queer oppornrnities and becoming licensed as a professional engineer. Obviously such assistance should not come under misleading or deceptive circumstances. Engineers have an ethical obligation to be honest and objective in their professional reports, and such repofts include wrirten assessments of the qualifications and abilities ofengineers and others under their direct supervision. Engineers that are not in a position to offer an evaluation ofthe qualifications and abilities ofother individuals should not provide such evaluations or prepare repofts that imply that they are providing such evaluations. Claiming to be in responsible charge of another engineer without actually having direct conuol or personal supervision over that engineer is inconsistent with the lemer and the spirit of the NSPE Code. By providing the report in the manner described, the Board believes Engineer A is sending the right message to Engineer B about what will be expected of Engineer B and his colleagues as professional engineers. Clearly, Engineer B desired the letter ofreference ftom EngineerA, a principal in a consulting firm, in order to improve his chances to become licensed as a professional engineer, and Engineer B is taking conscientious action to address the reguest. Professional engineers must always be mindfirl that their conduct and actions as professional
122
Cneprsn5 ExcnvnnnrNcErgrcs engineers set an example for other engineers, particularly those that are beginning their professional careers and who are looking for models and mentors upon which to build their professional identities. A professional engineer providing such a lener of reference should demonstrate that the author has obtained sufficient information about the candidate to write a letter
of substance and detail the individualt technical abilities
as
well as the individual's character.
A letter of recommendation for engineering licensure generally requires r:he recommending professional engineer to state in detail that the candidate possesses legitimate and progressive engineering work enperience. The Board is ofthe view thar an ahernative approach could have been for Engineer A to refer Engineer B back rc the engineer in the firm that was in responsible charge ofengineering for the lefter of recommendation. Howwer, the lecer provided by Engineer A was just as adequate and ethical.
Conclusion was ethical for Engineer A to provide the letter ofreference for Engineer B testi$'ing as to Engineer B's engineering o
It
Board of Ethical Review LorryT.
Bannes, P.E., NSPE E. Dave Dorchester, P.E., NSPE john Vl Gregorits, P.E., NSPE Paul E. Pritzker, P.E., NSPE Richard Simberg P.E., NSPE Harold E. S7illiamson, P.E., NSPE C. Allen \fordey, P.E., NSPE, Chair i':-
ENcTNEERING
Pnnt
Concrpts Evrnv
,t* : ...: '.':,,.
.".
:
., u;.
i l:i;;';
..r.':; j
..
I
F uNoAMENTALS
EUcTNEER
SHouro |(llow
:
i
Engineers are probfem solvers. Successfulengineers possess good communication skills =and are tearn phyers. Successful engineers have a good grasp of FUNDAMENTALS, which
they can use to
UNDERSTAND and SOLVE many
different problems. In Part Two of this
book, we willfocus on these engineering fundamentals. THESE ARE C0NCEPTS THAT EVERY ENGINEER, regardless
of his or her area of specialization, SHOULD KN0W. From
our observation of our surroundings, we have learned that we need only a few physical quantities to describe fully events and our surroundings. These guantities are length, time, mass, force, temperature, mole, and electric current. There are also many design variables that are related to these fundamental quantities. For example, length dimension is needed to describe how tall, how long, or how wide something is. The
fundamental dimension length and length-related variables, such as area and volume, play important roles in engineering design. To become a successful engineer, you need
to first fully understand these fundamentaland related variables. Then, it is important for you to know how these variables are measured, approximated, calculated, or used in engineering formulas.
Chapter 6
Fundamental Dlmensions and Units
Chapter 7
Length and Length-Related Parameters
Ghapter 8
Tlme and Tlme-Related Parameters
Chapter 9
Mass and Mass-Related Parameters
Chapter
l0
Force and Force-Related Parameters
Chapter
ll
Temperature and Temperature-Related Parameters
Chapter 12
Electrlc Current and Related Parameters
Chapter 13
Energy and Power
CHAPTER
6
FuNDAMENTAL D rnnENs roNs AND Uxrrs group of engineers preparing for the test flight of the Pathfinder, a solar-powered, remote-piloted
aircraft. The Pathfinder was constructed of composite materials, plastics, and foam to create a lightweight aircraft. The
aircraft has a wingspan of 30 m and
a
gross mass ot221kg.lt is capable of
moving at a speed of 7 m/s and is powered by a solar array that runs 6 electric motors which turn the propellers. Among other activities, NASA is
interested in using the aircraft
to study the upper atmosphere without disturbing it.
!-:'.U:jli!l1ir.!::-----r:{1.:!ilAr-----.!!:t1si!a:!-----
126
---rrir.!:!:l
6.1 In this chapter,
we
ExcrNrsRrNc PnoslErvrs eNo FuNpr,unNrel DrMsNsroxs
will ex?hinfundammtal
127
engineering dimensions, such as lmgth
and time, and their units, such at rneter and second, and their role in mgineering analysis and design. an engineering studant, and later as praaicing engi.neer, '4s whm perforrntng an analysis, you will fnd a need to conuert frorn one systern of units to anotlrer. Ve will expkin the stelts necessar! to conuert informationfrorn one slsteln ofunits to annther corre$b, In this chaptn, we will also emphasize thefact thatyou musr always show the a?praprinte anits that go with yur calcuhtions. Finally, we will explain what is rneant by an mgineering rystem and an mgineering corn?onent,
6.1
Engineering Problems and Fundamental Dimensions The eyolution of the human intellect has taken shape over a period of thousands ofyears. Men and women all over the wodd observed and learned from their surroundings. They used the knowledge gained from their observations ofnature to design, dwelop, test, and fabricate tools, shelter, wenpons, water transportation, and means to cultivate and produce more food. Moreover, they realized that they needed only a few physical quantities to firlly describe natural wents and their surroundings. For erample, the length dimension was needed to describe how tall or how long or how wide something was. They also learned that some things are heavier than other things, so there was a need for another physical quantiry to describe that observadon: the concept of mass and weight. Early humans did not firlly understand the concept of gravitf consequendy, the correct disdncdon between mass and weight was made later. Time was another phpical dimension that humans needed to understand in order ro be able to enplain their surroundings and to be able to answer questions such as: "How old are you?' "How long does it take to go from here to there?" "How long does it take to cook this food over fire?" The response to these questions in those early days may have been something like this: "I am many many Moons old," or "It takes a couple of Moons to go from our village to the other village on the other side of mountain." Moreover, to describe how cold or hor something was, humans needed yet another physical quantity, or phpical dimension, that we now refer to as temperanrre. By now, you understand why we need to formally define physical variables. The other impomant fact you need to realize is that early humans needed not only physical dimensions to describe their surroundings but also sorne way to scale or divide these physical dimensions. This realization led to the concept of unis. For example, time is considered a physical dimension, but it can be divided into both small and large ponions, such as seconds, minutes, hours, da1a, years, decades, centuries, millennia and so on. Today, when someone asks you how old you are, you reply by sa;nng, "I am 19 years old." You don't say that you are approximately 170,000 hours old or 612,000,000 seconds old, even though these statements may very well be true at that instant! Or to describe the distance berween two cities, we may say that they are 2000 kilometers aparq we dorft say the cities are 2,000,000,000 millimeters aparn The point of these examples is that we use appropriate divisions of physical dimensions to keep numbers manageable. Ve hane learned to create an appropriate scale for these fundamental dimensions and divide them properly so ttrat we c:rn describe particular events, the size of an object, the
128
Cnrprrn
6
Furp.ervrnmlr- DnasNsroNs eNp Uxrrs
thermal state ofan object, or its interaction with the surroundings correcdy, and do so without much &fficulty. Today, based on what we know about our physical world, we need sevenfunda.mental or base dimensions to arrectly express what we know of the natural wodd. They are ltngtb, mass, timo, tsrnperatuNe, elzctric current, rt nount of substance, and lwninous intensity. With the help of these base dimensions we can derive all other necessary physical quantities that describe how nature worlis.
6.2
Systems of Units Throughout the world, there are several qystems of units in use today. The most cornmon systems of unis are International System (SI), British Gravitational (BG), and the U.S. Customary units, which we will discuss next.
International System (Sl) of Units 'We
begin our discussion of qrstems of units with the International System (SI) of units, because SI is the most cornmon system of units used in the world. The origin of the present day International System of units can be traced back to 1799 with meter and kilogram as the first two base units. By promoting the use of the second as a base unit of time in 1832, Carl Friedrich Ganss (1777-1855), an imponant figure in mathematics and physics, including magnetism and astronomy, had a great impact in many areas of science and engineering. It was not until 1946 rhat the proposal for rhe ampere as a base unit for electric crrrrent was approved by the General Conference on Weights and Measures (CGPM). In 1954, CGPM induded amPere, kelvin, and candela as base units. The mole was added as a base unit by the 14th CGPM in 1971. A list of SI base (fundamental) units is given in Table 6.1.
;:'
m-2.0m
,',
.'.t..':,, ,
'r
-
---
---
t-tff il *o*r2okg
-:--_
r:i
,ffiiffiff*canfun kM ery
in aPProximatPl
*oil
to lo.seconds
:.]
=
-
,
,
,.. ;. fRrF=-1.25amas
.
,i:
in
rool:-":-"-
Comforttble kelvin nroruru roon tempera$re: 295 kervin 238
197 heavi€Et t:l atonrsknovm
;'
fr l2(-
:: '+4,
I+ '
:s
+.One of the
Common Carbon is
usedasa$andafil used as i Lightest arom
A caldle csnrlle bas hrc luminous hrminrurs A
'i
intensiry of approximately 1 candela
lz9
130
Cner'rsn
6
FuNpeunrvrar DrurNsroNs eNo UNrrs Listed below are formal definitions of base units as provided by the Bureau International des Poids et Mesures.
The meter is the length of the path traveled by light in a vacuum during a time interval of 11299,792'458 of a second. The bihgram is the unit of mass; it is equal to the mass of the international prototype of dre kilogram. The second is the duration of 9,192,631,770 periods of the radiation corresponding to the transirion between the two hyperfine levels of the ground state of the cesium 133 atom. The ampere is that constant current which, if maintained in two suaight parallel conductors of infinite length, of negligible circular cross section, and placed I meter apafi in a vacuum, would produce between these conductors a force equal to 2 x l0-7 nerwon per meter of length. The bebi.n" a unit of thermodynamic temperature, is the fraction | | 27 3 .16 of the thermodynamic remperature of the uiple point of water (a point at which ice, liquid water, and warer vapor coexist). The unit of Kelvin is related to the degree celsius, according to
K: "C + 273.16,
The mole is the amount of substance of a system that contains as many elementary entities as there are atoms in 0.012 kilogram of carbon 12. 'When the mole is used, the elementary entities must be specified and may be atoms, molecules, ions, electrons, other panicles, or specified groups ofsuch pardcles. The candela. is the luminous intensity, in a given direction, of a source that emits monochromatic radiation of frequency 540 x 7012 hertz and that has a radiant intensity in that direction of I1683 waft per steradian. You need not memorize the definitions of these units. From your weryday life experiences you have a pretty good idea about some of them. For example, you know how short a time period a second is, or how long a period a year is. However, you may need to dwelop a "feel" for some of rhe other base units. For example, How long is a meter? How tall are you? Under 2 meters or perhaps above 5 meters? Most adult peoplet height is approximately between 1.6 meters and 2 meters. There are exceptions of course. \fhat is your mass in kilograms? Developing a "feel" for units will make you a bemer engineer. For example, assume you are designing and sizing a new type ofhand-held tool, and based on your stress calculation, you arrive at an average thickness of 1 meter. Having a "feel" for these unia, you will be alarmed by the value of the thickness and realize that somewhere in your calculations you must haye made a mistake. \7e will discuss in detail the role of the base dimensions and other derived units in the upcoming chapters in this book. The CGPM in 1960 adapted the first series of prefixes and symbols of decimal multiples of SI units. Over the years, the list has been extended to include those listed in Thble 6.2. SI is the most common system of unia used in the world. The units for other physical quantities used in engineering can be derived from the base units. For example, the unit for force is rhe newton. It is derived from Newon's second laur of motion. One newton is defined as a magnirude of a force that 'qrhen applied to 1 kilogram of mass, will accelerate t}re mass at a rate of 1 meter per second squared (m/s2). That is:
1N = (1kCXlm/s2). Examples of commonly derived SI units used by engineers are shown in Thble 6.3. The physical quantities shown in Table 6.3, will be discussed in detail in rhe following chapters of
6.2 Sysrnnns or UNrrs
131
inerE;"i Multiplication Factors
=
Prefir
SI Symbol
t02a
yotta
Y
1021
ze$a
Z
1018
exa
E
l0r5
Pera
P
lo12
teta
1,000,000,000:
10e
ge
G
t,000,000:
106
mega
M
1000: 100 =
103
kilo
k
102
hecto
h
10r
da
lO-t
deka deci
10-2
centi
1,000,000,000,000,000,000,000,000 1,000,000,000,000,000,000,000 1,000,000,000,000,000,000 1,000,000,000,000,000 1'ooo'ooo'ooo'ooo
: : :
=
10:
0.1 0.01 0.001 0.000,001 0.000,000,001 0.000,000,000,001 0.000,000,000,000,001 0.000,000,000,000,000,001 0.000,000,000,000,000,000,0Qt 0.000,000,000,000,000,000,000,001
: : :
=
:
=
10-3 10-6 10-e
10-i2
I
d
milli
m
micro llano
tL P
n
: : :
10-r5
Pi@ femto
10-18
atto
1g-zt
zepto
z
=
10-24
yoclo
v
f
a,
this book. Saning in Chapter 7, we will discuss their physical meaning their significance and relevance in engineering, and their use in engineering analysis.
British Gravitational (BG) System In the British Gravitational (BG) qystem of unim, the unit oflength is a foot (ft), which is equal to 0.3048 meter; the unit of time is a second (s); and the unit of force is a pound 0b), which is equal to 4.448 newton. Note that in the BG system, a pound force is considered a primary unit and rhe unit of mass, while the slugis deived&om Newton's second laqr.'W'hen one slug is subjeced to one pound force, it will accelerate at a rate of 1 foot per second squared (ft/s2). That is: 116 : (1 slug)(lft/s'z). In the British Graviational s;rctem, the unit of temperature is orpressed in degree Fahrenheit fF) or in terms of absolute temperature degree Rankine ('R). The relationship berween the degree Fahrenheit and degree Rankine is given by:
"R:
"F
+
459.67
The relationship between degree Fahrenheit and degree Celsius is given by: g
z("F):;r("c)+32
132
Cneprrn
6
FunoeuBxrar. DnvmNsroNs eNo Uxrrs
Physical Qrotrtity
Name of SI Unit
Symbol for SI Unit
Expresion i't lbrms of Base Units
mR
Acceleration
A"g1.
Radian
rad
,;adl{
Angular acceleration Angular velocity
rad/s
#
Area
Densiry
I
Energy, work, heat
Joule
Force Frequency
Newton
N
Hetz
Hz
Impulse Moment or torque Momentum Power Pressure, $tress
b/m' N.m
or kg.mz/s'
k'*/t' s
N.S or kg.m/s
N.m or kg.mzls2 kg.m/s
'Vatt
\p
J/s or
Pascal
Pa
N/m2 or kglm.s2
N. m/s or kg.
m2/s3
m/s
Velocity Volume Elecuic charge Electric potential Electric resistance
Coulomb
C
A's
Volt Ohm
o
V
J1C or m2.kg/(s3.A2)
Electric conductance
Siemens
Elecric capacilance
Farad
Magnetic flux densiry Magnetic flux
'W'eber
Inductance Absorbed dose of radration
m3
Tesla
s F
T wb
Y/Aor m2.kg/(s3.,f)
llAorss.L2l(*'.kg) C/V or sa.A27(r''kg) V.s/m2 orkg/(f .A) V.s or m2.kg/(s2.A)
Henry
H
V.s/A
Gtf
Gy
Jlkgor r&l&
And the relationship between the degree Rankine and the Kelvin by:
Z('R)
:
g
art*>
\7e will enplore these relationships funher in Chapter 11.
U.S. Customary Units In the United States, most engineers still use the U.S. Customary qystem of units. The unit of length is a foot (ft), rvhich is equal to 0.3048 meter; the unit of mass is a pound mass (lb-), which is equal to 0.453592 kg; and the unit of time is a second (s). In U.S. Customary qrstem, the unit offorce is pound force (lbd and 1 lbsis defined as the weight ofan object having a mass of 1 lb- at sea level and at a latitude of 45o,where acceleration due to gravity is 32.2 ft1s2, One pound force is equal to 4.M8 newton (N). Be"ause the pound force is not defined formally
6.2
Camera
film
SvsrsMs
or Urvrrs
133
35 mm
Medication dose such
as
pills
100 mg, 250 mg,or 500 mg
Spons: 100 m breaststroke ot buterfly stroke 100 m, 200 m, 400 m, 5000 m, and so on
swunmrng
running Automobile engrne capacity
2.2L (liter),3.8 L, and
Ligbr bulbs
60
Electric consumption
kwh
Radio broadcasting sigrral ftequencies
V
100 \Pi
Police, 6re
153-159MHz
Global positioning sysrem signals
157 5.42MHz
US; Customary Unit Usage
broadcast band)
and 1227.60 MHz
U.S. Customary Units Used
20 galloas or2.67ff
Fuel tank capaciry
(l Sports (length ofa foorball
on
88-108 MHz (FM broadcast band)
A.54-l.6MHz (AM
F..amFlec of
so
or 150'!7
fiel$
Power capacity of an automobile
7.48 goJlons)
100 yd or 300
150 hp or
(l Distance beween two cities
ff :
hP
ft
S2500lb.ftls
= 5501b'ft/')
100 miles
(l
mile
=
5280 ft)
using Newtorfs second law and instead is defined at a specific location, a correction factor must be used in engineering fornulas when using U.S. Customary units. The units of temperature 'in the U.S. Customary system are identical to the BG q/stem, whidr we discussed earlier.
Finally, by comparing the units of mass in the BG $'stem (that is one slug) to the U.S. b. slighdy massive qfstem pound mass, we note 1 slug - 32.2Lb^. For those of us who (overweight), it might be wiser to o(press our mass in slugs rather"tight than in pound mass or kilogram. For enample, a person who has a mass of 200 lb. or 90.7 kg would appear skinny if 6.2 sl'rgs, and he were instead to express his mass as 6.2 slugs. Note that 200 lb. : 9O.7 4: therefore, he is telling the truth about his mass! So you dont hare to lie about your mass; knowledge of units can bring about instant results without any exercise or diet! Examples of SI and U.S. Customary uni$ used in our weryday lives are shown in Thbles 6.4 and 6.5, respecdvely.
134
Cneprsn
6
Fuvnervrnxter. DrlrsxsroNs enro Urcrrs
6.3 [fnit Conversion Some of you mqy recall that not too long ago NASA lost a spacecraft called, Mars Climate Orbiterbecause two groups ofengineers working on the project neglected rc communicate correcdy their calculations with appropriate units. According to an internal rwiew conduced by NASA's Jet propulsion laboratory "a failure to recognize and correct an error in a uansfer of information between rhe Mars Climate Orbiter spacecraft team in Colorado and the mission navigation team in California led to the loss of the spacecraft." The peer review findings indicated that one team used English units (e.g., foot and pound) while the other used SI units (e.g., meter and kilogram) for a key spacecraft operation. According to NASA, the information enchanged benryeen the teams was cridcal to the maneuvers required to place the spacecraft in the proper Mars orbit. An overview of the Mars polar lander mission is given in Figure 6.1.
Cfr@
. . .
IIL Entry, Descenf Landing
II. Cruise
L Launch
DeltaII7425 Launched January 3, 1999
. .
Launch mass: 574 kilograms
Thruster attitude control Four trajectory-correction maneuvers CICM); siteadjusfuent maneuver September
. .
1,t999;
contingency 5th TCM at entry -24 hours Eleven monftr cruise Near-simultaneous fracking with Mars Climate Orbiter orMars Global Survevor
dwing approach
I
figure
O.l
}tars polar landel mission overyiew.
S ourc e : Cottttesy of
NASd
. .
Arrival: December 3" 1999
.
.
Jettison cruise stage;
microprobes separate from cruise stage
. .
fV. Landed Operaffons
Hypersonic entry Parachute descent;
propulsive landing Descent imaging
landing site
of
Lands in Martian spring at 76 degrees South latitude, 195 degrees West
longitude (76os, 1 95"\ry)
. 90-day landed mission . Meteorology, imagrng, .
soil analysis, frenching Data relay via Mars
Climate Orbiter, Mars Global Surveyor, or directto-Earth high-gain antenna
6.3
UNrr Coxwnsrou
135
As you can see, as engineering studen$, and later as practicing engineers, when performing analysis, you will find a need to conven from one system of units to another. It is very imporrant foryou at this stage in your education to learn to conven information from one system of units to another correcdy. It is also imponant for you to understand and to remember to show all your calculadons with proper units. This point cannot be emphasized enough!Always show the appropriate units that go with your calculations. Example 6.1 shows the steps that you need to take to conven from one system ofunits to another.
A person who is 6 feet and 1 inch tall and weiglx 185 pound force (lb) is driving a
car at a
hour over a distance of 25 miles. The outside air temperature is 80oF and has a density of 0.0735 pound mass per cubic foot (lb-/ft'). Convert all of the values given in this example from U.S. Customary Unis to SI units. Conversion tables are given on the front speed of 65 miles per
and back end covers ofthis book.
H
Persoris heifitt,
)f ,ft ))(o.ao
t: vt
11: (r.854 Person's
weight,
=
1gJ.4cm
IZ
/ 4.443 N \ : h,)(-ff/ rrr.tr *
Y6r: (r85 Speed
/ too.-\
-)[ff/
ofthe car, S
s:
(e:
:
*)(ffi)(tiff-)
r04,607m/h
=
104.607
or
s: (ro<,eo'tX;#): Distance trareled,
D
rr.or,
^,,
D
*)(' : (25-r*r(P*)(o'911 rmile/\ lft
/\looo-/
1
Temperature of
ajLr,
T
\
r('c):;lT("F)-321 7('C)
:
1
;(so
-
*
32)
:26.7"C
\ = 4o'233knt
kmth
136
Cneprnn
6
FuNoeueNrer DrurNsroNs ervo Urrrs Densityof air, p
(
P
I ,_ .,,- ,..,,.,.
tb^)
= (0'0735 rr I f r
- --. r'i,r:lt::i.,.i.i:--.--iiil,rilr:
_.-l_.-_'
t'?l r
- .rrii!..:rlL:rr-
u*)
: t.t76kstm3
lb- / f=l+-) \0.3048 m/ ---,ti.rl"..rill!ltrl
--.:i.[:
For the following problems, use the conversion factors given on the front and back end covers of this book
a)
Convert the given value of area, A, from cm2 to m2.
A:
100
cr*
(
tm \2 : n: (roo.-')(.ffiJ o.ol m2 b)
Conven the given value ofvolume,
Z:
ftommm3 to m3.
1000 mm'
7: (rooo *-,)(,ofifu)' c)
V
=
ro-u
-,
Conven the given value of atmospheric pressure,4 from N/m2 to lb/in2.
?:
105
N/m2
, lP:) (o,zs4n\ -P: (ro'+)f 1in. ) \--m'/\4.448N/\ d)
:
kA\/ llb-
-'l=-r --lr----:r'-:--'---
6.4
lb-/ff.
1000 kglm3
/ \/rm\3 e: (rooo=J\"^*r)\ffi) --t-lt---"l1ll-
Msrbtn2
Conven the given value of the density of water, p, from kglm3 to
p
t
:
-----t'
:62'5 rb-/rt'
t- ---.:---t.----::---I-'--'---:-lt--:r::'::.-::--t-r--:--I
Dimensional Homogeneity Another important concept that you need to understand completely is that all formulas used in engineering analysis must be dimensionally homogeneous. W'hat do we mean by "dimensionally homogeneous"? Can you, say, add someone's height who is 6 feet tall to his weight of '$7hat would be the result of such a 1 85 lb6 and his body temperature of 98'F? Of course notl calculation? Therefore, if we were to use the formula L : a * b * c, inwhich the variable Z on the left-hand side of the equation has a dimension of length, then the variables 4 b, atd, c on the right-hand side of equation should also have dimensions of length. Otherwise, ifvariables
4 b, andchaddtmensions
such as length, weight, and temperature, respectively, the given
6.4 DrIvreNsroNAL
HotvroesNrrrY
137
formula would be inhomogeneous, which would be like adding someone's height to his weight and body temperature! Example 6.3 shows how to check for homogeneiry of dimensions in an engineering formula. (a) $Zhen a constant load is applied to a bar ofconstant cross section, as shown in Figure 6.2, the amount by which the end of the bar will defect can be determined ftom the following
relationship:
.-,PL AE
(6r)
d=-
where
d
of the bar in meter (m)
P = applied load in newton (N) Z = length of the bar in meter (m)
P
t
= end deflection
.,4
rigurc ce
Tie barin tmmple 6.3.
=
E:
cross-sectional atea of the bar in m2
modulus of elasticity of the material
V'hat are the units for modulus of elasticity? For Equation (6.1) m be dimensionally homogeneous, the units on the left-hand side of the equation must equal the units on the right-hand side. This equality requires the modulus of elasticity to have the unis of N/m2, as follows:
, PL
-a:- AE
(NX-) -
m2E
Solving for the units ofEleads to N/m2, newton per squared meter, or force per unit atea. (b) The heat cransfer rate drough a solid material is govemed by Fouriert
q
=
la:vi':
T,_7"
hA_tT-
rc.21
where
q
= hex transfer rate
/: A: T,
- Tz: Z
:
thermal conducdvity of the solid material in watts per meter degree Celsius,
'!7/m."C
areainn:.2 temperaflrre difference,
oC
thickness of the material, m
\Zhat is tJre appropriate unit for the heat transfer rarc q? Substituting for the unirc of &, A, 71, 72, and 4 we have q
=
. T,- T,: / Ut \.(_,)( -./.C\ = v kA__," / [*..i
From this you c:m -t_
._-.
see
-r:
that the appropriate SI unit for the heat transfer rate is the watt. r_i:-- l:lii
t38
Crngmn
6
ftnroenmr.r"rar. DrMsNsroNs axo Uivrrs
6.5
fit|umerical versus Symbolic Solutions '$7'hen
you take your engineering classes, you need to be arare ofwo imporant things: (1) understanding the basic concepts and principles associated with that class, and (2) how to apply them to solve real physical problems Gituations). In order m gain an understanding of the basic concepts, you need to study carefirlly the statement ofgoverning larrys and rhe derivations of engineering formulas and their limitations. After you have studied the undedying concepts, you then need to apply them to physical situations by solving problems. After snrdying a certain concept initially, you may think that you completely understand the concept, but it is .hto"gh the application of the concept (by doing the homework problems) thar you really can test yoru understanding. Moreover, homework problems in engineering rypicalty require either a numerical or a qfmbolic solution. For problems that require numerical solution, data is given. In contrast, in the symbolic solution, the steps and the final answer are presented with variables that could be substituted with data, if necessary. The following example will demonstrate the difference between numerical and symbolic solutions.
Determine the load that can be lifted by the hydraulic qFstem shown. All of the necessary information is shown in Figure 6.3. The general relationship among force, pressure, and area is explained in desil in Chapter 10. At this time, dorit worry about understanding these relationships. The purpose of this example is to demonsmate the difference berween a numerical and a symbolic solution. The concepts that are used to solve this problem are: F1 : zlttg, F2: m2g, and Fr: (r42lAt)fi, where Fdenotes forcs., m is mass, gis acceleration due to gravity (g= 9.8t m/sz), andr4 tePresen$ afea-
Numerical Solution 'We start by
making use of the given data and substituting tlem into appropriate equations
follows.
Ft: mrT: (1oo ks)(g.at
Rr:5cm
I Ite
riqure6.3 hydraulic systen of
Erample 6.4.
rn/s2)
: 9315
Rz:15cm
as
6.6 F,
A. : 't,Fr:
F2
:
8829
N
T(0.15 m\2
,
ao.o#(981
:
(*rkg)
(9.81
N)
:
rnls'z)
8829
:+
SrcNrrrceNr Drcrts (Frcunss)
139
N rnz
:
900 kg
Symbolic Solution For this problem, we could stan with the equation that relates F2to Fp and then simplify the as Tr and gin dre following manner:
similar quantities such
- Az- m8= n(R)', \ ft: n1^ryt*r9) 4rr: (15 cm)2 ,.^^ (R)' --+ ks) = 900 kB *t = nz = ffimr T.#(100 Often, this approach is preferred over the direct substitution ofvalues into the equation right array because it allows us ro change a value of a variable such as m1 or dte areas and see what happens to the result. For example, using the symbolic approach, we can see clearly that rn1 is increased to a value of 200 kg, then m2 changes to 1800 kg.
f
--.-iiiFdlilli.;ii.,lrii:l l--
irii
6.6
- ----.r.----.-. -- ---
Significant Digits (Figures) Engineers make measurements and c.ury out calculations. Engineers then record the resula of measuremen$ and calculations using numbers. Significant digi6 (figure$ represent and convey the extent to which recorded or computed data is dependable. For example, consider the instruments shown in Figure 6.4.'We are interested in measuring the temperature of room air using a thermometer, the dimensions of a credit card using an engineering nrler, and the pressure of a fluid in a line using the pressure gage shown. As you can see ftom these examples, the measuremenr readings fall between the smallest scale division of each instrument. In order to take the guess work out of the reading and for consistency, we record the measurement to one half of the smallest scale division of the measuring instrument. One hdf of the smallest scale division commonly is called the least count of rhe measuring instfl,rment. For enample, refering to Figure 6.4, it should be clear that the least count for the thermometer is l"F (the smallest division is 2"F), for the ruler is 0.05 in., and for the pressure g€e is 0.5 inches ofwater. There-
fore, using the given thermometer, it would be incorrect to record the air temperanue as 7L25"F and later use this value to c:rry out other calculations. Instqd, it should be recorded asTl +' 1"F. This way, you are telling the reader or the user ofyour measurement that the temperanrre reading falls between 70'F and 72oF. Note the correct way of recording the dependability of a measurement using the + sign and the least count value. As stated earlier, significant digir (figures) represent and convey the extent to which recorded or computed data is dependable. Signifi"sng digits are numbers zero through nine. However, when zeros are used to show the position of a decimal point, they are not considered significant digits. For example, each of the following numbers 175, 25.5, 1.85, and 0.00125
r40
Cnarrsn
6
FuNo,nrvrsNrAr-
DrmrNsrons exo UNns
b)3.35
a)
7l
=
* 0.05incfies
l't
c) 7.5
M
figure
6.4
+
0.5 inches of tlater
txanples of recorded measurements.
Note r}re zeros in number 0.00125 are not considered as sigrificant digits, since they are used to show the position of the decimal point. As another example, the number of significant digis for the number 1500 is not clear. It could be inteqpreted as having fwo, three, or four significant digits based on what the role of the zeros is. In this case, if the number 1500 was expressed by 1.5 X 103, 15 X 102, or 0.015 X 105, it would be clear that it has nro significant digits, By expressing the number using the power of ten, we c;ln make its has three significant digits.
6.7 ENcrNssRrNc CorvrpoxsNTs ANo Svstnts
141
accuracy more clear. Howsver, if the number was initially expressed as 1500.0, then it has four significant digits and would imply that the accuracy of the number is known to 1/10000. Addition and Subtraction
Rules When adding or subtracting numbers, the result of the calculation
should be recorded such tiat the last significant digit in the resuk is determined by the position of the last column of digits common to all of the numbers being added or subtracted. For example,
152.47
+ or
3.9
132.853
-
5
156.37
t?:7.853
(Your calculator will display.)
156.3
127
(However, the result should be recorded this way.)
The numbers L52.47 and.3.9 have five and nvo significant digits, respectively. IThen we add these rwo numbers, the calculator will display 756.37; however, since the first column after the decimal point is common to these numbers, the result should be recorded as 156.3. Rules 1$7hen multiplying or dividing numbers, the result of the calculation should be recorded with the least number of significant digits given by any of the numbers used in the calculation. For example, Multhlication and Ditlision
152.47
x or
3.9 594.633 5.9 X 102
152.47
+
3.9
39.0948717949 (Your calculator will display.) (Howwer, the result should be recorded this way.) 39
tJre above example, the number 152.47 has five significant digits and the number 3.9 has rwo significant digia. Therefore, the result of the calculadons should be recorded with two significanr digits, because the number 3.9 used in the calculations has the leasc significant digits. Finally, ir is wonh noting rhat in many engineering calculations it may be sufficient to record the results of a calculation to a fewer number of significant digits than obtained following the nrles explained previously. In this booh we present the results of example problems with mro or tfuee decimal points.
In
6.7
Engineering gomponents and Systems Every engineered product is made of components. Let us start with a simple example to demonsrate rrhat we mean by an engineering system and i$ componenm. Most of us own a winrer coat, which can be looked at as a q6tem. First, the coat serves a purpose. h primary function is to offer additional insulation for our bodies so that our body heat does not es@pe as quickly and as freely as it would without protecdve covering. The coat may be divided into smaller componenr: the fabric comprising the main body of the coat, insulating material, a liner, dreads, zipper(s), and bunons. Moreover, each component may be furher subdivided into smaller componenrs. For enample, the main body of the jacket may be divided into sleeves,
142
I
Crrer"rsR
6
FuNpalrsNrrAl DrunNsrous
ero Uwrrs
rigure6.t
A simple system and
its
components.
collar, pockets, t-he chest section, and the back seccion (see Figure 6.5).Each component serves a pu{pose: The pockets were designed to hold thiogs, the slewes cover oru arms, and so on. The main function of the zipper is to allow us to open and close the front of the jacket freely. It too consists of smaller components. Think once more about the overall purpose of the coat and the function of each component. A well-designed coat not only loola appealing to the eyes but also has functional pockets and keeps us warm during the winter. Engineering systems are similar to a winter coat. Any given engineered product or engineering system can be divided into smaller, manageable subqrstems, and each subsystem can be funher divided into smaller and smaller componen$. The components of a well-designed engineering sFstem should function and fit well together so that the primary puqpose of the product is attained. Let us consider another common example. The primary funcdon of a car is to move us from one place to another in a reasonable amount of dme. The car must provide a comfonable area for us to sit within. Furthermore, it must shelter us and provide some protection from the outside elements, such as harsh weather and harmfirl objects ouside. The automobile consists of thousands of parts. \Vhen viewed in its entirety, it is a complicated qlstem. Thousands of engineers have contributed to the design, development, testing, and supervision of the manufacnrre of an automobile. These include electrical engineers, electronic engineers, combustion engineers, materials engineers, aerodynamics er(perts, vibration and control expefts, air-conditioning specialists, manufacturing engineers, and indusnial engineers. \Vhen viewed as a q/srcm, the car may be divided into major subsystems or units, such as elecuical, body, chassis, power train, and ur conditioning (see Figure 6,6), Each major com' ponent can be further subdivided into smaller subsystems and their componen$. For example, the main body of the car consists of doors, hinges, locks, windows, and so on. The windows are controlled by mechanisms that are activated by hand or motors. And the electrical system of a car consists of a battery a starter, * ,lt.rn"tor, wiring, lights, switches, radio, .rri"roprocessors, and so on. The cart air-conditioning system consists of componen* such as a fan, ducs, diffirsers, compressor, evaporator, and condenser. Again, each of these componenB can be funher divided into yet smaller components. For example, the fan consists of an impeller, a
6.8
Prrysrcer Lews aNp OsssnvATroNs rN ENcrNBnRrNc
143
E rigure6.6 An engineering systen and
its
main components.
moror, and a casing. From these examples it should be clear that in order to understand a qrstem, we must first firlly understand the role and funcdon of its components. During the next four or five years you will take a number of engineering classes that will focus on specific topics. You may ake a satics class, rvhich deals with the equilibrium of objects at rest. You will leam abour the role ofexternal forces, internal forces, and reaction forces and their interactions. Iarer, you will learn the undedying concepts and equilibrium conditions for designing parts. You will also learn about other physical lalvs, principles, mathematics, and correlations that will allow you to analyze, desrgn, dwelop, and test various components that make up a system. It is imperative that during the next four or five years you firlly understand these larrs and principles so that you can design components that fit well together and work in harmony to firlfill the ultimate goal of a given system. Thus, you c:rn see the imporrance of learning the fundamentals. Ifyou don! you are likely to design poor components that, when put together, will result in an even poorer system!
6.E
Physical Laws and 0bservations in Enginecring As stated earlier, engineers apply physical and chemical lars and principles dongwith ma*rematics to design, develop, test, and mass-produce products and services r:hat we use in our everyday lives. The key concepts that you need to keep in the back of your mind are the phys-
ical and chemical laws and principles and mathematics. Hardnghad ahighschool education, you haveaprettygood ideaofwhatwemean bymath'Well, the universe, induding the earth that we ematics. But what do we mean by physical laws? live on, was created a cerain way. There are differing opinions as to the origin ofthe universe. Was it put together by God, or did it stan with a big bang?'We won't get into that discussion here. Bur we have learned .ht"gh observation and by the collecdve effort of those before us that things work a cerain way in nature. For example, ifyou let go of something that you are holding in your hand, it will fall to the ground. That is an observation that we all agree upon.
144
CneprsR
6
FuNoervrsNTAL DnvrnNsroNs ervo UNrrs
Ve
can use words to explain our observadons or use another language, such as mathematics, to
Newton (1642-1727) form,iarcd that observation inro a usefirl mathematical expression that we know as the universal lann of granitational attracdon. An imponant point to remember is that the physical larrs are based on observations. Moreover, we use mathematics and basic physical quantities to eKpress our observations in the form of a larv. Even so, to this day we may not firlly understand why nature worls rhe way it 'We just know it works. There are physicis* who spend their lives ff)4ng to understand does. on a more fundamental basis why nature behaves the way it does. Some engineers may focus on investigating the fundamentals, but most engineers use fundamental laq/s to design things. As another example, when you place some hot ob.iect in contact with a cold object, the hot object cools down n'hile the cold object warms up until they both reach an equilibrium temperature somewhere between the two initial temperatures. From your we ryday expertenc,e, yov know that the cold object does not get colder while the hot object gea hotter! \Vh.y is that? 'Well, it is just the way things work in nature! The second larr of thermodynamics, which is based on this observation, simply states that heat fl.ows spontaneously from a high-temperaflue region to a low-temperature region. The object with the higher temperature (more energetic) transfers some of its enerry to the low-temperafllre (less-energetic) object. \Vhen you put some ice cubes in a glass of warm soda, the soda cools down while the ice warms up and wentually melts away. You may call this "sharing resources." Unfornrnately, we as people do not follow this larnr closely when it comes to social issues. To bemer undersand the second law of thermodynamics, consider another example. Some ofyou may have young children or young brothers and sisters. Ifyou placed the child with some toys in a room that is tidied up and orderly, let the child play with the toys for a while, tJren came back in a few minutes, you would find toys scattered all over the room in a disorderly way. 'Why worft you find toys put auray nicely? \7ell, tlat is because tfrings work sponaneously in a express our findings. Sir Isaac
6vatvitartiovr arftratc*iorr coql be embarwassihg i* lrot pvogevly r^hAerstooA!
a
6.8
Prrrsrclr. Levs aNo Onsrrvetloxs rN ENcrNrsRrNc
145
certain direction in nature. These two examples demonsuate the second lav of thermodynamics. Thingr in nature work in a cerain direction by themselves. Engineers are also good bookkeepers.'What do we mean by this? Any of us with a checli ing account knows the imporance of accurate record keeping. In order to avoid problems, mosr of us keep track of the transactions in terms of payments (debits) and deposits (credits). Good bookkeepers can tell you instandy what the balance in their account is. They know they need to add to the recorded balance whenever they deposit some money and subtract from the balance with wery wirhdrawal from the account. Engineers, like weryone else, need to keep track of their accounts. Moreover, similar to bookkeeping a checking account' engineers keep track of (bookkeep) physical quantities when analyzing an engineering problem. To berer undersand rhis concept, consider the air inside a car tire. If there are no lealis, the mass of air inside the tire remains constant. This is a statement expressingcoweruation of mass, whichis based on our observations. If the tire develops a leak, then you know from your experience trhat the a$rounr of air wirhin the dre will decrease until you have a flat tire. Furrhermore, you know tfie air that escaped from the tire was not desuoyed; it simply became part of the surrounding atmosphere. The conservation of mass statement is similar to a bookkeeping method that allows us ro account for what happens to the mass in an engineering problem. 'What happens if we try to pump some air into the tire that has a hole?'Well, it all depends on the size of the hole and the pressure and flow rate of the pressurized air arailable rc us. If the hole is small, we may be able to infate the tire temporarily. Or the hole may be so large that the same amount of air thar we put into the tire comes right back out. To completely describe all possible siruations penaining to this tire problem, we can express the conservation of mass as the ratu at uhich air enters the tire rninus the rate at which the air baues the tire should be equal to the rate of accurnuktion or dep/ztion of ai.r insidz the tire, Of course, we will use the physical quantity mass along with mathematics to express this statement. We will discuss the conservation of mass in more detail in Chapter 9. There are other physical laws based on our observations that we use to analyze engineering problems. Con^senation of merg is another good example. It is again similar to a bookkeeping method that allows us to keep track of various forms of enerry and how they may '\7e will spend more time discussing the conservation of enchange from one form to another. ergy in Chaprcr 13.
Another important larv that all ofyou hare heard about is Newtonk second la t ofnoti.on If you place a book on a smooth table and push it hard enough, it will move. This is simply the way things work. Newton observed this and formulated his observation into what we call Newtont second larv of motion. This is not to say that other people had not made this simple observation before, but Newton took it a few steps funher. He nodced that as he increased the mass of the object being pushed, while keeping the magnitude of the force constant (pushing with rhe same effon), the objecr did not move as qui"kly. Moreover, he noticed that there was a direct relationship between the push, the mass of the object being pushed, and the acceleraalso noticed that there rras a direct relationship between the direction of the force and the direction of the acceleration. Newton expressed his observations using mathematics, but simply enpressed, this larv states that unbalanced force is equd to mass times acceleration. You will hare the oppornrnity to take physics classes that will allow you to srudy and explore Nevttorfs second lanr of motion further. Some of you may wen take a dynamics class that will focus in gtearer detail on motion and forces and their relationship. Dont lose sight of
tion of the objecr. He
the main idea" Phvsical lavs are based on observadons.
Crreprsn
146
6
FuNoervreNrer DruBNsroNs axo Uxrrs
)')
€
Another important idea to keep in mind is that a physical larr may not fully describe all possible situations. Statements of physical laws have limiations because we may not firlly undersand how nature works, and thus we may fail to account for all variables that can affect the behavior of things within our natural world. Some natural laws are stated in a panicular way to keep the mathematical expressions describing the observations simple. Often, we r€som to experimental work dealing with specific engineering applications. For example, to understand bener t{re aerodynamics ofa car, we place it inside awind tunnel to measure the drag force acting on the car. Ve may represent our experimental findings in the form of a chart or a correlation that can be used for design purposes over a predetermined range. The main difference bemreen laws and other forms of experimental findings is that the lavrs represent the results of a much broader observation of nature, and almost everything that we know in our physical world obeys these laws. The engineering correlarions, on the other hand, apply over a very limited and specific range ofvariables.
SUMMARY Now that you hane reached this point in the te:c
'
' ' '
You should understand the importance of fundamental dimensions in engineering analysis. You should also understand what is meant by an engineering q/stem and an engineering component. You should also realize that physical larvs are based on observation and experimentation. l6u should know the mosr common systems of units: SI, BG, and U.S. Customary. You should know how to convert values from one qFstem ofunirs to another. You should understand the difFerence between numerical and symbolic solutions.
Pnosr.EMs
61.
G
Examples 6.1and6.2.
64. Conven ftom SI Units
To U.S. Customarv Units
l20kmlh
miles/h and ft/s
w
80 ks
Btu/h and ho I( tb-
1000 ks/m3
lb-/fc
1000
100 m3
N
900
6.2.
lbcl':rr,2
9.81 m/sz
ftls2
6.3.
:
shear modulus of the material
e.
.'7n--7-
6-
F: rn : a: I/: R: r:
V)
t')
force (N)
(k$
mass
accnleraiton (m/sz)
velociry (m/s) radius (m)
time (s)
6.5. Determine the number of significant digits for the
65 miles/h
km/hr and m/s
60,000 Btu/h
w
120Lb^lFf
ke.lm3
286.5
30lbFllu;rz
kPa
2.2
200lb-
ks
2200
200 lbf
N
0.0286
following numbers. Number of Significant Digits
6.6"
X
tO2
Present the results of the following operations using the proper number of significant.ligits.
,TL
Your Calculator
a= lG
Dirpl"y.
+
r.2856
where
r55
d:
the angle of mist in radians
T
applied torque
:
-
where
To Sl Units
torque can be expressed by the following eguation:
m(Vz
(Vr- V)
r:-
Customarv Units
The angle of narist for a shaft subjected to twisting
(N'm)
I = length of the shaft in meter (m)
(N/-t)
'What is the appropriate unit for rI, if the preceding equation is to be homogeneous in units? 'Which one ofthe following equations is dimensionally homogeneous? Show your proof.
F(t, - tr) = ". d. F:nV
Conven the information given in the accompanying table &om U.S. Customary unia ro SI units. Refer to conversion ables on the inside front and back covers of this book. Show all steps of your soludons. See Examples 6.1and,6.2. Conven from U.S.
shaft's polar
u F=ma yz b. F: rn- ^ K
lbt
100 kPa
moment of inenia (measure of resistance to nristing)
,/:
Convert the information given in the accompanytng table ftom SI units to U.S. Customary units. Refer to the conversion tables on the inside front and back covers ofthis book. Show all steps ofyour solutions. See
147
155
10.1
:
- 0.52 = 0.52r
1558
X
3.585
+
12
=
1.2
=
:
Should Be Reconded As
Cneprsn
148
6 Fuloevnqrel DrunnsroNs
enp Uxrrs
following qFsterns, identi$ the major components, and briefly explain the function or the role of each component: a drcss shift, pants, a skin, shoes, a bicycle, roller blades.
61. For the
6.8.
For the following
systems,
d. a building e. a hot water heater
the following
Fouriert laur
Newton's larr of viscosity d. Newton's law of cooling
e. Coulombt larv Ohmt lar g. the ideal gas larr h. Hooket larv
f, i. j.
Fickt lav
k.
Faraday's lavr
the first law of thermodynamics
'Write Investigate t-he operation of various turbines. a briefreporc explaining the operation ofsteam rurbines, hydraulic turbines, gas turbines, and wind turbines. In a brief report, discuss why we need various modes of transponation. How did they wolve? Discuss the
role of public transpoftation, water ffanspoftation, highway transportation, railroad transportarion, and air transportation. 6.lL
Identify the major componen$ of a computer, and briefly explain the function or the role of each component.
6.13.
\tr7hich one
ofthe following equations is dimensionally homogenous? Show your proof,
a.
F(*
1l
r,)
:
!*V? 2'2 -
!*VZ
F: :mVZ 2'2 -:mV? c. F(V - Z) : !**l - !* 2'2 d. F(tt - tt) = mV2 - mV1 b.
6.15.
Express
car has a mass of 1500 kg. Express the mass and the weight of the car using BG and U.S. Customary units. Show the conversion steps.
the kinetic energy
](mass)(speed)2
of
a
with a mass of I2AA kg and moving at speed of 100 km/h using SI, BG, and U.S. Customary units.
lavs
c.
6.t!.
A
car
b. Darcy's lavr
6J0.
6J4.
a toaster
g. an airplane 69. Investigate what observations describe: a.
F': force (N) r = distance (m) ze : mass (kg) 7: velocity (m/s) r : time (s)
identi$' the major
comPonen$: a. ahousehold refrigerator b. a computer c. the human body
f,
where
Show the conversion steps. 6t6.
A machine shop
has a rectangular
foor
shape
with di-
mensions of 30 ft by 50 ft. Express the area ofthe floor in m2, in2, and cm2. Show the conversion steps. 617. A trunk of a car has a listed I'rggage capaciry of 18 Express the capacity in in3, m3, and cm3. Show rhe
f(,
ff.
conversion steps. 6.t8.
The 2005 Acura RL has a listed 300
horsepower,
3.5 liter engine. Express the engine size in both
k\f
and in3. Show the conversion steps. 6t9. The densiry of air at standard room conditions is l.2kglm3. Express rhe density in BG and U.S. Customary units. Show the conversion sreps. 6.20. On a summer day in Phoenix, fuizona, the inside room temperau[e is maintained at 68"F while the outdoor air temperature is a sizding 110oF. \fhat is the outdoor-indoor temperature difference in: (a) degrees Fahrenheit, (b) degrees Rankine, (c) degrees Celsius, and (d) Kelvin? Is one degree temperature difference in Celsius equal to one temperarure difference in Kelvin, and is one degree temperature difference in Fahrenheit equal to one degree temperature difference in Rankine? 6.21"
If
so, why?
A cantilwer beam shown in the accompanlng figure is used to support a load acting on a balcony. The defection of the centerline of the beam is given by the following equation -wX2, a t: ffiGz
'r 41"'c
+ 6I])
where
: zll : 7
defection at a given x location, (m) distributed load
Pnonrxms
149
Problem 6.21 A cantilever beam.
-E:
modulus of elasdciry
ing equation may be used to determine the temperature of a thin piece of marerial after some time t
N/-1
= second momenr of area (ma) r = disrance from the support as shown (m) 1
Z
:
T_T _ _entrcmmt Tnaa- L"i-*-,
length of the beam (m)
\J(hat is the appropriate unit for ar, if the preceding equation is to be homogenous in units? Show all steps
where
7:
ofyour work.
:
: 1:
'
where
g
=
/= 6;
=
7:
stopping sight distance (ft)
initial
speed
6.24.
(ftA)
acceleration due to gravity, 32.2
ftl?
coefficient of friction between tires and roadway grade ofroad
driver reaction time (s)
\7hat are the appropriate units for f and. G if the preceding equation is to be homogenous in units? Show all steps of your work 6,23. In an annealing process-a process wherein materials such as glass and metal are heated to high temperazures and then cooled slowly to toughen them-the follow-
plxe thickness (m)
r = dme (s) Those of you who will pursue
.c" zsuft G) + ry Z:
specific heat (J/kS . K)
c
TP
=
temperature ('C)
h -- hqt transfer coefficient p density (k/-1
6.n. A model knovrn u stopping sight disance is used by civil engineers to design roadways. This simple model estimates the distance a driver needs in order to stop his car while uaveling at a certain speed after detecting a hazard. The model proposed by the American Association ofstate Highway Officials (AASHO) is given by
S
*,(-#')
aerospace' chemical,
mechanical, or materials engineering will learn abour the underlying concepts that lead to the solution in your heat-transfer class. W'hat is the appropriate unir for / if the preceding equation is to be homogenous in units? Show dl steps of your work The air resistance to the motion of a vehicle is something important that engineers investigate. As you may also know, the drag force acting on a car is determined experimentally by placing the car in awind tunnel. For a given car, the experimental daa is generally represented by a single coefficient that is called the drag coefificient. It is given by the following relationship:
ud-
Fd
Ipv'A
where Ca
= &a9coefficient
.Fz :
measured drag force (lb)
150
CHaprsn
6
Furvoannnllrar, DrvmxsroNs.qNp UNrrs
p=
ir
V: A:
ur speed inside the wind runnel (ft/s)
density Glugs/fd)
6.29.
ftontalarea of the car (#)
C1 if the preceding equation is to be homogenous in units? Show all steps
Sir Isaac Newton discovered rhatany 2 masses zz1 and rn2 atrtac. each other with a force that is equal ir *"gnitude and acts in the opposite direction, according ro the following relationship:
Vhat is the appropriare unit for
r' -
ofyourwork or entended surFaces, commonly are used in a variety of engineering applications to enhance cooling. Common examples indude a motorrycle engine head, a la$rn mower engine head, heat sinks used in electronic eguipment, and finned rube heat exchangers in room heating and cooling applications. For long fins, ttre temperature distribution along the fin is given by
6.25. Fins,
Grn,m.
r-
where
F= G: 'rml :
ruz:
r:
the exponential relationship:
attractive force,
N
Universal Gravitational Constant mass
of panicle-l, kg
mass
of parricle-2, kg
distance between the center ofeach
particle, m
T- T*6*, = (fb-* - T^bi.or){* where 7"
=
*:\
6.30.
requested units. Show all the conversion steps. (a) 14.7
temperature (K)
lb/in2 to lb/ni, 6) 14.7lbli# to Pa (c) 14.7lblirf to kPa, (d) L4.7 Lbli# to bar (1 bar : 100 kPa).
t-Ei hA
h : the heat cansfer coefficient (W/m2. g; p : perimeter of the fin (m) .r4 : cross-sectional area of the fin (m2) / = thermal conducdvity ofrhe fin material
x=
distance from the base of the fin (m)
'\?'hat
6.26.
is the appropriate unit for h rf the preceding equation is to be homogenous in units? Show all steps of your worlc A person who is 180 cm tall and weights 750 newtons is driving a car at a speed of 90 kilometers per hour over a distance of 80 kilometers. The ouride air temperature is 30"C and has a densiq' of 1. 2 (l
grams per cubic meter. Convert all of the values given in this example from SI units to U.S. Customary units. 6.n. Use the conversion factors given on the &ont and back end covers ofthis text and converr the given values. (a) Value of arcaA: l6fuf to f/, (b) Value ofvolume
V: 64in3 to ff, (c) Value of area momenr I: 27.3'na to ft4.
6.28.
of inertia
The acceleration due to gravitygis 9.81 m/s2. Express the value gin U.S. Customary and B.G. units. Show all conversion steps.
What is the appropriate unir for G if rhe above equation is to be homogenous in units? Conven the atmospheric pressure in the given units to
6.3r.
For gases under cenain conditions, there is a relationship becween the pressure ofthe gas, its volurne, and its temperature as given by what is commonly called the ideal gas law. The ideal gas law is given by
PV:
mRT
where
P: absolute pressure ofthe gas, Pa Z: volume of the gas, m3 m: massrkg R: g* constant 7: absolute temperature, Kelvin 'What
is the appropriate unit for .& if the above equa' to be homogenous in units? 6.32. The amount of radiant enerry emitted by a surface is given bythe equation q : eaATft,where 4represents the rate ofthermal energy, per unit time, emitted by the surface in watts; e is the emissivity of the surface 0 ( e ( I and is untiless; o is the Stefan-Bolzman consranr (o : 5.67 X I 0-8); I represents the area ofthe surface in m2; and l5is the surhce temperaflue ofthe object expressed in Kelvin. $7'hat is the appropriate un it for a , if the above equation is to be homogenous in units?
tion
is
Pnosr-EMs 6.33.
A person's body temperarure is conrolled by (1) con-
6.36.
vecfive and radiative heat transfer to the surroundings, (2) sweating, (3) respiration by breathing surrounding air and exhaling it at approximately body temperattue, (4) blood circulation near the surflace ofthe skin, and (5) meabolic rate. Metabolic rate determines the rate
.Express the denThe density of water is 1000 kgl ^' sity of water in lb/fC and lblgallon (7.48 gdlons : 1
6.37.
A unit that is generally used to
6.38.
Start
ff).
of conversion of chemical to tlermal energy within a person's body. The metabolic rate depends on the person's activity level. A unit commonly used to a(Press
('F.fi3'hrlBtu). with I lbf 'ft /s and convert it to N'm/s
that
lb6'ft/s
1
is equal
lb6'ft/s is equal to
tary conditions, per unir surhce area, is cdled rnet;
is that?
pated by an average adult person sleeping for 8 hours, if he or she generates 0.7 mets and has a body surface area
of 19.6 fii. Expres your results in Btu and Joules
(1 Btu :
6.35.
express the insulating value of clothing is called clo. I clo is equal to 0.155 m'. "C /'W: Express tlis vdue in U.S. Customary units
the metabolic rate for an average person under seden1 met is equal to 58.2Y/1m2. Convert dris value to Sru/hr ff. Also, calculate the amount of energy dissi-
6.34.
151
1055 Joules).
The calorieis defined as the amount of heat required to raise the temperanre of I gram of water by l"C. Also, the energy content of food is rypically expressed in Calnies which is equal to 1000 calories. Convert the results of the previous problem to Calories (1 Btu = 252 calories). Convert the suength ofselected marerials given in the accompanfng table from MPa to ksi, where 1000
lb/in2
=
1 ksi.
Material
I-Iltimate
Ultimate
srength
strength
(MPa)
(Lsi)
6.39.
t
horsepower, how many kilowatts
Viscosity of fluid play significant role in the analyses of many fuid dynamics problems. The viscosity ofwater can be determined from the following correladon: /c.\ l-l
p: cJD\r-
caJ
where g,
:
I:
viscosiry (N/s'm2) temperature (K)
4=2.414x
l0-5
247.8 (9 cz: I40 (9
q:
\V'hat is the appropriate unit for c1, if the above equation is to be homogenous in units?
100-550
Aluminum alloys Concrete (compressio n)
10-70
Steel
550-860
Machine Spring
700-1,900 400-1,000
Stainlecs
Tool
900
Strucnual Steel
340-830
Titanium allovs
900-1,200
Vood (Bending) Douglas
fir
50-80 50-100 50-100
Oak Southem pine
"Nothing is too wonderfirl to be true if it be consistent with the lars of nature.'' Faraday Q 79 I - 1 867)
-Michael
and show
to 1 .36 watts. IGowing that 550
CHAPTER
LnNcrH AND LnncrH-RELATED PanaMETERs he important dimensions of a BMW 23 Roadster are shown in
the
illushation. The fundamental dimension length and other lengthrelated variables, such as area and volume (e.9., seating or trunk capacity), play important roles in engineering design. As a good engineer you will develop a "feel" for the relative
magnitude of various length units, area units, and volume units. lt is also
important for you to know how to measure, how to calculate, and how to
approximate length, area, and volume. Source: 23 Roadster 2.5il3.0i schematics reproduced with permision of BM\f AG
Munich.
152
All dimensions shown in inches
7.1 Lrxcnr
es.r, FuxpnrvrsNTAr. DrMsNsrox
153
'When
you becorne a practicing engineer, Jlou willfnd out that lou donl stop leaming new thingt eaen arter obtainingyour enginening fugu. For exampb, lou ma! worh on a projea in which the noise of a machine is A concem, and yoa rnay be asked to corne ap whh ways to redwce the leuel ofnoise. It may be the cate that during the four or fue years ofyour engineering education, you did not take a class in noise controL Considering your kck of understanding and bachground in noise re' ductioq you rnayfirst try tof,nd sorneone who speci.alizcs i.n noise contol who could solue the probbm for yoa. But your supmtisor may tell you that because of budga constraints and because this is a one-tirne prrject, you lnust corne up with a re*sonable solution yourself Therefora you will haue to leam somalting new and leam
farn lfyou haue a good grasp nfundeilying mginening concepts andfundammtals, the learning process coald be fun and quick. The point of this story is that during the nextfour years lou need to mahe sure that you leam the fundamenals well. an mgineering srudzn; you abo need to deuelop a hem awareness ofyour '4s surroundi.ngs. In this chaltter, we uill inuestigate the role of length, areq and uol' ume alongwith othn length-related aariabks in engineeriugapplications. You will leam how thaephysicaluariables affect mgineningfutiso dccisions. The tapics innoduced in this chapter are fundammtal in content, so dcuelopi.ng a good grasp of them will make you a better mgineer. You rnay see sorne of the concepts and ifuas innoduced hne in some other forrn in the engineering classes that you will take Iater. The main purpose of introducing these concepts here is ta help yoa become Aware of their importance and leam to looh for their rektion to other mgineering pararil.eters in yourfuture cla.sses when you study a specifc topic in deaiL
7.1 Length as a Fundamental
Dimension
'\?hen
you walk down a hallway, can you estirhate how tall the ceiling is? Or how wide a door Or how long the hallway is? You should develop this ability because having a "feel" for dimensions *ilt h.lp you become a better engineer. Ifyou decide to become a design engineer, you will find out that size and cost are imporrant design parameters. Hardng dweloped a "feel" is?
for the size ofobjects in your surroundings enables you to have a good idea about the acceptable range ofvalues when you design something. As you know, wery physical object has a size. Some things are biger than others. Some thinp are wider or taller than others. These are some common wa1's of expressing the relative size of objects. As discussed in Chapter 6, through their observation of nature people recognized the need for a physical quantity or a physical dimension (which today we call' bngth) so that they could describe their surroundings beter. They also realized that having a conrmon definition for a physical quantity, such as length, makes communication easier. Earlier humans may hane used a finger, arm, stride, sdck, or rope to meastue the size of an object. Chapter 6 also emphasized the need for harving scales or divisions for the dimension length so that numbers could be kept simple and manageable. Today, we call these divisions or scales Estems ofunits.
154
CHeprsR
7
LBNcrn arco I.sNcTH-REL^lrto Pl,nanrmreRs
In this chapter, we will focus our attention on lengdr and such length-related derived quantities as area and volume. Length is one of the swen fundamental or base dimensions that we use to properly express what we know of our natural world. In todayt globally driven economy, where products are made in one place and assembled somewhere else, there exists an even greater need for a uni-
form and consistent way of communicating information about the fundamenal dimension length and other relatedlength variables ,o Ih". parts manufacnred in one place can easily be combined on an assembly line with pars made in other places. An automobile is a good example of this concept. It has literally thousands of parts that are manufacnrred byvarious companies in different parts of the world. As sxplained in Chapter 6, we have learned &om our surroundinp and formulated our observations into larvs and principles.'We use these laqrs and physical principles to design, develop, and test products and services. fue you observing your surroundings carefirlly? fue you learning from your everyday observations? Here are some questioru to consider: Hane you thought about the size of a soda can? What are its dimensions? !?hat do you think are the important design factors? Most of you drink a soda wery day, so you know dut it fits in your hand. You also know that the soda can is made from aluminum, so it is lightweight. \7hat do you think are some of its other design factors? SThat are important considerations when designing signs for a highwa/ How wide should a hallway be? \D7hen designing a supermarket, howwide should an aisle be? Most ofyou have been going to class for at least 12 years, but have you thoughr about classroom seating amangemenm? For example, how far apafi are the desks? Or how far above the foor is the presentation board? For those of you who may take a bus to school, how wide are the seats in a bus? How wide is a highway lane? Wrat do you think are the imponant factors when determining the size of a car seat? You can also look around home to think about the dimension lengdr. Start with your bed: What are is dimensions, how far above the floor is it? !7hat is a typical standard height for steps in a stairway? V'hen you tell someone that you ovrn a 20-inch TY to what dimension afe you referring? How high offthe foor are a doorknob, showerhead, sink, lighr switch, and so on? You are beginning to see t}rat length is a very imporant fundamenal dimension, and it is rhus commonly used in engineering producr. Coordinate systems are examples of another ap-
plication where lengh pl"ys an imporant role. Coordinate qfstems are used to locate things with respect ro a known origin. In fact, you use coordinare sfstems eve{F day, even t}rough you dont think about it. \7hen you go from your home to school or a grocery store or to meer a friend for lunch, you use coordinate systems. The use of coordinate qrstems is almost second nanue to you. Lett say you live dowrrtovrn, and your school is located on the nonheast side of torvn. You know the exact location of the school with respect to your home. You know which streets to take for what distance and in which direcdons to move to get to school. You have been using a coordinate $rstem to locate places and things most ofyour life, even ifyou did not know it. You also know the specific location of objects at home reladve to other objects or to your-
TV is located relative to your sofa or bedroom. There are different tlpes of coordinate qFstems such as rectangular, cylindrical, spherical, and so on, as shown in Figure 7.1 . Based on the narure of a particular problem, we may use one or another. The most common coordinate sFstem is the recangular, or Cartesian, coordinate qystem (Figure 7.1).'\fhen you are going to school from home or meeting a friend for lunch you nse the rectangular coordinate q/stem. But you may not call it that; you may use the direffions nofth, east, west, or south to get where you are going, You can think of the axes of a self. You know where your
7.1 LnxcrnesaFur,roausNTAlDlMrNsroN
155
Cylindrical system
(r,0, z)
To locate an object at pointz4, with respect to the origin (point 0) of the Cartesian rystem,
Spherical system
(a o,4;
you move along the r axis by r amount (or steps) and then you move along the dashed line parallel to they axis byy amount. Findly, you move along the dashed line parallel to the z direction by z amount. How would you get to point,4 using the Cylindrical or Spherical qntem? Explain.
I
rigure?.t
Eranples of coordinate systems.
rectangular coordinate system as aligning with, for example, the east and nonh direcdon. People who are blind are expert users of redangular coordinate systems. Because they cannot rely on their visual perception, people with a vision disability know how many steps to ake and in which direction to move to go from one location to another. So to beter understand coor'!7hile at home, close your eyes for a few dinate systems, perform the following experimenr. minutes and try to go ftom your bedroom to the bathroom. Note the number of steps you had to take and in which directions you had to move. Think about it! All of us, at one dme or another, hane experienced not kno$'ing where we are going. In other words, we were lost! The smarter ones among us use a map or stop and ask someone for directions and dissnces (x and 7 coordinates) to the desired place. An o
156
b
Cnaprsn
7
LrNcrrr axp lxNcrrr-Rpranpo PanAMsrERs
tigure?.Z
An erample demoashatinq the use
of a coordinate systen.
Unie of Length
ir
Increasing
I
angstrom
I
micrometer or 1 micron
Order
I X l0-ro meter (m) 1 X 10-6 meter (m), l/1000 inch = 254' x 3.514598 x lO-a (m) 1/1000 meter (m) 4.217 ntllimeter (mm) 1/100 meter (m)
lmil I point (printer's) I millirnerer 1 pica (printer's)
I
Equivalent Value
cenrimeter
I hch
2.54 centimeter (cm)
I foot
12 3 feet (ft)
inc.hes
I yr"d
meter (m)
'l
1.1 yar4 (/O 1000 meter , ', 1.6093 kilometer (km; = 5250 ftet (ft)
I meter I kilometer
l mile
l
l0*t
1.O936yard-
(m)
:
prefixes in Chapter 6 thatwe use these multiplication prefixes to keep the numbers manageable. The International System of Units is used almost universally, except in the United States. The unit of length in the U.S. Customary qystem is foot (ft). The relation between foot and meter
unis
by 1 ft
:
0.3048 m. Thble 7.1 shows other commonly used units and their equivalent values in an increasing order and indudes both SI and U.S. Customary units to give you a sense oftheir relative magnitude. Some interesting dimensions in the narural world are is given
The highesr mounnin peak (Everest): Pacific
Ocean:
8848 m (29,O28
feei
Average depth: 4028 m (L3,2L5 feet) Greatest knovrn depth: 11,033 m (36,198 feet)
7.2 MresunrursNT
7.2
or l;encrn
157
Measurcment of LenEth Early humans may have used finger length, arm span, suide length (step length), a stich rope, chains, and so on to merrure the size or displacement of an object. Today, depending on how accurate the measurement needs to be and the size of the object being measured, we use other measuring devices, such as a nrler, a yardstick, and a steel tape. All of us have used a nrler or tape m€asr$e to measure a distance or the size of an object. These dwices are based on internationally defined and accepted units such as millimeters, centimeters, or meters or inches, feer, or yards. For more accurine measuremeffs of small objects, we hare developed measurement tools such as the micrometer or the Vernier caliper, which allow us to mensure dimensions within 1/1000 of an inch. In fact, machinists use micrometers and Vernier calipers -..rery day.
On a larger scale, you have seen the milepost markers along interstate highways. Some people acnrally use the mileposts to check the accuracy of their car's odometer. By measuring the time between two markers, you can also check the accuracyofthe car's speedometer. In the last few decades, electronic distance measuring instruments (EDMI) have been developed thar allow us to measure disances from a few feer to many miles with reasonable accuracy. These electronic distance measuring devices are used quite commonly for surveying purposes in civil engineering applications. The instrument sends out a light beam that is refected by a system of refectors located ar the unknown distance. The instrument and the reflector qfstem are situated such that the reflected light beam is intercepted by the instrument. The instrument then interprers the information to determine the distance between the instrument and the refector. The Global Positioning System (GPS) is another orample ofrecent advances in locating objects on the surFace of the earth with good accuracy. As of the year 2000, radio signals were sent from approximately 24 aftfrcrd satellites orbiting the eanh. Tracking stations are located around the world to receive and interpret the signals sent from the sarellites. Although originally the GPS
A Vernier callper and a
mlcrometer.
158
Cnerrrn
7
LsNcrH ervo I;sNcrH-RnLuso PenaMsrERs
An example of an electronic
distance measuring
instrument used in surveying. was funded and controlled by the U.S. Department of Defense for
military applications, it now
has millions of users. GPS navigation receivers are now cotrunon in airplanes, automobiles, buses, and hand-held receivers used by hikers. Sometimes, distances or dimensions are determined indirecdy using trigonometric principles. For example, let us say r:har we want ro determine the height of a building similar to the one shown in Figure 7.3.but do not have access to accurate measuring devices. Vith a drinking straw, a protractor, and a steel crpe we can determine a reasonable value for the height of the building by measuring the angle a and the dimensions d and h1. The analysis is shown in Figre 7 .3, Note that /1 represents the distance from the ground to the eye of a person looking through the stravr. The angle a (alpha) is the angle between the stramr, which is focused on the edge of the roof,, and a horizontal line. The protractor is used to measure the angle that the
srarv makes with the horizonml line. It is very likely that you have used uigonomeuic tools to analyze some problems in the past. It is also possible that you may have not used them recendy. If that is the case, then the tools hane been sitting in your mental toolbox for a while and quite possibly hane collected some rust in your head! Or you may have forgotten altogether how to properly use the tools. Because of their importance, let us review some of these basic relationships and definitions. For a right triangle, the Pythagorean reladon may be expressed by
a2+F:f
7.2
Mresunnrunr'mor Lrncrn
r59
Promctor
H=ht* hz=ht* dtzna EAiln
IaIn [trDn
nnEa
nnmm
il
Hgure?.3
l{easuring the height of a buildlng indhectly.
the rigpt tri@rgh shown, th" *gl" facing side a is denoted by a (alpha)' and the angle facing side bby B beta). Ttie sine, cosine, and the tangent of an angle are defined by
In
.
sln
(l :
srnp:
opposite
a
hyftotenuse
c
adjacent b sina opposite cosc:-:tanc.: cos a adjacent hypotenuse c
opposite b ^ hlpotenuse c ---r
adjacent a hypotenuse c
Thecosinerule:
4-: sina -\: sinp -t= sin9 8: F + * - 2bc(cosa) F:8+;-2ac(asB) P:3+F-2ba(aso)
ot
cos' :
8+f-* 2b,
^ ;+e-F cosp: 2* cos
o
:
/+b2-l --:;- zoa
b
sinB opposite b -^ *tF adjacent a
The sine and the cosine law (rule) for an a.rbinarily sbaped *iangle\s
Thesinenrls
d
r60
Crnprun
7
LsxcrH eNp Lsxcnr-Rrransp PanlvstrRs Some other usefirl trigonometry identities are
sin2e
* cos2a:
1
sin2a:2sinacosa
l-2sirlo. "os2o-sin2c:Zco*a - 1sin(-a) : -sin ct cos(-o) : cos a sin(a * B) : sinacosB * sinBcosc sin(a - B) : sin acosB - sinBcosa cos(a * B) : cosacosB - sinasinp cos(a - B) : cosacosB -| sinocosB cos2q:
7.3 Nominal Sizcs vcrsus Actual Sizes You have all seen or used 0,2 X 4 piece of lumber. If you were to measrue the dimensions of the cross section of a 2 X 4, you would find that the actual width is less than 2 inches (approximately 1.5 inches) and the height is less than 4 inches (approximately 3.5 inches). Then why is it referred to as a "2 by 4? Manufacnrrers of engineering parts (See Figure 7.4) we round numbers so that it is easier for people to remember the size and thus more easily refer to a specific pan. The 2 x 4 rs called the norninal size of the lumber. If you were to investigate other structural members, such as l-beams, you would also note that the nominal size given by che manufacurer is different from the actual size. You will find a similar situation for pipes, rubes, screws, wires, and many other engineering parts. Agreed-upon standards are followed by manufacnrers when providing information about the size of the parts that they make. But manufacnrrers do provide actual sizes of parts in addition to nominal sizes. This fact is important because, as you will learn in your furure engineering classes, you need the acrual sizn of parm for various engineering calculations. Examples of nominal sizes versus actual sizes of some engineering parts are given in Table7.2.
:,......,..:
I
tigure
-
7.1
.::i..,. . ;... .,-i-.. ...
}lanufacturen provide actual and romilal sizes for many parts and structural menbe$.
I
Noninal
Outide
Innide
Ifall
Slz€ (ln.)
Dismetff (in)
Dtsmeter (in )
Thickneec (in")
0.375 0.5
0.305
0.035
3t8
0.402
0.a49
1tJ
o.62t
0.5n
0.049
518
0.75
0.652
0.049
314
0.875
a.745
0.065
I
0.995 r.245
0.065
t7
r.125 r.375
l*
'{
r.62,5
1.481
0.072
2.12:5
1.959
0.083
2l
2.625
2.435
3
3.r25
2.907
0.095 0.109
3i
3.625
3.385
0.120
4
4.r25 5.t25
3.857 4.805
0.r34
u4
.,
r1
il
) -:,-.
i: ,r-
.,1:r.--
0.065
0.160
(inl
0.073 0.127 0.218 0.354 0.416 0.778
1.22 r,72 3.01 4.66 6.64 9.00 rr.7 19.1
r,l
-
Incide Diameter (tn.)
Ifa[ Thicknesr (inJ
v4
0.375
0.315
3le
0.5
0.4
u2 5/8
0.625 0.75
4l/L
a.875
0.545 0.666 0.785
a.042 0.045
1 rl
1.72' r.175
1.025
0.065
r.265
0.050
l*l
r,325
2
2,125
.L
2.625
0.060 0.070
3 3+
3.125 3.625
r.505 r.985 2.465 2.945
4
4.r25
3.425 3.905
5
5.r25_
4.875
i
Flow Cross Section
0.030 0.035
0.M0
0.055
0.080 0.090 0.100 0.110
FlowC,roos Section
(in2)
0.078
oJ45
a.n3
0.334 0.4M 0,825 1.26 L.78 3.09
4.77
6.81 9.2r 11.0 18.1
i i,
i
i: j
i t:: :
I
i l i 'i
,,
t::
"
) |
!
'l
i
i i i
l ,
l
i
i I
t,
i
----. ':-:r: . - -Continued
16l
r.. ISorribal Size (in.)
,
Outdde Piametet (ll.)
3/8
0.5
112 314
0.625 0.875
1
t.125
r+
1.375
ri ,,
4
Inside
Dismeter
r'
--l:--':,
FIow Cross Secdon (rnl
,
(in)
0.450
4.025.
0,569
0,028.,1
0.159 0.254
0.811 1,055
o.ot?
a.5v
o.w'
r.625
r.291 r.52V
0.049:
1.83
2.125-
2.449
0,018, 4.065,;
3,17 4.89 6.98 9.40
,
i
0.035,
,L
0,874 1.31
2.625
2495
i4
t.125
2.981
o.w2
3.625
0.083
4.t25
3.459 3.935
0,095i
122
5
5,725
4'e!7
0.r09
18.9
3
,
,
Deqgnation
Depth (mm)
i
x u3*
vaeo
w4l0 x 85 iI v360x57 i v20a x K.r l-.
r
-
,.-.
*Nominal depth
(-.4)
-l
-. .r-: -
FlangeVidth
(-'n)
Area (mm'z)
463
280
L4,/100
4t7
181
10,800'
358 203
172
7.230
243
.-
5,890 .,--
.---
and mass &g) per one rnen"r lengtb.
De3igneti$n
v
lwl
X
!116 x \tr14 X
w8.x
76* :,,
57 38 31
18.21
I
t6.43
'
I
:
14.10
,8.00
,
:
L,::.r:.:it::lr:.hlr r ---..-,-. ,.-,- .:,, . .-.r..-..., , ,-- .,-.-.:.t,-,.--,-= tNominal depth (io.) and weight 0b) per one foot length. 162
11.03 7.r2 6.77 7.95
.
ti.t,
22,3
t
r6.8 7t.2 9.1
-.r.: :..
7.6
L
AREA
163
fi9ure7.5
Ihe relationsiip anong arc length, radius, and angle.
7.4
Radians as a Ratio of Two Lengths Consider the circular arc shown in Figure 7.5.The relationship among the arc length, radius ofthe arc,4 and the angle in radians, 0, is given by
s,s
a---:_--_1
"
- &-
n,
$
0.1)
Note that radians represents the ratio of two lengths and thus is unidess. As you will learn later in your phpics and dynamics classes, you can use Eguation (7.1) as the basis for esablishing a relationship benareen the uanslational speed of a point on an object and its roational speed. Also note 2zr radians is equal to 360 degrees, and 1 radian is equal to 57.30 degrees.
7.5
Strain as a Ratio of Two Lengths material (e.g., in dre shape of a recangular bar) is subjected to a teruile load (pulling loao, the material will deform. The deformation, A4 divided by dre original length, I is called narrnal strai.n, shown in Figure 7.6. In Chapter 10, we will discuss the concepts of ^s stress and strain in more detail.'We will also explain horr stress and suain information is used in engineering analysis. But for now, remember that suain is the ratio of deformation length to original length and thus is unidess.
When
L I
:: jiaz _{_
a
,AL =J-
stram
Load
f, tigure7.6
trt.zI
Note that strain is another length-related engineering variable.
A bar subjected to a pulling
load.
7.6
Area fuea is a derived, or secondary, physical quantiry. It plays a significant role in many engineering problems. For example, the rate of heat transfer ftom a surface is direcdy proportional to the exposed surface area. That is why a motorcycle engine or a larnrn mower engine has extended surfaces, or fins as shown in Figre7,7.If you look closely inside buildings around your c:rmpus, you will also see heat exchangers or radiators with extended surfaces under windows and
164 CrreprsRT Lsr.rcrgANp LrNcrE-RELArun PenervrsrERs
E
rigure?.?
The important role of area in lhe
design of various systems.
against some walls. As you know, these heat exchangers or radiators supply heat to rooms and hallways during the winter. For another example, have you wer thought about why crushed ice cools a drink faster than ice cubes? It is because given the same amount of ice, the crushed ice has more surface exposed to the liquid. You may have also noticed that given the same amount of meat, it takes longer for a roast of beef to cook than it takes stew. Again, it is because the stew has more surface area exposed ro the liquid in which it is being cooked. So next time you are planning to make some mashed potatoes, make sure you first cut the potafoes into smaller pieces. The smaller the pieces, the soonet they will cook Of course, the reverse is also true. That is, ifyou want to reduce the heat loss from something one way of doing this would be to reduce the exposed surface area For example, when we feel cold we naturally try to curl up, which reduces the surface area exposed to the cold surroundings. You can tell by observing narure that trees take advantage of the effect and importance of surface area.'Why do trees have lots of leaves rather than one big leaf? It is because they can absorb more solar radiation that way. Surface area is also imporant in engineering problems related to mass transfer. Your parents or grandparents perhaps recall when they used to hang out clothes to dry. The clothes were hung over the clothesline in such way that they had a maximum area exposed to the air. Bedsheets, for example, were stretched out fiilly across the clothesline. Let us now investigate the relationship between a given volume and er<posed surface area.
Considera 1m
X I m X l mcube. W'hatisthevolume? 1m3.
lThatistheexposedsurface
Ifw" divide each dimension of this cube by half, we get 8 smaller cubes with the dimensions of 0.5 m X 0.5 m X 0.5 m, as shown in Figure 7.8. \Vh.at is the rctal volume ofthe 8 smaller cubes? It is still I m3. S7har is the total exposed surface area of the cubes? area of this cube? 6 m2.
Each cube now has an exposed suface area of L.5 face area of 12 sf .
fi],
which amounts to a total exposed sur-
Letusproceedwithdividingthedimensions of our 1m X 1m X
l
moriginalcubeinto
m X 0.25 m X 0,25 m. 'We will now have 64 smaller cubes, and we note that the total volume of the cubes is still 1 m3. Howwer, the surface area of each cube is 0,375 n],leading to a total surface au.e.s.of 24 m2. Thus, the same even smaller cubes vrith the dimensions of 0.25
cube divided into 64 smaller cubes now has an exposed surface area that is four times the original surface area.
7.6
il
AREA
r65
rigurctS
The relationship between the area
and volume of a cube.
This mental exercise gives you a good idea why, given the same amount of ice, the crushed ice cools a drink faster than ice cubes do. You may not know it, but you abeady own the best heat and mass exchanger in the world. Your lungs! Human lungs are the best heat and mass er
drag force
: ]{.i,
a..,.iv)(air speed)2(ftontal
area)
The frontal area represen$ the fronal projection of the vehicle's area and could be approximated simply by calculating 0.85 timCI the width and the height of a recangle that oudines the front of the vehicle. This is dhe area that you see when you view the car or truck from a direction normal to the front grill. l.ater in your engineerhg education, some of you may take a class in the physics of flight, fluid mechanics, or aerodynamics, where you will learn that dre lift force acdng on tlre wings of a plane is proponional to the planform area of the wing. The planform area is the area that you would see if you were to look from above at the wing from a direction normal to the wing. Cross-sectional area also plays an imporant role in disuibuting a force over an area- Foundations of buildings, hy&arlic systems, and cutdng tools (see Flgare 7 .9) are examples of objeca for which the role of area is important. For enample, hare you ever thought about why the edge of a shaqp knife cuts well? \fhat do we mean by a "shaqp" knife? A good sharp knife is one that has a cross-sectional area as smdl as possible along its cutting edge. The pressure along the cutting edge of a knife is simply determined by Pressure at dre
cuningsurface
:
force cross-sectional area at the curdng edge
(7.3)
166
E
Cnar"rnn
7
LsNcrH nNp LsNcrg-Rrr,etro PanavprsRs
Figure7.9
Area plays an important lole in
tie
design of cutting tools.
lbu
can see from Equation Q.3) than for the same force (push) on the knife, you can increase rhe cutting pressure by decreasing the cross-sectional area. fu you can also tell from Equation (7.3), we can also reduce the pressure by increasing the area. In skiing we use the area to our advantage and distribute our weight over a bigger surface so we won't sink into the snow. Next time you go skiing think about this relationship. Equadon (7.:) also makes clear why highheeled shoes are designed poorly as compared to walking shoes. The purpose ofthese examples is to make you realize that area is an important parameter in engineering design. During your
will learn many new concepts and laws that are either direcdy or inversely proportional to the area. So keep a close watch out for area as you study various engineering education, you engineering topics.
7.6
Units ofArea in Increasing Order
EquivalentValue
1 mm2
1
1cm2
I X lO-am2:
1 inz
645.16mn*
rfe
X
AREA
167
10-6 m2 100 mm2
lUi*
rv&
etr
lmz I acre I kmz I square mile
43,560 1,000,000
r.196y&
t{
2.59 Ur*
ri = 247.1 acres : 640 acres
Now that you understand the significance of area in engineering analysis, let us look at its 'We units. The unit of area in SI is m2. can also use the multiples and dre submultiples (see Thble 6.2) of SI fundamenml units to form other appropriate units for area, such as mm2, cm2,kmz, and so on. Remember, the reason we use these unirs is to keep numbers manageable. The common unit of area in rhe U.S. Customary qrstem is fi3. Table7.3 shows other units commonly used in engineering practice today and their equivalent values.
Area Calculations The areas of common shapes, such as a triangle, a circle, and a rectangle, can be obtained using the area formulas shown in Table7.4.It is a common practice to refer to these sirnple aras as priwitiae areas. Many composite surhces with regular boundaries can be divided into primitive areas. To determine the total area of a composite surface, such as the one shown in Figure 7.10, we fust divide the surface into the simpler primitive areas that make it up, and then we sum the values of these areas to obtain the total area of the composite sudace. Examples of the more usefirl area formulas are shown in Table7.4.
Approximation of Planar Areas There are many practical engineering problems that require calculation of planar areas of inegular shapes. If the irregularities of the boundaries are such that they will not allow for the irregular shape to be represented by a sum of primitive shapes, then we need to resort to an approximation method. For these situations, you may approximate planar areas using any of the procedures discussed next. The Trapezoidal Rule You can approximate the planar areas of an irregular shape with reasonably good accuracy using the trapezoidal nrle. Consider the planar area shown in Figure 7.11. To determine rhe total area of the shape shown in Figure 7.LI,we use the trapezoidal approximation. We begin by dividing the total area into small trapezoids of equal height b, as depicted in Figure 7.11.'We then sum the areas of the trapezoids. Thus, we begin with the equation
AoAr+Az+,L+"'+4
fl.4)
CHemsRT LrNcrHANp I;excrH-RBr-arBD Paneurmns
Triangle
e=|w F__t_____+l Recangle
T
A=
h
I
bh
l-b--*+| Paralldogram
A:
bh
I
ffij F--b*---4 F-4-_{
Trapezoid
a:
)tn
+
r--T7T
b)h
\\__/_J /t *-b---1
n siild.Polygon
^: (t)r*,(ry) Circle
A:
168
nRz
7.6
AREA
Ellipse
A= nab
Cy'indet
4:2nRh A.p: r{booo. : rP Ar*t= 4* A.p * luo*-
T h
i
Right Circrlar C,one
4: rtr: oR.. Fl ,] 4-.,". : nP 4*6:
A\+ &tu^
Sphere
A:4rR2
I
IigureT.l0
A composite surface (sudace ot a
|teat sinlr)
tlat
may be dtuided
into Ddmitiue anas.
169
170 Crilprsn7
LsNcrsANoLsNcrH-RELATEoPenelrsrsns
T
Yo
Q
Figure?.lt
Approrimation of a planar area by the hapezoidal rule.
ln
i
I
Substituting for the values ofeach trapezoid,
h. + + h, + ,.h, a: r0o t) 20, D + ;Ur+ t) + "' + !0,,-, + t,)
0.5)
and simplifying Equation (7.5) leads to
a:n\;h+yr+72+ "'rlo-r*rr-r*!rr,) '
tl
t
(7.6)
Equation (7.6) is known as the trapezoidal nrle. Also note that the acrl')tacy of the area apptoximation can be improved by using more ttapezoids. This approach will reduce the value of h ar,d thus imprwe the accuracy of the approximation. Counting the Squares There are otherwa)n to approximate t"he surFace areas ofirregular shapes. One such approach is to divide a given area into small squares of known size and then count the number ofsquares. This approach is depicted in Figute7.l2. You then need to add to the areas ofthesmall squares tfreleftoyer areas, whichyou mayapproximate bythe areas ofsmall triangles.
Subtracting Unwanted Areas Sometimes, it may be advantageous to first fit large primitive area(s) around the unknown shape and then approximate and subtract the unwanted smaller areas. An example of such a situation is shown in Figure 7.13. Ako keep in mind that for symmetrical areas you may make use of the symmetry of the shape. Approximate oily Il2, Ll4, or 1/8 of the total area, and then multiply the answer by the appropriate factor.
L-J Figuru7.t2 Approrination of a planar area using small squares.
7.5
I
AREA
171
rigurul.l3
Approilmation of a planar area using a rectangular primitive and small squares and triangles.
Weighing the
Area Another approximation procedure requires the use of an accurate
"nalytr-
cal balance from a chemistry lab. Assuming the profile of the area to be determined can be
drawnonasrr
x tt sheetofpaper,firstweighablanks| x tl
sheetofpaper,thenusingthe
analytical balance, weigh the sheet and record its weight. Next, &arv the boundaries of the unknown area on the blankpaper, and then cut around the boundary of thatara..Determine the werghr of the piece of paper that has the area drawn on it. Now, by comparing the weights of the blank sheet of paper to the weight of the paper with the profile, you c:m determine the area of the given profile. In using this approximation method, we assumed that the paper has uniform thickness and densiw. Using the trapezoidal nrle, determine the total ground-conad area of the athletic shoe shown
inFigre7.74.
l-{
tigure
7.14
lhe shoe profile for Emlnple
7.1.
'W'e have
divided the profile into two parts and each part heights of 1.0 in. Applyt"g the rapezoidal nrle, we have
.(r A- h\tyor
!,,+
y2r"'*
I \
y,-,+ l"-t+,J/")
into L2 trapezaids of
equal
172
Cneprrn
7
LnNcrn exp Lwcrs-Rnrarso Penqrvrcrnns
e,:1r.0;f,10; + r.t2 + +
4-
2.12
1r.01|,1107
+
,'',*'...
+
2.00
1.37
+ 2.t2 + 1.50
+
+
t.52
2.00
1.62
+
+
+
t.25 + 1.00+0.87+1.12+1.75
+
1.87
+
1.l
;(0)
1.75 + 1.52
t.25 +
+
l- L6.6il 1.75
+
2.00
+ 2.12 +
2.12
]ror]: rgsi#
Then the total area is given by
fuot-
-
i:tl--
16.6
+
t9.3
-
ls.g
fi
.
7J
Volume Volume is another imponant physical quantiry, or phpical variable, that does not get enough respect. W'e live in a dree-dimensional world, so it is only natural that volume would be an impomant player in how things are shaped or how things work" Let us begin by considering the role ofvolume in our daily liva. Today you may have treated yourself to a can of soda, which on average contains 12 fluid ounces or 355 milliliters of your favorite beverage. You may have driven a car whose engine size is rated in liters. For example, if you own a Buick ParkAvenue, its engine size is 3.8 liters. Depending on the size of your car, it is also safe to say that in order 'W'e to fill the gas tank you need to put in about 15 to 20 gailons (57 to77 liters) of gasoline. also express th" g* consumption rate of a car in terms of so many miles per gallon of gasoline. Doctors tell us that we need to drink at least 8 glasses of water (approximately 2.5 to 3liters) a 'W'e day. breathe in orygen ar a rare of approximately 1.6 fC lh (0.0453 m3/h). Of course, as you would er
7.7
Activity Exhausting effort Suenuous work or sports Moderate exercise
Mild
exercise; light
work
173
of
Consumption
Catbon Dioxide Production
Breathing
(t€th)
(ff/h)
(tr/h)
4.44 2.96
3.8
146 97
L55
Oxygen I-evel ofPhyuical
Vor,tnvrs
6.6
t<
Standing; deskwork
1.84 1.10
Sedentary, at ease
0.74
0.93 0.62
Reclining at rest
0.3
a.55
Rate
64
40 24
r6
t2
Source: Ameisnsociery of Heating Refrigemting and Air-Conditioning Engineers, Inc.
fluid pressure distribution acting over the submerged surface of an object is the buoyancy force. The magnitude of the buoyanry force is equal to the weight of the volume of the fuid displaced. It is given by
f,s =
pVg
tlll
where ,ft is the buoyancy force (N), p represents the density of the fuid (kg/-'), and gis acceleration due to gravity (9.S1 m/s2). Ifyou were to firlly submerge an object with a volume % as shown in Figure 7.l5,you would see that an equal volume of fluid has to be displaced to make room for the volume of the object. In fact, you c:rn use this principle to measure the unknourn volume of an object. This idea is demonstrated in Example 7.2. NASA astronaub also make use of buoyancy and train underwater to prepare for the in-orbit repair of satellites. This type of uaining mkes place in the Underwater Neutral Buoyancy Simulator, shorrn in the accompanying photograph. The changes in the apparent weight of the astronaut allows him or her to prepare to work under near-zero-Cra/ity (weighdess) conditions. Now that you understand the significance of volume in the analysis of engineering problems, let us look at some of the more common units in use. These units are shovrn nTable7.6.
F7
:'
f, rigure?.t5 Buoyancy lorce acting (ln a submerged sl|Iface.
174
Cne,prsn7 Lrrcrn
eNp LsNcrs-Rsr-arEo PenervrsrpRs
Volume Units in Increasing
CapacityOrder I milliliter 1 teaspoon
l/1000 liter (ap)
4.928 milliliter
1 tablespoon (tbsp)
3 tsp
I fluid I cup
2
ounce
l
NASA Astronauts
training
in the Underwater Neutral Buoyancy Simulator to prepare for the in-orbit Hubbell Space Telescope repair.
174
milliliter
(
I
:
/1000
F
16 tbsp
16ounces:2cups 1000 cm3 o 4.2 cups 4 quala 7.4805 gallons 1000 liters 264 gallons
gallon
quan
-
2 pins
I cubic foot I cubic meter I
tbsP
8 ounces
I pint I quart I liter
*l
EquivalentValue
(
:
1 teaspoon
I liter
(
I
(
gallon
1 mblespoon
{
< I cubicfoot
(
I 0uidounce
(
1 cubic meter.
1 cup
:
<
353 f€ 1
pint
<
7.7
Vor.rnas
175
As you go overTab!e,7.6, try to develop a lfeel" for the order ofvolume quantity. For example' askyourselfwhether a pint or a liter is the larger quantity, and so on.
Volume ealculations The volume of simple shapes, such as a cylinder, a @ne, or a sphere, may b€ obtained using volume formulas as shown in Thble 7.7.Indfuwt and direct estimation of volumes of objects are demonstrated in Examples 7 .2 ard7.3.
Clinder
V: nPh
Right Grcular C,one
v=
loer,
Section of a Cone 1
V::zrh(R?+ n7+ &Rz)
Sphene 4,
V: VtF Section 1
ofa Sphere
V:7nh(3a2+3b2+ h')
176
CneersR
7
I;BNcur eNo LENcTH-Rnr-emp Pnnnrvrcrsns Determine the exterior volume of rhe object shown inFigure7.l6. For this example, we will use the buoyancy effect to measure the exterior volume of the .We object shown. will consider trvo procedures. First, we obain a large conainer that can accommodate the object. \7e will then fill the container completely to its rim with water and place the container inside a dqy, empty tub. V'e next submerge the object with the unknown volume into the container until its top surface is just below the surface of the water. This will displace some volume of water, which is equd to the volume of the object The water that overflowed and'n'as collected in the tub can then be poured into a graduated cylinder to mensure the volume of the object. This procedure is shown in Figure 7.17.
t
figureT.l6
The object in Eranple 7.2.
E
risum?.|7
Using a displaced volume of water to neasure the volume of
the qiven object.
The second procedure makes direct use ofthe buoyancyforce. IZe first suspend the object in air from a spring scale to obtain its weight. W'e then place the object, still suspended from the spring into acontainerfilledwithwater. Next, we record rhe apparentweightoftheobjec. The difference between the actual weight of the object and the apparent weight of the object in water is the buoyancy force. Knowing the magnitude of the buoyancy force and using Equation (7.7),we can thendetermine thevolume ofthe object. This procedureis depicted in Figure7.18.
I
FigureT.l8
Using apparent neight (actual
weight minus the buoyancy
fone) to determine tie volme
7.8
SscoND MorrasNTs or AnsAs
177
Estimate the inside volume of a soda can. W'e hane used a nrler to measure the heiglrt and the diameter of the can, as shown in Figure 7.19, W'e may approximate the inside volume of the soda can using the volume of a cylinder of equal dimensions:
diameter = 6.3 cm
cm) : V: rrPh = (3.tat5)[6'',t* . , )'(rr.o
/
12cm
,/
374 cmr
:
374 mL
Compared to the 355-mL value shown on a typical soda can, the approximated value seems reasonable. The difference between the approximated value and the indicated value may be explained in a number ofwaln. First, the soda container does not represent a perfect cylinder. Ifyou were to look dosely at the can you would note that the diameter of the can reduces at the top. This could explain our overesdmation ofvolume. Second, we measured dre outside diameter of r:he can, not the inside dimensions. However, this approach will introduce smaller inaccuracies because of the small thickness of the can. 'We could have measured the insidevolume ofthe can byfilling the canwithwater and then pouring the water into a graduated cylinder or beaker to obtain a direct reading of the volume.
_1
I
riqurcl.l9
Soda can in txample 7.3.
-t- -
-..-._-
__.
Finallp it is wonh nodng here that numerical solid modeling is an engineering topic that with computer generation of the surface ares and volumes of an acnral object. Solidmodeling software programs are becoming quite common in engineering practice. Computergenerated solid models provide not only great visual images but also such information as magnitude of the area and the volume of the model. To generate numerical solid models of simple shapes, area and volume primitives are used. Other means of generating surhces include dragging a line along a path or rotating a line about an axis, and, as with areas, you can also generate volumes by dragging or sweeping an area along a path or by roating an area about a line. deals
Ve will
7.8
discuss computer solid modeling ideas
in more detail in Chapter 16.
Second Moments of Areas In this section, we will consider a properry of an area known as rhe second rnoment tareaThe second moment of arca, also known as the area rnornmt of inertiq is an imporrant properry of an area that provides information on how hard it is to bend something. Next time you walk by a construction site, mke a closer look at the cross-sectional area ofthe support beams, and notice how the beams are laid out. Pay close attention to the orientation ofthe cross-secdonal area of an I'beam with respect to the directions of expected loads. Are the beams laid out in the ori-
entation shown in Figute 7 .20(a) or in Figure 7 .20(b)? Steel I-beams, which are commonly used as strucnual members to suppoftvarious loads, offer good resistance to bending and yet they use much less material than beams with recangular cross secdons. You will find I'beams supponing guard raits and I-beams used as bridge cross members and also as roofand flooring members. The answer to dre question about the orientation of I-beams is that they are oriented vrith respect to the loads in the configuration shown in Figue7.20(a). The reason for having I-beams support loads in that configuration is that about
178
Cneprnn
7
LsNcrH errrp Lrxcrn-Rrranso Penenarruns I Direction of exlected load fi
+
fin
l!:#fl f
u--t
riqureT.zo
ltlhich way
h
an l'bean oriented
(b)
with ruspect to loading?
H
t FigureLzt
v
Bend the rod and yardstick in the
(a)
dhections shown.
|
the a-aaxis shourn, the value of the area moment of inenia of the I-beam is higher for configuration (a) than it is for configuration (b). To better understand this imponant properrF of an area and the role of the second moment of area in offering a measure of resistance to bending, try the following experiment. Obtain a thin wooden rod and a yardstick. First try to bend the rod in the directions shown in Figare 7 .21 .If you were to repoft your findings, you would note that the circular cross secdon of t.he rod offers the same resistance to bending regardless of the direction of loading. This is because the circular cross section has the same distribution ofarea about an axis going through the center of the area. Note that we are concerned with bending a member, not twisting it! Now, try bending the yardstick in the directions shown in Figare 7 .2I. \07'hich way is it harder to bend the yardstick? Of course, it is much harder to bend the yardstick in the direction shown in Figure 7.2I(a). funn, that is because in the orientation shown in Figure 7.21(a), the second moment ofarea about the centroidal axis is higher. Most of you will take a statics class, where you will learn more in depth about the formal definition and formulation of the second moment of area, or area moment of inertia, and its role in the design of strucnrres. But for now, let us consider the simple siruations shown in Figare 7 .22, For a small area element A, located, at a distance r from the axis z-4 rhe arca moment of inertia is defined by
frgurel.Z2
Small area element located at distance r ftom the
(b)
rz
aris.
Int: /A
7.8
Srcoxn MourNrs orAnns
179
I I
D
rigure7.Z3
Second monent ol area for three
!"-Flo, l-."---5'" tF-r3---flo, I
small area elements.
Now let us expand this problem to include more small area elements, as shown inFiyre7.23, The area moment of inenia for the system of discrete areas shown about the z-z axis is now
I,-": 4Ar + &r,4, + luA,
ff2'l
Similady, we cnn obtain the second moment of area for a cross-sectional area such as a rectangle or a circle by summing the area moment of inertia of all the litde area elements that make up the cross section. As you take calculus classes, you will leam that you can use integrals instead of summing the fAterms to evaluate the area moment of inertia of a continuous cross-sectional area. After all, the integral sign, J, is nothing but a big "S" sign, indicating summation.
t-": lff dA
[/lo]
Also note that the reason this propercy of an area is called'second moment of area" is that osecond the definition contains the product of disunce squared and an area, hence the name moment of area." In Chapter 10, we will discuss the proper definition of a moment and how ir is used in relation to the tendency of unbalanced forces to rotate things. As you will learn later, the magnirude of a moment of a force about a point is determined by the product of the perpendicular distance from dre point about which the moment is taken to the line of action of the force and the magnitude of that force. You have to pay attention to what is meant by'a moment of a force about a point or an axiso and the way the term lnornenr is incorporated into the name "the second moment of area" or "the area moment of inenia." Because the distance term is multiplied by another quantiry (area), the word "moment" appears in the name of this propeffy ofan area. You can obtain the area moment of inertia of any geometric shape by perForming the integration given by Equation (7.10). You will be able to perform the integration and better understand what these terms mean in another semester or two. Keep a close watch for them in the upcoming semesters. For now, we will give you the formulas for area moment of inenia without proof. Examples of area moment of inenia formulas for some common geomeffic shapes are given nexr.
CseprenT lrxcrrenp
180
LaNcrH-RELArno Panar,rstrRs
Bedangle
T
l,
I
I
4': |u'h3
I
( *-
,,!
4,: |h*u
:
----*-t---
(r.il)
!
t,
,l
I
I
_t_
I
v
tircle
I-*: Ir! = In*
(7J2)
Thevalues ofthe second moment ofarea forsrandardwide-fange beams are shown in Thble 7.8.
Dep&(mm) widtl(-m)
x 113 x 85 ',, v36o x 57 \1460
463
i -1rr410
4r7
i
lvzoo x
+e.r
|
,
',358 2A3 .i --- r--
Area
280 181,'
r4,400
r
10,800
L72 ,,l
7,230 rl8e0
to1
-',i,
(mrnl
Inr(mm4) 554 x to6 it6 x rou reo x r01,,
45.S
x i06:
,, .
;l''
:'
63.3 x,106 l:7.9 X rc6
ll.t x to6 15.4 x to6
.
Continued
7.8
SscoNo MourNrs oFAREAs
7.8 Examples of the Second Moment of Area of Standard Beams (confinued) Cusnrnary umion) ft.15.
181
TABLE
r'al
Designation
r
wr8 x
wr6x X \tr8 X
Ttr14
Depth (inJ
Width (inJ
76
lB.2l
11.03
57 38
r6.43
71)
14.10 8.00
7.95
31
I-"(h)
tuea (in1
'r)
rrr(i#)
4
r330
r52
16.8
43,r
LL.L
758 385
o1
110
37.r
25.7
In Chapter 9, we will look at another similarly defined property of an object, mass moment of inertia, which provides a measure of resistance to rotational modon.
Calculate the second moment of area for a 2
X 4 stud with the
acrual dimensions shown on
Fig.rc7.24. I I I
I
T
'i" '-
I
I I I
v
F-3.5
fl
riqure?.24
I-":
r fiwh'
4,: j* L
in.-____----+l
/r\
:
0.98 ina
: (#),t'5 in')(3'5 in')3 :
5'36ina
:
( ,, /(r.5
in.) (t.5 in.)3
Note that the 2 X 4 lumber will show more than five times more resistance to bending about they*y axis than it does about the r-r axis.
1--a:r:-i!:ra!l:r:!r.j.:ariri.iiliir
i-,r.,r.!-r.r:rii-------I]I]1- .
182
CrrlprsnT LexcrHaNp LsNcru-Rprarnp PanervrsrsRs Calculate the second moment of area for a 5-cm-diameter shaft about the
shown in
Figte7.25.
x-r
and ttre
!-!
a;ries
v
E
Figure
7.25
The shaft in txample 7.5.
r-*: rr! = I,*
ii 'ilt
=
|lu.rn
r)(+) :
30.7
c'f
i1
i
-:l:rl.
t:r.,1:.ir;it :r-.rl
--,,1
Finally, it is imponant to emphasize the fact that all physical variables discussed in this chapter are based on the fundamental dimension length. For example, area has a dimension of (length)2, volume has a dimension of (length)3, and tlle second moment of area has a dimension of (length)4. In Chapter 8, we will look at time- and lengch-related parameters in engineering.
SUMMARY Now that you have reached this point in the text
. "
. .
.
You should understand the significant role the fundamental dimension length plays in engineering problems. You should also realize the imporance of area and volume in engineering
applications. You should have developed a "feel" for the relative magnitude of various length units, area units, and volume unia. You should know the difference between the acual size and a nominal size of objects. You should know how to measr[e areas and to approximate planar surfaces using the trapezoidal nrle. You should have a good undersmnding of the physical significance of the second moment of area in structural analysis.
Pnosr,EMs
183
NM'IE 71. Seasoned engineers are good at estimating phpical
the dimensions of the given objects. Next measure,
value without using tools. Therefore, you need to begin nfeel" for the sizes of various physical dweloping a guantities. This enercise is intended to help you develop this ability. Using the table below, fust esdmate
or look up, the actual dimensions of the objects, and compare them to your estimated values. How close are your esdmations? Do you hane a "feel" for units of length yet?
Estimated Values
Difference
Measured Values
(cm)
(in)
(.-)
(in.)
(cm)
(inJ
The distance ftom home to school Use your car's odometer to mquure the acnral dismnce
(k-)
(miles)
(km)
(miles)
(km)
(mile')
Adotlar bill The height ofyour engineering buildins Vingspan of aBoeng747
(cm)
(in)
(m)
(ft)
(m)
(ft)
(m)
(ft)
(m)
(ft)
Obiect This book A oen or a oencil A laptop computer (dosed) A.L2.fl.-oz. so&can
1.2.
The following exercises are designed to help you become aware of the significance of various dimensions around you. You see these dimensions every day, but perhaps you never looked at them with the eyes ofan engineer. Measure and discuss the significance of the dimensions of the following items a. The dimensions of your bedroom or living room b. The dimensions of the hallway c. The window dimensions d. The width, height, and thickness of your apartment doors or dormitory doors e. The distance &om the floor to the doorknob f. The disance &om the foor to the light switches g. The dimensions ofyour desk h, The dimensions ofyour bed i. The distance from the floor to the bathroom sink
j. 7.3.
The distance from the tub surFace to the showerhead
In this exercise, you are to explore the size ofyour classroom, the seating arr.rngemen$, the location of the
chalkboard (or the whiteboard) with respect to the classroomt main entrance, and the size ofthe board relative to classroom size. Discuss your findings in a brief report to your instructor. a. What are the dimensions of the classroom? b. How far apart are the seats placed? Is this a comforable arrangement? lVhy or why not? c. How far above the floor is the chalkboard (or whiteboard) placed? Howvride is it? How tall is it?
'Vhat
is its relative placement in the
classroom?
Can someone sitting in the back corner of the room see
the boardwithout too much difficulty?
1.4. This is a sports-related assignment, First look up the dimensions, and then show the dimensions on a diagram. You may need to do some research ro obtain the information required here. a. A basketball courr
b. A tennis coun c. A foo$all field
Cneprrn
184
d. A
soccer
7
LsricrH exo l;eNcrn-Rnransp PenarvrsrERs highway is, Do all interstate highway lanes have the
field
e. Avolleyball court 7.5.
same width?
c. Find out how tall the Hoover Dam is. How wide is '\tr7hat lhe Hoover Dam? is the area of the dam
These dimensions concern ffansportation systems. Look up the dimensions of the following vehicles; give the model year and t}re source of your information.
d. Howwide is the driver's seat in each car mentioned
f.
enposed to upstream water? Discuss how and why the thickness of the dam varies with its heighu How tall are the interstate bridges so that an average truck can go under them? How far above the road are the highway signs
in (a) through e. Acitybus
g.
placed? '$?'hat is the average height of a runnel?
a.
A car ofyour choice
b.
BM\tr760 Li Hon&Accord
c.
d.
(c)?
f.
How wide are the bus seats? g. A high-speed passenger train h. How wide are the sain seats in the coach section? 7.6. This is a bioengineering assignment. Investigate the following dimensions and write a brief repon to your instructor about your findings. a. What is the average diameter of a healthy red blood cell?
b. '!7hat is the average diameter of awhite blood cell?
c.
Measure and record the length of each finger of ten male adults in your class. Also, measure and record the length of each finger of ten female adults. Compute rhe anerages for males and females, and compare the female results to the male results. Pre-
sent the results in both SI and U.S. Customary units.
d. Measure the average
e.
surface area of an adult male arm by covering the arm with paper and then measuring the area of the paper. Measure the surFace area of an adult male leg. How much plaster would be required for a plaster cast around an aver€e adult male leg, assuming 2-mm thickness? Vhat is the volume of the plaster needed? Estimate the average surface (skin) area of an adult male.
f.
\Vhat are the lengths of human small and large intestines?
7.7. This
assignment is related co civil engineering. is each lane of a street in your neighborhood? Talk to yo:or ciry engineer to obtain infor-
a. Howwide mation.
b. Visit the U.S. Department of liansportation \Zeb site to find out how wide each lane of an interstate
area
Vhat
is the
ofa tunnel at the enfance?
W'rite a brief repon discussing your findings. Investigate the size of the main waterline in your neighborhood. What are the inside and ouaide diameters? Vhat is the nominal size of the pipe? lfhat is the cross-sectional area of water fow? Investigate the size of piping used in your home. '!?'rite a brief memo to your instructor discussing your findings. 7.9. Iook up the length of the Alaska pipeline. \Vhat are the inside and outside diameters of the pipes used in ffansporcing oil? How far apan are the booster pump stations? How thick is the pipeline? $7hat is the oosssectional area of tle pipe? \7hat is the nominal size of 'Write a brief memo to your instructor disthe pipe? cussing your findings. 710. Investigate the size ofpipes used in transporting natural gas to yonr state. Vhat are the inside and outside diameters of the pipe? \IZhat is the distance between boosting stadons? \Vhat is the cross-sectional area of the pipe? '$7riee a brief memo to Four instrucor discussing your findings. 7.n. Investigate the diameter of the electrical wire used in your home. How thick are the interstate transmission lines, and whar is their cross-sectional area? Wrire a brief memo to your instructor discussing your findings. 712. ![hat is the operating wavelength range for dre following icems? 7.8.
a. Acellularphone b. FM radio transmissions
c.
TV broadcasting brief memo to your instructor discussing your findings. Satellite
Write
a
PnoslErvls 7.13. Derive the formula given for t}re area of a uapezoid. Start by dividing the area into trvo uiangular areas and one rectangular area. 7.14. Trace on a white sheet of paper the boundaries of the
of the United States shown in the accompanylng diagram. a. Use the uapezoidal nrle to determine the total area. b. Approximate the total area by breaking it into small squares. Count the number oftotal squares and add what you think is an appropriate value for r:he remaining area, Compare your findings to pan (a). c. Use an analytical balance from a chemistry lab to weigh an 81-x-11 sheet of paper. Record the dimensions of the paper. Draw the boundaries of the area shown on the accompanying figure and cut around the boundary of the area. Determine the
area
for each shoe, and assuming the woman who wears
these shoes weighs 120 lb, calculate the average pressure
at the bomom of the shoe for each shoe sryle. \Vhat are your findings? \fhat are your recornmendations? Recall that
area
weight of the piece ofpaper that has the area drawn on it. Compare the weights and determine the area of the given profiles. I7.hat assumptions did you make to arrive at your solution? Compare the area computed in this manner to your results in parrs (a) and (b). fue there any other ways that you could have determined the area of the profile? Explain.
185
force
'Pfessufe = 716.
area
-
Obain two different brands of cross-country skis. Make imprints of the expeced contact surface of the skies with snow. Obtain a value for the average pressure for each ski, assuming the person wearing the ski
711.
weighs 1801b. Estimate the anerage pressure enerted on the road by the following.
a. Afamilycar b. Atruck c. Abulldozer Discuss your findings. 718.
Using area as yotu variable, suggest ways
to
cool
freshly baked cookies faster.
il9. Estimate hornr much material
is
needed
to make
00,000 stop signs. Investigate the size ofone side ofa stop sign and the kind of material it is made from. 'S?'rite a brief memo to your instructor discussing your findings. Estimate how much material is needed to make 100,000 traffic yield sigrs. Investigate the size of one side of a leld sign and the kind of material it is made from.'Write a briefmemo to your instructor discussing your findings. As explained in the chapter, the air resistance to motion of a vehicle is determined experimentdly by placing it in a wind runnel. The air speed inside the unnel is changed, and the drag force acdng on the vehicle is measured. The experimental data is normally represented by a single coefficient that is cdled I
1.20.
t.?1.
rhe drag cofficimt.
It is defined
by the following
reladonshio: 7.15.
Obtain a woman's high-heeled dress shoe and
a
woman's athletic walking shoe. For each shoe, make imprints of the floor contact surfaces. Determine the total
drag
coefficient
-
drag force
I
i(air
densiV) (air speed)'z(frontal area)
185 Cnapren7
Lnr,vcrn aNn LeNcrrr-Rnr-erno PenenrsrrRs
or, in a mathematical form,
Collect information on the standard sizes of automobile tires. Create a list of these dimensions. lfhat do the numbers indicated on a tire mean?'W'rite a brief memo to your instructor discussing your findings. n.n. Visit a hardware store and obtain information on the standard size of the following items. 7.26.
F.
'd'
Cr:
;ov'za I.
*
"h;
explained in this chapter that the frontal
arca Ais the frontal projection of the area, and could be approximated simply by multiplying 0.85 times the width and the height of a recangle that outlines the front of a vehicle when you view it from a direction
normal to the front windshield. The 0.85 facror is to adjust for rounded corners, open space below the bumper, and so on. Typi"d drag coefficient values for sports cars are between 0.27 and 0.38, and for sedans the values are between 0.34 and 0.5. This assignment
a. Screws b. Plywood sheets c. PVC pipes d. Lumber Create a list with both the actual and the nominal sizes. Using just a yardstick or a measuring upe, open r:he classroom door 35 degrees. Yoa cannotuse a?rotrdL-tor. '[.29. This is a group project. Determine the area of each class membert right hand. You can do that by tracing 7.28.
the profile of weryonet fingers on a white sheet of paper and using any of the techniques discussed in this chapter to compute the area of each hand. Compile a data bank containing the areas of the right hands ofeach person in class. Based on some average, classify the data into small, medium, and large. Estimare how much material (outer leather and inner lining) should be ordered to make 100,000 gloves. How many spools of thread should be ordered? Describe and/or &aw a diagram of r.he besr way to cut the hand profiles from sheets of leather to minimize wasted
requires you to actually measure the frontal area of a car. Thpe a nrler or a yardstick to the bumper of your car. The ruler will serve as a scale. Thke a photograph of the car and use any of the methods discussed in this chapter to compute the frontal area of the car.
t.u.
A
machinist
in an engineering machine
shop
has
ordered a sheet of plastic with dimensions of 10 ft X L2 ft x 1 in. width, length, and thickness, respectively. Can the machinist get the sheet of plastic inside
the machine shop provided that the door dimen-
ft X 8 ft? Give the maximum dimensions of a sheet that can be moved inside the shop and sions are 6
explain how. 7.8. Investigate the volume c,apacity of a barrel of oil in gallons, cubic feet, and cubic meters. Also, determine the volume capacity of a bushel of agricultural products in cubic inches, cubic feet, and cubic meters. Write a brief memo to your instructor discussing your
materials. 7.30.
land, how many crus can be parked safely there? Determine the appropriate spacing between cars and the
width of drive lanes. Prepare a diagram showing your solution, and write a brief report to your instructor 7.31.
findings. 1"24.
7.25.
Measure the outside diameter of a fagpole or a streetlight pole by first wrapping a piece of sning around t}re
object to determine its circumference. Wh.y do you think the flagpole or the streedight pole is designed to be thicker at the bomom near the ground than at the top? Explain your answer. Measure the width and the length of a readmill machine. Discuss what some of the design factors are that determine the appropriateness of the values you measured.
This is a group project. Given a square kilometer of
7.32. 7.33.
7.34. 7.35.
discussing your assumptions and findings. Estimate the length of tubing used to make a birycle rack on your clmpus. Determine the base area of an eledric steam iron. Determine t-he area of an incandescent 100-'ttr( light bulb. Obtain a trash bag and verifr its listed volume. Discuss the size of a facial, tissug a paper towel, and a sheet
1.36.
oftoilet
paper.
Estimate how much silver is needed to make a fourpiece set of silverware containing teaspoons, tablespoons, forls, and knives (16 pieces in all). Discuss your assumptions and findings.
PnosLEMs Calculare the second moment
of
area
diameter shaft about the r-r utdH
of z
Calculate the second moment of area for a2 X 6 piece of lumber about the r-r andy-y axer, as shown. Visit a hardware store and measrue the actual dimensions of a typical2 X 6 piece of lumber.
2-in.-
axes, as shown.
v I
v I
T
i"-
I Problem ?.3?
187
hoblen 7.30
CnepraR
7
LnNcrn eNp Le,Ncrrr-RBr-rtsu Paneurmns
Objective: To build a boat from a 6" aluminum foil, which
x
6" sheet of
will carry as nrany pennies
as
minute. Thirry minutes will be allowed for preparation.
possible. It should float for at least
1
,::.
"Teamwork divides the task and doubles the success." Camegr.e (1 835 -1 9 I 9)
-Andrew
188
- -l
..':_'-,
;WaterTunneINo.3*:.
Source Redravtfrom art bv Don Folev/National Geoeraohic lmaee Collecd6n, Redmwir with
p".dirsion.
@
Nadond Geographic
Sociery.
*Materials were adapted with permission of the New York City Department of Environmental Protection. 189
190
Crraprsn
7
LsNcrH eNo LeNcrrr-R-elArso PenaivrsrnRs
In
1954, New York City recognized the need for a third water tunnel to meet the growing demand on its more than 15O-year-old water supply system. Planning for City Tunnel No. 3 began in the eady 1960s, and acnral construction commenced nearly a decade later in 1970. The New York City !7ater Tunnel No. 3, because of its size, length, controlling devices, and depth of excavation, represents one of the most complex and challenging engineering projects in the world today. The New York City'W'ater Tunnel No. 3 also represents the largest capiml improvement project in the city's history. Tunnel No. 3, when complete, is projected to cost nearly $6 billion dollars. Constructed by the New York Cicy Department of Environmenal
Protection (DEP) employees, engineers, and underground construction workers (known as will eventually span more than 60 miles. The tunnel is expected to be complete u;.2020. Since 1970, when construcrion on the runnel begau, a total of 24 people
sandhogs), the tunnel have died
in consuuction-related accidents.
I and No. 2, it will enhance and improve the adequary and dependabiliry of the entire water supply system as well as improve service and pressure to oudying areas of the city. It will also allow rhe DEP to inspect and repair City Tunnels No. 1 and No. 2 for the first time since they were put into operadon in 7917 and 1936, respecdv+ Ahhough City Tunnel No. 3 will not replace City Tunnels No.
The City Tunnel No. 3 project is to be completed in four stages. Stage I of the project has already been completed. Similar to Ciry Tunnels No. I and 2, srage 1 of Ciry Tunnel No. 3 begins at the Hillview Reservoir in Yonkers. It is constructed in bedrock 250 to 800 ft below the suface and runs 13 mi, extending across Central Park until about Fifth Avenue and 78th Street, then stretching eastward under the East River and Roosevelt Island into Astoria, Queens. The first srage of the tunnel, which cost $ 1 billion to construcr, consists of a 24-ft-dnnreter concrete-lined pressure tunnel that steps down in diameter to 20 ft. 'W'ater travels along this route and rises from the tunnel via 14 supply shafts, or "risers,' and feeds into the cityt water disribution system. Currendy, Ciry Tirnnel No. 3 is serving the Upper Easr and Upper'West Sides of Manhattan, Roosevelt Island, and many neighborhoods in the Bronx west of the Brorx River. There are four unique valve chambers that will allow stage L to connec to furure ponions of the tunnel without disrupting rhe flow ofwater service. Each of the existing valve chambers contains a series of 96-in.-diameter conduits with valves and fowmeters to direct, control, and measure the flow ofwater in sections ofthe tunnel. The largest of the valve chambers is in Van Conlandt Park in the Bronx. Built 250 ft below the surface, this valve chamber will conuol the daily flow ofwater from the Catskill and Delarvare water supply qrstems into Tunnel No. 3. In a desrgn departrue from the two enisting tunnels, valves that control the fow of water in Tunnel No. 3 will be housed in large underground valve chambers, makhg them accessible for maintenalce and repair. The valves for City Tunnels No. I and 2 are at the tunnel lwel and thus inaccessible when the tunnels are in service. Three of r:hese four unique subsurface valve chambers have already been built to allow t}le connection of fumre sages of the tunnel without removing the water or aking any other stage of the tunnel out of service. These tluee valve chambers are located in the Bronx at Van Corclandt Park (Shaft 2B), in Manhaman at Central Park (Shaft 13B), and Roosevelt Island (Shaft 15B). Stage 2 of City Tunnel No. 3 construction is comprised of a two-leg Brooklyn/Queens secdon and a section in Manhactan. Stage 2 is scheduled to be completed in 2009 at an approximate cost of $1.5 billion. The combination of stages 1 and 2 will provide the system with
the ability to bypass one or bot'h of City Tunnels No.
I
and2.
The New York City Water Tunnel No. 3
191
Excavation of stage 1 of City Water Tunnel No. 3 looked like this in 1972. Today, water filling this portion of the tunnel is serving areas of the Bronx and Manhattan.
Stage 3 involves rhe construction of a 16-mi-long section ortending from *re Keruico Reservoir to the Valve Chamber in the Bronx, which contains water &om the Catskill and Delaware sysrems. \7'hen stage 3 is completed, City Tunnel No. 3 will operate at greater pressrue' induced by the high elevadon of Kensico Reservoir. k will also provide an additional aqueduct to supply water to the city that will parallel the Delarrare and Catskill aqueducts. Stage 4, rrhich will be 14 mi long, will navel from \6.n Cordandt Park under the East River into Woodside, Queens.
Consrruction of the remaining three stages of City Tunnel No. 3 is being accelerated by the use of a mechanical rock excavacor called a runnel-boring machine (TBM). This machine, which is lowered in sections and assembled on the tunnel floor, chips offsections of bedrock through t-he continuous rotation ofa series of steel cutting teeth. The TBM, which replaces the conventional drilling and blasting methods used during the construction of stage 1, allows for faster and safer excavation. A photo summary illusuating various aspects of Tunnel No. 3 is shown in the accompanying photos.*
*Photos courtesy ofNew York City Department ofEnvironmental Protection.
A form is constructed
to
prepare the tunnel to be lined with concrete.
This finished section of the tunnel, which is Z0 feet in
diameter, is under Shaft l3B, near the Central Park Reservoir.
192
The valve chamber in Van
Cortlandt Park was completed in the early 1990s.
The tunnel boring machine (TBM), which was lowered in
sections and assembled on
the tunnel floor, chips off sections of bedrock in stage 2 of City t{ater Tunnel N0.3. The TBM's rotating steel cutting teeth replace the dynamite needed for earlier excavation.
193
CneprER
7
LsNcrH eNp LrNcrrr-Rsr.Arnp PNaMsrERs
A close-up view of the tunnel boring machine (TBM)
PR0BtEil4 1. Estimate the volume of the
earth that has to be removed to make room for the 60-mi-long, 24-ft-dtatneter, concrete-lined water tunnel. Also, investigate the capaciry of typical dump trucks used in removal of eanh materials.
CHAPTER
B
Trnan AND TIrvIE-RELATED PnneMETERS ood engineers recognize the role of
time in their lives and in calculating speed, acceleration. and flow of
traffic, as well as flow of materials and substances. They know what is meant by
frequency and period and understand the difference between a steady and
a
transient process. Good engineers have a comfortable grasp of rotational motion and understand how it differs from
translational motion.
195
196
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8
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In
nsu
PARAMntnRs
the preuious chapter, we considered the role
of ltngth and hngth-rektedpararneters in engineering. In tltis chapur, we will inuestigate the role oftirne as afundarnentaldirnension and other tirne-related engineeringparametns, such wfrequency and pniod. Ve will first discass why the ti.rne uariable is needcd to describe euents, processes, and othn occalrences in our pbysical surroundings. Ve will then mplain the role of periods and frequency in recurring or periodic, eaents (euents that repeat thernselues). A brief innod.uction to paraTneters fustibing nffic flo* it also gium in this chapter. Finally, we will look at mgineering uariables that inuolae lmgth and tirne, including linear uelocities and accelerations and uolurne flow rate offlwid,s. Rotational rnotion is also introduced at the md of this chapten
E.n
Time as a Fundamental Dimension 'W'e
live in a dynamic world. Everything in the univene is in a constanr state of motion. Think about it! Everything in the universe is continuously moving. The eanh and everything associated with it moves around the sun. All of the solar planets, and everytJring that comprises rhem, are moving around the sun. We know that every*ring oumide our solar q/stem is moving too. From our everyday observation we also know that some things move faster than others. For example, people can move faster than ants, or a rabbic can move faster tiran a tunle. A jet plane in fight moves faster than a car on a highway. Time is an important parameter in describing motion. How long does it take to cover a cenain distance? A long time ago, humans learned that by defining a parameter calle d tirne rhey could use it to describe the occurrences ofevents in dreir surroundings. Think abour the questions &equendy asked in your everyday lifel How old are you? How long does it take to go from here to there? How long does it ake to cook this food? How late are you open? How long is 'We your vacation? use the parameter of time to express how old we are, \7e have also associated time with natural occurrences in our lives, for example, to s(press the relative position of the earth with respect to the sun, we use day, night,2:00 e.la. or 3:00 n,u., or May 30. The parameter time has been conveniendy divided into smaller and larger intervals, such as seconds, minutes, hours, days, months, years, centuries, and millennia. 'W'e are condnuously learning more and more about our surroundings in terms of how nature was put together and how it worls, thus we need shoner and shorter time divisions, such as microseconds and nanoseconds. For example, with the advent of high-speed communication lines, the time that it takes elecffons to move between short distances can be measured in nanoseconds. We have also learned from our observation of the world around us that rve can combine the parameter time with the parameter length to describe how fast something is moving. \f.hen we ask how fast, we should be careful to state with respect to what. Remember, everyrhing in the universe is moving. Before discussing the role of time in engineering analysis, let us focus on the role of time in our lives-our limited time budget. Todqy, we can safely assrune that the ulerage life expectancy of a person living in the'Western world is around 75 years. Let us use this number and
8.1 Trur es a Furvorvrrn'u'r Drurusrox
197
perform some simple arithmetic operations to illustrate some interesting points, Converdng the 75 years to hours, we have: (Z
5 years)(165.25 daysl year) (24 hours/day)
=
657,450 hours
On an average basis, we spend about 1/3 of our lives sleeping; this leaves us with 438,300 waking hours. Considering that traditional college freshmen are 18 years old, you have 333'108 waking hours still available to you ifyou live to the age of 75 years. Think about this for awhile. If you were given only $333,108 for the rest of your life, would you throw arvay a dollar here and a dollar there as you were strolling through life? Perhaps not, especially knowing that you will not get a$y more money. Life is sho*! Make good use ofyour time, and at the same time enjoy your life. Now let us look at the role of time in engineering problems and solutions. Most engineering problems may be divided into two broad arex; steady and unstea.dy. The problem is said to be steadJt when the value of a physical quandry under investigation does not change over time. For example, the length and width of your credit card doesrit change with time if it is not subjected to a rempemrure change or a load. If the value of a physical quantity changes with time, t-hen the problem is said to be unsteady or transient A good example of an unsrcady situation that is familiar to everyone is the rate of a person's physical growrh. From the time you were born until you reached your late teens or early twendes, your physical dimensions changed with time. You became taller, your arms grew longer, your shoulders got wider. Of cowse, in this example not only the dimension length changes with time but also your mass changes with time. Another common example of an unsteady event that you are familiar with deals with the physical variable temperature. S(h.en you lay fteshly baked cookies out to coo[, the temperature of the cookies decreases with dme until they reach the air temperature in the kitchen. There are many engineering problems that are
/
HogePully
lwill
harVe
art I earst a*\ol+\ev 300 A00 hoqrs ,wr'rilarble to Ae +o Ao gool. thiugs ir li€e!
There is still time for you to do some good things in life.
198
Cneprsn
8
TrtvrsANo Trur-Rnr,errp Penervrnrsns
Once I reach the age of 20,
does my physical dimension
really become steady? What if lwere to gain mass?
-/ou
will find examples of unsteady and steady processes in the cooling of electronic equipment, biomedical applications, combustion, materials casting, materials processing, plastic forming, building heating or cooling applicadons, and in food processing. The also unsteady.
transient response of a mechanical or structural s)rstem to a suddenly applied force is another instance of an unsteady engineering problem; for example, the response of a cart suspension system as you drive through a pothole, or rhe response ofa building to an earrhquake.
8.2
Measurement of Time Early humans relied on the relative position of the earth with respect ro rhe sun, moon, srars, or other planets to keep track of time. The lunar calendar was used by many eady civilizations. These celestial calendars were usefirl in keeping rack of long periods of time, but humans needed to devise a means to keep track of shoner dme intervals, such as what today we call an hour. This need led to the development of clocks (see Figure 8.1). Sun clocls, also known as shadow clocls or sundials, were used to divide a given day into smaller periods. The moving shadow of the did marked the dme intervals. Like other humaa-invented instruments, the sundial evolved over dme into elaborate instruments that accounted for the shoftness of rhe day during the winter as compared to the summer to provide for a bemer year-round accuracy. Sand glasses (glass containers filled rvith sand) and water clocks were among t},e first timemeasuring devices that did not make use of the relative position of the eanh wirh respect to rhe sun or other celestial bodies. Most of you have seen a sand glass (sometimes referred ro as an hourglCIs); the water clocks were basically made of a graduated container with a small hole near the bonom. The container held water and was tilted so that the water would drip out of the hole slowly. Graduated cylindrical containers, into which water dripped at a constant rare, were also used to me:$ure the passage of an hour. Over the years, the design of water clocls was a-[so modified. The next revolution in timekeeping came during the 14th cenrury when weight-driven mechanical clocks were used in Europe. Lacer, during the 15th century, came
8.2
I
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or Trur
199
rigureo.t
An we rcally measufing time?
springJoaded clocls. The spring mechanism design wentually led to smaller clocks and to watches. The oscillation of a pendulum was the next advancement in the design of clocks. Quarz clocks wentually replaced the mechanical clocks around the middle of the 20th century. A quarz clock or watch makes use of the piezoelectric propefty of quarz crystal. A quara crysal, when subjected to a mechanical pressrue, creates an electric field. The inverse is also true-thar is ro say, the shape of the crystal changes when it is subjected to an electric field. These principles are used to design clocks that make the crysnl vibrate and generate an elecuic signal of constant frequency. As we stated in Chapter 6, the natural frequenry of the cesium atom was adopted as the new standard unit of time. The unit of a second is now formally defined as the duration of
9,192,631,770 periods of the radiation corresponding to the transition between the flvo hyperfine levels of the ground state of the cesium 133 atom.
The Need for Time Zoncs You know that the earth rotates about an axis that runs from the South Pole to the North Pole, and it akes rhe earh 24 hours to complere one revolution about this axis. Moreover, from
studying globes and maps, you may have noticed that the earth is divided into 360 circular arcs thar are equally spaced from east to west; these arcs are call ed longitudcs. The zero longitude was arbitrarily assigned to rhe arc that passes doo"gh Greenwich, England. Because it takes the earth24 hours to complete one revolution about its axis, every 15 degrees longitude corresponds to t hour (360 degrees/Z4 hovs = 15 degrees Per one hour). For example, someone exacdy 15 degrees west of Chicago will see the sun in the same exact position one hour later as it was observed by another person in Chicago one hour before. The eanh is also divided into latitudes, which measure the angle formed by the line connecting the center of the earth to the specific location on the surface of tJre earth and the equatorial plane, as shown in Figure 8.2. The latitude varies from 0 (equatorial plane) degrees to 90 degrees nomh (Nonh Pole) and from 0 degrees to 90 degrees south (Soudr Pole).
200
CnarrsR
8
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exo
Trtvrs-RELATEp Per,arvrBrnns
N
I
figure8.Z
The longitudes and latitudes of
the earth.
The need for time zones was not redized until the laner parr of the 19th centurywhen the railroad companies were expanding. The railroad companies realized a need for standardizing rheir schedules. After all, 8:00 A.M. in New York Ciry did nor correspond to 8:00 a.u. in Denver, Colorado. Thus, a need for a uniform means to keep track of time and its relationship to other locations on the eanh was born. It was railroad scheduling and commerce that wentually brought nations together m define dme zones. The standard time zones are shown in Figure 8.3.
Daylight Saving Time Daylight saving time was originally put into place to save fuel (energy) during hard times such as'W'orld !?'ar I and \forld'War II. The idea is simple; by setting the clock forward in the spring and keeping it that way during the summer and the early fall, we extend the daylight hours and, consequendp we save enerry. For example, on a certain day in the spring without daylight saring time, it would get dark at 8:00 p.u., but with the clocls set forward by one hour, it would then get dark at 9:00 p.u. So we nrn our lights on an hour later. According to a U.S. Deparrment ofTransportation study, daylight saving time saves energy because we tend to spend more time outside of our homes engaging in outdoor activities. Moreover, because more people drive during the daylight hours, daylight saving could also reduce the number ofautomobile acciden$, consequendy saving many lives. In 7966, the U.S. Congress passed the Uniform Time Act to establish a system of
uniform time. Moreovet in 1986, Congress changed tlre initiation of daylight saving time from the last Sunday in April to rhe first Sunday in April. In 2005, Congress approved an amendment to hane d"yligh. saving time start on r:he second Sunday in March and end on the fust Sunday in November, commencing in2007, Today, most countries around the world follow some type of daylight saving schedule. In the European Union countries, the daylight saving begins on the last Sunday in March and continues through the lasr Sunday in October.
8.3 Frnrons exo Fnrqunrcrrs
201
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STANDARD TIME ZONES Coreted to April2@1 tuIM6m@ll@ Dayligh Slviog Tine (S@/ fme), Nally oB hM io sdrc of S@dad Tiebkarir sreFl@ ,l,fqodlJ.ft
Cryil'il
@ Mo@ain Elsh Maps br HM N@icdl Ah@ Afr.e
120"w
il
Fgure8.3
SMdqd Time = Univemd Ti@ + ralue frcm eble !n he hE
90:w 60:w 30:w
Q'
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2
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E
S
6
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Standardtinezones.
Source: Map oudine @
8.3
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Mountain High Maps. Data supplied by HM Nautical Almanac Office
Periods and Frequencies For periodic evens , a periodis the dme that it takes for the event to repe.rt itself. For example, every 365.24 days the eanh lines up in exacdy the same position with respect to the sun. The orbit of the eanh around the sun is said to be periodic because this event rePeats itself. The inverse of a period is c,alled a frequency. For example, the frequency at which t.he earth goes around the sun is once a year. Ler us now use other simple examples to explain the difference between period and frequency. It is safe to assume thar most of you do laundry once a week or buy groceries once a week. In this case, the frequenry of your doing laundry or buying groceries is once a week Or you may go see your dentist once everF six months. That is the frequency of your dentist visits. Therefore, frequency is a measure of horr frequendy an event or a process occurs' and period is the time rhat it akes for that event to complete one rycle. Some engineering examples that include periodic motion are oscillatory systems such as shakers, mixers, and vibrators. The piston inside a cart engine rylinder is another good example of periodic motion. Your car's suspension system, the wings of a plane in a turbulent flight, or a building being shaken by strong
I
rigure8.4
A spring-mass system.
winds may also show some component of periodic motion. Consider a simple spring-mass sy$tem, as shovrn in Figure 8.4. The spring-mass sy$em shown could represent a very simple model for a vibratory slstem, such as a shaker or a
20?
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Application
Frequenry
Alternating curent (US$
60Hz
AM radio
540 kHz-1.5
Television (channels 2 -6) FM radio Emergenry, fire, police Television (channels 7-13) Penonal computer clocls (as ofyear 2000) Cordless phone (as ofyeat 2004)
54*88MHz
MHz
88-108 MHz 153-r59 MHz
174-216MHz
uptolGHz 5.8 GHz
i ,
'
l
vibrator. \7hat happens if you were to push down on the mass and then let go of it? The mass will oscillate in a manner that manifests itself by an up-and-down motion. If you study mechanical vibration, you will learn that the natural undamped frequency of the qrstem is given by
f = 1 Ja 2n
L
(8J)
7n
wheref
is the natural frequency of the sJrstem in cycles per second, or Herz (Hz), & represents the stiftress of the spring or an elastic member (N/m), and rn ls the mass of the qystem (kg). The period of oscillation, 7l for the given qfstem-or in otherwords, the time rhat it takes for the mass to complece one cycle-is given by
-t-f
| $.21 Jn
Most ofyou have seen oscillating pendulums in clocla. The pendulum is another good example of a periodic system. The period of oscillation for a pendulum is given by
f
: Zo.f
(8.3)
I is the length of the pendulum (m) andgis the acceleration due to gravity (m/s2). Note that the period of oscillation is independent of the mass of rhe pendulum. Nor too long ago, oil companies used measured changes in the period of an oscillating pendulum to detect variations in acceleration due to gravity that could indicate an underground oil
where
reservoir.
An understanding of periods and frequencies is also importanr in the design of electrical and electronic components. In general, excited mechanical systems have much lower frequencies than electrical/elecronic qFstems. Examples of frequencies of various electrical and electronic systems are given in Thble 8.1.
8.4 Frov or Tnarrrc
203
@Determinethenaturalfrequencyofthesimplespring_masssystemshowninFigure8.5. ,,,
'r'l
ft=5N/mrn=5000N/m
,1,
ri
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8.5
A simple spring-mass systetn.
Using Equation (8.1), we have
i'i
ti1
F:I Jn 2n \lm 2r
i'l i,i,
li----:
Figure
_-- -_.-
-
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-- -- l":-,lh-----
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----
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8.4 Flow of Traffic Those of you who live in a big ciry know what we mean by uaffic. A branch of civil engineering deals with the design and layout of highways, roads, and streets and the location and timing of raffic control devices that move vehides efficiendy. In this secdon, we will provide a
A congested flow!
204
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brief overview of some elementary concepts related to trafifi.c engineering. These variables are time related. Let us begin by defining what we mean by traffic flow. In civil engineering, traffic flow is formally defined by the following relationship:
q:
3600n {8.4}
T
In Equation (8.4), 4 represents the ffamc flow in terms of number of vehides per hour, a is the number ofvehicles passing a known location during a time duration ?-in secon&. Another usefirl variable of traffic information is densiry-how many cars occupy a stretch of a highway. Density is defined by
h-
1000n CI.51
where / is density and represents tJre number of vehicles per kilometer, and n is the number of vehicles on a stretch of highway, f, measured in meters. Knowing the average speed of cars also provides valuable information for the design of road layouts and the location and timing of traffic control devices. The average speed (7) of cars is determined from
a- +2",
(8"6)
In Equation (8.6), uiis the speed of individual cars (we will discuss tle definition of speed in more detail in Section 8.5.), and z represents the total number of cars. There is a relationship among the raffic parameters-narneh the fow of traffic, densrty, and the average speedaccording to
q:
(8.71
Ea
where q, h, and,T.were defined earlier by Equations (8.4) through (8.6). The relationship among flow, density, and average speed is shown in Figures 8.6 and 8.7. Figure 8.6 shows that when the average speed ofmoving cars is high, the value ofdensity is near zero, implying that there are not that many c{us on that stretch of highway. As you may exped,
Uncougested
o
C)
c)
Density (ft)
n
tigure
8.6
The relationship between speed and density.
flow
Flow (4)
a
Figute
8.?
The relationship between speed and
florl.
8.5
PenemstrRs
Ixvorvrrc LsNcrH axp Tnns
205
of density increases (the number of vehicles per kilometer), the average speed of vehicles decreases and will wentually reach a zero value, meaning bumper-to-bumper ffamc. Figure 8.7 shows the relationship between the average speed of vehides and the flow of traffic. For bumper-to-bumper traffc, with no cius moving, the fow of traffic stops, and thus q has a value of zero. This is the beginning of the congested region shown in Figure 8.7. fu the average speed of the vehides increases, so does dre flow of traffic, eyentually reaching a maximum value. As the average speed ofvehicles continues to increase, the fow of traffic (the number of as the value
vehicles per hour) decreases. This region is marked by an uncongested fow region. Finalln uaffic engineers use various measurement devices and rcchniques to obtain realtime dam on the flow of uaffic. They use the collected information to make improvements to movevehicles moreefficiendy. You mayhave seen er
Show that Equation (8.7) is dimensionally homogeneous by carrying out the appropriate
unia
for each term in that equation.
\-/kilometer\ 4\lvehicla\,1 h-" / = u\ttt,r,'.* )'\ h"* ) vehicles
r-r-:-: -:-
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rri r
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8.5 Parameters Involving Length and Time In this
section, we will consider dzriued. pbysical quantities thar are based on the fundamentd dimensions of length and rime. Wb will first discuss the concepts of linear speed and acc.eleration and then define volumeuic flow rate.
Linear Velocities V'ewill
begrn by explaining the concept of linear speed. Knowledge of linear speed and acceleration is important to engineers when designing conveyer belts that are used to load luggage into aiqplanes and product assembly lines, ueadmills, elevators, automatic walkrarays, escalators, water or gas fow inside pipes, space probes, roller coasters, transportation q/stems (such as cars, boats, aiqplanes, and rockem), snow removal equipment, baclup computer tape drives, and so on. Civil engineers are also concerned with velocities, pardcularly wind velocities, when designing strucnu€s. They need to account for the wind speed and its direction in their calculations when sizing strucnrral members. Let us now take a close look at what we mean by linear speed and linear velocity. All of you ,ue familiar with a car speedometer. It shows the instantaneous speed of a car. Before we explain in more detail what we mean by the term iwananeous Eeed" let us define a physical variable that is more easily understood, the aaaage geed" which is defined CI average speed
:
disance traveled ume
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Note that the fundamental (base) dimensions oflength and time are used in the definition of the anerage speed. The ar/erage speed is called a deiued pltysical quanfiry because its definition is based on the fundamental dimensions of length and time. The SI unit for speed is m/s, although for fast-moving objects km/h is also commonly used. In U.S. Customary units, ft/s and miles/h (mph) are used to quantify the magnitude of a moving object. Now let us look at what we mean by the instantaneous speed and see how it is related to the average speed. To understand the difference between irverage and instantaneous speed, consider the fol-
loroirg mental exercise. Imagine that you are going from NewYork City to Boston, a distance of 220 miles CI54 kn.). Let us say that it took you 4.5 hours to go from the outskirts of New York City to the edge of Boston. From Equation (8.8), you can determine your average velocity, which is 49 mph (79 kmlh). You mayhave made a rest stop somewhere to get a cup of coffee. Additionally, the posted highway speed limit may have varied from 55 mph (88 km/h) to 65 mph (105 km/h), depending on the stretch of highway. Based on the posted speed limits and other road conditions, and how you felt, you may have driven the car faster during some stretches, and you may have gone slower during other suetches. These conditions led to an average speed of 49 mph (79 km/h). Let us also imagine that you recorded the speed of your car as indicated by the speedometer every second. The acnral speed of rhe car ar any given instant while you were driving it is called the instantaneous Eteed" To better understand the difference between the average speed and the instantaneous speed, ask yourself the following question. If you needed to locate the car, would you be able to locate the car knowing just dre average speed ofthe car? The knowledge ofthe average speed of the car would not be sufficient. To know where the car is at all dmes, you need more informadon, such as the instantaneous speed of the car and the drrection in which it is traveling. This means you must know the instantaneous velocity of the car. Note that when we say veIocity ofa car, we not only refer to the speed ofthe car but also the direction in which it moves. Physical quandties that possess both a magnitude and a direction are called oectors.You will learn more about vectors in your calculus, physics, and mechanics classes. For now, just remember the simple definition of a vector quantity-a quantity *rat has both magnitude and direction. A physical quantiry that is described only by a magnitude is called a scalar quarntity, Examples of scalar quantities include temperaffe, volume, and mass. Examples of the range of speed ofvarious objects are given in Thble 8.2.
r:
.
m/s
I i ; '; I i i
Average speed of a persou
ri'alking
, l
The fastest runner in dhe world (100 m) Professional tennis player serving a ball Top speed ofa spons car 777 Boengatqplane (cruhe speed) Orbital speed of t}re space shutde Avetage orbital speed of the eanh
l,,j 10.2
,
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209
o/
24r
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,
,
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Linear Aeeeleratisns Acceleration provides a measure ofhow velocity changes with time. Something that moves with a constant velocity has azero acreleraion. Because aehcity is a ueaor quantity and hat both rnagnitudc and. d.irection, aqr change in either the di.reaion or the rnagnindc ofuebcity cAn c?ente accel,eration. For example, a car moving at a constant speed following a circular path has an acceleration component due to the change in the direction of the velocity vector, as shown in Figure 8.8. Here we will focus on an obiect moving along a straight line. The a/erage acceleration is defined as average acceleration
:
change in velocity
(8.e)
ume
Again note that acceleration uses only the dimensions of length and time. Acceleration represents lhe rate at which the velocity of a moving object changes with time. Therefore, acceleration is the time rate of change of velocity. The SI unit for acceleration is m/s2 and in U.S. Customary units ft/s2. The difference benareen instantaneous acceleration and average acceleration is similar to the difference between instantaneous velociry and average velociry. The instantaneous acceleration can be obtained from Equation (8.9) by making the time interval smaller and smaller. That is to say, insananeous acceleration shows how the velocity of a moving object changes at any instant. Let us now turn our attention to acceleration due to gravity. Acceleration due to gravity plqrs an important role in our weryday lives in the weight of objecrs and in the design of projectiles. \(hat happens when you let go ofan object in your hand? It falls to the ground. Sir Isaac Newton discovered that two masses attract each other according to
F:#Gm,ra f
0.10)
where Fis the attractive force bet"neen the masses (N), Grepresents the universal graviutional atrracdon and is equal to 6.7 10-11 m3/k. &, rq ar..d m2are the mass of each particle (kg), and r denotes the distance between the center of each particle. Using Equation (8.10), we qrn determine the weight of an object having a mass m (on Farth) by substituting m fot mp substituting for m, the mass of the Earth, and using the radius of the earth as the distance r between the center of m1 and m,. At the surface of dre Earth, the atractive force .Fis called the
x
f
rigun8.B
Ihe acceleration of a car moving at a constant speed following a circular path.
208
CneprEn
8
Tur'rB.lNp TrMr-RELArno PenarrsrpRs
weight of an object, and the acceleration created by the force is referred to as the accelaa.tion At the surface of the Eanh, the value ofg is equal to 9.81 m/s2, or 32,2 ft1s2. In general, acceleration due m gravity is a funcdon of latitude and longitude, but for most practical engineering applications near the surface of the Earth the values of 9.81 m/s2 or 32.2 ftlsz due to graui4t (re).
are sufficient.
Thble 8.3 shows speed, acceleration, and the distance uaveled by a falling object from the 'We roof of a high-rise building. have negleced air resistance in our calculadons. Note how the drstance traveled by the falling object and its velocity change with time.
rnmur a.s,
Acceleration
Time of the Object (seconds) (m/s'z) 0 1
2 3 L
5
6
9.81 9.81 9.81 9.81 9.81 9.81 9.81
Speed of
the
Distance Travelqd (m)
Fa[ingobject (a=, "\ (m/s) (Ir= gq \ 'tC' ) 0
0
4.90
9.81 19.62 29.43 39.24 49.05
4/t.14 78.48 r22.62
58.86
176.58
19.62
8.5
Panerr,rnrnns lNvor,vrNc
Ixucrn exp Trun
.
209
accelerates to a speed of 100 km/h in 15s. The acceleration during this period is constant. For the next 30 minutes, the car moves with a constant speed of 100 km/h. At this time the driver ofthe car applies the brake and the car decelerates to a firll stop in 10s. The variation of the speed of the car with time is shown in Figure 8.9. \ffe are interested in determining the total distance traveled by the car and its average speed over this distance.
A car sarts from rest and
g C)
o
(n
Ttme (s)
fil
figure
8.9
The variation of the speed
0l the car with time for Eromple 8.3.
During the inidal 15 seconds, the speed of the car increases linearly from a value of zero to 100 kmih. Therefore, ttre average speed of the car is 50 km/h during this period. The distance traneled during this period is
7,
:
(time)( ayetaFespeed)
:
1r:
,y
(:o$) f "]5) ( toqo m \ \ n/\aeoosl\ tk^ )=
During the next 30 minutes the car moves with uaveled during this period is
a constant speed
208'3m
of 100 km/h and the distance
/
r dz:(r8o0s)\.l0c,!a\ f -r-!-\ (teeqh /\aeoos,/\ 1 k^ ):50000m Because the car decelerates at a constant rate from a speed of 100 km/h to 0, the average speed of the car during the last 10 seconds is also.50 km/h, and the distance traveled during this
period is
, (ro.)(:o;7 ,<^,/-^k*\ ( rt \/rooo*r d,:
t_.;/ tffi,) : r38.e m
The totd distance uaveled by the car is
d: dr+ dz+ fu:208.3 + 50000 + 138.9:50347.2m= 50.3472knr And finally, the average speed of the car for tfie entire duration of travel is &stancf traveled Vnog. =-.1--
-
--
---,-.----li-riti:irrlillill
dme
=
to3-!:' : 1825
27.56 mts
:
99.2kmlh _-.
--__---_--__i
210
Cnesrsn
8
Tr*rsANo Trvrr-RBr,.lrsp PnnaurrsRs
Volume Flow Rate Engineers design flow-measuring devices co determine the amount of a material or a substance fowing through a pipeline in a processing plant. Volume flow rate measuremenm are necessary in many industrial processes to keep rack of the amount of material being transponed from one point to the next point in a plant. Additionally, knowing the flow rate of a material, engineers crul determine the consumption rate so that they can provide the necessary supply for a steady state operation. Flovr-measuring devices are also employed to determine the amount of water or natural gas being used by us in our homes during a specific period of time. City engineers need to know the daily or monttrly volumetric water consumption rates in order to pro-
vide an adequate supply of water to our homes and commercial plants. Many homes in the United States, and around the worl4 use natural gas for cooking puqposes or for space heating. For example, for a gas firrnace to heat cold air, natural gas is burned inside a heat exchanger that transfers the heat of combustion to the cold air supply. Companies providing the naturd gas need to know how much fuel-how many cubic meters of natural gas-are burned wery day, or every month, by each home so that they can correcdy charge their customers. In sizing the heating or cooling units for buildings, the volumetric fow ofwarm or cool air must be determined to adequately compensate for the heat loss or heat gain for a given building. \Gntilation rates dealing with the introducrion of &esh air into a building are also expressed in cubic feet per minute (cfm) and cubic meter per minute or per hour. On a smaller scale, to keep the microprocessor inside your computer operating at a safe temperafllre, the volumetric fl.ow of air must also be determined.
Another example thatyou maybe familiarwith is the f.owrate of awater-antGeezemixture necessary to cool your car's engine. Engineers who design the cooling sJrstems for a car engine need to determine the volumeuic flovr rate of the water- antifreeze mixnrre (gallons of water-antifreeze mixture per minutQ through the car's engine block and tluough the car's radiator to keep the engine block at safe rcmperaftres. Now that you have an idea what we mean by the term uolame*icfaw rate,let us formally define it. The volume fow rate is simply defined by the volume of a given substance that flows through something per unit time. volume flow rate
:
volume
CIJl)
ume
Some of the more common units forvolume flowrate include, m3ls, m3lh, L/s, ml/s, ffls (cft), gal /min (gpm), or gal /day (gpd). Note that in the definition given by Equation (8.1 1), the fun-
damental dimensions of length (length cubed) and dme are used.
trpefiment for You to Gonduct LJse a stopwatch and an empty Pepsi or Coke can to determine the volumetric flow rate of water coming out of a drinking fountain. Present your results in liters per second and in gallons per minute. For fluids flowing through pipes, condui$, or nozzles, there enists a relationship between the volumetric flow rate, the average velocity of the flowing fluid, and t"he cross-sectional area of the flow, according to A Fun
volumeuic flow rate
:
(average velocity)(cross-sectional area
of the
flow)
(812)
8.5
PanevntsRs Ixvor.vnvc I;sNcrH aNo Trrvre
211
flow-measuring devic€s make use of Equadon (8.12) to determine the volume fow rate of a fluid by 6rst measuring the fluid's aver€e velocity and the cross-sectional area ofthe fow. Finally, note that in Chapter 9 we will explain another closely defined variable, mass flow rate of a fowing material, which provides a measure of time rate of mass flow through pipes or other carrying conduits.
In fact, sorne
W
Consider the piping system shovrn in Figure 8.10. The average speed ofwater flowing *oo"gh rhe l2-inch-diameter secdon of the piping system in 5 ft/s. !?hat is the volume fow rate of water in the piping qrstem? Express the volume flow rate in ff/s, gpm, and L/s. For the case of steady fow of water through the piping qystem, what is the average speed of water in tlle 6-inch-diameter section of the system?
j) tl
ij 'tii
i:t
.
Figme8.t0 'W'e
Thepipingsystemof Example8.4.
can determine the volume flow rate of water through the piping qfstem using Equa-
tion (8.12). volume flow rate
e:
:
(average velocity)(cross-sectional area of
volumeflowrate
:
(s fi/s)
(DU
ft)2
q: (3.e26ro(+#X#) : q:
Q.e26
:
fow)
3.926ff ts
tT6zgpm
/ 28.3r!r-BL) n:/,)( 1tr l = rr r.2Lts
For the steady flow of water through the piping system the volume flow rate is constant. This fact allows us to calculate the speed ofwater in the 6-inch section of the pipe.
Q:
3.926
fflr = (average velocity)(cross-sectional
3.e26 f€ts= (averase **o) avel. gespeed
:
(;)(
area
of flow)
0.5 ft)z
20ft|s
that qrhen you reduce the pipe diameter by afactor of 2, the water speed in the reduced section of the pipe increases by afaaor of 4.
From the results of this example you
------:ftifr.-_
.
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[-.
czln see
-:
--
..ia,-----=-1.
----.. . -..-- ----..
-..-. ..-. --
---- -.llil
212
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8.6
Angular Motion In the next two sections, we erations of rotating objects.
will
discuss angular motion, including angrlar speeds and accel-
Angular (Rotational) Speeds In the previous section, we explained the concept of linear speed and acceleration. Now we will consider variables that define angular motion. Rotational motion is also quite common in engineering applications. Examples of engineering components wir:h rotationd modon include shafts, wheels: g€?rs, drills, pulleys, fan or pump impellers, helicopter blades, hard drives, CD drives, Zip drives, and so on.
The average angrlar speed of a line segment located on a rotating object such as a shaft is defined as the change in its angular position (*gul* displacement) over the time that it took the line to go through the angular displacement.
,:E
L0
A stroboscope and a tachometer.
(8r3)
8.5
ANcur-an MorroN
213
anerage angrlar speed in radians per second, A0 is the angular displacemenr (radians), and Aris the time interval in seconds. Similar to the definition of insantaneous velocity given earlier, the instantaneous angular speed is defined by making
In Equation (8.13), tlr represents the
the time increment smaller and smaller. Ag"itt, when we speak of angular velocity, we not only refer to the speed of rotation but also to the direction of rotation. It is a common practice to express the angular speed of rotating objects in revoludons per minute (qpm) instead of radians per second (radA). For example, the rotational speed of a pump impeller may be expressed as 1600 rpm. To conveft the angular speed value from rpm to rad/s, we make the appropriate
conversion substitutions.
- rad -^^( revolutions\ ( 2tr radnns \ / t minute \ : ,.r67't; 1600(' /\t'*"r,*o,'/\u.'*"'a" ) -
practice, the angular speed of rotating objects is measured using a stroboscope or a tachometer. To get some idea how fast some common objecs rotate, consider the following examples: a dentist's drill runs at 400,000 rpm; a current state-of-rhe-aft computer hard drive runs at 7200 rpm; the earth goes through one complete revolution in 24 hours, thus the rotational speed of earth is L5 degrees per hour or I degree every 4 minutes. There exists a relationship between linear and angular velocities of objects that not only rotate but also nanslate as well. For example, a car wheel, when not slipping will not only rotate but also ranslate. To establish the relationship between rotational speed and the ranslational speed we begin with
In
AS:
(8J4)
rAd
dividing both sides by time inoement
AS At
At
L0
(8.r5)
A,t
and making use of the definitions of linear and angular velocities, sre have
V-- ra
(8.r6)
For example, the acnral linear velocity of a particle located 0.1 m away from the center of shaft that is rotating at an angular speed of 1000 rpm (104.7 radls) is approximately 10.5 m/s.
Determine the rotational speed of a car wheel 55 mph. The radius of the wheel is 12.5 in.
if the car is uanslating along at a speed of
Using Equation (8.16), we have
V
a:-: r
("+)(#)(#) : (t2.5in.)
(;*) .
77.4 rad/s
=
739 rpm
214
Cnal'rpn
8 Trurexo TrMs-REr.arsp
PaneMsrsRs
Angular Accelerations speed and ir irnportance in engineering applications. We now define angular acceleration in terms of the rate of change of angular velocity. Angular acceleration is a vector quantity. Note also that similar to dre reladonship be*reen the a',erage acceleration and insantaneous acceleration, we could first define an average an$rlar
In the previous section, we defined angrlar
acceleration as
angular acceleradon
ry
:
change in angular speed
and then define the instananeous angular acceleration by making the time interval smaller and smaller in Equation (8.17).
It
takes 5 s for a shaft of a motor to go from ?rro to 1600 rpm. Assuming constant angular acceleration, what is the value of the angular acceleration of the shaft? First, we conven the final angular speed ofthe shaft from qpm to radls.
ii
I
(8r7)
2z'radians \/ t mi""t. ,Uoo(revolutions)( \ minutes Z t r r*ot,r.iorr/\at *-"a"
\ : rad t67'5; I
215
PnosLEMs
Then we use Equadon (8.17) to calculate dre angular acceleration. angular acceleration
-
changes in angular sPeed
time
-
(L67'5
:, 0) ndls = 5s
33.5
rad
7
itaritr-ii+Ei,ltrllr!,flati'li.rrrn
Finally, remember that all the parameters discussed in this chapter involved dme, or length and dme.
SUMMARY Now that you have reached this point in the text
" You should have a good grasp ofyet
' ' ' '
8.1.
another fundamental dimension in engineering, namely time, and im role in engineering analysis. You should also recognize the role of time in calculating speed, acceleration, and fow of traffic, as well as fow of marcrials and subsances. You should know what we mean by frequency and a period, and be able to give some ex-
amples of mechanical and electrical q/stems with frequency and periods. You should know how to define average and instantaneous velocity. You should also be able to define average acceleration and instantaneous acceleration. You should have a comforable grasp of a rotational motion and how it differs from a trans-
lation motion. You should know how to define angular velociry and acceleration. You should understand the significance of volume fow rate in our weryday lives and in engineering applications. You should also have a good grasp of what is meant by volume flow rate.
Create a calendar showing the beginning and the end
Daylighr SavingTime Year
Begins at 2:00
e.u.
Daylight Saving Ends at 2:00
a-v.
2009
8.3.
2010
20ll 2012 2013 2014 2015
20r6 8.2.
According to the Depanment of Transportation analysis of energyconsumption figures for 1974and1975,
obseryation of daylight saving time in the month of years of 1974 and L97 5 saved the country an estimated energy equivalent of 10,000 barrels of oil each day. Estimate how much energy is saved in the United Sates or yoru country due to observing daylight sardng time in the current year. Is there a need for a country near the equator to observe daylight saving time? Explain. BesidCI energy savings, what are other advantages that observation of daylight saving time may bring? Every year all around the world we celebrate certain cultural events dealing with our past. For enample, in Christianiry, Easter is celebrated in dre spring and Christmas is celebrated in December; in the Jewish calendar, Yom Kippur is celebrated in October; and Ramadan, the time offasting, is celebrated by Muslims accordingto alunar calendar. Briefly discuss thebasis of the Christian, Jewish, Muslim, and Chinese calendars.
April in the
of dq/ight saning time for the years 2009-2016.
8.4.
8.5.
216 8.6.
8.7.
Cneprrn8 TrugeNpTrMr-R-sr-drnoPer.qurrnns
In this problem you are asked to investigate how much watera leakyfaucetwastes in oneweeh one month, and one year. Perform an experiment byplacing a container under a leaky faucet and acnrally measure the amount of water accumulated in an hour (you can simulate a leakyfaucet by just panially closing the fauce$. You are to design the o
Speed
8.9.
8.r0.
8i1.
Speed
Limit
Speed
Limit
l)
30 35 40
4) )> 65 70
c.
at/e'?lge
d. e.
average yearly basis
monthly basis
oYer aperiod of ten years Fxpress your results in gallons and liters.
812. Calculate the speed of sound for the U.S. standard atmosphere using, a
: \/ h-RT,where c represents the
air (h = L.4), and R is the gas constant for the air (R = 286.9 Jlkg. K) Trepresents the temperature of "nd the air in Kelvin. The speed of sound in atmosphere is the speed at which sound propagates through the air. You may use Excel to solve this problem. speed of sound in m/s, & is the specific heat rado for
Altitude
Air Gmperature
Speed
(m)
(K)
Sound (m/s)
fountain.
500
284.9
1000
28r.7
2000
275.2
5000
255.7
10,000
223.3
15,000
216.7
20,000
216.7
40,000
250.4
50,000
270.7
maF use Microsoft Excel to solve this problem. Most car owners drive their cars an a\rerage of 12,000
8J3.
miles a year. Assuming a 20 miles/gallon gas consumption rate, determine the amount of fuel consumed by 150 million crr owners on the following time basis:
8.14.
a. average dailybasis b. average weekly basis
Umit
(m/s)
tank needs to be refilled. State your assumptions. You are to investigate the water consumption in your house, apartment, or dormitory-whichever is applicable. For example, to determine bathroomwater consumption, time how long it normdly takes you to shower. Then place a bucket under the showerhead for a known period of time. Determine the total volume ofwater used when you take a typical shorver. Estimate how much water you use during the course of ayear just by showering. Identifr other activities where you use water and estimate your amount of use. Using the concepts discussed in this chapter, measure the volumeuic flow rate of water out of a drinking
Convert the following speed limits from miles per hour (mph) to kilometers per hour (km/h) and from feet per second (ft/s) to meters per second (mls). Think about the relative magnitude ofvalues as you go from mph to ft/s and as you go from km/h to m/s. You
Speed
(ft4)
(km/h)
25
borhood gas station. Btimate how often the storage 8.8.
Limit
(mph)
of
Speed
of
Sound (km/h)
Express Equation (8.5), the uaffic density, in terms of number ofvehicles per mile. Express the angular speed of the Earth in rad/s and
rPm. 815.
8J6.
\?hat is the magnitude of the speed of a person at the equator due to rotational speed ofthe Earth. Calculate the average speed of the gasoline o
Pnosr.Ervrs don, measure the volumeuic flow rate of the gas fust, and then measure the diameter of the nozzle. Use Equation (8.12) to calculate the average speed of the gasoline coming out of the nozzle. Hint:Firstmeuwe the time that it takes to put so many gallons of gasoline into the gas tank! Measure the volumetric flow rate ofwater coming out a of faucet. Also determine the average velocity of
8J7"
8.27.
8.t9.
8.20.
Derermine the narural &equency
8.2t.
of a
Consider the duct system sho.vvn in the accompanying
figure. Air fows through two 8-in.-by-10-in. ducr that merge into a 18-in.-by-14-in. duct. The {verage speed of the air in each of the 8 X 10 ducts is 30 ft/s. What is the volume fow rate of air in the t8 X 14 duc? Express the volume flow rate in ffls, ff lmin, and m3is. \0hat is the average speed of air in the 18 X 14 duct?
water leaving the faucet. Determine the speed of a point on the Eanht surface in ft/s, m/s, mph, km/h. Into how many time zones are the United States and iu territories divided? Btimate the rotational speed of your car wheels when you are traveling at 60 mph.
8J8.
217
8"X
18'x l4'
pendulum
whose length is 5 m. 8.n. Determine the spring constant for Example 8.1 if the q/stem is to oscillate with a naturd frequency of 5 Hz. 8.23. Determine the traffic flow if 100 cars pass a known
location during 10s. A conveyer belt runs on 3-in. drums that are driven by a motor. If it takes 6 s for the belt to go from zero to the speed of 3 ft/s, calculate the final *ngtl"r speed of the drum and its angular acceleration. Assume con-
8.24.
stant acceleration.
Chinook, a military helicopter, has rwo thee-blade rotor systems, each rurning in opposite direction. Each blade has a diameter of approximaterly 4l ft. The blades can spin ar an5rlar speeds of up to 225 qpm. Determine the translational speed of a partide locarcd at the tip of a blade. Express your answer in ft/s, mph,
8.25.
m/s, andkm/h. Consider the piping system shown in the accornpanying figure. The speed of water flowing through the 4-in.-diameter section of the piping system is 3 ft/s. Vhat is the volume flow rate of water in the piping qntem? Express the volume flow rate in ff/s, gpm, and L/s. For the case of steady Ilow of water .t"o"gh the piping sJntem, what is the speed of water in the 3-in.-diameter section of the svstem?
8.26.
T_I l \ Y t)
( |\-/-T
4in.
Problen 8.26
->3ffls
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 Mass-RELATED 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 mass-related 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 rnass-related 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 day-to-day 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 Mess-Rnr-rrso PARAMsTEns
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,
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6.941
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al
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21
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Sc
Ti
Fe
Co
Ni
44.956
47.88
50.942
51,W6
54.938
55,M5
58.933
58.69
s9
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41
42
4{l
44
,+5
-.s
Z
zg
27
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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
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92.W
95.94
98
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102.906
106.-42
107.868
t12.4t1
72
73
74
76
T7
78
80
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Hf
Th
w
Re
Os
Ir
ft
79
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Hg
138_906
178.49
180.948
183,M
186,2t7
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195 nfl
196.967
200J9
89
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105
107
108
109
110
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Ac
Rf
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Sg
Bh
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Mt
Uun
261
262
x3
262
zo
zffi
269
62
63
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50-36
t'r.9a
Cs 132-q]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
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,n
fia&
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1
u
ffi
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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 mass-related 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 Mess-RBr.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 water@4"C
(e.3)
9.3 Dwsrrr, Sprcrrrc VoluuE, AND Spscrrrc
Gnevrrr
223
It is imponant to note that specific g-raviqf 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
1790@t00"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)
n0a@Lal'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
ta00@4"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 Mess-RsLATEp 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 t-he 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
=
um€
/, ,/volume\ : \oensrry;l --:\umel I
(9'6]
(densiry)(volume fow rate)
The mass flow rate calculation is also important in excavation or tunnel-digging 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
lt|ass 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 z-z axis is now
f-u=
/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 p€e.
Cneprsn 9
Mess
lNp Mess-Rrrersp
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
volume censlry
:
(nl4)(d2)(leng&)
I*,
9.6
:
0.01571 m3
mass
*ffi;
123.51<S
To calculate the Equation
btlL)(OJ m)2(z m)
volume
7360kg/m3: rnass :
:
mass moment
(l.tt1. tl = )mRz 22'
:
of inenia of the shaft abour its longitudinal anis, we
:(t23.5 kg)(0.05 m)t
:
use
0.154 kg. m2
Momentum Mass also plrys an important role in problems dealing with moving objects. Monerztt?n rs a physical variable that is defined as the product of mass and velocity. j
L: mV
(9.14)
228
Crraprrn
9
Mass lr.lp Mass-R-ElAruo PlnervrsrERs
wihAshielA! Thart
liftle pebble awe
a, I av ge mowrehtqm b ec oP
its velatfive velocity!
av1
\( >J
In Equation (9.14), Z represents momentum vectot, ln is mass, and Vis the velocityvecor. Because the velocity of the moving object has a direction, we associate a direction with momentum as well. The momenfl.rm's direction is the same as the direction of the velocity vector or the moving object. So a 1000-kg car moving north at a rare of 20 mls has a momentum rrith a magnitude of 20,000 b.-h in the nonh direcdon. Momentum is one of those physical concepts that are commonly abused by sports broadcasters when discussing sporting wents. During a timeout period when a*rletes are standing still and listening to their coaches, the broadcaster may say, "Bob, clearly the momentum has shifted, and team B has more momentum now going into the founh quarter." Using the preceding definition of momenfllm, now you know that relative to eafih, the momentum of an object or a person at rest is zero! Because the magnitude oflinear momentum is simplymass times velociry somedringwith reladvely small mass could have a large momentum value, depending on irc velocity. For example, a bullet with a relatively small mass shot out of a gun can do los of harm and penetrate a surface because ofits high velociry. The magnitude of the momentum associated with the bullet could be relatively large. You have seen stunt actors falling from the top oftall buildings. As the stunt performer approaches the ground, he or she has a relatively large momenrum, so how are injuries avoided? Of course, the stunt performer falls onto an air bag and some soft materials that increase the time of contact to reduce the forces that act on his or her body. We will discuss the reladonship berween linear momentum and linear impulse (force acting over time) in Chapter 10 after we discuss the concept of force.
9.7 CoNssnvarrox or Mess
229
Determine the linear momennrm of a person whose mass is 80 kg and who is running at a rate of 3 m/s. Compare it to the momentum of a car t}rar has a mass of 2000 kg *d is moving at a rate of 30 m/s in the same direction as the person. Ve will use Equation (9.14) to arNwer the questions.
tl:!----
-i'i1i't!?,:-iiri!{:iti"-ill
person:
i:
Car:
7, --
*i = (s0 kg)(3 m/s) =
: *T:
(2000 kgx30 m/s)
iiTi1}rlrIi----,---t'ir'
,:*.:-E"i-
240k4.mts
:
60,000
l€.
m/s
-.-,:.{iLli::1tr,'i'l;iiltj
9:l Conservation of lvlass Recall that in Chaprcr 6 we mentioned that engineers are good bookkeepers. In the analysis of engineering work we need to keep track of physical quantities such as mass, energy, momentum, and so on. Let us now look at how engineers may go about keeping track of mass and the associated bookkeepingprocedure (see Figure9.4). Simplystated, the conservation ofmass says that we cannot creaJe or desuoy mass. Consider the following example. You are uking a shower in your bathtub. You turn on the water and begin to wash yourself. Let us focus our atention on the rub. \7ith the drain open and clear of any hair and dead skin tissue, the rate at which water comes to the tub from rhe showerhead is equd to the rate at which water leaves the tub. This is a statement of conservation of mass applied to the water inside the tube. Now what happens if the drain becomes plugged up by hair or dead skin tissue? Many of us have experienced this
at one time or another. Liquid Drdno time! The rate at which water comes to the tub now is not exacdy equal to the rate at which water leaves tire tub, which is why the water level in the tub begins to rise. Howwould you use the conservation of mass principle to describe this situation? You cln express it this way: The rate at which water cornes to the ab is equal to the rate at which the utatn lzaoes the tub plus the tirne rate ofaccumulation ofthe mass ofwater within the nb. !f'h.at happens if you were taking a bath and you had the tub filled with water? Let us say that after you are done taking the bath you open the drain, and the water begins to leave. Being r"he impatient person that you are, you slowly rurn on the shower as the tub is being
Control volume
Irigurct.l Ihe rate at whicfi rrlater enters the container minus the rate at which water leaves tfte container should be equal to the rate of
acumulation or depletion of the mass of water within the container.
230
Crlar"ren
9
M.lss aNo M,lss-Rsr-mpp Pen-o,MrrsRs drained. Now you norice that the water level is going down but at a slow rate. The rate the water is coming into the rub minus the rate at which water is leaving the tub is equd to the rate ofwater depletion (reduction) within the tub. Let us now turn our attention to an engineering presentation of conservation of mass. In engineering we refer to the tub as a control volume, because we focus our attention on a specific object occupying a cerain volume in space. The control volume could represent the flow boundaries in a pump, or a section of a pipe, or the inside volume of a tank, the flow passage in a compressor or water heater or the boundaries of a river.'We can use the bathtub o
ahe vatle art which warteY the tqb is eqqarl fo lhe vatte art which wartev learves the tqb plns the tiu,re varte o0 acct^mhlritiou oP the t^ratss o€ urartev within the ttab. comes ihlo
Ithink l'll invite my class here tomonow to demonstrate the consentation of masst
9.7 ConsrnvarroN
or
Mrc,ss
23r
How much water is stored after 5 minutes in each of the tanla shosrn in Figure 9.5? How long will it take to fill the tanla completely provided that the volume of each tank is 12 m3? Assume dre density ofwater is 1000
E
kg/^'.
rigurc 9.5
(b)
(a)
Ihe tanks of tnmple 9.3.
We
will
(the
to solve this problem.
use the conservation of mass statement
(th" rateatwhichthefuid
rateatwhichfluid _
enters a control
volume)
leaves rhe
control
volume)
(the rate of accumulation or depletion ofthe mass of fluid within the given contol volume)
Realizing that no water leaves tank (a) we have
(zkel')
- (0) =
changes of mass inside the
control volume
change in time
After 5 minutes,
r\t'
t/
eOs
' (2 ksAX5 ."i")(.ffi changeofmassinsidethecontrolvolume:'^'
\
)
:
600kg
To determine how long it will take to fill the tank, first we will make use of rhe relationship among mass, density, and volume-rnass = (density)(volume)-to compute how much mass each tank can hold. mass
:
(looo kg/m3) (rz m3)
:
12,ooo kg
By rearranging terms in the conservation of mass equation we can no'w solve for the time that is required to fill the ank time required to fill dre tank
:
?ffi
:
(6000 s)
:
100 min
\(ater enters mnk (b) at 2 kg/s and leaves the tank at I kgA. Applying conservation of mass to tank (b) we have
(zkst')
- (l ksh) :
changes
of mass inside the control volume change rn trme
23?
Cnngrnn
9
Mass eNo Mess-Rrr.ltnp Pan*rsrERs After 5 minutes, change of mass inside the control volum
" -- l(2r.g/r)
:
-
(JU) (f fu/s)](5 ,ri"; ___/\lmin)
=
12000 s
300 kg
and the time that is required to fill the tank (b)
time required to fill the tank .:.r-. --r..-i
.
:
12000 kg
l(zkgl')
-
(t tgl')l
= 200 min
:.1:..::::.:l-::-l:
SUMMARY Now that you have reached this point in the text
'
You should have a good undersanding of what is meant by mass and know about the impoftant roles of mass in engineering applications and analysis. o You should know that in engineering to show how light or heavy materids are, we use propercies such as density, specific volume, specific weight, and specific glalvity, You should also know the formal definition of these propenies. ' You should hare a good understanding of what is meant by mass fow rate and how it is related to volume flow rare. ' You should realize that mass provides a measrue of resistance to uanslational motion. You should also know that if you want to rotate something, the distribudon of mass about the center of rotation also plErs an imponant role. The further the mass from a center of rotation, the more resistance mass offers to rotational motion. " You should know how we define momentum for a moving object. You should also know that something with a relatively small mass could have a reladvely large momentum if it has a Iarge velocity.
9J.
9.2.
look up the mass of the following a. recent model of an automobile b. the earh c. aBoeingTTT fitJly loaded d. anant
objects:
The density of standard air is a function of temperature and may be approximated using the ideal gas lav, according to
p:fr
where
p = density (kg/-')
?
:
standard atmospheric pressure (101.3 kPa)
R
=
gas
conssurt; its value for air is 286.9
rl-) \kc.K/
P
T
=
ir
temperauue in Kelvin
Pnorr.sMs Create a able that shows the densiry of air as a function of temperature in the range of 0"C Q73.15 K) to 50"C (323.15 K) in increments of 5"C. Also, create a graph showing the value of densiry as a function of 9.3.
9.4.
9.5.
ternperanue. Determine the specific grarrity of the following materials: gold @: 1208lblf€), platinum (p: 1340lblf€), silver (p = 654lblff), sand (p :94.6lblf1), frahly fallen snow (p : 3l lb/#), ar (p = 75 lb/#), hard rubber (p :74.4lblf1). Show that the specific weight and density are related according to specific weight : (density)(acceleration due to gravity). Compute the values of momentum for the following
911. The use of ceiling fans ro circularc air has become quite common. Suggest ways to correc[ the rotation of a wobbling fan using the concept of mass moment of inenia. fu a starting point, you can use pocket change such as dimes, nickels, and quaners and chewinggum. How would you stop the fan &om wobbling? Exercise 9.12.
9.13.
9J4.
situations:
a. a9}-kgfoo$all
player running ar 6 m/s a 1500-kg car moving at a rate of 100 km/h
b. c. a 200,000-kg Boeing 777 moing at a speed of 500 km/h
d.
9J5.
a bullet with a mass of 15 gtraveling at a speed
of
500 mls
e. a l40-gbaseball traveling at 120 km/h f. an 80-kg stunt performer falling off a ten srory 9.6"
9.1.
building reaching a speed of 30 m/s Investigate the mass flow rate of blood through your heart, and write a brief report to your instructor discussing your findings. Investigate the mass fow rate of oil inside the Alaskan
pipeline, and write a brief report to your instructor discussing your 6ndings. 9.8. Investigate rhe mass flow rate of water through the Mississippi River during a normal year, and write a brief repon discussing your findings. Explain flooding using the conservation of mass statement. 99. Determine the mass fow rate offuel from the gasoline tank to the car's fuel injection system. Assume that the gasoline consumpdon ofthe caris 20 mpgwhen dre car is moving at 60 mph. Use the specific gravityvalue 910.
of 0.72 for the gasoline. Determine the mass moment of inenia for the follovring objects: a thin dish circular c/inder, and a sphere. Refer to Equations (9.10) through (9.13) for appropriate relationships. Ifyou were to place these objects alongside of each other on an inclined surhce, which one of the ob.iects would get to the bottom first, provided that th"y hane the same mass and diameter?
"ll
233
9.r6.
9J7.
caution! Determine the mass moment of inerda of a steel shaft that is I m long and has a diameter of 5 cm. Determine t}re mass of the shaft using the density information provided in Table 9.1.
Determine the mass moment of inertia of steel balls used in ball bearings. Use a drameter of 2 cm. Determine the mass moment of inenia of the earth about its axis ofrotation, going through the poles. Assume the shape of the eanh to be spherical. Iook up information such as the mass of the eanh and the radius ofthe eamh at the equatorial plane. Next time you put gasoline in your car, measure the mass flow rate (kg/s) of gasoline at the pumping scrtion. Record the amount of gas in gallons (or liters if you are doing the experiment ou$ide the United Sates) that you placed in your car" gar tank and the time that it took to do so. Make the appropriate conversion from volume fow rate to mass fow rate using the density of gasoline. Measure rhe mass flow rate of water coming out of a drinking fountain by placing a cup under the running water and bymeasuring the time that it took to fill the cup. Measure the mass of the water by subtracting the total mass from the mass of the cup. Obtain a graduated beaker and accurate scale from a chemistry lab and measure the densiry of the following
liquids:
a. a cooking oil b. SAE 10Sf40 engine oil c. water d. rnilk e. ethylene glycol, antGeeze Express your findinp in kg/m3. Also determine the 9J8.
specific gravit)'of each liquid. Obtain piecrs of steel, r rood, and concrete of kno,nrn volume. Find pieces with simple shapes so thar you can rne:$ure the dimensions and calculate the volume quickly. Determine their mass by placing each on an accurate scale, and calculate their densities.
234
CnarreR
9
Mrcs nNo Mess-Rsr-erso
PARAMgTERs
Take a 500-sheet ream of computer paper as it comes wrapped, Unwrap it and measure the height, width, and length of the stack. Determine the volume, and measure the mass of the ream, ald obtain the density. Determine how many rsrms come in a standard box. Estimate the total mass of the box. Discuss your assumptions and estimadon procedure. 9.20. In this assignment you will investigate how much water may be wasted by a leaky faucet. Place a large cup under a leaky faucet Ifyou dorit have a leakF faucet at home, open the faucet so that it just drips into the cup. Record the dme that you stared the experiment. Allow t.he water to drip into the cup for about an hour or two. Record the dme when you remove the cup from under the faucet. Determine the mass flow rate. Estimate the water wasted by 100,000 people with lealqf faucets during a period ofone year. 9.21. Obtain a graduated beaker and accurate scale from a chemistrylab and measure the density ofSAE 10W40 engine oil. Repeat the experiment ten dmes. Determine the mean, variance, and smndard deviation for your density measurement. 9.n. A hydrometer is a device which uses the principle of buoyancy to measure the Specific Graviry (S.G.) of a
liquid. Daign
9.t9.
a
pocket-sized hydrometer thar can be
to measure S.G. for liqui& having 1.2 < S.G. < 1.5. Specifr impoftant dimensions and the mass of the device. S?'rite a brief report to your insuuctor explaining how you arived at the final design. Also include a used
sketch showing the calibrated scale from which one would read S.G. In addition, construct a model and attach the scale to the working model. To keep things simple, assume that the hydrometer has a cylindrical shape. 9.23.
9.24.
Referring to Figure 9.5, how much water is stored after 20 minutes in each of the tanks? How long will it take to fill the tanks completely provided that the volume of tank (a) is 24 m3 and tank (b) has a volume of 36 m3? Assume the densiw of water is 1000 kg/m3. Investigate the size ofthe storage tank in a gas stadon.
Apply the conservation of mass statement
'Education is what remains after one has forgotten everydring he learned in school."
-
Albm Einstein
(187 9
-I9 5 5)
to
the
gasoline fow in ttre gas station. Draw a control volume showing appropriate components of the gasoline flow q/stem. Estimate how much gasoline is removed from the storage tanh per day. How often does the storage tank need to be filled? Is this a sreadyprocess?
0biective: Given the following materials, design a vehicle that will transport a penny. The vehicle must remain in contact with the penny and the ground at all dmes. The design that travels a penny the funhest wins. Each desim is allowed one practice run. OL
PrOvided
Materials: Construction paper (one sheet*
g' x 12' and one shssl-ll" x l8'), I
balloon,
"If you study to remember, you will forget, but, if you study to understand, you will remember.' -Unknoum 235
cHAPTER
10
FoRcE AND FoRcE-RnrarED PeneMETERS o surface, air is blown into the
blast tank of a submarine to push
water out of the tank to make the submarine lighter-the submarine floats
to surface.
To dive,
the blast tank is
opened to let water in and to push out
the air. The submarine sinks below the surface. The diving and surfacing of the submarine demonstrate the relationship between the weight of the submarine and
the buoyancy force acting on it.
236
10.1 lf'narI(s MraN BvFoncr
237
The objectiues ofthis chapter are to innoduce the concept offorce, its uarious types,
and other force rekted aariables such as pressare and stress. Ve will discuss Newton's laws in this chapter. Newtonls lawsform thefoundation ofrnechanics and analysis and design of rnany engineering problems including strac-tures, airpkne airframes (fusekge and wings), car frames, medical irnplants for hips and othn joint repkceruents, machine parts, and orbit of satellites. We will ako expkin the tendmcies of unbalanced mechanicalforces, which are to translate and rotate ob-
will carcsider sarne ofthe rnechanicalproperti.es ofmatniak that show hou stif orfexible a rnaterial is when subjected to aforce. Ve will then explain the ffttt of nforce acting at a distance in terrns of creattng a rnzrnent about a point; the ffict of a force acting ouer a distance, what is forrnally defned as rnechanical worh; and the ffio of nforce acting ouer a period oftirne in terms ofwhat
jects. Moreoner, u)e
is
gmerally refenedto as linear iwpulse.
10.1 What
We Mean By Force
What is force? The simplest form of a force that represents the interacdon of nvo objects is a push or a pull. lfhen you push or pull on a lavn mower or on a vacuum cleaner, that interaction between your hand and the lawn mower or the vacuum cleaner is call edforce.Vhenan automobile pulls a U-Haul trailer, a force is exened by the bumper hitch on the trailer. The interaction representing the trailer being pulled by the bumper hitch is represented by a force. In these examples, dre force is exemed by one body on anorher body by dired contact. Not all forces result from direcr contact. For example, gravitational and magnetic forces are not exerted by direct contact. If you hold your book, say, 3 feet above the ground and let it go, what happens? It falls; that is due to gravitadonal forcewhich is exerted bythe earth on the book. The gravitational atuactiveforces act at a distance. A satellite orbiting the earth is continuously being pulled by the earth toward the center of the earth, and this allows the satellite to maintain its orbit. Ve will discuss the universal lavr ofgravitadonal atraction in detail later. All forces, whether they represent the interaction ofnvo bodies in direct contact or the interaction oftrro bodies at a distance (gravitational force), are defined by their magnitudes, their direcdons, and the points of application. The natural tendency of a force acting on an object, if unbalanced, will be to uanslate t}re object (move it along) and ro rotate it. Moreover, forces acting on an object could squeeze or shorten, elongate, bend, or twist the object. The amount bywhich the objectwill elongate, shoften, bend, or twist will depend on the type of material the object is made from. Machine components, tools, pans of the human body, and strucurd members are generally subject to push-pull, bending or wisting types of loading. Those of you who are studfng to become aerospace, civil, manufacnrring or mechanical engineers will take classes in basic mechanics and mechanics of materials in which you will explore forces, stresses, and materids behanior in more depth.
Units of Foree One newton is defined as the force that will accelerate 1 kilogram of mass at a mte of This relationship is based on Newton's law of motion. 1
newton
:
(1
kgxl m/s2) or lN : llrg.m/s2
I
m/s2. (r0J)
238
f
Cneprsn^L0 ForcsaxoFoncE-RErArroPanelrsrERs
Thin plate acting as a spring
Torsional spring
riguret0.l
Eramples of different types of
springs.
In comparison, 1lb6 And
1
:
1
pound force will accelerate 1 slug at a rate of 1 ft/s2,
as
given by
(1 slugxl ft/s'z)
pound force is equal to 4.48 newtons (1
(10.2)
lb6:
4.448 N).
Spring Forces and Hooke's Law Most ofyou have seen springs that are used in cars, locomotives, clothespins, weighing
scales,
and clips for potato chip bags. Springs are also used in medical equipment, electronics equipment such as printers and copiers, and in many restoring mechanisms (a mechanism that returns a component to its original position) covering a wide range of applications. There are different types of springs, including extension or compression and mrsional springs. Examples of different types of springs are shown in Figure 10.1. Hooket law (named after Robert Hooke, an English scientist who proposed the lar) states
that over the elastic range the deformation of a spring is direcdy proponional to the applied force, according to
F:
hx
{r0.3)
where
F = applied force (N or lb)
: r: &
spring constant (N/mm or N/cm or lb/in.) deformation of the spring (mm or cm or
in.-u5g units that are consistentwith
A)
By elastic rangewe mean the range overwhich ifthe applied force is removed the internal spring force will return the spring to irc original unsuetched shape and size and do so with no permanent deformation. Note the spring force is equal to the applied force. The value of the spring constant depends on the type of material used to make the spring. Moreover, the shape and winding of the spring will also affect its k value. The spring constant can be determined experimentally.
For a given spring, in order to determine the value of the spring constant, we halre atached dead weights to one end of the spring, as shown in Figure 10.2. \Ze have measured and recorded the defection by the corresponding weights as given in Thble 10.1. 'I7hat is "aused
f0.f
wirerVs
MnaN svFoncs
239
the value of the spring consant? W'e have plotted the results of the experiment using Excel (shown in Figure 10.3). The spring constant A is determined by calculating the slope of a force-defection line (recall that the slope of a line is determined from slope : rise/run, and for this problem, the slope : change in force/change in deflection). This approach leads to a value of b : 0.54 N/mm.
I
Often, when connecdng experimenal force-deflecdon points, you may not obtain a straight line that goes through each experimental point. In this case, you will try to come up with the best fit to the data poins. There are mathematical procedures (including least-squares rechniques) that allowyou to find the best fit to a set of data points. 'We will discuss curve fitting using Excel in Section 14.8, but for now just drav a line that you think best fits the daa poins. As an example, we have shown a set of data points in Table 10.1(a) and a good corresponding fit in Figure 10.4.
tigurelo.Z
Ihe spring setup.
Force-deflection
-f----r-
--i-
^24 z 415
./-/
o10 () Weight (N)
The Deflection of the Spring (mm)
4.9
9
9.8
l8
L+,/
1'7
19.6
36
rY
)
010203040 Deflection (mm)
t
tigurel0.S
The force-deflection diagram for the spring in Erample 10.1.
Force-deflection curve 25
I
nX
z
als ei l0
The Defection
Weight (N)
ts
rY
)
of the Spring (mm)
5.0 10.0 15.0 20.0
9
0102030
1'7
Deflection (mm)
29 35
lil
Figurelo.4
A good :r: -:i--::::---,::-:i:
:
;il
- -1 --l:l-
-:.-:
,
----- :--
-:1
fit to
a set 0f force'detlection data points.
240
Cneprsn
10
Foncr errro Foncr-Rnr.erno Penel.srnRs
Friction
t*
force
I
Friction force
Normal force
Maximum static friction force
Dynamic friction force
Applied force
E
figure
10.5
The relationship between applied force and the
friction force.
Frictisn Forees-Dry Friction and Viseous Friction There are basically tvvo tFpes offrictional forces that are important in engineering design: dry and. uiscous ftictioz (or the fuid friction). Let us first rake a closer look at dry friction, which allows us to walk or to drive our cars. Remember what happens when frictional forces get relatively small, as is the case when you try to walk on ice or drive your car on a sheet of ice. Stated in a simple way, dry &iction er
fri.ctionalforces
F-o: FN
fl0.4)
Nis the normal force-exerted by the table on the book to prevent it from f*llirgand, in the case of a book resting on the table, l/is equal to the weight of the book, and pc is r:he coefficient of static fricdon for the two surfaces involved, the book and the desk surface. Figure 10.5 also shows that once the book is set in motion, the magnitude of the fricdon force drops to a value called the dynamic, orl
lO.2 NevroN's
Lnvs rN MrcneNrcs
241
side by side on an inclined sufice, which of the two liquids will fow easier? You know the answer: The water will fow down the inclined surface faster because it has less viscosity. Fluids with relatively smaller viscosiry require less energy to be transponed in pipelines. For example, it would require less energy to ffansport water in a pipe than it does to transport motor oil or
glycerin. The viscosity of fuids is a function of temperature. In general, the viscosity of gases increases with increasing temperanrre, and the viscosity of liquids generally decreases with increasing temperature. Knowledge of the viscosiry of a liquid and how it may change with temperarure also helps when selecting lubricants to reduce wear and friction between moving parrs.
The coefficient of static friction between a book and a desk surFace is 0.6. The book weighs 20 N. If a horizontal force of 10 N is applied to the booh would the book move? And if not, what is the magnitude of the friction force?'!7hat should be the magnitude of the horizontal force to set the book in motion? The maximum friction force is given by
4no
= pll:
(0.6)(20)
:
12 newrons
Since the magnitude of the horizontal force (10 N) is less than the maximum friction force (12 N), rhe bookwill not move, and hence the friction force will equal the applied force. A horizontal force having a magnitude greater than 12 N will be required to set the book in modon.
,i_ l
-iitrt
r1:1 11crt1i:
..,!:li4-alCLi.r'.r:i.-rr
10.2
-----l'a!-l----
--
__
_---
__:____-___-____.__-__
-
------:-_...
-_
.-_ _-__
---_--l:. ..____lglill
F'lewton's Laws in Mechanics in Chapter 6, physical laws are based on observations. In this secdon, we will briefy discuss Newton's larvs, which form the foundation of mechanics. The design and analysis of many engineering problems, including strucnrres, machine parts' and the orbit ofsatellites, begins with the application of Newton's laws. Most ofyou will have t}re oppornrniry to take a phpics, statics, or dynamics class thatwill explore Newtont laws and their applications further.
As we mentioned
hlewton's First Law given object is at rest, and ifthere are no unbalanced forces acting on it, the object will then remain at rest. If the object is moving with a constant speed in a certain direction, and if there are no unbalanced forces acting on it, the object will continue to move with its constant speed and in the same direcdon. Newton's first law is quite obvious and should be intuitive. For example, you know from your everyday experiences that if a book is resting on a table and you don't push, pull or lift it, then the book will lie on that table in that position until it is disturbed
Ifa
with an unbalanced force.
hlewton's Second Law \7e briefy explained this la'r in Chapter 6, where we said that ifyou place a book on a smooth table and push it hard enough, it will move. Newton observed this and formulated his observation into what is called Newtont second law ofmotion. Newton observed that as he increased
242
10
Crnrrnn
Foncn eNp Foncs-RELATEo Pan-ervrrreRs Weight
of
the book
Force from the table
supporting the book
5
Figure
tO.6
The lorces of action and reaction; they have the same lnagritude ard the same line of action,
but they act in opposite dheclions.
the mass of the object being pushed, urhile keeping the magnitude of the force consant, the object did not move as quickly and had a smaller rate of change of speed. Thus, Newton noticed that there was a direct relationship berween the push (the force), the mass of the object being pushed, and the acceleration of the object. He also noticed that there was a direct relationship between the direction of the force and the direction ofthe acceleration. Newton's second larv of motion simply states that dre net effect of unbalanced forces is equal to mass times acceleration ofthe object, which is given by the expression
>7: *Z
fl0.5)
where the syrnbol ) Gigma) means summation, F represents the forces in unit of newton, m is the mass of the object in kg, *d a is rhe resulting acceleration of the mass center of the object in m/s2. In Equation (10.5), a summation of forces is used to allow for application of more than one force on an object.
Newton's Third Law Returning to oru example of a book resting on a desk, as shown in Figure 10.6, because the
nz
mt \PQ> ( 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
l0-11 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 Foncs-REr,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^* _le-ezz
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
10.3
-lii
- ------..,{lil
-.-:-lii!i--
---.r.----.1-:ir--: - -iilili1li1;rilrr-rilar[
:ii]llLlrriiTlllitrJi.,.{:l1
L-_-j
Pressure and Stress-Force Acting Over an Area Prcssureproides a measure of intensity of a force acting over an aren. It can be defined as the ratio offorce over the contact surface area: Dressure
'
:
force fi0.8)
areg.
-
To better understand what the magnitude of a pressure represents, consider the situations shown in Figure 10.8. Let us first look at the situation depicted in Figure 10.8(a), in which we lay a solid brich in the form of a rectangular prism with dimensions of 2l .6 x 6.4 x 10.2 cm (A\ x Zi X 4 in.) that weighs 28 N (6.4 lb), flat on its face. Using Equation (10.8) for this orientation, the pressure at the contact surface is
-pressufe
force atez
28
N
(0.2r6 m)(o.r02 m)
: l27r*: m-
1271 Pa
f0.3
I
Pnrssunr.l,lqp Srnnss-FoRcEAcrING OvsnANAREA
245
Figurel0.S
An erperiment denonstrating
tie
concept of pressure. (a) A solid
bdck resting on
ih
face. (b) A
solid brick resting on its end. In position (b), the block creates a
(a)
higher pressure on tfte surface.
Note that one nevnon per squared meter is called one pascal (1 N/l m2 : I Pa)' Now, if we were ro Iay the brick on its end as depicted in Figure 10.8(b), ttre pressure due to the weight of the brick becomes
pressure
^
:
force areit
-:
28N (0.064 m)(o.toz m)
:
42894
: 42s9 P,
is 28 N, regardless of how it is laid. contact surface depends on the magnitude of the contact surface area. The smaller the contact area, the larger the pressure created by the same force. You already know this from your everyday experiences-vrhich situation would create more pain, pushing on someone's arm with a finger or a thumbtack? In Chapter 7, when we were discussing the imporance of area and its role in engineering applications, we briefy mentioned the imponance of understanding pressure in situations such as foundations of buildings, hydrar:lic systems, and cutting tools. W'e also said that in order to design a shalp knife, one must design the cutting tool such that it creates large pressures along the cuning edge. This is achieyed by reducing the contact surface area along the cutdng edge. Understanding fluid pressure distributions is very important in engineering problems in hydrosatics, hydrodynamics, and aerodynamics. Hydrostatic refers to water at rest and its studp howwer, the understanding of other fuids is also considered in hydrostatics. A good example of a hydrostatic problem is calculating the water pressure acting on the surfacr of a dam and how it varies along the height of the dam. Understanding pressrue is also important in hydrodynamics srudies, which deal with understanding the motion of water and other fluids such as those encountered in the fow of oil or water in pipelines or the flow of water around the hult of a ship or a submarine. Air pressure distribudons play imponznt roles in the analysis of air resistance to the movement of vehicles or in creating lift forces over the winp of an airplane. Aerodynamics deals with understanding the motion of air around and over
It
is
imporant to note here that the weight of the brick
But the pressure that is created at
surfaces.
t.he
246
Cnapren
10
Foncnaxo Foncr-Rmerrp
PanervrsrnRs
Common Units of Pressure In the International
System ofunits, pressure units are expressed in pascal. One pascal is the presslrre created by one newton force acting over a suface area of I m2:
t
p":
lT
flo.t)
m-
In U.S. Customary unit, pressure is commonly expressed in pounds per square inch or pounds per square fooq one lb/inz (t psi) represents dre pressure created by a l-pound force over an area of 1 in2, and I lblfi represents the pressure created by a l-pound force over an area of I f?. To convert the pressure &om lb/ff to lb/in2, we take the following step:
tr* \ : ,/.q\/ -\ffl\a?) "ra\ \i,,,./
fl0i0)
I lblin2 is usually referred to as I psi, which reads one pound per square inch. Many other units are also commonly used for pressure. For example, the atmospheric pressure is generallygiven in inches of mercury (in. Hg) or millimeters of mercury (mm. Hg). In airconfitioning applications, often inches of water are used to express the magnitude of air pressure. These unis and their relationships with Pa and psi will be explained further in Example 10.6, after we explain how the pressure of a fluid at rest changes with ia depth. For fuids at rest, there are two basic larvs: (1) Pascalt law, which explains that the pressure of a fluid at a point is the same in all directions, and (2) the pressure of fluid increases with its depth. Even though these lars are simple, they are verypowerfi.rl tools for analyzrngvarious enNote that
gineering problems.
Pascal's Law Pascalt law states that for a fuid at test, ?retsure at a point is the same in all directions. Note carefully that we are discussing pressure at a point. 'Io demonstrate this laur, consider the ves-
,_4. {}' a
in Figure 10.9; the pressrre atpointAis the same in all directions. Another simple, yet imponant concept for a fuid at rest states that pressure increases with the depth of fluid. So the hull of a submarine is subjeced to more water pressure when cruising at 300 m than at 100 m. Some ofyou hare direcdy experienced the variation of pressure with depthwhen you have gone scuba diving or swimmingin a lake. You recall fromyour experience
figureto.t
that as you go deeper, you feel higher pressure on your body. Referring to Figure 1 0. 1 0, dre relationship beween the gauge pressure (the pressure that is above atmospheric pressure) and the height of a fluid column above it is given by
sel shown
'tb
I
In a static fluid, pressure
at a paintisthe sane in all dheclions.
P:
pgh
(1o.il)
where
P: p: g: /:
is the fluid pressure at point B (see Figure 10.10) (in Pa or is the density of the
fuid
is the acceleration due
is
lblff)
(in kg/m3 or slugs/ff)
to gradty
(g:
9.8 m/s2 org
:
32.2 ft1l2)
tle height of fuid column (in m or ft)
Also shown in Figure 10.10 is the force balance between the pressure acting on the bottom surface of a fluid column and its weight. Equation (10.1 1) is derived under the assumption of constant fluid density.
10.3
PnsssuRn alro
Srnrss-Foncs AcrINc Own
eN
Arnl
247
PatrnosohericA
TJ
of
Weight the column of fluid
|,I I
lll lf l Itl ?
PsA
E
figurct0.l0
I L I I
B
&A=rv"orumn+PuooA
I
PBA=-#+Pa'#A
Pta=Fffis+Pun4 ""=pgh+P6a, pgh Ps(sage) --
\;i,,r;:,:::::,':'i;',':,1,1,:,',.,::l:,ll,l''
,r.tfii,iiilij:i'
i
.:ri
j
Thevariationof pressurewithdepth.
Another concept, which is closely related to fluid pressure distribution is buoymcy. Aswe discussed in Chapter / , buoyancy is the force that a fluid exefts on a submerged objecl The net upward buoyancyforce arises from the fact that the fluid exerts ahigherpressure at the bottom surfaces ofr:he object than it does on the top surfaces ofthe object. Thus, the net effect of fl.uid pressure distribution acdng over the submerged surface of an object is the buoyancy force. The magnitude of the buoyancy force is equal to the weight of the volume of the fuid displaced, which is given here again for convenience.
&:
pvg
where.Fs is the buoyancy force (N), p represents the densiry of the fluid ation due to gravity (9.8 m/s2), and Vis the volume of the object (l).
(kg/*'),
g is acceler-
Most of you may have seen water towers in and around small towns. The function of a water tower is ro create a desirable municipal water pressure for household and other usage in a town. To achieve this purpose, water is stored in large quantities in elevated tanlis. You may also have noticed that sometimes the water towers are placed on top of hills or at high points in a town. The municipal water pressure may vary from town to town, but it generally falls somewhere benveen 50 psi and 80 psi.'We are interested in dweloping a able that shows the relationship between the height of water aboveground in the water tower and the water Pressure in a pipeline located at the base of the water tower, as shown in Figure
1
0.
1
1. !tre
in psi, in a pipe located at the base of the water tower as we vary the height of water in incre'\7e will create a table showing our results up to ments of l0 ft. will
calculate the water pressure,
100 psi. And using rhe height-pressure table, we will answer the quesdon, "W'hat should be the water lwel in the water tower to create a 50-psi water pressure in a pipe at the base of the water tower?" The relationship between the height of water aboveground in the water tower and the water pressure in a pipeline located at the base of the water tower is given by Equation (10.11):
P:
pgh
I
Figureto.tt
The water tower of Eranple
t0.5.
?48
CHepreR
10
Foncs AND Force-Rrr"arso PenerusrsRs where
P=
p: g=
/ :
is the water pressure at the base of the warer mwer is the density
(lb/ft)
ofwater (equal to 1.94 slugsfff)
is the acceleration due to gravity
(g:
32.2 ftls2)
is the height ofwater aboveground (in ft)
Substituting the known values into Equarion (10.11) leads to
/lb\ n\F
)
/slues\/ : 1.e4(t'
rt\ )ltz.zi )th(ft)]
Recall that to conveft the pressure from
tr \ : -o/lL\/ '"f-!L\ \i"'/ \tC ) \r44in2 )
lb/ff
to lb/in2, we rake the following srep:
And performing the conversion,
/tU\ : r.e4(? /slues\/ +\
"(#/
/
)\zz.z;rtr(ft)l(
rn?
\
*#):
(0.4i38)th(ft)l
Note that the facor 0.4338 has the proper units that give the pressure in lb/in2 when the value
/
is
input in feet. Nort, we generate Thble 10.2 by substituting for
i
in increments of 10 ft
in the preceding equation. From Thble 10.2, we c:rn see that in order to create a 50-psi water pressrue in a pipeline at the base of the tower, dhe water level in the water tower should be approximately 120 ft.
Vater level in theTbwer (ft)
110 t20
:: '40
3a
I lrl
l::
l90 , i ,
the
130
8.7 13.0
140 150
'7
150 170
26.A
180
11
3A.4
190
34.7
20Q
39.4
210 220 230 240
ll0
43.4 47.7
l2O
52.0
100
Tiver
47
17,3
150 t60 '70 igo
io (ft)
Vater Lwel
Water Prescure flb/in'?) 55.4 60.7 55.0 69.4
73.7 78.1 82.4 86.8
9r.r 95.4 99.8
t04.1
f03
PnsssuRE ANp
Srnpss-FoncsAcruYc OvsnAN
AREA
249
Atrnospherie Pressure Eafth's atmospheric pressure is due to the weight of t}re air in the atmosphere above the surface of the eanh. It is the weight of the column of air (extending all the way to the outer edge of the atmosphere) divided by a unit area at the base of the air column. Standard atmosphere at sea level has a value of 101.325 kPa- From the definition of atmospheric pressure you should realize that it is a function of altitude. The variation of standard atmospheric pressure and air density with ahitude is given in Thble 10.3. For example, the atmospheric pressure and air density respectively. The at the top of Mount Everest is approximately 31 kPa and 0.467 (exacdy 8848 m), and thus less air lies on elwation ofMounr Everest is approximately 9000 m
k/-',
top of Mounr Everest than there is at sea level. Moreover, the air pressure at the top of Mount Everest is 30o/o of the value of atmospheric pressrue ar sea level, and the air density at its top is only 38o/a of air density at sea level. New commercial planes have a cruising akitude capacity of approximately 11,000 m. Referring to Thble 10.3, you see the atmospheric pressure at that altitude is approximately one-fifth of the sea-level value, so there is a need for pressurizing the cabin. Also, because of the lower air density at that aldtude, the power required to overcome air resistance (to move the plane through atmosphere) is not.as much as it would be at lower
Atmospheric Altitude (m)
Pressure (kPa)
0 (sea level)
t01.325
1.225
95.46 89.87 84.55
r.167
500 1000 1500 2000 2500
3000
1.058 1.006
0.957
6r.66
0.909 0.863 0.819
/-/)
0.777
54.05
0.736 0.660 0.590
3500
70.11 65.87
4000 4500
)
5000 6000
47.22
7000
4t,1r
8000
35.56 30.80 26.50
0.526 0.457
22.70 t9.4A r6.58
a365 0.3t2
r4.17
0.228 0.195
9000 10,000 11,000 12,000
73,000 14,000 15,000 Source:
79.50 74.70
t.Llz
D*a fromU,S.
12.11 Standard Atmosphere (1962).
0.4r3
0.266
250
Cneprpn
10
FoncsANo Foncs-RELAruo Pl-n.avprnns altitudes. Atmospheric pressure is commonly expressed in one of the following units: pascals, pound per square inch, millimeters of mercury and inches of mercury. Their values are
I atm :
: : 1 atm l atm : 1
atm
101325 Pa
=
101.3 kPa
l4.69lblin2 760 mm. Hg
29.92in.H9
Study Example 10.6 to
see
how these and other units of atmospheric pressrue are related.
Staning with an armospheric pressrue of 101,325 Pa, express the magnitude of the pressure in the following units: (a) millimeters of mercury (mm. Hg), (b) inches of mercury (in . Hg), (c) meters ofwater, and (d) feet ofwater. The densities ofwater and merorfil arc ps.e : 1,000 kglm3 and pns: 13,550 kglm3 respectively. (a) \[e stan with the relationship between the height of a fluid column and the pressure at the base of the column, which is
P=
psh: rorr25(+)
:,r,,ro(*)[r
"(?)]r*,
And solving for h,weharc
h:
0.76 m
=
760 mm
Therefore, the pressure due to a standard atmosphere is equal to the pressure created at a base of a760-mm-tall column of mercury. Stated another way, 1 atm : 760 mm. Hg. (b) Next, we will convert the magnitude of pressure from millimeters of mercury to units that express it in inches of mercury. 760
a') (mm.ug(wzz lmm \
/
: ,o i,.Hs
(c) To express the magnitude of pressure in meters of warer, as we did in pan (a), we begin with the relationship between the height of a fluid column and the pressure at the base of
the column, which is
P: psh: ,o,,rr:({)
\m',/
: ,,ooo($)|-r.rt(+)lrt,,l \m',/L
\s',/l
And then solve for /. which resuks in h
l
:
10.328 m
Therefore, the pressure due to a standard atmosphere is equal to t.he pressure created at a base of a 10.328-m-tall column of water, that is, I atn : 10.328 m. H2O. (d) Finally we convert the units of pressure from meters of water to feet ofwater.
ti
t0.328(m. H2o) ii,,,-.,,,".,,*.,,r'.,.'.,.u,ui,t4l,ill...l
,, -riilirlrr,.1:.li:ii-i:iiir
(=#) = 33.87ft . H2o
iiilili:litri1i.'liii,:iii-
riL:ri,iliiiii.:i
I'iilllill-:El:iriil;ll
.TilEi:iilr.lil.i--.-.l.tl
10.3 Pnrssuns.ero Srnnss-Foncs Acnxc Own.lx
A-nne
251
Absolute Pressure and Gauge Pressure a gas or a liquid relative to the local atmospheric pressrre. For example, when a tire pressue gauge shows 32 psi, this means the pressure of air inside the tire is 32 psi above the local atmospheric pressure. In general, we can express the relationship between the absolute and the gurge pressure by
Most pressure gauges show the magnitude of the pressure of
Pnb*rur"
:
Pg"rg"
* ?"m*phoi"
fl012)
Vacuum refers to pressures below atmospheric level. Thus, negative gauge pressure readings indicate vacuum. Some of you, in your high school phpics class, may have seen demonstrations dealing wirh an enclosed container connected to a vacuum pump. As the vacuum pump draws air out of the container, thus creating a vacuum, the atmospheric pressure acting on the outside of the container made the container collapse. As more air is drawn, the container continued to collapse. An absolute zero pressure reading in a container indicates absolute v:Lcutun' meaning there is no more air left in tfie container. In practice, achieving absolute zero pressure is
not possible, meaning at least a litde air will remain in the container.
We have used a tire gauge and measured the air pressure inside a car tire to be 35.0 psi. \7hat is the absolute pressr.ue of the air inside the tire, if the car is located in (a) a city located at sea level, (b) a city located in Colorado with an elevation of 1500 m? Express your results in both
unia ofpsi and
pascals.
The absolute pressrue is related to the gauge pressure according to
Pnb-lo*:
Pgurg"
+
P*orph"ri"
(a) For a city located at sea level, Patm*phei" P"b-ru,"
:
P"b-ru*
: :
(b)
.W'e
35.0 psi 4e.7
i
14.7 psi
:
:
14,69 psi
:
14.7 psi,
49.7 psi
(prt)(q21}) lpsi \
/
342,680Pa: 342.68kPa
can use Thble 10.3 to look up the standard atmospheric pressure for a city located at an elevation of 1500 m 84.55 kPa.
in Colorado with an elevation of 1500 m. The atmospheric pressure is
P*o*5*i. Ps*s,
P"b-lur"
P"b*ro,"
:
=
35.0
(psi)f
\
q=+) : lpsi /
z4r,325Pa
:241,325 + 84,550 325,875Pa =
:
325.875 Wa
:
325,875(P",
(#)
:
47.3 psi -ri.-- -.l]
252
Cneprrn
l0
Foncnarvo Foncn-Rsremo PenanrsruRs
A vacuum pressrue
gauge monitoring the pressure inside a container reads 200
mm. Hg
vacuum. What are the gauge and absolute pressures? Express results in pascals. The stated vacuum pressure is the gauge pressrue, and to convert its units from millimeters ofmercury to pascal, we use the conversion factor I mm . Hg: 133.3 Pa. This results in
P*w
:
-2oo (mm. Hd
(#ft
)
:
-ru,uuo
r,
The absolute pressure is related to the gauge pressure according to
P"b*lo*: P** + P"b*ro,"
:
?"*oqph"ri"
-26,660
+ I0I,325 :74,665Pa
Note that the negative gauge pressure indicates vacuurn, or ril':iiJiii
i.rtr
i,lliia;lliilltl,iL,filiiili.i.,rirlrFii,.lti.tl:.
a
pressure below atmospheric level. -- r-----illlill
---..1
Vapor Fressure of a tiquid You may already know from your everyday experience that under the same surrounding condidons some fluids evaporate faster than others do. For example, if you were to leane a pan of water and a pan of alcohol side by side in a room, the alcohol would evaporate and leave the pan long before you would notice any changes in dre level of water. That is because the alcohol has a higher ua.Por pressure than water. Thus, under the same conditions, fluids with low vapor pressure, such as glycerin, will not evaporate as quickly as those with high values of vapor pressure. Stated another way at a given temperaflre the fuids with low vapor pressure require reladvely smaller surrounding pressrue at their free surface to prevent them &om
waporating.
Measuring blood pressure.
10.3 Pnrssunr anp Srntss-Foncs AcrrNc OvsnANAREA
253
Pressure W'e all hare been to a doctor's office. One of the first things that a doctor or a nurse will mquure is our blood pressure. Blood pressure readings are normally given by rwo numbers, for example ll5l75 (-m. HS).The fust, or the higher, number corresponds to systolic pressure, which is the maximum pressure exemed when our heart contracrs. The second, or r:he lower number, me{Nures the diastolic blood pressure, which is the pressure in the arteries when the heart is at rest. The pressure unit commonly used to represent blood pressure is millimeters of mercury (mm. Hg). You may already know that blood pressure may change
Blood
with level of activity, diet, temperature, emotional state, and so on.
Flydraulic Systems Hydranlic dtd"Cmachines and bulldozers perform asla in hours that used to take da)n or weels. Today, hydraulic systems are readily used in many applications, including the braking and steering systems in cars, the aileron or flap control system ofan aiqplane, and as jacls to lift cars. To understand hy&aulic qnterns, you must first firlly understand the concept of pressure. In an enclosed fluid system, when a force is applied at one point, creadng a pressure at that point, that force is transmitted doo"Sh the fluid to other poins, provided the fluid is incompressible. When you press your car's brake pedal, tJrat aftion creates a force and subsequendy a pressure in the brake master cylinder. The force is then transmitted through the hydraulic fluid from the master cylinder to the wheel cylinders vrhere the pistons in the wheel cylinder push the brake shoes against the brake drums (rotator). The mechanical power is converted to hydraulic power and back to mechanical power. \7e will discuss the idea ofwhat we me.n by power later in this chapter and in Chapter 13. The rypi""l hydraulic fluid pressure in a car's hydraulic brake lines is approximately 10 MPa (1500 psi). Nowlet us formulate a general relationship among force, pressute, and area in a hydraulic qrstem such as the simple hydraulic system shown in Figure 10.12. Because the pressure is nearly constant in the hydraulic qzstem shown, the relationship benreen the forces fi and F2 could be esablished in the following manner: P\
_FL
(10J3)
A1
P2
_F2
Pr
-
(r0.14)
4
Fr:
Pz=
F,
-Ft 44,
(1015)
A" fl0.r6)
iF,
F-k-4 E
figurel0.lZ
An exanple of a simple hydraulic
system.
254
Cseprrn
l0
FoncneNp Foncs-Rgr-areo PlnevsrERs
It
imporant to realize that when the force
F1 is applied, the drive piston (piston 1) moves by while the driven piston (piston 2) moves by a shoner distance, 12. However, the volume of the fuid displaced is constant, as shown in Figure 10,12. Knowing that in an enclosed qrstem, the volume of the fuid displaced remains @nstant, we c:m then determine the disance l2by tvhich piston 2 moves, provided the displacement of piston 1, 21, is known. is
a distance 11,
fluid volume displaced
O
= ArLt
: 412
00J7)
A,
=;t,
fl0.18)
Moreover, a relationship between the speed of the drive piston and the driven piston can be formulated by dividing both sides of Equation (10.18) by the time that it takes to move each piston. This leads to the following equation:
A, V::V '42 where
(
00.re)
represents the speed ofthe drive piston, and 72 is the speed arwhich the driven piston
moves.
No
as
much
fuid
as
needed to extend the driven piston to any desired level.
Release
valve (closed) Hand-activated pumP
High pressure check valve (open) (a)
I
Figure
10.13
valve (closed)
(b)
Two eramples of hydraulh systems. (a) A land-activated systen. (b) A qear 0r rotary pulnp systen.
L0.3
Pnrssuns ANo SrnEss-FoncpAc"rrr.sc Own aN AnsA
255
The hydraulic system shown in Figure 10.13(b) replaces the hand-activated pump by a gear or a rotary pump that creates the necessaqr pressue in the line. As the control handle is moved up, it opens the passage that allows the hydraulic fluid to be pushed into the load cylin'When the control handle is pushed down, the hydraulic fluid is routed back to the reserder. voir, as shown. The reliefvalves shown in Figure 10.13 could be set at desired pressure levels to control the fluid pressure in the lines by allowing the fluid to retum to the reservoir as shown.
Determine the load that can be lifted by the hydraulic system shown in Figure 10.14. Use g : 9.81 m/s2. W'e use Equation (10.16) to solve this problem.
Ft: mtg: (t00kg)(g.St m/s2) : 9915 n(0.15 m)2. A" F,: ,::;-+(981 N) = 8829 N ' -]4: At n(0.05 m)" Fz: 8829N : (rr,rkg)(9.81 rr/s'?) => n2:
9OOl,g
Note rhat for this problem we could have smred with the equation that relates F2to F1, and then simplified the similar quantities such as zr andg in the following manner:
T(&)' . = 7n8: n1^Sl**)
- A2 F,=iF ,nr:
(R)'
cm)2 : (15 (100 ks) = 900 k (5;F ffiw'
100 kg
Rz=15cm
Rt=5cm
E
figuret0.l4
Thehydraulicsystem0f Emmplel0.9.
This approach is preferred over the direct substitution ofvalues into the equation right arvay because it allows us to change the value of a variable, such as rq or the dimensions of the pistons, and see what happens to the result. For e:
256
Cneprnn
10
Foncs ANo Foncn-RsrArso Pa-narvrnrsRs Normal component of the force
Force
/
I An
Figurel0.l5 anchoring plate subjected to a
cornplessil|e and a shearing force.
Stress Stress provides a measure of the intensity of a force acting over an area. Consider the situation shown in Figure 10.15. The plate shovrn is subjected to the compressive force. Because the force is applied at an angle, it has two componenc: a horizontal and a vertical component. The tendenry of the horizonal component of the force is to shear the plate, and the tendency of the vertical component is to compress the plate. The ratio of the normal (vertical) component of the force to the area is called the normal strasg and the ratio of the horizontal component of the force (the component of the force that is parallel to the plate sudace) to the area is called sltear stress. The normal sness component is
n0.4
often called pressure.
Modulus of Elasticity, Modulus of Rigidity, and Bulk Modulus of Compressibility In this section, we will discuss some important propenies of materials. Engineers, when designing products and srucnrral member, need to know how a seleced material behaves under applied forces or hors well the material conducs thermal energy or electricity. Here we will look at ways of quantifying (measuring) the response of solid materials to pulling or pushing forces or twisting torques. W'e will also investigate the behavior of fuids in response to applied pressures in terms of bulk modulus of compressibiliry. Tensile tests are performed to measure the modulus of elasticity and strength of solid materials. A test specimen with a size in compliance with the American Society for Testing and Materials (ASTM) standards is made and placed in a tensile testing machine.'S?hen a material specimen is tested for its strength, the applied tensile load is increased slowly. In the very beginning of the test, the material will deform elastically, meaning that if the load is removed, dre material will return to its original size and shape without any permanent deformation. The point to which the material exhibits this elastic behavior is called elastic point As the material
LO.A Mouur,us or Er.asrrcrrr, Mopulus or Rrcrprrv,.nxo Bur,r Mopurus or ColrpnssslBrlrrY
Crt= 63
r
-T
60
I
i/
50
of =47 !1
'r4
o a I
257
I
I
I
I
t/
(o"), = 38
]!-
(or)1= 36
ce =35
rn
30
-i
20
I
I t
--.1---
I
10
l
i I I
0.050
0.
0
0.30 /' o.40
0.20
er = o'030
e/ = o'380
Shain (in./itr.)
f
Figure
10.16
Shess-shain diagram for a mild steel sanple (ksi
Uperyield stress, (o1)7 Source :
k
=
loweryield sttess,
o,
:
:
ultimate stress, dld
1000 lb/in?,
oy:
oe
:
elastic stress,
(oy)r:
fmctule stress).
C. Hlibbler, Mechanics ofMareiah.
suetches, rhe normal suess (the normal force divided by the cross-sectional area) is plotted versus the strain (deformation divided by the original length of the specimen). An orample ofsuch test results for a steel sample is shown in Figure 10.16. The modulus of elmticity, or Young's modulus, is computed by calculating the slope of a suess-suain diagram over the elastic region. T-lte modulus of elasthityisa measure of how easily a material will stretch when pulled (subject to a tensile force) or how well the material will shorten when pushed (subject to a compressive force). The larger the value of the modulus of
elasticity, the larger the force required to stretch or shonen the material by a cerain amo{rnt. In order to befter understand what the value of the modulus of elasticity represents, consider the following enample: Given a piece of rubber, a piece of aluminum, and a piece of steel, all having the same recta:ngular shape, crossRubber Aluminum Steel sectional area, and original length as shown in Figure 10.17, which piece will stretch more when subjected to the same force F? As you &dy know from your experience, it will require much less effon to elongate the piece of rubber than it does to elongate the aluminum piece or the steel bar. In fact, the steel barwill require the greatest force to be elongated by the same amount when compared to the other samples. This is because steel has the highest modulus of elasdcFigurel0.l? l{hensubjectedtothesameforce, ity value of the three samples. The results of which piece of material will stretch more?
I
?58
Crrerrpn
l0
An example of a tensile
test
Foncs elvp Foncr-Rruaruo PanervrBlrns
machine.
our observatioll-in lsuns of rhe example cired-are o<pressed, in a general form that applies to all solid materials, by Hooket laqr. W'e discussed Hooket law earlier when vre were explaining springs. Hooket larr also applies to stretching or compressing a solid piece of material. However, for solid pieces of material, Hooket law is elpressed in terms of stress and strain according to
C:
EA
fl0.20)
where
o = normal .g
=
a
:
stress
(N/mz or lb/inz)
modulus of elasdcity (or sometimes called Young's modulus), a properry of material (Nl*' or lb/inz) normal strain, the racio of change in length to original length (dimensionless)
fu
we explained earlier, the slope of a stress-strain diagram is used to compute the value of the modulus of elasticity for a solid specimen. Equation (1,0.20) defines the equadon of the line
that relates the stress and strain on the stress-strain diagram.
lO.4 Mopur,us or Elesncrrr,
Mopur.us or Rrcrorry, rurp Bur.r Moowus or CoupnrssrBrlrrr
259
As an alternative approach, we will next explain a simple procedure that can be used to E Although rhe following procedure is not the formal way to obain the modulus of elasticity value for a material, it is a simple procedure that provides additional insight into the modulus of elasticiry. Consider a given piece of rectangular*haped material with an original length I, and cross-sectional area l, as shown in Figure 10.18. When subjected to a known farce 4 tle bar will suetch; by measuring the amount the bar stretched (final length of the bar minus the original lengrl), we can determine the modulus of elasticity of the material in the following manner. Staning with Hooket larr, o = Ee, and, substituting for a and the strain e, in terms of their elementary definitions, we have measure
F
E A
ti$rel0.ls
rcctangular bar suljected to a
tensile load.
F -x A -L
(r0.zr)
Rearranging Equation (10.21) to solve for modulus of elasticity
p: lL
4
we have 00.2a
Atc
where
: F: L: I : r: E
modulus of elasdcity (N/mt) the applied force (N) origSnal length
of the bar (m)
cross-sectional area of the bar (m2)
elongation, final length mimrs original length (m)
Note that the physical variables involved include the fundamenal base dimensions length and arcaftom Chapter 7 andthe physical variable force from this chapter. Note also from Equation (10.22) that the modulus of elasticiry, ,fl, is inversely proportional to elongation *; the more the material stretches, the smaller the value of .E Values of the modulus of elasticity E for selected materials are shown in Thble 10.4. Another important mechanical propertF of material 's rhe modubs of rigidity ot sbea.r maduha. The modulus of rigidity is a measure of how easily a material can be twisted or sheared. The value of the modulus of rigidity shows the resistance of a given material to shear deformation. Engineers consider the value of shear modulus when selecting materials fot shafts or rods that are subjected to twisting torques. For example, the modulus of rigidiry or shear modulus for aluminum alloys is in the range of 26 to 36 GPa, whereas the shear modulus for steel is in the mnge of 75 to 80 GPa. Therefore, steel is approximately duee times more rigid in shear when compared to aluminum. The shear modulus is measured using a torsional test machine. A din&ical specimen of known dimensions is twisted with a known torque. The angle of twist is measured and is used to determine the value of the shear modulus (see E:rample 10.11). The values of shear modulus for various solid materials are given in Table 10.4. Based on their mechanical behardor, solid materials are commonly classified as either dactile or brittb. A ductile material, when subjected to a tensile load, will go through significant permanent deformation before it breaks. Srcel and aluminum are good examples of ducdle materials. On the other hand, a britde material shows litde or no permanent deformation before it rupnues. Glass and concrete are examples of britde materials.
264
Cnamsn
10
FoncsANo Foncn-Rnr.erBo Peneuprens
i
l\[aterial
Modulusof
Shear
Elasticitr (cPa)
Modulur (GPa)
Aluminum alloys
70-79
Brass
96-1 10
96-l2o
Bronze C-ast
26-10 36-41 36-44 32-69
,
83*i70
iron
Concrete (compression) Copper alloys
17-31
:
110.120
40-47
Glass
48-83
Magnesium alloys
4l'45
19-35 15-17
Nickel
210
80
Plasdcs
2.1-3.4
Nylon Polyethylene Rock (compression)
Granite, marble, quare, Limestone, sandstone
''
0.7*t.4 40*100
20-70 0.0007*0.004
Rubbsr
0.0002-0.00r
Steel
l9A:210
75-80
Tianium alloi's
rQa-t2a 340-380
140-160
Tungsten \Food (bending)
..
Oak
t1-r3' l1-I2
Southem pine
11-r4
Douglas
fir
39-U
Source: Adapted fromJ. M, G*e, Mecbania of Mate/iab, Reprioted a division of Thomson Leaming: www.thomsonrights.com
with permission of Nelson,
The tensile and compressive suength are other imponant propenies of materials. To prestress calculations that are compared to the tensile and compressive suength of materials. The tensile srength of a piece of material is determined by measuring the maximum tensile load a material specimen in a shape of a rectangular bar or cylinder can carrywirhout failure. The tensib strengtbor ultimate strength ofa material is expressed as the maximum tensile force per unit of the original cross-sectional area of the specimen. As we mentioned earlier, when a material specimen is tested for its suength, the applied tensile load is increased slowly so that one can determine the modulus of elasticity and the elastic region. In the very beginning of the test, the material will deform elasticalln meaning that if the load is remove4 the material will return to its original size and shape without any permanent deformarion. The point to which the material exhibir this elastic behanior is the elastic point. The elasdc srength represents the maximum load that the material can carrywithout any permanent deformation. In many engineering design applications, the elastic suength or yield strength (the yield vdue is very close to the elastic strength value) is used as the tensile strength. Sorne materials are stronger in compression than they are in tension; concrete is a good example. The compression strength of a piece of material is determined by measuring the
dict failure, engineers perform
lO.4
Mopur-us or Eresrtqrv, Mopur-us or Rlcturrr, aNp BuLK MoDULUs or
XeldStength (MPa) Aluminum alloys Brass
Btonze
,
Cast iron (tension) Cast iroq (compression) Conirete (compression)
:
Copper
allop
Gtnss
:
ColrpnsssrBrlrrr
261
Ultimite Strength (MPa) , 100-550
35-500 70-550 82-690 lao-29A
2A0-62A
200-830
69-#0 M04,4A0
1010 230-830
55-76A
30-1,ooo
i
70
Plate glaq Glass fibers Mgnesium allqrs
7,000-20,000
Nickel
80-280
r40-340
rca-620
310)760
Plastis
40-80
Nyl"n
7-28
Polyethylene Rock (compression)
50-280 20-2A0
Granite, mar-ble, quar,tz Lirnestone, sandstone
7-20
1"-7
Rubber, Steel
High*crength
340-1,000
550
Machine
340-700
550*860 700*1,900
Spring Stainless
400-1,600 280-700
,
l
Tool
1,200
400=1,000 900
52A
280-1,000
550-1,4t0
Strucnrral steel
2A0 7A0
,40-830
l ltailuh
760*1,000
900-1,200 1,400-4,000
Steel
wiie auo)'s
Tungsten
Vood (bendind Doqgla$,fu
30-50
50*s0
Oak
40-60 40-60
50-100 50*100
l
Southern Pine '\Food (compression parallel to grain) Douglas fir
I
Oak Southernpine
:
30-50
4Ou7A
30*4A
30-50 4A-74
30*50
Sozrca: Reprinted with permission of Nelson, a division of Thornson www.rJromsondghts,com
lemiag:
mar
Thble 10.5.
262
Cnaprnn
10
FoncraNo Foncr-Rnarpo PanetvrsrsRs Asuuccural memberwith a recangular cross section, as shown in Figure 10.19, is used to support a load of 4000 N disuibuted uniformlyover the coss-sectional area of the member. $?hat type of material should be used to carry the load safely? Material selection for structural members depends on a number of facors, including the density of the material, its strength, its toughness, its reaction to t}re surrounding environmenr, and its appearance. In this example, we will only consider the strength of dre material as the design factor. The average normal stress in the member is given by
o':
4000
N
(0.05 m)(0.005 m)
:
16 MPa
Aluminum alloy or structural steel material with the yield strength of 50 MPa and 200 MPa,
I Figurel0.tg
respectively,couldcarrytheloadsafely.
ll lhestructuml nemb$ ol
lj
Eramplel0.l0.
I One of the goals of most structural analysis is to check for failure. The prediction of failure is quite complex in nature; conseguendy, many investigators have been srudying this topic. In engineering desrgn, to compensate for what we do not know about the exact behavior of material and/or to account for future loading for which we may have not accounted but to which someone may subject the pan or the strucnrral member, we introduce afaqor of safety (F.S.), which is defined as
F's':
Pk
flo.23)
P* is the load that can cause failure. For cerrain situations, it is also custornary to defne the factor of safery in terms of the ratio of maximum stress that causes failure to the allowable stresses ifthe applied loads are linearly related to the stresses. The factor ofsafety for Example 10. l0 using an alumnium alloy with a yield suength of 50 MPa is 3. I .
where
A setup similar to the one shown in Figure 10.20 is commonly used to measure tlre shear modulus, G. A specimen of known length Z and diameter D is placed in the test machine. A known
I
Figure
10.20
A setup
to measure the shear noduhs of a naterial.
t0.4
Moowus or Er-nsrrcrrr, Moour.us oF RrcrDrrr, er*o Brrrr MoDt torque
Iis
LUs
or CoupnrssrBrlrrr
applied to the specimen and the angle of twist @ is measured. The shear modulus
is then cdculated
from
32TL
Q:
263
(t0.24)
rTD4+
Using Equation (70,24), calculate rhe shear modulus for a $ven specimen and test results T: 3450 N. m, Z : 20 cm, D : 5 cm,6 :0.015 rad.
Q,l
32TL
32Q45a N'm)(0.2 m)
rrD44
flo.os m)a(o.ot5 rad)
:75
of
GPA
i,ilrrfili'a--
Bulk Modulus of Compressibility Most of you have pumped air into a bicycle dre at one time or another. From this and other experiences you know that gases are more ezsily compressed than liquids. In engineering, to see how compressible a fuid is, we look up the value of a bulk modulus of compressibility of rhe fluid. The value of fluid bulk modulus shows how easily the volume of the fluid can be reduced when the pressure acting on it is increased. The bulk modulus of compressibility is formally defined
o___ increaseinptessure
_ decrease in volume original volume
as
increaseinpressure increase in densiry
fl0.25)
original densiry
Equation (10.25) expressed mathematically is
Ev=
dP
_dv V
:-
dP
(10.26)
dp
7
where
Ey
:
modulus of compressibilitf
dP
=
change
in pressure (increase in pressure) (N/m1
change
in volume (decrease in volume, negative value) (m3)
: I/ : dp : p:
dV
N/m1
origind volume (m3) changein densiry (increase in density)
original density
(kS/-')
(kg/.1
!7hen the presure is inoeased (apositive change in value), thevolume of the fluid is decreased (a negative change in value) , and thus the need for a minus sign in front o f the dV wm to make Eya positive value. In Equation (L0.26), also note that when the pressure is increased (a posi tive change in value), the density ofthe fluid is also increased (a positive change in value), and thus there is no need for a minus sign in ftont of the dp term to make Ey a positive value. The compressibility values for some common fluids are given in Thble 10.6.
264
Cneprsn.
10
Foncs nxo Forcs-REr-arEo Panerrarrnns
Compressibility Modulus qNlut)
nuid hhyl alcohol
1.06 1.3
Gasoline
Glycerin MercurF SAE 30 oil 'Water
x
Compressibility Modulw flbr/in'z)
10e
L54X
i I
rc5
X 10e 4.52x roe 2.85 X l0t0
4.74
1.5 x loe 2.24 x tA,
2.2X t0, 3,25 X t05
X 10: 6.56 X 10) 1.9
x
LA6
'lChen
we were defining the bulk modulus of compressibilitywe said rhat as the pressure is inoffluid is decreased, and consequendy the density ofthe fluid is increased. 'Why does the density of a fluid increase when its yolume is decreased? creased, the volume
Recall that density is defined as the ratio of mass to volume, as given by densrty
mass
volume
From examining this equation, for
a given mass,
you ca:r
see
that as the volume is reduced; the
density is increased.
For warer, sardng with I m3 of water in a container, what is the pressure required to decrease the volume ofwater by 1olo?
From Thble 10.6, the compressibility modulus for water k 2,24 x 10e N/m2. It would require a pressure of 2.24 x 107 N/m2 to decrease the original 1 m3 volume of water to a final volume of 0,99 m3, or, said another way, by 17o. It would require an equivalent pressure of 221, atm Q.24 X 107 Nlml to reduce the pressure of a unit yolume of water by to/o.
ll
I
,la,=,,a,jaa::I
-::-:-
10.5
Moment, Torque-Force Acting at a Distance As we mendoned earlier, the two tendencies of an unbalanced force acting on an object are to translate the object (i.e., to move it in the direction of the unbalanced force) and to rotate or bend or nvist the object. In this section, we will focus our attention on understanding the rcndency of an unbalanced force to rotate, bend, or twist objects. Being able to calculate momens created by forces about various points and axes is important in many engineering analyses. For example, all of you have noticed that the sueet light poles or the traffic light poles are thicker near the ground than they are on the top. One of the main reasons for this is that wind loading creates a bending moment, which has a maximum value about the base of the pole. Thus, a bigger section is needed at the base to prevent failure. Nature undersands the concept ofbending
10,5
I
Figure
10.21
MotvtsNT,
Tonqur-Foncr Acrrxc
an a DrsteNcs
265
iloment of the force created by someone opening or closirg a door.
I
d;"
I
ftgure 10.22
The moment
ofthe force
about points l|,
f,
f
and d.
moments well; that is why trees have big trunks near the gound to support rhe bending moment created by the weight of the branches and the wind loading. Understanding moments, or torques, is also imponant when desigrung objects that rotate, such as a shaft, gear, or wheel. To betrer gasp the concept of moment, let us consider a simple example. Vhen you open a door you apply a pulling or a pushing force on the doorknob (or handle). The application of this force will make the door roate about its hinges. In mechanics, this tendency of force is measured in terms of a rnoment of aforceabout an axis or a point. Mommt has both direction and rnagnitudr.In our example, the direction is defined by the sense of rotadon of the door. As shown in Figure 10.21, looking at the door from the top, when you open the door by applying the force shown, the direction of moment is clochrise, and when you close the same door, the force creates a countercloclcrrise moment. The magnitude of the moment is obtained from the product of the moment arm times the force. Moment arm represents the perpendicular distance beE'reen the line ofacdon ofthe force and the point about which the object could rotate. In this book, we will only consider the moment of a force about a point. Most ofyou will take a mechanics class, where you will learn how to calculate the magnitude and the diredion ofrnoment ofa force about an arbitrary axis. For an object that is subject to force 4 the relationship berween the line of action of the force and moment arm is shown in Figure 10.22. The magnitude of moment about arbitrary points l, B, and C is given by
My: d1F)
CI0.2n
?66
Cnepren
10
Foncs ANp Foncs-RETATEp Penerrarrnns
Mp: drFl Mc: 0
00.28)
fl0.29)
Note that the moment of the force Fabout point C is zero because the line of action of the force goes thtough point C and thus the value of moment arm is zero. \7hen calculating the sum of momen* creifed by many forces about a certain point, make sure you use the correct rnornent arm for each force, and also make sure to ask yourselfwhether the tendency ofa given force is to rotate the object clockrrise or counterclockwise. fu a bookkeeping procedure, you may assign a positive value to clockwise rotations and negative values to counterclockwise rotations or vice versa- The overall tendency of the forces could then be determined from the final sign of the sum of the momen$. This approach is demonsrated in Example 10.14. Determine the sum of the moment of the forces about point O shown in Figure 10.23. 0.1 cos 35
100N
t
figure
10.23
lhe dhgram for Eranrple
^2Mo: :
(50 NX0.05 13.55
10.14.
m) +
(50 NX0.07 cos 35
m) + (100 NXO.I
cos 35
m)
N. m
Note that in calculating the moment of each force, the peqpendicular distance berween the line of action of each force and the point O is used. Moreover, for book keeping sake, we assigned a positive sigr to countercloclcvise rotation,
:.-l:::;]i
:::
rlr-:::,,:-:rn::::ri
,:-::.-:1.r.-{
10.5
MowTBNT,
Tonqur-Foncr Ac:rlvc rr
A DrsraNcE
261
Determine the moment of the two forces, shown in Flgue L0.24, about points 24, B, C, and. D. can see, the forces hare equal magnitudes and acr in opposite directions to one another.
fu you
R ZMn:
(100 NXO)
R2M":
(looNXrom) + (looNXo):1oN.m
R2Mr:
(100Nx0.25m)
-
(100Nx0.15m)
R2Mo:
(100N)(0.35m)
-
(100Nx0.25m) = 10N.m
+
(100 NX0.1
m) =
10
N. m
:10N.m
100N
tmN
[
lj li '1, '!1 [,:-:a,,a-:-=,'.,.=.tlllllg l
figure0.el
AscftematicdiagranforEranplel0.l5.
Referring to Example 10.15, note that rwo forces that are equal in magnitude and opposite in direction (not having the same line of action) constitute a couple. As you can see from .he resuks of this example, the moment created by a couple is equal to the magnitude of either f"rce times the pelpendicular distance between the line of action of the forces involved. Also note for this problem, we assigned a positive sigo to dochrise rotation. Since the srr- of the has a positive vdue, the overall tendency of forces is to tum the object clocl(rise.
^ments
a]:at:..:i-:T.l-l'-:-i::ra:i::--.-L t--.r--r =.---:-:-,-.
-'.r,:.-:-:
r-.:-----::-ir::r :.-:-.:-
268
I
Cnaprsn
10
ForcB.eNo Foncs'RELATEp Penarvmruns
Figuml0.Z5
wo* done on the cal by tie force fh equal to
The
4_,
:
lFcos OXd).
10.6 Work-Force Acting
Over a Distance
'\ilVhen
you push on a car that has run out of gas and move it through a distance, you perform mechanical work. When you push on a larvn mower and move it in a certain direction, you are doing work. Mechanical work is done when the applied force moves the object through a distance. Simply stated, mechanicaluorhis defined as the component of the force that moves the object times the distance the object moyes. Consider the car shown in Figure 10.25. The work done by the pushing force moving the car from position I to position 2 is given by
Vt-z: @cose)@)
00.30)
Note from examining Equation (10.30) that the normal component of the force does not perform mechanical work. That is because the car is not moving in a direction normal to ttre ground. So next time you are mowing the lawn, ask yourself if you should push on the lavrn mower horizontally or at an angle? Another point you should remember is that, if you were to push hard against an object and were not able to move it, then by definition, you are not doitg *y mechanical work, wen though this action could make you tired. For example, if you were to push against a rigid wall in a building, regardless of how hard you were to push with your hands, you wouldn't be able to move the wall, and thus, by definition, you are not doing any mechanical
work*
Determine the work required to shown in Figure 10.26.
lY il
lift
Y:
a box that weighs 100
(100 NX1.5
m) =
150
N,
1.5 m above the ground,
as
N. m
T-*
1.5 m
il
I
I
riguretO.zO
The bor in trample 10.16.
NI
Ll;i:.-iiirrt!,ll1iati
r,it
:..-----l.:i:i-
*Of coune, your eflon
goes into deforming the wall on a very very small scale thar cannor be dereced by the naked eye. The forces created by your push deform the wall material, and your effort is stored in the material in the form of what is called the stain enngl
lO.7 Lnrpan Iupursr-FoncrAcrnc Own Tnur
269
The SI unit for work is N. m, which is called joale; thus the unit joule (J) is defined .hr""gh a distance of I m.
as
the work done by a 1-N force
l joule : (t N)(t m) Srudents often confirse the concept of work with power. Power represents how fast you want to do the work. It is the time rate of doing work, or said anorher way, it is the work done divided by the time that it took to perform it. For instance, in Example 10.16, if we want to lift the box in 3 seconds, then the power required is
power
work 150I = ,i*" :;;::
50Jls:
50watts
Note that I J/s is called I watt (W). If you wanted to lift the box in 1.5 seconds, then the re'Wl quired power to do this task is: l5O J ll.5 s or 100 Thus, rwice as much power is required. It is imponant to note that the work done in each case is the same; howwer, dre power requirement is different. Remember if you want to do work in a shorter time, then you are going to need more power. The shorter the time, the more power required to do that work !?'e will discuss power and its units in much more detail in Chapter 13.
X0.7 linear lmpulse-Force Acting 0ver Time Up to this point, we
have defined the effecs of a force acting at a distance in terms of creating moment, and through a distance as doing work Now we will consider force acting over a period of time. Understanding linear impulse and impulsive forces is important in the design of produca such as air bags and spon helmets to prevent injuries. Understanding impulsive forces also helps with daigning cushion materials to prevent damage to products when dropped or when subjected to impact. On TV or in the movies you have seen a stuntman jumping offa roof of a multistory building onto an air mat on the ground and not getting hun. Had he jumped onto concrete pavement, the same stuntman would most likely be killed. \(hy is that? 'Well, by using an infated air mat, the stuntman is increasing his time of contact with the ground through staying in touch with the air mat for a long time before he is completely stopped. This statement will make more sense after we show the relationship between the linear impulse and linear momentum.
Linear impuberepresents the net effect of a force acting over a period of time. There is a relationship berween the linear impulse and the linear momentum. 'We explained what we mesn by linear momentum in Section 9.6. A force acting on an object over a period of time creates a linear impulse that bringp about a change in the linear momenftlm of the object, according to
i*",oa;:
&u
- A1
(10.3r)
where
Fo*e, =
average magnitude
At = time period
of the force acting on the object (N)
over which the force acts on the object (s)
ln: mass of the object (Lg) Vr: finalvelocity of the object (m/s) '14 : initid velociry of the object (m/s)
274
Crnr"rnn l0
Foncr exo Foncr-Rrr-lrno Pan-avBtnns It is important to note that Equation (10.31) is avectoral relationship, meaningit has both magnitude and direction. Using Equation (10.31), let us now explain why the stuntman does not huft himselfwhen he jumps onto the air mat. For rhe sake of demonstration, let us assume that the snrnunan is jumping offa ten-story building where the.werage height of each foor is 15 ft, Moreover, let us assume that he jumps onto an inflated air mat that is 15 ft tall. Neglecting air resistance during his jump to make the calculation simpler, the stuntman's velocity right before he hits the air mat could be determined &om
r(: t/Ifi
fi0.32)
where I{ represents the velocity of the stuntman right before he hits the air mat, g is the acceleration due to gravity (g: 32.2 ft/$, and / is the height of the building minus the height of the air mat (h : 135 ft). Substituting for gand 6 in Equation (10.32) leads to an initial velocity of V,= 93.2 ft/s. Also, we realize that the air mat reduces the velocity of t*re stuntman to a final velocity of zero (Vr: 0). Now we can solve for .fu"*o from Equation (10.31), assuming different values for Aa and assuming a mass of 4.65 slugs (150 lb-). The results of these calculations are summarized and given in Thble 10.7. It is important to note that .Foo"o represenn
Time of Contact
fall.
Average Reaction Force (lbs)
0.5
4334 867
0.1
Two stuntmen practice a
(s)
1.0
433
2.0
)17
5.0 10.0
43
87
PnosLEMs
27l
the force drat the $untman exer$ on the air mat and subsequendy through the air to the ground. But you recall from Newtorfs drird law that for every action there is a reaction equd in magnitude and opposite in direction. Therefore, the anerage force exened by the ground on the stuntman is equal in magnitude to 4"".ge.Also note that the reaction force will be disuibuted over the back surface of the stuntman's body and thus create a relarively small pressure disribution on his back You can see from these calculations that the longer the time of conac, the lower the reaction force, and consequendy the lonrer the pressure on the stuntman's back
SUMMARY Now that you have reached this point in the
.
n
"
o
.
You should have a good understanding ofwhat is meant by force and is common units. You should also knorq that there are different types offorce. You should understand the tendenry of an unbalanced force, which is to translate and/or rotate objects. The application of forces can also shoften, elongate, bend, and twist objects. You should be familiarwith the Neftorls larvs used in mechanics. These lalvs are the basis of analysis in many engineering problems. You should undersand what is meant by pressure. For static fluids, you should understand Pascal's lar, which states that fuid pressure at a point is the same in all directions. You should also remember the relationship between the fluid pressure and the depth of the fluid. You should understand what the fluid propenies such as viscosity and bulk modulus of compressibility mean. You should also have a good idea of how hydraulic systems work You should realize that engineers, when designing products, need to know how a selected material behaves under applied forces. You should knowwhat material properties mean, such as modulus of elasdcity, shear modulus, and tensile and compressive suength. You should understand that when a force is applied to an object, it can affect the objectt state in a number ofways. In engineering, these effects are measured in terms of sffess, moment, worh and linear impulse. Remember that the intensity of a force acting within the materid, in ferms of pulling it apart, compressing, or shearing it, is measured in terms of suess; the effect of a force acting at a distance is measured in terms of creating a moment about a point or an axis; the effect ofa force acting over a distance is represented by mechanical work; and the effect of a force acting over a period of time is measured in terms of a
r0l.
r0.2.
te:
linear impulse.
Dotgr a mass-spring qfstem that can be aken to Mars to measure the acceleration due to gradty at the surface of Mars. Explain the basis of your design and how it should be calibrated and used. In the past, scientists and engineers hare used pendulums to measure the value of g at a location. Design a
pendulum drar can be used to measure the value ofg at your location. The formula to use to measure the acceleration due to gravity is
f = 2n.1ft
272
Cneprun
L0
FonceAND Foncs-REr,ATEp PanavrrsRs
where ?"is the period of oscillation of the pendulum, the time that it takes the pendulum to complete one rycle. The distance between the pivot point and the
n0.8.
center of the mass of the suspended deadweight is represented by Z.
Designers often learn from their narural surroundings; do the bigger animals have bigger feet? Calculate the pressure exerted by water on the hull of a submarine that is cruising at a depth of 500 ft below ocean level. fusume the densiry of the ocean water is 1025 kg/*'.
p:
Investigate
10.r0.
devices such as manometers, Bourden tubes, and diaphragm pressure sensors. W'rite a brief repon discussing your findings. Conven the following pressure readings from inches of mercury unia to psi and Pascal units: a- 28.5 in. Hg b. 30.5 in. Hg
1o.fl.
If
hoblem 10.2
r0.3.
An asuonaut has a mass of 70 fu.'What is the weight of the astronaut on Eanh at sea level? What are the mass and the weight of the astronaut on the Moon,
10J2.
[0.5.
n0.6.
Basketball player Shaquille O'Neal weighs approximately 335 lb and wears a sizr 23 shoe. Estimate the pressure he enens on a floor. SThat pressure would he exert if his shoe size were 13? State all your assump-
Psl pascal
feet ofwater feet of mercury
An air compressor intakes atmospheric ir x I4.7 psi and increases its pressure to 175 psi (g*rg"). The pressurized air is stored in a horizontal tank V'hat is the absolute pressure answers
ofair inside the tank? Express your
in
a. psi b. kPa
tions, and show how you arrived at a reasonable
c.
solution.
d. standard atm (recall that
Investigate the reladonship between pressure for the following situations: a. when you are standing up barefooted b. when you are lying on your back c. when you are sitting on a tall chair with your feet offthe ground and the chair Investigate the pressure created on a surFace by the fol-
10L325l<J:a) Calculate the pressure exened by water on a scuba diver who is swimming at a depth of 50 ft below the water surface. Investigate the grpical opentiagpressure range in ttre hydraulic sFstems used in the following applications. '\?'rite a brief repon to discuss your findings.
1013.
10.t4.
b, a car c. a bicycle d. a pair of cross-countrf
bar
1
atm
:
L4.7
psi:
a. controlling wing faps in airplanes b. your car's steering qystem c. jacls used to lift can in a service station
lowing objects: a. abulldozer
t0.7.
pressure-measuring
a pressure gauge on a compressed air tank reads 120 psi, assuming standard atmospheric pressure at sea level, what is the absolute pressure of air in
a. b. c. d.
and on Mars? !7hat is the ratio of the pressure exened by the astronaut's shoe on Earth to Mars? n0"4.
the operation of
n0.9.
t0J5.
skis.
Do you think there is relationship berween a person's height, weight, and foot size? Ifso, verifyyour answer.
Using the information given in Thble 10.3, determine the ratio of local pressure and density to sea-level values. Estimate the value of air density at the cruising dtitude of most commercial airliners.
Pnosl.Ervrs 273
Rz=20cm
1000 N
Problem 10.19
p Altitude (m)
Peloet
P
P*
lut"l
XOJE.
10.19.
0 (sea level) 1000
3000
r0"20.
5000 8000
0,000 2,000
4,000 5,000
Convert the 130/85 systolic/diastolic pressures given Pa, psi, and in. H2O. Determine the load that can be lifted by the hydraulic qrstem shown in the figure above. All of the necessary information is shown in the figure.
in millimeters of Hg to
Determine the pressure required to decrease the volume of the follorring fluids by 2o/ot a. water b. glycerin
c.
10.21.
SAE motor oil Compute the defection of a structural member made of aluminum if a load of 500 lb is applied to the bar. The member has a uniform recangular cross section
with the dimensions of { in. X 4 in., t0.16.
t0J7.
Bourdon-type pressure gauges are used in thousands of applications. A deadweight tester is a device rlat is used to calibrate pressure gauges. Investigate the operation of a deadweight pressure tester. Write a brief repon to discuss your findings. Investigate what the rypiol range of pressure is in the following applications: a birycle tire, home water line, natural gas line, and refrigerant in your re&igeratort evaporator and condenser lines. Present your findings in a brief repon.
accompanying figure.
L='l
F= 500Ib
Problem 10.21
as
shown in the
274 1l0.n.
Cnar"rrn
10
Foncr eNo Fox.cs-RELATEp Pan qMETERs
A strucrural member with as shown
Fo-l
a rectangular cross section
in the accompanying figure is used
to support aload. of 2500 N. \Vhat type of material do you recommend be used to carry the load safel)r? Base your calculations on the yield strength and a factor of
It.
safew of 2.0.
Problem 10.24
n0.25.
10
[0.23.
cm
Problem 10.22
F'ol
The tire wrench shown in the accompanying figure is used to tighten the bolt on a wheel. Given the information on the diagram, determine the moment about point O for the two loading situations shown: a. pushing peqpendicular to the wrench arm
b. pushing
Determine the moment created by the weight of the lamp about point O. The dimensions of the street light post and arm are shown in the accompanyrng figure. The lamp weighs 20 lb.
at a75" angle, as shown
Problem 10.25
10.26.
Determine the moment created by the weight of the uaffic light about point O. The dimensions of the uaffic light post and arm are shown in the accompanying figure. The uaffic light weighs 40 lb.
Problem 10.23
tr024.
Determine the moment created by the weight of the suspended sign about point O. Dimensions of the sign and the support are shown in the accompanyitlg fig,rt.. The sign is 2 mm thick and is made of aluminum.
Proilem 10.26
PnoslErvrs 275
100
N
Problem 10.27
10.27.
r0.28.
t0.29.
changed
forces shown in the accompanying figure about points
shown in the accompanying table.
A, B, C, andD, Determine the amount of work done by you and a friend when each ofyou push on a car that has run out of gas. Assume you and your friend each push with a horizontal force with a magnitude of200 N, a distance of 100 m. Also, suggest ways of acnrally measuring the magnitude and the direction of the pushing force. Determine the work done by an electric motor lifting an elevator weighing 1000 lb .loough five floors; assume a distance of L5 ftbetween each foor. Compute the power requirements for the motor to go from the first to the fifth foor in
a.
Express your results
in
horsepower (1
hp :
reaction force from the
is
Reaction Force (N)
0.05 0.1 1.0
2.0
10.31.
10.32.
foor if the time of contact
as
0.01
550
ft'lb/s). If a laptop computer weighing 22 N is dropped from a distance of 1 m onto a foor, determine the anerage
materials,
Average
Time of Contact (s)
5s
b.8 t0.30.
by using different cushion
Determine the sum of the moments created by the
Obain the values of vapor pressures of alcohol, water, and glycerin at a room temperanue of 20"C. '!?'hen learning to play some sports, such as tennis, golf;, or baseball, often you are told to follow through with your swing. Using Equation ( I 0. 3 I ), explain why follow-drough
10.33"
is important. In many applications, calibrated springs are commonly used to me:$ure the magnitude of a force. Investigate
276
CneprsR
10
Foncraxo Forcr-Rrr.ersD Peneurmns 70 tb
80 tb Proilem 10.34
how ty'pical force-measuring devices work W'rite a briefrepon to discuss your findingr.
ro.il. Calculate the moment created by the forces shovm in the accompanying figure about point O.
t0.s. \Fe have used an experimental setup similar to Example 10.1 to determine the value of a spring constant. The defection caused by the corresponding weighs are given in the accompan)ang table.'What is the vdue of the spring constant?
10.36.
If an astronarr and her space suit weight
250 lb on
Earth, urhat should be the volume of her suit if she is to practice for weighdess conditions in an underwater neutral buoyant simulator, similar to ttre one used by NASA as shown in Chapter 7?
2n
Caterpillar 797 Mining Truck
*
The Cateqpillar@ 797,which features many patented innovations, was developed using a cleansheet approach. But the design also draws on the experience garned from hundreds ofthousands of operating hours accumulated by more than 35,000 Caterpillar construction and mining tmcks working worldwide. The 797 builds on such field-proven designs as the mechanical power train, elecuonics that manage and monitor all qrstems, and structures that provide durability and long life. Engineers from various fields, induding mechanical, electrical, strucnral, hydraulio and mining, collaborated to design the Cateqpillar 797 mining truck
*Materiats were adapted with permission from Caterpillar@ documens.
278
CersRPru-A,R 797
Mnrnrc Tnucr
279
Cateqpillar dwelopedtheT97 in response to mining companies that were seeking a means to reduce cost per ton in large-scale operations. Therefore, $e797 is sized to work efficiendy with loading shovels used in large mining operadons, and Careqpillar matched the body design to the material being hauled to optimize payloads. The new 797 mining truck is the largest of the extensive Caterpillar oFhighway truck line, and it is the largest mining truck wer constructed. The Caterpillat 797 mintng truck has a payload capacity of 360 tons (326 metric tons) and design operating mass of 1,230,000 lb (557,820 kg). Th" Cateqpillar 797 MiningTruck specificadons include:
F.ngine modeft 35248 high displacement Gross power rpm: 3400 hp Q537 klV) ^tL75O Transmission: 7-speed planetary power shift
Top speed:40 mph (64 km/h) Operating massr 610,082 lb (1,345,000 k&) Nominal payload capacity: 360 tons (326 metric tons) Body capacity (SAE 2:1)r 290 y& (220 m3) Tire sizq 58/80R63
A summary of some information about various componens of the 797 truck follows.
Station The 797 cab is designed to reduce operator fatigue, enhance operator performance, and promote safe operation. The controls and layout provide greater operator comforc alongwith an automotive feel, while enhancing functionality and durability. The cab and frame design meets SAE standards for rollover and falling-object
Comfortable and Efficient 0perator
protecdon. The truck and cab design provides exceptional all-around visibility. For example, the clean deck to the right of the cab improves sight linCI. The spacious cab includes nvo firll-sized air suspension seats, which allow a trainer to work with the operator. The steering column tilts and telescopes so the operator can adjust it for comfon and opdmum conrol. The Vital Information Management System (VIMS) display and keypad provide precise machine status information. The new hoist control is fingenip actuated, allowing the operator to easily and precisely control hoist functions-raise, hold, foat, and lower. Elecrically operated windours and standard air conditioning add to operator comfon. The cab is resiliendy mounted to dampen noise and vibration, and sound-absorbing material in the doors and side panels cuts noise further. Cast
Frame Mild steel castings comprise the entire load-bearing ftame for durability and reto impact loads. The nine major castings are machined for precise fit before being
sistance
joined using a robotic welding technology rhat ensures firll-peneration welds. The frame design reduces the number of weld joints and ensures a durable foundation fot rhe797. The suspension system, which uses oil-over-nitrogen struts similar to other Caterpillar mining trucks, is designed to dissipate haul road and loading impacts. Electronically Controlled Engine The new Cat 35248 high-displacement diesel engine produces 3400 gross hp (2537kV).It is turbocharged and aftercooled, and features electronic unit injecdon (EUI) technology, which helps the engine meet year 2000 emissions regulations. The engine uses two in-line blocls linked by an innovative coupler and precisely controlled by electronic conrollers. These electronic conuollers integrate engine information with
280
Cneprrn
10
Foncn eNp Foncs-RELATno PenAtr smRs mechanical power train information to optimize uuck performance, ortend component life, and improve operator coffirt. The engine uses a hydraulically driven fan for efficient cooling. The fan design and operadon also reduce fuel consumption and noise levels. The engine and its components are designed to minimize service time, which helps keep availability of the truck high. Efficient Mechanical Power Train The 797 power train includes a new torque converrer with lockup clutch that delivers high mechanical efficienry. The new automatic-shift transmission features seven speeds forward and one speed reverse. Electronic clutch pressure control (ECPC) technologr smoothes shifts, reduces wear, and increases reliability. Large clutch discs give the transmission high torque c.apacity and extend transmission life. The transmission and torque converter enable the truck to maintain good speed up grades and to reach a top speed of40 mph (64 km/h). The differential is rear mounted, which improves access for maintenance. The differential 'S7ide is pressure lubricated, thus promoting greater efficiency and long life. wheel-bearing spread reduces bearing loads and helps ensure durability. A hydraulically driven lube and cooling qrstem operates independendy ofground speed and pumps a continuous supply offiltered oil to each final drive. Oil-cooled, multiple &isc brakes provide fade-resistant braking and retarding. The electronically managed automatic retarder conuol (ARC) is an integral pan of the intelligent power train. ARC controls the brakes on grade to maintain optimum engine rpm and oil cooling. Automatic electronic tracdon aid (AETA) uses the rear brakes to optimize uaction. A combination of constant-displacement and variable-displacement pumps delivers a regulated flow of brake-cooling oil for consant retarding capability and peak tmck perFormance on downhill grades.
Electrohydraulic Hoist Control New hoist hydraulics include an elecronic hoist conuol, independent metering valve (IMV), and a large hoist pump. These fearures allow automatic body snubbing for reduced impact on the frame, hoist cylinders, and operator. This desigo also allows the operaror to modulate flow and conuol overcentering when dumping.
Serviceability Cateqpillar designed the 797 to reduce service time thus ensuring maximum availabiliry. Routine maintenance points, such as fluid fill and check points, are close to ground level. Easily accessed connectors dlow technicians to download data and to calibrate machine functions, thus extending maintenance intervals which further increases arailability. The797 is also designed to make major components accessible, which effectively reduces removal, installation, and in-truck maintenance time for all drivetrain components. To allo$r technicians to raise the truck body inside a standard-height maintenance building, the canopy ponion ofthe body is hinged so that it can be folded back. In the folded and raised position, rhe797 body height is no higher than the 793, Cateryrllar's 24D-ton-capacity (218-metric ton) mining truck.
Testing Severd 797 minngtrucfts were tested at the Caterpillar proving groun&. Mine evaluation performances were also done during 1999 -2000. The test results are used to correct unexpected problems and fine-tune various components of the truck The797 has been available commercially woddwide since January 2001.
Cernnprr-r-en 797 MrNrxc
PROBLE]tlS 1. Calculate the braking force required to bring the truck from its top
Tnucr
281
speed to firll stop in a of 250 ft. distance 2 Calculate the linear momentum of the truck when frrlly loaded and cruising near its top speed. Calculate t}re force required to bring the truck from its top speed to a firll stop in 5 s. 3. Esdmate the rotational speed of the tires when the truck is moving along near its top speed. 4. Estimate the net torque output of the engine from the following equation:
power: Ial where power is in lb6'ft/s
I:
torque (lbr'ftA)
r,r :
angular speed (rad/s)
CHAPTER
Tn TPERATURE AND ThvTPERATURE-RELATED PannMETERs olten steel is being poured by a ladle into ingot molds for
transport to another processing plant to make steel products. Depending on the carbon content of steel, the
temperature of molten steel can exceed 1300'C.
282
11
ll.f
TieDTpeRATURE es
e Furoervrnnrer, DrupnsroN
283
In this chapter, we will expkin what telnperature, thennal mergt, and heat transfer rnean, We will tahe a close looh at the role of ternperatt re and heat tranrfer in mgineering d.esign and examine how they play hiddcn roles in our euayday liues. Ve will discuss temperatare and its aarious scales, inclu.ding Cekius, Fahrmheit, Ranhine, and Keluin. We will ako erplain a number of ternpnature-related properties ofmaterials, sucb as specifc heat, thermal expansion, and thermal conductiu-
&u
tt"dyi"S this chapter, you will haae lzamed that thermal energ nansfer occurs whmeuer there exists a telnperatare diffnmce whhin an objea or wbenarcr there is a telnperatare dffirence betwem tuto bodies or a body and its sumoundings. Ve wi.ll bri.fly discus tbe uarious rnodes of heat transfCI, what the R-ualue ofinsulation FneAng and what the terrn metabolic rate rneAn* We will ako explain the heating ualues offueh. In order to see how this chapter fts into what you haue bem studying so far, recallfrom our discussion in Chaptn 6 that based on what we know about our pblsical world today, we need seven fundamental or base dimensions to cowectl! ex?ress our natural world, The seuen fundamental dirnmsions are length, mass, time, temperature, elecric current, amount of substance, andluminous intensity. Recall that with the help ofthese base dimensions, u)e can explain all other necesary physical quantities that fustibe ltow nature works. In the preuious cbaptns, you studied the roles of length, mass, and time in mgineering applications and in your eueryday liues. In this chapter, we will discuss the role oftemperature, anotlter fundamenta[ or base, dimension, in mgineering analysi* iry.
11.1 Temperature as a Fundamental Dimension Understanding what temperature means and what its magnitude or value represents is very important in understanding our surroundings. Recall from our discussion in Chapter 6 that in order to describe how cold or hot something is, humans needed a physical quantiry, or physical dimension, which we now refer to as temperature. Think about the important role of temperantre in your weryday lives in describing various states of things. Do you know the answer to some of these questions: \?hat is your body temperature?'What is the room air temperarure? 'lVhat is the temperature of the water that you used this morning to take a shower? What is the remperanue of the air inside your refrigerator that kept the milk cold overnight? Vhat is the remperatrue inside the fregzer section of your refrigerator? \7'hat is the temperature of the air coming out ofyour hair &yer? lVhat is the surface temperanrre ofyour stove's heating element when set on high? lVhat is the surface tempeftture of a 100-V light bulb? !7hat is the average operating temperarure of the elecuonic chips inside your TV or your computer? Vhat is the remperanrre of combustion producs coming out ofyour cart engine? Once you start thinling about the role of temperature in quantifring what goes on in our suroundings, you realize that you could ask hundreds of similar questions.
284
Cnrr.rsRll
TEMpSRATURE^lNoTBupnneruns-RELATEoPanerusrgRs
Regardless of which engineering discipline you are planning to pursue, you need to develop a good undersanding of what is meant by temperature and how it is quantified. Figure 11.1 illusuates some of the systems for which this understanding is imporant. Electronic and computer engineers, when designing computers, televisions, or any electronic equipment, are concernedwith keeping the temperature ofvarious electronic components at a reasonable operating level so that the electronic componen* will function properlp In fact, they use heat sinls (fins) and fans to cool the elecronic chips. Civil engineers need to hane a good understanding of temperature when they design panement, bridges, and other strucnues. They must design the strucnues in such a wty as to allow for expansion and contraction of
I
tigurett.l
Exanples of engineeilng systens
for which understanding of tempemture and heat hansfel is important.
1
1.1
TieMpsRATURE as
285
e FuNpervrENTAL DrMENsroN
1.0 0.8 0.6
o.4 0.2 0.1
Density of air as a function of temperature
:-.ll,
-l--.---
irll \i
bI)
,v
N
0.08 0.06
3
0.M
h
a o a 0.02
F
0.01 0.008
0.006
\: \{
h a
0 5 10t52025
0.004
0.00r
-20
30354045
20 40 60 80
0
I
Figurell.Z
100
120
Temperature (oC)
Temperature ("C)
B
figurell.3
Density of ah at atmospheric pressure as a function of
Vhcosity of SAt 10ltl,
temperature.
temperature.
SAE
101{'30, and SAE 30 oil as a function of
materials, such as concrete and steel, that occur due to changes in the surrounding temperatures. Mechanical engineers design headng ventilating, and air-condidoning (HVAC) equipment to create the comfonable environment in which we rest, *oth *d play. Th"y need to understand heat transfer processes and the propenies of air, including its temperature and moisture content, when designing rhis equipment. Automotive engineers need to hane a good understanding of temperature and heat uansfer rates when designing the cooling system of an engine. Engineers working in a food processing plant need to monitor temperanre of the drying and cooling processes. Materials engineers need to have a good grasp of temperature and heat transfer in order to create materials with desirable propenies. Material propenies are a function of temperature. Physicd and thermal propefties of solids, liquids, and gases vary with temperaffe. For example, as you know, cold air is denser than warm air. The air resistance to your car's motion is greater in winter than it is in summer, provided the car is moving at the same speed. Most of you who live in a cold climate know it is harder to staft your car in the moming in the winter. As you may know, starting difficulty is the result of the viscosity of oil increasing as the temperanre decreases. The variation in density ofairwith temperature is shown in Figure 11.2. The temperature dependence ofSAE 10W SAE 10'!f-30, and SAE 30 oil is shown in Figure 1 1 .3. These are but a few examples of why as engineers you need to hane a good understanding of temperature and its role in design.
286
Cnepmn
11
Tinupnnanunr aNo Tien.prn-lrtrns-Rsr.ATEp Pen-cMnruRs Let us norv examine more closely what we mean by temperature. Temperature provides a me:uure of molecular activity and the internal enerry of an object. Recall that all objecs and living things are made of matter, and matter itself is made up of atoms or chemical elements. Moreover, atoms are combined nanrrally, or in a laboratory sefting, to create molecules. For example, as you already know, water molecules are made of two atoms of hydrogen and one arom of oxygen. Temperature represents the level ofmolecular activity of a substance. The molecules of a substance at a high temperanue are more active than at a lower temperaffe. Perhaps a simple way to visualize this is to imagine the molecules of a gas as being the popcorn in a popcorn PoPPer; the molecules that are at a higher temperanre move, rotate, and bounce around faster than the colder ones in the popper. Therefore, temperature quantifies or provides a measure of how active these molecules are on a microscopic level. For er
11.2 Measuremcnt of Temperature and lts Units Early humans relied on the sense of touch or vision to measrue how cold or how warm something was. In fact, we sdll reh on touch today. \il7hen you are planning to ake a bath, you first turn the hot and cold water on and let the bathtub fill with water. Before you enter the rub, however, you first touch the water to feel how warm it is. Basically, you are using your sense of touch to get an indication of the temperantre. Of course, using touch alone, you cant quantify tlre temperature of water accurately. You cannot say, for example, that the water is x 2I.5"C,
I \-/
ll.2
MresunsMENT or ThlrpsRATURE.ervo
r r , ,
Dark blood red, black red Dark red, blood re4 low red Dark cherry red Medium cherry red Cherry, frrll red Light cherry light red Orange
Light orange
990 1050
rt75 1250 r375
r550 r650 1725 1825
Yellow
Light yellow
r975
$fhite
2200
Soarce:T.Baumeister,
287
Temperature (oF)
Color
' I I
Its UNrrs
a al, Marh's Handbooh.
So note rhe need for a more precise way of quantirying what the temperanrre of something is. Moreover, when we er(press the temperature of water, we need to use a number that is understood by all. In otherwords, we need to establish and use the same units and scales that are un-
derstood by weryone.
Another example of how people relied on their senses to quantify temperature is the way blaclsmiths used to use their eyes to estimate how hot a fire was. They judged the temperature by the color of the burning fuel before they placed the horseshoe or an iron piece in the fire. In fact, this relationship bemreen the color ofheated iron and ia actual temperature has been rneasured and established. Thble 1 1. I shows this relationship. From these examples, you see that our senses are usefirl in judgrng how cold or how hot something is, but they are limited in accuracy and cannot quandry a value for a temperature.
Thus, we need a measuring device that can provide information about the temperature of something more accurately and effectively. This need led to the development oft{rermometers, which are based on thermal expansion or contraction of a fuid, such as alcohol, or a liquid metal, such as mercury. All of you know that almost everything will expand and is length increase when you increase its temperature, and it will contract and i$ length decrease when you decrease its temperature. We will discuss thermal expansion of material in more detail later in tlis chapter. But for now, remember that a mercury thermometer is a temperanrre sensor that works on the principle of expansion or conffaction of mercurywhen its temperanue is changed. Most of you hare seen a thermometer, a graduated glass rod that is filled with mercury. On the Celsius scale, under standard atmospheric conditions, the value zero was arbitrarily assigned to the temperature at which water freezes, and the value of 100 was assigned to the temperature at which water boils. This procedure is called calibration of an instrument and is depicted in Figure lL.4.It is important for you to understand that the numbers were assigned arbiuarilp had someone decided to assign a value of 100 to the ice water temperanue and a value of 1000 to boiling water, we would hare had a very different type oftemperaturescale today! In fact, on a Fahrenheit temperature scale, under standard atmospheric conditions, the temperature at which water freezes is assigned a
288
I
Cner.rrn
ll
TntvIpSRATUREaNp TieLrpsRATUnn-RsLArBp
Penanvrersns
riguren.l
Cali[ration of a mercury thernometer.
va\ue of 32" , and the temperature at which the water boils is assigned a value of 2l2o . The relationship benveen the rwo temperanre scales is given by
rfc)
: 9'llryr1 - n1
g I('r; : ;lreql )-
+
32
flt.0
01.2)
As with other instruments, thermometers evolved over time into rcdayt accurate instruments that can measure temperanrre to 1/100"C increments.
Todaywe also use other changes in properties of mafter, such as electrical resistance or optical or emf (elecuomotive force) changes to measure temperature. These properry changes occur within matter when we change its temperature. Thermocouples and thermistors, shown in Figure 11.5, are examples of temperature-measuring devices that use these propenies. A thermocouple consists of two dissimilar metals. A relatively small volmge oulput is created when a difference in temperature between the two junctions of a thermocouple exists. The small voltage oufput is proponional rc rhe difference in temperature between the two junctions. Two of t}re common combinations of dissimilar metals used in thermocouple wires include iron/ constantan (J-tfp") and copper/constantan (T-gp"). (Th. J and the T are the American Nationd Standards Institute (ANSD symbols used to refer to these tirermocouple wires.) A thermistor is a temperature-sensing dwice composed of a semiconductor material with such propenies that a small temperature change creates large changes in the electrical resistance of the material. Therefore, the elecrical resistance of a thermistor is correlated to a temperature value.
ll.2
MsAsuRErvrENT
or
TrurpsRATuRE ano
Irs UNrrs
289
TCopper
\Constantan
,ffi({[ fl 3 \\
ilflil
A{/fl
G=W
T-type Thermocouple
il
figuretl.5
Eramples of temperature'
Thermistor
lneasuring devices.
Absolute Zero Temperature Because both the Celsius and the Fahrenheit scales are arbitrarily defined, as we hare explained, scientists recognized a need for a better temperaflre scale. This need led to the definition of an absolute scale, the Kelvin and Rankine scales, which are based on the behardor of an ideal gas.
You have observed what happens to the pressure inside your cart tires during a cold winter day, or what happens ro the air pressure inside a basketball if it is left ouside during a cold night. At a given pressure, as the temperature of an ideal gas is decreased, its volumewill also decrease. W'ell, for gases under cerain conditions, there is a relationship between the pressure of the gas, its volume, and its remperaft.re as given by what is commonly called the idzal gas law.The ideal gas larv is given by
PV=
rnRT
CI|.3)
where
: 7: rn : P
R
:
7:
absolute pressure of the gas (Pa or
volume of the gas (m3 or mass (kg gas
lb/ff)
ff)
or lb.)
consranr
/ | .r ft.lb\ \-, - ru-.RT
absolute temperature (Kelvin or Rankine, which we will explain soon)
Consider the following experiment. Imagine that we have filled a rigid container (capsule) with a gas. The container is conneced to a pressure gauge that reads the absolute pressure of
294
I
CrHprun
ll Truprurunr nNo TienapsRATuRn-Rrl-trrsp P$erusrERs
rigmell.6
The capsule used in the absolute
zero temperature eranple.
,h.
g
inside the container, as shourn in Figure 11.6. Moreover, imagine thar we immerse the in a surroundingwhose temperature we can lower. Of course, ifwe allow enough time, the temperature of the gas inside the container vrill reach the rcmperanue of its surroundings. Also keep in mind that because the gas is contained inside a rigid container, it has a constanr volume and a constant mass. Now, what happens to the pressure of the gas inside the container as indicated by the pressure gauge as we decrease the surrounding temperature? The pressure will decrease as the temperature of the gas is decreased" Ve can determine the relationship berween the pressrue and the remperaflue using rhe ideal gas larv, Equation (11.3). Now, let us proceed to run a series of enperiments. Saning with the surrounding temper' ature equal to some reference temperanue-say, f-and allowing enough time for the container to reach equilibrium with the surroundings, we then record the corresponding pressure ofthe gas, {. The pressure and temperarure of the gas are related according to the ideal gas law: capsule
P.V
t'-
flr.4)
*R
Now imagine that we lower the surrounding temperature to 21, once equilibrium is reached, recording the pressure of the gas and denoting the reading by P1. Because the temperature of th. g.. was lowered, the pressure would be lowered as well.
P,V
tl-
Dividing Equation (t1.5) bf Equation (IL.4), we T1
flr.s)
*R
_
T
have
lv mR
Py
flr"6)
mR
And after canceling our the ra the R, and the I{ we have
Tr:p,
T1
(1r.7)
ll.2 Tt=
MgAsuREMTNT
or TslrpnRATuREeNp Irs Uurs
291
(11.8)
'(+)
Equadon (11.8) establishes a reladonship benveen the temperature of the gas, its pressure, and the reference pressure and temperature. If we were to proceed by lowering the surrounding temperanue, a lower gas pressrue would result, and if we were to extrapolate the resuls of our er
r(K)
: rfc) + 273.15
fll.e)
The relationship berween degree Rankine ('R) and degree Fahrenheit ("F) in U.S. Customary units is
ZfR)
:
7(.F) +
459.67
fi1.10)
Note that the orperiment qurnot be completely carried out, because as the temperature of decreases, it reaches a point where it will liquify and thus the ideal gas lar will not be valid. This is the reason for extrapolating the result to obtain a theoretical absolute zero temperanrre. It is also important to note that while there is a limit as to how cold something can be, there is no theoretical upper limit as to how hot something can be. Finally, we citn also esablish a relationship bet'reen the degree Rankine and the Kelvin by
th"
g*
?"(K)
:2r('R)
(ll.ll)
9
Equadons (11.9) and (1 1.10), unless you are dealing with very precise experiments, when convening from degree Celsius to Kelvin, you can round down the 273,15 to 273. The same is true when convening from degree Fahrenheit to Rankine; round up the 459,67 to 460.
In
is the equivalent value of T : 50oC in degrees Fahrenheit, Rankine, and Kelvin? Wb can use Equadons (11.2), (11.9), and (11.10):
'Vhat
c) /c)\ : iT("C\ * 32 : t ; l(50) I 32 : r22"F Ifn; = f('F) + 460: r22 t 460:582'R f(K) : fCC) + 273: 50 + 273:323K
f('F)
Note that we also could hare convened the 582oR to Kelvin direcdy using Equation (l
r(K)
t ::r1"n; 9
: (i),'', --_ - -
----:-
:323K . ----t-.t-
--
1.1
l):
292
Crreprrn
ll
T&rpnnAruns aNo ThrapsRATUns-RELArnp Panervnruns On a summer day, in Phoenix, Arnona, the inside room temperature is maintained at 68oF while the outdoor air temperanue is a sizzling 110"F. lZhat is the outdoor-indoor remperarure difference in (a) degree Fahrenheit, (b) degree Rankine, G) degre" Celsius, and (d) Kelvin? Is a 1o temperature dlfference in Celsius equal to a 1o temperature difference in Kelvin, and is a
Lo temperature dlfference
If
so, why?
in Fahrenheit equal to a 1o temperature difference in
Rankine?
I7e will first answer these questions the long way, and then we will discuss the short way.
(*)
4ora-,
-
4oo*,
:
110'F
-
68"F
=
(b) 4"a-('R) : 4o,a*.(oF) + 460 : Z-u*,fR) 4urd-,
-
-
7La*,fF)
4na*,
:
*
570'R
460
-
:
42"F 110
68
* 460:
+ 460:
528"R:
570"R
528'R
42"R
Note that the temperature difference expressed in degrees Fahrenheit is equal to the temperature difference expressed in degrees Rankine. ll (.) 4",a*,('C) :;(4,,a*,("F)
Z-a"*("C) 4o.d*,
-
(d) 4*a*(K)
- 32):;(tto -
32):43.t"a
55 :;(Z.a*,('F) - 32):;(eS - 32):29"a Tiod*"
:
:
43,3"C
- zoc :
23.3C
4u.a-('C) + 273 : 43.3 + 273: 316.3K
Z-a*$) = 4.a"-(.C) + 273 : 20 + 273 : 293K 4,od*, - I-d*, : 316.3K - 293K: 23.3K
It should be clear by now that a 1o temperanue difference in Celsius is equal to a 1o temPeraflrre difference in Kelvin, and a 1o temperature &fference in Fahrenheit is equal to a lo temperature difference in Rankine. Of course, this reladonship is true because when you are compudng the difference between two temperatures and converting to the absolute temperature scale, you are adding the same base value to each temperature. For example, to compute the temperature difference in degrees Rankine between two temperatures 7, and 72 given in degrees Fahrenheit, you first add,460 to each temperature to converr T1 and T2 ftom degrees Fahrenheit to degrees Rankine. This step is shown next.
n
("R)
4fR)-4(.R)=tffir
4 fR)
-tffir
=4fF)+460_72(F)_46a
11.3
TieMpnRATURE
DtrrsRsxcsAND Hner TneNsmn
293
And simpliSing this relationship leads to
4fR)
- AfR):4fF) -
rr("F)
Also note that you could have convened the temperature difference in degrees Rankine to Kelvin direcdy in the following manner:
Ar(K)
: llrfn) :
G)ro
11.3 Temperature Difference and
:23.3K Fleat Transfen
Thermal energy uansfer occurs whenever there exists a temPerature difference within an object, or whenwer rhere is a remperature difference between two bodies, or a temperature difference berween a body and its surroundings. This form of energy uansfer that occurs between bodies of different temperaflues is called heat nanrfer. Additionally, heat dways fows from a high-temperature region to a low-temperature region. This statement can be confirmed by observation of our surroundings. When hot coffee in a cup is left in a surrounding such as a room with a lower temperature, the coffee cools down. Thermal energy transfer akes place from the hot coffee through the cup and from its open surface to the surrounding room air. The thermal enerry uansfer occurs as long as there is a temperature difference between the coffee and its surroundings. At this point, make sure you underssnd the difference between ternpnature atd heat lJrat is a form of energy that is uansferred &om one region to the nent region as a result of a temperature difference between the regions, whereas temperanue represents on a macroscopic level, by a single number, the level of microscopic molecular movement in a region. There are three units that are commonly used to quantify thermal energ;r: (1) the British thermal unit (Btu), (2) the calorie, and (3) the joule. Ttrc Britishtbermalmit(Bn) is defined as the amount of hear required to raise the temperature of I pound mass (1 lb-) of water by 1 degree Fahrenheit (1'F). The calorie is defined as the amount of heat required to raise the remperature of 1 gam ofwater by 1"C. Note, howsver, that the energy c,ontent of food is typically expressedin Calories,which is equal to 1000 calories. In SI units, no distinction is made between the units of thermal energ;r and mechanical energy and therefore energ;r is defined in terms of the fundamental dimensions of mass, length, and time.'We will discuss this in more deail in Chapter 13. In the SI Sptem of Unis, thejouhis the unit of energy and is defined as 1
joule
:
1
N.m :
1
kg.m2ls2
The conversion factors among various units of heat are given in Thble 11.2.
Relationship Between the Units of Thermal Enerry
t Btu
:
Relationship Between the Unito
I W= I \7'=
1055J
lBtu = 252 cal 1 cal = 4.186J
of
Thermal Enerry per Unit Time @ower)
l
1Jls 3.4123 Btu/h
cal/s = 4.186.W
294
CHer"rsR
1l
T$TpSRATURE eNo TielrpsRATunp-RslArso PelervrsrERs
Using the units of energy and time, show that t wam
inThble 11.2.
@
is equal to 3.4123 Btu/h, as shown
,w:'(lX##X#) :34rn+ ,i
--'1-n!iEh,i?Iir?!,Iir.i1ra?r
--ia-li:aillrrii--------.---:l
-
In many parts of *re United
States, in order to keep a house warm in the winter months, a gas a gas furnace puts out 60,000 Btu/h to compensare for heat loss from a house, what is the equivalent value of the thermal power (energ;r per unit time) ouqut of the
fiunace is used.
If
firrnace in watts? From Thble 1I.2, you know that:
1
Btu
:
1055 J; you also know that
t h : 3600
s.
Sub-
stituting for these values,
-= 17'583\;/I\ : a=60.000(nru)/t'o::l)f-i!-) h r nru \ /\
Or you could
a
)
./\1,600.7
have used the direct conversion factor benveen
: 6o,00o(+)
(#b)
- --.---,riE
-- -
-liiriTlili-l:,ltriil
:,,,r,usr
:
77'583'v
Btu/h andlM
17 583
as
:
r7'583kv
shown:
kw
li!-ariir'lllililll?iri'aiAti :-... ---.-llrt::iiU,
li----------:--------j.rij
Now that you know heat transfer or thermal energy tansfer occurs as a result of temperature difference in an object or between objects, let us look at different modes ofheat transfer. There are tfuee different mechanisms by which enerry is transferred from a high-temperature region to a low-temperanr.re region. These are referred to as the rnodzs of hex transfer. The three modes ofheat transfer are conduction, convection. and radiadon.
eonduction Condactianrefers to that mode ofheat transfer that occurs when
a
temperanue difference (gra'
dient) exisr in a medium. The energy is transponed within t}re medium from the region with more-energetic molecules to the region with less-energetic molecules. Of course, it is the interaction of the molecules with their neighbors that makes the transfer of enerry possible. To bemer demonstrate the idea of molecular interactions, consider the following example of conduction heat transfer. All of you hane experienced what happens when you heat up some soup in an aluminum container on a stove. \Vh.y do the handles or the lid of the soup conrainer ger hot, wen .h"ugh the handles and the lid are not in direct contact with the heating element? \?'ell, let us examine what is happening.
11.3
TslrprRATURE DrrrsRENcsANp
Hrer Tnensrnn
295
lleat transfer
Cold eir
il
rigurell.T
Conduction heat hansfer tftrough a glass window.
Because
of the energy tansfer from the heating element, the molecules of the conainer
in the region near the heating element are more energetic than those molecules further away. The more energetic molecules share or transfer some of their enerry to the neighboring regions, and the neighboring regions do the same thing, until the energy transfer wentually reaches the handles and the lid of the container. The energ;r is transponed from the hightemperarrue region to the low-temperature region by molecular activity. Th. rate of heat transfer by conduction is given by Fouri.erb laut,which states that the rate of heat transfer thtotgh a material is proponional to the temperatue difference, normal $ea A.htongh which heat
ffansfer occrus, and the rype of material involved. The laqr also states that the heat transfer rate is inversely proportional to the material thickness over which the temperature difference exists. For example, referring to Figure 11.7, we can write the Fouriert lanr for a single-glass
window
as
q:
T,_7" u--T-
flr.r2)
where
q: z
:
./ : Tt
- Tr:
hffitransfer rate (in \7'or Btu/h) thermal
cond uctivitY
( v \ - . .c -
Btu \ h. ft. .F /
cross-sectional area normal to heat
fow (m2 or fl)
temPerature dlfference across the materid
Z = material thickness (m or ft)
ofZ thickness ('C or'F)
?96
Crrl'r.rsnll
TISMpSRATUREeNoTsMpsRATuRE-RrrArsoPenelrsrERs
Material
Air
Thetmal Conductivity (V/m . k)
(at atmospheric pressure)
Aluminum (pure) Aluminum alloy'2\24:T6 (4.5o/o apper, 1.5% magnesium,
0.0263 237 177
0.60lo manganese)
Asphah
0.a62
Brorvs (90o/o copper,
100/o
aluminum)
<,,
Brass (709o copper, 3\o/o zinc)
n0
Brick (fue clay)
1.0
Concrete Copper (pure)
t.4 40r
Glass
r.4
Gold
317 4.2
Human fat layer Human muscle Human skin
o.4l 0.37
Iron (pure) Stainless steels
(AISI 302,304, 316,347)
Lead Paper
Platinum (pure)
80.2 15,1, L4.9,13,4, r4,2 35.3 0.18
71.6 0.27
Sand
Silicon
1,48
Silver
429
Znc
116 0.61
Warer (liquid)
The temperature difference T,
-
temPelahtre gradi"nr. Again, keep
T, over material thickness is commonly reGrred to as the in mind that a temperaflre gradient must erdsr in order for
heat transfer to occur.
Thermal conductivity is a property of materials that shows how good the material is in uansferring thermal energy (heat) from a high-temperature region to a lowtemperantre region within the material. In general, solids have a higher thermal conductiviry than liquids, and liquids have a higher thermal conductivity than gases. The thermal conduc-
tivity of some materials
is given
in Table 11.3.
Calculate the heat uansfer rate from a single-pane glass window with an inside sufice temPeratue of approximately 20"C and an outside surface temperature of 5"C. The glass is I m tall' 1.8 m wide, and 8 mm fii"k, a shown in Figure 11.8. The thermal conductivity of the glass is approximately b : L.4 V/m. K
z:s(mm)( ' '\
t*
):0.008m
1000 mm,/
f
rs"C \ \ n\.
I lass I I II lrm II
tl
ll
\--\Li-d|| )b-.-
ill
figure
n.O
13
ThrvrpnncJruRr DrrrnnnNcs er.ro
A: (l m)(1.4 m) : Tr- Tr:2ooc -
5"c =
Hner
TneNsrun
297
1.8 m2
15"c:
15K
Note, as we orplained before, al5"C temperaure drfference is equal to a 15 K temperature difference. Substituting for the values of h, A, (\ - Tr), and Z in Equation (11.12), we have
..r,- r, . ..( rv \.. ^ ,,( r5K \ q: *T:0.4)(;kjr"*')(ffi):+tzsw (Be carefirl with lowercase &, denoting thermal conductivity, and the capital Kelvin, an absolute temperature scde.)
K
representing
li The single'pane llindow of
ij Erample
11.5.
ili---.-..ill!i----ri:rl!i1,rr:l=.-it]lir.r:Ih---ri.ll-
Thermal Resistanee In this secdon, we will explain whar the R-values of insulating materials mean. Most ofyou understand dre imporance of having a well-insulated house because the better insulated a house is, the less the heating or cooling cost of the house. For example, you may have heard that in order to reduce heat loss through the attic, some people add enough insulation to their attic so that the R-value of insulation is 40. But what does the R-value of 40 mean, and what does the R-value of an insulating material mean in general? Let us stan by rearranging Equation (11.12) in the following manner. Starting with Equation (ll.l2),
T;_7. q: kA_-T-
(flr2)
and rearranging it, we hane
q--- Tt-L T, :
u
temperafure difference
(ixJ3)
thermal resistance
and thermal resistance : L/hA Figure 11.9 depica the idea of thermal resistance and how it is related to the materialt thickness, area, and thermal conductivity. When examining Equation (11.13), you should note the following: (1) The heat transfer (flow) rate is direcdy proponional to the temperarure difference; (2) the hear fow rate is inversely proportional to the thermal resistance-the higher the value of thermal resi$ance, the lower the heat tansfer rate will be. \Vhen expressing Fourier's law in the form of Equation (11.13), we are making an analory between the flow of hear and the fow of elecricity in awire. Ohm's lont, which relates the voltage V to crurent .I and the elecmical resistance R", is analogous to heat fow. Ohm's law is expressed as
V:
R"I
(11.X4)
298
Cnnprnn
1l
TEMpsRAruRE aNo TTMpsRATuRB-RBr.ersn PanevrrsRs
Tr qAAAr'<-
T,t
L KA
I
rigumll.g
A slab of
naterial and its thermal resistance.
or
t:! & Ve will
flr.r5)
Ohmt laq/ in more detail in Chapter 12. Comparing Equarion (11.13) to 1.15)' note that the heat flow is analogous to electric currenr, the temperature difference to voltage, and dre thermal resistance to electrical resistance. Now turning our attention back to thermal resistance and Equarion (11.13), we realize that the thermal resistance for a unit area of a material is defined as discuss
Equation
(1
R,:*
(fl.r6)
or oF. h/Bru fR. h/Btu). Vhen Equation (11.1@ is orpressed per unit area of the marerial, it is rderred to as the R-value or the R-facor.
R'
has the
units of
'C/\7 (K/W
L
ft:
0r.r7)
b
where R has the units
m2.oC
of
/*t.r\
\v/
or
#.
"F
Btu h
/ ul."* \ lB*l \h
/
Note that neither R' nor R is dimensionless and sometimes the R-values are orpressed per unit thickness. The .R-value or R-factor of a material provides a mensure of resisance ro heat flow: The higher the value, the more resistance to heat flow the material offers. Finally, when the materials used for insulation pulposes consist of various components, rhe total R-value of the composite material is the sum of resistance offered by the various components.
11.3 Thlrprneruns
299
DFrsRENcs ANo Hnar Tnerrrsrsn
Determine the thermal resistance ft' and the R-value for the glass window of Example 11.5. The thermal resistance R' and the R-value of the window can be determined from Equations (11.10 and (11.17), respecdv+
R'
0.008 m : L kA: .1.4[ ^/ \r \*
K
:
o'oo317t
-/(1.8 m')
And the R-value or the R-facor for the given glass pane is
L *:7:., lili---.----lirlil.itil,lirir,ii]aiilii
0.008
m :o'0057m2. K \?
rl-t-\ t'*\*'K/ - -.---r ---
-. :liirll,,i
For Example 11.6, convert the thermal resistance R' and R-factor results from SI units to U.S. Customary units.
R'
R
=
:
ooo3L't*)(;,;*)F) :
+--
o
oosz(g')
5'\''!x
1o-4
"R Btu h
(rfu X+) (+*)'
"R.
= 0.01
ff
B* h
Or, if the R-value is e4pressed in terms of in2,
a
: !h: o ooiT(#l
(r+X+) (t*t)' :, 43H
ir A double-pane glass window consists of rwo pieces of glass, each having a thickness of 8 mm, with a thermal conductivity of h : 1.4 !7/m . IC The nvo glass panes are separated by an g"p of 10 mm, as shown in Figure 11.10. Assuming the thermal conducdvity of air to be '\V'lm. "ir h : 0.025 K determine the total R-value for this window. The toal thermal resistance of the window is obained by adding the resistance offered by each pane of glass and the air gap in the following nanner: R,ool
= Rrt^ f R6 .| R4* =
Ld^
, L*, zr*
h*- h*- h*
300
Cnerrsnll
TrMpsRATr,rRBaNpTh:,llrpsRATrrRn-RnanspPenanasrERs
I
Figurc
ll.t0
lhe double-pane glass window of Eralnple
11.8.
substituting for Lg"o, hgoo, La,, han we hane
R..
:
. i#S\ i##\ -#(+T. \.*'K/ \-.K/ [m'K-/ 1'4
0'025
:'*
1'4
(#)
Note the units of R-value. The R-value for the double-pane glass window in U.S. Customary units is
R.-
:
" (#)(rr+X+)(Tf)'
:
"H
fu you can see from the results of Examples I 1.6 and I 1.8, ordinary glass windows do not offer much resistance to heat fow. To increase the R-value of windows, some manufacnrrers make windows that use triple glass panes and fill the spacing between the glass panes with Argon gas. See Example 11.11 for a sample calculation for total thermal resistance of a typical exterior frame wall of a house consisting of siding, sheathing insulation material, and gypsum wallboard (drywall).
eonvection Conaecti.onhat transfer occur when
a
fuid
(a gas or a
liquid) in motion comes into contact
with a solid surface whose temperature differs from the moving fuid. For example, on a hot summer day when you sit in front of a fan to cool down, the heat transfer rate that occurs from your warm body to the cooler moving air is by convection. Or when you are cooling a hot food,
11.3 Thuprnerunr DrrrnnsxcnaNo Hnar Turvsrun
CPU being cooled by a
301
fan.
such as freshly baked cookies, by blowing on it, you are using the principles ofconvection heat transfer. The cooling of computer chips by blowing air across them is anotler example of cooling something by convection heat transfer. There are two broad areas of convection heat transferforced canaec'tion andfree (nawral) conaec'tion. Forced convecdon refers to situations where
the flow offluid is caused or forced by a fan or a pump. Free convection, on the other hand, refers to situations where the flow of fluid occurs naturally due to densityvariation in the fluid. Of course, the densiry variadon is caused by the temperature distribution within the fuid. \9hen you learre a hot pie to cool on the kitchen counter, the heat transfer is by natural convection. The heat loss from t*re exterior surfaces ofthe hot oven is also by nanrral convection. To cite another example of natural convection, the large electrical transformers that sit outdoors at power substations are cooled by natural convection (on a calm day) as well. For both the forced and the ftee convection situations, the overall heat transfer rate between the fluid and the surface is governed by Nemon's law of cooling, which is grven by
q:
hA(T"
- Ti
(1118)
'Wm2
. K (or Btu/hr . fi3 . "R), .4 is the area of the wherc h is the heat transfer coefficient in exposed surface in m2 (fC), T"is the surface temperature in "C ("F), and Irepresents the temperaflue of moving fluid in 'C fF). The value of the heat transfer coefficient for a panicular situation is determined from experimental correlation; these values are available in many books about heat uansfer. At this suge ofyour education, you need not be concerned about how to obtain the numerical values of heat transfer coefficients. However, it is imporrant for you to know that the value ofZ is higher for forced convection than it is for free convection. Of course, you already know this! \7hen you are trying to cool down rapidly, do you sit in front ofa fan or do you sit in an area of the room where the air is still? Moreover, the heat uansfer coef6cient E is higher for liquids than it is for gases. Have you noticed that you can walk around comfonably in a T'shin when the outdoor air temperature is 70oF, but if you went into a swimming
302
CruprBnl1 ThlrpsRATUREeNpThtrpsRATURE-Rsr-amoPenarvrsruRs
Heat Transfer Coefficient,
i(W/mz. K)
Conrrection Tnrc
Heat Transfer Coefficient,
h (Btru,tu,.
F.'R)
Free Conaetion
2 to25 50 to 1000
Gases
Liquids
0.35 to 4.4 8.8 to 175
Forced Canuection
25 to250
Gases
Liquids
100 to 20,000
,4.4to 44 17.6 to 35A0
pool whose water temperanue was 70oR you would feel cold? That is because the liquid water has a higher heat transfer coefficient than does air, and therefore, according to Newton's larp of cooling Equation ( I 1 . I 8) , water removes more heat from your body. The typical range of heat transfer coefficient values is given in Thble 11.4. In the field ofheat transfer, it is also common to define a resistance term for the convecdon process, similar to the.R-value in conduction, The thermal convection resistance is defined as:
R,--h
(r1J9)
oF. or h/Bru fR. h/Btu). Equation (11.19) is commonly expressed per unit area of solid surface exposure and, is called f/m resistance or filrn
Again, R' has the units
ofoC/If (Kl\V)
cofficient
*:zI
flr.20)
where R has the units
m2."C
of
/*'.r\
,u, \or/
or
rl."p /ry'..R\ Btu I Bru I
h \r'/
It is important to reaJize oncs again that neither.R' nor R is dimensionless, and rhey provide a measure of resistance to heat fow; the higher the values of.R, the more resistance to heat fow to or from the surrounding fluid.
Determine the heat transfer rate by convection from an electronic chip whose surhce temperature is 35"C and has an exposed surface area of 9 The temperature of the surounding "m2. air is 20"C. The heat transfer coefficient for this situadon is h : 40 W/m2 . IC
A: er"-,1(n#*;) : o.ooou*, f"
- T:
35C
-
20"C
=
15'C
:
15
K
11.3
TprrpsRATrrRE DrrrpnrNcs AND
Hnar TneNsrsn
303
Note again that the 15oC temperature difference is equal to a 15 K temperature difference, from the chip by substituting for the value s of h, A, and (7"- T) in Equation (11.18), which results in
i"
\?'e can determine the heat transfer rate
il tj
4*nv"don = hA(T"
- r) :40(;xR),0.000e
#)(t5K)
:
0.54v
i1 'tl,
t-'
-: ...I
-t - . :t----:-
:: _:,ii::-
f-:ar-:--- -- - 1.-'r'r--,.--a-: --
: r;:::--;i:-_-:
l
:n.. --:,-...:
Calculate the R-factor (film resistance) for the following situations: (a) wind blowing over
h
"F . fe, and (b) still air inside a room ne:u awall, b For the situation where wind is blowing over a wall:
= 5.88 Btu/h.
ft: h
5'88
Bat
^:
I.47
a
wall,
Btu/h.'F. ff.
: 0.17h. ff. "F Btu
h. n3. .F
And for still air inside a room, near
t -i:
:
_ Bru "" h.ff..F 147-
a
wall:
: 0.68h.#.'F f-'ru
A typical exterior frame wall (made up of 2 X 4 studs) of a house contains the materials shown in Thble 1 1.5 and in Figure 1 1. 1 1. For most residential buildings, the inside room temperature is kept around 70oF. fusuming an outside temperaffe of 20'F and an exposed area of I50 ft, we are interested in determining the heat loss through the wall. In general, the heat loss through the walls, windows, doors, or roof of a building occurs due to conducsion heat losses through the building materials-including siding, insulation
Thermal Resistance
(h.
P.
oF/Btu)
I. Outside film resistance
o.L7
(winter, 15 mphwind) 1 Siding, wood (1,12x 8l*pp"O
0.81
4
4.
(ll2 n. regular) Insulation batr $ - t in) Sheathing
Gypsum wallboard (1/2 in.) 6. Inside film resistance (winter)
1.32 11.0 0.45 0.68
lL
rigurett.tl
l{all layen to accompany
Table 11.5.
304
Crr,lr"rsR
11
PNlvsrERs
TSMppRATURE eNp TsMppRATtrRE-RsLArso
material, gypsum wallboard (drywatl), glass, and so on-and convection losses tfuough the wall to the indoor warm air and the outdoor cold air. The totd resistance to heat fow is the sum of resistances offered by each component in the path of heat flow. For a plane wall we can write: surfaces exposed
: x' ': 4*ia"
- 4rtia" (Z*t* -
T""or*)A
>"
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 Stefan-Bolv,mann constaint (c = 5.67 X . 10-8 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 in-depth understanding of the underlying concep$ and geometry of the problem.
306
Cner.ren
1l
TSMpTRATuRE eNp
Truprnarunr-Rrr,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
Surfac€
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 Surfrc€ (w)
' q ='esAT: (0.9)(5.67 (0.9)(5.67 (0.9)(5.67 (0.9)(5.67
x 10-8X400X32r4 x 10-)(400)(313)a : x rO 1(400X303)a : x 10-8X400X298)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 air-conditioning (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
TyP-c
'.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 Air-Conditioning Engineen.
€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 €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
TieMpBRATURxeNoTnMpsRATURE-Rrr.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, short-sleeve shirt Trousers, shon-sleeve shift Tiousers, long-sleeve shin
0.36 0.57
Trousers, long-sleeve shirt, plus suit jacker
0.96
Same as above, plus vest and
0.51
l.t4
T-shirt
Sweatpants, sweatshirt
I I
KneeJength skin, short-slewe shin, panty hose, sandals Knee-lengh skirt, short-sleeve shirt, full slip, panty hose KneeJength skirt, long-sleeve shin, halfslip, panty hose, long-slewe 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 Air-Conditioning 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 TgMpsn-erunn-Rnr,rrno MersRrAL PnopsRTlss
309
Some Temperature-Related Material Properties
Thermal Expansion As we mentioned earlier in this chapter, accounting for thermal e-Fansion 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 expand-increase in
length-and if the temperature of the material is decreased, it will contract-decrease 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
ThtvrpnnerunsaNpTielrpsRATuRE-RnranspPenanusrsRs
(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 (hard-rolled) Steel (soft-rolle0 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 10-6
2.9 5.5 3.3
5.9 X 10-6 8.0 X 10-6 5.0 x 10-6 1.8 X 10-6 4,5 x l}-G
L0-6-5.0
x 2.9 x
13.4
X
x X
10-6 10-6 10-6
4.4x rc-6 2.8
X
10-6
x 10-6 2.5 x 10-6 t.4x 10-6-2.8 x 1.0
x 10-6
7.4X
LO-6
10-6
1.0-6
x 10-6 x 10-6 x 10-6 L.7 x LO-6 1.5 x 10-6 1.7 x LO-6
l}-G
1.6
5.6 X 10-6 6.3 x 10-e 3.0 X 10-6 2J X 10-6 3,0 X 10-6
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 1000-ft-long 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 10-6 lfF. Using Equation (11.25), we have
AL: arl 67: l!uru,=.,,,,-,..-
,,.
liiiiiaiiitrltlii,iilli''
:!.
(2.9
1
X
10-6 1l'F)(1000 ft)(100"F)
=
0.29
ft :
-- --l:..:,
3.48 in. ---a.r:'rr,r!. ,i.r,i-iiL,:iir,::i:
Speeifie Heat Have you noticed t-hat 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 ThupBnerunp-Rsr^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(7|66- 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).
i-i
i,
l-:_.::...:.
i
Spt$fic lleat (Y[S. IO
Maretial
1
eir(atatrngsptrgricpressure) Alqgrinuq (pure) Alumiirum'alloy'2024.T6 (4.5% copper,l.5 7o Mg; 0.6% Mn) ' r AsPhali ',, |
,
,
i
i
903
)
875
l
,
j
l
I
Bronze (907o
spper,
1
0olo.
Brass (707o coppetl 30o/o
aluminutn)
tnc) ,
,.
380
:
Brick (fue clay)
Concrercl
l"
l.
:
r-oPp€r (P$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:V-ater(iquid)
:"
arso
i 1
447
48b,477,46t,n
,
, r
r
"
312
CnarrsR
1l
Truprnarunp ANo Thtvrppneruns-Rr,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 l0-5 m3)(zZoo kg/*') :0.191kg
Volume
F,h*,r:
mc(7|66
-
X t0-5
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 Air-Conditioning 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
-o-p*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 by-products 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 TnMpsRATURE-Rrranro Panaurrsns Calculate the total amount of thermal energy released when 60 is burned inside a gas fumace.
Eh*"r Ii .*liiGi*!i?ir,I-r:--:-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 R-value or R-factor 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" R-type
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 loose-fill cellulose insulating material provides a table showing the relationship between _
R-value
(units?)
R40 R-32
R-24
R-r9 R.T3
Thickness
ll 9 6.5 5.25 3.5
:__ _
(in.)
i
',
316 Cner"rnnll
Thupnn-rrtrnnnNpTieMpnRATunr-Rnr-etnoPenelrsrERs
the thickness of the material and its R-value. 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 R-values 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.
R-20 Calculate the R-value for the following materials:
a. 4-in.-thick brick
b. lO-cm-thick brick c. l2-in.-thick concrete slab d. 20-cm-thick concrete slab e. l-cm-thick 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 5-mJong 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 2.5-cm 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 TetvrprRATunr-Rrrerro 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 aver€e per kVh usage of electricity, If a hot-air 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 i|r
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 length-related parameters such as area and volume, the fundamental dimension mass and mass-related parameters, the fundamental dimension time and time-related parameters, and the fundamental dimension temperature and temperature-related 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 10-7 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 opposite-charge 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 high-pressure region to the lower-pressure 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 high-temperature 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 nickel-cadmium (NiCd), and nickel-metal 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
f-a
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.5-volt 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 hand-held 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 su-ris 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 power-generating 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 low-pressure 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,rc-rnrc Cunnnnr aNo RBr-lrBp 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 long-distance 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
Determine the value of current i3 in the circuit shon'n in Figure 12.6.
I
figuru12.6 ThechcuitforErarnphlz.l.
Ve can solve this simple problem by applying Kirchhofft cunent law to nodelV This leads
to:
*
i1:
t2
4=
24
i3
Cnarrrn
326
12
Er,Bctnrc CunnrNr nxp Rrr,nrso PenarvtrsRs
Residential Power Distributisn example of a typical residential power distribution system is shown in Figure 12.7, which gives examples of amperage requirements for oudets, ligh*, kitchen appliances, and cenual air conditioning. To wire a building, an electrical plan for *re building must first be dweloped. In the plan the location and the types of switches and oudets, including oudets for the range and dryer, must be specified.'W'e will drscuss engineering symbols in more detail in Chapter 16. For now, examples of elecuical symbols used in ahouse plan are shown in Figure 12.8. Moreover, an example of an elecrical plan for a residential building is shown in Figure 12.9.
An
12.4 Electrical Circuits and Components An elzc*i.cal circuit refers to the combination of various electrical components that
are
connected together. Examples of electrical componens include wires (conductors), switches, oudets, resistors, and capacitors. First, let us mke a closer look at electrical wires. In a wire the resistance to electrical crurent depends on the material from which the wire is made and its length, diameter, and temperature. Materials show varying amount of resistance to the flow of electric current. Resistittity is a measure of resistance of a piece of material to electric crurent.
Pool outlets
Conduit to meter
Basement fighting and outlets
First floor lighting
Garage outlets Furnace
Service cable to house
Meter Shop equipment
Circuit breaker Livingifamily room Bathroom fighting and outlets Bedroom outlets Kitchen appliances
200-amp main circuit breaker
and distribution panel
204
?TA
15A
20A
A
?TA
15A
20A
15A
30A
Kitchen lighting
t
Figure
S ource
326
12.7
Oven-range Dishwasher
204 20A. 15A
An erample of an elechical dhhibution system for a residential building.
: AAVIM, Ehctrica I Wring.
Utility room
Sink disposal unit
Kitchen appliances
Kitchen outlets
Utilitv room
20A
50A
Water heater
Cenfral air conditioner
Common Electrical Symbols gsiringoutlet
O
serieennmcepanel
-O n vPS
Wail outlet
S
Single-poleswitch
Ceiling outlet with pull
Sa
Double-pole swirch
_^ vPs
Walloudetwithpull
33
3-way switch
€ -ra €wP
Duplex convenience
54
4-wayswitch
Sp
Switch with pilot light
switch switch
ooti"t Weatherprmf cnnvenience orrllet Convenience outlet
_6 -'vl'3 l=sinsle 3=triole
|T|
Pushbutton
€"
Rangeoutlet
Eb
B"lorchim*s
_.a --vs
Convenieneoutlet with switch
l(
Telephone
€o
Dryeroutlet
El
€
sPlit-wtuedduPlex
O E
I
YZ
s'--Jwitcnwfing
Special-purposeoutlet
1------7 Fluorescentceiling
Ebctricdooropener
4)v 4)
figure
Souree:
Televisionoutlet
12.8
rrnre
Fluorescentwall qiYhrrA
tranples of electrical symhols in a house plan.
AA\IM, E bctrical Vlring.
General-Purpose Circuits Circuit no. 2
Circuit no.
1 from Service Entrance Panel (SEP)
ftom
Circuit no. 3 from SEP
I
figure
S o urce :
12.9
Circuit no.4 fromSEP
An erample of an electrical plan for a house.
AAWM, Ebaical Wii ng
Circuit no. 5 Aom SEP
328
CrHprsR
12
Er,scrRrc Cunnnrvrexp
Rrratro
Pana.tvrereRs
Resistance
l
O)
l
(ohms,
Met4l Aluminum
v.,41
Brass
66.80
Copper
r0.37
Gold Iron
14.7 59.9 132.0
Lead
Nickel Platinum Silver
50.8 63.8 9.8
Tin
7A.0
Tungsten
33.2
Znc
35.58
The resistivity values are usually measured using samples made of a l-cm cube or a rylinder having a diameter of I mil and len$h of I ft. The resistance of the sample is then given by R
o4
- --
(12.21
where p is Le resistivity, { is the length of the sample, and a is the cross-sectional area of the sample. The resistance for l-ft-long wire hardng a diameter of 1 mil made of various materials is given in Thble 12.1. The electrical resistance of a material varies with temperanre. In general, the resistance of conductors (er
0hm's Law Obm's law describes the relationship among voluge,
I{
resistance,
4
and crurent,
.f,
according to
V:
RI
flz.3)
Note from Ohm's law, Equation (12.3), that crutent is direcdy proporrional to voltage and inversely proportional to resistance. As electric potential is increased, so is the currenq and ifthe resistance is increased, the current will decrease. The electric resistance is measured in unis of ohms (O). An element with I ohm resistance allows a current flow of I amp when ttrere exists a potential of 1 volt across the element. Stated another way, when there exists an elecrical potential of I volt across a conductor wirir a resisance of I ohm, then L ampere of electric current will flow through the conductor.
L2.4
Er,scrRrcAr- Crncurrs eNp ColrpoNsNTs
329
The Ameriean Wire Gage (AWG) Elecuical wires are typically made of copper or aluminum. The actual size ofwires is commonly expressed in rerms ofgage number. TheAmerican \07ire Gage (AVG) is based on successive gage numbers having a constant ratio of approximateh 1.12 betvreen their diameters. For example, the ratio of the diameter of No. l AVG wire to No.2 is 1.12 (289 l;tisl258 mils : 1.12), or, as another example, the ratio of diameter of No. 0000 to No.000 is 1.12 (460 mils/ 410 mils : l.I2). Moreover, the ratio of cross sections of successive gage numbers is approximately (1.12)2 : I.25.The ratio ofthe diameter ofwires dlffering by 6 gage numbers is approximately2.0. For example, the ratio of diameters ofwire No. I to No. 7 is2 Q89 mils/ 144 mils - 2.0), or the ratio ofdiameters ofNo. 30 to No.36is2 (10.0 mils/5.0 mils : 2,0).Table 12.2 shows the gage number, the diameter, and the resistance for copper wires. lfhen examining Table I2.2, note that the smaller the gage number, the bigger the wire diameter. Also note from data given in Thble I2.2 rhatthe electric resistance of awire is increased as its diameter is decreased.
The National Electrical Code, published by the Fire Protection Association, contains specific information on the type of wires used for general wiring. The code describes the wire
AmericanWire
(AT[G) Number
Gauge
0000 000 00
I 2 5
6
7
l0 t2 r4 r6 18
Diameter (mils) 460.0 410.0 365.0 289.0 258.0 182.0
r62.0 r44.0 91.0 81.0 64,0 51.0 40.0
20 22
32.4
24
20.1 15.9 12.6 10.0 8.0 6.3
26 28 30 32 34 36 38 40
25.3
5.0
Reistance per f 0O0 (ohms) @7fF 0.0500 0.0630 0.0795 0.126 0.159 0.319
0.403 0.508
t.28 r.62 2.58
4.09
6.5r 70.4 16.5
25.2
4r.6 66.2
r33 167
265 107
4.0
673
3.r
ro70
ft
330
Crr.n'prsn
12
Elncrnrc CunnsNT lNo Rrrarpo
Pan-eMsrgRs
types, maximum operating temperanues, insulating materials, outer cover sheaths, the type usage, and the specific locadon where a wire should be used.
Electric Power The elecuic pov/er consumption of various electrical components can be determined using the following power formula:
P: W
t12.4)
where P is power in watts, Zis the volage, and ,I is the current in amps. The kilowaft hour units are used in measuring the rate of consumption of elecricity by homes and business. One kilowatt hour represents the amount of power consumed during t hour by a device that uses one kilowatt (k\0, or 1000 joules per second.
The electric resistance of a light bulb is 145 O. Determine the value of current fowing through the lamp when it is connected to a 120-volt source. Using Ohmt law, Equation (I2.3),we have
V: [4,4,t,a4,,,4,4,,,,4--,,,,,',-'-
l
RI
v t20 o'83 A *: r45: ': :-.tlttil ii -t-r::1: t--
.::
-
:
r---'r
--
-
- -:----r - -':--:-:t::-
Assuming that your elecric power company is charging you 10 cenc for each k'$f'h usage, estimate the cost of leaving five 100-\f [gh bulbs on from 6 p.vr. until 1 1 p.v. every night for 30 nights.
(5 light bulbs)
/
(
:
750 cents
:
$7.5a
t00
v
\/
lish; burb /
lkw \/ (-tooo
*:/
(I
h \
-;
,)rl0
nigh,')
/
( tt
o.'a\
ffi/
Series Cireuit Elecuical components can be conneced in either a series or a parallel arrangement. As we mentioned earlier in this chapter, there are many different types of electrical components. Here in this section we will only look at resistors and capacitors. Aresistorls anelectrical component that resists the fow of either direct or alternating current. Resistors are commonly used to protect sensitive components or to control the fow of cuffent in a circuit. Moreover, a resistor can be used to divide or control voltages in a circuit. There are two broad groups of resistors: fixed-value resistors and variable resistors. As the name implies, the fixed-value resistor has a fixed value. On the other hand, the variable resistors, sometimes called potmtiolnetels, can be adjusted to a desired value. The variable resistors are used in various circuits to adjust the current in the circuit. For example, variable resistors are used in light switches that allow you to dim the light. They are also used to adjust the
12.4
Er,ncrnrcer. Crncuns nNu ColrpoNsNTs
331
volume in a radio. A resistor dissipates heat as elecric current flows through it. The amount of the heat dissipated depends on the magnitude of the resistor and the amount ofthe current that is passing through the resistor. Similar to a constant fow of water in a series of pipes of varying size, the electric current flowing through a series of elements in an electric circuit is the same (constant). Moreover, for a circuit that has elements in a series arrangement, the following is true:
'
The volage drop across each element can be determined using Ohmt la\il. across each element is equal to the total voltage supplied to the
n The sum of the voltage drop
'
clrclut. The total resistance is the sum of resistance in the circuit.
For resistors in series, the total equivalent resistance is equal to the sum of the individual resistors, as shown in Figure 12.10.
R.ol:
Rt
+
R2
+ Rr + R4+ R5
fi2.5)
One of the problems with a circuit that has elements in a series arrangement is that if one of the elements fails, R5 Ra that failure prevents the current from fowing through tigurelZ.l0 other elements in the circuit; thus the entire circuit fails. fiesistors in series, You may have experienced this problem with a series of ftotar = f,r + nz+ h+ /.4+ /'5. light bulbs in a string of Christmas tree lights. In a series arrangement, when one light fails, the current to other lights stops, resulting in all of the lights
!
being off.
Determine the total resistance and the current fowing in the circuit shown in Figure 12.11.
Rr=3Sl
Rz=7
E
figurel2.ll
A
ThecircuitforEranplel2.4.
The light bulbs are conneced in a series arrangement; therefore, the
toel
grven by
R,oot: Rl W'e can now use
+R2*Rr:3O+7A+ Ohmt
lanrr
9O:190
to determine the current fowing thtough the circuit.
r:+:L:0.47A-0.5A R.ool 79
resistance is
332
CHnprnn
12
Er.Bc"rnrc Cunnprirr exo RsLArEo PanelrersRs 'We
v_z: R,1:
,j,
t:
i 'il,
iiu.,,,.,.,.r,,* .,rrrii,.l- ---i
can also obtain the voltage drop across each lamp using Ohm's lam
: Vz_t = R I : (z)(0.47) : Vt-q = R /: (l)(0.42) :
i
(3)(0.47)
t.4IV 3.29v
4.ruu
Note that, neglecting rounding-offerrors, tle sum of the volage drops across each light bulb should be 9 volts.
- lltil-
-.- -
=-
.'lr,i'ni.iiirirlrili'i-:lrialilr,rr.l.
Parallel Circuit Consider the circuit shown in Figare 12.12. The resistive elements in the given circuit are connected in a
parallel arrangement. For this situation, the electric current is divided among each branch. For the parallel arrangement shown in Figure 12.12, the electric potential, or the voltage, across each branch is the Figurel2.l2 same. Moreover, the sum of the current in each flesistors in panllel, branch is equal to the total current fowing in the circuit. The current fow in each branch can be derermined using Ohmt lav. Note that unlike a series circuit, ifone branch fails in a parallel circuit, depending on the failure mode, the other branches could still remain operational. This is the reason parallel circuit arrangement is usedwhen wiring different zones ofa building. Howwer, it is wonh noting that ifone branch fails in a parallel circuit, it could result in an increased current fow in other branches, which could be undesirable.
il
llllll +-+-+-+-
fffiP,knxn4P{"
The light bulbs in the circuit shown in Example 12.4 arc placed in a parallel arangement, as shown in Figure 12.13. Determine the current fow through each branch. Also compute the total resistance offered by a[ light bulbs to current fow.
ftt=3O
f,
tigurel2.l3
Rs=9O
lhechcuitforEramplel2.5.
light bulbs are conneced in a parallel arangement, the vohage drop across light bulb is equal to 9 volts. W'e use Ohm's lar to determine the current in each branch
Because the each
in the following manner:
V:R{1 =+ 9:3It :+ 1r:3.04 V= R2I2 + 9:71, + Ir: l.3A V-- RJ3 + 9:9$ =+ ,I3 : 1.04
12.4
Er,sc'rRrcAr, Crnctnrs eNo ColrponsNTs
333
The total crutent dravrn by the circuit is
ii i!
I,oot: It + Iz+ \:
i'1,
't,,
3.0
+
1.3
+
1.0
:
5.34
The total resistance is given by
'il;
+
+:***+*=1*l*1 R.dRlR2R3379
i",
r1
i
R,o,,1: 1.7o
Note that we could have obtained the total current dravn by the circuit using the total resistance and the Ohm! lalv in the following manner:
'il'l ri
V:
'ii
i.:lt:t-1. - -.: --'l-.
R,o6,I-o1
+
9V:
(1.7
OX/."J =+
.I.o1
:5.3A
.,r.1...--r."-f:
Determine the total current dravrn by the circuit shown in Figure 12.14.
il
figurelZ.l4
Thecircuitfortmmplel2.6.
The circuit given in this example has components that are in both series and parallel \[e will first combine the light bulbs in the parallel branches into one equiva-
arrangements.
lent resistance.
; I :|**:l*1
R"qoi*rl*, R2 R3 7
9
+
R*.i,*r...:3.9f) 'Y'
Next, we add the equivalent resistance to R1, noting that the two resistors are in series now.
f R-uiu.r*r:3 * 3.9:6.9Q 7: R,oor"(o,a + 9V: (6.9 OX/."r) +
R.o,rt: ii
R1
.::'l:-::-t--l -.
:--------:.:
'--t-
--t-:
,I,oo1
-: ':-=-:: r':
= 1.3A
- -, :-
-- - -!
Capacitors Capacimrs are electrical components that store electrical energil/. A capacitor has rwo oppositely charged electrodes with a dielecuic material insened between the electrodes. Dielecric material is a poor conductor of electricity. Capacitors are used in many applications to serve as frlters to protect sensitive components in electrical circuits against power surges. They are also used in large computer memories to store information during a temporary loss of electric powet to the computer. You will also find capacitors in tuned circuis for radio receivers, audio filtering applications (e.g., bass and treble controls), timing elements for suobe lights, and interminent wipers in automobiles.
334
Cnaprnn
12
Er-ncrnrc CunnnNraNo RsLArrp Panavrrrns
Thefarad (F)
is the basic unit used to desrgnate the size of a capacitor. One farad is equal per volt. Because the farad is a relatively large unit, many capacitor sizes are expressed in microfarad (,u,F : 10-6 F) or picofarad (pF : tO-12 F).
to 1 coulomb
12.5 Electric
Motors
As engineering studen$, most of you will ake at least one class in basic electrical circuits wherein you will learn about circuit theory and various elements, such as resistors, capacitors, inductors, and uansformers. You will also be introduced to different types of motors and their operations. Even if you are studying to be a civil engineeg you should pay special amemion to this class in electrical circuits, especially the part about motors, because motors drive many devices and equipment that make our lives comfortable and less laborious. To understand the significant role motors play in our weryday lives, look around you. You will 6nd motors running all types of equipment in homes, commercial buildings, hospitals, recreational equipment, automobiles, computers, printers, copiers, and so on. For example, in a typical home you can identify a large number of motors operating quiedy all around you that you normally dorit think about. Here, we have identified a few household appliances with motors:
. Refrigerator: compressor motor, fan motor ' Garbage disposer . Microwave with a turning tray . Stove hood with a fan . Exhaust fan in the baduoom ' Room ceiling fan . Thpe player in a VCR . Hand-held power screwdriver or hand-held . Heating, ventilating, or cooling qystem fan . Vacuumcleaner . Hair dryer . Electric shaver . Computer: cooling fan, hard drive
drill
Some of the factors that engineers consider when selecting a motor for an application arer (a) motor type, (b) motor speed (rpm), (c) motor performance in terms of torque ourput, (d) efficiency, (e) duty cycles, (f) cost, G) life expectanry, (h) noise level, (i) maintenance and senrice requirements. Most of these are self-explanatory. \0?'e discussed speed and torque in previous chapters; next we will discuss motor types and duty cycles.
lvlotor Type Selecting a motor for an application depends on a number of factors, induding the speed of the motor (in rpm), the power requirement, and the type of load. Some applications deal with difficult starcing loads, such as conveyors, while others deat with easy saning loads, such as a fan. For this reason, there are many different types of motors. Here we will not discuss the principles of how various motors work Instead, we will provide examples of various motors and their applications, as given in Table I2.3. Most of you will take some classes later that will be
devoted to the theory and operation of motors.
12,5
Er,rc"rnrc Morons
335
Schemadc Diagrams forAC Motors Repulsion-start induction
Splirphase
Capacitor-start
Capacitor-motor
Repulsion-induction
qc*-ffisili rh Fswltclj \Y
0 Electrical Power
Reqniremenr
I
i
Ioad
MotorType
Typ"
(1) Shaded-pole
induction
StartingAbilit/ Cforque)
(2)
Verylow ll2 to 1 times
Sta4ingCurrent (3)
Low
Phaoe
hp (4'
Size
6)
Vottage (6)
.035-2 krtr (1t20-r14Lp)
SinCL
Usudly 120
.035-.56 k\r',
Si"gl"
Usually 120
Single
Single voftage
nuuring torque Split-phase
tt
:
Ear! Sarti.ng Loads
low I to lj
High times
6 to 8 times
running torque
running curent
reilnanent-sPut,
Verylow
Low
capacitor-
1/2 to I tims running torque
inductiorr
Low
running torque
running current
Capacitor-sfart,
Hish
Medium
induction-nrn
3 tg 4 times
3 to 6 times
runuing torque
running current
Repulsion-sart,
Hish
Iow
induction:run
4 times
2! to 3 rimes
numhg torque
running current
Dfficuh
Capacitor-stan,
Medium
Sarting
capacitor-run
Hish 3| to 4| times
InaA Repulsion-san, capacitor-run Three-phase, general-puqpose
.035-.8
(1/20-l
Verylorr 1/2 to I time"s
Soft-start
(rt20-3t4hp)
2to2L, dinta
3 to 5 times
running torque
running current
High
low
4 times nrnnhrg torque
2] to 3 times
tuledium 2 to 3 times running torque
High
k\r
5.6'25kIff (4*so hp) .14-8
120 or240
hp)
k\r
Si"Cl*
240
Single
120-240
Si"gl"
r20-240
Si"CL
r20-240
Si"Cl"
Usually
(116-10 hp)
.14-15kV (116-20 hp)
.4^2Aklr (It2-25hp) .8-i2 kr?' (1-15 hp)
240
i
running current 3 to 6 times
rulningflrrrent
.4-300
krr
(rl2-40ahp)
Three
720-240; 240-480 or higher
i i
i
i
I
Source:
AAYM, Ehctrie Moar.
Continued
335
335
CHemsR
12
Er:ecrnrc Cunnmr:rexo Rnnrpo PenarvrBrsRs
Sctremafic Diagrams forAC Motors Tlvo-value capacitor
Universal motor
Shaded pole
Repulsion motor
TW
Sp€edRange
V)
Reyersible 'Relsdr€
(8)
900, 1200,
OtherrCha$cteridcs
(10) . :,
C.osr(g)
Verylow
Light dury low in
Iow
Simple
lyplcal
efficiency
Small fans, freezer blowers, arc welder blower, hair dryers
3@0 Yes
consmrcdon
Fans, fiunace blowers,
1800,
3600,
Low
Variable 1800,
lathes, small shop tools,
^ Usually
900-1800
Ueec
'{tl)
'
:
1800,
900, 1200,
squirel cage
o
.-€ Load
3-phase
jet pumps
custom-designed
Small compressors, fans
for special application Yes
High
i
3600
normaily nrheq
Used in motor sizes served by 3-ptrase power
Cenuifirgal pumps, crop dtyer &ns, feed giuder
lphase power not available 900,
tks
Moderate
Iong
service,
low,
,1,''
.
,'V'ater system!'air
'
'
'1, maintenance, very popular
1200,
compressors, ventillating
l'
1800,
fan6, grinders, blowers
3600 1200,
Yes
1800,
Moderate
Handles large load variations
Grinders, deep-well pumps,
to high
with lide variation in current
srlo unloaderq grains
3600 900, 1240,
demand Yes
Moderate
conveyors, baro cleaners
Good staning ability and
frrllload
effiiiettcy ,
,, Flrmps, air compressors,
3600 1200, 1800,
.', Yes
service than
\y'ery
Very simple constructioq,
3600
900, 1200, 1800, 36A0
Source:
336
AAYIM, Ebctrit Moar.
.
Moderate to high
tlrying fans, large conveyors,
:,
1800,
t-ilgh efhclency, reqlures more mo$ motors
,,fitdffi :
'
Conveyon, deepwell pumps, feed mills, silo unloaders
''t.,, Yes
low
dd""ddfu*,;*io-fr*,
Conveyors, dryers, elevatcirs, hoists, inigation pumps
12,5
Ersc"rRrc Morons
337
Duty Cycle Manufacnrrers of morors classifr them according to the amount of time the motor needs to be operated. The motors are generally dassified as continwus daty or intermittent daty. The continuous duty motors are used in applications where the motor is o
J. |)uncan Glover,
Ph.D.,
SUil{lvlARY Now that you have reached this point in the text
"
You should understand the imponance of elecnicity and elecuical power in our everyday lives.
" You should knowwhat
is meant
and altemating current.
byelectric current and the difference between direct current
338
Cneprsn
12
Er.rc"rnrc Cunnelrr AND Rrlerso PlnnrvrersRs
. You should hane an idea about various gources ofelecricity. . You should understand the concept ofelectrical resistance and ia unit. " You should be familiar with the AITG wire designations. . You should be familiar with the role of elecuic morors in our evervdav lives.
t lIfiliTHtr{ llJ.
_
I
How are bareries conneced in the following products: a hand-held calculator such as a
TI
85, a flashlight, and a porable radio? fue the baneries conneced
in a series arrangement or in a parallel arrangement? of batteries used in the following
n2.8.
12.2. Identify the types
products: a- a laptop computer b. an elecuicshaver c. acordless drill d. a camcorder e. a cellular phone
f.
n2.9.
XZJ0.
relative resistance of l-ftJong wires having diameters of 1 mil made of the metals given in Thble 12.1 to l-ftJong copper with a diameter of 1 mil. For example, using the data given in Thble 12.1, the relative
aflashlight
g. a wrisrwatch h. a camera 12.3. Investigate the size and the material used for heating elements in the following products: a toaster, hair dryer, pressing iron, coffeemaker, and an elecuic stovetop. 12.4. As explained in this chapter potentiomerers (rheostats) are used in applications to adust the electric current in a circuit. Investigate the different types ofresisrance element used for various applications. The resisance element of the rheostat is typically made from metal wires or ribbons, carbon, or a conducting liquid. For example, in applications where the current in the circuit is relatively small, the carbon type is used. !?'rite a brief report discussing your findings. 12.5. Elearical furnaces are used in the production of steel that is consumed in the strucrurd, automotive, tool, and aircraft indusuies in the United States. Elecuical furnaces are typically dassified into resistance firnaces, arc fi.rrnaces, and induction fumaces. Investigate the 'W'rite operation ofthese dree types offumaces. a brief report describing your findings. X2.6. Identify examples ofmotors used in a new automobile. For er
building belonging to someone you know. If possible dran' a diagram showing the power distribution, similar toFigarc I2.7. Vhat is the current that fows through each of the following light bulbs: 40 W 60 W 75 W 100 \7? Each lighl ls conneced to a 120-V line. If a 1500-V hair dryer is connected to a 120-V line, what is the maximum current drarvn? Refering to Thble 12.1, crate a table that shows the
resistance
of l-foot-long I'mildiameter aluminum l-mil copper wire 1s 1.64 (17.01 Ql
wire to l-ftJong,
r0.37dl:
r.64).
12.11.
Investigate how an alkaline battery and an automobile
12.12.
maintenance-free battery (ead acid, gel cell) work W'rite a brief repon discussing your findings. When subjected to pressrue, certain materials create
a relatively small voltage. Materials that behare in this manner are called piezoelecuics. Investigate the applications in which piezoelecrics are used. Write a brief repon discussing your findings. 12J3. Prepare an elecuical circuit plan similar to the one shown in Figure 12.9 for your apaftmenr, your home, or a section of your dormitory. 12J4.
The National Elecrical Code (NEC) covers the safe and proper installation of wiring, electrical devices, and equipment in private and public buildings. NEC is published by the Nadonal Fire Protection Association (NFPA) eveq/ three years. As an orample of an NEC provision, the receptacle oudets in a room in a dwelling should be placed such that no point on the wall space is more than 6 ft *"y from the oudet in order to minimize the use of extension cords. After performing a W'eb search or obtaining a copy of tlre NEC handbook, give at least duee other
PnosLEMs examples
of National Electric
Codes
for a
fam-ily
n2.20:
dwelling. 1215.
339
Use Kirchhoff's cufrent laqr to determine the missing current in the circuit shown in the accompanyrng
fig*".
Obain a multimeter (voltmeter, ohm, and current of the headng element in the back window of a car.
meter) and me:$ure the resistance and the voltage Determine the power output of the heater. t2J6.
Visit a hardware store and obtain information on the sizes of heating elements used in home hot water heaters. If a hot water heater is conneced to a line with 240
V
determine the current &av"n by the hot water
heater and its power consumption. 12fi.
n218.
fu
explained in this chapter, the National Elecuical Code gives specific information on the type of wires used for general wiring. PerForm a search and obtain ffirmation on wire types, maximum operating temperatures, insulating materials, outer cover sheaths, and usage types. Prepare a table that shows examples of these codes; show wire size, temperaflrre radng, application provisions, insuladon, and outer covering. Determine the total resistance and the current fow for the circuit shown in the accompanying figure.
Problem 12.20
Obtain a flashlight and avoltmeter. Measure the resisance of the light bulb used in the flashlight. Draw the electrical circuit for the flashlight. Estimate the current drarn by the light when the flashlight is on. 12.2. Contact your elecuic utility company and obtain information on what the company charges for each kilowa$ hour usage. Estimate the cost of your electric power consumption for a typical day. Make a list of your activities for the dry *d estimate the power con'W'rite a brief sumption for the dwices that you used. report discussing your firdi"gr. X2.23. For Example 12.6, we computed the toal current dravn by the circuit. Determine the current in each branch of the circuit given in Example 12.6. n2.24. pq1 an 2utomobile banery investigate what is meant by the following terms: arnpere-hour rating cold-nanhing rating and resente capacity rating. Vrite a brief repon
n2.21.
Rz=9 A
t2Y
Problem lZJB
12.19.
Determine the totd resistance and the cuffent flow in each branch for the circuic shown in the accompanying figure.
Rt=6f,)
t2Y
discussing your findings. 12.?5.
12.26. Problem 12.19
For a primary cell such as an alkaline battery what does the term amp-hourrepresent? Gather information on the amp-hour rating for some alkaline batteries. V'rite a one-page summary of your findings. As you know, a fire is a safety device that is commonly placed in an electrical circuit to protect the circuit
340
CHer"rsR
12
Erxcrmc CunnrNraNo RBr.emo
PanarvrsrERs
againster
CHAPTER
T3
ElwRGY AND PowER owerful cranes lift and lower structural members in place. We need energy to build shelter, to
cultivate and process food, to make goods, and to maintain our living places
at comfortable settings. To quantify the requirements to move objects, to lift objects, or to heat or cool something, energy is defined and classified into
different categories. Power is the time rate of doing work. The value of power required to perform a task represents how fast you want the task done. lf you
want a task done in a shorter period of time, then more power is required.
341
342
CHaprrn
13
ENsncyANo Povpn
ofthis chapar are to introduce the concq)t ofenngt, the uarious types of energt, and what is rneant b the tewn power. W will expkin aarious mechanical fomns ofmerg including kinetic energ/, ?otmti.al mergt, and elastic merg. Ve will abo reuisit the def.nition of therrnal enng forms from Chapter 11, including heat andinternalmngl Next, wewillpresentconseroAtionofmerg andits applications. Ve will defne Power as the rate ofdoing work and explain in detail the dffirence betwem work/energt and?ower. our discussion, these dffirmces should be clear '4fter t0 )/oa. The cornrnon units ofpower, wans and horsepower, are also explained. Once you haue agoodgrasp ofthe concepts ofwork, energ1 andpower, thenyoa can bener understand the rnanufacturer?s power ratings of engina and motors. Moreouer, in this chapter we will ako explain what is rneant b! the rmn fficimcy and laoh at the fficiencies ofpower plants, intemal cornbustion mgines (car mgines), elecnic motors, parn?s, and heating air-conditioning, and rfiigeration sJ/stems. Tbe obiectiues
13.1 Work, Mechanical Energy, Thermal EnerEy in Chapter 10, mechanical work is performed when a force moves an object through a distance. But what is energy? Energy is one of those abstract terms ttrat you already
As we o
hare a good feel for. For insance, you already know that we need energy to creare goods, to build shelter, to culdvate and process food, and to maintain our living places at comfonable temperature and humidity settings. But what you may not know is that energy can have different forms. Recall that scientists and engineers define terms and concepts to explain various physical phe-
nomena that govem nanue. To beter explain quandatively the requiremen$ to move objects, to lift things, to heat or cool objecs, or to sretch materials, energT is defined and classified into different categories. Let us begin with the definition of binetit @qgy, When work is done on or against an object, it changes the kinetic energy ofdhe object (see Figure 13.1). In fact, as you will learn in more detail in your physics or dynamics class, mechanical work perFormed on an object brings about a change in the kinetic energy ofthe object according to
workl-,
r1^ : i*V/ - i*V?
(l3l)
where rni, A. ,1"r, of th"'obj""t and. V2are the speed of the object at positions I ^nd V and 2, respectively. To bener demonstrate Equation (13.1), consider the follorving example. Ifhen you push on a lawn mower, which is initially at rest, you perform mechanical work on the larvn mower and move it, consequendy changing its kinetic energy from a zero value ro some nonzero value.
I
Figuret3.l
Tle relationsfiip Ietween work altd change in fiinetic energy.
13.1
343
Worur, MecHANTcAL ENsncy, TnSRMAL ENsncv
Kinetic Energy An object having amass rz and moving widr kinetic energy
:
a speed
Vhas akinetic energy, which is equal to
1
(13.2)
;mVz
The SI unit for kineti"lrr"rgy i. the joule, which is a derived unit. The unit ofjoule is obtained by substiruting kg for the units of mass, and m/s for the units ofvelocity, as shown here.
N
'
kinetic energ'y
:
1
: (l -, ::(k")fs)'=;7+i(-) (kC) = N.m: joure: \;i \ s, ./ .
mv2
,*V"
J
j factor in the kinetic energf equadon is unidess. As we discussed in the previous secrion, it is the change in kinetic energy (AKE), that is used in engineering analysis as grven by Equation (13.1). The change in the kinetic energy is grven by Note that the
11
AKE: 2'2 ^*Vl-)*V1
fl3.3)
Determine t*re net force needed to bring a car that is uaveling at 90 km/h to a fi.rll stop in a distance of 100 m. The mass of the car is 1400 kg. \Ufe begin the analysis by first changing the units of speed to m/s and then use Equation (13.1) to analyze the problem.
( rn \/tooom\:25; ^-m /\**/\ , r.* i
V:
--lk*\ v.^a: ro( ,
V:
Vna:
:
o
:
I*rl - I*rl l)' (force)(100 m) = 0 ]t,noo ks)(zs workl-2
(force)(distance)
&
force :
,
- . ...
:'
ir-liiillli{
\
-/
-4375N
----------i.Eatllrri-
-:---A:-lf-.- --a-----------l[]""'lr.Al':
Note that in Example 13.1, change in the kinetic energ;r' was used in the analysis. Also note, the negative value of force indicates that the force must be applied in opposite direction of modon, as expected.
fotential Energy lift an object with a mass rn by a venical drstance Ai is called graoiuti.onal potential energ, It is the mechanical work that must be performed to overcome the gravitational pull of the earth on the object (see Figure 13.2).The change in the potential energy of the object when its elevation is changed is given by
The work required to
change in potential energy
:
APE
:
rng
A,h
03.4)
34
Crrer"rsn
L3
ErvsncyANo Povsn where
Fn: g
=
A/ :
mass
of the object (Lg)
acceleration due to graviry (9.4t
mlsl
change in the elwation (m)
The SI unit for potential energy is also joule, a derived unit, and is obtained by substituting kg for the units of mass, and mls2 for the units of acceleration due to gravity, and m for the elevation change:
N
E figurel3.Z Cftange in potential energy
of
potendalenerry
:
msh=
(kc)($)(*) : N.m :
an object.
J
As in the case with kinetic energy, keep in mind that it is the change in the potential energy that is of significance in engineering calculations. For example, the energy required to lift an elevator from the first floor to the second floor is the same as lifting the elevator from the third foor to the founh foor, provided that the disance between each floor is the same. Cdculate the energT required to lift an elevator and its occupant with a mass of 2000 kg for the following situations: (a) between the first and the second foor, (b) berween the third and the fourth floor, (c) bemreen the first
and the fourth
foor.
(See
Figure 13.3.) The vertical disance
between each floor is 4.5 m. We can use Equation (13.4)
M=4.5m
(") change in potential energy
= M=
(2ooo
(b) change in potential energy
4.5 m
ar;ralyze
rng
: (2ooo b)(r.sr
:
this problem; the
:
M
m)
mg
$)*.r
(c) change in potential energy
4.5 m
:
kl(r.sr $)*.r
: (2ooo k)(r.ar M=
to
energy reguired to lift the elevator is equal to the change in its potential energy, starting with
J
M
m)
rng
: 88,2e0
: 88,2eoJ
Lh
: $)f tr.r m)
264,8701
Note that the amount of energy required to lift the elevaror from the first to the second floor and from the third to the
E
fourth floor is the same. Also realize that we have neglected any frictional effect in our analysis. The actual energy requirement would be greater in the presence of frictional effect. FiguretS.S
A schematic diagram
iilr-f
for Eranple 13.2.
13.1 'Wonr, MncneNrcel ENnnc! Tnrnv.nr. Eurnct
345
Elastic Energy a spring is stretched or compressed from its unstretched position, elasti.c enngis stored in the spring, energy that will be released when the spring is allowed to rerurn to its unstretched position (see Figure t3.4) . The elastic energ;r stored in a spring when stretched by a distance x or compressed is given by
\?hen
elastic energy
I : ;h*' L
fl3.5)
where
: rc : A
spring consant (N/m) deflection of spring from its unstretched position (m)
The SI unit for elastic energy is also the joule. It is obtained by substituting N/m for the units of spring constant and m for the units of deflecdon, as shown: elastic energy
/N\ : I : (.; jt*l' : N. m : ;b'
J
j
factor in the elastic energyequation is unidess. Let us now consider the spring shown in Figure 13.5; the spring is stretched by 11 to position I and then stretched by x2to position 2.The elastic energy stored in the spring in position I is given by
Note once again that the
elastic energy
=
elastic energy
:
I*l f,*l
change in elastic energy
:
AEE
:
|U - )*l
(13.6)
---T;, Unsrebhed
___l_'
--
x2
position Position 2
Stretched
position Force
Force
E
E
figurel3.{
The elastic energy of a
sprirg.
rigurel3.5
The change in the elastic energy of a spring.
346
CHaprBn
13
ENrncyaNo Porrrn Determine the change in the elastic enerry of the spring shown in Figure 13.6 when
it
is
stretched from: (a) position I to position 2, (b) position 2 to position 3, and position I to position 3. The spring consant is A 100 N/cm. Additional information is shown in Figure 13.6.
:
-1 x3=7 cm
-3 Force
E tigurc13.6 TfiespringinExamplel3.3. \7b begin by converting the units of the spring constant from N/cm to N/m in the following manner: (tOO N/cm)(100
cm/m)
:
10,000
N/m
Using Equadon (13.6), we qrn now answer the questions:
:
change in elasticenergy
AEE
:
I*, - l*,
(b) AEE
:
)*l - **r:
AEE
:
**Z -
G)
G)
I-----lilir .li--
l*l
AEE
: I*l - *l
:j{,o,ooo-*,m)(o.o:),
]{to,ooo
N/m)(0.02),
:J{to,oooN/m)(o.oz),
-
0
: 12.51
- }{to,ooo N/m)(0.05)2 :
-
0
:
721
24.51
-
eonservation of Mechanical Energy In the
absence of heat transfer, and assuming negligible losses and no
work, the conseraation
ofmechaaical energtstetes that the toml mechanicd energy of a system is constant. Stated anor:her way, the change in the kinetic energy of the object, plus the change in the elastic enerry,
13.f
'Wonr, MscHAMc,AL Errwncv, Tnrmrel Ewrrciv
3ql
plus the change in the potential energ:r of the object is zero. This statement is represented mathematically as follows.
AKE+APE+AEE=0 Equation (13.7) stares
tlat
flg.n
energy @ntent of a qrstem can change form, but the
toal
energy
content ofthe qntem is constant
In a marurfacnrring process, cars are rolling down an inclined sufice, as shon'n in F rgure L3.7 . fron which the cart must be released so that, as it reaches pointr4, the can has avelocity of 2.5 mls. Neglect rolling fricdon. Estirnate the height
I
Figurcl3.? 'W'e
Aschenaticdiagramfor0nmplelS.4.
can solve this problem using Equation (13.7)
AKE+APE+AEE:0 AtrCE
:
ltl
t*V/ - i*r?:
,m(2.5
m/s)z
-
o
AEE=0 APE
:
rngah:
f,*tz.srn/s)2
+
-*(o.rr3)"
-*(r.trt)":
o
And solving for F/we have
Ff
=
0.318 m
Thermal Energy Units ln Chapter 11, we explained that thermal energt/ transfer occurs whenever there e*ists a temperature difference within an object, or whenever there is a temperature difference between two bodies, or a temperaf,ure difference between a body and its surroundings. This form of energy
348
CHeprsR
13
ENsncyANp Porwpn transfer is call ed heat Remember the fact that heat always fows from a high-temperature region to the low-temperature region. Moreover, we discussed the tluee different modes ofheat transfen conduction, convecdon, and radiation. W'e also discussed *ree units that are commonly used to quantify thermal energy (1) the British thermal unit, (2) the calorie, and (3) the joule. As we explained the Btu (British thermal unit) in Chapter 11, one Bru is formally defined as the amount of thermd energy needed to raise the temperature of 1 lb. ofwater by loF. The calorie is defined as the amount of heat required to raise the temperature of 1 g ofwater by loC. And as you may also recall from our discussion in Chapter 11, in SI units no distinction is made between the uni* of thermal energy and mechanical energy, and therefore the units of thermal energy are defrned in terms of fundamental dimensions ofmass, length, and time. In the SI system of units, the joule is the unit of energlr and is defined as
l joule: 1N.m = l kg.nfls2 The U.S. Customary unit of thermal energy is related to mechanical enerry *o"ugh
: 1 Btu : 1 Btu
7781b,ft 1055 J
Finally, internal energy is a measure of the molecular activiry of a substance and is related to the temperature of a substance. As we explained in Chapter 11, the higher the temperature of an object, the higher its molecular activity and thus its internal enerry.
13.2 Conservation of Energy-First
Law of Thermodynamics
Earlier in this chapter, we discussed the conservation of mechanical energy. W'e stated that in the absence of heat ffansfer, and assuming negligible losses and no work, the conservation of mechanical energy states that the total mechanical energy of the qatem is constant. In this section, we will discuss the effects of heat and work in conservation of energy. There are a number of different wErs that we can describe the general form of the conservation of energ;r, or the first law of thermodynamics. Expressed simply, $ef.rst laat of thnmodlmamics states that energT is conserved. It cannot be created or destroyed; energy crn only change forms. Another more elaborate statement of the first lavr says that for a sysrcm having a fixed mass, the net heat transfer to r:he qystem minus the work done by the system is equal to the change in total energy ofthe system (see Figure 13.8) according to
Q- V=
A,E
fl3.8)
where
= tt"t heat transfer into the system € Qr" - > a,J in joules (J) ilZ = net work done by the system EW, - >V.) in joula (J) AE = net change in total enerry of the system in joules (), where Erepresents Q
the sum of the internal enerry, kinedc energfr potential enerry, elastic enerry, and other forms of energy of the qrstem.
13.2 Coxsrnvarron or ENnncv-Frnsr Lev or Tnx,nuoovNevrcs 349
t The
Figurel3.S
fint
law of thermodynamics
for a system having a fired mass.
There is a sign convention associated with Equation (13.8) that must be followed carefirlly. Heat transfer into the system, or the work done by the system, is considered a positive quantity, whereas the heat transfer out of the qystem, or the work done on the qrstem, is a negative quantity. The reason for this type of sign convention is to show that work done on the system increases the total energ;r of the qystem, while work done by the qatem decreases the total energy of the qTstem. Also note that using this sign convention, hear transfer into the system increases the toal enerry of tfre system, while heat transfer out of the system decreases its total enerry. 'When it You may also think of the fust lau' of thermodynamics in the following manner: energy out of qrscomes to enerry, the best you can do is to break even. You cannot get more a tem than the amount you put into it. For example, ifyou put 100 J into a q6tem as work, you can get 100 J out ofthe system in the form ofchange in internal, kinetic, or potential energy of the system. As you learn more about energyyou will also learn that according to the second lariy of thermodynamics, unfornrnately you cannot even break even, because there are always 'W'e losses associated with processes. will discuss t'he effect of losses in terms ofperformance and efficienry ofvarious qystems later in this chapter.
350
CHaprrn
13
ExsncyANp Poven Determine the change in the total energy of the system shown in Figure 13.9. The heater puts 150 \f (J/s) into the water pot. The heat loss from the water pot to the atmosphere is 60'Wi Calculate the change in total energy ofwater in the pot after 5 minutes. 60 watts
I
Figurcl3.9 'W'e
AsclematicdiagramforEramplel3.g.
can use Equation (13.8) to solve this problem.
Q_ W: AE 'V= 0 (150J/s)(300s)
A,E:
-
(60Jls)(300s)
:
A,E
27 kJ
:r.r.,:,..::..1.:r,,i.ii::rl rr:':::rrlr..l.-i:-:..Ltrtri
13.3 Understanding
rr-,4-ii.-J.l.i
ai., ::.:..:-,..Li.!
What We Mean by Power
In Secdon 13.1, we reviewed the concept of work as presented in Chapter 10 and explained the different forms of energy. Vb now consider what is meant by the rcrm power. Poueris formally defined as the time rate of doing work, or sated simply, the required work, or enerry, divided by the time required to perform the task
-power =
Power
:
work
(force)(distance)
trme
ume
03.9)
energT
ume
From Equation (13.9), the definitionofpower, itshould be clearthat thevalue ofpowerrequired to perform a task represenrc how fast you want the task done. Ifyou want a task done in a shorter time, then more por^rer is required. For the sake of demonstrating this point better, imagine that in order to perform a task 3600 J are required. The nent question then becomes, how fast do we want this ask done? If we want the task done in L second, 3600 Jls power is required; ifwe want the task done in I minute, then 60 J/s power is needed; and ifwe want the
13.4 \7errs
eNp Honsnpownn
351
in t hour, then the required power is 1 J/s. From this simple example, you should clearly that to perForm the same task in a shofter period of time, more power is required. More power means more energy expenditure per second. Another example that you have a direct experience with is the following situadon: !7h.ich requires more power, to walk up flight " of stairs or to run up the stairs? Of course, as you already know, it requires more pov/er to run up the stairs, because as compared to walking, when running up the stairs, you perform the same amount of work in a shorter time period. Many engineering managers understand the concept of power well, for they undersand the benefit of teamwork. In order to finish a prgt ect in a shorter period of time, instead of assigning a task to an individud, the task is divided among sweral team members. More usefirl energ;r expenditure per &y is expected from a team than from a single person, thus the project or the task can be done in less time, task done see
13.4 Watts and Horsepower fu we explained in the prwious section, power is defined
as the
time rate of doing work, or
stated anorher way, work or energy divided by time. The unis of power in SI units are defined
in the following manner: power -
work : -:trme
(force)(distance)
=
_N.m_J_* w ---
(r3.r0)
ss
ume
Notethefollowing: lN.miscalledl joule(J),""dlJ/siscalledlwattffi.InU.S.Customary unie, the units ofpower power -
: work . : trme
are expressed
in lbs. ft/s and horsepower (hp), in the following manner:
(force)(distance)
lbf.ft
ttme
s
03J1)
and
t hp :
lb,. ft 55A
The U.S. a*.** ing manner:
1lb.. ft : s
---:-
03.t2)
*ts I.3558
of power are related to the SI unit of power, watt (.W), in the follow-
\7 =
\7
1.36
(r3r3)
t hp : 745.69W = 746W
(r3.r4)
'Wi
Remember that 1(1b6. ftA) is slighdy greater in magnitude than 1 Also keep in mind that t hp is slighdy smaller than 1 k\il Another unit that is sometimes confirsed for the unit of power is kilowatt hour, used in measuring the consumption of electricity by homes and the manu"€acturing sector. First, kilowan hour (kVh) is a unit of energy-not power. One kilowatt hour represents the amount of energy consumed during t hour by a device that uses one kilowatt (k\f) or 1000 joules per second (J/s). Therefore,
: I klfh : I kW
1000'W'
:
1000 J/s
(1000J/s)(3600 s)
1k\fh :3.6MJ
:
3,600,000J
:
3.6 MJ
352
Cneprnn
13
ENrncyeNp Porw'rn In heating, vendladng, and air-conditioning (HVAC) applications, Btu per hour (Btu/h) is used to represent the heat loss from a building during cold months and the heat gained by the building during summer months. The units of Btu/h are related to the unit ofwatt in the
following manner: 1
Btu/h
:
1055
'W:
1.055
k\7
Anodrer common unit used in dre United States in air-conditioning and refrigeration systems is ton of rfiigeration or cooling One ton of refrigeration represents the capacity of a refrigeration system to freeze 2000 lb- or 1 ton of liquid water at 32"F into 32oF ice in 24 hours. It is
I ton ofre&igeration
:
12,000 Btu/h
In the case ofan air-conditioning unit, one ton ofcooling represents the capacity ofthe aircondirioning system to remove 12,000 Btu thermal energy from a building in t hour. Clearly, the capacity of a residendal air-conditioning system depends on the size of the building, its construction, shading the orientation of its windows, and its climatic location. Residential airconditioning units generally have a 1- to 5-ton capacity. The sizes of home gas furnaces in the United States are also expressed in units of Btu/h. The size of a typical single-family-home gas furnace used in moderate winter conditions is 60,000 Bru/h. lb get a feel for ttre relative magnirudes that watt and horsepower physically represent, consider the following examples.
Determine the power required to move 30 people, with an average mass of 61 kg (135 lb^) per person, berween two foors of a building, a vertical distance of 5 m (16 ft) in 2 s. The required power is determined by
/
work
_ Power: at*.: _
rao
p***l(e
r
ks,
\'/
\
#,)(g.sr $)(: -) = 45,000 \s zt
The minimum energy requirement for this task is equivalent to providing electricity to fifteen 100-\[light bulbs for 1 minute (90,000 ]. Next time you feellazy and.ate t]rinking about taking the elevator to go up one floor, reconsider and think about the total amount of energy that could be saved if people would take the sairs instead of taking the elevator to go up one foor. As an example, if 1 million people decided to take the stairs on a daily basis, the minimum amount of energy saved during a year, based on an estimate of 220 working days in a year,
would be energy savings
fr----- - i:.::::r: : l
I-::-::-:::..
:i--_ ::- -::::
'We
:
/
go,oooJ
\/ r \,.t,000,000 persons) (220 days)
ril.rr._,/ ( *r, = 66 x 10eJ : 660 GJ
::t-l-
|'
:
:- :
:::
(
.
:i:i;:i:::-;:liri-
will revisit this problem, after we discuss efEcienry, to determine the amount of fuel needed in a power plant to provide the amount of energy that we just calculated in Example 13.6.
13.4 Werrs nNo HonsspowEn
353
?20 jouvrAs ot Aislatr\ce of Z.tfeetirr I secoh2.
rhar#s I
horsegower!
Determine the power required to move a person who weighs 220 lb a vertical distance of
2.5ftin1s. -power
:
work
(22otb)(2.5 ft)
flme
1s
:
550
lb.. ft ----j: t hP
Therefore, I honepower represents the power required to lift a person weightng22}lb6, a distance of 2.5 ft in 1 second. There are a number of other ways to think about what t horsepower physically represents. It could also be interpreted as the power required to lift an object weighting 100 lbsa distance of 5.5 ft in I second. How powerfi:I are you?
Determine ttre power required to move an object thatweighs 800 N (179.85 lbe), avenical distenc.r of 4 m (L3.12 ft) in 2 s. The power requirement eirpressed in SI units is given by Power
work (800 N)( = .i_. : ---t-:
m)
1600'\7
And using U.S. Customary Units, the power requirement
is
lbr'ft work (179.851brx13.12ft) --^^ = 1180 s Power: ,i*. : -----;The power expressed in horsepower is
power
:'so(!re)(+) : z.t4hp \ r '\55o +/
354
L-
CHepreR
13
ENsRcvANo Powsn In this example, think about the relationships among various units of power needed to perform the same amount ofwork in the same amount of time.
a:], .--
a:,,
-.. t -a1,,-.
i:-
,-litr:: r:l :1:-- t::ttr::: l::: -:-l:!-l-:r1
--t1
t-:-::_a:::i
Most ofyou ha\re seen car advenisements where the car manufacturer brags about how fast one of its cars can go from 0 to 60 mph. According to the manufacturer, the BM$7' 750iL 2001 model can go from 0 to 60 mph in 6.7 seconds. This performance is usually measured on a test uack The engine in the car is rated at 326 hp at 5000 rpm. The car has a reponed weight of 4597 lbf.Is the claim by the manufacurer justifiable? 'Well, to answer this question you may agree it would be more fun to drive to a BM!7 dealership and take the car for a test run on a racing traclc But let us answer this question with the knowledge you have gained so far in this course. W'e need to make some assumptions first; we cnn assume that the driver weighs 180 lb6 and the reported weight of the car includes enough gasoline for this test. Next, we convert the speed and the mass values into appropriate units.
fr
V: Vaoa: o; vz= vs*t= werght o b
/
'ru\/tn)(rzson\ (uo.r.r\ruoos/\ I *i /
:
ft
ss.=
(4597 + 180) lbf :jffi:148slugs
32.2 ,
s-
Using Equation (13.1), we c:rn determine the required work to go from 0 to 60 mph.
work,-":!*V3-!*v? 2'2 workl-,
1 / ff\2 :1!+aslucs)(s8-:,
- 0:
573,0561br.ft
The power requirement to perform this work in 6.7 seconds is Power
573,056Lbf . ft
work : -:-: ume
6.7
Ib,.ft :85,530L s
s
The power o<pressed in horsepower is
power
:,rr.r ro : 85,530(l!dl(+) J \
,\:foy;l
Keeping in mind that power is needed to overcome air resistance and that there are always additional mechanical losses in the car, it is still safe to say the claim is good. . : - .-
--:':: :--:':]--:--:':'l:
--
.i---.il
rtl - :i
I3.5 ErrrcrrNsr 13.5
355
Efficiency As we mentioned earlier, there is alwal's some loss associated with a dynamic system. In engineering, when we wish to show hovr well a machine or a qfstem is functioning, we express its efficiency. In general, the overall efficiency of a system is defined as: acnral output
efficiency
requrred
fl3'15)
input
All
machines and engineering sJrstems require more input than what they put out. In the nent few secdons, we will look at the efficiencies of common engineering components and qrstems.
Fourer Plant Effieiency 'Water is used
in all steam power-generating plants to produce electricity. A simple schematic of a power plant is shown in Figure 13.10. Fuel is bumed in a boiler to generate heat, which in turn is added to liquid water to change its phase to steam; steam passes *uough turbine blades, turning the blades, which in effect runs the generator connected to the turbine, creating elecuiciry. The low-pressure steam liquefies in a condenser and is pumped through the boiler again, completing a cycle, as shown in Figure 13.10. The overall efficiency of a steam power plant is defined as power plant efficiency
:
energy generated (t316)
energy input from fuel
The efficiency of todayt power plants where a fossil fuel (oil, gas, coal) is burned in the boiler is nan 40o/o, and for the nuclear power plants the overall efficienry is nay'ly 34o/o. Electricityis also generatedbyliquidwaterstoredbehinddams. Thewateris guidedinto water turbines located in hydroelectric power plants housed within the dam to generate electricity. Transmission Iine
-E l,q
IA ttl
Disposal to storage pond or landfill
River
I
tigurc
13.10
A sciematic diagram of a steam water plant.
Source: C-owcsy ){cel
Enog.
356
CnqPTEn
13
Exnncvaxo Povsn The potential energy of the water stored behind the dam is converted to kinetic energy as the water fows through the turbine and consequendy spins the rurbine, which turns the generator.
In Example 13.6, we determined the power required to move 30 people, with an average mass of 61 kg (135 1b-) perperson, beween nvo foors of abuilding: avertical distance of 5 m (16 ft) in 2 s. The energy and the power requirements were 90,000 J and 45,000 S respectively. Moreover, we estimated savings in energT if 1 million people decided to walk up a foor instead of taking the elwator on a daily basis. The minimum amount of energy saned during a year, based on an estimate of 220 working days in a year, would be 660 GJ. Let us now estimate the amount of fuel, such as coal, that can be saned in a power plant, assuming a 38olo overall efficiency for the power plant and a heating value of approximately 7.5 MJ/kg for coal.
power plant efficiency
0.38
:
:
energT generated energy input from fuel
660 GJ energy input from fuel
amount ofcoal required
-
l'74
+ x
energyinputfromfuel
7ot2l
:
:
L.74
x
1012J
:1.74TJ
232,000 kg(sLt,A12 lb^)
7.5Xrc6ft As you crrn see, the amount of coal that could be saved is quite large! Before you get on an elevator nort time, think about the amount of fuel-not to mention the pollution-that can be saned if people just walk up a foor!
Internal eombustion Engine Efficiency The thermal efficiency of a tlpi."l gasoline engine is approximately 25 to 30o/o and for a diesel engine is 35 to 4lo/o.T)he thermal efficiency of an internal combustion engine is defined as thermal efRciencv
:
Power ouqput heat power input as fuel is bumed
Keep in mind thatwhen expressing the overall efficiency of
chanical losses
as
ac*,
ffifl one must account for the me-
well.
Motor and Pump Efficiency As we explained in Chapter 12, motors run many dwices and equipment that make our lives comfonable and less laborious. As an example, we identified a large number of motors in var-
ious devices at home, including motors that run the compressor of your refrigerator, garbage disposer, exhaust fans, tape player in a VC& vacuum cleaner, turntable of a microwave, hair dryer, electric shaver, computer fan, and computer hard drive. W'h.en selecting motors for these
tt.5
producs, engineers consider the efficiency of the motor efficiency of an elecuic motor c:rn be simply defined
efficiency:
as
EprrcrsNcY
357
one of the design criteria- The
as
power input to the device being driven electric power input to the motor
fi3.r8)
The efficiency of motors is a function of load and speed. The electric motor manu.facnrrers provide perFormance curves and tables for their products that show, among other information, the
efficienq of the motor, You find pumps in hydraulic qystems, the fuel system ofyour car, and qystems that deliver water to a city piping network. Pumps are also used in food processing and in petrochemical plants. The function of a pump is to increase the pressure of a liquid entering the pump. The pressure rise in the fluid is used to overcome pipe friction and losses in fiaings and valves and
358
Cnaprrn
13
ENsncvANo Poven to transpoft the liquid to a higher elevation. Pumps themselves are driven by motors and engines. The efficiency of a pump is defined by efificienry
:
power input to the
fuid by the pump
power input to the pump by the motor
(r3.r9)
The efficienry of a pump, at a given operating speed, is a function of flow rate and the pressure rise (heao of the pump. Manufacturers of pumps provide perFormance curves or tables that show, among other performance data, the efficiency of the pump.
Efficiency of Heating, Cooling, and Refrigeration Systems Before we discuss the efficienry of heating, cooling, and refrigeration systems, let us briefly explain the main components of these systems and how they operate. \(e begin with the cooling and refrigeration systems because their designs and operations are similar. Most of todayt airconditioning and refrigeration systems are designed according to a vapor-compression cycle. A schemadc diagram ofa simple vapor-compression cycle is shown in Figure 13.1 l. The main components ofthevapor-compression cycle include a condenser, an evaporator, a compressor, and a tluotding device, such as an expansion valve or a capillary tube, as shown in Figure I 3. 1 I . Refrigerant is the fluid that transports thermal energy from the waporator, where thermal energy or heat is absorbed, to the condenser, where the thermal energy is ejected to the surroundings. Referring to Figure 13.11, at state 1, the refrigerant exists as a mixture of liquid and gas. As the refrigerant fows through t-he evaporator, its phase completely changes to vapor. The phase change occurs becarrse ofthe heat transfer from the surroundings to the evaporator and consequendy to the refrigerant. The refrigerant enters dre evaporator rube in a liquid/vapor mixture phase at very low temperature and pressure. The temperature of the air surrounding the evaporator is higher than the temperature of the waporator, and thus heat transfer occurs from the surrounding air to the evaporator, changing the refrigerant's phase into vapor. The evaporator in a refrigerator is a coil that is made from series of rubes that are located inside the freezer section of the refrigerator (see Figure 13.12). In an air-conditioning unit,
Condenser
Expansion valve or capillary tube
J
FigurelsJl
A schematic diagram of
a vap0r-cotnpression cycle.
Condenser
I
Figurc
13.12
The
typhal locations of lfte evaporator and the condenser ir a housefiold relrigerator.
359
360
Crreprrn
13
ExBncveNo Poven the evaporator coil is located in the ducnvorh near the fan-furnace unit inside the house. After leaning the waporator, the refrigerant enters the compressor, where the temperature and the pressure of the refrigerant is raised. The discharge side of the compressor is connected to the inlet side ofthe condenser, where the refrigerant enters the condenser in gaseous phase at a high temperature and pressure. Because the refrigerant in the condenser has a higher temperature than the surrounding air, heat transfer to the surrounding air occurs, and conse' quendy thermal energy is ejected to the surroundings. Like an eyaporator, a condenser is also made of a series of rubes with good rhermd conductivity. As the refrigerant flows through the condenser, more and more heat is removed (or transferred to dre sumoun&t$)t consequendy, the refrigerant changes phase from gas to liquid and it leaves the condenser coil in liquid phase. In the older version of a household refrigerator, the condenser is the series of black tubes located on the back ofthe refrigerator. In an air-conditioning unit, the condenser is located outside the house in a housing that also contains the compressor and a fan tfrat forces air over the condenser. After leaving the condenser, the liquid refrigerant flows through an expansion valve or a long capillary tube, which makes the re&igerant expand. The expansion is followed by a drop in the refrigerantt temperature and pressure. The refrigerant leaves the er<pansion valve or the capillary tube and flows into the waporator to complete the cycle shown in Figure 13.11. Another point worth mentioning is that as the warm air flows over the evaporator section of an air-conditioning unit, and as the air cools, the moisture (the water vapor) in the air condenses on the outside of the waporator coil. The condensation forming on the ouaide of the waporator is wentually drained. Thus, the evaporator acts as a dehumidification device as well. This process is similar to what happens when hot, humid air comes into contacr with a glass of ice water. You have all seen the condensation that forms on the outside surface of a glass of ice 'We will discuss what we mean by absolute water as the neighboring hot, humid air cools down. and relative humidiry in Chapter 17. The efficiency of a refrigeration qystem or an air-conditioning unit is given by the coefficient of performance (COP) which is defined as
COP:
heat removal from the weporator energy input to the compressor
fl3.20)
You should use consistent units to calculate the coefficient ofperFormance. The COP ofmost va-
porcompression units is 2.9 to4.9. IntheUnitedSates, itis customaryto enpressthecoefficient of performance of a refrigeration or an air-conditioning qFstem using mixed SI and U.S. units. Quite often, the coefficienc ofperformance is called energy efficiency ratio (EER) or the seasonal energ;r efificiency ratio (SEER). In such cases, the heat removal is expressed in Btu, and the en'Wh : ergy input to the compressor is orpressed in wan-hours, and because 1 3.412 Btu, EER or SEERvalues of greater than 10 are obtained for the coefficient of performance. Therefore, keep in mind that in the United States, the EER is defined in the following manner:
EER
=
heat removal from the evaporator (Btu) energ;r
input to the compressor (Vh)
03.21)
The reason for using the units of I7h for energy input to the compressor is that compressors are powered by electricity, and electricity consumption is measured (even in the United States)
13.5 ErrrcrnNcv
351
in k\Ch. Today's air-confitioning units have SEERvalues that range from approximately 10 to 17 . ln fact , all new air-conditioning units sold in the United States must have a SEER value of at least l}.In 1992, the United States goveflrment established the minimum standard efficiencies for various appliances, including air-conditioning units and gas furnaces. In a gas fi.unace, natural gas is burned, and as a result the hot combustion producs go through the inside of a heat exchanger where thermal energy is transponed to cold indoor air that is passing over the heat exchanger. The warm air is distributed to various pars ofthe house through conduits. As mentioned, in 1992 the United States government established a minimum annual fuel utilization efficienry (AFUE) rating of 78o/o for firrnaces installed in new homes, so manufacnrrers must design their gas firrnaces to adhere to this standard. Today, most high-efficienry firrnaces offerAFUE ratings in the range of 80 to 96010.
An air-conditioning unit has a cooling capacity of 24,000 Btu/h. If the unit has a rated energF efficiency ratio (EER) of 10, how much electrical energy is consumed by the unit in I h? If a power company charges 12 cents per k\Vh usage, how much would it cost to run the air conditioning unit for a month (30 day9, assuming the unit runs 10 hours a day? lChat is the coefficient of performance (COP) for the given air-conditioning unit? 'W'e can compute the energy consumpdon of the given air-conditioning unit using Equation (13.21).
EER:
10:
heat removal from the waporator (Btu) energy input to the compressot
(\71)
24,000 (Bru) energy input to the compressor (\71r)
Energy input to the unit for t h of operation : 2400 Vh : 2.41i\7h. The cost to run the unit for a month for a period of 10 hours a day is calculated in the following manner: cost to operate the
unit
: (iPX#)
(t#)ur
d"y,)
: $s6.40
The coefficient of performance (COP) is calculated from Equation (L3,21)z
COP =
heat removal from the evaporator energy input to the
compresso,
24,000 Btu
:2,9
1z4oo*r(i#)
Note the relationship between the EER and COP:
EED
COp:-llll:4:2.9l0 3.412 3.412 ::_
r--:
---
-__J
,I .l
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','i
Pllolrfr*e Engineering has alwaya been a way of to one fl was a member of the Hybrid Electric Vehicle thinking for me. I remember growing up Team). I wish I had the time to create another technol-
lq
completely fascinared by science, math, ory and have it patented before graduadng college. I and with TV programs like Mr. Wizard had so many ideas on paper that never came to fruition,
# hrff',*ffiill
!::ffi: i,r ffiHfnl;lfro1,m*n*ffi'tr il;
funny because I was given an
opportuniry
the ideas on paper to life. It's not a toal wash because there is dways_grad-uate school, which does provide the centure (my current employer). One major up-close resource,s needed for research to make ideas happen. and personal chance to pursue my passion as an engi- The undergraduate curiculum is such thar most proneer c:rme when I entered high school and was able fessors arent allowed to offer sruden$ exciting experito do more hands.on experiments in programs like ments and projects becaus€ they have to meet requireBEAMS (Becoming EnthusiasticAbout Math and Sci- menu for accreditation. ence), sponsored by the Thomas Jefferson National The thing I like most about engineering is that itt laboratory (a Department of Energy Sponsored Re- wideopen. Icanrememberjudgingatechnologyfuturesearch Facility). While in that internship, I was able m city competition some months ago, which included acnrally *."i S..r.,*ry O'L€ary, who served one term middle *d .I.-.rrory smdents. Sorn. of tlese ideas under President Clinton. \)9'e took several picnrres, were just amazing. Today, engineering is the backbone which still hang proudly at my mothert home in New- of any technolory or innovation. Engineering is the port News, VA My excitement later turned into a se- track and the wheels needed to maLe ideas move. I like iies of internships with Thomas Jefferson National its diveniry because I've found myself speaking to Laboratory, NASA, and the Newpon News Shipbuild- people with all sores of accents, religion, and gender in ing which all colleccively helped to solidify my com- our aftempt to solve complex problems by sharing our to work with PBS on one of mymar^r)' projects with
Ac-
ri
and the real hand.s-on (life-changing) intemsf,ips I sued and won, all before and during The icing on the cake was that I was always
college.
phy.d ;y
;Gt;;;sf;h"Jry ;-p'"d,
had internshipsin the field of engineering and
tons ofscholarship
money.
pur-
I sdll dorft think my most iqteresdng project has come yet. But it's fi.rnny because all of the projeca I've enjoyed have been simple hands-on projects. I havent
em,j**' ;;;i.-;;;"sh;;i;;fi;d-;;*d";;-
received ing forthe
past c3;rple lears. Most ofmywork b€ei lris along the path ofbusiness process design oversight and
ThebiggestchaIlengeIfacedasastudentryTtry-sofrwareimplemenarion.MyworkwiththeHybrid ing to manage my dme with srudies and e'xua-surricu- Elecuic Vehicle (where auromobiles use fuel cells and tar aai"itie. there really wasrit enough dme for me to have a bi-product of water) project was probably the excel in any outside acdvities the way I dreamed. As an most rewarding because it still has reallife applicadon. enginecr, you really have to commit l00o/o of your aca- Of course, we-can all look around tod"y see how demic regiment to leaming a ton of information in a gas prices have affected our choice to look"lri for altemashon peiiod of time. Iooking back on my four-year iu. .t rgy sources. My excitement really is for the fuspurt in engineering I wish that I could have done ture of the technology and in knovring that my ideas *stuff" thar more of the rurned me on to engineering may one day make their way to the show-room floor of such as joining two mechanical auto teatru as opposed .some sr dealership or household.
364
CHrprsR 13
ENsrcvANo Povsn
SUMMARV Now that you have reached this point in the text
. . . ' .
You should understand how different forms of mechanical energy are defined. You should know how and when to use kinetic energy, potendal enerry, and elastic energy in engineer-
ing analyses. You should know what is meant by intemal energ;r and heat. You should know how and when to use conservation ofenergy to solve engineering problems. You should cleady understand the definition of power, its common units, and how it is re-
lated to work and energy. You should know the basic definition of efficiency and be familiarwith its various forms, including the definitions of thermal efficienry, SEER" and AFUE, which are commonly used to express the efficiencies of components such as pumps, compressors, and motors, as well as
complete qFstems including heating, vendlating, and air-conditioning (HVAC) equipment and household appliances.
t3J.
t3.2.
Identify ways that you c:m sa\re enerry. For example, walking up a floor instead of taking the elwator, or walking or riding your bike an hour a day instead of taking the car. Estimate t-he amount of energy that you could save every year with your proposal. Also, estimate the amount offuel that can be saved in the same maruler. Sate your assumptions, and present your detailed analysis in a report. Iook up the manufacnrrer's data for the most recent year of the following cars: a. Toyota C"*ry
t3.4.
13.5.
Power input to the pump by the motor (k!7): 0,5,0.7,0.9, l.o, 1.2 Power input to the fuid by the pump (kV): 0.3, 0.55, 0.7, o.g, 1.0
b. HondaAccord c. BMV750 Li
Plot the efficiency curve, The efficiency of a pump is a function of the fow rate. Assume that the flowrate readings corresponding to power data points are
You can visit the cars.com'W'eb site to gather information. For each car, perform calculations similar to Example 13.9 to determine the power required to accel-
erate the car ftom
0 to 60 mph.
Smte
all of your
equally spaced. t3.6.
assumptions. x3.3.
An elevator has a rared capacity of 2000 lb. It
can
transport people at the rated capaciry benveen the first and the ftfth foors, with a verrical distance of 15 ft be*reen each foor, in 7 s. Estimate the power reouirement for such an elwaror.
364
Determine the gross force needed to bring a car that is traveling at 1 10 km/h to a firll stop in a distance of 100 m. The mass of the car is 2100 kg.'What happens to the initial kinetic energy?'!7here does it go or to what form ofenergy does the kinetic energy convert? A centrifirgal pump is driven by a motor. The perFormance of the pump reveals the following information:
t3.7.
A power plant has an overall efficienry of 30%o. The plant generates 20 MI7' of elecricity, and uses coal from Montana (see Thble 11.12) as fuel. Determine how much coal must be burned to sustain the generation of 20 M\7 of electriciry. Estimate the amount of gasoline that could be saved
if all of the passenger cars in the United
States were
PnosLEMs
r3.8.
driven 1000 miles less each year. Sate your assumptions and write a brief repon discussing your findings. Investigate the typical power consumption range of the following producs:
t3Jl. Investigate
a. home refrigerator
nfl. An air-conditioning unit
the size of a gas furnace used in a rypical single-family dwelling in upstate New York, and compare ttrat size to the firrnaces used in Minnesota and in Kansas.
has a cooling capacity of 18,000 Btu/h. If the unit has a rated energy efficiency ratio (EER) of 11, how much electrical energ;r is consumed by the unit in t h? If a power company charges 14 cents per k\Vh usage, how much would it cost to
b. 25-inch television set c. clothes washer d. elecric clothes dryer e. vacuum cleaner
f
hair dryer
run the air-conditioning unit for a month (31 days), assuming the unit runs 8 h a day? What is the coefficient of performance (COP) for the given air-
Discuss your findings in a brief repon. R.9.
Investigate the typical power consumption range the following products: a. personal computer with a l9-inch monitor b. bser printer
of
size of the airconditioning unit in your oriln home or apartment.
conditioning unit? Visit a store that sells window-mount air-conditioning units. Obain information on their rated cooling capacities and EER values. Contact your local power company and determine the cost of electricity in your area- Estimate how much it will cost to run the airconditioning unit during dre summer. 'Vrite a brief report to your instructor discussing your findings and
Investigate the SEER and the AFUE of the units.
assumptions.
t3.t3.
c. cellularphone d.
p"l-
calculator
Discuss your findings in a brief repon. 8.t0.
365
look up the firrnace size and the
For Problams 13.14 througlt 13,20 use the datafrom the accompanying table shown bebw.
Electricity Generation by Fuel, 1980-2030 (billion kilowam-hours)-Data from U.S. Depanment of Energy -
r i
i,, i ;,
:_
___ _-.._
___
Co4t
Petrole'-
1980
116r.562
245.9942
346.2399
25r.1.156
284,6883 actual values
L990 zooo 2005 2010 2020 2030
1594.0r1:
125.62rr Lrr.221
372.7652
576.8617
357.2381acqral values
501.0382 75r.8189 773.8234
753.8929 774.0726
356.478rt acnral values 375.8663 actual values 47 5,7 432 proj eaed values 5 I5.L 523 projected values 5 59.1 335 projected values
1956.26' 2A40.9r3 2217.555
r15.4264 104.8r82
25:A4.786
106.,6799
3380.674
1t4.674r
Natural Gas
tro2.762 992.7706
Nudear
808.6948 870.698
870.5909
Renewable/Other
l
plot the percentage of each fuel used in generaring elecuicity for each year shown in the table.
1314. Calculate and
n.l5. Calculate and plot the percentage of increase consumpdon for the data shown in the able.
in
coal
365
1316.
Convert the data given in the able from kilowatt-
hours to Btu. nJ7. Assuming an average 35Vo effiaency for power plants and a heating value of approximately 7.5 Mllkg, celculate the amount of co"l (i" kg) required for generating electricity for each year shovrn in the table. BJ8. How many kilograms of coal could be saved ifwe were to increase the average efficienry of power plants by Io/o
366
to
360/o?
t319.
Assuming an average 35o/o efficiency for power plants
and a heating value of approximately 1000 Btu/ff (22,000 Btu/lbJ, calculate the amount of natural gas required in ff and lb- for generadng electricity for each year shown t3.20.
in rhe able.
How many ft' and pounds of natural gas could be sayed if we were to increase the average efficiency of power plants by |o/o to 360/o?
Hoover Dam*
Hoover Dam is one of the Bureau of Reclamations' multipulpose projects on the Colorado River. These ptor""t conuol floods; they store water for irrigation, municipal, and industrial use; and they provide hydroelectric pov/er, recreation, and fish and wildlife habit*. The Hoover Dam is a concrete arch-gravity type of dam, in which the water load is carried by both gravity action and horizontal arch action. The first concrere for the dam was placed on June 6, 1933, and the last concrere was placed in the dam on May 29, 1935.The following is a summary of some facts abour the Hoover Dam. *Materials wete adapted ftom U.S. 3u€au sf'Reclamarion.
368
Cner"rsR
13
ENsncvAND
Pown
Dam rfimensions: Heighc 726.4ft; length at crest: I244ft; width at topt 45 ft; width basq 660
at
ft
Weighc 6.6 million tons Reserwoir statistics: Capacity: 28,537,000 acre-feet; length: 110 mi; shoreline: 550 mi; max depth: 500 ft; surface area:157,000 acres Quantities of matedals used in proiecc Conoete: 4,440,000 yd3; explosives: 6,500,000 lb; plate steel and oudet pipes: 88,000,000Ib; pipe and finings: 6,700,000Ib (840 mi); reinforcement steel 45,000,000 lb; concrete mix proportions: cemenfi 1.00 part, sand: 2.45 parts, fine gravel 1.75 para,intermediate gravel 1.46 parts, coarse gravel: 1.66 parts, cobbles (3 to 9 in.): 2.18 paru; water: 0.54 parts The dam was built in blocfts, or vertical columns, varying in size from approximately 60 ft square ar the upsueam face of the dam to tbout25 ft square at r:he downstream face. Adjacent columns were locked together by a s;ntem of vertical keys on the radial joints and horizontal keys on the circumferential joints. A-fter the concrete was cooled, a cement and water mixnue called grout was forced into the spaces created between the columns by the contraction of the cooled concrete to form a monolithic (one-piece) sffudure.
Hoover Dam itself contains 3.25 million yd3 of concrete. Aftogether, there are 4,360,000 y& of concrete in the dam, power plant, and appunenant works. This much concrete would build a monument 100 ft square and 2-ll2 mi high; would rise higher than the Empire State Building (which is 1250 ft) ifplaced on an ordinary ciry block; or would pave a standard highway, 16 ft wide, from San Francisco to New York City. The Reservoir At elevadon 1221.4, l,ake Mead, the largest man-made lake in the United States, contai ns 28,537 ,000 acre-feet (an acre-foot is the amount of water required to cover I acre to a depth of 1 foot). This reservoir will store the entire ar/erage flow of the river for two years. That much water would cover the entire state of Pennsylvania ro a depth of 1 ft. lake Mead extends approximately 110 mi upsfteam toward the Grand Canyon and approximately 35 mi up the Virgin River. The wi&h of Lake Mead varies &om several hundred feet in the canyons to a maximum of 8 mi. The reservoir covers about 157,900 acres, or
247 square miles. Recreation, although a by-product of this project, constitutes a major use of the lakes and controlled fows created by Hoover and other dams on the lower Colorado River today. l,ake Mead is one ofAmerica's most popular recreation areas, with a l2-month season that attracts more than 9 million visitors each year for swimming, boating, skiing, and fishing. The lake and surrounding area are administered by the National Park Service as pam of the Lake Mead National Recreation fuea, which also includes lake Mohave downstream from Hoover Dam. The Power Plant There are 17 main turbines in Hoover Power plant. The original turbines were all replaced through an upgrading program between 1986 and 1993. With a rated capacity of 2,991,000 hp, and two station-service units rated at3500 hp each, for a plant totd of 2,998,000 hp, the plant has a nameplate capaaty of 2,074,000 kW This includes the two station-service units, which are rated at 2400 kV each. Hoover Dam generates low-cost hydroelecftic power for use in Nevada, fuizona, and California. Hoover Dam alone generates more than 4 billion k\Vh ayeut-enough to serve 1.3 mil-
Hoown
oervr
369
lion people. From 1939 to 7949, Hoover Power plant was the worldt largest hydroelectric installation; with an instelled capacity of 2.08 million kW it is still one of this countryt largest. The $165 million cost of Hoover Dam has been repaid, with interest, to the federal treasury through the sale of its power. Hoover Dam energy is marketed by the'WesternArea Power Administration to 15 entities in fuizona California, and Nevada under conffacrs thar expire in 2017 . More than half, 560/o, goes to southern California users; Arizona contractors receive l9o/o, and Nevada users get 25o/o. The revenues from the sale of this power now pay for the dam's operation and mainrenance. The power conrractors also paid for the uprating of rhe power plant's nameplate capacityfrom 1.3 million to over 2.0 million k\M
PROBLEMS
1.
2.
3. 4.
Calculate the water pressure at the bottom of the dam when the water level is rwo- thirds of the height of the dam. Express your result in lb/in2 and Pa. Also, calculate the magnitude of the force due to water pressure acting on a narrow suip (1 ft by 100 ft wide) located at the base of the dam. As mentioned in the article describing Hoover Dam, lake Mead contains 28,537,000 acrefeet ofwater (an acre-foot is the amount ofwater required to cover I acre to a depth of I ft). Express t-his water volume in gallons and m3. Hoover Dam generates more than 4 billion kVh a year. How many 100-watt light bulbs could be powered every hour by the Hoover Damt power plant? How much coal must be burned in a steam power plant with a thermal efficienry of 34o/o to generate enough power to equal the 4 billion I(S7h a year generated by Hoover Dam? Assume the coal comes from Montana (see Chapter 8 for heating value data).
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Microsoft Excet and
MATLAB,
two
used commonly by engineers to solve engineering
problems. These computationaltools are used to record, organize, analyze data using formulas, and present the results of an analysis in chart forms. MATLAB is also versatile enough that you can use it to write your own program to solve complex problems.
--
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..
.
-
Electronlc $preadsheets .
MATLAB
CHAPTE,R
T4
ErncrRoNrc SpnnADsHEETs n electronic spreadsheet is a tool
that can be used to solve an engineering problem. Spreadsheets are commonly used to record, organize, and analyze data using formulas. Spreadsheets are also used to present
the results of an analysis in chart form. Although engineers still write computer programs to solve complex engineering problems, simpler problems can be solved
with the help of a spreadsheet.
372
l4.l
Mrcnosorr F,:rcrr,-Besrc IpBes
373
In this chaprcr, we will discuss the use of spread.sheex in soluing mgineering problems. Before the innodaction of electronic spreadslteets, engineers wrote tlteir own cornpater prograrn* Computn prograrns were typically utritten for problerns where rnore tltan a feu., hand calculations were required. FORTMN uas A cornrnon prograrwning langwage that was used by many mginens to perform nurnerical computations. Ahhough mgineCIs still write computerprograms to solue comltlex engineeringproblerns, simpler problerns can be solued with the help ofa spreadsheet. Compared to writing a com?ater program and debuging it, spreadsheets are much easier to use, to record, organize, and analyze data asingformulas tltat are tnput the user. Spreadsheex are also used to show the resalts of an analysi.s in theforrn of charts. Because of their ease of use, spreadsheets ar€ cornrnon in many other disciplines, including business, rnarketing and accounting This cbapter begins by discussing the basic makeap ofMiuosoft Excel, a cornmon spreadsheet. Ve will explain how a spreadsheet is diuidzd into rows and colurnns, and how to input data or aforrnwla into an actiae cell. We will ako explain the use of other tools suclt as Excel's mathernatical, statistical, and logicalfunctions.
b
Plotting the results of an engineering analysis using Excel is also presented.
14.1 Microsoft Excel-Basic ldeas \(e will
begrn by explaining the basic components of Excel; then once you have a good understanding of these concepts, we will use Excel to solve some engineering problems. As is the case with any new areas you explore, the spreadsheet has ir own terminology. Therefore, make sure you spend a litde time at the beginning to familiarize yourself with the rcrminology, so you can follow the examples later. A tfpi.rl Excel window is shown in Figure 14.1. The main components of the Excel window, which are marked by arrows and numbered as shown in Figure 14.1 are
l. Tide bar: Contains the name of the current active workbook. 2. Menu bar: Contains the commands used by Excel to perform certain tasks. 3. Toolbar buttons: Contains push bumons (icons) that er(ecute commands used by Excel. 4. Active cell A worksheet is divided into rows and columns. A cell is the box that you see as the result ofthe intersection ofa column and acell,.Aaiue cellrefers to aspecificselected cell. Formula bar: Shows the data or the formula used in the active cell. 5. 6. Name box Contains the address of the active cellColrimn header: Aworksheet is divided into rows and columns. The columns are marked byA, B, C, D, and so on. 8. Row header: The ror'\r's are identified by numbers 1,2,3,4, and so on. 9. Worksheet tabs: Allow you to move from one sheet to anot-her sheet. fu you will learn later, you can name these worksheers.
10. Status bar: Gives information about the command mode. For example, "Ready'' indicates the program is ready to accept input for a cell or "Edit" indicates Excel is in an edit mode.
374
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6.le that you create and save. A workbook could consist of many worksheets and charts. A utorhsbeet represen* the rows and columns where you input information such as data, formulas, and the result of various calculadons. As you will see soon, you may also include charts as a part of a given worksheet as well.
A utorhbooh is the spreadsheet
Naming Worksheets To name a worksheet, double-click the sheet ab to be named, qpe the desired name, and hit the enter key. You can use the Edit or the Insert menu to delete or insen a worksheet. You can move (or change the posidon of) a worlsheet in the workbook by selecting the sheet tab and while holding dovrn the left button on the mouse move the tab to the desired position among other shees.
14.2
Cells and Their Addresses As shown in Figure I4.I, a worlsheet is divided into rovrs and columns. The columns are marked byA, B, C, D, and so on, vrhile the rows are identified by numbers I,2,3,4, and so on. A cell rcpresena the box that one sees as the result of the intersecdon of a row and a column. You can input (enter) various entities in a cell. For example, you c:rn type in words or enter numbers or a formula. To enter words or a number in a cell, simply choose the cell rvhere you want to enter the information, rype the information, and then hit the Enter key on your keyboard. Perhaps the simplest and the easiest way to move around in a worlsheet is to use a mouse. For example, ifyou want to move from cellA5 to cell C8, move the mouse such that the mouse
14.2
CsLLs ANo
TnnnAppnrssns
375
pointer is in the desired cell and then click the left mouse button. To edit the content of a cell, choose the cell, double-click the left mouse bufton, and then similar to editing aword processing document, use any combination of delete, baclspace, or arrow keys to edit the content of the cell. fu an alternative to double clicking, you c:rn use the F2 key to select the edit mode. Keep in mind that as you become more proficient in using Excel, you will learn that for certain tasks there is more than one way to do something. In this chaprcr, we will explain one of the ways, which can be followed easily.
A Range As you will soon see when formatting, analyzing, or plotdng daa, it is often convenient to select a number of cells simultaneously. The cells that are selected simultaneously are called a range.To define a range, begin with the first cell that you want included in the range and then drag the mouse (while pressing down the left button) to the last cell that should be included in the range. An example of selecdng a range is shown in Figure 14.2. Note that in spreadsheet language, a range is defined by the cell address ofthe top-left selected cell in the range followed by a colon, :, and ends with the address of the bottom-right cell in the range. For example, to select cells A3 through B10, we first select A3 and then drag the mouse diagonally to B10. In spreadsheet language, this range is specified in the following manner-A3:B10. There are situations where you may want to select a number of cells that are not side by side. In such cases, you must first select the contiguous cells, and then while holding (pressing) the Cul key select the other noncondguous cells by dt"ggirg the mouse button. Excel allows the user to assign names to a range (selected cells). To name a range, first select the range as just described, and dren click on the Name box in the Formula bar and type
t
figure
14.2
An erample showing the selection of a range of cells.
376
Cner"rsn
14
Erec"rnoNrc Spnrepsnssts in the name you want to assrgn to the range. You can use upper- or lowercase letters alongwith numbers, but no spaces are allowed between the characters or the numbers. For example, as shown in Figure l4.2,wehure grouped the measured voltages and the resistance into one range, which we have called rneAsuretnetuts. You can tlen use the name in formulas or in plotting data.
Inserting eells, eolunnns, and Rows After entering data into a spreadsheet, you may realize that you should hane entered some additiond data in between two cells, columns, or rows that you hare just created. In such a case, you c:rn always insert new cells, column(s), or row(s) among already existing data cells, columns, and rows in aworlsheet. To insert new cell(s) benveen other enisting cells, you must fust select the cell(s) where the new cell(s) are to be inserted. Next, from the Insert menu choose *1s Qells sption. Indicate whether you want the selected cells to be shifted to the right or down. For example, lett say we want to insen duee new cells in the location E8 through El I (E8:E1 1) 'We and shift the existing content of E8:E1 1 down. first select cells E8:E1 1; then from the Insert menuwe choose the Cells option, and then choose the Shift cell down option. To insen a column, click on t}re column indicator button to the right ofwhere you would like to have the new column inserted. From the Tnsert menu, choose Columns. The procedure is similar for insening
a
new row among already existing rows. For example, ifyou would like to insen a new
column between columns D and E, you must first select column E, and then from the Tnsert menu choose Columns; the new column will be inserted to the left of column E. To insen more than one column or row simultaneously, you should select as many column indicator buttons as necessary to the right of where you would like to have the columns insened. For example, ifyou would like to insert three new columns between columns D and E, then you must first select columns E, F, G; then from the Insert menu choose Colunns, and duee new columns will be insened to the left of column E.
14.3
Creating Formulas in Excel By now, you know that engineers use formulas that represent physical and chemical lanrs governing our surroundings to analyze various problems. You can use Excel to input engineering formulas and compute the resuks. In Excel, a formula always begins with an egual sign, =.
Fxompler and A6
Operation Addition Subtraction
Muldplication Division Raised to a power
Symbol
+ * | A
C€llsAi C€llA7 contains the the result of the
conain
I0
andz'
formulagiven respectively inthe example ,
values
:A5+46+20 32 8 =,A.5-A6 =(A5*AO+9 29 6 --(L5t2.5)+A6 :(A5r,{gag.5 t0
l
',
i
1
f
I
f
I4.3
3n
CnperrNc Fonuur-e,s ru Excsl
To enter a formula, select t}re cell where you want the result of the formula to be displayed. In the Formula bar, then type the equal sign and the formula. Remember when typing your formula to use parentheses to dictate the order of operation. For example, if you were to type : 100 * 5*2, Excel will perform the multiplication first, which resuhs in a value of 10, and then this result is added to 100, which yields an overall value of 110 for the formula. If however, you wanted Excel to add the 100 to 5 first and then multiply the resuldng 105 by 2, you should have placed parentheses around the 100 and 5 in the following manner: :(100t5)*2, which results in a value of 2L0 . The basic Excel arithmetic operations are shown in Tirble 1 4. I .
As we explained in the previous chapters, thermophysical properties of a substance, including density, viscosity, thermal conductivity, and heat capacity, play a key role in engineering calculations. As discussed, the thermophysical properry values represent information such as how compact the material is for a given volume (density), or how easily a fuid fows (viscosiry), or how good a materid is in conducting heat (thermal conductivity), or how good the material is
in
storing thermal energy (heat capacity). The values of thermophysical propenies
are
commonly measured in laboratories at given conditions. Moreover, the values of thermophysical propenies of a substance generally change with temperature. The following example will show how the density of standard air changes with temperature. The densiry of sandard air is a function of temperature and mqy be approximated using the ideal gas larv according to
P p=fr
where
fip$ii&it'iffii|Wi
id:
D6ffilsmryl
: R:
P
l:F ,,iufl:l
&liffi " ,...P .:,lt.tl
u
-A=--i-f-; Deqs-lt1t I
S1,S1LSs
u a tunc1lo1 of tempqratule
,D,,.;
i
=
ir
value for air is
air temperatrue in Kelvin (K)
(323.L5 K) in increments of 5oC. Refer to the F.:
&
s
1 I
'18
-
. In cell Al , type Density
of air
as a
function
of temperature.
I
PIHW.U ':!*.r*
/ | \ 286'e (G'K,)
shows the density of air as a function of temperature in the range of OoC (273.15 K) to 50'C
Qenqltv ftsfnl-l
4
11
g"t constant and
Using Excel, we want to create a able that
L
Ie-$r-er!qF-JQ)
T
sandard atmospheric pressrue (101.3 kPa)
2. In
cells A3 and 83, type Temperature (C), Density (kg/m3), respectively.
I
D
3. In
cells
tivelv.
A5 and A6, type 0 and 5,
respec-
Cnaprun
378
14
ElBcrnoNrc Spnneosnrurs Pick cells A.5 and A6 and use the fill command with the * handle to copy the pattern into cells A7 to A15.
In cell 85, type the formula :(101300)/ Jiffi fa* q*t ra
IJ
bw
Frse*
ia*"t
i*t
6mlsRvl,*rumd
'qg
{li+_* __.to rln ,{ Ul '-SrAel=('101300)ffl2BE.Bl*(A^5+273)) Es ,-l Ft+tn ti Densitv of air as a functisn sf temFerature :fi
lI
:''9.:i
iDensity
Temperature {C)
4:
5
t!
5 10 15
lu uti TT
?5 3u
fts/ru)
|
(Q8 6.9)* (A5 + 27 3)), as shown.
6; :lFl-
u;i; ud;;;
&l
T
" ,ii:
14.3
CnEATTNc Fonuur-e,s nrr
E:rcrr.
379
6.
Use the
7.
Use the Format menu and the Cdls command to change t*re number of decimal digits, as shown,
Edit menu and the FiIl command to copy the formula into cells B6 to B15.
D6F{lgEtriehBd
ry"i
E5
1:
i-,rg:.1
u -r glel::.&.rf,..,"
e | =f ,l B 1 30t1}{(286.9fi45+2/3}}
[- ll
A]:
B p"el_qill.slair,,ec_qfu 4r!o_a"oJJff lSEtgUlE Temperature
(CJ
iDensity [kg/m3J
4
1 1
ft
13 14
15
[l*[tit9Hl"$F*a"d:ffi,1 J*t J--.---;-ffil*,I - "- ""i -
i ; lsffi*a'nrn-ruq$:
.
i
380
Cneprrn
14
Er-ecrRoNrc SpnnepsnsETs
I4.l are shown in Figure 14.3. The cell conten$ were centered using the center button (icon) from the Toolbar. The 6nal results for Example
I
Figurel{.3
Thefinal resultlorEmmplel4.l.
r,-:ilt*ll------.:----------
:r,r.iri.:L-
....ll'alln:iir.iitin,::;li-r,l;trl|-:rli.I;l
::iilr',ri;.-iriti:l!-i]:l::-rL
Ahsolute eell Reference, Relative eell Reference, and Mixed Cell Reference When creating formulas you have to be carefi.rl howyou refer to the address of a cell, especially ifyou are planning to use the FiIl command to copy the pattern of formulas, in the other cells. There are three ways that you can refer to a cell address in a formula: absolate, rebtiae, and rnixed refermca
To better undersand the differences among the absolute, relative, and mixed reference, consider the examples shown in Figure 14.4. fu the name implies, absolua refnmce s absolute, meaning it does not change when the FiIl command is used m copy the formula into other cells. Absolute reference to a cell is made by $columnJetter$row-number. For orample, $A$3 will always refer to the content of cell A3, regardless of how the formula is copied. In the example shown, cell'A3 contains the value 1000, and if we were to input the formula :0.06* $A$3 in cell 83, the result would be 60. Now ifwe were to use the FiIl command and
14.3
n6H,
$tu .6-A* ,.r'l I---
HtutJ B3
CnsATrNc Fonrvrur.es rx Excrr-
38t
''
*:l=0.t16*SA$3
IIY
A
B
:,,',v
uollaf Amounl lnt8r€d Rate 0.06i 6fl1
1000 125{l 150{l 1750 ?000
3
T. Fi
6
E[]
ml 6B
ffi
s
60 6fl
??50
g
m
1U
611
iui
$ 4
sl
#
m l'liHhshmtl
qtBBtl
lll f
-'lE
:i:r)i (b)
il
tigme
14.4
Examples showing the difference between the results 0f
I
formula when absolute and
relatitte cell references are made in the forlnula.
copy the formula down in cells 84 ttoo"gh 811, this would resuh in a value of 50 appearing in cells 84 through 811, as shown in Figarc 14.4(a). On *re other hand, if we were to make arelatioe refnance to !3, that would change the formula when the Fill command is used to copy the formula into other cells. To make a relative reference to a cell, a special character, such as $, is not needed. You simply refer to the cell address. For example, if we were to input the formula :0.06*43 in cell 83, the resuh would be 60; and ifwe use the Fill command to copy the formula into cell 84, the,{3 in the formula will automatically be substituted by M, resulting in a value 75. Note that the formula in cell 84 now becomes :0.06*A4. The result ofapplying the Fill command to cells 84 through 811 is shown in Figure 14.4(b). The mi.xed cell reference cotid be done in one of rwo wqrs: (1) You can keep the column as absolute (unchanged) and have arelative rovr', or Q) you can keep the row as absolute and have a relative column. For example, if you were to use $A3 in a formula it would mean that column A remains absolute and unchanged, but row 3 is a reference row and changes as the formula is copied into other cells. On the other hand, A$3 means row 3 remains absolute while column A changes as the formula is copied into other cells. The use of mixed cell reference is demonsuated in the following example.
Using Excel, create a table that shows the reladonship between the interest earned and the amount deposited, as shown inTable I4.2. In order to create the able for Example 74.2, using Excel we will first create the dollar amounr column and the interest row, as shown in Figure 14.5. Next we will type into cell B3 'We can now use the Fill command to copy the formula in other the formula :SA3*B$2. cells, resulting in the able shown in Figure 14.5. Note that the dollar sign before A3 means
382
CHar"reR
14
F.r.ncrnoNlc Spnr^epsgrurs
lntetst
Rate
Dollar Amount
f
1000
60
7A
/)
75'
t25A
87.5
93.75
1500
90
175A
105
t22.5
rr2,5. 13t,25
2000
120
L4A
150,
?250
135
168.75 2A6.25
71\
105
25AA
150
?:750
rc5
r57,5 175 r92.5
3000
180
21,A
Figurel4.5
1
r87.5
80
r00
I
i
t20 L4A 160 180 l 204 tl
j
i,
',,
n0') 240
l
Ercelspreadsheetfortramplel4.2.
column A is to remain unchanged in the calculations when the formula is copied into other cells. Also note that the dollar sign before 2 means that row 2 is to remain unchanged in calculations when the FiIl command is used. i!,a.,,,,,r,l'lt
a..,,-.]ru*
r,
,.':i.,-:'i:rt--:-:i,:rrl:!r:i-i.iij
-:-.,rri]ulrltflf?ilr.ui?lE;,i
,:.i:.--:.i,.-
-.-
---:!:trLr:l'it: l
14,4 Usrxc E:rcsl Fuxcrrors
14.4
383
Using Excel Functions F.xcel offers a large selection of built-in funcdons that you can use to analyze data. By built-in funcdons we mean standard functions such as the sine or cosine of an angle as well as formulas that calculate the mtal value, the average value, or standard deviation of a set of data poina. The E:rcel functions are grouped into various categories, including mathematical and trigonometric, satistical, financid, and logical functions. In this chapter, we will discuss some of the cornmon functions that you may use during your engineering education or later as a practicing engineer. You can enter a function in any cell by simply typing the name of the function ifyou know it. If you do not know the name of the function, then you can press the Paste "lt""dy Function fuf) button on the toolbar, and then from the menu select the Function category and the Function name. There is also a Help button, marked by a question mark, ?, on the lower left corner of the Paste Function menu, which once activated and followed leads to information about what the function computes and how the funcdon is to be used. Some examples of commonly used Excel functions, along with their proper use and descriptions, are shown in Thble 14.3. Refer to Example 14.3 ao.d Figure 14.6 when studying
TableI4.3. A set ofvalues is given in the wodaheet shown in Figure 14.6.Falnfialize yourselfwith some of Excel's built-in functions, as described in Thble 14.3. !7h.en studying Table 14.3, note that columns A and B contain the data range, which we
EdG{lghryl SB@dlo'"*"1%lffi' ' -, io_:ln"r Ul=-g.Af:nrrat .
have named ualues; cell D 1 contains the angle I 80.
Also note that the functions were typed in the cells El through E14; consequendy, the resuks ofthe executed Excel functions are shown in those cells.
t'}
E14
,'lr
sl=SIN{RADIANSIDIII
At,"s Iu
c'r
Fr
{r .ll "0,
1m
"ql
U
10
q
n
a
,3:
E:
I
E
1il
-" -*--Bl EI
a::;.,
2u
ii
1u
:lil
b
1.1n5[125U3
i\,
E
3.1415S2E54
B
4l
184
8.2
Bl
s Y
'tt,i :.it!,
lBBl
ti
,
q
1.5707s6327[ li
:
B
3:i4ibetcs4_-n
E
11
6.125-/4E-17
1a
-1
{?
l
I 1.22515E-161 T
J-thmeJ rljiiL i'1,
A
Rd I
ii,
,i
s#wtl
I
figure14.6 lhe hcelworksheetforEnmplel4.3. -,r-
-----'i-
-.i-llri,iri!:ij,. --
-
-ililt:r.iii-fiiif
---.- -- -
-.ii --:ir;
,ttl
384
Cnepren
14
ErncrnoNrc Spng'e-psHrsts
-
Decdption ofthe Function It
t*"tni''":;:rlr'";r :SUM(AI:B10) of
in the given range.
sums the values
164
:Iuv(uauo) It
calculates the average value data in the given range,
:AVERAGE(AI:810)
ofthe
or =AVERAGE(vatues):, ;,,
It counts the number ofvalues in the
=COUNT (A1:B10)
given range.
or :COUNT(values)
It determines the
largest value in the
=MAX(AI;Bl0)i,
grven range.
or :MAX(values)
It determines the smallestvalue in the
:MIN(A1:B10)
glven range.
?r----.
10
6
.
=MIN(values) It calculates *re standard deviation for Se
:STDEV(AI:810)
values in the given range.
or =STDEV(values)
It renrrns the value of zr, 3.141519265358979, acntrare to 1 5 digits.
:PIO
DEGREES
It converu
:DEGREES(PI0)
, i):,t
radians to
RADIANS
It
:.:,
";'
,.
.qo,
the value
3.141519265355979
d.S..r. RADIANS(90) of
cosine value ofthe argument. The argument must be in radians.
Ir returns the sine value of r-he argument. The argument must be in radians.
180
j "
L57479
:RJdIDIAI{S(Dl)
3.r4r59
:CoS(PI0/2)
0
.
It returns the
I
li
.l
in the cell from
converm the value &om degrees to radians.
1;105
or
:CoS(RADIANS(D1)) :SIN(PIO/2)
I
or
=51tr{(trApIANS(Dl)) -.,--.-:-,,,
--, ----- -,-- -l:!--,1'...- - ---
0
...-.....'
More examples of Excel's funcdons are shown nTable I4.4.
The Now and Today Functions 'When
you work on an imporant Excel document, it is a good idea to indicate when the document vras last modified. In one of the top cells, you may want to type "1o"11\d6diffed:", and in the adjacent cell, you can use the =now( ) or =today( ) function. Then, each time you access the Excel document, the now( ) function will automatically update the date and the time the file was last used. So, when you print the sheet, it will show the date and the time. If you use the today( ) function, it will only up&te the date.
14.4
SQRI(x) FACTk)
'
,
UsrNc E:rcnr. FuncrroNs
Returns the square root ofvalue
r. ,
385
,
Returns the yalue of the factorial of r. For eample, FACif (5) will return: (5X4X3)(2X1)= 1ZO,
'
Trigo norne*i c Fanctio ns
TAN(x)
:
DEGREES (x)
Retuins the value for the grngent of x. The argurnent must be
'
rafians. .
Coaverts theivalue
ofrfrom
radians to degrees.
ofrin degtees. ACOS(x)'
,,'
',.
:
in
'.
It
returns the value
l
This is the inverse cosine funcdon of r, It is used to &termine the value of an angle when its cosine value is known. It renuns the angle value rn radians, wheo the value ofcoslne benveeol :- I and 1 ' is used for argqment rc ,:
ASIN(x)
This is the iqverse sine function of .rc It is used to determine the value of an angle when its sine yalue is known. It returns the angle value'in radians when the value of sine falls between - I and 1.
ATAN(x)
This is the inverse tangent of the *function, It is used to determine the value ofan angle when its tangent value ii known.
Exponmtial and Logarithmh Fwnaions
EXP(X)
'
t*t
Rqrurnsthevalueofat
,,,
.
Returns the value of the natural logqdthm
.t
lo.G.lil
greaqer
i.**,
thanO.
'
ofr.
Note thqt rmust be
i
rtre value of the common
hgpithm of r,
..-......--..'.
Using Excel, compute the average (arithmetic mean) and the standard deviation of the density of water data given in Thble 14.5. Refer to Chapter 19, Section 19.5, to learn about what the value ofstandard deriation for a set ofdata points represents. Refer to Figure 14.7 when following the steps.
1. ln cell 81, rype Group A findings, and in cell Cl rype Group B findings.
of cells 83 and C3, type Density (kg/-'). Highlight the 3 in the kg/m3, and use following command to make 3 a superscript. Choose Cells from the Format, Cells the menu, and turn on the superscript toggle switch. In cells 85 to C14, rype densityvalues for Group A and Group B. 3. Next, we want to compute the arithmetic means for the Group A and Group B data, but first we need m crerite a tide for this compuadon. Because we are calculating the average, we might as well just use the word AVERAGE for the tide of our calculations, thus in cell
2. In
each
815 typeA\IERAGE:.
Cnlr'rrn 14
Er,Bcrnor.rrc SpnBeosnrprs
Group A Findingg
p (kll'"-3) r020 1015
990 1060 1030
L
Figurel{.?
Tfte Ercel spreadsheet
trample 14.4.
for
.
Group B Findings p
(Whlf) 950 940 890 1080 1120
954
900
975 1020
1040 I 150
980
910
960
r020
I
14,5
UsrNc E:
387
4. In order to hare Excel compute the average, we use theAVERAGE function in the following manner. In cell 816, we rFpe :A\IERAGE(85:814), and similarly in cell C16, we type :A\/ERAGE(C5:C14).
5. Next, we will make a tide for the standard dwiation calculation by simply ryping in cell B 1 8 SIAI\ID. DEV. 6. To compute the standard deviation for the Group A findings, in cell B19 type :SIDEV(85:814), and similarly to calculate the standard deviation for the Group B findings, in cell C19 type =SIDEV(C5rCL4). Note that we used the funcdon STDEV and the appropriate data range.
The final results for Example L4.4 are shown in Figare 147.
i-', ,,a,,
-,
-.
14.5
r
: --= '-t -:l'r-
Using Excel Logical Functions In this secdon, we will look at some of Excelt logical functions. These are functions that allow you to test various conditions when programming formulas to analyze data- Excel's logical funcrions and their descripdons are shown in Thble 14.6. F-lrcel also offers relational or comparison operators that allow for testing of relative magnitude of various arguments. These relational operators are shown in Table L4.7. Ye will use Example 14.5 to demonstrate the use of Excel's logical funcdons and relational operators.
Iogical Functions
Description of rfi e Function
AND(logicl, logic2, logic3, . . .)
Returns tnre if all iugumen$ are ffue and returns false if any of the arguments are false,
False( )
Returns the logical value false.
IF (logical test,
value-if-true,
value-ilfalse)
It first
if true, then it retums waluadon of the logical test deems false, then it returns the value-if-false the
evaluates the logical test;
valutiltrue; if tlre
value,
of its argumerq returns true for a false argument and false for the true argument.
NOT(ogical)
Reverses the logic
OR(ogicall, logical2, . . .)
Returns TRUE if any argument is true and returns FAIJE if all arguments are false.
TRUEo
Returns the logical value TRUE.
388
Cnaprrn
14
ErpcrnoNrc SpnEADsHrsrs
Relational Operator
Description Le"ss
.
than
Less than
:
or equal to
Equal rc : Greater Greater than or equal to
than
Notetualtl
t ,l I 1
l
.,,, ,.,
,,il-
1
l
The pipeline shovrn in Figure 14.8 is connected to a control (check) valve that opens when the pressure in the line reaches 20 psi. Various readinp were taken at different dmes and recorded.
E Figurell.8 A schematic diagran
for
Erample 14.5.
t
Figurel4.g
The solution
to Enmple 14.5.
14.6 Pr.orrrxc vrrn Excsl
389
Using Excel's logical funcdons, create a list that shows the corresponding open and closed position of the checkvalve (see Figure 14.9). The solution to Example 14.5 is shown in Figure 14.9. The pressure readings were entered in column, - In cell 83, we r;pe the formula =IF(A3 )= 20,"OPEM,"CLOSED') and use the Fill command to copy the formula in cells 84 through Bl0. Note that we made use of the relational operator ): and relative reference in the IF function.
14.6 flotting with Excefi Todayt spreadsheen offer many choices when it comes to creating charts. You can create column charts (or histograms), pie chara, line charts, orry charts. As an engineering student, and later as a practicing engineer, most of the charts that you will create will be ofry-type chara. Therefore, next we will explain in detail how to create an ry chart. Excel offers Chart Mzard, which is a series of dialogue boxes that walks you tfuough the necessary steps to ctez.te chart. To create a chan using the Excel Chart \Vizard, follow the
^
procedure explained here.
o
. .
"
Select the data range as was explained earlier in this chapter.
Click the Chart'Wizard icon from the toolbar buttons. Select the XY (Scatter) Char rype. The XY Chan rype offers four Chan sub-rype options. (It is important to note here that the Line chan is often mistakenly used instead of XY (Scaaer)). tdata points connected by smooth lines" From the four Chan sub-type opdons, select the
Chart option. Click on the Next bunon. " o The selected data range will show in the Data Range box. " Click the Next bunon. In the Chan Option dialog boxes enter the Chart tide, Category (X) tide), Value (Y) axis (this is the y axis tide), and click the OK button. 'When
oi*
(this is the x axis
creating an engineering chart, whether you are using Excel or using freehand methods, you must include proper labels with proper units for each axis. The chart must also conain a figure number with a tide explaining what the chart represents. If more than one set of data is plotted on the same chan, the chart must also contain a legend or list showing symbols used for different data sem.
Using the resuhs of Example L4.1, create a graph showing the value of air density as a function of temperature.
1. First we will select the data as
range,
shown.
2. Next clic.k the Chan \Tizard icon from the toolbar buttons and select the XY chart type. Also select the Chart sub-type, as shown. Click the Next button.
390
14.6 Pr,ornxcwnnExcnr.
391
3. You will nonr see the Chart Source Daawindowwith the seleced das range in the Data Range box. Click the Next button.
4.
Novr enter the Chart tide, the X axis
tide, and the Y axis tide, Click the Next button.
lWlli
as shourn.
5. You now can display the chan as an object in the wodrsheet or as a new sheet.
Select the "As object irf button and clic.k Finish.
6. If for any reirson you want to modifr or edit the chart,
dick the Chart menu and choose the Chan Options.
Now you can modi& or edit the tide, axis labels, spacing among the gridlines, legend, and so on. Just choose the perdnent tab and proceed.
7. Finally, you can place
'u111*n;.j {ji i:'.*
392
the
chaft in an appropriate location, as shovrn.
14.6 Pror:rrxc wrrn
F,:rcsr.
393
It is worth noting that you can plot more than one set of data on ttre same chan. To do so, first pick the chan by clicking anywhere on the chart area, and then from the Chart menu use the Add Data command and follow the steps to plot the other data set to the charc
Plotting Two Sets of Data with Different Ranges on the Same Chart At times, it is convenient to show the plot of two variables
versus the same variable on a single
chart. For example, in Figure 14.10, we have shown how air temperanrre and wind speed changes with the same variable time. Using Example L4.7, we will show how you can plot rwo sets of data series with different ranges on the same chart.
L-J
Figure
14.10
Using trcel to plot two sets of data with different ranges.
394
Cueprrr'
14
Fr.nqln6rvrs SpnnaosnsBts Use the following empirical relationship to plot the fuel consumption in both miles per gallon and gallons per mile for a car for which the following relationship applies. Note: Vk the speed of the car in miles per hour and the given relarionship is valid for 20 s V < 75.
Fuel Consumption (Mriesper ---- r-- Gallon)
: :909 tr,y900 + fr.85
Refer to the Excel sheets shown in the accompanfing figures when following the steps.
1. First, using Excel and the given formula, we compute the fuel consumption in miles per gallon and gallons per mile for the given range of speeds.
Note that the values in cells C3 through CI4 are the inverse of cell values in 83 drough B14.
!*'
fffi
Fprnd I@b ghff
2. Whdow
:eP -_ I U.: @"3 ro.: q! SJ ,', q' dd ta ; . r,.rur. t r
\7'e plot the fuel consumption in miles per gallon versus speed, as shown.
3. Sfith the mouse pointer in the chart area, click the right mouse bufion and select dre $gurce
Data...,asshown.
L4.6 Pr.orr:ncwrrn 4. In
the Name box, type Miles Per Gallon and then press the
fr{os88 &osd{rytt
{dd button.
Excrr,
395
396
Crrll.ran
14
Ewc"rnorvrc Spnpeosnrsrs
5. In the Name box, qrye Gallons Per Mile. Next, in the X
Values box, click on the litde spreadsheet icon, then select cells A3 rtuough A14, and press rhe Enter key. Then, in the Y Values box, click on the litde spreadsheet icon, select cells 83 through B14 and press the Enter key. Because Gallons Per Mile values are relatively small, they will be displayed close
to the Speed axis,
as shown.
14.6 Pr-orrutc vrrn 6.
Excrr,
397
\fith
the mouse pointer over the Gallons Per Mile curve, double-click the left mouse button. From the Format Data Series window, pick theAxis tab, select the Secondary axis, as shown, and press OI(
7. Next, we will shon' how to get rid of the data markers and change the line type. Vith the mouse pointer over the Gallons per Mile curve, right-dick and select the ChartType. . ., as shown.
398
CrHprnn
14
F.r.pctr'RoNrc Spnrepssrurs
8. From the Chart sub-type box,
select the
'Scatter with data points connected by
smoothed Lines without markers,"
as
shornrn.
9.
\fith Per
the mouse cursor on the Gallons
Mile curve, right-click and select the
Fgflnat Dafa Series
...
option,
as
shown.
14.6 Prornrc vrrrr Excsl 10. From the Format Data
Series
399
window, pick the Patterns tab and change the line style and
weight, as shown.
The final results for Example 14.7
are
25
0.070 0.060
a2o
0.0s0 E 0.a4a b
8D 15
o
0.030
810 f 0
shown in Figure 14.11.
t-Mtt"..-rr"tl* l. --- Gallonspermite
| 020406080
0.020
€ F
|- 0.010 0.000
Speed (mph)
il
figurel4.ll
Tiefinalresultsfortranrplel4ll. ,:
t
,:r,:ar
.:l
.
aa:-,,::---, -: ;:- i-lr.--.-.
:,1
400
Crr.er.rrn
14
Er,scrnoNrc SpnneosnnrTs
14.7 Matrix Computation with
Excel
During your engineering education, you will learn about different types of physical variables. There are those that are identifiable by either a single value or the magnitude. For example, time can be described by a single value such as two hours. These types of physical variables, which are identifiable by a single value, are called scalars. Gmperature is anor:her example of a scalar variable. On the other hand, ifyou were to describe the velocity of a vehicle, you not only have to specifr how fast it is moving (speed), but also its direction. The physical variables that possess both magnitude and direction are called aec'tors.Tltere are also other quantities that in order to describe them accurately we need to specifr more than rwo pieces of information. For o
5 el [6 lrrl = lr26t4l L-5 8 ol
{L}:
fi)
Here, matrix [lV] is a dree by tfuee (or 3 X 3) matrix whose elements are numbers, and {Z} is a three by one matrix with its elements representing variables n y and a The [tr/] is called a square matrix , A square matrix has the same number of rows and columns. The element of a matrix is denoted by its location. For example, the element in the first row and the third column of matrix Fy'] is denoted,by ns, which has a value of nine. In this book, we denote the mauix by a boldface letter in brackets [] and {}, for enample: !Vl, [f], {F}, and the elements ofmauices are represented by regular lower case leners. The {} are used to distinguish a column matrix. A column matrix is defined as a matrix that has one column but could hare many rows. On the other hand, a row matrix is a mauix that has one row but could hane many columns.
U}:
{i}
are examples of
and{X}:
{i
column matrices, whereas
fcl = lS o z -3landlr] are examples
: ly yt tu)
of row matrices.
discuss matrix algebra in more detail. Ifyou do not have an adequate background in mauix algebra, you may want to read Section 18.5 before studying the
In Section 18.5, we will
fo[owing examples.
14.7 Mernrx
CorvrputnrroN vrrH Excnr
401
Matrix Algebra Using E:
ry
=
Givenmatrice,,,r,
[l i i),"t
:li
-_il,^{c}=
:
{-i},-"
Excel ro perForm the following operations.
:
(a) L4l + lrl (b) [,{] - [a]
k) lA)[B]: ($ [/]{c} =
:
?
I
? 1
If you do not have any background in mauix algebra you may want to study Section 18.5 to learn about matrix operation nrles. The manual hand calculations for this example problem is also shown in that section.
Refer
to the
Excel sheets shown
in the
accompan)nng figures when following the
srePs.
1. In the cells shown, type the appropriate characers and values. Font option to create the boldface variables,
as
shown.
Use the Format Cells and
4A?
Cneprnn
14
F.r.ncrnomc SpnEADsHEErs
+ IB) =, and using the left mouse bumon, pick cells 89 through D11, as shown in the accompa-
2. In cell A10, tFpe lAl
nying figure.
3. Next, in the Formula bar, type = 83: D5 + G3{5 and, while holding down the Ctrl and the Shift kqrs, press the Enter i ,A: i::i
E
r
c-l
a
;TT:;!
HI
I
r,
Note that using the mouse, you could B3:D5 or G3:I5 instead of ryping them. This sequence of operations will create the result shown. You follow a procedure similar to the one outlined in step 2 to perform [A] - lBl, exkey.
M:-
also pick the ranges of
0
r Et
;i
I
4
6
-2
3
4
'l
g $
at* tBlr
15 10
A
Al.fBl' l:!atuir
l8
I
i^/"
'11
-z
5
10
cept in the Formula bar type
G3rI5.
a
i, 7
2
.;,
4 13
1
f
.,
ls TF
= B3:D5
-
L4.7 Marnor CoupurerroNwrrg
Excnl
403
4. To cany out the matrix multiplication, fint type lAllBl = in cell A18, as shown. Then pi& cells 817 through D19.
5.
In the Formula
bar, type =MMULI
(83:D5,G3:I5), and, while holding down the Ctrl and the Shift keys, press the EnA:
ni
]
ts
x
.t
I z
o
{
lAl
q
"
7
s
SE
6 a
a 1lI
{i
:
A]+ IFI=
4
1
1
4
rfr l!
l8
E
60
lAl{q}a
33
tailttt
jl
75
19
n x
*
type =MMULT(83:D5, I3rL5), and, while holdingdoqrn the Ctrl and the Shift kqrs, press the Enter key. This sequence of operations will create the result shown in the accompanying figure.
tg
[Al-lBl'
key. Similarly, you can perform the ma-
trix operation lAllcl. First you pick cells 821 through B,23, n the Formula bar,
10 it
""
..t
404
Cne'prrn
14
ErrcrnoNrc SpnplosnsErs The formulation of many engineering problems leads to a system of algebraic equations. As you will learn later in your math and engineering classes, drere are a number ways that you c:ln use to solve a set of linear equations. Using Example 14.9, we will show how you crul use Excel to obtain a solution to a simulsneous set of linear equations.
Consider the following three linear equations with three unknowns: xy x2, ar,d x3. Our intenr here is to show you how to use Excel to solve a set oflinear equations.
2x1*x214=13 3x1 * 2x, * 44: 32 5x1 - x2 * 3xs: 17 x2 :
The solution to this problem is discussed in detail in Chapter 18 and is given by: x1
:
5, and e3 = 4.The solution easily can be verified by substituting the resuhs (the values xp x2, and 4) back into the three linear equarions.
2,
of
2(2)+5t4:t3 3(2) + 2(5) + 4(4) -- 32
5(2)-5+3(4):17
Q.ED. Now the Excel solution. Refer to the Excel
sheets shown
in the accompanying
figures
when following the steps.
1. In the
cells shown, type the appropriate characters and values. Use the Format Cells and
Font option to create the bold and subscript variables.
14,7 Mernpr CouprrrerroNvrrlr EncsL 405 2. In
cell A10, type [A]-1
=, and using the left mouse bufton, pick
cells 89 through
Dl1
as
shown.
3. Next, in the formula bar, type =MIN\IERSE (83:D5) and, while holding down the Ctrl and the Shift keys, press the Enter key. This sequence of operations will create the result shown. The inverse of mauix [r4] is computed.
406
Cnerrrn
14
Er,ncrnorrc Spnnensnwrs 4. Type the information shovrn in in the accompanying figure.
cells
A14, B13 through Bl5, CI4, D14, and E14,
as
shown
5. Then, pick cells F13 through F15; in the formula bar, type =MMULT (89:D11,
K3:K5); and, while holding down the Ctrl and the Shift keys, press the Enter key. This sequence of operations will create the result shown in the accompanymg figure. The value of calculated.
x1, x2,
and x3 are now
L4.8
14.8
Cunvn
Frrrrxcwrrn Excsl
407
Curve Fitting with Excel Curve fining deals with finding an equation that best fim a set of daa. There are a number of techniques that you can use to determine these functions. You will learn about them in your numerical methods and other future engineering classes. The purpose of this section is to demonstrate how to use Excel to find an equation that best fits a set of data which you have plotted. V'e examples.
will demonstrate the curve-fitting capabilities of Excel using the following
In Chapter 10, we discussed linear springs. IZe will revisit Example 10.1 to show how you caa to obtain an equation that best fits a set offorce-deflection data for a linear spring. For a given spring, in order to determine the value of the spring constant, we atnched dead weights to one end of the spring, as shown in Figure l4.l2.Ve have mer*ured and recorded the defection .aused by the corresponding weights, as given in Thble 14.8. Vhat is the value use Excel
ofthe spring constant? Recall, the spring constant
I
Figurel4.l2
The spring setup
Erample 14.10.
for
2 is determined by calculating the slope of a forcedeflection line (i.e., slope : change in force/change in deflection). !7'e have used Excel to plot the defection-load resuls of the experiment using the XY (Scarter) without t}re data points conneced, as shown in Figure l4.l3.If youwere to connect the experimental forcedeflection points, you would not obain a suaight line that goes through each experimental point. In this case, you will ffy to come up with the best fit to the data points. There are mathematical procedures (including least squares techniques) that allow you to find the best fit to a set ofdata poina; Excel makes use ofsuch techniques. To add the trendline or the best fit, with the mouse pointer over a data point, click the
?\s .3
10
The Defection
ofthe $priag Weight(N) 5.0
(mm)
15 20
?5
q
Deflection (mm)
9
l0
T7
r5.0
29
E
35
Ihe IY (Scatter) plol of erperimental data points without the
2Q.0
figuret4.t3
data points connected.
408
CHer"rsR
14
F-r.nqlReNrs Spnneosgrsrs
t
Figurel4.l4
Add Trendline.. . .
L
Figurelll5
TlreAddTrendlinedialog box-Type.
right button and choose Add Trendline . . . , rs shown in Figwe 14.14. Nent, from the Add Trendfine dialog box, under Trend/Regression type, select Linear, as shown in Frgue 14.15. Then click on the Options tab of the Add Tiendline dialog box and toggle on the Set intercept= and the Display equation on chart, as shown in Figure 14.16. After you press OBl, you should see the eguation y : 0.5542x on the chart, as shown in Figarc I4.l7. Ve have edited the variables of the equadon to reflect the experimental variables as shown in Figure 14.18. To edit the equation, left-click on the equation $ : A.5542x) and change it to read F : 0.5542 x, where F : load (N), and x : defection (mm), as shown in Figure 14.18. In order to examine how good the linear equation F : 0.5542x fits dre dat& we compare the force resula obtained from the equation to the actual daa points as shown in Thble 14.9. As you can see rhe equation fics the data reasonably well.
14.8
CunYs FrrrrNcwrrn E:rcer,
449
y = 0.55 42x
?s
-2
910 5 0 4A
Deflection (mm)
bd
Figuru
14.16
F
:
The Add Trendline dialog
0.5542 x, where F
:
I
ior-0ptions.
Ihe linear fit to data of Enmple
14.10.
load (N),
-2.
The Measured
q10
the Sptiog' r (-n)
Defection of
5
010203040 Deflection (mm)
I
l4.l?
andx:deflection(mm)
4rs
0
tigure
figurc
tl.t8
The edited linear
fit to
data of trample 14.10.
,9 ;17 :29 '35
Measured Spring Force (N)
The Predicted Force (N) using
F = O.5542
5.0 10 15.0
16.1
20.0
19.4
*
5.0
9.4
I
Find the equation that best fits the following set of data points in Tirble 14.10. 'We fust plot the data poina using the )ff (Scater) without the data points connected
as
shown in Figure 14.19. Right-click on anyof the data points to add a trendline. From the plot of the data points, it should be obvious that an equation describing the relationship between r andT is nonlinear. Select a polynomial of second order (Order: 2), as shown in Figure 14.20. Then click on the
Options tab oftheAddTiendline dialogbox and toggle on the Disptayequation on chartand
410
CneprsR
14
Er.ecrRoNrc Spnpeosnnsrs
1.50
,X i
r r I i, 1
0.50
0.00 0.50 1.00 1.50
2.00 2.50 3.00
2.00
0.00
4.75 0.00
-0.50 L-
'0.25
0.00 0.50
0.75
2.00
1.00 1.50 2.A0
I
figurel{.|9
The XY (Scatter) plot of data points
il
2s0 3.00 3.50
x
0.00
for Emmple l4Jl.
tigurc14.20 lheAddlrendlinedialogbor-Type.
the Displey [-squared value on chart, as shown in Figue 14.2I. After you press OK, you should see the equation y : * - 3x * 2and R2 : 1 on the chart, as shown nEigare 14.22. The R2 is called the coefficient of determination, and its value provides an indication of how good the 6t is. R2 : 1 indicates perfect fit, and R2 values that are near zero indicate extremely poor fits. The comparison between the acrual and predicted y values (using equation y : x2 - 3x * 2) is shown in Thble I4.II.
Suvruanv
I
Figurel4.2l
411
TheAddTrendlinedialogbor-0ptions.
I.50
x 0.50
0.00 0.50
0.00
-0.50 0.00 0.50 1.00 1.50 2.00 2.50 3.00 3.50
I
figurelL22
2.00
2.00
1.00
a.75 0.00
4.75 0.00
r.50
-0.25
-0.25
2.00 2.50
0.00 0.75
0.00 0.75
3.00
2.40
2.AO
TieresultsforExamplel4.ll.
SUMMARY Now that you have reached this point in the text
'
You should know that a spreadsheet is a tool that can be used to solve an engineering problem. Spreadshee* are commonly used to record, organize, andanalyze data using formulas. Moreover, you can use a spreadsheet such as Excel to present the resulrc ofan analysis in chaft
Crurrrn 14
412
Er.pcrnorvrc Spnu,osnnsTs form. You can input your own formulas or use the built-in functions provided by the spreadsheet.
" You should know how to move around in a workbook and input data into different cells. You should also know how to edit the content ofa cell. n You should know how to select muldple cells and cre:fie a range. You should also realize that you c:m name a range and use the name in your formulas or in ploming data. " You should underssnd how to refer to a cell by address. You should also knovr the differences among a cell's relative address, absolute address, and mixed address, and remember to use the proper address when creating formulas. You should be familiar with Excelt built-in funcdons.
ir
.
n You should know . You should know . You should know . You should know
how how how how
to insert cells, columns, and rows in an existing worlsheet. to create a proper engineering chan using Excel. to perform matrix operations with Excel. to perForm curve fitting with Excel.
l_-ll x4J.
Using the Excel Help menu, discuss how the following functions are used. Create a simple example and demonstrate the proper use of the function.
a. TRUNC(number, num-digits) b. ROUND(number, num-digits)
-
c. COMBIN(number, number-chosen) d. DEGREES("ngI") e. SLOPE(known-y's, known-x's)
f. x4.2.
14.3.
of water above ground in the water tower and the water pressure in a pipeline located at the base ofthe water tolver. The relationship is given by
P: where
P:
CEILING(number, significance)
In Chaprcr 20, we will
cover engineering economics. For now, using the Excel Help menu, familiarize yourself with the following functions. Create a simple example and demonstrate the proper use of the function. a. FV(rate, nper, pmr, pv, type) b. IPMT(rate, per, nper, pv, fv, type) c. NPER(rater pmtl pv, fr, .yp")
d. PV(rate, nper, pmt, f",.fp.) In Chapter 10, we discussed fuid pressure and the role of water towers in small towns. Recall that the function of a water tower is to create a desirable municipal water pressr[e for household and other usage in a town. To achieve this purpose, water is stored in large quantities in elevated tanlis. Also recall that the municipal water pressure may vaty from town to town, but it generally falls somewhere between 50 and S0 lb/in'z (psi). In this assignment, use Excel to create a table that shows the relationship between the height
pgb
p
:
: h:
g
the water pressure at the base of the warer tower in pounds per square foot (lb/fl)
the density ofwater in sl'r$ per cubic foot,
p:
I.gfslugs/ft'
the acceleration due to gravity, g
= 32.2 ftl*
the height ofwater above ground in feet (ft)
Create a table that shows the water pressure in lb/in2 in a pipe located at the base of the water tower as you
vary the height of water in increments of 10 ft. Also plot water pressure (lb/in2) vs. the height of water in feet. What should be the water level in the water tower to create 80 psi water pressure in a pipe at the base of the water tower? X4.4. As we o
PnosLEMs fluid dynamics problems. The viscosity ofwater can be determined from the following correlation: 11'
-
6r!QatT-c)
where
p=
viscosity (N/s. m2)
7= rcmperature (K) 4=2.4t4 x 10-5 (N/s.m2) cz= 247.8K
ct= l40K
t4.5.
14.6.
Using Excel, crate a uble that shows the viscosity of water as a function of temperature in the range of 0"C (273.15 K) to 100'C (373.L5 K) in increments of 5oC. Also create a graph showing the value ofviscosity as a function of temperature. Using Excel, create a table that shows the relationship between the units of temperature in degrees Celsius and Fahrenheit in the range of -50o to 150'C. Use increments of 10'C. Using Excel, crqrte a able that shows the relationship among the unis of height of people in centimeters, inches, and feet in the range of 150 cm to 2 m. Use increments of 5 cm.
Problen
l4.ll
413
Using Excel, create a table that shows the relationship among the units of mass to describe people's mass in kilogram, slugs, and pound mass in the range of 20 kg to 120 kg. use incremens of 5 kg. 14.8. Using Excel, create a table that shows the relationship among the units of pressure in P4 psi, and inches of water in the range of 1000 to 10,000 Pa" Use increments of 500 Pa. 14.9. Using Excel, create a table that shows r:he relationship between the units ofpressure in Pa and psi in the range of 10 kPa to 100 kPa. Use increments of 5 kPa. 14.t0. Using Excel, cr@te a able that shows the reladonship between the units ofpower in watts and horsepower in the range of 100 V to 10,000 'Wi Use smaller increments of 100 \tr up ro 1000 I7, and then use increments of 1000 \f a[ the way up to 10,000 W t4lt. As we explained in Chapter 7, the xt resistance to the motion of a vehicle is something imporant thar engineers investigate. As you may also know, the drag force acting on a car is determined experimentally by plac' ing the car in a wind tunnel. The air speed inside the tunnel is changed, and the drag force acting on the car is measured. For a given car, the experimental dara is generally represented by a single coefficient that is 141.
414
Crrnpren
14
called,the drag
ElncrnomcSpnsADsHEEts
cofficimt lt
is defined by the
Discuss your findings in terms of power consumption
following
relationship: Fd
L;:-
l_ :oVzA 2',
where Ca
=
: : p 7: A:
-F6
function of speed and air temperature. The cantilevered beam shown in the accompanying figure is used to support a load acting on a balcony. The deflection ofthe centerline ofthe beam is given by the following equation: as a
1412.
,, -IrN2 r j + 6L2) - 4Lx t: ffi@2
drag coefficient (unidess) measured drag force (N or lb)
where
ab density (kg/m3 or slugs/ff) air speed inside the wind tunnel (m/s or ft/s)
frontalarea of the ,-ar (m2 or f?)
The fronal arear4 represents the frontal projecdon of the car's area and could be approximated simply by multiplying 0.85 times the wi&h and the height of a recangle that oudines the front of a car. This is the area that you see when you view the car from a direction normal to the front grills. The 0.85 factor is used to adjust for rounded corners, open space below the
7: ea : E: .[ : r : Z:
defection at a given
r location (m)
disuibuted load (N/m) modulus of elasticity
(N/*1
second moment of area (ma) distance from the suppoft as shown (x)
len$h of the beam (m)
bumper, and so on. To give you some idea, typical drag coefficient values for sports cars are benveen 0.27 to 0.38, and for sedans are between A34 rc 0,5. The power requirement to overcorne air resisance is computed by
P:
Fav
where
P
:
t
horsepower (hp)
power (watts or
:
ft.lbA) 550 ft.lbA
Pmblem 14.12
Using Excel, plot the defection of a beam whose length is 5 m with the modulus of elasticity of E =
and
t
horsepower (hp)
= 746W
The purpose of this
exercise
-I: 99.1 x 106 mma. The beam is decarry a load of 10,000 N/m. V'hat is the maximum deflection of the beam? 1413. Fins, or enended surfaces, are commonly used in a variety of engineering applications to enhance cooling. Common examples indude a motorcycle or lalrtn mower engine head, extended surfaces used in electronic equipment, and finned tube heat exchangers in 200 GPa and
is to
see how the
power requirement changes with the car speed and the air temperature. Determine the power requirement to overcome air resistance for a car that has a listed drag coefficient of 0.4 and widrh of 74.4 in. and height of 57.4 in. Vary the air speed in the range of 15 m/s ( V < 35 m/s, and change the air density range of 1.1 1 kg/m3 1 p 1 L29 kgl m3.The given air density range corresponds to 0o to 45oC. You may use the ideal g"s lry to relate tle density of the air to its temperature. Present your findings in both kilowatts and horsepower as shown in the accompanying spreadsheet.
signed
to
room heating and cooling applications. Consider a rectangular profile of aluminum fins shown in the accompanying figure, which are used to remove heat from a surhce whose temperature is 100"C (4* = 100'C). The temperarure of the ambient air is 20"C.
PnosLEMs 'We
in determining how the temperanrre varies along its length and ploming this temperantre variation. For long fins, the remperanre distribution along the fin is given by are interested
of the fin
415
Using Excel, creare a table that shows the relationship between the circular and the recangular duct, similar to the one shown in the accompanying able.
T- T^bi"nt: (4* - T^ti*r)e-^
Length of One Side of Rectangular Duct (length a), mm
where Lene$ b 400
207
450
: p: .r4 A: h
the heat uansfer coefficient (\V'/.n'
. K)
500
550
perimeter of the fin 2*(a*b), (m) cross-sectional areaof the
fin (a*b),
600
(m2)
thermal conductiviw of the fin material
(Wm '
14.15.
19
Plot the temperarure distribution along the fin using
the following data: h: 168 '!7/m . K l, L2Wlnr2 . K a = 0.05 m, b : 0.01 m. Varyrfrom
A Pitot tube is a device commonly used in a wind tunnelto measure the speed ofthe airflowing overamodel. The air speed is measured from the following equation: 2Pa V_
0 to 0.1 m in increments of 0.01 m.
p
where
V: Pa : p:
I
b
T 14.16.
Problem 14.13
ab speed (m/s) dynamic pressure (Pa) density of
ur (1.23 t g/-u)
Using Excel, create a table that shows the air speed for the range of dynamic pressure of 500 to 800 Pa. Use increments of 50 Pa. Use Excel to solve F.xample 7.1. Recall we applied the trapezoidal rule to determine the area of the shape given.
14Jt. 14.14.
A person by the name of Huebscher dweloped a relationship between the equivalent size of round ducts and recangular ducts according ro (
\Fe will discuss engineering economics in Chapter 20. Using Excel, create a table that can be used to look up monthly payments on a c:u loan for a period of five years. The mondrlypayrnents are calculated from
ab\0.62s
D: t.3;r-: . (a * b)o'25
A_
where
D
-
diameter of equivalent circular duct (mm)
a
=
dimension of one side of the recangular
duct (mm) b
=
the other dimension of the rectangular ducr
(-nt)
\'*
l+
5 1200
1200 OU
/
T 1
where
: P: I:
z4
monthlypaymenr in dollars the loan in dollars interest f?t€, €.g.r 7,7.5, , . .
,9
416
CH*run
14
loan ffi
Errc"rnoNrc Spnr,eosasrts Create a table that shows the viscosity of air as a function of temperature in the range of 0"C (273.15 K) to I00"C (373.L5 K in increments of 5"C. Also cr€rfie a graph showing the value of viscosity as a funcdon of temperafue as shon'n in the accompanyrng spread-
Interest Rate
10,000 15,000
sheet.
20,000 t4J9.
25,000
In Chapter 11, we orplained the concept ofrrindchill
Ife said that the heat transfer rates from your body to the surroundings increase on a cold, windy day. Simply stated, you lose more body heat on the cold, windy day than you do on a calm day. The windchill index accounts for the combined effect ofwind speed and the air temperature. It accounts for the additiond body heatloss that occurs on a cold, windyday. Thewindchill factors.
14.X8.
A person by the name of Sutterland
has dweloped a
correlation that can be used to evaluate the viscosiw air as a function of temperature. It is given by
l":
c,7fr5
t*; n
values are determined empirically, and a common correlation used to determine the windchill index is
where ;r,
= viscosity (NA .
7=
-')
temperature (K)
q = t.458 X cz= II0'4K
of
10-6
r
: (10.45 - v + ro\/-v)Q3 - r^)
where
iks)
(
wcr
*'s'K1i2,/
VCI: I/: 4:
windchill index (kcal/m2 . h) wind speed (m/s) ambient air temperature (oC)
Problem 14.18
PnonleMs and the value 33 is the body surface remperature in The more common equivalentwindchill temperaQ*,6. ("C) is given by
ture
:
Create a table that shows the windchill temperafor the range of ambient air temperature -30'C < T, < 10"C and wind speed of20 km/h < y< 80 km/h as shown in the accompanying spread-
tures
degrees Celsius.
4qoh"r"nt
417
0.045(5.27vo'5
+
x (T^-
33
33) +
10.45
- 0.28v)
sheet,
Note that Zis expressed in km/h.
Problem 14.19
Use the data given in Figure 14.10 and duplicate rhe chan shown there. 14.4. Use Excel to plot the following data. Use two different -/ axes. Use a scale of zero to 100"F for temperature, and zero to 12 mph for wind speed. 14.20.
Time Temperature Vhd (p.r"rJ CF) Speed (mph) ,1 ,2 t4
3
,l
,6 t7 d
14.2. Use Excel to plot the following &ta for a pump. Use two differentT axes. Use a scale of zero to 140 ft for the head and zero to 100 for efficiencv.
FlowRate Headof r (cPM) Pump (ft)
Efficiency(o/d
l
0
120
2
t19.2
10
30 50 70 80
0
75 80 6/, 82
4
4
116.8
5
6
ttz.8
8
t07.2
)
10
78
5
t1
ql ,
79
75
zl
I4
80.8
'7.)
70 68
5
r6
68.8
50
18
55,2
3A
I
,
100
418 n4.23.
Cs,lprrn
14
Fr.ncrRoNrc SpnslosrrsErs
Use the following empirical relationship to plot the fuel consumption in both miles per gallon and gallons per mile for a car for which the following relationship applies. Note: I/is the speed of the car in miles per hour and the given relationship is valid for 30 < V < 70.
14.24.
1050xy + y1'88
g10
Startingwith a 10 cm X 10 cm sheet of paper, what is the largest volume you c:m create by cutting out r cm X x cm from each corner ofthe sheet and then folding up the sides. Use Excel to obtain r:he solution. Hint : The volume created by cucing out * cm X x cm from each corner ofthe 10 cm X 10 cm sheet ofpaper is given by V: (10 - 2tX10 - L'c)x.
n4.25.
l, i il{t} l-',:: o I
L
lAl:17
lt, 3 -tl3l,
f 'l
[C] : \-2 ], p.tfor* the following
-r;r
-!0,,
:
:r lta
-0.76 0.85r -o.oer l{a} -ooer -1:z;)l?)
:1:l
-1|
[r:.aalrzoll
lo.qz)(zq) 14.28.
Find the equation that best fits the following set ofdata points. Compare the acrual and predictedT values.
14.29.
Find the equation that best fits the following set ofdaa points. Compare the acnral and predictedT values.
14.30.
Find the equation that best fits the following set ofdara points. Compare the actual and predictedy values.
-7 l,lBl:15 lq 5 -7) Lr -5 3)
and
:{f}
the following set of equations using Excel.
: :
Given matrices:
l+ zo tf
Solve the following set of equations using Excel.
[4.27. Solve
Fuel Connmption(Miles per Gallon)
_
14.26.
opera-
t4J
tions using Excel.
lAl+[r] :l b.[/] -[r] :r a.
c. 3lAl:7 d. [/]lrl e. [/]{c} :
:
?
?
CHAPTER
15
MATLAB ATLAB@
is a mathematical software
that can be used to solve a wide range of engineering
problems.
MATLAB, wrote their own computer
Beforethe introduction of engineers
programs to solve engineering problems. Even though engineers
still
computer codes to solve
write
complex
problems, they take advantage of
built-in
i
i
i
i
i,
i
i,
',
I
functions of the software. MATLAB is also versatile enough that you can use it write your own
to
i
programs. i
I
419
4?0
Cner.rsn
15
MAILAB
In
this chaptn, we will discuss the use of MATLAB in soluing mgineeringproblems. MATLAB is a rnathernatical sortware aaailable in rnost of the uniuersity cornputational labs today. MATLAB is a aery powerful too[ egecially for manipalatingmanices; infact, it originally was fusignedfor thatparpose. Thne are many good textbooks that discuss the capabilities rfMATLAB to solue afull range ofproblems. Hne, our intent is to innoduce on$r some basic ifuas so that lou can perforrn you continue your engineering education iru other sorne essential operations. '4s yu will leam more about how to use MATLAB ffictiuely to solue a widz clz.sses, range of engineering problems. we discussed in Chapter 14, before the into'4s of electonic spreadsheets and mathernaical nrtware such as MATLAB, duction engi.neers wrnte their own cornputer prlgrarns to solue mgineering problems, Eum though mgineers still write cornputer cod,es to solue complex problerns, they tahe adaantage of buih-infunctions ofsoftwares that are readily auailable with computational tools, such as MATLAB. MATLAB is also aersatile enough that you cAn ase
it
to write yoar own programs. Thi.s chapter btsr^ by discussing the basic make-up of
MATLAB.
We
will
xplain how to input data or a forrnula in MATLAB and how to carryt out sorne typical mgineering compatations, Ve will also explain the wse ofMATLAB's mathelnatical, statistical, and logical functions. Plotting the resubs of an mgineering analysis using MATLAB is ako presmted in this chapter. Finally, we will discuss britfl! the curue ftting and symbolic capabilities of MATLAB. $trnbolic rnathernatics refns to tbe use of syrnbols rather tltan numbns to set a? ?roblems. Morenuer, oue will use exarnplesfrom Chapter 14 (Exce} to emphasize that MATLAB is just another tool that you cAn use to solue a uariery of engineering problems.
15.1 IVIATLAB-Basic ldeas 'W'e
begin by explaining some basic ideas; then once you have a good understanding of these concepts, we will use MAII,AB to solve some engineering problems. As is the case with any new software you explore, MA|LAB has its own synux and terminology. A typicd windoq/ is shown in Figure 15.1. The main components of the MAILAB window in the default mode are marked by arrows and numbered, as shown in Figure 15.1.
MltfI/'B
1. Menu bar: Contains the commands you can
use
to perform cenain tasfts, for example, to
sane your workspace or to change the View settings.
2. Current Directoryr Shows the active directory but you c:m also use it to
change the
drrectory.
3. Current Directorywindow:
Shows all files, their types, sizes, and descriptions in Current
Directory.
4. Moving window outside the desktop: Clicking on this icon will move the window ourcide the desktop.
15.f MAILAB-Besrc
I
Figurel5.l
lonm
421
Thedesktoplayoutforlr|AT|AB.
5. Command Wlindow: This is where you enter variables and issue MAILAB commands. 6. Command Histoty lfrindow: Shows the time and the date at which commands were issued during r:he previous M.{|LAB sessions. It also shows the history of commands in the current ("aiv.) session. As shown in Figure 15.1, the MATLABT desktop layout, in deftult mode, is divided into *uee
windows: the Current Directory the Command'!7indow, and the Command History. You qpe (enter) commands in the Command Window. For example, you can assign values to yariables or plot a set of variables. The Command History window shows the time and the date of the commands you issued during prwious MATLAB sessions. It also shows the history of commands in the current ("cti"") session. You can also transfer old commands, which you issued during previous sessions, from the Command Historywindow to the Command !?indow. To do this, move your mouse pointer over the command you want to move then click the left bufton, and, rrhile holding down the bufton, drag the old command line into the Command 'Window. Alternatively, when you strike the up-arrow key, the previously ercecuted command will appear again. You can also copy and paste corunands &om the current Command '$7indow, edit them, and use them again. To clear the contents of the Command 'S7indow, type cJ.c. Once in the
MAII,AB environment, you c:m assign values to a variable or define the elemen$ of a matrix. For example, as shown in Figure 15.2, to assign a value of 5 to the variable a in the Command Vindow after the prompt sign >> you simply type x = 5. The basic MAILAB scalar (arithmetic) operations are shown in Thble 15.1. As we explained in Section 74.7, duingyoru engineering education, you will learn about different types ofphysical variables. There are those that are identifiable by a single value or by
422
il
Crul"rrn
15
MAILAB
rigurctt.z
Eramples of assigning ualues or
defining elements of a mahir in lrlATLAB.
Erample Operation Addition
Symbol
x=5 andy=3
T
x+y
Subtraction
Multiplication
I
Division Raised to a polver
Result 8
,
x-y *Y
tl
(x+y)12 x^2
25
4
magnitude. For enample, time can be described by a single value such as two hours. These types of physical variables which are identifiable by a single value are called scalars. Temperature is another example of a scalar variable. On the other hand, ifyou were to describe the velocity of a vehicle, you not only have to specifr how fast it is moving (speed), but also its direction. The physical variables that possess both magnitude and direction are called aec'tots. There are also other quantities that require specifying more than two pieces of information. For example, if you were to describe the location of a car parked in a multi-story gamge (with respect to the garage entrance), you would need to specify the floor (the s coordinate), and then the location of the car on that foor (randycoordinxa). Amatrix is often used to describe situations that require manyvalues. In general, a matrix is an array of numbers, variables,
or mathematical you
type
terms. To define the elements of a matrix, for example Vl :
[r 5 ol
a Z 7 l, L6 2 9 ) I
A=[1-50i837;629] Note that in MAil-AB, the elements of the matrix are enclosed in brackets
t I and iue sepai as shown in
rated by blank space, and the elements of each rols are separated by a semicolon
Figure 15.2.
15.1 MAftAB-Besrc
loBas
423
How the Result of
ll,IATTllB Command
x
=
format shorL
E4ilanation
213 is Displayed
0.6667
Shows four decimal
digitsdefailt format.
format long format rat
0.66666666666667
Shows 14 decimal digits.
2t3
Shows as fractions
ofwhole
numbers.
formal bank format short
0.57
e
Shows rwo decimal digits.
6.6667e-001
Shows scientific notation
with four decimal digits. fnmal-
1nna
a
6.666666666666666e-00
I
Shows scientific notation
wfth 14 decimd digia.
format hex format + format
3fe5555555555555
Shows hexadecimal output.
+
Shows or blank based on whether the number is positive, negative, or zero.
*, -,
compact.
lt
suppresses the blank lines
in the output.
In Section 18.5, we will discuss mauix algebra in more detail. If you do not have an adequate background in matrix algebra, you may want to read Section 18.5 before studying the MAtrIAB enamples that ded with matrices.
The
formaL, disp,
MA|L,AB offers
and
fprinLf
eommands
number of commands that you clrn use to display the resula of your calculation. The MA[-{B fomat command allows you to displqT values in cerain wap. For example, if you define x = 2/3, then MA|LAB will displqyx = 0.6667. By default, MAILAB will display four do;ttt;[ digrts. Ifyou want more decimal digits displayed, then you type fo:mat long. Now, if you type x again, the value of r is displayed with 14 decimal digits, that is x = 0 . 55666666666661 . You can control theway the value ofxis displayed in a number of other walzs, as shown in Table I5.2. It is impomanr ro note that the f ormat, command does not affect the number of digis maintainedwhen calculations are carried out by a
MAILAB. It only affects the way
the values are displayed.
command is used to dislay t€xr or values. For example, given x = lL 2 3 4 5J , then the command dtsp (x) will display 1 2 3 4 5. Or the command disp ( 'Result = ' ) will display ResulL =. Note that the text that you want displayed must be endosed within a set of single quotation marks. The fprlntf, is a print (display) command that offers a great deal of flexibility. You can use it to print text and/or values with a desired number of digits. You can also use special formatting characters, such as \n and \t, to produce linefeed and tabs. The following example will demonstrate the use of the fprintf command.
The
dlgp
424
Cuaprrn
15 MAftAB In the MAII,AB Command Vindow, type the following commands
as
shown.
x=10
fprLntf ('The vaLue of x l-s
2"9
\n', x)
The MAII-AB will display
llhe value of x is A
10
screen c:rpntre of the Command'STindow
ro
I
llir
f,!'rf.asf{'rhr lalqe d x lo $g\n,, x} Figurct5.3
ral[e
for F,xample 15.1 is shown in Figure 15.3.
of,
x ts
llj l,l
10
The screen capture of
lhe Command Window
for
Example 15.1.
Note that the text and formatting code are enclosed within a set of single quotation marls. Also note that the o/og is the number format character and is replaced by x value, 10. It is also important to mention that Mlflf-AB will not produce an output until it encounters the \n. Additional capabilides of disp and fprintf commands will be demonstrated using other examples later. :lLr!:il:l!
r1_-!
rrt
::lr
.i.l..l.t-:rr= - - -----.,: -illl:..------:-:- -.--
:1..
Saving Your MATLAB Workspace You can save the worlspace to a file by issuing the command: gave your-f 5.lena.me. The your-filename is the name that you would like to assign rc the worlispace. Later, you can load the file from the disk to memory by issuing the command: Load your-f LLenane. Over command rc list the content of the time you may create many files; then you can use the directory. For simple operations you c:rn use the Command\?indow to enter variables and issue MAil.AB commands. However, when you write a program that is longer than a few lines, you use an M-fiIe. later in this chapter, we will explain how to create, edit, run, and debug an
dlr
M-file.
Generating a Range of Values 'When
creating, analyzing, or plotting data, it is often convenient to create a range of numbers. To create a range of daa or a row matrix, you only need to specify the starting number, the increment, and the end number. For example, to generate a set ofrvalues in the range of zero to 100 in increments of 25 (i.e.,0 25 50 75 100), in the Command'lfindow, type
x= 0:25:100
f
5.1 MAILAB-Be,src
F.xamde x
Addition
Symbol T
Subtraction
Multiplication Division Raised to a power
425
."
r[tsi#int.
Operation
Ipna.s
I
=
andy=2
a the Result of the Formula 10 Given in the FramFle
z:x*y*20 z:x-Y
4-)
z=(x*y)+9 s=(x12.5)*y
29
2=(x^y)^0.5
l0
B
5
Note that in MA|L-AB language, the range is defined by a starting value, followed by a colon !, the increment followed by another colon, and the end value. As another example, if you were to tyPe
Countdown = 5:-1:0 then the countdown row matrix will consist of values 5 4 3 2 1 0.
ereating Formulas in MATLAB As we have said before, engineers use formulas that represent physical and chemical lanvs gov-
erning our surroundings to analy'zr various problems. You can use MAILAB to input engineering formulas and compute the resula. $7hen typing your formula, use parendleses to dictate the order of operation. For example, in MAII,ABI Command\Vindow, if you were to type count=100+5*2, MATI-AB will perform the multiplication first, which resuhs in a value of 10, and then this result is added to 100, which yields an overall value of 110 for the variable count. If, however, you wanted MAI*I-A3 to add the 100 to 5 first and then multiply the resulting 105 by 2, you should have placed parentheses around the 100 and 5 in the following manner sssag=(100 +5) *2, which resula in a value of 2lA. The basic MAII-{B arithmetic operations are shown in Table 15.3.
Element by Element 0peration In addition to basic scalar (arithmetic) operations, MATI-A.B provides element-by-element operations and mauix operations. Mlffl-{Bt symbols for element-by-element operations are shown in Thble 15.4. To beter understand their use, suppose you hane measured and recorded the mass (LS) *d speed (mls) of 5 runners who are running along a straight path: m = 160 55 70 68 721and s = [4 4.5 3.8 3.6 3.I1. Note that dre mass m array and speed s array each have 5 elements. Now suppose you are interested in determining the magnirude of each runnert momentum. We can calculate the momentum for each runner using MAILAB's element-by-element multiplication operation in the following manner: momentum = m.*s resulting in momentum = lzm 247 .5 266 2U.8 223.21 . Note the multiplicarion qymbol (.*) for element-by-element operation. To improve your understanding of the element-by-element operations, in ttre MAIABt Command Vindowtype a = V 43 -ll and b = [1r 3,5r 71^od then rrythe following operations.
426
Cnlpren
L5
MAILAB
Equivalent Element by Element Symbol for Operations the Operation
Aritfmatic
Operation
+
Addition Subuaction
:
Multiplication Division Raised to a power
>>a+b
ans=8786 >>a-b
ans=61-2-8 *a ans = 2L >>3 . *a >>3
L2
9-3
ans =
21, L2 Y -5 >>a. *b
ans=7L2
1"5
-7
ans=7L2
1-5
-7
>>b. *a
>>3. ^a
ans = 1.0e+003* 2.L870 0.0810 0.0270 0.0003 >>a. ^b
ans = 7 54 243 -L >>b. ^a
ans = 1.0000 8l-.0000 125.0000 0.1429 Tty
"ko
the following example on your own.
This example will show how the density of standard air changes with temperature. It also makes use of MATJ,ABI element by element operation. The density of standard air is a function of temperature and may be approximated using the ided gas larv according to
p:fr P
where
: R: T: P
standard atmospheric pressure (101.3 kPa)
g"" constant and its value for ur is 286.9
ur temperattue in Kelvin (K)
(#)
15.1 MAILAB-Besrc
IosAs
427
Using MAILAB, we want to creirte a table that shows the density of air as a function of temperature in the range of 0"C Q73.15 K) to 50"C (323.15 K) in increments of 5oC. In the M,{IT-AB's Command'S7indow, we ty?e the following commands.
femperature = 0:5:50; Density = 101300 ./ ((286.9)*(Temperature+273)) t >> fprJ-ntf (' \n\n' ) ;dlgp(' Tenperature(C) Denelty (kglm^3 ) t ) r dLsp ( lTenperature r , DeoEj-tsy' l ) >> >>
In the commands just shown, the semicolon i suppresses MATI-AB's automatic display action. Ifyou type Tenperature = 0 :5:50 without the semicolon at the end, MATIAB will display the values of temperature in a row. It will show rFamnarafil16
-
0510r_5 The . /
20
25
30
35
is a special element by element division operation that tells
40 MlfflAB
45
50
to carry the
division operation for each temperature value. In the dLsp command, the prime or the single quotation mark over the variables
Telnperature'
and
Density'
will change the
values of the temperature and the density,
which are stored in rovrs, to column format before they are displayed. In matrix operations, the process of changing the rows into columns is called the Tiranspose of the matrix. The final results for Example 14.1 (Rwisited) are shown in Figure 15.4. Norc that the values of the temperanue and the density are shown in columns. Also note the use of the fprint f and disp commands. F{e Rfit vrew
y** sfr Y ?io,
-** ndl+r{r@'.(.
d
Figure
15.4
lhe result of
Example 14.1. (Revisited).
fnilztltlmirir&1lt,.0
Matrix 0perations MAII/.B
offers many tools for mauix operations and manipulations. Thble 15.5 shows exMl$TABt matrix operations in more deail in
amples of these capabilities. S?'e will explain Section 15.5.
428
Cner"rpn
15
MAILAB
---:Frornple:AandBare Marricec WhichYou Synbols or Com-*nds
Operation Addition
,
:
Subuaction
*
Mukiplication
Determinant Eigenvalues
A+B
A_B A*B
maffix narn/
/t'
inv{na*hname} det(matix name) eig(watth ndrile)
mv(n/
d"(A)
\
See Example 15.5
Transpose
Inveme
Have Deffned
Matrix left division
eig(A)
(uses Garss elimination to solve a set oflinear
equations)
Using MAILAB, create a table that sholvs the relationship berween the interest eamed and the amount deposited, as shown in Thble 15.6. To create a able that is similar to Thble 15.6, we tlpe the following commands.
fomat, bank >>.Amount = 1000:250:3000; >>
IntereEt_Rate = 0.06;0. 01:0.08; Interest_Earaed. = (.Anount' ) * (IatereEt_Rate), fprl-nrf (' \n\n\r\L\r\r\r\r\r rnrereEr Ratet ) ;fprLntf (' \a\t .Anount\t\t' ) I . .. fprLatf ( ' \t\t e"g[' , Interest-Rate) ; fprl-ntf ( ' \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.r-nr 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 built-in functions that you qln use to analyze data. fu we
discussed in the prwious &apter, by built-in 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. MA|I,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 MAlf-ABt built-in 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 giv-eqrena)r It sorts the values in the give$' ,,, , array in ascending order. It returns the value of zr,
i
r min ,l
sEq
:
joir; I,:
3.14151.926535897.. i
'
,.,'i;
.
It retums tangent value of the
9@l
argument. The argument must be
in radians,
cos
It
renrrns cosinevalue ofthe argument. The argument must be
in
It
returns sine value of the argument.
in radians, s
,',,,,,
frg*su**l**rh,h:ifT:,
;l
sqrt (x) f actorial (x)
Retums the square root ofvalue.t
ky"^ *:.:*e of factorial of*. For example, factorial(5) wlll return: (5)(4)(3)(2)(r) : 120.
Trigonornzt ix Functions (x) This is the inverse cosine function ofx. It is used to determine the value of an angle when its cosine value is kno\trn. asin (x ) This is the inverse sine function of*. It is used to determine the value of an angle when its sine value is known. atan (x) This is the inverse tangent function of r. It is used to determine the value :. of an angle rylieg its glgent value is knor,rn. '',1 ,,
acos
,
',
'
kponcntial and Ingarithmic Functiot s exp
(x)
los (x) 1og10 (x) 1og2 (x)
Retums the value of a". Retums rhe value of the natural logarithm of *. Note that
* must
greater than 0.
Returns the value ofthe common (base 10) logarithm of.r. Returns the value of the base 2 loSanthm of x.-
be
15,2 Usnc MAIIAB Bur.r-nv FuNcrroxs
431
Usiag MAil.AB, compure the aver4ge (ari hmecic mean) and the sandard deviation of the density of water data given in Thble 15.9. Refer to Chapter 19, Section t9.4, to learn about what the vdue of the standard deviation for a set of data points represents.
1,,
Group-AFindingq
GroupB Flndings
1.,
ffi* The final resuks for Example 14.4 (Revisite$ are shown in Figure 15.6. The commands leading to these results are:
MAILAB
>>DensJ.ty-A = [1020 1015 990 1060 1030 950 975 1020 980 950] t >>DeasLty-B = [950 9AO 890 1080 1120 900 ].040 1150 910 10201; >> DensLty_A_Avela![e = nean (Denstty_A) Density-A_Av€rdg€ = 1000.00 >> DenEl.ty_B_Aver?![€ = mean (DensLty_B) Densi-Ly-B-Av€rdg€ = 1000.00 > > Standard._Xlevlat lon_For_croup_l\ = std ( Dengity_A) Standard-Deviat 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&'ce@-*l&jt
I
figumls.O
Command l{indow
i|ATIAB'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 = sLart-value : incremenL : end-va1ue 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 MlftI-AB 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 MAII-A,B Bunr-rN FuNc"rrors
433
Note that in the preceding example, the index is the integer i and its start-value is 1, it is incremented by a value of 1, and ia end-value 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 1i-ne or a block of vour comouLer
code
end 'With the while command,
as
long
as
the controlling-expression 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 <= 24.00 Y=x^2+10;
disp ( lx' ,y'l x = x + 0.5;
)
end In the preceding example the ( = symbol denotes less than or equal to and is called a relational or comparison operator.'We will explain MlffLABt logical and relational operators next.
ftfsing MATTAB Logical and Relational Operators In this section, we will look at some of MAlf-AB's logical operators. These are operators that allow you to test various conditions. MIIILABI logical operators and their descriptions are shown in Thble 15.10.
'*rrli
s$iil#jo;;iii;f ffi iffi
L"d""l
Symbol
Operator
6. I
and
or noI exclusive
xor
or
-:i'ri,iln$iililiffi
Description TRUE, If both conditions are true TRUE, If either or bo*r conditions are ffue FALSE, If the result of a condition is TRUE, and TRUE, If the result of a condition is FAI-SE FAISE, If both conditions are TRUE or borh conditions are FAI-SE, and TRUE, If either condition is
all
l
'i; l l :
true
TRUE, If any element of vector is nonzero TRUE, If all elements ofvector is nolrzero
any
ffi
"
I I
MAn-48 also offers relational or comparison operators that allow for the tesdng of relative magnitude of various arguments. These relational operators are shown in Thble 15.11.
The Conditional '\Vhen
Statements' if, else
compurer program, sometimes it becomes necessary to execure a line or a block of code based on whether a condition or set of conditions is met (true). MAn"$ provides z/ and eke cnmmands for such situations.
writing
a
434
Cneprpn
15 M/IIUB
Ids Meaning Less
than
Less than
or equal to
Equal to Greater than Greater than or equal to
Not equal to
The
lTStatement The l/statement is
d1e simplest
form of a conditional control. Using the
,/
statement, you crn o(ecute a line or a block ofyour program as long as the expression following the i/statement is true. The qfntax for the l/statement is
if expression a line or a block of your computer code end For example, suppose we have a set of scores for an enam: 85, 92, 50, 77, 80, 59, 65, 97 , 72, 'We 40. are interested in writing a code that shows that scores below 60 indicate failing. The MAILAB code for this example then could hane the following form:
scores = [85 92 50 77 80 59 65 97 72 40]; tnrl_l.t.ttl
if scores (i) <60 fprintf('\t ?g \t\t\t\t\t
FAILfNG\nt, scores (i))
end
end 4 e/se Statement The ekestatement allows us to enecute other line(s) of computer code(s) if the expression following the l/statement is not tnre. For enample, suppose we are interested in showing not only the scores that indicate failing but also the scores that show passing. !7'e can then modifr our code in the following manner: The
scores = 185 92 50 77 8A 59 65 97 72 401; t^r
1-
| . | . I Il
if scores (i) fprintf ('\t
>=60
%g
\L\L\L\t\t. PASSING\n', scores (i)
%g
\t.\t\t\t\t
else
fprintf('\t
FAILING\nr, scores (i))
end end In the MAILAB Command lfindow, try the following eremples on yorr orrn.
);
15.2
UsrNc MAILAB Brnrr-rx FuxcrroNs
435
qB also provides the ekeifcnmmand that could be used with dre ifartd else statements. It is left for you to learn about it Try the MAIIAB help menu: rype help ekeiftolam about the elseif satement.
MAn
The pipeline shown in Figure 15.7 is connected to a control (check) valve that opens when the pressure in the line reaches 20 psi. Various readings were taken at different times and recorded. Using MAILAB's logical funcdons, create a list that shows the corresponding open and closed posidons of the checkvalve.
I
cter&vdveg$&f;ffiffi
Figuret5.l
A schematic diagram
for
tmmple 14.5 (Revisihd).
The solution to Example 14.5 (Revisired) is shown in Figure 15.8. The commands leading to the solution are:
>>pressur6=[20 tS 22 26 L9 1,9 2L L27i >>fprlotf('\t Line Pressure (pgt) \t VaLve FosLtlon\n\n');
for L=1:8 if pressure(t) >=20 fPrLntf (' \t e"9 \t\t\ts\t\t el-se
fpriatf ( '\t
end end
I
rigurel5.8
The solution
ol Ennple
14.5 (Revhited).
?"9
\t\t\t\t\t
OFEN\n'rBreEsure(L) CITOSED\n'
)
,pregsure (L) )
436
Cneprsn
15
MAILAB
The M-file As explained previously, for simple operations you c:m use M,{'l,AB's Command lTindow to enter variables and issue commands. However, when you write a program that is more than a few lines long, you use an M-frle. It is called an M-file because of its .zz extension. You can cre-
M-file using any text editor or using MAIT,ABI Editor/Debugger. To crqrte an M-fiIe, open the M-file Editor and MlfIl,AB opens a new window in which you can type your program. fu you type your program, you will notice that MAILAB assigns line numbers in the left column of the window. The line numbers are quite useftrl for debugging your program. To save the file, simply click File -* Save and type in the file-name. The name of your file must begin with a letter and may include other characters such as underscore and digits. Be carefirl not to name your file the same as a MAILAB command. To see if a file-name is used by a MATI-AB command, type exist ('file-namd) in the M,$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\ t-tt
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 twice-leading to the answer:
.
approach and came up with the formula
100(10 I )
--;:
n(n+ r)
-- ,
2-since
he wrote down the numbers
5050. l,ater, he generalized his
15.2
Flle Edt
Tqt
Bdffitr ,
Cell T@h Dehrg De$ktop Wrdow
&EB-
UsrNc MAILAB Burr,t-rN Funcrrous
437
Ftalp
"leitmTl aelc*e-areeli:
F{iF
rtil 4.. ":,i :
g Ask the usar lo input the uFper vaLue ,, upper_val-ue=input('Please input th€ 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 lr|-file 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 M-file 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 For-Ioop-F,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
x-y
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 MAII-AB commands that lead to the solution are:
I The
>>x = 0:0.1:51 >>volune = (10-2*x) .*(10-2*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 II|AIIAB Command l|indow
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
r-mark
+
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 x-aris. The text that you enclose will be shown below the x-axis. Similady, the ylabeL ( 'text' ) command creates the tide for the y-axis. 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) double-click 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 double-clicking 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 daa-point 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 y-axis scales. For enample, the MAIT,AB's
Loglog(x,y)
usesthebase-l0logarithmicscalesforr-andy-axes.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 base-l0 logarithmic scales for either only dre r-axis ory-axis. 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 free-hand 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 $lindow
for Erample t4.6 (Revhited).
15.4 Iuponrrxc E:rcrr,aNo Orrrrn Dere
L
Figure
15.?l
Frr.ss rNro
MAil,AB
445
Plot of density of air for Erample 14.6 (Revhited).
i] t1
IttA,.i,
r:ff1&1i.lilr,,t,t -
.i,-r"lllta?nji:11-lftlfiil!:
15.4
--ii,!liijitii!.'i iiirxlitiijliirrrtii
--
---ii-,rXitilrl!ii1-11ta-- lr,iilril'liTrlTi1.ir'rrira
l:
Jli:1ii-,1ir,irlrirl..- -.'lTlLr.li4lltl
Nmporting Exeel and Other Data Filcs into MATLAB At times, it might
be
convenient to impon data files tlat were generated by other programs, such
MAil-AB for additional analpis. To demonsrate how we go about imponing a daa file into MATLAB, consider the Fxcel file shown in Figure I5,22.The Excel file was created for Example 15.4 with two columru: the r values and the corresponding volume. To impon this file into MAII-AB, from the Menu bar, we select File *1sn Irnport Data . . . , t}ren go to the appropriate directory, and open the file we want. The Impon'Wizard window, as
Excel, into
in Figure L5.23,r ,il|appear next. Select the "Create vectors from each column using column nameso and the \Pizard will i-port the data and will sare them as x and voXu:ne
as shown
variables.
Now lett the
MIIIIAB
Figure 15.25.
'We
that we want to plot the volume as a function of r. then simply type commands that are shoiyn in.Figtire 15,24.Tlte resulting plot is shown in
say
-*r; ; .';I-"-**-*-"
t11
:A
,
,:.9.i,
:'l
x
:.t:,
2 0.1
0.0 9.6
A:
na
18.4
sl
0.3
20.5
6
9.4
l,1.9 40.5
8l
0 0.
116.€
0.€
58.4
r:,$
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
i|AIIABI lmport ltizad.
@:-..:
n aiI*, a
fo';
-4'g.i'1i;;**l.rryffii!
Ar Sat rtst d, el,el ..qdf|l!8 Ealtn trs l&r E ls &m $Frg 61ffid o!e*d @l,ablo, ln rhe ol@9 @tt!D€R. > al@ ta.wlwel > td.tla lr9ol@ e a 6Mt@ ot r'l I ilib€t (,a a@l'l > Itd!€r, (190r@ 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
Given the following matrices:
[o 5 ol l+ 6 -21 f-r'l f/l =18 3 zl,lnl:lt 2 3l,""a1ci:l 2fusingMArlAB,perform Le-zel
(5J
113-4)
:
:
the following operations. (d [,a] + [B] = ?, (b) t1l lBl ?, (c) 3[A]: ?, (d) La.l t8l ?, (e) Lal{C}: ? (f) determinate of lAl. The solution is shown in Figure l5.24,Yhen studying these examples, note that the response given byMAIT-AB is shonrn in regular typeface. Information that the user needs to type is shown in boldface.
-
To get startcd, seleet "IMAT'LAB Fflelp" f,rom the F{elp menu"
>) A=[0 5 0;8 3 7;9 -2
91
A-
0 50 8 37 9-29 >> 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
=
4It-2 1-5 5 10 1
10 5
>> A-B 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 MAIT-AB to solve a set of linear equations.
2qI x2* 4:13 3x1*2x2*4xu=i2 5q-x2t34:17 For this problem, the coefficient matrix [r4]
"nd
the right-hand 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. l----lr.-:--------- -- ---- ---- :-----il ------r.r:i-
15.6
-----"----..:--.------l
Curve Fitting with MATLAB In Section 14.8, we discussed the concept of curve fitti"g. MAII-A.B offers a variety of curve' fitting options. V'e will use Example L4.ll to show how you can also use MAlf-AB 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
*
c2xo-2
*
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 MAII-A,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 MA|I.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 MA|LAB'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 MAn-A.Bt qymbolic operations,
as
Thble 15.14.
f(*):*-5x+6 rtQ): *- t
: (x+ 5)2 fsk):5*-t-r2x-Y
fi(r)
I
Function
Description of the Function
ExamFle
It
Flx = qfm('x^2-5\+6') F2x = sfm('x-3')
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('51-y+2\-y')
Flx = x^2*5*x*6
F2x: x-3 = (x*5)^2
p31
F4x=
5\-y+2\-y
faao(Fxl)
(x-2).(x-3)
simplifr(Flx/F2x) expand(F3x)
x-2 x^2*10\*25
collea(F4x)
7*x-2*y
solve(Flx)
x=2andx=3
ezplot(F1x,0,2)
See
Figure 15.27
shown in
452
Cru.prnn
15
MAil-AB
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
MAII-ABI 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,Lon-3 = '5*x-y+3*z=L7' i >>[x,Yrz] = golve(equatl-on 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 MAIf-AB 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
Pnonr-eMs
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 buih-in funcdons provided by MAII,AB. know that can use
o know how to edit the content of a MIILAB file.
.
be familiarwith MAILABI built-in 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 MAII-A.8I curve-fiting 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 10-5NA.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 MAIT-A3, 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 MAII-AB, 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
P:
of water in dre range of 1000 Pa to 10000 Pa. Use increments of 500 Pa. Using MAILAB, create a table that shows the relationship benveen the units ofpressure in Pa and psi in the range of l0 kPa to 100 kPa. Use incremens of 0.5 kPa Using MAILAB, create a table that shows the relationship between the unis of power in W'atts and 'uf to 10000 W Use horsepower in the range of 100 smaller increments of 100 \7 up to 1000'!(, and then use increments of 1000 \f all the way up to 1 0000 \[
where
t
horsepower (hp)
:
horsepower (hp)
: 746lff
power (!7 or
ft. lb/s) 550
ft.lbA
The purpose of this problem is to see how the power requirement changes with the car speed and the air temperature. Determine the power requirement to overcome the air resistance for a car that has a listed drag coefficient of 0.4 and width of 74,4 inches and herght of 57 .4 inches. Vary the air speed in the range of 15 m/s < V < 35 mls, and change the air density range of 1. 1 I kg/or' < p < 7.29 kC l t f .The given air density range corresponds to 0"C to 45"C.You may use the ideal gas lalv to relate the density of the air to its temperature. Present your findings in both kilowatts and horsepower. Discuss your findings in terms of
tunnel. The air speed inside the tunnel is changed, and the drag force acting on the car is measured. For a given car, the experimental data generally is represented by a single coefficient that is cailed &ag coefficient. It is defined by the following relationship:
l5.ll.
,PV"A
:
I
in earlier chapters, the air resistance to the motion of a vehicle is something imponant that engineers investigate. The drag force acting on a car is determined experimenally by placing the car in a wind
F,
P
and
As we explained
: ----:3C)-1
F6v
porilrer consumption as a funcdon of speed and air temPeranue. The cantilevered beam shown in r:he accompanlng figure is used to support a load acting on a balcony. The deflection of the centedine ofthe beam is given by the following equation
where Ca
: drg coefficient
F4
=
(unidess)
fteasured drag force (N or lb)
: it density (kg/-t or slugs/ff) 7: air speed inside the wind tunnel (m/s or ft/s) ,4 : frontal area of the car (mz ot #) p
-MX2, , t=ffiQr-4Ix+6L2) where
-/
= defection
za :
at a given
r location (m)
distributed load (N/m)
PnoslElvrs
E:
modulus of elasticity
(N/-t)
1=
second moment of area (ma)
r:
distance from the suppoft as shown (m)
:
Z
,4
=
*:
457
cross-sectional area ofthe 6n (a',b) (m2)
thermal conductivitv of rhe fin material
(Wim. K) Plot the temperame distribution along the fin
length of the beam (m)
'Wm.K h : I2Vlm2.K a:0.05 m, and b :0.01 m. Yuy,
using the following dan:
t513.
h --
158
from 0 to 0.1 m in increments of 0.01 m. A person by the name of Huebscher developed a relationship between the equivalent size of round ducs and rectangular ducts according to
D:1.3
(ab\o'ezs
t or'la -a,vrii
where
: a:
D Problem 15.ll
&
x
Fins, or extended surfaces, commonly are used in a variety of engineering applications to enhance cooling. Common oramples include a motorcycle engine head,
-
Tn*i^r = (7b*"
-
T*fi,r
r)t-*
Length of One Side of Rectangular Duct (length a), mm
Len$h
&
40a 450 500
550 600 15.14.
A Pitot tube is a device commonly
hA
perimetell*(a
m
\l-----:
YP
where
dte heat uansfer coefficient
*
(!f/m2. K)
b) of rhefin (m)
used
in a wind
runnel to measure the speed of the air flowing over a model. The air speed is measured from the following
V:
hp
and
b: p:
table.
equation:
where
7/t:
dimension of the other side of the rectangular duct (mm)
in
electronic equipment, and finned tube heat exchangers in room heating and cooling applications. Consider aluminum fins of a rectangular profile shown in Problem I4.l3,whrch are used to remove heat from a surface whose temper:rtue is 100'C. The temperature of the ambient air is 20"C.'We are interested in determining how the temperature of the fin varies along its length and plotting this temperature variation. For long fins, the temperature distribution along the fin is given by
T
:
Using MAIT.AB, ceate a table that shows the relationship between the circular and the rectangular duct, similar to the one shown in the accompanying
maximum defection of the beam?
a lanrn mower engine head, extended surfaces used
dimension of one side of the rectangular
duct (mm)
Using MIIII-AB, plot the deflecdon of a beam whose lengh is 5 m with the modulus of elasticity of 106 mma. The beam is E = 2O0 GPa and .I = 99. I designed to qrry a load of 10000 N/m. !?hat is the 15.12.
diameter of equivalent circular duct (mm)
V: Pa : p:
ar
speed (m/s)
dynamic pressure (Pa) density of
ur (L.23 kS/*')
Crnrrsn
458
15 MAILAB Create a able that shows the viscosity of air as a function of temperature in the range of 0'C (273.15 K) to 100"C (373.15 K) in increments of 5oC. Also, create a graph showing the value ofviscosity as a func-
Using MIIILAB, create a table that shows the air for the range ofdynamic pressure of500 Pa to 800 Pa lJse increments of 50 PaUse MIIILAB to solve Example 7.1. Recall that we applied the trapezoidal nrle to determine the area of the shape given. Create an Excel file with the given data, and then impon the file into MlfIf.AB. speed
15"r5.
tion of temperature. In Chapter 11, we explained the concept of windchill factors. The windchill values are determined empirically, and the common equivalent windchill temperanue Tr"quhalot fC) is given by
will discuss engineering economics in Chapter 20. Using MIIILAB, create a table that can be used to look up mondrly payments on a car loan for a period of five years. The mondrly payments are calculated from W'e
T"q,*)*,= 0.045(5.27V0'5 + .(7"- 33) + 33
,:{ 15J9.
0.28V)
=
loan in dollan
/
:
interest rate; e.9., 7,7.5, . . .
Ll -5
,9
t'l f {C}: \-2I,
Interest Rate
ffi
Given the matrices:
2 -t1 14 2 rf [r 3 :17 :15 o 3l,and -71, [B] t/l
monthly payments in dollars
P
Loan
-
Crate a able that shows the windchill temperatures for the range of ambient air temperature of -30'C < T^ < l0oC and wind speeds of 5 m/s ( V120 mls.
where
,4 :
10.45
3)
t4 5 -7)
perform the following operations
lq)
usingMAIT,AS.
10,000 15,000
20,000
a.
25,0Q0
15.t7.
b.
A person by the name of Sutter/and
has dweloped a correlation that can be used to evaluate the viscosiw of afu as afirnction of temperarure. It is given by
lL=
c.
3lAl:7
d.
IAJIB):
?
e.
lAl{c}:
?
c,To5
,*i
15.20.
Ct
ro ol
viscosity (NA.
*1
temperature (K)
cr: 1.458x cz:770'4K
Given the following matrices:
lz lz [/] :l16 6 r4l,ana[r] :l 4 Lrz -4 181 lrz
where
p: 7:
+ 7n1= 7 Pl - [r1 :1
L'al
10-6
ut---)
r\m's'Kt/2/
10
ol
20
01,
-4
the determinant of tAl and tBl Mltfi-A3. Vhich mauix is singular? n5.21. Solve the following set of equations calculate
MATLAB.
r8_J
using using
Pnosr.EMs roaztooo
I
-
-
1812500
18i2500 6343750
-^rrrrrzf{::r\ 4531250
-4531250
L
459
)
\uq)
: {,,1} 15.2?.
Solve the following set of equations using MATLA.B.
ll{''.| l:; L-a , ,)(:l: ll;l |'ul
15.23.
Solve the following set of equations using MATI-AB.
|
-":: -!o,u : : I I'll: ;'' o -0.76 o.8ir -o.oel r il
I
L 15.24.
-i;;
: : 0 0
105
0
-r.49 -1.49
o I -1:2i)17) l{
-1',)
Solve the following set of equations using
7.2
15.25.
-o oel
ssxzorl
}
[.,n,],^,j
M.{fl,AB.
000 7.2 0 -4.22 o 8.44 0 *4.22 0 4.22 -r.49 -4.22 0 0 -1.49 0
-r.49 -r.49
-r.49 -r.49
-4.22
0
0
0
5.7r r.49
r.49
Find the equadon that best fits the following set of data points. Compare the actual and predictedy values. Plot the data firsr.
r.49
{il{ Find the equation that best fits the following set of data points. Compare the acual and predictedy values. Plot the data first.
Find the equation that best fits the following set of daa points. Compare the actual and predictedy values.
Plot the data first.
i 1
i
T-r
Pn r
/-
-.f,NGINEERING bRAPHICAL n i UOMMUNICATION I
l
4
i
i CoNttEyrno It'lF0RtvtATT0lt T0 0TttER EtleTNEERS, i lv|ncHrltrsTs, TrcHltrcrANs, nnu MInAGERS
, !
'l
_
_ltltw:_n
1.
14@1ryq@ry1
Engineers use techfical drawings to convey useful information to others in a standard
=
manner:-An engineering drawing provides information, such as the shape of a product,
its dimensions, materials from which to fabricate the product, and assemhly steps. Some engineering drawings are specific to a particular discipline. For example, civil
engineers deal with land or boundary topographic, construction, and route survey drawings. Electricaland electronic engineers, on the other hand, could dealwith printed
circuit'board assembly drawings, printed circuit-board drillplans, and wiring diagrams. Engineers also use special symbols and signs to convey their ideas, analyses, and
solutions to problems. In Part Four of this book, we will introduce you to the principles and rules of engineering graphical communication and engineering symbols. A good grasp of these principles will enable you to convey and understand information
effectively.
ii:lll'l
Englneering Drawings and Symbols
CHAPTER
ENcTNEERTNG
L6
DnmrrNcs
AND SvnnBols ngineering drawings, such as the cellular phone schematics here, are important in conveying useful
information to other engineers or machinists in a standard manner that allows for visualization of the proposed
"'*\
r-'---r --1
\l
'-
i--
product. lmportant information, such as the shape of the product, its size, type of material used, and the assembly steps required, are provided by these drawings. Source : F rcm Pro / ENGINEEP
2000i2, W louis Gary Lamit, pp. L10-19 and L10-20, Brooks/Cole, 2001. Reprinted with permission of Brooks/Cole, an imprint of the Wadsworth Group, a division ofThornson Leaming. Fax: 800-73O-2215,
462
+I
I
-_T-.1I-T
r
lii'"lt -l-ltl-t-' ril +'ll I
I
J
t
f6.1
lrvrponrexcn on ENerxeeRrNc Dnewrxc
463
Engineers use special kind ofdrawings, calhd mgineeringdrawings, t0 conaEltheir
idtas and dzsign inforrnation about produas. These drawings paway uital information, such as the shape of the product, its size, rype of rnateriak used, and assetnbly steps. Moreouer, machinisn use the information prouided by mgineers or drafts l,ersons on the mgineering drawings to make the pats. For complicated. systerns madz ofuarioar ?afts, the drawings ako smte as a ltotu-to-assernble guide, showing how the uarious parts fit ngethn. Most ofyou will eumtually take a semester-long class in engineeringdrawingwhne you will leam in rnuch rnore dztail how to neate such drawings. For now, the following sections prouidz a brief innoduction to engineering graphical communication principles. We will discuss wlry mgineering drawings are i.mportant, haw they are drawn, and wbat rales must be followed, to crea.te such drawings. Engineering symbols and signs abo prouide ualuable i.nformation. These symbob are a "knguage" used hy mgineers to conae! their idcas, solutions n problems, or analyses of certain siruailons. In this chEtter, we will also discuss the needfor conuentional mgineering symbob and uill show sorne cornrnon syrnbok used in ciuil, electrical, and mechanical mgineering
16.tr lmportanec 0f Engineering Drawing Hane you wer had an idea about a new product that could make a certain task easier? How did you get your idea across to other people? What were the first thingp you did to make your idea clearly known to your audience? Imagine you are ha.ring a cup of coffee with a friend, and you decide to share your idea about a product with her. After alking about the idea for a while, to clarifr your idea, you will naturally dra$r a picnrre or a diagram to show what the product would look like. You have heard the saying "a picrure is wonh a thousand words"; well, in engineering a good draring is worth even more words! Technical drarings or engineering drarrings are important in conveying usefi.rl information to other engineers or machinists in a
464
Cnlpren
16
ENcrNrsRrNc Dnemrrcs,errro Sv:vrnor.s standard acceptable manner to allow the readers of these drarings to visualize what the proposed product would look like. More significandy, information such as the dimensions of the proposed product, or what it would look like when viewed from the top or from the side or the front is provided. The dranings will also specifrwhat type of material is to be used to make this product. In order to drar or read engineering drarrings, you must first leam a set of sundard nrles that are followed by all engineers, drafispersons, and machinists. In the nent sections, we
will briefv
discuss these nrles.
16.2 OrthoEraphic
Views
(dragrams) shorv what an object's projection lools liLe when seen from the top, the &ont, or the side. To better understand what we mean by onhographic views, imagine rhat you hare placed the object shorun in Figure 16.1 in the center of a glass box.
Ortbogrqbic vieus
fl
tigurel6.l
The orthograpIic projection
of an oliect into the hoftontal,
ueilical, and profile planes.
16.2 OnnroenapnrcVrs\rs
t-,'
F-; t"l B..k - ---l L------J
t-
f
I
lL"ftl
l-'"f
t-"-l
I
465
F*t-l
r------l I nieht
I
[ll;t f
figure
16.2
Tie relative locations of the top, bottom, front, lack, right-side, and left-side views.
Now ifyou were to drar perpendicular lines from the corners of the object into the faces of the glass box, you would see the oudines shown in Figure 16.1. The oudines are called the orthographic projection of the object into the hoizpntal, uertical, andtheproflc plznes. Now imagine that you open up or unfold the faces of the glass box that have the projecrions of the object. The unfolding of the glass faces will result in the layout shov*n in Figure 16.2. Note the relative locations of the top view, the bottom view, the fronr view, the backview, the right-side view, and r:he left-side view. At this point, you mxy rcalize that the top view is similar to the bottom view, the &ont view is similar to the back view, and the right-side view is similar to the left-side view. Therefore, you nodce some redundancy in the information provided by these six views (di"gr"-r). Therefore, you conclude that you do not need to drar all six views to describe rhis object. In fact, the number ofviews needed to describe an object depends on how complex the shape of an object is. So the question is, then, how many views are needed to complerely describe the object. For the object shown in Figure 16.1, tfuee views are sufficient to firlly describe the object, because only three principle planes ofprojection are needed to show the object. For the example shown in Figure 16.1, we may decide to use the top, r:he front, and the righcside views to describe the object completely. In fact, the top, the fronr, and the righr-side views are the most commonly created views to describe most objects. These views are shown in Figure 16.3. From examining Figure 16.3, we should also note that three differenc types oflines are used in the orthographic views to describe the objeaz solid lines, hiddcn or dashed lines, and cmterlines, The solid lines on the orthographic views represent the visible edges of planes or the intersection of two planes. The dashed lines (hidden lines), on the other hand, represent an edge of a plane or the extreme limits ofa cylindrical hole inside the object, or the intersection ofnvo planes thar are not visible from the direction you are looking. In other words, dashed lines are used when some material exists between the observer (from where he/she is looking) and the actual location of the edge. Referring to Figure 16.3, when you view the object from the right side, irs
466
Cneprrn
16
ExcrNrunrNc Dnsrrxcs.ero Snvrnors
t-rlTr---------t Irit I
rop lllil litlll
I I
l--!!-r-J---------------
t@l Front I
t'
G
I
Figureto.3
The top,
front,
and the right-side views of an object.
projection contains the limia of the hole within the object. The right-side projection of the object also contains the intersection oftwo planes that are located on the object, Therefore, the solid and dashed lines are used to show t-hese edges and limits. The third rype ofline that is employed in orthographic projections is the centerline, or the line of symmetry which shows where the center ofholes or the center of rylinders are. Pay close attention to the difference in the line pafierns between a dashed line and a centerhne. Examples of solid lines, dashed lines, and lines of symmetry are shown in Figure 16.3. As we said earlier, the number of views that you should drarv to represent an object will depend on how complex the object is. For example, if you want to show a bolt washer or a gasket, you need to draw only a single top view and specify the thickness ofthe washer or the gasket. For other objecs, such as bolts, we may draw only nvo views. Examples of objeca requiring one or two views are shown in Figure 16.4.
I
tigure
16.l
Examples of objects requiring one 0r tlvo uiews.
16.3 DnvrsNsroNrNcewoTor,nnencrrc Drar the orthographic
I
views of the object shown in Figure
467
t6.56).
Figurel6.S
An object and
ils
orthograpiic views: (a) the object, (b) the top view, (c) the front
vieq
and
(c)
(d) the side view.
ir--.--
-- --------i- ---:
(d) ------..
16.3
t
.-r.illiiL!dt------.i
i,i
--,L--j
Dimensioning and Tolerancing Engineering dmidngs provide information about shape, size, and material of a product. In the previous section, we discussed how to draw the orthographic views.'We did not say anything about how to show the actual size of the object on the drarrings. The American National Standards Institute (ANSD sets the standards for the dimensioning and tolerancing practices for engineering drawings. Eveqy engineering draring must indude dimensions, tolerances, r:he materials from which the product will be made, the finished surfaces marked, and other notes such as part numbers. Providing this information on the diagrams is important for many reasons. A machinist must be able rc make the part from the detailed drarrings witlout needing to go back to the engineer or the draftsperson who drew the dran'inp to ask questions regarding the size, or the tolerances, or what type of material the pan should be made from. There are
468
f
CHepren
16
ENcTNEERTNc
DnevrNcs eNp Sylrsor.s
Figurel6.6
The basics of dimensioning
pmcticss: (l) dimension line, (2) extension line, (3) centerline, and (4) leader.
basically rwo concep$ that you need to keep in mind when specifring dimensions in an engineering draring: sizc and, bcation As shown in Figure 16.6, not only do you need to specifr how wide or how long an object is but you must also specify the location of the center of a hole or center of a fillet in the pan. Moreover, a draring is dimensioned with the ud of dirnmsion lines, qtewion lines, cmteilines, and, leaders. Dirnension lines proide information on the size of the objecq for example, how wide it is and how long it is. You need to shovr the overall dimensions of the object because the machinist can then determine the overall size of the stock material from which to make the piece. As the name implies, qtension lines ate those lines that extend from the points to which the dimension or location is to be specified. Extension lines are drarvn pardlel to each other, and the dimension lines are placed between them, as shown in Figure 16.6,The had,ats are the arrows
that point to a circle or a fillet for the purpose of specifying their sizes. Often the draningr are shown Not To Sca& (NTS), and therefore a scaling factor for the draring must also be specified. In addirion to dimensions, all engineering drau'ings must also contain an information box with the following items: name of r:he person who prepared the drawing, tide of the draring date, scale, sheet number, and draring number. This information is normally shown on the upperor lower-right corner ofa draring. An example ofan information box is shown in Figure 16.7. Let us now say a few wor& about filles, which are often overlooked in engineering dravdngs, a shortcoming that could lead to problems. Fillet rders to the rounded edges of an
R 1.25 cm
The sides of the middle hexagon are 9 rnm long.
Drawn by:9.M.
1.5 cm
The sides of the comer hexagons
arc7 mmlong. The wrench is 3 mm thick.
-J' I2.5 cm
E
Figuru
16.7
An example
ol engineedng drawing with an infomation [or.
Title: Eike wrench
)oale: NT9 Date:6 Oct.2OO1 Drawing number:2
16.4
IsoMerRrc VrEw
469
object; their sizes, the radius of roundness, must be specified in all dravings. If the size of fillets is not specified in a drawing, the machinist may not round dre edges; consequendy, the absence of fillets could create problems or failure in pars. As some of you will learn later in your mechanics of materials class, mechani""l p* with sharp edges or a sudden reduction in their cross-sectional areas could fail when subjeced to loads because ofhigh stress concentradons near the sharp regions. As you will leam later, a simple way of reducing the stress in r:hese regions is by rounding the edges and creating a gradual reduction in cross-sectional areas. Engineered products generally consist of many parts. In todayt globally driven economy, some of the para made for a product in one place must be easily assembled with parts made elsewhere. When you specify a dimension on a &er,ring-sty,2.5O gsntim$g1-horn' dose does the acnral dimension of the machined pan need to be to the specified}.50 cmfor the pan '\[ould to 6t properly with other parts in the product? werphing fit conecdy if the acual dimension of the machine paft were 2.49 cm or 2.51 cm? If so, then you must specifr a tolerance of -10.01 cm on your drarving regarding this dimension. Tolerancing is a broad subject with its own nrles and symbols, and as we mentioned earlier, the American National Standard Institute sets the tolerencing standards that must be followed by those creating or reading engineering drawings. Here, we have briefy introduced these ideas; you must consult the sandards if you are planning to prepare an acrud engineering draning. Shon' the dimensions of the object in Figure 16.8 on its onhographic views.
Q 0.s". ---J-
0.5" )r _T g/
,.}.--.-
tM
F1" 0.5'
I n
Figute
-T t'
_l_ 0.5'
l6J
An obiect and
its dimensions.
lrlrtlril',:r'li:r
::lr11l:-::::t:l
:.r 1lt
r::
r
:-:.-i!Il;::r rI::r
:
r
::u:'r
-:rr:
16.4 lsometric View '!7hen
it
is
difficult to visualize an object using only its orthographic views, an isometric sketch Tlte isonetric drauingshows the three dimensions of an object in a single view.
is also drarvn.
The isometric drarvings are sometimes referred to
as
technical illusradons and are used to show
470
Cneprsn
16
ENcrNsERrNc Dnevnrcs ar.ro Snunors
11.0'(280 mm)
LI.L'
14.4' (365
[
figure16.9
nm)
Eramplesofhomehicdranings.
what parts or products look like in pans manuals, repair manuals, and product catalogs. F,xamples of isometric drarings are shovm in Figure 16.9. As was the case with onhographic views, there are specific nrles that one must follow to drarv the isometric view of an object. W'e will use the object shown in Figure 16.10 to demonsuate the steps that you need to follow to draw the isometric view of an object.
l: Draw the width, height, and the depth axes, as shown in Figure 16.11(a). Note that the isomeuic grid consisu of the width and the depth axes; they form a 30o angle with a horizontal line. Also note rhat the height axis makes a 90" angle with a horizontal line and a 60o angle with each of the depth and width axes. Step 2: Measure and drav the toul width, height, and depth of the object. Hence, drarv lines l-2, l-3, and t-4 as shown in Figure 16.1 1(b).
Step
16.4 IsorrsrRrc
I
VrEw
471
Figurel6.l0
The object used in demonstrating
the steps in creating an isomehic view.
Step (a)
I
I
Step 2 (b)
Figurel6.ll
Steps in creating an isometdc
(l) Create the isometric ares and grid; (2) mark the
view:
height, widti, and depth ofthe object on tfte isometric grid; (3) create the front, the top, and the side work faces; and
(4) complete the drawing.
Step 3: Create the front, the top, and the side work faces. Drar line 2-5 parallel to line l-3; draw line 4-6 parallel to line 1-3; dralv line 3-5 parallel to line 1-2; draw line 3-6 parallel to line 1-4; draw line 5-7 parallel to line 3-6; and draq/ line 6-7 parallel to line 3-5 u shown in Figure 16.11(c). Step 4: Complete the drarn'ing as marked by the remairiing line numbers, and remove the unwanted lines 3-6, 3-8,3-72,6-13, aad.6-9, Next, we will use an enample to demonstrate tte steps in creating an isometric view of an object.
472
Cnerrrn
16
ExcrxnsRrNc Dnerrnqcs er.ro Snvrnols Draw the isometric view of the object shown in Figwe 16.12.
Step
I
I
Step 2
Figurel6.l2
I
The orthogmphic and
isonehic views of the object
Step 4
Step 3
for Erample 16.3. ll:j11?.11l1[&3ATn.T,[ill.{ii?'4il1'J{ltliili1.'!i.il"iii!iii,rl.iii!lir-.
..4ir.?iiir.i;!il,liiuli&l
16.5 Sectional
-..,.rrtltit'ilirllttllii:l!
----,,
- - _-.
..1i
Views
You recall from Section 16.2 thatyou use the dashed (hidden) lines to represent the edge ofa plane or intersection of two planes or limits of a hole that are not visible from the direction you are looking. Of course, as mentioned eadier, this happens when some material ercists benveen you and the edge. For objects with a complen inl6des-fe1 erample, with many interior holes or edges-the use of dashed lines on orthographic views could resuh in a confirsing draring. The dashed lines could make the draring difficult to read and thus difficuft for the reader to
16.5 SscrroNALVrEvs
473
Top view
Ii--tj-r-r-r-iT11Ii-i;;:;--,-l-l j
I
tit lii ii|i l--ur-r-F-L;+-.:!-.i I
I
LjJJ-d-LLJ
Front view
l I
Sectional view Figurel6.l3
An object
witi
a compler interior.
a visual image ofwhat the inside of the object lools like. An example of an object with a comple,r interior is shown in Figure 16.13. For objects with complen interiors, sectiondl aierrrs are used. Sectional views reveal the inside ofthe object. A sectional view is created by making an imaginary cut duough the object, in a cerain direction, to reveal its interior. The sectional
form
views are drawn to show clearly the solid portions and the voids within the object. Let us now look at the procedure that you need to follow to create a sectional view. The first step in creating a sectional view involves defining the curting plane and the direction ofthe sight. The direcdon of the sight is marked using direcdonal arrows, as shown in Figure 16.14.
Cutting plane
f,
figurel6.l4
A sectional view of an object.
Crosshatch patterns also indicate the
$pe of matedal that parts
made from.
are
Cross-harching
474
Cnepren
16
ENcrNeERrNc DnavrNcs aNo Snvrsors
Moreover, identifyingletters are usedwith the directional arrows to name rhe section. The next step involves identifying and showing on the sectional view which ponion of the object is made ofsolid material and which portion has the voids. The solid section of the viewis then marked by parallel inclined lines. This method of marking the solid portion of the view is called crosshatching An orample of a cutdng plane, irs directional arrow, its idendfying letter, and crosshatching is shown in Figure 16.14. Based on how complex the inside of an object is, different methods are used to show secdonal views. Some of the common section types indudez full seaioa half seaioa brohm-out sectiou rotnted section, and remoaed sec'tion.
o Full-section tieus
.
"
te
created when the
cuning plane
passes
through the object completely,
in Figure 16.14. Half-seaianal dews are used for symmeuical objects. For such objects, it is customary to drar half of the object in sectional view and the other half as exterior view. The main
as shown
advanage of half-sectional views is that they show the interior and exterior of the object using one view. An example of a half-section view is shovrn in Figure 16.15. Rotaad section d.eat may be used when the object has a uniform cross section with a shape that is difficult to visualize. In such cases, the cross section is rotated by 90" and is shown in the plane ofview. An example of a rotated section is shown in Figure 16.16.
I
figurut6.t5
f
Figure
16.t6
Anerampleof ahalf-sectionalview.
An eranple
ol a rotated section view.
16.5 Srcrroxlr. Vrgws
475
B_B
@ il
Figurel6.lT
An eranple of an object
witi
removed sections.
"
Remoaed sections are similar to rotated sections, except instead of drawing the ros.ted view on the view itsel{, removed sections are shown adjacent to the view. They may be used for objects with a variable cross section, and generally many cuts ,lo"rgh the section are shown. It is important to note that the cutting planes must be properly marked, as shown in Figure 16.17.
Dran' the sectional view of rhe object shown in Figure 16.18,
f
as
marked by the cutting plane.
Figurel6.tg
The object used in Emmple 16.4.
.ftllillTri0JEi?tr1ilt@.[l?E?r?[
--@d,Wf&{uff1&]-ffi,ftI
476
Crupren
16
Excnrrnnnrc Dnewrxcs exo Svunor,s
(b) Highwaymap
(a) Highwaymap consmrction lines
il
Figuru
16.19
tramples ol drawings used
16.6 Civil, Electrical,
ir civil
engineering.
and Electronic Drawings
In addition to the type ofdrarings that we have discussed so far, there are also discipline-specific drawings. For o
n6.T Solid hlodeling In recent years, the use of solid modeling software as a design tool has grown dramatically. Easyto-use packages, such as AutoCAD, IDEAS, and Pro-E, have become corntnon tools in the hands of engineers. With t{rese software tools you c:rn create models of objecs with surfaces and yolumes that look almosr indistinguishable from the actual objects. These solid models provide great visual aids for what the para that make up a product look like before they are manufacnrred. The solid modeling software also allows for experimenting on a computer screen with the assembly of parts to eramine any unforeseen problems before the parts are ac-
16.7 Sorro Moonrrxc
LTR
OTY SIZE
NONI
't6
A
4
ts
.J
.04t) .063 .LZJ
477
REMARKS
PLATEDTHRU PLATEDTHRU
(a) A printed circuit board drill plan
f
Figurct6.20
Eramples of drawings used in
electrical and electronis
(b) A wiring diagram
engineedng.
rually made and assembled. Moreover, changes to the shape and the size of a part can be made quickly with such software. Once the final design is agreed upon, the computer-generated drawings can be sent direcdy to computer numerically controlled (CNC) machines to make the parts.
Solid modeling software is also used by architects and engineers to present concepts. For example, an architect uses such software to show a client a model ofwhat the exterior or interior of a proposed building would look like. Dotgt engineers employ the solid modeling software to show concepts for shapes of cars, boats, computers, and so on. The computer-generated
478
Cnrr.rrn
16
ErrrcrNEgRlyc DnewrNcs exp Snvmors models save time and monqy. Moreover, there is additional software that makes use of these solid models to perForm additional engineering analysis, such as stress calculations or temperature distribution calculations for products subjected to loads and/or heat transfer. Examples of solid models generated by commonly used software are shown in Figure 16.21. Let w now briefly look at how solid modeling software generates solid models, There are two ways to creare a solid model of an obj ecr;. bofiom-ap modelingandnp-doun modelingVrth bonom-up modeling you start by defining keypoints first, then lines, areas, and volumes in terms of the defined keypoins. Keypoints are used to define the venices of an object. Lines, next in the hierarchy of bottom-up modeling, are used to represent the edges ofan object. You can then use the created lines to generate a surface. For example, to create a rectangle, you first
I
riguret6.Zt
Eramples of conputer'generated
solid models.
L6.7
Sor.ro MoosuNc
479
Line Rotating lines
Dragging lines
al@ Skinning (area "skinned"
over lines
E
tigum16.2
Ll, L2, andL3)
Offsetting areas (cylindrical area *infl ated' and "defl ated")
Additionalarea-gererationmetltods.
define the comer points by four keypoints, next you connect the keypoints to define four lines, and then you define the area of the rectangle by the four lines that enclose the area- There are additional ways to creirte areas: (l) dragging a line along a pa&' (2) roating a line about an axis, (3) creating an area fillet, (4) skinning a set oflines, and (5) oftetting areas.'With the area-fillet operation, you c:m crqlte a cqnstant-radius frllet tangent to rwo other areas. You c,rn generate a smooth surface over a set of lines by using the skinning operation. Using the area offset command, you can generate an ar@, by offsetting an enisting area. These operations are all shown in Figare L6.22, The created areas then may be put together to enclose and create a volume. As with areas, you can also generate volumes by dragging or €$ruding an area along a line (path) or by rotating an area about a line (axis of roadon). Examples of these volume-generadng operations are shorvn in Figure 16.23. lCith top-down modeling, you can create surfaces or three-dimensional solid objects using area andvolumeprimitiues. Primitives are simple geometricshapes. Two-dimensional primitives indude redangles, circles, and polygons, and tluee-dimensional volume primitives include blocks, prisms, cylinders, cones, and spheres, as shown nFiguue L6.24. Regardless ofhow you generate areas or volumes, you can use Boolean operations to add (union) or subffact entities to create a solid model. Examples of Boolean operations are shown in Figure 16.25.
f
Figmel6.u:t
Erampler of Yolume-generating
opentions.
$Dexamples
!
ngunu.a
fiamples of
ho'
and three'
dimensional primithes.
/il80
2-D examples
16.7
Sor.ro
Moorr.ruc 481
CylinderB
O I
ll\-J
Fi9mel6.25
ANB
Eramples of Boolean union (add)
(Intersection operation)
and subhact operations.
Discuss how to create the solid model of the objects shon'n in Figrure L6.26, using the operadons discussed in this section.
(a)
t
A microprocasgel
foeaf
sink
(b) Abracket
Figure16.26 lheobjectsforExamplel6.S.
(a) In order to create the solid model of the heat sink shown in Figure 16.26(a), we draw its front view first, as shovrn in Figarc L6.27.
482
Crrlpren 15 Ercmwrnruc Dn-avnres .elro Sntaors
tl ilnnriFil
f
S'" il
Figure16.27
Front view.
Ll
L]Ll,uLl\
'We
then extrude this profile in the normal direction, which leads to the solid model of the heat sink as depicted in Figure 16.26(a). (b) Similarly for the bracket given in Figare 16.26(b), we begin by creating the profile shown in Figure 16.28.
H
figurel6.28
Partial Drofile of
t[e bncket.
Next, we will extrude the profile in the normal direction, as shov,'n nEigure 16.29. W'e then create the block and the holes. In order to cr@te the holes, fust we cr@te two solid cylinders and then use the Boolean operation to subtract the cylinders from the block Finally, we add the new blockvolume to the volume, which we created by the entrusion method, as
I
in Figure 16.30. Block
I
i
ii
i
I
L ii.i---.----
Figurul6.29
ErtrudedProfile. -. -- -L- ----
-a
I
Figure16.30 Thesolidmodeloftlebracket.
--.I.--,lir]i-...'-------
.-.--i.-
.,--.
---- -- . --.-
-
16.8 'Wrrr Do
16.8
Wn
Nrrp EncrNnrnnvc
Svrvrsols?
483
Why Do We Need Engineering Symbols? The symbols in Figure 16.3I are a "languageo used by engineers to convey their ideas, their solutions to problems, or analyses of cenain situations. As engineering srudents you will learn about the graphical waln by which engineers communicate among themselves. In this section, we will discuss the need for conventional engineering symbols as a means to convey informa'We will begin by explaining why there don and to effectively communicate to other engineers. erdst$ a need for engineering signs and syrnbols. \7e will then discuss some cornmon symbols used in civil, electrical, and mechanical engineering.
-/\AAz\resistor ---.1
e-
capacitor
{1 speaker
logic symbot (a nd)
+----
r>-
op-amp
logic symbol (or)
v
gt,,^ 90"
natural gas line
elbow
45'etbow
+D
_IXF globe valve
Y check valve
L
{-} --1___J
Fi9urel6.3l
tngineering symbols and signs provide valuable information.
-G -E
-
electric service
-
o
Beginni ns (PoB)
l'.7
71 l"l l. i l4 I
concrete
glass
484
Crraprsn
16
ENcrNrnnnc DnarxrrNcs.lNo Snr,rsor,s
It E
w@w
figure16.32
Eramples of traffic signs.
Almost all of you hane a driver's license by now; perhaps you eyen oriln a car. If this is the you are already familiar with conventional traffic signs that provide valuable information to drivers. F.xamples of traffic signs are shown in Figure 16.32. These signs are designed and developed, based on some acceptable national and international standar&, to convey information not only effectively but also quickly. For example, a stop sign, which has an octagon shape with a red background, tells you to bring your ciu to a complete stop. case, then
\ carYe+rrl! \ voaA. atheadl
---t-
16.9
Exerupr.rs
or CorvruoN
Syrvrsor.s IN Crvrr, Er-scrRrcAr" ar,vu
MrcruNrcar, ENcnrEERrNc
485
A sign that signals a possibly icy road ahead is another example that warns you to slow down because the road conditions may be such that you may end up in a hazardous situation. That same information could have been conveyed to you in other wa)'s. In place of the sigrr indicating dippery road, the highway depanment could have posted the following words on a " board: "Hey you, be carefirl. The road is slippery and you could end up in a ditch!" Or they could have installed a loudspeaker warning drivers: "Hey, be carefi.rl, slippery road ahead. You could end up in a ditch." Which is the most effective, efficient, and least orpensive way of conv#ng the inforrnation? You understand the point of these examples and the question. Road and traffic information can be conveyed inexpensively, quickly, and effecdvely using signs and symbols. Of course, to understand the meaning of the signs, you had to study them and learn what they mean before you could take your driving test, Perhaps in the future the new advancements in technology will reach a sage that road and uaffic information can be conveyed firecdy in a wireless digital format to a computer in your car to allow your c:r to respond accordingly to the given information! In that case, would you still need to know the meaning of the signs, and would the highway departmenr need to post them?
16.9
Examples of eommon Symbols ln eivil, Electricafl, and Mechanical Engineering As the examples in the previous section demonstrated, valuable information can be provided in a number ofways: through a long written sentence, orally, graphically or symbolically, or any logical combination of these. But which is the rnore effective way? As you study various engineering topics, not only will you learn many new concepts but you will also learn about the graphical way that engineers communicate among tihemselves. You will leam about engineering signs and qymbols that provide valuable information and sarre time, money, and space. These symbols and signs are like a language that engineers use to convry their ideas, their solutions to a problem, or their analyses of certain situations. For example, electrical engineers use various sfmbols to represent the componens that make up an elecuical or an elecrronic s;rstem, such as a television set, a cellular phone, or a computer. Examples of engineering rymbols are shown in Thble 16.1. Mechanical engineers use diagrams to show the layout of piping networls in buildings or to show the placement of air supply ducts, air return ducts, and fans in a heating or cooling qrstem.
For more detailed information on symbols, see the following documents: Graphic ElectricalSymbokforAir-Conditioningand Rfiigeration Equipmmtby frRI (ARI 13088). Graphic Symbob for Ebctrical and Electronic Diagrams by IEEE (ANSI/IEEE 315-
7975). Gra?hic Synbok for Pipe Fi.nings, Values, and Pipi"S by ASME (ANSIiASME ASME Y32.2.2.3 -r9 49 (R I 9 8 8)). Symbob for Mechanical and Acoustical Elzmmx as Ued in Schernatic Diagrams by ASME (ANSr /ASME Y32.r8-r97 2 (R 1 985)).
Electrtcal Svmbols
{-{&WWr\-
-*>F
:+ F-
*-J *::=
tr.
I
e l'lrl,Variable resistor
Battery
Lamp
Photo resistor
C"
Diode
\\ ,
Solar qell
Tl
Tt
\.af/-
S#ater
o,pamp
l
Single-phase
motor
:;.
OR
--{-\,_
-t1
Outputif any one of the control inputs is
on
,'
g
D
-b.D/
'Outputifsingle s_ignal
is
off
NOR
ID>
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*ou-o
i^
r-=:l
il--
Oval batltub
X Iloor sink
Showersall
fr&iirlpool bath
t,,,
iH
':,, Tows-
iry
r
'45"
Elbow
Chillpddrinking ril€tsr t:
f
90o
supply
Elbow
drinking waterretqr,ni
Chill€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
,,,
Straight-size
'.i
gfoss
-rD<]f-
,
Snaight-pi2e
-',
@',,.
1.
Gate
valve
+xF Globe valve
Pnoar,r*rs
487
SUMMARY Now that you have reached this point in the rext
o You should have a good undentanding of the imponance of engineering drawings in con-
. . . "
. . .
veying information to other engineers, machinists, and assembly personnel. You should undersand what is meant by onhographic views, isomeuic drawing and solid modeling. You should undentand the basic nrles required for an engineering draring including showing dimensions, specifring material size, and indicating finished surfaces. You should know when to use isometric views and sectional views. You should be familiar wirh the different types of sectional views. You should understand the importance of solid modeling in conveying concepts and examining para for their abiliry to fit with other parts. qFmbols to communicate among You should know why we need and use ourselves. You should be familiar with some of the common civil, elecrical, and mechanical engineering symbols so that you can understand and interpret a technical drawing.
For Probltms 16,1 throaglt 16.10, draw the np, thefrona and tbe right-side ortbographh aieus oftbe objec* are shoun. Indican when an ohject needs on$ one or two ui.ews to b,
16.3.
f"h
dzsctibed. t6.t.
Problen 16.4
488
CneprsR
16
EncNEERT{c Dnevrncs eNo SyMsoLs
16.5.
Itt.
Problen 16.9
For Probbms 16.11 through 16.15, use the catting plznes sbown to drau tlte sectional aieus.
Problem 16.8
Pnonr.sMs
489
For Probbms 16. 16 through I 5.20, asing the nrbs discassed this chapter, show the dirnmsions ofthe uiews shown.
1616. 0
tl
234 ttl
1
5cm I
t6.t4.
|
n6.17 0
rlrrrl
Top view
Front view Problem 16.15
zln.
in
490
CnerrBn
16
ENcrNsERrNc DnerrrNcs er,ro Syunors
16.t8.
16.20.
0
3
I
I
o
1
1,,,,rr,,,1
in.
2345 tltl
6cm I
Problem 16.18
[6.f9.0 r tltll
2
3
4
5cm I
Problem 16.20
For Probbms 16.21 through 16.26, draw the isorne*ic uiew of the folhwing objeas. Mahe the necessa?y rneasurenzents or estimatiotu of dirnmsions.
16.23.
A television set. A telephone.
f6.25.
A
f6.21.
chair.
tr6.24.
A compurer. A razor.
tr6.26.
Acar.
n6.U.
For Problems 16,27 through 16,31,
d.iscass how yoa would cleate the solid rnodcl of the gium objectt. See Een?b 16.5 to better undarstand what you are being ashed m do.
r=
0.125 (holes)
Pnosr..EMs
16.28.
Awheel.
'
Asocket.
16.31.
A heat exchanger. The pipes pass duough all ofthe fins.
Dimensions are in inches. Proilen
n6.29.
16.30.
16.28
Apipe.
Platethickness=
! 1.6
rt=
2.O0
tn
rz=2.25'rn t =0.75 in.
H =0.75in.
io.
492 16.32.
CnaprsR
16
ENcrNsrRrNc DnevrNcs.aNp Sytvrsor-s
Using Table 16.1, identifr the engineering
e
shown in the accompanying
figure.
grmbols
16.33.
UsingTable 16.l,identifythecomponenaofthelogic s)rctem shown in the figure
-_+|\\
D
iF
*T*
D DF
-DWSProblem 16.32 Problem 16.33
Boeing TTT
*
Commercial Airplane 199 ft 11 In. (60.e m) 70
ft
7.5 tn.
-T 60ft9in.i (18.5
li ft 70
m)
,,
I
7.5 In,
(27.5
n)
i
209 ft 1 ln. (63.7 m)
Overview TheBoengTTT is the first commercialairplane tharwas fullydesigned usingthree-dimensional igial solid modeling technology. The core of the design group consisted, of 238 teams ttrat included engineen of various backgrounds. A number of intemational aerospace companies,
.f
*Marerials were adaprcd with pernission from Boeing documents.
493
494
CHer'rsR
16
ENcrNssRrarc DnevtNcs eryp SvMsoLs
from Europe, Canada, andAsia/Pacific, contributed to the design and producdon of rhe777. The Japanese aerospace industry was among dre largest of the overseas participants. Representatives from airline customers such as Nippon Airways and British Airways also provided input m the design of777. Throughout the design process, the different components of the airplane were designed, tested, assembled (to ensure proper fifting), and disassembled on a nenvork of computers. Approximately 1700 individual worlistations and 4 IBM mainframe computers were used. The use of the computers and the engineering software eliminated the need for the dwelopment of a cosdy, firll-scale prototype. The digital solid modeling technology allowed the engineers to improve the quality ofwork, experimentwith various design concepts, and reduce changes and errors, all ofwhich resulted rn lower costs and increased efficiency in building and installing various parts and components. The engineers used, among other software, CATIA (Computer-Aided Three-Dimensional Interactive Application) and ELFINI (Finite Element Analysis System), both dweloped by Dassault Systems ofFrance and licensed in the United States *rough IBM. Designers also used EPIC (Elecronic Preassembly Integration on CAIIA) and other digital preassembly applications developed by Boeing. The 777 series, the world's largest rwinjet, is available in three models: the inidd model, 777-200;rhe777-200ER (Extended hS") model; and the Iaryer777-300 model. The777300 is suetched33 ft (10 m) from the initi^777-200 model to atotal,of 242 ft4in. Q39 m). In an dl-economylayout, rhe777-300 can accorilnodate as many as 550 passengers. However, it m4y be configured for 368 to 386 passengers in three classes to provide more comfort. In terms of range capabiliry, the 777-300 can serve routes up to 6450 statute miles (L0,370 km). The 777-300 has nearly the same passenger capacity and range capability as the 747-1001-200 models but burns one-third less fuel and has 40o/olower maintenance costs. Of course, this resule in a lower operating cost. Baseline maximum takeoffmass for the 777-300 is 580,000Ib (263,080 kg); the highest maximum takeoffweight being offered is 660,000 pounds Q9937A k$. Maximum fuel capacrty is 45,220 gal (17l,160 L). The 777-300 has a total available cargo volume of 7080 ff (200.5 m3). Satellite communication and global positioning systems are basic to rhe airplane, The777 wing uses the most aerodynamically efficient airfoil ever developed for subsonic commercial ariation. The wing has a span of 199 ft 11 in. (60.9 m). The advanced wing design enhances the ailplane's abiliry to climb quickly and cruise at higher altitudes than its predecessor airplanes. The wing design also allows the aiqplane to carry firll passenger payloads out of many high-elevation, high-temperature airfields. Fuel is stored entirehwithin the wing and its strucmral center section. The longer-range model and the777-300 model can carry up to 45,220 gaJ.
(r7r,155 L).
The Boeing Company, upon request from airplane buyers, can install engines from three leading engine manufacnrrers, namely, Pratt & \Vhimey, General Electric, and Rolls-Royce. These engines are rated in the 74,000 to 77,000-pound thrust class. For the longer-range model and the777-300,these engines will be capable ofthrust ratings in the 84,000 to 98,000pound category. New, lrghtweight, cost-effective strucnrral materials are used in several 777 applicacions, For example, an improved aluminum alloy,7O55, is used in the upperwing skin and stringers. This alloy offers greater compression strength than prwious alloys, enabling designers to save weight and also improve resistance to corrosion and fatigue. Lightweight composites are found
BorrNc
777
Corqvrnncrer ArnprnNs
495
in the venical and horizontal tails. The floor beams of the passenger cabin also are made of advanced composite materials.
The principal fight, navigation, and engine information is all presented on six large, liquidcr;rstal, flat-panel displays. In addition to saving space, the new displays weigh less and require less power, and because they generate less heat, less cooling is required compared to the older conventional cathode-ray-tube screens. The flat-panel displays remain clearlyvisible in all con-
ditions, wen direct sunlight. The Boeing 777 uses an Integrated Airplane Information Management System that provides fight and maintenance crews all pertinent information concerning the overall condition of the aiqplane, its maintenance requirements, and its key operating functions, including flight, thrust, and communication management. The flight crew transmits control and maneuvering commands through electrical wires, augmented by computers, direcdy to hydraulic actuators for the elwators, rudder, ailerons, and other controls surfaces. The tfuee-axis, "fy-by-wire" fight-control sys-tem saves weight, simplifies factory assembly compared to conventional mechanical systems relying on steel cables, and requires fewer spares and less maintenance in airline service. A key pan of the 777 system is a Boeing-patented, two-way digital daa bus, which has been adopted as a new industry standard: ARINC 629.It permia airplane systems and their computers to communicate with one another thto"gh a common wire path (a misted pair of wires) instead of through separate one-way wire connections. This further simplifies assembly
and saves weight, while increasing reliability through a reducdon in the amount of wires and connectors. There are 11 of these ARINC 629 padrways in rhe777. The interior of the Boeing 777 is one of the most spacious passenger cabins ever devel' oped; the777 interior offers configuration flexibiliry. Flexibiliry zones have been designed into the cabin areas specified by the aidines, primarily at the airplane's doors. In f -in. incrementsi galleys and lavatories can be positioned anywhere within these zones, which are preengineered to accommodate wiring plumbing, and atachment fr.xtures. Passenger service units overhead storage compartmenc are designed for quick removal without disturbing ceiling panels, airconditioning ducts, or suppoft struccute. A typical 777 configtraion change is expected to take as litde as 72 hours, while such a change might take two to tfuee weelis on other aircraft. For improved, more efficient, in-flight service, the 777 is equipped with an advanced cabin man€ement qFstem. Linked to a computerized control console, the cabin management qfstem assists cabin crews with many tasks and allows aidines to provide new services of passengers, including a digrtal sound system comparable to the most state-of-the-art home stereo or compact disc players. The main landing gear for the 777'sin a standard two-post arrangement but features sixwheel trucls, instead ofthe conventional four-wheel units. This provides the main landing gear with a total of 12 wheels for better weight distribution on nrnways and taxi areas and avoids the need for a supplemental nvo-wheel gear under the center of the firselage. Another advantage is that the six-wheel trucks allow for a more economical brake design. The 777 landing gear is the largest ever incorporated in[o a commercial aiqplane. The Boeing-United Airlines 1000-rycle fight tests for the Pran & t07h.itney engine were completed on May 22, 1.995.In addition, engine makers and the many suppliers of parts for the aiqplane intensified their own development and testing efform to ensure that their products met aidine requirements. This thorough test program demonstrated the design features needed to obtain approval for extended-range nvin-engine operations (ETOPS). A'll777s are
496
CH.cr,rER
16
ExcrNnrnrxc Dnewnrcs axo Syrusors
DestgnVariable
7774M
Seating
305 ro 320 passengers
tluee
m-300 in
368 to 386 passengers in three classes
classes
209 ft 1 n. (63.7 m)
Length
ft
242 fr.4 in. (73.9 m)
ft
Vingspan
199
liril
60 ft 9 in. (18.5 m)
60
Engines
Pratt &'\Thitney 4000 General Elecuic GE90 Rolls-Royce Tient 800
Pratt & tU7himey 4000 General Electric GE90 Rolls-Royce Tient 800
Maximum takeoffmass
506,000Ib
580,000Ib (263,080 ks)
height
11 in. (60.9 m)
199
Q29,52Okd Fuel capacity
ft
8
11 in. (60.9 m)
in. (18.5 m)
45,200 U.S. gallons (171,160 L)
31,000 U.S. gallons
(rr7,335L) ft (1r,975 m)
ft
Akitude capability
39,300
Cruise speed
555 mph (893 km/h)
555 mph (893
Mach 0.84
Mach 0.84
Cargo capacity
5656fe (160 m3)
7552ff Qr4m3)
Maximum range
5150 nautical miles; 5952 starute miles; 9525Uf,la
5600 nautical miles; 6450statute miles; 10,370 km
36,400
(1 1,095
m)
kn/h)
ETOPS-capable, as part of the basic doig".To ensure reliabiliry, the777 with Pratt &'!7hitney engines was tested and flown under all appropriate conditions to prove it is capable offying ETOPS missions. A summary of Boeing 777 spectficxions is shown in Tlble l.
PROBLEMS 1. Using the data given for the Boeing 777, london.
esrimate the
fight time from New York City to
2. Estimate the mass ofpassengers, fuel, and cargo for a firll Ilighc 3. Using the maximum range and fuel capacity data, estimate the fuel consumption of the Boeing 777 on a per hour and per mile basis. Calculate the linear momentum of the Boeing 777 at crwse speed and at two-thirds maxi-
mum takeoffmass.
5. Calculate the Mach number of the Boeing 777 at cruise Mach =
cruise speed
\/ hRT
speed using
BosrNc
77
7 C,ov.unncr,tr. fu RPLANT
Avl
where
h: R: T:
speitficheatratio air gas constant
=
=
1.4
287
Jll
ab temperature at cruising altitude (IQ
Compare your Mach number to the one given in the Boeing data table.
6. As mentioned previously, the flight crew transmits control and maneuvering commands *oough elecnical wires, augmented by computen, direcdy to hydraulic actuators for the ebaa.tors, rudder, ai/erons, and other controls surfaces. The elevators, rudder, and ailerons for a smdl plane are shown in the accompanying figure. In a small plane the ailerons are moved by turning the control wheel in the cockpit. Vhen the wheel is rurned left, the left aileron moves up and the right aileron moves down. This is how the pilos starts a turn to the left. In a small plane the rudder is operated by the pilott feet. !7hen the pilot presses the left rudder pedal, the nose of the plane moves left. \7hen the right rudder pedal is pressed the nose moves right. The elevators make the nose of the plane move up and down. 'Vhen the pilot pulls back on the control wheel in the cockpit, the nose of the plane moves up. \f'hen the control wheel is pushed forward, the nose moves down.
Investigate the aerodynamics of maneuvering flight in more detail. Explain what happens to air pressure distribution over these surfaces as their orientations are changed. lfhat are the direcions ofthe resulting force due to the pressure distributions over these surfaces?
'Write a brief repon explaining your findings.
I ,|
l
^E,NGINEERING
i MATERTAL SnrpcrroN i
j AN ltupontEnr Drston Druston I
I
you are designing a machine part, a toy, a frame of a car, or a
of materials is an important design decision. In this part of the book, we will look more closely at materials such as metals and their alloys, plastics, glass, wood, composites, and concrete that commonly are used in various engineering
applications. We will also discuss some of the basic characteristics of the materials that are considered in design.
I
:,
,',
Englneerlng ifaterlals .
CHAPTER.
ENCINEERING METERIALS ecause pure aluminum is soft and has a relatively smalltensile
strength, it is alloyed with other metals to make it stronger, easier to weld, and to increase its resistance to corrosive
environments. Lightweight metals, such as alloyed aluminum, are used in many
structural and aerospace applications because of their small densities (relative
to steel) and high strength-to-weight ratios.
500
T7
I7.l
MarsRrALSsLEcrroN
501
As we discussed in Chapter 1, mgineers dcsisn millions ofproducts and seraices that we ase in owr ruelyda! liues: products such as cArs, cornpaters, aircrart, cbthing toys, home appliances, surgical equiprnent, heating and cooling equiprnmt, heabh care deuices, took and machines that rnahe uarious products, and so on. Engineers ako design and superaise the constru.ction of bwildings, dams, highwalts, Power
plants, and mass nansit
systerns.
a machine part, A to!, a frame for a car, or a stntcture, the selection of rnaterial is an imporwnt design decision. There are a nurnber offactors that engineers consider whm selecting a rnatnialfor a specifc application. For example, thq consid.er properties ofmaterial such as density, ubirnate snmgth, fiexibility, rnachinability, durability, therznal expansion, electrical and therrnal conductiuity, and resistance to corrosion. They also consider the cost of the rnaterial and how easily it can be repaired, Enginens are always searchingfor wals to use aduanced rnaterials to mahe products lightt, and snonger ,4s dzsign mgineers, whether yoa are designing
for dffirmt app lications. In this chaltter, we utill look rnore chwly at waterials that are comrnonly wsed in aarious mgirceering applications. Ve will also discus sorne of the basic physical characteri.stics ofmateriak that are considered in daign, We will examine solid rnaterials sucb as rnetals and their alloys, plastia, glass, wood, and those that solidifii ruer tirne, such as concrete. Ve will also inuestigate in more daail basicfluids, such as ai.r and water, that not only are needrd to sustain life but abo pky important roles in engineering. Did yow eaer stop to think about the irnportant role that air plays in food processing driuing power took, or in your car's tire to prouidz a cusltiony rid,e? You mE not think of water as an mgineeri.ngrnaterial either, but ute not only need uAter to liue, we also need water to generate elecnicity in stearn and hydroelectric power plants, and we wse high-pressurized water, which funaions lihe a sAu), to cut rnaterials,
17.1 MaterialSelection Dobt engineers, when faced with selecting materials for their products, often ask questions such as: How strong will the material be when subjected to an expected load? \fould it fail, and if not, how safely would the material carry the load? How would the material behave if its temperanrre r /ere changed? \Mould the material remain as sffong as it would under normal conditions if its temperaflre is increased? How much would it expand when its temperature is increased? How heavy and fexible is the material? Vhat are its energy-absorbing propenies? \trould the material corrode? How would it react in the presence of some chemicals? How
Vould it dissipate heat effectively?'lfould the material ductor or as an insulator to the fow ofelectricity? expensive is the material?
act as a con-
502
Cneprsn
17
ExcrNmnrxc MarsRrAr^s It is imponant to note here that we have only posed a ferv generic questions; we could have asked additional questions had we considered the specifics of the application. For example, when selecting materials for implants in bioengineering applications, one must consider many additional fadors, including: Is the material toxic to the body? Can the material be sterilized? \?hen the material comes into contact with body fluid will it corrode or deteriorate? Because the human body is a dynamic qrstem, we should also aslc How would the material react to mechanical shock and fadgue? fue the mechanical propenies of the implant material compatible with those ofbone to ensure appropriate stress distributions at contact surfaces? These are examples of additional specific questions that one could ask m find suitable material for a specific application. By now it should be clear that material properties and material cost are imporant design factors. However, in order to berter understand material propenies, we must first undersand the phases of a substance. We discussed the phases of mater in Chapter 9; as a rwiew and for the sake of continuity and convenience, we wrll briefly present the phases of matter again here.
The Phases of Matter: Solids, Liquids, Gases, and Plasma in Chapter 9, when you look around, 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. Ve also explained that all solid objects, liquids, gases, and living things are made ofmatter, and maaer itself is made up of atoms or chemical elements. There are 106 known chemical elements to date. Atoms ofsimilar characteristics are grouped together and shown in a table-the periodic table of chemical elements. Atoms are made up of even smaller panicles we call electrons, protons, and. neutrorar. In your first chemistry class, you will study these ideas in more detail, ifyou have not yet done so. Some ofyou may decide to study chemical engineering in which case you will spend a geat deal of time srudying chemistry. But for now, remember that atoms are the basic building blocla 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 nvo atoms of hydrogen and one atom of onygen. A glass of water is made of billions and billions of homogeneous water molecules. A molecule is the smallest ponion of a given matter that still possesses its microscopic characteristic propenies. Matter can exist in four states, dgpsnding on its own and che surrounding conditions: solid, liquid, gaseous, or plasma. Let us consider the water that 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, water e*ists in a solid form as long its temperature is kept under 0oC. Under sandard atmospheric pressrue, ifyou were to heat the ice and consequently change its temperature, dre ice would melt and change into a liquid form. Under standard pressure at sea level, the water remains liquid up to a temperatue of l00oc as you condnue heating the water. If you were to carry out this experiment furher by adding more heat to the liquid water, eventually the liquid water changes its phase from liquid into a gas. This phase of water we commonly refer to as stearn. If you had the means to heat the water to eyen higher temperatures, temperatures exceeding 2000"C, you would find that you can break up the water molecules into their atoms, and wentually the atoms break up into free electrons and nuclei that we callphsma.
As we discussed
17.2 Er-rc"rnlceq In
MscHANrcAL, arvo TnrnrvropuysrcAl Propanrtes
or MarBnrers
503
general, dre mechanical and thermophysical propenies of a material depend on its
phase. For example, as you know from your everyday experience, the density of ice is different from liquid water (ice cubes float in liquid water), and the density of liquid water is different
from that of steam. Moreover, the properties of a material in a single phase could depend on its temperature and the surrounding pressure. For example, if you were to look up the density of liquid water in t.he temperature range of, s*y,4" to 100"C, under standard atmospheric pressure, you would find that its density decreases with increasing temperature in that range. Therefore, properties of materials depend not only on their phase but also on their temperature and pressrue. This is another important fact to keep in mind when selecdng materials.
flZ.2 Electrical,
Mechanical, and Thermophysical Properties of Materials
As we hane been explaining up to this point, when selecting a material for an application, as an engineer you need to consider a number of material properties. In general, the properties of a material maybe divided into three groups: electrical, mechanical, and thermal. In electrical and
elecuonic applications, for orample, the electrical resistivity of materials is important. How much resistance to the fow of elecuicity does the material offer? In many mechanical, civil, and aerospace engineering applications, the mechanical propenies of materials are imporant. These propenies indude modulus of elasticity, modulus of rigidity, tensile strength, compression strength, the strength-to-weight ratio, modulus of resilience, and modulus of toughness. In applications dealing with fluids (liquids and gases), thermophysical properties such as thermal conductivity, heat capacity, viscosity, vapor pressure, and compressibility are imporrant propenies. Thermal expansion of a material, whether solid or fuid, is also an imponant design factor. Resistance to corrosion is another imporant factor that mu$ be considered when selecting materials.
Material properties depend on many factors, including how the material was processed, its age, its exact chemical composition, and any nonhomogeneity or defect within the material. Material properties also change with temperature and time as ttre material ages. Most companies that sell materials will provide upon request information on the important properdes of their manufactured marerials. Keep in mind that when practicing as an engineer, you should use the manufacnrrer's material properry values in your daign catculations. The properry values given in this and other textbooks should be used as rypical values-not as exacr values. In the prwious chapters, we hare explained what some propenies of materials mean. The meaning of those propenies and other propenies that we have not explained alrady are summarized next. Electrical Resistivity The value of electrical resistivity is a measure of the resisrance of material to the flow of electricity. For er
504
Cneprsn
17
ENcnvErRrNc MrreRrArs
Density Density is defined as mass per unit volume; it is a measrue of how compact ttre material is for a given volume. For example, the average density of aluminum alloys 's 2700 kgl m3 and compared to steel density of 7850 kg/m3, aluminum has a density that is approximately one-third the density ofsteel. Modulus of Elasticity (Young's Modulus) Modulus of elasdcity is a measrue of how easily a material will stretch when pulled (subject to a tensile force) or how well the material will shonen when pushed (subject to a compressive force). The larger the value of the modulus of elasdcity is, the larger the required force would be to suetch or shonen the material. For example, the modulus of elasticity of aluminum alloy is in the range of 70 to 79 GPa, whereas steel has a modulus of elasticity in the range of 190 to 210 GPa; therefore, steel is approximately 3 times stiffer than aluminum allovs.
Modulus) Modulus of rigidiry is a measure of how easily a material can be twisted or sheared. The value of modulus of rigidiry, also called shear modulus, shows the resistance of a given material to shear deformation. Engineers consider the value of shear modulus when selecting materials for shafis and rods that are subjected to rwisting torques. For orample, the modulus of rigidity or shear modulus for aluminum alloys is in the range of 26 to 36 GPu whereas the shear modulus for steel is in the range of 75 to 80 GPa. Therefore, steel Modulus of Rigidity (Shear
is approximately three times more
rigid in shear than aluminum.
Strength The tensile suength of a piece of material is determined by measuring the maximum tensile load a material specimen in the shape of a rectangular bar or cylinder can carrywithout failure. The tensile strength or ultimate suength of a material is expressed as the maximum tensile force per unit cross-sectional area of the specimen. '$7hen a material specimen is tested for its strength, the applied tensile load is increased slowly. In the very beginning of the test, the material will deform elastically, meaning that if the load is removed, the material will re&rn to its original size and shape without anypermanent deformation. The point to which the material exhibia this elastic behardor is cdled, yield point The yield strength represents the maximum load that the material can carry without any permanent deformation. In cerain engineering design applications (especially involving britde materials), the yield Tensile
strength is used as the tensile strength. eompression Strength Some materials are snonger in compression than they are in tension; concrete is a good example. The compression snength of a piece of material is determined by measuring the maximum compressive load a material specimen in the shape of cylinder or cube can carry without failure. The ultimate compressive strength of a material is expressed as the maximum compressive force per unit cross-sectional area of the specimen. Concrete has a compressive stren$h in the range of 10 m 70 MPa. Modulus of Resilience Modulus of resilience is a mechanical properry of a material that shows how effective the matedal is in absorbing mechanical energy without sustaining any permanent
damage.
17.2 Er-ecrnrcaq Mncuexrcero
eNo TneRMopHysrcAL
Pnopnnns or Mrrunrars
Toughness Modulus of toughness is a mechanical propeny of ability of the material to handle overloading before it fracnres.
Modulus of cates the
a
505
material that indi-
Strength'to'Weight Ratio As the term implies, this is the ratio of the strength of the material to its specific weight (weight of the material per unit volume). Based on the application, engineers use either the yield or the ultimate suength of the material when determining the suength-to-weight ratio of a material. Thermal Expansion The coefficient of linear expansion can be used to determine the change in the length (per original length) of a material that would occur if the temperature of the material were changed. This is an important material propefty to consider when designing producs and structures that are expected to o
temperature region within the material. Heat Capacity Some materials are better than others in storing thermal energy. The value of heat capacity represents lhe amount of thermal energy required to raise the temperature of I kilogram mass of a material by 1'C, or, using U.S. Customary Units, the amount of thermal energy required ro raise one pound mass of a materid by 1'F. Materials with large heat capacity values are good at storing thermal energy. Viscosity, vapor pressure, and bulk modulus of compressibility are additional fuid propenies that engineers consider in design.
Viscosity The value ofviscosity of a fluid represents a measure of how easily the given fluid can flow. The higher the viscosity value is, the more resistance the fuid offers to flow. For example, it would require less energ;r to transport water in a pipe than it would to transport motor oil or glycerin. Vapor Pressure Under the same conditions, fluids with low vapor pressure values will not evaporate as quickly as those with high values of vapor pressure. For example, if you were to leave a pan of water and a pan of glycerin side by side in a room, the water will waporate and leave the pan long before you would notice any changes in the level ofglycerin. Bulk Modulus of Compressibility A fluid bulk modulus represen* how compressible the fluid is. How easily c:rn one reduce the volume of the fuid when the fluid pressure is increased? For example, as we discussed in Chapter 10, it would uke a pressure of 2.24 X 107 N/m2 to re-
duce
I
m3 volume
ofwater by lo/o or, said another way, to
a
final volume of 0.99 m3.
In this section, we explained the meaning and significance of some of the phpical propenies of materials, Thbles 17.1 through 17.4 show some properties of the solid materials. In the following secdons, we will examine the application and chemical compositionlof some common engineering materials.
Aluminum alqrs
l'.L Cast iron Concrete (compression) Copper allop Glass
Magnesium allrys
Nickel Plastics
Nylon Polyethylene Rock (comprecsion)
Granite, marble, quare Limestone, saodstone
Rubber "r. t,""1 ' 'ri :l
:
Titardurrr alla;d
Tungpten
,
,,
Wbod (bending)
Douglasdr
..
I
O& "', r'
l-13
tl-12
i-.,-.-:1.1tn1*lr"=- .r-." *...
11-14 .
Source : Adaptd ftom J. M. G*e, Me cbania a division of Thomson l-eaming: www.thomsonrights.com
nfate,rial
Mass Density
;r**t;""t;
(kg/mu)
G$Vhj)
Brass
2600-2800 8400-8500
Bronze
s200-8800
Qast iron Cooat"t
7000=74a0
Aluminundllays
Plain Reinforced
Ughtweight
'il
Copp.t GIe$s
,
Magnesium
aIIgrc ' Continued
Mcerial
I ; i i
Nickel
1
Rock Granite, marble, quartz
i i i i i
Masr Density
Specific\Feight
(kg/*o)
(kl{lnt)
8800
86.3
Plastics
880-t 100 960-r40a
Nylon Polyethylene
8.6-10.8
9.4-r3.7
26A0-2900 2000-2900 950-1300
25.5-28.A
77.4
Titanium alloys
7850 4500
Tungsten
1900
18.6
Limesgone, sandstone
Rubber Steel
r9,6-25.4
9.4-rzJ 44.1
1 \food ("itdty)
I
ii
Douglas
fu
ouk Southern pine
S*ro
*o
48A-.560
4./_>.'
640-72A 560-64A
6.3-7.r
i.
5"5-6.3
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)
35-500 70-554 82-69A
na-290 55-760
Gtass Plate glass Glass fibers
Magnesium allqrs
Nickel
s0-2sp 100-620
Plastics
Polyethylene Rock (compression)
l l
Granite, marble, quaru Limestone, sandstone Rubber
100-550 2AA-520
l
11
,
l
200-830
69-480 340-L400 lO-70 230-830 30-1000 70: 7000-20,000 1404/+A 310-760
|
I
l
r
I
r
i
l
t.
Nylon
'
of
40-80 7-28
l
i
50-280 2o-2AA 7-20
l
'
l
Continued
Matetial Steel
Yteld Strength
@a)
Illtimate Strength (MPa)
,
High-srength
340'1000
550-tzAO
Machine
550-860
Spti"e
34A-7oA 400-1600
Sainless
280-700
700-1900 400-1000
520
900
Tool Steel wire
280-1000
Smrctural steel
?00-700
Titanium alloys
760-1000
550-1400 340*830 900-1200 1400*4000
Tungorcn Wood (bending) Douglas
fu
30-50
Oak Southern pine \Food (compresion parallel to gain) Douglas
fir
Oak
50-80 50-100
4A-60
50=100
,30-50
4A-70 30-50 40-70
30*40 30-50
Southern plne Source: Adapted fromJ.
40-60
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.t-21.2
Brass
:,
Coefficient ofThernal
i0.6-11.8
T8_21
9.9-11.6
9.9-r2
5.5-5.6
7-r4
4-8
16.6-r7.6
9.2-9.8
5 -11 26.L-28.8
L4.5-16.A
3-6
l3
aa
Plastics
Nylon Polyethylene
Rock:
70-L4A
40-80
140-290
80-150
5-9
3-5 70-l 10
,
130*200
Rubber
l0-t8
5.5-9.9
T4
8.0
Strucnual
t2
9.6 6.5
I ltailum aroys Tungsten
4.3
Steel
r,
High-strengti Stainless
*Note
thr you nust muhiply the coefficients of thermal expansion.
8.
co"ma"""
l-r I
giri i" .hi"
4.5-6.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 one-third 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 strength-to-weight-ratio. 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 products-from high' performance engines to soda cans.
as
510
CnrprsR
17
Enqcrxrrmxc
Memnns
Because of their excellent strength-to-weight ratio and resistance to corrosion, titanium alloys are used to produce a range of products-from 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 one-half that of steel. Because of theit relatively high strength-to-weight 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) low-alloy steels having a maximum of elements, and (3) high-alloy 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 low-alloy steels have good strength and are machine or tool pars and as strucural members. The high-alloy 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
applications-from 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 (water-to-cement 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 load-bearing 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 strengt-h, 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 direction-dependent. 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 load-carrying 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 groc€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 high-density 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 soda-lime-silica glass. The materials used in making soda-lime-silica 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 light-dispersion 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 high-temperature 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*€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 light-emitting 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 Memnrer-s
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, fast-food 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) polymer-matrix composites, (2) metal-matrix composites, and (3) ceramic-matrix 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 t-hat 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 two-thirds 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 str€,!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 polver-generating 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 low-pressure 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. High-pressure 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 red-cedar 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 ofR-value. 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. Cobalt-chromium 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 strength-to-weight
ratio of material is an imponant criterion when selecting material for aerospace applications. Calculate the average strengtfr-to-weight 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 fiber-optic 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
wet-inking
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$on and its typical properties. Discuss your findings in a brief repon.
524 17.28.
Csepren
17
ENcrNnnnrNc MereRrALs
As most of you know, commercial uansport planes cruise at an altirude of approximately 10,000 m (-33,000 ft). The power required to maintain
fight depends on air drag, or resistance, at that altitude, which may be estimated by the following level
reladonship:
Power
17.30.
I : |o*coulu
where p,;" is alrrri.y of
the given akitude, Cp rep"i, ", resenr the drag coefficient of the plane, r4 is the planform area, and U represens the cruising speed of the plane. Assume that a plane is moving at constant speed
and, Cp remaining constant, determine the ratio of power that would be required
if the plane is cruising
at 8000 m and when the plane may be cruising at 11,000m. 17.29.
Investigate the average daily water consumption per capia in t}re United States. Discuss the personal and public needs in a brief report. Also discuss facors such as geographical location, time of the year, time of &" d"y, and cost of consumption pafterns. For example, more water is consumed during the early morning hours. Civil engineers need to consider all
these factors when designing water systerns for cities. Assuming that the life expectancy of people has increased by five years over the past decades, how much additional water is needed to sustain the lives of 50 million people? Vhen a ring gets stuck on a finger, most people reson to water and soap as a lubricant to get t-he ring off. In earlier times, animal fat was a common lubricant used in wheel axles. Moving pans in machinery the piston inside your car's engine, and bearingr are examples of mechanical componen$ 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 petroleum-based lubricants in reducing wear and friction in today's mechanical componens. Vrite a brief repon discussing the application and characteristics
poins
of liquid-petroleum-based 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, 9-foot-diameter 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 4-bedroom 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.
l--rrTiili: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 engineering-design 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
)<)
58.7
309
s:-* Zq(t'r
zza
1;
, ,. .|:
go
Lrr,?
g4L
Ezr700 li3 u,600 liE 5(s00 f*
c)
where
i,H 400 41 f-
S:
stopping sight distance (ft)
[:
initial speed (ft/s)
g:
acceleration due to gravity, 32.2ltlsz
f:
3( i90 300
:&z20olliBu100 l* iv) i
014 0
i-_---.-,___
coefficient of friction between tires and roadway
6:
/ grade of road
or^\
( j00 J
I: driver reaction time 5:12
(s)
(ft)
66.0
1000 to10( ie sre00 Tl* i9 sr800 l*
yz
Distance
/r5 '
Highway and Transportation 0fficials (AASHT0) is given by
Stopping Sight
7tr3' 436 :, '.55 80.7 , 508 60 88.0 584 65 95.3 666 ':" 7A rc2.7 : 753 'tt 75 : : 11o.o t g44
\\
36.7
t,l:,:,:
't 50 ,
78
u.o
,
Sp€€d Sp€4 (mph) (ft/s)
Distance
known as stapping sight disfanceis used by civil engineers to design roadways.
t,
,
__
-__r _---=, - ____--
18.1
MernsMATrcAL Symsols eNo
GnBxAr,prrABET
533
In general, engineering problems are mathernatical mod.els of physical situations. Mathematical modeb of mginening problems may haue rnany differmt forms. Sorne engineering problerns lead to linear mofub, whereas others resub in nonlinear rnodek. Some mgineeringproblems areformulated in theform of diffumtial equatiow and sorne in tbe form of integrab. Forwulation of maryt mgineering problems resubs in a set of linear algebraic equations that are solued simultaneously. A good understanding ofmatix algebra is essmtial in theforrnulation and solution ofthese problens. Thnefore, in this chapnr, we will discus uarious rnathemati.cal rnodcb that commonly are used to solue engineningproblerns. We begin our discussion with an expknation of the needfor conaentional math syrnbok as a rneAns to conae! information and to ,ffitti*h cornmunicate to other mgineers, Examplcs of math syrnbob will ako be giuen. Next, we discuss the irnportance of knowing the Greeh alphabet and its use in engineeringforrnuke and engineeringdrawings. This sectinn will be followed with a discussion of simph linear and nonlinear mod.ek. Next, we introduce rilatrix algebra, with its own teTwtinolog and ruleg which we will defne and expkin. Ve will then bri$y discuss calculus and its irnportance in soluing engineeringproblerns. Calcalus comrnonly is diai.ded into two areas: dffirmtial and integral calculus. Finally, the role ofdiffomtial equations informulating mgineeringproblens and their solwtions is presmted. It is importam to heep in rnind that the pwrpose of this chapter is n focus on irnportant rnathematical concepts and. to point out why rnathematics is so important in your mgineering education. Thefocus ofthis chapter is not to explain rnathematical concepts in deail, that will take place later in your math classes,
18.1 Mathematical Symbols and Greek Alphabet As you already know, mathematics is a language that has its own qymbols and terminolory. In elementary school, you learned about the arithmetic operational symbols, such as plus, minus, division, and multiplication. Later, you learned about degree symbols, trigonometry qymbols, and so on. In the next four years, you will learn additional mathematical qymbols and their meanings. Make sure that you understand what they mean and use tiem propedy rvhen communicating with other students or with your instructor. Examples of some math symbols are
shovm in Thble 18.1.
The Grcek Alphabet and Roman Numcrals As you take more and more mathematics and engineering classes, you will see that the Greek alphabedc characters quite commonlyare used to express angles, dimensions, and phlnical variables in drawings and in mathematical equations and expressions. Thke a few moments to learn
534
CuerrrsR
18
MenrBrvnrrcs rN Encnrnnnnrc
I
Plu or positive
l
i-
t-F
I
xor.
i
Mrrltiplicuion
+ orl
Ratio Les than Grearer than Muchless than Much greatel.rhan
r< 1> :
i<<
r)> .: i:
Equal
iz iH lo
rI iK IA iM
5t:5x4x3x2xl
A
Delta indicating difference Pardat
g
Pi, iuvalue 3.1415926; . ,
,:
oo
Infinity
o |l IJ
Parenthess Brackets
{}
Braces
Degree
Identical ncith Equal to or less dran Equal to orgreater than
i>
iE
Integral Facorialr forexample,
!
Not equal to
i<
iB' ir ia
F*&i. Summation
Approximately equal to
J
rA
to
d
Alpha
P "t
Beta
N
Gamma
s
v
no.-,1,,ii,,
.'t
o II
5
Xi'
o
Omicr<m
P
p
rr
.
0
Y
a
Upsilon;,,'
L
Iota
o
a
K
Kappa
Phi Chi or khi
/\
Iambda
tL
Mu
6
rl
i
s T X
v o
a
r
x '!a,
-
Pi .t,.:;: Rho ::
Delta Epsilon Tnta Eta Theta
t
r
Proportional to
I
Division
l
Absolute value of
lxl 0(
Minus or negative Plus or minus
Siglna
Tau , Psi
'
r,.
,
Om"ga ..,,..1,
and memorize these characters. Knowing these qymbols will sane you time in the long run when communicating with other students or when asking a question of your professor. You don't want to refer to @t^) as that "curly thing" when speaking to your professor! You may also find Roman numerals in use to some extent in science and engineering. The Greek alphabet and the Roman numerals are shown in Thbles 18.2 and 18.3, respectively.
t
18.2
)o( )oo(
=20
-a
-j
XL
=40 =50 =60 =70
iI
lu
IIU nn 6t1Y ,Y lvI i YII r VItr lD(
ix ix
L
-8 -9
XC
:10 = 1l =12
i )0r )ofl ilav ixr i xvl lxm i :wm
:13
=14 =15 =16 =17
:18
i)(Ix
18.2
-4 :5 -(
=19
IX
IJO( I)OO(
c cc ccc CCCCoTCD
D DC DCC DCCC
CM
M MM
Ln*EARMoosLs
535
-30
:80 =90 = 100 = 200 = 300 = 400 = 500 = 600 = 70a = 800 = 900 = 1000 = 2000
Linear Models Linear models are the simplest form of equations commonly used to describe a wide range of engineering situations. In this section, we first discuss some examples of engineering problems where linear mathematical models are found. Ve then explain the basic characteristics of linear models.
Spring In Chapter 10, we discussed Hoote's law, which srates that, over the elastic range, the deformation of a spring is direcdy proportional to the applied force, and consequendy to the internal force developed in the spring according to A Linear
F:b
fl8.t)
where
.F:
spring force (N or lb)
: r:
spring constant (N/mm or N/cm or lb/in.)
A
It
deformation of the spring (mm or cm or in.) (use units that are consistentwith
f)
is clear from oramining Equation (18.1) that the spring force Fdepends on how much the spring is stretched or compressed. In mathematics, .Fis called e depmdent uariable.The sping force is called the dependent variable because its value depends on the deformation of rhe
536
Crr^q.prrn
L8
Memrr,ulrrcs
rN ENGTNEERTNG
,^25
F:2,t
a
&20 r(mm)
^F
00 510 10 l5 20
E
Figure
N)
20 30 40
r
l8.l
x (mm)
A linear model for spring force-deflection relatiorship.
springx Considertheforceinaspringwithastiffnessof h:2N/mm,asshowninFigure 18.1.
:
For the linear model describing the behavior of this spring, constant P 2 N/mm represen$ the slope of the line. The value of 2 N/mm tells us that each time the spring is suetched or compressed by an additional 1 mm; as a result, the spring force will be changed by 2 N. Moreover, note that for r 0, the spring force F 0. Not all springs exhibit linear behaviors. In fact, you find many springs in engineering praffice whose behaviors are described by nonlinear models.
:
:
Temperature Distribution Across a Plain Wall Temperarure disribution across a plain wall is another example where a linear mathematical model describes how temperature varies along rhe wall. Under a steady-state assumption, the temperarure disuibution-how temperature varies across the thickness of the wall-is given by
r(x)
: G, - n)1+ r,
(18.?)
where
: [ : [ : rr :
T(*)
or'C) or'C) I ("F or'C)
temperanue distribution ('F temperature at surface 2 ("F temperature at surface distance from surface
I (ft or m)
Z = wall thickness (ft or m) For this linear model, Iis the dependent variable, and r is rhe independent variable. The variable r is called an independznt uariable, because the position r is not dependent on temperature. Now, let us consider a situation for which I : 68"F, Tz: 38"F, and, L = 0.5 ft. For these
L8,2 Lrxrm.Mooprs
T: -6 0r+68
4 = 68'F Tz=
L= r
s37
38"F 0.5
ft
(ft)
50
T(r)
T- qo t!
30
68 62
0 0.1 0.2 0.3 0.4 0.5
E
56 50
0L
44
0.0
38
Figunl8.2
0.1
0.3
0.5
0.6
r (ft)
Temperaturedhhibutionalongawall.
conditions, the slope of the linear model is given by (Tz- T)lL = -60oF/ft, as shovvn in Figure 18.2. Note that for the given conditions, the line that describes the reladonship betrveen the temperature and position intercepts the temperarure axis at the value of 68 (i.e., at r : 0,
I:68'F). \V'e can describe many other engineering situadons for which linear relationships enist berween dependent and independent variables. For example, as we explained in Chapter 12, resistiuity is a measure of the resistance of a piece of marerial to elecuic current. The resistiviry values usually are measured using samples made of a centimeter cube or a cylinder having a diameter of 1 mil and length of 1 ft. The resistance of the sample is then given by
n:4 4
where p is the resistivity, { is the length of the sample, and a is t}re cross-sectional area of the sample. As you c:ul see, for constant values ofp and a rherc exists a linear relationship between R and the length {. The relationship amongvarious systems of unis is also linear. Let us demonstrate this fact using an example dealing with temperature scales. In Chapter I 1, we discussed the relationship berween the two temperatrre scales Fahrenheit and Celsius, rvhich is given by
Z('F) =
I =T("C) )
+
32
fl8.3)
'We
have ploned the relationship benveen the Fahrenheit and Celsius scales for the temperaffie 1.8, range shown in Figure 18.3. Note the slope of the line describing the relationship is 915 and the line intercepts the Fahrenheit axis at 32 (i.e., I("C) = O, T ("F) 321.
:
-
538
CHnprsR
,-30 :-22 -25 -r3 *20 -4
18
Mnrrrrue.rrcs nr Eucnrrrnnvc
35 4a 45 50
95' LoA
rrg 122
))
/
6A 65
1@
70 7J
;
80
tx
85
7lo
/
60
-{
--
-J
90 95
_/
-4SF -N
tm
m&
-n
80
T("C)
f
figure
I8.3
lhe relationship beheen Fahrenheit and Celsius scales.
linear Equations and Slopes Now that you realize the irnporunce of linear models in describing engineering situations, let us consider some of &e basic characeristics of a linear model. As you know, the basic form of a line equation is given by
y: ax* b wtere
,
CI0.4)
Ay:
: l-
4:
Srope
b:
yintercept
^
Ar
change
nyvalue
change in xvalue
(rlevalueofyatr:
0)
Equation (18.4) is plotted and shown in Figure 18.4. Note positive values were assumed for the.y-intercept and the slope in Figure 18.4. The slope of a linear model shovrs by how much
E
FigmlS.4
Alinearmodel.
18.2
539
LrNEARMopsLs
the dependent variable 7 changes each time a change in the independent variable duced. Moreover, for a linear model, the value of the slope is always constant.
r
is intro-
Comparing our previous models of examples of engineering situations to Equation (18.4): for the spring example, the slope has a value of 2 and drey-intercept is zero. As we mentioned before, t}re slope value 2 N/mm conveys that each time we stretch the spring by an additional one unir, as a result the spring force will increase by 2 N. For the temperature-distribution model, the slope has a value of -60'F/ft, and they-intercept is given by the value of 68oF. For the temperature-scale er
a:
EloPe
=
Ay
chanseinyvalue
Ar
change in
Affi
rvalue (-35) - (-40) --
4L-32:_:_ 2r2-203 100-95 5-0
9 5
Linear models could have different forms with different characteristics. \U?'e have summarized the characteristics of various linear models in Thble 18.4. Make sure to studv them carefirlly.
Systems of Linear Eguations At times, the formulation of an engineering problem leads to a set of linear equations that must be solved simultaneously. In Section 18.5, we will discuss the general form for such problems and the procedure for obtaining a solution. Here, we will discuss a simple graphical method that you can use to obain the solution for a model that has two equations with
naro
unknowns. For example, consider the following equations with
r
andT
as
unknown
variables.
* 41: 19 4xI Y:6
2x
(18.5a) (18.5b)
Equations (18.5a) and (18.5b) are plomed and shown in Figure 18.5. The intersection of rhe nvo lines represents the r solution, which is given by * : 1; because as you c:m see at Jd : 1, both equations have the same.l value. V'e then subsdtute for r into either Equation (18.5a) or Equation (18.5b) and solve forn which yields a value of I : 2.
LinearModel
Charactetistice
!:4x+b
Slope a and y-intercepr b.
!:
Slope zero,y-intercept b, and
b
horizontal line going
rt"o"sh
point b ony axis.
x:
Undefined slope, .n-intercept c, and venical line going tlro"gh
c
point con r axis.
qx+c2y=ct
General form with r- and y-intercepts, slope - qlq,
v ca
q,
yintercept
cal
x-interc-epn
cal c1
c2, and
ca c1
4x + Y:6 """ -2x+4y=19
I
FigurelS.S
The plot of Eguations
(t8.5a) and (10.5b).
w
f
18.3
8.3
NoNr,rNsAR
Moonrs
541
Nonlinear Models For many engineering situations, nonlinear models are used to describe the relationships berween dependent and independent vadables beca"*e they predict the actual relationships more accurately than linear models do. In this section, we fust discuss some examples of engineering situations where nonlinear mathematical models are found. We then explain some of the basic characteristics of nonlinear models.
Folynomial Functions Pipe For tlose ofyouwho are planning to become an aerospace, chemical, civil, or mechanical engineer, later in your studies you will take a fluid mechanics class. In that class, among other topics, you will learn about the fow of fluids in pipes and conduits. For a laminar fow, the velocity distribution-how fuid velociry changes at a given qls5s-5sctieninside a pipe is given by
Laminar Fluid Velocity Inside a
=
a(r)
nlt -
(;)'l
fl8.6)
where
u(r) Z"
:
g64r"rocity
=
center line velocity (m/s)
r: R
at the radial distance r (m/s)
radial distance measured from the center of the pipe (m)
=
radius of the pipe (m)
The velocity distribution for a situation where V": 0.5 m/s and R = 0.1 m is shown in Figure 18.6. From Figure 18.6, it is evident that the velocity equation is a nonlinear second-order polynomial and the slope of this type of model is not constant (it changes with r). For the given enample, for any 0.01 m change in r, the dependent variable zr changes by different amounts depending on where in the pipe you waluate the change. _0.095
-0.005
slope
:
change in velocityvalue change in position
_
value
(0.495) (0.01
-
-\- 0.01
(0.5) 0)
, 0-
0.095
0.1
-
-/-
0.01
0.09
Stopping Sight Distance A model known as aopping sight di.sance is used by civil engineers to design roadways. This simple model estimates the distance a driver needs in order to stop his car while travelling at a certain speed after detecting a hazard. The model proposed by the American Association of State Highway and Tfansponation Officials (AASHTO) is given by
t:*#")+rv
fl8J)
542
Cneprr.n
18
Mernnrvrerrcs
urr
ENcrNssR[\rc
---.--;--- - .----. --i
r
u{r)
-#r -0.1
0 ,:i
j
0.6
I
.
*0.09 -0.0s
ooe5
0"18
ly
I
l
0.5
-0.07
.. '. 0.3? 0.t75 .i -0.0t 0.42 -0.04 i 0.455 -0.03' 0.48 -0,o2 0.495 -0.01 00.5 0.0r 0.495 o.o2 . o.4g 0,03' a.455 'i o.M a.42 o.g5 0375 0.06 . 0.32 .l 0.07 0.255 0.08 o.rs . j 0"09 . 0.095
-0.06
.tlope
..1
^
j i
/
0.4
I
4 0.3
l
I
o
1
E
|
t
o2
I
1
ti
0.1
0.0
oj ,.:_._i
-0.15 -0.10 -0.05 0
I
Figure
18.6
\ \ \ Slope
\
I
.1
f
\
0.05
t\
0.10
Radial distance ftom tle center of pipe, r (m)
An eranple of fluid velocity dishibution inside a pipe.
where
S: Z: g: / :
stopping sight distance (ft)
initid
speed (ft/s)
acceleration due
to g;rrty,32.2
ftl*
coeffi"i"nt of ftiction berween tires and roadway
/lL)
G:
grade
I:
driver reaction time G)
ofroad
[,ooz
lnitialspeed = V
:i,r,:
In the
't
above equation, the typical value for the coefficienr of friction berween tires and roadway/is 0.33, the driver reaction time varies between 0.6 to L.2 seconds; however, when designing roadways, a conservarive value of 2.5 second commonly is used- In the denominator
18.3 Noxr.rxran Mopnrs Sp""d
Speed
(-ph)
(fth)
I
0.0
0
)
47
L5
7.3 14,7 22.0
2A
29.3
tt4
25
36.7 44.0
r55
.g 600
201
5r.3
)<1
E
30 35 40 45
2t 78
58.7
309
66.0
370
50
73.3
436
)>
80,7
508
60
88.0
)64r
65
753
/)
95.3 102.7 110.0
80
rr7.3
94r
7A
f
Stopping Sight Distance (ft)
0
10
Figure
18.7
543
/
X
zoo
/
s00
b0
./)
/
ao
zl00
R
300
v) 200
666
0rz
844
40.0 60.0
0.0
80.0
r20.0
Speed (ft/s)
The stopping sight distance for a car haveling speeds of up to 80 mph.
of Equation (18.7), plus (*) indicates upgrade, whereas minus (-) is for downgrade. A graph showing the stopping sight distance for a flat roadway as a function of initid speed is shown in Figure 18.7 (G : 0,f : 0.33, andT: 2.5 seconds were used to generate this graph). This is another example where a second-order polynomial describes an engineering siruation. Again, note the slope of this model is not constant, that is, for any 5 mph change in speed, dre dependent variable S changes by &fferent arnounts based on where in the speed range you inuoduce the change. 26
shown: slope
:
change in stopping distance S change in speed
l/
* (2r) , 508 - 436 - 55-* \--_-,5)
(47)
=- (ro -
5 mph (7.4 ftls)
,ior,-rzr,,o
'We
can describe many other engineering situations with second-order polpromials. The trajectory of a projectile under a constaot deceleration, power consumption for a resistive element, drag force, or the air resistance to the motion of a vehicle are represented by secondorder models. Deflection 0f a Eeam The defecdon of a cantilwer beam is an example of an engineering situation where a higher-order polynomial model is used. For example, the cantilever beam shown in Figure 18.8 is used to support aload acting on a balcony. The deflection ofthe centedine of the beam is given by the following fourth-order polynomial equation.
t: ffif*
- 4Lx + 6L2)
(r8"8)
544
CHaPrsR
18
Mrtrrrruarrcs rx ExcnrsERn{c
I
FigurelS.S
Acantileverbeam.
where
y: a; : E: / : x: I :
deflection at a given
r location (m)
disuibuted load (N/m) modulus of elasticity
(N/-')
second moment of area (ma) distance from the suppoft as shovm (m)
length of the beam (m)
The deflecdon of
a beam
with
a
lengrll of 5 m with the modulus of elasticity of E
I = 99.1X 106 mma, and for a load of 10000 N/m
is shown
:
200 GPa,
in Figure 1S.9.
Now that you realize the importance of polynomial models in describing engineering
.t(m)
0.2 "' 1 1.2
y(m) -0.00012
-0.Q0275
-0.00386
t'a',:,:li*gdffi
2 2.2 I
*0,00959 -0.01128 -0.01305
Hguret8.9
0.0m
'-'l--l'*t-
2.6 -0,01489 2.8 . -0;01,678 3 :0.0187j 3.2 -U,A2072 , 3.4 ':0,02274 u0,82478 ', 3.9, ;0.02685 ., 3.8 4 -0.a2s93 4,2 -0.03102 4.4 -0.03311 4.6 -0.03521 4.8 -A.$732 5 -0.03942
lhedsflectionof acanlileverbeam.
-0.005
-0.010 rl
-0.015 e
-.Ga\
\ \ \
-0.020
\
A -0.025 t)
o o
\ \
-0.030
\
-0.035
\
-0.M0 -0.045
23 Distance from support,r (m)
18.3 Noxr.rrnenMoorrs
54s
flr) I
Slope
" f(x)=f-6*+3*.l.rO |
-21
|_, I -r .0 rl '2 t3 :4 r) t5
I
SJope-
-80 -28
-20
0
r
/
\ -l Slope
4
I
10 8
I
0
-8 -10
-80
0 28
Figure
I
-100
-4
18.10 Tie real roots of a third'order
-3
-2
-1
polynornial tunctior.
situations, let us consider some of the basic characteristics of polynomial models. The general form of a polynomial function (model) is given by
l: fQ) :
do
* ap * ar* )- auf l "' *
a,x"
{r8.e)
where as, dy . . . 2 a, are coefficien$ tfiat could ake on different values, and z is a positive integer defining the order of the po\momial. For the laminar fluid velocity and the stopping sight distance examples, n is 2, and. the defection of the beam was represented by a fourth-order
polynomial. Unlike linear models, second- and higher-order polynomials have variable slopes; meaning each time you inuoduce a change in the value of the independent variable r, the corresponding change in the dependent variable.2 will depend on where in the x range the change is inuoduced. To becer visuelize the slope ar a certain .r value, drarr a tangent line to the curve at the correspondingrvalue, as shown in Figure 18.10. Another imporant characteristic of a polynomial function is that the dependent vafiable y has a zero vdue at points where it intersects the n axis. For example, for the laminar fuid velocity situation shown in Figure 18.6, the dependent variable, velocity 6has zrro values at r = 0.1m and r : -0.1 m. As another example, consider the third-order polynomialT = f@) = f - 6* * 3x * L0, as shown in Figure 18.10. This function intersects the r axis at x = -1, rc : 2, and x : 5. These poina are called the real roots of the polynomial function. Not all polynomial funcdons have real roots. For example, the function/(r) : * + 4 does not have a real root. As shown in Figure t 8. I 1, the function does not intersect the r axis. This also should be obvious, because if you were to solveflr) : x2 + 4 : 0, you would frndxz : -4.Even though this function does not have a red root, it still possesses imaginary roots. You will learn about imaginaqr roots in your advanced math and engineering classes.
Nort, we consider other forms of nonlinear engineering models.
546
Cneprnn
18
MefirsMAflcs n* ENcrxssRrNc flx)
T4
I
\
I
\ f(*)=*++
\
13
-3 -2
I
8l
-l
\
/
/
4l
0
I 2
13)
3
U
18.4
FiguretS.ll
0
-4
*3
-2
_I 1
Afunctionwithinaginaryroots.
Exponential and Logarithmic Models In this section, we will
discuss exponential and
logarithmic models and their basic character-
istics. The Cooling of Steel Plates In an anneding process*a process wherein materials such as glass and metal are heated to high temperanrres and then cooled slowly to toughen them-thin steel
plates([=thermalconductivity:40$7./m'Kp:density:7800kg/-',andc:specific
:
40OJlkl. K) are heated to temperanres of 900'C and then cooled in an environment with temperature of 35'C and a heat transfer coefficient of h : 25 \M/mz'IC Each plate has a thickness of L : 5 cm. W'e are interested in determining what the temperaflue of the plate is heat
after one hour. Those of you who will pusue aerospace, chemical, mechanical, or materials engineering will learn about the underlying concepts that lead to the solution in your heat-uansfer class. For now, in order to determine the temperature of a plate after one hour, we use the following exponential equation.
'7_',r I I oviromor 7i"r,a
* 4nri-**,
=*r(#,)
(t8J0)
represents the temperature of the plate at time t Using Equation (18.10), we hane calculated the temperature of the plate after each 12 minute (0.2 hd interval. The corresponding temperature distribution is shown in Figure 18.12. From examining Figure l8.l2,wenote that the temperature of the plate after one hour is 308'C. Moreover, Figure 18.12 shows that, during the first hour, the temperature of the plate
In Equation (18.10), 7
18.4 Time (hr)
0 0.2 0.4 0.6 0.8 I 1.2 t.4 1.6 r.8 2 2.2 2.4 2,6 2.8 362 3.2 3.4 3.6 3.8 444 4.2 4.4 4.6 4.8 538
E:rpoNsNTrALexo Locemrnurc Moonr,s
547
Temperature
fc) 900 722 580
468 379
t
308 252 207
172 143 121 103
89 78 69
\ O
sao
H -
aso
Euo 19 soo
57
\
52 49
\
46 42 40
I
r20
\ \
aa
39 38 J
Time (hr)
t
figurelS.l2 lhecoolingofapieceof
metal.
drops from 900'C to 308'C (a temperanre drop of 592"C). During the second hour, the temperatrue ofthe plate drops from 308oC to l2l"C (a temperarure drop of 18/C), and during the third hour, a temperaflrre drop of 59"C occurs. As you ciln see, the cooling rate is much higher at the beginning and much lower at the end. You should also note that the temperatr[e begins to lwel offtoward the end of the cooling process (after about 4.6 hours). This is the most important characeristic of an exponential function. That is to say, the value of the dependent variable begins to lEvel offas the value of the independent variable gets larger and larger. Now, let us look at what is meant by an exponential function. The simplest form of an exponential function is given by f(*) = e", where a is an irrational number with an approximate value of e : 2.778281. To provide you with a visual aid for better understanding of e* and for comparison, we hane plotted functions 2', e*, arrd3' and shown the results in Figure 18.13. Also note that in Figure 18.13, we have shown how the value of a is determined. Spend a few minutes studying Figure 18.13 and note trends and characteristics of the functions.
548
Cruprnn
18
Marnruerrcs rx ENcnwERrNc
" (,.;)' 2.000000 2.250000
ilt,
2.48$2A 2.593742 2.653298 2.691588 2.704814
il0 20
,50 r , i r r I
100
211r5r7
200
2.715569 2.716924 2.717603 2.718010 2.718146
500 1000 2000 5000 10000
As n
x
-+
f(r)=z* 1,
0 0.2
i
l
1.15
a.4
1.32
0.6 0.8
r.52 r.74
I t4
2.00 2.30 2.64
l.o
3.AE
1.8
3,48 4.00 4.59
t.2
,,
',
I
'l
|
2.2 2.4 2.6 2.8
,
l
I
|
i
l
i
2.718281.8285. .
.
.
5.28 6.A6
f@)=e* f(4=er 1.00 1.22 1.49 1.82
2.23 2.72 3.32 4.06 4.95
1.00
1?< r.93 2.41
3.00
474 4.66 5.80
6.05
all
7.39 9.03
9.00
1r.02
Lt.2l 13.97 17.40 21.6V 27.00
3
8.00
13.46 16.44 20.09
3.2 3.4 3.6
9.19
24.53
33.53
r0.56
29.96
4r.90
12.13
36.60
3.8
13.91
44.70
4
16.00
54.60
52.20 65.02 81.00
6.96
l---r,l l-erl
I
......3'
L
60 lt ^50 \40
!'/
30
v
20 10
./
a-----'-
U
2
x
L
Figuru
f8.13
The comparison of functions
?r, d,and
3'.
The exponential functions have imporant characteristics, as demonstrated by examples shown in Thble 18.5. Example I shows the changes that occur in a o<ponential model when the growth rate of an exponential function increa.ses. Example 2 demonsuates similar changes for a decaying exponential function. Note these important effects as you study Thble 18.5. A good understanding of these concepts will be beneficial when you take future engineering classes.
18.4
-__ - . - t - -
-
...,-:. . - -
549
..
Form of the E4ronential Function
fk)=rt*
ExporvsrvflAr. AND LocARrrHMrc MoDELs
Characte.risticc
aoea'
(Example 1)
fo*
fk)
= rt
-
aaa-w frt fo>
ao
no
(Exarnple 2)
:la
fo-
ao
Another interesting form of an exponential firnction isflr) : {r.Youwill find this qpe of orponential funcdon in expressing probability distributions. Ve will discuss probability distributions in more detail in Chapter 19. For comparison, we hare ploned the functions f(x) : 2-" and"f(x) : d'" and shon'n them in Figure 18.14. Note the bell shape of these firnctions.
Logarithmie Functions In this section, we will
discuss logarithmic functions. In order to show the importance of logarithmic functions, we will rwisit the cooling of steel plates example, and ask a different
question.
(Revisited) In an annealing process, thin steel plares (A = thermal conductivity:40!P'lm.K-p= density:7800kglmt,anda:specificheat:400Jlkg.Ig are heated to temperaffies of 900oC and then cooled in an erivironment with temperature of 35oCandaheattransfercoefficientof h:25lff /m2. K Eachplatehasathicknessof L:5 crl.. The Cooling of Steel Plates
'We
are norr interested
in determining how long it would take for
a
plate to reach a temperature
550
CrLcl'rER
18
MrrnEMATrcs rN
ENGTNEERTNG
r-:-:
!= 2-n'
tc
-'2.6
0.009
-2.4 -2.2
0.018
0.003
0.035
0.008
0.063
0.106
0.018 0.039
-1.6
0.r70
0.077
-1 ItT
0.257
0.141
4369
0.237
0.500 0.642
0.368 0.527 0.698
-2 - 1.8 /,
' t.z *1 -0.8 -0.6 -u.4
--4.2 t0 0.2
,0.4 0.6 0.8
0.779 0.895 0.973 1.000 0.973 0.895 0.779
0.001
t.2
4369
I.rt
0.257
,1.6
4.r70
0.237 0.141 0.077
.I.8
0.106
.4,
0.063
0.009
FigurelS.f4 Theplotof fA)
t'I
{
I
I
ti
,
i1 t T I
!l
0.698 0.368
0.035 0.018
;-r
0.951 0.852
0.500
2.4 2.6
)'r'
0.961 1.000
0.527
nn
tr
1.0
0.852
o.M2
I
I
d"
fi
;t
I I
I I t tl ll ll
I
\
0.039 0.018
0.008
/
0.003 0.001
:
2-"
0.0
"
and
f{xl
:
\
-2.0 -1.3 -0.6 0.1 0.8
-2. 7
ta
1.5
x
e-r'.
of 50"C. To determine the time that it akes for following logarithmic equation.
a plate
to reach a temperaure of 50"C, we will
use the
r
:
pcL,
*
tn
Ti- Ty , - ,r: =
(zaoot
=
3.5 hr
But what is meant by a logarithmic function? The logarithmic functions are defined for the ease of computations. For example, ifwe let 10" : !, then we define logy : r. The sFmbol "log" reads logarithm to the base-l0 or common logarithm. For enample, you know any number raised to the power of znro has a value of 1 (that is 100 : 1); then using the definidon of the corntnon logarithm then log 1 : 0, for 101 : 10 then log 10 : 1, for 102 : 100 then log 100 - 2, and so on. On the other hand, if we let a' : 7, then we define ln jt : 4 and the qymbol ln tads the logarithm to the base-a (or natural logarithm). Moreover, the reladonship between the natural logarithm and the common logarithm is given by
18.5 MarnurArcsBRA pPa
ln x
140 dB
:
(ln
10) (log
also prove the 100,000,000
13O
Th'
551
r) :
2.3O2585 logx. Using the logarithmic definitions, \Me can followingidentides, whichyouwill find usefi.rlwhen simplifying en-
gineering relationships.
logxy: logr* 10,000,000
logy
lu; :
logx
- log!
logx":
Scale In engineering, the loudness of sound is rypically
nlogx in
11.0
Decibel
100
unit called decibel (dB). The threshold of hearing (that is the softest sound that a healthy human can hear) is 20 p,Pa or 20 X 10-6 Pa. Note this is caused by a
-
a
very very small air pressure change. Amazing! At the other end of hearing range
X 106 pPa of pressure change. To keep the numbers manageable in the range of hearing (from 20 ltPa to 100,000,000 pPa) the decibel scale is defined by dB :
1,000,000
lies the threshold of hearing pain, which is caused by approximately 100
90
20 log (1120);r,P4 where ,Irepresents the pressure change (in g,Pa) created by the sound source. For example, a sound created by a moving car (creating a pressure change of200,000 p,Pa) has a comesponding decibel rating of20lo gQ00000 p,Pal 20) p,Pa: 80 dB.The decibel rating of common sounds is shown in the accompanying figure. Finally, the following are some additional mathem*ical relationships that you may find h.lpfirl during your engineering education.
50
xft34m
4,0
expressed
-
=
,ntm
I x-' :'-;
1,000
x-
30
(*!)n =
,"!"
(r')- : *"t
r0
: 1
(x
+
0)
xo / *\' :v j:x"-\r)
xo
10
0
Thresholdofhearing
18.5 Matrix
Algebra
As you will leam later during your engineering education, formulation of many engineering problems, such as the vibration of machines, airplanes, and stnrcnrres; joint deflections of suucnrral qystems; cunent flow through branches of electrical circuits; and the fluid fow in pipe nenvorlis lead to a set of linear algebraic equadons that are solved simultaneously. A good understanding of matrix algebra is essential in the formulation and solution of these models. As is the case with any topic, matrix algebra has its own terminology and follows a set of rules. Ife will provide a brief overview of matrix terminology and matrix algebra in this section.
Basie Definitions During your engineering education, you will learn abour different types of physical variables. There are those that are identifiable by a single value or magnitude. For example, dme can be described by a single value such as two hours. These types ofphysical variables which are
552
Cner"rsn
18
MATHEMATTcs rN ENcrNsERrNc
identifiable by a single value arc caTled scalan Temperature is another example of a scalar variable. On the other hand, if you were to describe the velocity of a vehicle, you not only have to specifr how fast it is moving (speed), but also ia direction. The physical variables that possess both magnitude and direction are czlledaec'tors.There are also other quantities that require specfing more than two pieces of information to describe them accurately. For example, if you were to describe the location of a car parked in a multi-story garage (with respect to the gar€e enuance), you would need to specifr the floor (s coordinate), and then the location of the car on that floor (r andy coordinates). A matrix is often used to describe situations that require many values. A rnafuix is an array of numbers, variables, or mar-hematical terms. The numbers or the variables that make up the matrix are called rhe elements ofrnanix.The size of a maffix is defined by its number of rows and columns. A mauix may consist of m rcws and n columns. For enample,
,',:
[_f i
';] -:{i}
matrix [.lf] 's a3by 3 (or 3 X 3) mauix whose elements are numbers, and {Z} is a 3 by 1 mauix with its elements representing variables x, !, ar'd e, The [N] is called a square matrix. A squtre mau:txhas the same number of rows and columns. The element of a matrix is denoted by its location. For example, the element in the first row and the third column of a mauix [N] is denoted by n6 (it reads n sub 1 3) , which has a value of 9 . In this book, we denore the mauix by a boldface letter in brackea like [] and {}, for example: [N], [T], {F}, and r:he elements of mauices are represented by regular lowercase letters. The {} is used to distinguish a column matrix. A column matrix is defined as a matrix that has one column but could have many rov/s. On the other hand, a rols matrix is a mauix that has one row but could have many columns. F-xamples of column and row matrices follow.
-:{i} are oramples
fcl
and {x}
tt)
of column mauices, whereas
: [5 0 2 *3]
are examples
:
and lY7:
lt, lz tl
of row matrices.
ftlahices A diagonal matrix is one that only has elements along its principal diagonal; the elements ate zoro everywhere else. An example ofa4X4diagonalmatrix Diagonaland Unit
follows.
,[ii:,i]
18.5 M.*rnocAleesRA
553
The diagonal along which values 5, 7, 4, and lL lies is called rhe principal diagonal An idmtity or unit matix s a diagonal matrix whose elements consist of a value of 1. An example of an identity matrix follows.
l/l :
100 010 001
.00 .00 .00
;;; 000
10 0l
Matrix Addition or Subtraction Two mauices can be added together or subffacted from each other provided that they of the same
are
size-ech matrix must have
the same number of rows and columns. We can add matix lAf .yo of dimension rn by n (having rn rows and a columns) to matrix lBl-r, of the same dimension by adding the like elements. Matrix subtracdon follows a similar nrle, as shown.
lto
[,4]
:l::: 2
t z)
:l: [{ro
|
(5rr)
Lo
fvlu
x. q)
trl
lI
r [r] 151 Lg
lvlatrix
sl
3
,rl
(3 t 12) (1 a7)
(2! (0t :Ll :
(2
+
55)
:
:,:
ltipl ieation
In this section we will
discuss the rules for multiplying a mauix by a scalar quanrity and by
another mauix.
0uanti$ '$Vhen a mauix p{l is muldplied by a scalar quantiry with a magnitude such as 5, the operation results in a matrix of the same size, whose elements are the product of elements in the original mauix and the scalar quantity. For example, when we multiply mauix [.{ of size 3 X 3by ascalar quantity 5, this operation results in another mauix of Multiplying a lr|atrix by a Scalar
CHapmn
18
Marrrra,rerrcs rN ENGTNEERTNc
3 X 3 whose elemen$ are computed by multiplying each element of matrix [A] by 5 shown here. size
5lAl:
as
t.l ,o o 5l | -24 o e 2 l: || -ro 45 ro LiTrol Lzs35 5oJ
5l
I
t|ahh
'\Thereas
any size matrix can be multiplied by a scalar quantity, mauix muldplication can be performed only when the number of columns in the premuhiplier matrixis equal to the number ofrows in the2osnnabipliermatrix. For example, matrix [r4] ofsize m X. n canbe premultiplied by matrix [.8] of size n X p because the number of columns a in matrix [.r4] is equal to number of rows z in matrix [B]. Moreover, rhe multiplication resula in another matrix, say [C] of size m X p. Mauix muldplication is carried out according to the following nrle. Consider the multiplication ofthe following 3by 3lA) and the 3 by2lBl matrices. Multiplying a }|ahir by Another
t,4ltal :l
I z 4 11 lt nf L 6 5l ltt el :lc)y, L-z a slr"rlte lrla*z
Note the number of columns in matrix L4] is equal to number of rows in matrix [.8], and the multiplication will result in a matrix of size 3 by 2.The elements in the fust column of the resulting [C] matrix are computed from
7 12 16<
: tz
r,til
n1
''' L-l : :]Ll ,?l (z)(z) + (rz)(4) + (r6xr) | rc ',,1 : || (7Xr) + (r2X6) + (re)(5) ',,1 ',,1: I rst ,,1 L(7)(-2) + (12X3) + (16Xs) ctz) Ll50
ctz-l
and the elements in the second column of the [C] matrix are 23
g tt+--l
4 tfl 7 l'l I zt ',,1 sllrz tl:lrl czzl
lz e [c] :l 1
L-2
3
lrc
+ (ex4) + (23)(1)+(ex6)+(11)(5) (23)(-2) + (ex3) + (11x8).1
:115e
L 150
8-l L
16
(23)(2)
tl
L
r:o
,zz)
(lrxl)l I zr %1 l: lr5e B2l Lr:o
6e)
If you are dealing with larger matrices, the elements in the other columns are computed similar manner. Also, when multiplying matrices, keep in mind that matrix muhiplication is not commutative except for very special cases. That is in
a
lAllul + lBllAl
1S.5 MernmAressRA 555 This may be a good place to point out that if [.I] is an identity mauix and uix of matching size, then it can be readily shovm that the product of
!{l
is a square ma-
ll)lAl: lAllll: lAl
(ls.l3)
@Givenmarrices'|,]:i:z]|,V]:|:i_1rand{c}:{-;},perform il
th" following operations. (a)
(b)
[,{ + [,tr]: t4 - [B] :?
?
:
(c) 3[.4]
?
:
(d) [^4ur] ? (e) L{{C}: I
(f) Showthat tlltr{l 'We
will
: [/]lI):
use the operation
[A)
des
discussed
in the preceding
sections
to
answer [hese
questions.
(a) [.,4]
+ [8] :
?
lAl + lBl
(b) t
{l - [B]
[o z5ol l+6-21 :ls , 3l zl+lz
-2 e) [r 3 -4) [(o + 4) (i + 6) (o + (-2))] | + 1r -21 :l(8+7) (3+2) (7+3) l=lr i rol 5J L(g + r) (-2 + 3) (e + (-4))_l Lro I Lp
:7
rol 14u-r1 [o 371-17r 3l lA) - lBl: l8
L9 -2 eJ Ll 3 -4) [(o
- 4)
(5
- 6)
(o
- (-2))l | -4 -1
(3-2) (7-3) l=l 1 I l(8-7) L(e - 1) 1-z - z) (e - (-4))l L a -5 (c)
3lAl
:
21
4l B)
t
3lAl
o (3X5) ol Io ti ol 5 ol fo s z zl: Ilta)ta) (3X3) (3X7) | :lz+ e 2rl (3)(-2) (3Xe)l eJ
:31
Lg
-2
L(3Xe)
lzz -6
27 J
556
CneprsR
18
Mnrnnmerrcs rN ENGTNEERTNG
(d)tAltBl:
?
[o 5ol[46-zfjl felfrl: ls t tllt z Lg -2 sllr 3 -4) (oxa)
+
(5x7) + (0x1) (0)(6)
+
(5x2) + (0x3) (0x-2)
+
(5x3) + (0)(-4)l
f :l(a)ta) + Q)Q)+(7)(1) (sx6)+ (3)(2)+(7x3) (8x-2)+ (3x3)+(7)(-4)l L(px<) + (-2)(7) + (exr) (ex6) + (-2)(2) + (r)(:) (ex-2) + (-2x3) + (e)(-4)l
lts 10 15.l :l6o 75 451 tl 131 77 -60l
(e)
L4{c}: i
s o'l[-rl [(o)t-r) + (5X2) + (0Xi)) [to] 3 7ll 2l:\ (8x-1) + (3)(2) +(7x5)}:1331 -2 e)L 5) ((ex-r) + (-2)(2) + (e)(5) ) lzz) (f) Show that [f]L4l = Vt]l[ : lAl
[o f/l{c}: l8 Lg
5 o'l [o 5 ol [r o ollo :loroll 8 3rl:18 3?l lrltel Lo o lll9 -2 el Lg -2 9l : olfr o ol [o 5 ol [o [.4][/]:18 3 Tllo I ol:18 3 7l Ls -2 gllo o ll Lr -2 el
and
Transpose of a klatrix fu you will
see in the dasses which you will ake later, the formuladon aad solution of engineering problems lend themselves to situations wherein it is desirable to reanange the rows of a matrix into the columns of another matrix. In general, to obtain the uanspose of a matrix [B] of size rnx n, the first row of the given matrix becomes the first column of fBfT, the second row of [,8] becomes the second column of lBfr, and so on, leading to the mth row of [B] becoming rhe rnth column of t.All and r resulting in a matrix with the size of n X m, The matrix [,B] reads as a uanspose of the [.8]
matrix. Sometimes, in order to save space, we write the soludon matrices, which are column maas row matrices using the transpose of the solution-2n61hg1 use for the ranspose of a maffix. For example, we represent the solution given by the [/matrix.
tices,
18.5 MernmArcrsRA
557
174e6L2) This is a good place to define a symmetric matrix. A rymrnetric rnatix is a square maffix whose elements are s)'mmetrical with respect to its principal diagonal. An example of a qfmmetric matrix follou/s.
[r
4
2
5 t5 ;Z] f,{l:l 4 -3 | 215 L-5 20 8 ;j lt
Giventhefollowingmauices: following operations: (a)
5ol 146-11 7 | nd lB) : | 7 2 3 l,performthe el Lt 3 -4) :
[o [/] : | 8 3 Lg -z
@r =
7
and (b)
lBlr
?
(a) As explained earlier, the fust, second, third, . . . , and zth rows ofa mauix become the first, second, thir4 . . . , and rnth column of rhe uanspose marrix, respecrively.
Vf
el =15 -21 107 el
[08
3
(b) Similarly,
l4z lalr=l 6 2
l-, 3
11
3l
-4)
l-rrrl]:tl-liaril.L-*l:a.L}.?
Determinant of a Matrix Up to this point, we have defined essential mauix terminolory and discussed basic matrix operations. In this section, we will define what is meant by a determinant of a matrix. Let us consider the solution to the following set of simultaneous equations:
/l1f\* Aptcz= 4 aa\ t tt22x2: b2
(lEJ4a) fl8.r4b)
558
Cneprsn
18
Mernruerrcs
rN ENcrxssRrNc
Expressing Equations (18.14a) and (1 8. 14b) in a matrix form, we have
-g_
: {r1\ ln" azz) lxz) Latt ^,1{^\ lbt) To solve for the unl<nowns 11 and x22 we tnaf first solve for x2 in terms of x1, using Equation (18.14 b), and then substitute that relationship into Equation (18.14a). These steps are shown next.
b2-a21x1 *r: -- nrr- =+ Solving for
^l
'
--
(h-arrx-t\ : t nrr\ a )
,
Ot
rc1:
b1a22
-
apb2
anhz
-
dv4r
After we substitute for
X2:
411x1
ank an4z2
q
(l8l5a)
in either Equation (18.14a) or (18.14b), we get
- h4t -
(18.X5b)
aI2e4r
Refering to the solutions given by Equations (18.15a) and (18.15b), we see that the denominators in these equations represent the product of coefficients in the main diagonal minus the product of the coefficient in the other diagonal of che [.4] matrix. The a11a22 - apa21 is the d*errninant of the 2 X 2lA] matru and is represented in one of following ways:
Det f./] or det [.4]
orlo" lar
n"l hzl
=
d,&zz
- dtzhr
fls.16)
Only the determinant of a square mauix is defined. Moreover, keep in mind that the determinant of the L{] matrix is a single number. That is, after we substitute for the values of ay, a22, de, alad a2linto asa72- 412421t we get a single number. Let us now consider the determinant of a3 bv 3 matrix such as
[*tcrz truf czz czz| lCl:Itt Lczr ca2 %) which is computed in the following manner:
a, crt t"'"1 : lty czz czZl cr^l
I
b^
c3z
tl
\1c22ca3
* cpc4cal *
q3c21ca2
-
cgC22Q1
-
C1C23C32
-
Cpc21Q3
(r8.r7)
There is a simple procedure called direa expansion that you can use to obtain the results given by Equation ( I 8 . 1 7) . Direct expansion proceeds in the following manner. First we repeat and place the first and the second columns of the matrix I C] next to the third column, as shown
18.5
\\\,/.
i"u'i
C1. Ctt Crt Cro * "\'-.>< -X.',"
Ctr
Ctt Cr Ctt '..' C..1X.;,\ "\ -
.azi ,r\
Ctr
\
,r'
Ctt -Czt Cta Ct '-y' "/ "\ -'r
(a)
f
Czt
"'r
(b)
tigurelS.lS
Direcherpansion procedure for computing the detenninant of (a) 2 2 mahir, and (b) 3 3 lnatrh.
x
x
f,-1.x1xN
,..6,..2'..)
J
elements lying on the dashed arrows. This procedure shown in Figure 1 8.15 results in the determinant value given by Equation (13.17).
The direct-expansion procedure c:mnot be used to obtain higherorder determinants. Instead, we resoft to a method that first reduces the order of the determinant-to what is called a minor-and then evaluates the lower-order determinants. You will learn about minors later in your other classes.
trf
: j f
Z
Le
ll,catculate the d.eterminant of
-z e)
!41.
As explained earlier, using the direct-expansion method, we repeat and place the first and the second column ofthe matrix nerft ro the third column as shown, and compute the
products of the elements along the solid arrows, then subtract them from the products of elements along the dashed arrows, as shown in Figure 18.16. Use of this method resula in the following solution.
'Xu\'-..
Figurcl8.l6
Ir 35 ol7l : (1x3xe) + (5)(7)(6) + (oX8X-2) ls rc -2 el _(5xsxe) - (1)(7)(-2) - (ox3x6) : -roe
Ihe dhed erpansion method for Erample 18.3.
't------
559
in Figure 18.15. Then we add the products of the diagonal elemena lying on the solid arrorvs and subtract them from the products of the diagonal
Given dre followingmatrix,
\1-\-><-\('.' -i-\o'' I" s"
M,c,rRD(ALcEBRA
....
---
....-.--
--i:--
\Vhen the determinant of a mauix is zero, the matrix is called a singular, A singular matrix results when the elements in two or more rr"iit" n*n mauix are idendcal. For example, consider the following matrix:
lA)
As shown next, the determinant of
:
[.{
12
|
Ll 3
4
l,
whose rows one and two are identical.
5l
is zero.
lzt4l lz t 4l: (2X1X5)+(rX4X1) +(4)(z)(3) 11
3 5t - (1X2Xj) - (2)(4)(3)- (4XrXl) :
0
Matrix singularity can also orcur vrhen the elements in two or more rows of a matrix are linearly dependent. For example, if we multiply the elements of the second row of matrix [r4] by
lz |
41
[/] : | ,n 7 28 lis singular because rows one and two are now tinearly dependent. As shown l.:, ;. a;5,J*i"*, of the new
a scalar factor such as 7, then rhe resulting matrix
!{]
matrix is zero.
4l lz 7t 281:(2)(7)(5) +(1x28xr) +(4)(14)(3)
lr4 | 1 3 5t -
(rX14)(5)
-
(2)(28)(3)- (4X7X1)
:
0
560
Cnar.rsR
18
Manrnuerrcs
tN ENcrNsBRrr\c
Solutions of Simultaneous Linear Equations fu
we discussed earlier, the formulation of many engineering problems leads to a qystem of algebraic equations. As you will learn later in your math and engineering classes, there are a number ways that we c:rn use to solve a set of linear equations. In the section that follows, we will discuss one of these metho& that you can use to obtain solutions to a set oflinear eguations.
Ve will begin our discussion by demonstrating the Gauss elimination method using an example. Consider the following three linear equations with three unknowns:
Gaus Elimination lir|ethod
xy x2, afrd
x3,
*
* xu:
13
(l8.l8a)
* 4xu: 32 54 - x2 I 3xu: 17
(t8l8b)
2x1 3x1
Stepl: ,dr
*
x2
2x2
(18.18s)
W'e begin by dividing the first equation, Equation (8.18a), by 2: the coefficient of the term. This operation leads to
11t3 x1:xn*)x^:' 2' 2"
(1019)
2
Step2:'We multiply Equation (13.19) by 3: the coefficient
?,?.39 3x,+|n*ir=t
ofrl in Equation (18.18b). fl8.20)
'We
q
then subuact Equation (18.20) from Equation (18.18b). This step will eliminate from Equation (18.18b). This operation leads to
3x1* 2x2+ 44:32
/
-[r",
z ? 39\ +;xz+i":i)
| ,5
fl8.A)
25
i*r'i*t: T Stqr 3: Similarly, to eliminate r, from Equation (18.18c), we multiply Equation (18.19) by 5: the coefficient ofrcl in Equation (18.18c)
- +,5 *,5 in t,
5xr
.
=
65 =i
08.22)
'We
then subtract the above equation from Equation (18.18c), which will eliminate x1 from Equation (18.18c). This operation leads to
5\-&.*3xz:17
(. ,5 ,5 _65\ -\rrt +tx2+t*t:T) 7 -f,r : 37 --Xt
2'
2-
-X^
--
2
(18.23)
18.5 MnrnrxAr,crsRA
551
Let us summarize the results of the operations performed during steps 1 tluough 3. These operations eliminatedq from Equations (18.18b) and (1S.1Sc).
.1 I 13 \+ixz+ Z*u:T t5,25
t*r. r.t:
t
7 T.r -X^:
--
2'
(10.24b)
3l
2'
--Xt
08.24a)
08.24c)
2
Step 4: To eliminate 12 from Equation (18.24c), first we divide Equation (1S.24b) by rl: the coefficient ofx2.
x,
*
54 =
25
(|B.ZS)
ThenwemultiplyEquation (18.25) by j:rhecoefficientofx2inEquation (18.24c),and subtract that equation from Equation (I8.24c). These operations lead to
7 T,r -Xa:
3t
2' 2" ( 7 35 --Xt
--
2
-\-to-T*':- r75\ ,) r8r,:
Dividing both
sides
(x8.26)
?,
of Equation (18.26) by 18, we get
4:4 Summarizing the results of the previous steps, we have
,1
Xr '-l -Xt
' 2' xz
-T
I
13
2"
-2
-Xt:
fl8.2n
* 5xt: 25 xt: 4
08.28) (1s.29)
Step 5: Now we can use back substitution to compute the values of x2and.4. W'e substitute for r, in Equation (18.28) and solve for x2. x2
*
5(4) =
25 -+
Next we substitute for
, x, +
5
4 and x2 in Equation (18.2-7) and solve for .q.
1,-. +;\g , 1 ,,, :
,\5)
tcz:
13
2 -+ 4:2
Inverse of a Matrix In the prwious sections, we discussed matrix addition, subtraction and multiplicarion, butyou may hane noticed that we did not say anything about matrix division. That is be
56?
Cueprsn
18
MerrrsMATrcs tN ENcrxsERrNc
lA)-'lAl = l/l[/]-1
: [/]
(18.30)
In Equation (18.30), [r4]-t ir called the inverse of [,4]. Only has an inverse.
In the previous section, we explained the
matrix elimination method that you
a square and nonsingular
Gauss
can use to obtain solutions to a set of linear equations. Matrix inversion allows for yet another way of solving for the solutions of a set of linear equations. As you will learn later in your math and engineering classes, there are a number ofways to compute the inverse of a matrix.
18.6
Calculus Calculus commonly is divided into mro broad areas: differendal and integral calculus. In the following sections, we will explain some key concepts related to differential and integral calculus.
Differential Calculus A good understanding of differential calculus is necessary to determine the rate of change
n
engineering problems. The rate of change refers to how a dependent variable changes with respect to an independent variable. Lett imagine tfrat on a nice day, you decided to go for a ride. You get into your car and turn the engine on, and you staft on your way for a nice drive. Once you are cruising at a constant speed and enjoying the scenery your engineering curiosity kicks in and you ask yourself, how has the speed ofmy car been changing? In other words, you are interested in knowing the time ran of change of speed, or the tangendal acceleration of the car. As defined above, r.he rate of change shows how one variable changes with respect to another variable. In this example, speed is the dependent aariable and,trme is the indepmdent oariable. The speed is called the depmdent uariabb, because the speed of the car is a function of dme. On the other hand, the time variable is not dependent on the speed, and hence, it is cBlled an independent variable. Ifyou could define a function that closely described the speed in terms of time, then you would difomtiarr the function to obtain the acceleration. Related to the example above, there are many other questions that you could hane asked: W'hat is the time rate of fuel consumption (gallons per hour)? \7'hat is the distance rarc of fuel consumption (miles per gallon)? What is the dme rate of change of your position with respect to a known location (i.e., speed
ofthe
car)?
Engineers calculate the rate of change of variables to design products and services. The engineers who designed your car had to have a good grasp ofthe concept ofrate ofchange in order ro build a car with a predictable behavior. For example, manufacturers of cars make cenain information available, such as miles per gallon for city or highway driving conditions. Additional familiar examples dealingwith rates of change of variables include:
How does the temperature of the oven change with time, after it is tumed on? How does the temperature of a soft drink change over dme after it is placed in a refrigerator? Again, the engineers who designed the oven and refrigerator understood the rate ofchange concepr to design a product that functions according to established specificadons. Traffic flow and product movement on assembly lines are other examples where a detailed knowledge of the rate ofchange ofvariables are sought.
18.6 Celcur.us
I *ff*ia;l-**#;uiofn'itiilf$ln;1it*ffiiifll-i-l,, Definitions and Rules
7 2 3 4 5 6
1"'|-'"ilir'ilrr
563
,llilfli;rrl',,,
Explanation
df Iimit"# .. . f(* + h) - f(x) f'(*) : *: hc h'-+o h lf f(r) : constant then /'(r) : 0 If f(x) : x" then f'(x) : Txxn-l If f(r) : a' SQ) where ais a constant then/'(r) : a' g'(x) If f(x) : g(x) t- h(x) then f'(x) = g'(x) t H(x) lf f(x) : g(x) ' h(x) then f'(x) =
The definition of the derivative of the
functionflr). The derivative of a constant function is zero.
The Power Rule
(see
Example 18.5).
The rule for when a constant such
as
a is
multipliedbyafunction (seeExample 18.6). The nrle for when two funcions are added or subtracted
(see
The Product Rule
Example 18.7). (see
Example 18.8).
g'(x)' h(x) + g(x)' h'(x)
7
If
f(x) :
#n
The Quotient Rule
^
g'(x) .,, \ r\x):w h(x).
: f[SG)]: f(") where u: g(x) df{') : af!4 then/'(r) dtc du 'a! dx
9
If
10
f(x)
f(x) : lglx))' : u'where u: g(x) then/'(r) : o- u'-' .ft If
f(x) : ln lg(r)l
rhen
Example 18.9)
SQ). h'(r)
8
If
(see
The Chain Rule.
The Power Rule for
^
g(r)
(see
a general
function such
Example 18'10)'
The rule for natural logarithm functions
r'\*) /'(x) :
(see
F-xample 18'11)'
s@)
11
If f(x) :
exp(g(r)) or f(x)
thenf'(x) = g'(x) '
ed')
: ,s(x)
The rule for exponential functions (see Example 18.12).
During the nefi two years, as you take your calculus classes, you will learn many new concepts and nrles dealing with differential calculus. Make sure you uke the time to understand these concepts and nrles. In your calculus classes, you may not apply the concepts to acnral engineering problems, but be assured that you will use them eventually in your engineering classes. Some of these concepts and nrles are summarized in Thble 18.6. Examples that demonsffate how to apply these differentiation nrles follow. As you study these er
564
Crrepren
18
MerHsMATrcs m ENcrNnnrNc
Identify the dependent and independent variables for the following situations: water consumption and traffic fow. For the rrater consumption siruation, the mass or the volume is the dependent variable, with time the independent variable. For the uaffic Ilow problem, the number of cars is the dependent variable and time is the independent variable.
.iilt:r,liii!.IL:ll:,1,..i-:t
-- -- -llr. ---
f
Find the derivative
'We use nrles "f 3 and
(*)
:
ll,iti.lilii:-irrjr.l::l,r'
N3
-
5,f'(*) =
+
1012
i:-ririlii!.Tlir,ili
8.
nxo-|, from Thble 18.6 to solve this problem
as shovrn.
:3* - zox
f'(*)
.!.-iii'll-
-,lt{ -..rt'ti.tiail-i-lxr',ali
f;'a
-- .:,i
i:i:lrai!.lr,.i.iiirrif''i:--.:rrr.,:Li-
--
'i:].:irii
: 5(*3 - 10.12 + 8). : f'(x) a' g'(x), ftom Table 18.6 to solve this problem. For the given problem, rf a : 5 ar.d g(tc) : x1 - l0x2 + 8, then the derivative of f(x) is f'(x) = 5(3x2 - 20x) : l5x2 - I00x. Find the derivative of f(x)
\Fe use nle 4,
= 1xi - LOx2 t 8) -r (x5 + 5x). : Ve use rule 5,f'(r) g'k) ! h'(x), from Thble 18.6 to solve this problem as shown. f(*) : Q* - 20x) -r (sx4 + s)
Find rhe derivative of f(x)
,-,.:.,.:-
-,..r.:rl:,,t
,1.-..,1.:-.L.:'lliila.,!.ll'iltirll'aL...-..{:11:llt,ll
-------'liill
Find the derivative of f(x) : (tc' - lox2 + 8)(tc5 - 5x). \?b use nie 6,f'(x) : g'(x). h(x) + g(x) . h'(x), from Thble 18.6 to solve this problem shown. For this problem, h(") : (*5 - 5r) andg(r) : (x3 - 10*2 + 8).
f'(r) : (3r' -
:
8x7
20x)(x5
-
5x)
- 7ox6 + Aoxa -
Findthederivative of f(x): (x3 'W'e use
problem
as
nie
7,
f'(*) :
-
- loxz + 8)(5tc4 zo; + $o* - 40 +
(x3
1012
[h(x)'S'@)
-
shown. For this problem, h(x)
f'(r) :
1x5
- sx)(l* -
20x)
+ e)l(r5 =
-
5)
5r).
Sk)' h'(x))l[h(x)]2 from Thble 18.6 to solve this (xs
- (f -
(x5
as
5x)z
-
5x) and g(x)
rox2
*
8)(5'c4
- 1ot
-
5)
1012
+
8).
18.6
- rc* + q4. Thble 18.6 to solvethis problem 9, f '(r) : n. tt'-' ,(,fro dJc
Find dre derivative of f(x) 'Weuserule this problem, u
f'(r) :
Cer.culus
:
4(*u
=
(x3
-
-
loxz
(x3
t0r2 +
+
55s
asshown. For
a).
8)3 (3tc2
-
2ox)
Find the derivative offlr) = Iol*t - 1012 + al. \7e use rule 10,/(*) : g'(x)lg(x), from Thble 18.6 to solve this problem thisproblem, g(x) = 13 - 1012 + 8.
as
shown. For
(3x2-zox)
.,t\ I\x):7-r0S*t ,ii -.--
:,-:-
,-l
Find the derivative of f(x) : ror-lo''?+8). .W'e use rule ll,f'(x) : d@) . tdv), from Tirble 18.6 to solve this problem this problem, g(x)
:
f,(*) : (3*, -
x3
-
10x'
+
2ox) 8' -to*'
i*,,,,u,,o,
as
-,I:iIl
shown. For
8. +a1
.--
----
--.:
. _-_-_
Integral ealculus Integral calculus plays a vial role in the formulation and solution of engineering problems. To demonstrate the role of integrals, consider rhe following examples.
Recall in Chapter 7, we discussed a properuy of an area known as the second moment of arq. The second moment of arca,, also known as the area moment of inenia, is an imporant propemy of an area that provides information on how hard it is to bend something and therefore
il
Figuret8.l?
Small area element located
at a distance
y-y
axis.
r
from the
566
Cnaprrn
18
M.lrnruerrcs
rN ENcrNsERrnrc
playr an imporant role in design of structrues. V'e explained that, for a small area elementr4 located at a distance r from the axk y-y, as shown in Figure 18.17, the area moment of inenia is defined by
I-r= iA
fl8.31)
'We
also included more small area element, as shown in Figure 18.18. The area moment of inenia for the system of discrete areas shown about they-1r axis is now.
v
I
l---"-fl,, 3
l--"---fl,, Figurel8.t8
--ffi,,
Second moment ol area for three
small alea elements.
v
I.r: *A + ,7,4, + *!4
fl8.32)
Similarly, we qln obmin the second moment of area for a cross-sectional area, such as a recrangle or a cirde, by summing the area moment of inertia of all the litde area elements that makes up the cross-section. However, for a continuous cross-sectional atea, we use integrals instead of summingthe x2Aterms to evaluate the area moment of inenia. After all, the integral sign, J, is nothing but a big "S" sign, indicating summation.
, | xz",.dA Irr: I
fl8.33)
'W'e
can obtain the area moment of inenia of any geometric shape by performing the integration given by Equation (18.33). For example, let us derive a formula for a rectangular cross-section about the77 axes. step
I
step
2
steP
3
sep 4
;;-;;!!rn,: J| *' at= J| *'hd*: h J| *2 dx: *r*u 12 'ul2
-wl2
-ul2
Step 1: The second moment of the rectangular cross-sectional area is equal to the sum (integral) of liale rectangles. Step 2: \7e substitute for dA = hdx (See Figure 18.19) 'We simplify by taking out & (constant) outside the integral. Step 3: solution.'We will discuss the integration des later. 4: The Step
18.6
Cnr.cur.us
561
--'-'dO
fl
FiguretS.lg
0iffenntial element used to calculate the second moment of area. lrlTft{i'Ei'tlffiiitl1i!11[+iEllJnTiii."t
civil engineer, you may be assigned the task of determining the force enened by water that is stored behind a dam.'We discussed the concept of hydrosatic pressue in Chapter 10 and sated that, for fluid at test, the pressure increases with the depth of fluid as shown in FigAs a
ure 18.20 and according to
P= pg!
fi8.34)
where
P
:
: g:
p
7
I
:
fluid pressure at a point located a distancey below the water surface (Pa or lb I f?) density of the fluid
(kgl-'
acceleration due to gravity
g:32.2ftK)
or slugrlft3)
(g:
9.gt
mls2 or
disrance of the point below the fluid surface (m or ft)
Since the force due to the water pressure varies with depth, we need to add the pressure exerted on arqu at various depths to obain the net force. Consider fie force acdng at figure10.20 T[evariaiionof pressurewitldepth. depthT over a small aren" dA, as shown in Figure 18.21. The procedure for computing the total force is dernonstrated using the following steps. Note these steps make use ofintegrals. $ep
I
step
2
step
3
step
4
step
5
step 6
HHHHH
.. t oaaa= .. og | ydA: pg, | t4 = | --" - : f .- f Ne: rps,oH" JOr= J J J )
NetForce
00000
Step 1 : The net force is equal to the sum (integral) ofall the litde forces acting at different deptbs. Step 2: Ife substitute for dF: ?&4 (recall, force is equal to pressure times area).
558
CHer"rsR
18
Marrruruerrcs w ExcuvssRrNc
X
t8.2t
Figure
The forces due to pressute acting on a veftical surface.
Step 3:'We make use of the relationship berween the
fuid
pressure and depth
ofthe fluid, that
.liJ
'We
could come up with many more enamples to emphasize the role of integrals in engineering applications. During the next two years, as you take your calculus classes, you will learn many new concepts and rules dealingwith integral calculus. Make sure you take the time to undErsand these concepb and nrles. Some of these integral concepts and nrles are summarized in Thble 18.7. F-lramples rhat demonstrate how to apply some of these nrles follow. As you study the examples, keep in mind again that our intent is to familiarize you with these nrles, not to provide a detailed coverage.
Evaluate 'W'e
t. l(Z* -
20x)dx.
use nrles 2 and 6
from Thble 18.7 to solve this problem,
as shown.
I
irttff .JJJJJ
I
Qi -
20x)dx: l3xzdx + l -zoxdx = 3l
x2dx
-
20l xdx
u
I r I I ^l : zl:i-*ul zol:1:*'lI + c L1+1 l2*r I :f-to*+c I
,
til ,),
ltwirillaqa,ta;ni.tilltiLtilt"llt
:.1:iiiiii.'ii-iLtliiiil-il ---
,iia--.r,-
--i-----
1s.6 Celcur.us
i
D"finitions and Rulec
Fxplanation
ax* C
The integral of a consrant a,
["*: | *" a": -J_,*,+t I 6 n* I
t,
Tinre
for
1
i
;3 I:*= /l I
i5 :
i6 I
atnlxl +
t8 :
t9 '
True for
u'(x)dx
lr^r''61a*=
'!?hen
I#-:
+c - l"(4li-' nt r s44
4-
r*
(see
Example 18.15).
0 (see Example 1S.18).
See
a
=
constant
(see E:
l8,l@,
Example 18.17.
The substitution rnethod.
s
The subsdturion method
c
hlu(x)l +
n* -l
The nrle for exponential function.
[vo t c(i]e: [row' [aa* [b(4]' J
j
c
Io*:L**c *: ^lro* ["'on
I
569
(see
Ixample 18.20).
The substirution method.
? t
Evaluate
l5(3*
J
-
We use nie 5, given problem, d
I
5(3*'
-
:
20x)dx.
Ia.f(x)dx = a[f(x)dx,
2ox)dx: 5(*'
I
Evaluare 'We
)l{l*t use nne6,
from Thble 18.7 to solve this problem. For the 20x, thenusing the resuhs of Example 18.15, we get 3x2 loxz + c1.
5 nd,f(x)
zox)
-
*
:
(5xa
[[f(x) !
-
- ,lar.
g(x)]d.x
= !f(r)dxx. JS@)dn from
Thble 18.7 to solve this
problem, as shown.
fff
llli JJJ tl
2ox)
+ (5*n - 5)ld'c=
l(t*, - zlx)dx+- l(5x4 -
5)d*
=(f-ro*+C)t(x5-5x+e)
iirakaar:ltilttltmammii,q&?i'i lalrill?t&iii-- --ili.iIii![]j,nrii,.ii,ili,ii:l;iiir!.- ---ril..-irir,:!,r.Lirri'lIr-i:I
l:,
570
Cnar"rrn
18
MernsMATIcs lN ENcrNsrRrNc
Evaruate
I**
\V'e use rule 3,
I alx dx
I** : l0lnlrl + i
: aln lxl *
C from Table 78.7 ro solve this problem, as shown.
C
_-i:::il
I
lf("- tX*t t'.
Evduate 'We
-
zx)ldx.
substitution method lnrle 7) from Thble 18.7 tosolve this problem, as shown. : tc2 - 2x and duldx: 2x - 2 -- 2(x - 1), and rearrange the terms as l)dx or d.ul2 : (x - l)da. Making these substitutions, we get
use the
For this problem, tt
du
:
f I
J
t_
r, .,-a--,,la.a---,--.., 4,t
,
-.-.-
-----
2(x
tk -=
-
- r)(*' :
-
.--
rf
zx))dx=
I"+
=f,\"^:i(t. 4=:|ry.4
----.-
-:.,1-!!i:-
I
(' rllt-'z0 lax. I| V" *"A. rub.dtutiori method (nrle 8), ! s"@ u' (x)dx = su(x) I C from Thble 18.7 to solve this problem, as shown. From the prwious example, ,!' = x2 - ?* and' dul2 : (x - l)dx' Eva\tar'e
,,
Making these substitutions, we get
,I
I [o
ag - *e-a)a* : *l un" : lrnt *C:!1$-ul'1 2'
I
------:-:--:..--.-l--'tl:-.1--:-..
-t--
---
:r::--,-,t-r-:i--,ll ---i
18.7 DifferentialEquations Many engineering problems are modeled using differential equations with a set of corresponding boundary and/or initial conditions. As the name implies, differential equations contain derivatives of functions or differential terms. Moreover, the differential equations are derived by applying the fundamental lars and principles of nature (some ofwhich we described earlier) ro a very small volume or a mass. These differential equations represent the balance of mass, force, enerry, and so on. Boundary conditions provide information about what is happening physically at the boundaries of a problem. Initial conditions tell us about the initial conditions of a system (at time t: 0), before a disturbance or a change is introduced. When possible, the exacr soludon of these equations renders detailed behrvior of the system under the given set
of
L8.7 Drrrrnrxrrel EquarroNs
5n
conditions. Examples of goveming equations, boundary confitions, inftial confitions, and solutions are shown in Thble 18.8. In Example t, the function lrepresents the defecdon of the beam at the location denoted by the position variable X As shown in Tirble 18.8, the variable Xvaries from zero to I the position of the beam. Note Xis measured &om the left-support point. The load or the force acting on the beam is represented by V/,The boundary conditions tell us rrhat is happening at
fifffit Croverning Solution, Boundary Conditionr or Inithl C.ondidons
Problem Tnrc Example 1: The deflecdon of a beam.
Solrrtion
^-d2Y wX(L- x) Erm=
Deflection of the beam
z
Il
as
dre function
of distanceX
Boundary cosditionsr
xtX:
0,
lt:
0 and
atX+ l.,Y=
F
y:;Er?x4+zrx3-Lux)
Q
r.-
Example 2: The oscillation of an elastic gntem.
fr * ,iyi- a
The position of the mass y, as the function of time: yG) ys cos o)kt
:
*hrrrrl= "rnL
Initial conditions: at time
arrimq
'__ _,_
_,1
Example 3; The tempeiature distribution along a long 6n.
rl2T
t=
0,
y:
ys atrd
&t
/= 0,7=
0
ho
fr- aQ- r,): Bounilary conditions:
uX= as
0,
T=
Tu*
I-+oo, T*T^i,
o
Temperamre distribution along the
fin
as
T=
the frrnction of
Ta,+
(4*
-
X:
_ t2_. T"i,),
\A
572
Crrerrrn
l8
M.emsMAncs rN ENcTNEERTNG tlre boundaries ofthe beam. For Example 1, at supports locatedxX:0 andX: Z, the defection of the beam l/is zero. l,ater, when you uke your differential equation class, you will learn hovr to obain rhe soludon for this problem, as given in Table 18.8. The solution shows, for a given load Uf, how the given beam defects at any locarionX Note that, ifyou substituteX = 0 or X = Z in the solution, the value of I/is zero. As expected, the solution satisfies the boundary conditions. The differential equation for Example 2 is derived by applying Newton's second law to the 0, we given mass. Moreover, for rhis problem, the inidal condidons tell us that, at time / pulled the mass upward by a distance of and then release it without giving the mass any initial velocity. The solution m Example 2 gives the position of the mass, as denoted by the vari-
:
I
able y, with respect ro time funcdon.
r.
It
shows the mass
will oscillate according to the given cosinusoidal
In Example 3, I represents the temperature of the fin at the location denoted by the position -{ which varies ftom zero to Z. Note that Xis measured from the base of the fin. The boundary conditions for this problem tell us that the temperature of the fin at its base is Z5*, and the remperanue of the tip of the fin will equal the air temperature, provided that the fin is very long. The soludon then shows how the temperafltre of the fin varies along the length of the fin. Finally, keep in mind *rat the purpose of this chapter was to focus on important mathemadcal models and concepts and to point out why mathematics is so impomant in your engineering education. Deailed coverage will be provided later in your math classes'
SUMMARY Now that you have rqched this point in the text you should:
.
be familiar
with the oramples of math qymbols given in this chapter and
be prepared to learn
new math rymbols as you take additional math and engineering rhe role of the Greek alphabet in engineering and its importance in terms that represent angles, dimensions, or a physical variable in a formula. You should also memorize the Greek alphabet so you €n communicate widr others effectively. understand the impomance of linear and nonlinear models in describing engineering problems and tieir solutions. You should also know the defining characceristics of these classes.
o understand
. .
models.
rer,lhe that the formulation of many engineering problems leads to a set of linear algebraic equations that are solved simultaneously. Therefore, a good understanding of matrix algebn is essential. o know that calculus is divided into two broad areas: differential and integral calculus. You
.
should also know that differential calculus deals with understanding the rate of changehow a variable may change with respect to another variable, and that integrd calculus is related to the summation or addition of things. understand that differentid equations contain derivatives of funcdons and represent the balance of mass, force, enerry, and so on; and boundary conditions provide information about what is happening physicdly at the boundaries of a problem. Initial conditions tell us about the initial condition ofa qystem before a disturbance or a change is introduced.
ProsLEMs
for tluee springs in the accomp"nfrg figure. !?hat is the
n8.1. The force-defection relationships
are shown
tr8.3. The equations describing the position of a water stream (with respect to time) coming out ofthe hose, shown in the accompanying figure, are given by
stiftress (.pting consant) of each spring? Which one of the springs is the stifi:sd
Spring
A
---
Spring C
Ezo o0
+ @)ot
!:
lo
*
(ur)ot
- )*t
these relationships,
9.81 m/s2, and ris time. Plot dre xand they position of the water stream as a function of time. Also, plot *re pa*r the water strenm will follow as a function of time. Compute and plot the components of velocity of the water strenm as a function of dme.
Bro
**
xo
nates, r0 andye are
.F
46
0
x:
x ar'd y are position coordiinitial coordinates of the tip of the hose, (ar)e and (ur)o are the initial velocities of water coming out of the hose in the randy directions, g : In
...... SpringB
6zo
573
Deflection (mm) Problem lE.l
18.2.
A is a linear
(Vy)o- r.5 m/s
spring andspringB is ahard spring, with characteristics that are described by the relationship F kxo. Deter-
t,
In the
accompanylng diagram, spring
:
mine the stiftress coefficient h for sach spring. What is the exponent n for the hard spring? In your own
lo:
words, also explain the relationship berween the spring force and the deflection for the hard spring and how it
(V')o
:
2'6 m/s
t'5 m
differs from the behavior of the linear spring.
Problem 18.3
18.4.
?zo o 9
820 oo
Eto
E 0
46 Deflection (mm)
Problem 18.2
10
In Chapter
12, we explained that the electric power consumption of various elecuical components can be determined using the fo[owing power formula: P: VI = RIz;where P is power in \fatts, Zis the volage, 1is the current in amps, and R is dre resistance of the component in ohms. Plot the poriler consumption of an elecuical component with a resistance of I45 ohms. Vary the value of the current from zero to 4 amps. Discuss and plot the change in power consumption as the funcdon of cuffent drarn *uough the component.
574 r8.5.
Cner.rrn
18
Mergruerrcs
rN ENcnrrsRrNc
where
The defection of a cantilevered beam supponing the weight of an advertising sign is given by
: measured drag force (N or lb) Ca : d.€ coefficient (unidess) p : ab density (kg/-'or slugs/ff) ,Q
t= -Vx2 Uu QL- *) .
V
where
!:
deflecaon
:
speed inside the
wind tunnel (m/s or
ftA)
n a gven xlocation (m)
A:
IZ= weight of the sign (N) E = modulus of elasticity (N/m1 .I = second moment of area (ma) r = distance from the support as shown (m) Z
= ur
ftontal.area of the car (m2 or
f/)
The power requirement to overcome the air resistance is computed by
P:
Fav
Plot the power requirement (in hp) to overcome the
length of the beam (m)
air resistance for a car with a frontal area of2800 in2,
a drag coefficient of 0.4, and for an air density of 0.00238 stugs/fd. \Ary *re speed from zero to I 10 ft/s (75 mph). Also, plot the rate of change of power ret8.7.
qurrement as a funcdon of speed. The cooling rate for three different materials is shown in the accompanyrng figure. The mathemadcal equation describing the cooling rate for each material is of the exponential form T(t) = Taue *. In this relationship, (r) is the temperature of material at time t and the coefficient r represents the thermal capacity
and resistance of the material. Determine the initial temperature and the a coefficient for each material. \7hich material cools the fastest, and what is the corresponding a value?
80
u".... Pmblem 18.5
9r
Plot the defection of a beam with a length of 3 m, the modulus of elasticity of E = 200 GPa, and 1 : E] 1.2 x 106 mm4 and for a sign weighing 1500 N. €40 !?'hat is the slope of the deflection of the beam at the k wall (x : 0) and at the end of the beam where it sup- 930 ,q n20 ports the sign (x : L). 10.6. As we explained in earlier chapters, the drag force acting on a car is determined experimentally by placing the car in a wind runnel. The drag force acting on the car is determined from
fo:
---
\ \". \' \
2
46 Time (min)
1
;CopVz,l
Material C
...... MaterialB
Problem 18.7
10
Pnosr.Errs n8.8"
As explained in earlier chapters, fins, or extended surfaces, commonly are used in a variety of engineering applications to enhance cooling. Common examples include a motorcycle engine head, a lawn mower engine head, heat sinks used in electronic equipment, and 6nned tube heat orchangers in room heating and cooling applications. For long fins, the temperature distribution along the fin is given by:
r8.9.
T^bi*, =
Use the graphical method discussed in this chapter to
obtain the solution to the following set of linear of equations.
xt 3y= 14 4x* Y= 1 18J0.
T-
575
(4* - T-ri*r)e *
in this chapter to obtain the soludon to the following set of linear of
Use the graphical method discussed equations.
-Zxt * 3x2:5
where
4lx2:I0
hp
ln=
hA
18.11"
h
: p:
(\fl/mt. K) perimeterof the frn2(a + /) (m)
.r4:
cross-sectional area of the fin (axb) (m1
*:
thermal conductiviry of the fin material
heat transfer coefficient
n8.r2.
(\7/m.K)
18.13.
I
1814.
b
T n8.r5. Problem 18.8 x8.x6.
Vhat are the dependent and independent
variables?
'!(ithout
using your calculator, answer the following guestion. If 6* is approximately equal to 1300, then what is the approximate vdue of 68? 'tilTithout using your calculator, answer the following questions. If log 8 - 0.9, then what are the values of 1o964, log 80, log 8000, and log 6400? Plot the functions,y : x, ! : l0', arrd,y: log r. Vary the r value from 1 to 3. Is rhe function 1 = log x a mirror image of y : 19' *th respect to y = n and if so, why? Explain. Using a sound meter, the following measuremen$ were made for the following soruces: Rock band at a concert (100 X 106 ltPa), aJack hammer (2 X 106 pPa), and a whisper (2000 p,Pa). Convert these readings to dB values. A jet plane taking offcreates a noise with a magnitude of approximately 125 dB. !7'hat is the magnitude of the pressure disturbance (in pPa)? Identify the size and the type of the given marices. Denote whether the matrix is a square, a column, a diagonal, a rolv, or a unit (identity).
Next, consider aluminum fins of a rectangular profile shown in the accompanying figure, which are used
to remove heat from a surface whose temperature
is
100'C. The temperature of tjhe ambient air is 20'C.
.l
[r2ol 4 5l
Plot the temperatue distribution along the fin '!7/m . using the following data: h : 180 K, 11 : lS.\ffln?.K a = 0.05 m, and b = 0.015 m. Yuy *
^.12
from zero to 0.015 m. S7'hat is the remperature of the dp ofthe fin? Plot the remperature of the tip as a funcaon of h. Vary the h value from 1 80 to 350 \7/m . IC
d.fl ! f !')
Lo56J
[,,
,l:11 . fi :l L"-l
"[;?:l Lo o
r.l
576
Cneprnn
L8
MernsMATrcs rN ENcrNseRrxc ffi.n" Solve the
r8.r7.
Givenmatrices:
"t:lj I z
tBl
j)
II
: [r 3 -tl
l'r'l 3 I' and {c} : I -2 f, Perform 15 | 4) L+ 5 -7)
ns.n.
a. lAl + [B] :? b. [e] - lBl =? c. 3lAl:? d. L{l[r]:? e. !41{C} =1
18.18.
g.
Show that
nant
[1][,a]
: lA)lll :
ro ol lz4 20 0 |
Itz -4
l,
X8.20. Solve
1 :1rnV2.
of .", with a mass of 1500 kg Plot the kineti" "n""rgy " the funcdon of its speed. Vary the speed from zero to 35 m/s (126 km/h). Determine the rate of change of
lAl
10 ol lz = 16 6 I
14
|
18J
calculate the determi-
181
expansion. \Vhich
kinetic energy of the car as function of speed and plot does this rate of change represent? In Chapter 13, we explained that when a spring is suetched or compressed from its unstretched position,
it,'V'hat m.23.
elastic energy is stored in the spring and that energy will be releasedwhen the springis allowed to return to its unstretched position. The elastic energ;r stored in a spring when stretched or compressed is determined from
['
set
of equations using the Gaussian
ElnsticEnerg: I FAc J0
method.
x't 3y 4x*y:l
114)
as
Lr2 -4
of [r4] *d t,Bl by direct
matrix is singular? 18J9. Solve the following
5)\xz)
As we explained in Chaprcr 13, An object having a mass za and moving with a speed /has a kinetic energy,
Kinetic Enagt
Giventhefollowingmatrices: tr4l
and [.Bl :
|
which is equal to
:?
lAl2
e) t I r1(x,) tt't |f 2 5 t11"l=115i
|L-3 t
the following operations.
f.
following set of equations using the Gaussian
method.
14
the following set of equations using the Gaussian
method.
-2x, I 3x2:
5
4*x2:L0
Obain expressions fot the elastic energy of a linear spring describ edby F : &r and a hard spring whose behavior is described by F: M. For Example I in Thble 18.8, verifr that the given solution satisfies the goveming differential equation and the boundary condidons. For Example 3 in Thble 18.8, verify that the given solution satisfies the governing differential equation and the initial conditions.
CHAPT'ER
PnoBABTLTTY AND IN ENCINEERING
L9
SreusTrcs
|I ngineers use statistical models to address quality control and fl E
reliability issues and to perform
failure analyses.
5n
578
CneprsR
19
PnosAB[JTyeNo Strrrsrrcs tN ENcrNsnRrNc by practicing mgineers to n perforzn fai.lure analyses. Ciuil engineas ase statistical rnodek to study the reliabiliry of consm.tction mzteriak and sm.rctures and to futtS"forfuod connol and water supply managemmt Electrical mgineers use statistical modzb for signal processing or for deuehping uoicerecognition software, Mechanical mgineers use statistics n srudy thefailure ofmateriab and machine parts and to futig" experiments. Manufauuring engineers use statistics for qaahA control assarance of the products they produce, These are but a few acanples of why an undnstandi.ng of stati.stical conce?ts and modtls is important in mgineering. W'e will begin by exphining some of the basic idzas in probability and statistics. \Ye will thm discussfrequmcy disnibutiont, measure ofcmnal tendenry ftnean and rnedian), rneasare of uariation within a daa set (swndard dcuiation), and norznal dis*i butions.
Statistical
rtuodals are being used
inneasingfu more often
address qaality connol and reliability issues and
19.n Probahility-Basic ldeas Ifyou were to ask your instructor how many students
are enrolled in your engineering class this
semester, she could give you an exact number: sxy 60. On the other hand, if you were to ask her how many students will be in the class next year, or the year after, she would not be able to give you an exacc number. She might have an estimate based on trends o! other pieces of information, but she cannot know exacdy how many students will be enrolled in the class next yer. The number of studenm in the class next yeat, or the year after , 's randon. There are many situations in engineering that deal with random phenomena- For example, as a civil engineer, you may design a bridge or a high*.f. It is impossible for you to predict exacdy how many cars will use the highway or go over the bridge on a cenain day. As a mechanical engineer, you may
design a heating, cooling and ventilating qrstem to maintain the indoor temperature of a building at a comfortable level. Ag"i., it is impossible to predict exacdy how much heating will be required on a future day inJanuary. As a computer engineer, you may design a nerwork for which you cannot predict its future usage exacdy. For these types of siruations, the best we can do is to predict outcomes using probability models. Probability has its own terminologlt therefore, it is a good idea to spend a litde time to familiarize yourself with it. In probability, each time you repe,rt an experiment is called a nial The result of an experiment is called an ntttcorne. A randorn experimmt is one that has random outcomes-random outcomes qurnot be predicted exacdy. To gain a better understanding of these terms, imagine a manufacnrring setting wherein cell phones are being assembled. You are
positioned at the end of the assembly line, and in order to perform a final qualiry check, you are asked to remove cell phones at random from the assembly line and turn them on and off, Each time you remove a cell phone and turn it on and ofi you are conducting a randnrn acperirnmt Each time you pick up a phone is a nial with a result that can be marked as a good phone or a bad phone. The result of each experiment is calledan outcome.Now, suppose in one day you check 200 phones, and out ofthese phones, you find five bad phones. T\en rhe rela' ti.ae ftequency of finding bad phones is given by 5 I 200 : 0 .025 . In general, if you were to repeat an experiment a times under the same conditions, with a certain outcome occurring
19.2
$a411s116s-Besrc lona,s
579
m dmes, the relative frequenry of the outcome is glenby m ln. As n ge$ larger, then the prababilfu p of a specific outcome is given by p = rnln, Each question on a muhiple-choice exam has five answers listed. Knowing that only one of the answers is correct, ifyou are unprepared for the exam, what is the probability that you pick the
correct answer? 1
P=5:0'2 For those of you who follow spons, you may have noticed that sometimes the probabfity of a certain outcome is expressed in terms of odds. For orample, the odds in faror ofyour team winning may be given as I to 2. !7hat does "odds in farror ofan event" mean? The odds in favor of an event occurring is defined by probability (orcuring)/probability (not occuring). Therefore, if the probability of your team winning is given by 0.33, then the odds in favor of your team winning is given by 0.3310.66 : 712 or 1 to 2. On the other hand, if the odds are e:rpressed as r toy, then the probability of a specific outcome is calculated fromxl(x * y). For this example, as expected,la = 1/(1 * 2) : 9.33. li,ii,ir,airio.,I..
- i,j!.IiiLrliiii.irirl!,81?.i:Ailitri
--.i'!.1!ri{ii __
--,D11{!,,irIiiitf
- --.r.--_-,1ll1ititr.lr'
fu you take advanced classes in engineering you will learn more about the mathematical models that provide probabilities of cerain outcomes. Our intent here is to make you avare of the imporance of probabiliry and statistics in engineering, not to provide detailed coverage
of
these topics.
19.2 Statistics-Basic
ldeas
Statistics is that area ofscience that deals with collection, organization, analysis, and inteqpretation of data. Statistics also deals with methods and techniques that can be used to drav conclusions about the characteristics ofsomethingwith a large number ofdata points-commonly
ulledapopuktioa-using
smaller portion ofthe entire data. For example, using statistics, we with two million registered voters, by gathering information only from 1000 persons about how they are planning to vote. As this example demonstrates, it is neither feasible nor praceical to contact two million people to find out how they are planning to vote. Howwer, the sample selected from a population must represent the characteristics of the population. It is important to note that, io statistics, population does not refer necessarily to people but to all of the data that pertain to a situation or a problem. For example, if a company is producing 15,000 screws a &y and they want to examine the quality of the manufacnrred screws, they may select only 500 screws randomly for a quality test. In this example, 15,000 screws is the population, and the 500 seleced screws represents the sample. Statistical models are becoming common tools in dre hands of practicing engineers to ada
can predict the outcome of an elecdon in a state, say
control and reliability issues and to perform ftilure analyses. At this stage of your education, it is imporant to realize that, in order to use statistical models, you need first to completely undersand the underlying conceps. The next secrions are dwoted to some of these dress qualiry
imponant concepts.
580
Crrer"rsn
19
PnosAgrl.rryaNp Srarrsrrcs rx ENcnwrnrrtc
19.3
Frequency Distributions As we have said repeatedly throughout the text, engineers are problem solvers. They apply physical and chemical laws and mathematics to design, develop, test, and supervise the manufacture of millions of products and services. Engineers perform tests to learn how things behave or how well they are made. As they perform experiments, they collect data thar can be used to explain cerain things better and to reveal informadon about the quality of produca and services they provide. In the previow section, we defined what we mean by population and samples. In general, any statistical analysis starts with identifying the population and the sample. Once we hare defined a sample ttrat represents the population and have collected information about the sample, then we need to organize the data in a cenain way such tfiat pertinent information and conclusions can be extracted. To shed light on this process, consider the
following example. The scores of a test for an inuoductory chemistry class of 26 smdents are shown here. Cer'We are interested in drawing some tainly, the scores of your class would be beter than these! conclusions about how good this class is. The scores of a test for Example 19.2: Scores: 58,95,80,75,68,97,60, 85, 75,88,90,78, 62, 83,73,7A,70,85, 65,75,53, 62,
56,72,79 As you can see from the way the data (scores) are represented, we cannot easily drav a conclusion about how good this chemistry class is. One simple way of organrzngthe data better would be to identify t"he lowest and the highest scores, and then group the data into equal intervals or rangesi say a range of size 10, as shown in Thble 19.1. !7'hen data are organized in the manner shown in Thble 19.1, it is commonly referred rc as a groapedfrequmcy distibation.
Range
Scores
59,53,56 68,6A,
A,65,62
75, 75, 78, 73, 70, 70, 75, 72, 79 80, 85, 88, 83,85,87
95,97,90
Frequency
50-59
3
60-69 70-79
5 9 6
80-89 90-99
3
The way the scores are now organized in Table 19.1 reveals some usefi:l information. For example, three students did poody and three performed admirably. Moreover, nine students received scores that were in the range of 70-79,which is considered an aver€e performance. These average scores also constitute the largest frequency in the given data set. Another usefirl piece of information, which is clear from examining Thble 1 9 . 1 , is that the frequency (the number ofscores in agiven range) increases from 3 to 5 to 9 and then decreases from 6 to 3. Another way of showing the range of scores and their frequency is by using a bar graph (what is
commonly called a bisngram). The height of the bars shows the frequency of the data within the given ranges. The histogram for Example 19.2 is shown in Figure 19.1.
L9.3 FnrqurNcr DrsrnrsurroNs
E The
rigurul9.t
50-59
iistoqmm for the scores
581
70-79 80-89 n-99
60-69
given in Table 19.1.
Scores
Cumulative Fregueney The data can be organized further by cdculatin gthe camulatiue frequenry. The cumulative frequency shows the cumulative number of students with scores up to and including those in the given range. \7'e have calculated the cumulative frequency for Example I9.2 and shown it in Table 19.2. For Example 19.2, eight scores fall in the range of 50 to 69, and 17 students' scores (the majority of the class) show an average or below-average perFormance. The cumulative frequency disuibudon can also be displayed using a histogram or a cumulatiae frequmE polygon, as shown in Figures 19.2 and. 19.3, respectively. These figures
Range
Frequency
5A-59
3
6a-69 70-79
5
80-89
6
90-99
3
Cumulative Frequency 3
3
3t5:8
8
3+5+9=L7or8*9=17 3+5+9+6:23or17 t6=23 3 + 5 + 9 + 6 + 3 = 26or23 * 3 = 26
9
>?5 o
=tn
trs E
E10
v5
fl
tigurelg.?
The cumulative.f requency
histogram for Enmple 19.2.
50-59
50-69
s0-79 s0-89 Scores
50-99
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
In this
tg.4
MeAsuRBs
or Crnrns. Trsorr.lcv.rNo Vmrerrox-MnAN, MuDrAN, exr Stexpano DsvIATroN 583
Group B nndings
GroupAFindiag;s
p (kglm3)
p fl,e/m?)
950 940 890
1020 1015 990 1060
1080
1030
1120
r
950 975
r040
1020
1150
p*s
900
980
910
960
1020
=
1000
Pn€
='1000
GroupA
Group B
(o'
@-o*) i i I i I I i
1020 1015
+20
20
-L 1<
15
990
-10
10
950 940 890
1060 1030
+60 +30
60
1080
30
lI20
+80 +120
950 971
-50 *25
50
900
-100 +40
+24
20
1040 1r 50
+t50
-20 -40
20 40
910
-90
I t0zo I sso i
o*)
960
X=o
,,
,<
1020
2=zgo
=50
-60
-
110
+20
E:0
l(p
- p*)l 50 60" ll0 80:
LZA 100 4D 150 90 20 E=820
1
:
i
i 'i
j I
I
i
pute horv much each reponed density deviates from the mean, add up all the deviations, and then take their anerage. Tahle 19.4 shows the deviation from the mean for each reported density. fu one can see, the sum ofthe deviations is zero for both groups. This is not a coincidence. In fact, rhe sum of deviations &om the mean for any given sample is always zero. This can be readily verified by considering the following.
,=+2., d;: where
4
(x;
fl9J)
- V)
(19.21
represents data points,
sents the deviation
from
anerage.
I
is the average, a is the number of data points, and, di repte-
584
Crreprra.
19
Pnorn-anrry.cNo SrArrsrrcs rrl ExcrNsERrNc r
- ); Zd,: )", i=l i=\ i=1
flg.3)
u
r& - r&:0 2d,: i=l
09.4)
Therefore, the average of the deviations from the mean of the d*a set cannot be used to measure the spread of a given data set.'!f'hat if one considers the absolute value of each dwiation from the meani Ve can then calculate the arerage of the absolute values of dwiations. The result of this approach is shown in the third column of Tirble I9.4.For group A, the mean dwiation is 29,.r,ghereas for group B the mean dwiation is 82. It is clear that the result provided by goup B is more scattered than the group A data Another common way of measuring the dispersion of data is by calculadngthe uriance. Instead of aking the absolute values of each deviation, one may simply square the deviations and compute their averages:
2e,-o)'
i=l
fie.5)
n- |
Notice, however, for the given example the variance yields units that are (Wl^')'. To remedy this problem, we can take the square root of the variance, which results in a number that is called,
sundad deaiation h
t=
- -*)' 2@' i4L n- I
(1e.6)
This may be an appropriate place to say a few words about why we use n - t nrher then a to obtain the standard deviation. This is done to obain conservative values because (as we have mendoned) generally the number of experimental trials are few and limited. Let us turn our aftention to the standard deviations computed for each group of densities in Table 19.5. Group A has a standard dwiation that is smaller than group B's. This shows the
:
1000 kg/*'), whereas the redensities reported by group A are bunched near the mean (p suls reponed by group B are more spread our. The sandard deviation can also provide information about the frequency of a given data set. For normal disuibution (discussed in section
19,5) of a data set, we will show that approximately 680/o of the &ta will fall in the interval s) to (mean * s), about 95o/o of the data should fall between (mean - 2s) to (mean f %), andalmost all data points must lie berween (mean - 3l) to (mean -f 3s). In Section I9.3, we discussed grouped ftequency disnibution. The mean for a grouped
of (mean -
disuibudon is calculated from
>Qn _ ,:--; where
r = midpoints of a given range / = frequency of occunence of daa in the range o : 2 f = total number of data points
fl9J)
19.4 Mresunss or CENTRAL
ThxosNcyANo VnnHrror.r*MnaN,
Mronn,
GroupA @
- o*)'
585
b - o*)t
400
2500
225
3600 12,100
3600
6400
900 2500
r4,40a
625
10,000 1600
400
22,504
400 1600
8100
E=
DsvrATrox
Group B
100
s
AND STANDARD
):
lo,75o
- 34.56 (ke/*1
s
=
400 81,600
95.72 (kg/m3)
The standard deviation for a grouped distribution is calculated from
fle.8)
Next, we demonstrate the use of these formulas.
For Example 19.2, using Equations (19.7) and (19.8), calculate the mean and standard devia-
don ofthe
class scores.
Consult Thbles 19.6, 19.7, and 19.8, respectively, while following the soludon. To calculate the mean, fust we need to evaluate the midpoins of data for each range and then evaluate the ) x/as shown.
Range
Frequency
50*59
I
60-69 70-79 80-89
90-99
)
I
6 3
586
Cneprsn
19
Pnonanrr.rrveNp Stlrrsrrcs rx ENcrNnsRrNc
i''ir.liffi
.r''l
50*59
J
60-69 70-79
5
80-89
90-99
t63.5
54.5 64.5
322.5
c)
/4,)
670.5
5 5
u4.)
507
94.5
283.5
n=2f:26
>
, ,,
,
*f: !?q
it' i*-,"
Frequency Midpoint Rarye
fn
50-59 6A-69
3
7A-79
80-89 9A-99
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(r-V)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: n-t: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
50-59
60-69 70-79 80-89
90-99
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 bell-shaped 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 F-xample 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 probability-distribution curve becomes smoother. A probability distribution that has a bell-shaped 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 normal-distribution 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:- x-V
fi9.e)
J
In Equation (19.9), arepresenc the number of sandard deviations from the mean. The mathematical function that describes a normal-distribution 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 normal-disaibudon 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.
z=1.@
Mean:0
z =3.N
z=2.@
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 2-09 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
PnosABrr-rrv.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(x-x)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
x-V 7-9.22 ,: --;-: --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: x-V 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
z--0.97
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
x-V ro-9.22 -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 cumulative-frequency 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 probability-distribution 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 probability-disuibution curve. Determine the average, variance, and standard dwi-
ation for the following parr. The measured values are given in the accompanying able.
Screw l-ength
(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.ments-Problems 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. Therefof€, 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 first-year 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, t-he 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 $41.83 for the same duration.
20.3
Future Worth of a Present Amount Now we will dwelop a general formula that you can use to compute the future value Fof any present amount (principd) 4 after nyears collecting zolo interest compounding annually. The cash
fow
di"S"-
for this situation is shorrn in Figure 20.3.In order to demonsffate, step-by-
step, the compounding effect of the interest each year, Table 20.2 has been developed. As shown
in Thble 20.2, sardng with the principal P, at the end of the first year we will hare P * Pi or P(l + i). During the second year, dre P(l * i) colleca interest in an amount ofP(l + i)i' au;.d. by adding the interest to the P(l * i) amount that we stamed with in the second year, we will have a toal amount of P(L + i) + P(l * i)i. Fae'oling out the P(1 * /) term, we will hare
2O.3 Ftmrnr Wonmr or e PnEseNT Arvrorlxr
I
601
Figure20.3
The cash flow diagram for future
wort[ of a deposit made in the bank today.
Balance at the
Year
Begimiryof
Interest for
theYear
the Year
i1
P
t3 :4 i-
P(l + r) P(t + ilz PU + i)3
ll ,A
5
Balance at the Eqd
P+(P)(i)=P(r+i) P0+i) +P(l + DQ)=P(L+i)z
(P)(t P(t + t)(i) P(r
+
P0 + il2 + P(l + i)2(t) : P(r + il3 P0 + il3 + I{1 + i)3(i): P(L + il4 P$+ 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 +
il-t(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 eight-year 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
(y€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 iirr-lfifilii.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 fust-year 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 casi-flory 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 user-friendly, 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-!-r---I
A
A
O;;l;;* 1t* o'''A A I + (r + i)',-t (l 4"1 I-l
(20.8)
Simplifying the right-hand side of Equadon (20.8) leads to the following relationship:
P(r+i)-P-A-
i
€0.8b)
(r + i)"
And after simplifying the left-hand 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 step-by+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 y@r_at the same frequency as the compounding interest-leading 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 five-year, $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 ):$20276
of Engineerinq Economies Analysis
The engineering economics formulas that we have dweloped so far are summarized in Thbles 20.6 and20.7. The definitions of the terms in the formulas are given here:
P: F:
cost-lump sum ($) future wofth, or fuguss ssst-lump 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 interest-time 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+;\t-1'1 (PtA,i.n):Ld;1
| A=Pl:--l
(Atp,,,o)=l#++]
': ^L---i-)
(FtA,i,n):lga#]
LQ+ir-rJ 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 ;\ou-1
-7--76-
\'*;)
|
-'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\einterest-timefaaor,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 interest-time fa*or (AIP, A n). As an otample' the numerical values of interest-time 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-
9-1216381t
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.olo
rc,rt5,t64n lt"o5rw849 11.15840601
11.25175334 1r.34979939
rlAU9994/*
12.67604964 0.07888893 M 13-69013361 0-0730453r 35, 14.78534429 0.06763454 36 15.96817 184 0.06262458 t7 17.24562558 0.05798572 38' 18.62527563 0.05369048 39 20.11529768 0.0497 1341
11,51388837 LL.5,8693367
3!:
.,,'
2r72452150
4L
0.04603093 23.46248322 0.04262123
42
2,5.31945187' O,A3g464tt
4'
27.36664042 0.03654084 29.55597166 0.033834rr
44
0.09983225
45.7619&30 50.42292144 55.45675516
1.00000000
a.48a7692s 030803351 0.22192080
0.17031528 0.15769934
,2 1r.73708300 0.08520005
40
1.08000000 1.00000000 0.56076923 2.08000000 0.38803351, 3.24ffi000i0 0.30192080 4.50611200 0.25045645 5.86660096 a.U63r539'7.335929Ai4 0.19207240 s.92280336 0.17401476 10.63662763 0.1600797r 12.48755784 0.14902949 14.48656247 0.14007634 16.64548746 0.13269502 18.97712646 0.12652181 21.49529658 0.12129685 24.21492030 0.1,1682954', :n J52tr393 0.11297687 30.32428304 0.10962943 33.7502256s 0.10670210 37.45024374 0.t04t2763 4r.4626324
0.08961854 103.96593622 0.08882743 rr3.28321trr
0.088r0728
123.34586800
j
|
0:00961854 0,04882743 0.00810728
0.08745081' 134:,21353744 0.08685163 145.95062044
0,00745091
0.08630411
0.0063M11,
i l
158.62667007
0.00685163,
,j
1r,65456822
0.08580326 172.31680368
0.00580326 li
rr.7L7Dn9
0.08534467 r87.10214797
a.0afia67:lt'l
lr.7:t5L7851
0.08492410 203.07031981
a,gg4gz|Ab li
I r.82886899
o.08453894 220.31594540
a,0a4fi894
1r"87858240 1r.92461333
0.a04rcfl3
1r.9672345:7
0.08418513 238,94r22W3 0.08386016 259,05651871 a.083561.49 280.78rc402r
12:006698,A 12.9432995r
0i.a$28684 304i-24352342 0.08303414 329.58300530
A;A$28684
n.a77w362
0.08280152 356.94y&572
0.00280152
l'l
0.003s6016 0.003561/Lg
0.a$$414 -l
Continucd
612
CrrerrsR
20
Encrxrrnrnc EcoNoutcs
:.rtl1
4>
46
+/
t48
'49 ,50
(F/4 i, n)
(P/4 i, n)
3r.92A44939 34.47408534 37.23201217 40.21057314 43.42741899
a.$132788
469016125r
0.02900730 0.02685861
0.024869A8 0.02302693 0.02132123
n)
(A/4 L n)
12.10840150 12.1374A580
0.0823899tr
(P/A, i,
12,1642574t 12.18913649
0.08258728
(F/4
i n)
0.08220799
386.50561738 4r8.{o60657V 452.9A015211
0.a82040n
490.t32r6428
12.2r2t634r
0.08188557
530.34273742
12.23348464
0.ast74286
src.n0r5642
(A/4
i
n)
0.44258728
0.aon899r a.a0220799 0.00204027 0.00188557 0.00174286
Additional values of interest-time factors for other interest rates can be created using Excel. Keep in mind that you can use those tables or other similar tables found in the back of most engineering economy text boolis to determine interest-time factors for interest rates that compound more frequendy than once a yeax. To do so, however, you must first divide the quoted nominal interest rxe i by the number of compounding frequency m and use the resuhing number to pick the appropriate interest able to use. You must then multiply the number ofyears z by ttre number of compounding ftequency rn and use the outc ome of n times rn as rhe period when looking up inrerest-time facors. For example, if a problem states an interest rate of 18o/o compounding mondrly for four years, you use the 1.57o interest table (18/12 : 1.5), and for the number of periods, you will use 48 (4 X L2 : 48).
is the equivalent present worth of the qsh fow given in Figure 20,5? Put another way, how much money do you need to deposit in the bank today in order to be able to make the withdrawals shown? The interest rate is 87o compounding annually. The present wonh (PW of the given cash flow is determined from
Vhar
PV = 1000(PlA,8o/o,4) + 3000(?lF,8o/o,5) +
500O(P|F 8o/o,7)
$5000
PW=
U
?
Figure20.5
Thecash flondiagmmfff Elalnple20.9.
2O.9 CnoosrNc rnr Bssr ALTERNeTnEs-DecrsroN MnrflNc
613
We can use Thble 20.8 to look up the interest-time factor values, which leads to
(PIA' 8o/o'
4) = 3.3lnZ6A4
(PIF, 8o/o,
5) = A.65058320
(P 14 8olo,
7)
:
p17
:
PW
= $827L32
(
0.58349040
1000)(3.31212684)
+
(3000X0.68058320)
+
(5000)(0.5 5349040)
Therefore, f rcday, you put aside $827I.32 in an account that paln 8%o interest, you can withdrarr $1000 in the next four years, and $3000 in five years, and $5000 in seven years. 1.1i,.,,--,t,..
---,.-a
-
rI'iI?41:ilIIiE't1;Al .iri!iliT..r-
20.9
--'--rn1l'
tiiT._-- :.ir:.jiiiiilli'iili
Choosing the Best Alternatives-Decision Making Up to this point, we have been discussing general relationships that deal with money, time, and interest rates. Let r$ now consider the application ofthese relationships in an engineering setting. Imagine that you are assigned the task of choosing which air-conditioning unit to purchase for your company. After an exhausrive search, you have narrowed your selection to nro alternatives, both of which have an anticipated 10 years of working life. fusuming an 8o/o interest rate, find the best alternative. Additional information is given in Thble 20.9. The cashflow diagrams for each alternative are shown in Figure 20.6. Here we will discuss three different methods that you cln use to choose the best economical alternarive from many options. The drree methods are commonly referred to as (1) present worth (PW) or present cost analysis, (2) annual wonh (A\[) or annual cost analysis, and (3) future worth (FIf) or future cost analysis. \[hen these methods are applied to a problem, thty.ll lead to the same conclusion. So in practice, you need only apply one of these methods to evaluate options; however, in order to show you the details of these procedures, we will apply all of these methods to the preceding problem. Present Worth or Present Cost Analysis Wirh this approach you compure the toal present worth or the present cost of each alternative and then pick the altemative with the lowest present cost or choose the alternative with the highest present worth or profit. To employ this
AltemativeA Initial cost Salvage value
$100,000
after 10 yean
Operaring cost per yqu Maintenance cost per year
AlternativeB
$10,000
$85,000 $5ooo
$2500 $1000
$3400 $1200
614
Cnerrrsn2O ENcnvsnRrNcEcoxourcs $10,000
$100,0m
AlternativeA
$85,000 Alt€rnative B
L
Figure
20.6
The
casi'flow diagmns l0r the erample problen.
method, you begin by calculating the equivalent present value of all cash fow. For the example
problem mentioned, the application of the present worth analysis leads to: Altenative A;
PV = -100,000
-
+
(zloo
The interest-dme facrors for
pv = -100,000 -
i:
1000XPi/4,87o, 10)
8o/o are gSven
(2500
+
+
10,000(P/4 8olo, 10)
in Thble 20.8.
1000x6.71008140)
+ (10,000x0.463t9349)
PV = -118,853.35 lltenative B:
PV= -85,ooo -
(3400
+ r2o})(PlA,8olo, 10) +
10)
'oo}(PlF,87o,
P'W': -85,ooo - (3400 + 1200x6.71008140) + 50A0(0.46319349)
PW: -rt3,550.4a
2O.9 CrroosrNc rns Brsr ALTERNarrvEs-DscrsroN
Merrxc
515
Note that we have determined the eguivalenr present wonh of all furure cash flow, including the yearly maintenance and operadng costs and the salvage value of the air-conditioning unit.
In the preceding ana\nis, the negative sign indicates cost, and because ahemative B has a lower present cost, we choose alternative B. Annual Worth or Annual eost
Analysis Using this approach, we compure the equivalent annual
worth or annual cost value of each alternative and then pick the alternative with the lowest annual cost or select ttre alternative with the highest annual worth or revenue. Appln"g the annual wonh analysis to our example problem, we have Altenative A:
AW: -(2500 + 1000) A!7 = -(2500 + 1000) -
+ 10,000(AlE8o/o,10) (100,000)(0.14902949) + (10,000)(0.06902949) 100,000(l148o/o,10)
AW = -17,712.65 Altenative B:
:
-Qqo0 + 1200)
A!7': -Q40A + 1200) AW
AV =
85,0o0(Alp,8olo,
t0) +
5000(AlE 8%o, 10)
+
5000(0.06902949)
85,000(0.14902949)
-16,922.35
Note that using this method, we have determined the equivalent annud wonh of all cash flow, and because alternative B has a lower annual cosf, we choose alternative B. Future Worth or Future Cost Analysis This approach is based on evaluating the future wonh or future cost of each alternative. Of course, you will then choose the alternative with the lorrest future cost or pick the alternative with the highest future worth of profit. The furure worth analysis of our example problem follows. Aftemative A:
FV
:
FruT: *10,000 F:V:
*10,000
100,000(FlP,8o/o,10) (100,000)(2.15892500)
*
+ 1000XF/4 8%,10) (2500 + 1000)(14.48656247)
(2500
-
-256,595.46
Altenative B:
- 85,000(F1p,870, 10) - (3400 + 1200)(FlA,8%o, 10) FW : *5000 - (85,000)(2.15892500) - (Z+00 + 1200)(14.48656247) FY,r : -245,146.81 F$7
= *5000
Because alternative B has a lower
future cost, again we choose alternative B. Note that regard-
ofwhich method we decide to use, alternative B is economically the better option. Moreover, for each alternative, all of the approaches discussed here are related to one another thro"gh the interest-time relationships (factors). For example, less
616
Cnapmn
20
ENcrxEEMNc EcoNolttcs Altenative A:
PV:
A\(P/r{,8olo, 10) :
(-
17,7 12.65) (6.71008
140)
: - 118,853.32
or
P\( = F\0(P/4
8%o,
10)
: (* 256,595.46)(0.46319349) : -
118,853.34
Altenative B:
PI7 = A\(P/r{,8ol0, 10) : (* r 6,922.35X6.71
p\7
:
F!((? tF, 8o/o, r0)
:
008 140
)
(-245,r46.81)(0.463t9349)
: - tr3,550.40 =-
113,550.40
Finally, it is worh noting that you can take semester-long classes in engineering economics. Some of you will eventually do so. You will learn more in depth about the principles of money-time reladonships, including rate-of-return analysis, benefit-cost ratio analysis, general price inflation, bonds, depreciation methods, evaluation ofalternatives on an after-tax basis, and risk and uncenainry in engineering economics. For now, our intent has been to introduce you to engineering economics, but keep in mind that we have just sctatched the sur'We face! cannot resist but to end this section with definition of some of these imponant concepts that you will learn more about them later.
Bonds States, counties, and cities issue bonds to raise money to pay for various projects, such
as
schools,
highways, convention centers, and stadiums. Corporations also issue bonds to raise money to expand or to modernize their facilities. There are many different types of bonds, but basically, they are loans that investors make to govemment or corporations in return for some gain. When a bond is issued, it will hare a rnahtrity date (a year or less to 30 years or I onger), par aalue (the amount originally paid for the bond and the amount that will be repaid at maturity date), and an intelest rate (percentage ofpar value that is paid to bond holder at regular intervals).
Depreciation fusets (such as machines, cars, and computers) lose their value over a period of time. For o
Life-Cyele Cost In
engineering, the rerm ltfr-"ytlt cost referc to the sum of all the cos$ that are associated a structure, a service, or a product during its life span. For example, if you are designing a bridge or a highway, you need to consider the costs that are related to the initial definition and assessment, concepnral design, deailed design, planning, construction, operation, maintenance and disposal of the project at the end of is liG span.
with
Pnonr,rus
617
SUMMARY Now that you have reached this point in the text
. .
z0J.
You should realize that economics plays an imponant role in engineering decision making. Moreover, a good understanding of the fundamentals of engineering economics could also benefit you in better managing your lifelong financial activities. You should knoqr the relationship among mon€y, time, and interest rate. You should be familiar with how these relationships were derived. Moreover, you should also know how to use the engineering economics formulas summarized in Thbles 20.6 and,20.7 to solve problems.
Compute the future value of the following deposits
20.9.
made today:
a. $10,000 at 6.750/o compounding annually for
monthly.
10 years
b. $10,000
x
6.750/o compounding quarterly for
x
6.750/o compounding mondrly for
20.10.
10 years 20.2.
Compute the interest eamed on the deposits made in
20.3.
How much money do you need to deposit in a bank todqy ifyou are planning to hane $5000 in four years
Problem 20.1.
20.4.
20.5.
by the time you get out of college? The bank offers a 6.750/o interest rate that compounds monthly. How much money do you need to deposit in a bank each month if you are planning to have $5000 in four years by the time you get out of college? The bank offers a 6,750/o interest rate that compounds monthly. Determine the effective rate corresponding to the following nominal rates:
a. b. 20.6.
6,250/o compounding
monthly
compounding monthly c, 16,90/o compounding mondrly Using Excel or a spreadsheet of your choice, create interest-time factor tables, similar to Thble 20.8, for
i:
9,25o/o
6.50/o and
i:
6.750/o.
201.
Using Excel or a spreadsheet of your choice, create interest-time facor tables, similar to Thble 20.8, for
20.8.
Using Excel or a spreadsheet of your choice, create interest-time factor tables, similar to Thble 20.8, for
i:7.5o/o
and
i:
7.75o/o.
i:8,5o/oandi:9.5o/o.
Most of you hane credit cards, so you already know that if you do not pay the balance on dme, the credit card issuer will charge you a cerain interest rate each
10 years
c. $10,000
Using Excel or a spreadsheet of your choice, create interest-time factor tables, similar to Thble 20.8, that can be used for i : 8.5o/o compounding
month. Assuming thar you are charged 1.25olo interest each month on your unpaid balance, nrhat are the nominal and effective interest rates? Also, determine the effecdve interest rate that yorr own credit card issuer charges you. 20.11. You hare accepted a loan in the amount of $15,000 for your new car. You have agreed to pay the loan back in four years. Vhat is your mon*rly payment if you agree to pay an interest rate of 9olo compounding
montfily? How much money will you have available to you after five years if you put aside $100.00 a month in an acconnt that gives you 6.750/o interst compounding mon*rly? 20J3. How long does it take to double a deposit of $1000 a. at a compound annual interest rate of 60/o b. at a compound annud interest rzrte of 7o/o c. at a compound annual interest rate of 87o d. Ifinstead of $1000 you deposit $5000, would the dme to double /our money be different in parts (a)-(d)? In otherwords, is the initial sum ofmoney a factor in determining how long it takes to double your money? Now use your answers to verifr a nrle of thumb that is commonly used by bankers to determine how long it 20.12.
618 Crreprnn20
ENcrNrsRrNcEcoNolucs
takes to double a sum of money. The nrle of thumb commonly used by bankers is grven by
time period to double a sum of 20J4.
mon.t
- i"*#;
Imagine that as an engineering intern you have been assigned the task of selecting a motor for a pump. After reviewing motor catalogs, you narrow your choice to two motors that are rated at 1.5 k]n Additional information collected is shown in an accompanying table. The pump is expected to run 4200 hours etrery yenL After checking with your electric utility company, you determine the average cost of electricity is about 11 cents per k\Vh. Based on the information given here, which one of the motors will you recommend to be purchased?
Criteria
lVlotorX
MotorY
Expected usefirl
life cost Efficiency at the
5 years
5 years
lnitial
$300
$400 0.85
0J5
operating point
Esdmated
$t2
per yeat
30J5.
?016.
20.17.
\V"hat is the equivalent present worth of the cash flow given in the accompanying figure? Assume i : 8o/o. \7hat is equivalent furure wonh of the cash fow given in the accompanyrng figure? Assume i = 8o/o. lVhat is the equivalenr annual worth ofthe cash flow given in the accompanying figure? Assume i = 8o/o, $3000
Problem20.15 0 $3000
Problem 20.16
$8000
s67 ttl ttl {._t___+ Problem 20.17
$10 peryear
maintenance cost
$zooo
ProsLEMs
$7000
Problen 20.18
$7000
2010.
Vhat
2019.
You are to consider the following prqects. \ftich prqect would you approve ifeach project creates the
are the equivalent present won:h, annual worth, and future wonh of the cash flow given in the accompanlng figure? Assume 8o/o.
i:
same income? Assume
I = 87o and a period of Project
X
15 years.
Project
Y
Initial cost
$55,000
$80,000
Annual operating cost
$15,000
$10,000
$6,ooo
$4,000
$10,000
$15,000
Annual maintenance cost Salvage value at the end
of 15 years
619
20.20.
In order to
purchase a new car, imagine rhat you recendy have borrowed $15,000 from a bank that charges you according to a nominal rate of 8%o. The loan is payable in 60 months. (a) Calculate the monthly paymenb. ft) Assume the bank charges a loan fee
of 4.5o/o of the loan amount payable at the time they give you the loan. \V'hat is the effective interest rate that you actually are being charged?
.Apps@ A Summary of Formulas Discussed in the Book Traffic flowr.
3600n
q:
aYerage speed
ume
aver€e acceleration
weighe
change in velocity
:
hydrostatic pressrue:
flme buoyancy:
volume
angularspeed: ar
:
censrty
'
59
_
=
change in angular speed
flow rate
T'-
knT
Newton's law of cooling: q
q:
Ee
density of a material density of water@4"C weiehr
-------Y-
F*:;
ume
Ohm's Law; V: RI
(densiry)(volume flow rate)
elecrricalpower:
spring force (Hooke's lau): F
hq'q'
kineticenergy: LrnV2 hx
Ts)
AL
P= W I
:
-
linear exPansron, ot: TLT
AV o": vLF Coulomb's law
linearmomen*,7 = *i
bA(T,
coefficient of thermal volumetric expansion:
volume
mass : -:--
:
T,
eoAT!
coefficient of thermal
mass fow rate :
620
Fourier'slawzq=
ume
volume : -;;
-
:
32)
,
specific weiEhr
lm)t o
r('F) : z("c; = ;g("F) i7fc)) + zz (K) = rec) + 273.15 i"("R) : ryF) + 459.67
radiation:
=
mass
pVg
volume
specific gravity
pgh
;
mass
sPeclncvolume
r-)
Temperature conversion:
ror
anSrlar acceleration
Grnlrn2
L0
the relationship benveen linear and angular speed:
V=
Fs:
P:
stress-strain relation (Hooke's
fime
F:
V: *g
volume
f.ow rate :
ma
Newton's law of gravitational atraction:
distance traveled
:
) F:
Newton's second law:
T
Apprxpnr change in potential enerry
elasdcenergy
:
A,PE:
mgAh
I ::-N
power
-
(force)(distance) : work --------:ttme ume
2
efficiency
:
conservadon of mechanical eners/:
acnral outout
--.--;jrequrreo mpu[
LKE+ LPE+ LEE=0 conservation of energT-first lariv of thermodynamics:
Q_
V:
snndard deviation: s =
A,E
The Greek Alphabet
A a B p y I A 6 E e ( Z H rt @ 0 I t K r< A I Mpmu
N
v
nu
€
xi
o
omicron
delta
o T
epsilon
P
p
rho
Cf
sigma
alpha beta gamma
z,*a
pi
eta
T
T
tlu
theta
Y
U
upsilon
iota
o
0
phi
kappa
X
X
chi
lambda
v
qt
psi
o
o)
omega
orkhi
Some Useful Trigonometric Relationships Ilthagorean relationr cos
a2+F:& stn ct
:
sinB: @sa =
oPPosrre
adjacent -I'^ : hypotenuse
c cos (t
opposite
4.
adracent
b
opposite
b
sin
c
oppostte
!
^ tanp=
B cos p
b
The sine
rule
adacent hyPotenuse
c
tana =
hypotenuse
ny?orenuse
4
c
sin
abc sina sinp
aojacent
sin0
or power
:
62l
energT
ume
622
AppnNour
t F) : rittacosB * sinBcosa sin(c - F) = titt6ycosB - sinBcosc cos(a * F) = *racosB - sincsinB cos(o - F) : -r a cos B * sin cr sin B
The cosine rule:
sin(c
= 8 t c2 - 2bc(asu) F: * + * - 2ac(asB)
az
f:*+F-2ba(aso) Some other usefirl trignometqy identities:
s,s'
Q--__:_-
Rt
sin2a*cos20:1
R,
irn2a:2sinacpsa cos2c: cos2d - sin2a = 2asza - | : | - 2sirlo sin(-o) : -sin a cos(-o) : cos (t Some Useful Mathematical Relationships
l0':!
rr:3.14L59... 2rr
l
=
360 degrees
radian
-
180 ,If
:
57.2958'
'tl
1 degree
tcox*
= lg0 :
: N'*-
(x")-: x*
01174533 rad
(q,)"
to:
: *"!' I (x+
0)
x -xo-. (z\":t x-,=L tco x\l / Jt' log
= 1o*t*^
logy=x
logl=0 Iog 10 : 1 log l0A = 2 log 1000 : lo9ry: logr* logy .x l"g Logrc - Iogt ), logx" : nlogx e: 2.71828 ...
ln = logarithm to the base e (natural logarithm)
f:! lo x
lo!:* =
(ln 10)(log x)
:
2.302585 log
r
to the base 10 (common logarhhm)
Some Useful Area Formulas 1
Triangh n: ;bh
Rectanfu
A:
bh
T h
r--
F--b
3
u
I
AppsNDD( 623
Paallelagran A:
///
Cylindar A:
bh
2nRh
/h
----77
T
F_r______+l
h
i
Trqenid e: lQ + t)t
WT \ l+a
/+
RigSt circula.r cow
A:
:rrRs=
*NF
+ l?
F_U___-.1 n-sided
polEson
^
: (1) U *r(ry) Spberc A:4nfr
Circle A=
nR2
Trdpenidalrulet
a= n(f,tr* tr* 1t2* "' * Jlo-r* lnr.i") Ellipse
A
:
rab
f,-
@
Yo
624
Appnnpur
Some Useful Uolume Formulas
Cyllndn V: n<
Rigpt
ciratln cmz
v: lr&h
Seaionufaconc V=
|nfini
+
R7
+ Rt&)
Spbne v=fnF Sedronofaqberc
V:loh\*
+ 38 +
fi)
i::t-'::Iir1:r':-:f{
This page constitutes an extension ofthe copyright page. !?'e have made every effort to trace the ownership ofall copyrighted material and to secure permission from copyright holders. In the event ofany question arising as to tfie use of any material, we will be pleased to make the necessary corrections in future printings. Thanks are due to the following authors, publishers, and agents for permission to use the material indicated. 2: NASA4: Gop) @ inagel00/Corbis; (bonom) @ # 391610Index Stoc.k Imagery Inc.; (rieht) NASA 13r Michad Newman/Photo Edit 16: Charles Thatcher/Stone/Getty Images 1Z @ Virgo Productions/zefalCorbis 21: @ Masterffle 26: Saeed Moaveni 27r Tom Rosenthal/Superstock 32r @ Kevin Radford/SuperStock 33l @ Ingram Publishing/SuperStock 35r Ed lGshi/Corbis/Magma 36 ASME w\Dw.asme.org AIChE; ASCE www.asce.ors SAE www .sae.org 4L Copynght @ Nokia, 2001 60r (right) Corbis Images/Picnre Quesq (eft) Digital Vsion/Picnue Quest 62r Underwriters Iaboratories Inc. 68:2001 PhotoDisg Inc. 70: $ottom left) @ ThinkStock / SupetStock; (top left) 2001 Photodisc, Inc; (right) Corbis Images/PicturQuest 75r Saeed Mo$reni 76r vrww.howstuffrorks.com 81: @ Image Source/Corbis IOZ Lester Lefkowitz/CORBIS 105r @ # 310825 Index Stock Imagery lnc, L24 @ Roger Ressmeyer/CORBIS 126r James Sugar/Corbis 140: Saeed Moaveni 157r Saeed Moarreni 158r @ Brian McEntire/Shutterstock 174r NASA (Marshall Space Flight Cented 188: Saeed Moaveni 195: @ Geray Sweeney/ CORBIS 203: @Aaron Kohr/Shuttentock2l2: @Alan Schein Photography/CorbislMagna2l2t Roger Ressmeyer/ Corbis/Magma 218: @ epalCorbis 235: Saeed Moaveni 236: U.S. Navy photo counesy of Electric Boat Div., General Dynamics C-nry. 252: @ StocLbyte/Corbis 258: Couttoy of MTS Systems Corporation 270: liahanat Stunt School2S2r @ Master6le 301r Photo by Stephen Engler 319: @ Roger Ressmeyer/CORBIS. 341: @ Stocl Image/ SuperStock 359: Counay ofRheem Corporation 370: Tonis Valing/Shuterstock 419: NASA 460: Harry Sieplinga/ HMS Imaga/Image BanklGetty Images 498: Pasquale Sorentino / Photo Researchers, Inc 5O0r Boeing/counesy photo 509r (eft) Cowesy Canoll Shelby Enterprises; (right) Steve Allen/Brand X Pictures/Picnue Quest 510r (top right) Couneqy of Seiko Corpor*ion ofAmerica; (bottom right) Courtesy Maui Jim, Inc.; (eft) Courtesy of Utespeed; 51fr (all) 2001 PhotoDisc, lnc,5L\ (right) 2O0t PhotoDisc Inc.; (bottom left) @ Image Source/Electra Yrsion/ Picnue Quest Gop l"ft) Image Farm Inc/Picnue Quest 513r (top left) Tony Freeman/Photo Edit; (bottom left) Image Ideasllndex Stock, (righd Corbis Images/PictureQuest 515: (top right) ZOOI PhotoDisc, Inc.; ftoaom right) 2001 PhotoDisc, Inc.; (left) Stockbyte 5L6t 2001 PhotoDisc, Inc. 51& Coming Cable Sptems 530r NASA 532r Saeed Moaveni 577r BanialTanlGetty Lnaga 592 Lightscapes Photography, Inc./CORBIS/Magma
625
ABEI raaAmditation Board for Enginaing md Tirhnolotr (ABE"[) Abslute all refsene, Exal, 380-381 Abrctute hmidity, 520
Amerim Vire Gage (A\fG), 329-330 Amomt of asubstane, 128, 130 bxe (fundmental) dimmion, o a t28
Abolute
pt*m,
Absolute
thmodynmic temlmnre,
251-252
291 Abrclute m tempemnre, 289 -293 Aedemic dishonsty, engiqqing ethic
md, 112
mole (mol), SI mit of, 130 Anpete (A), bas mit of dctdel cur-
mt,320
rcational
a\glx,2L4
rrnFlc
207
linlw,207-2Q9 pmac inrclvinglmgth
od
time,
81207-209,214 Arcdiation Bo{d for Fng,nsing md Tecfnolory (ABFD , 14-22. zl9 F nginqirg
ANSI, geAmerien Narional Staadads Imtirute (ANSQ
Arbimily
fua
mmtofam
See
Ara"
Dirciplinc enginqing teclnolory
progm, 21-
aimgle" I 59 mommt of inenia, Jer S@nd me shaped
outingsques,
pre
dngoeficient, 167 qamlle of, 17I-172 foreaangover m.2&-256
tiom,98-99, Active
100
EEel,393 EEcl,393 Add Tmdllne optios, Excel, 408-4f I Aucprc mginering prcfsioa o{, 17Air quality mdads, 68 -70, 70 -72 Anaio Sciety of Haing, Refrigqatiap ud Aironditioning Engi-
nm(ASHRAE),72 ClmAirAcL 65,68-69
ontamimt managmot,
methods
L7 0
-
mbmctingumedarca
170
hating ventilation, md air@ndftioningGryAC)
sy$m,
indoor air quality
oudor,68-70
[AQ,
70
Nr,5L9-520
mmt
Almim,509
(rc), 323 -326
(ANSr),462 509
(AsrM,61-62 Amaim Sciety ofHqtin& Reftigerrak
ing' md Airooditioaing Eagiaen (ASHRAE),72
308 Code of ethis for
w
enginm, 3, I 07-l
1I
study for, 120-122 fuadammel mons, 107 importane of,l Ncional Sciety of Prcfesional F'g.(NSPE), 3, 107, 120
-122
107
profsional obliguioro, 109-1 I 1
n:ls
integnl nrle* 569 Calibraion of insuunots for tmpemture,287-288
of pmctice, 107-109
361
Col"-" hedq, Exel,373
Co-qd
Hiaory windw, MAILAB,
Gndela, defined, 130
Comd
windw, MIIILAB,
Cap*itos,333-334
Commiqtion,
Calorie
(al),
qij
ef thermal energr, 293
flow diagnms, 'oh Cemdads,63
al
8l -104.
See
421 al:o
Dw-
ings; Enginering graphici PmrPoint prenwiom; Solucioq of eo-
598
Cxlls, Fxcel, 374 -37 6, t8O - r82 absolute referene, 380-381 add$sse$ 374 -376, 380-382
ginaingprcblem detailed techniel repom, 87-90
dwings, 100-101 enginering gaphio,
d&td"374
f 00-l0l s61iys srrnm{is, 8/
imening 376 mhedrefrrence 381
memos, 87 oml prmatiom, 90
mnges,375
PwaPoint pmtadon$ 9l-1 00 pmgffineports,87 skills md prsmtadon of enginering work 82
r/r
Arithmetic operatiom, 376, 42L- 422, 425
M.qIILAB,421-422,425
slan,MAn-48,421 de
nomalistion
(ATNOR),63 ASTM, wAmericar Sciety for Tsing adMaerials (A.9Inl) Atmospheric prsure, 249-250 Atomic structure of utsials, 502 Avemge mion force X9-271
Br
graphs 580-581
Brydimmioa.g saFu&maal dimmiom Brc uits, aalntmtional System (SI)
ofuie
Bar'tsis,S2l-322
homwork prmtarion, 85-86
reladrerefeme,38l Celsiu fC), SI mit of tempenture 291-293 C*nterlina, onhographic dwings, 465466
ad vaiation, mzuurc of,,582-586 Chan \fiza!d, EEel, 389-392 Chuical mg"ering profesion of, 18 Central tmdmcy
China State
Buruof QgalityadTirh-
oiql Supwision (CSBTS), 63 Ciratts,326,528-334
Amaia Vir
,
t20-t22
relghuq,
Nadonal Smdarrls Institute
Aaeria Scierf 6r Taing ed Mffii-
Cla (c1c) ommm4 Ml|ln AB, 421 Clothi'g (clo), imulatingvalue of, 307-
Fillcomd.380-382 Imetmu,376
Asiation Fmgais
7L-72 ndoor,70-72
6, 47 6
napeoidal nrle, 167, 170, 623 us or, lo/
Fscl,376
of,72
Amsim
t72 prinitive, 167
wndmomotof, 1//-182
18
Altmting
momot of inenia lZ-182 plam, apprcximadon of, | 67,
-l
Cods, serSundarls md cod6 Coeficimt of perfomc (COP), 360-
ntqnl565-570
fomul,m for, 167, 168 -169, 622' 623 imponana of in enginqin& 163-
5
prcfesion of, 15-16 Clean Air Acr, 65, 68-69
nes
dsivarive nrla, 563 di$erential, 562-565
165-166
16
AddDataom4 Add Dataomd"
sistor,330-331
prr-ble,
defined,163
ell, Exel,373
?s3-2u,505 matrrial pmperty of, 505 ma&rial sletion m4 51
Crlqlu,562-570
rc-stional,
22 prcgram requimots, 22 Acion butmns, PwaPoint prmm-
British Stmdads Institute (BSI), 63 British thmal mit (Bnr), 29 3 - 29 4, 352 pom,lryAC uits of (Bn/h),352 thernal oagr, mit of,293-294 Brinle msial behryior, 259 Bulk modulu of omprcibility, 51,
Buoyarcy, 172-173,247
170
22 function of, 14-15
potentiomeErs, 330 rcistivity, 326, 328
dwings for,476
masrrercnt offore of 263-264
-172, ln -L 82, 2M -246, 622-623. SeeaboPtwe 163
Ohm'slm,328 WallcS,332-333
Civil engin€ing, |
unia,131-132
Animatio, PwsPoint pmutiom, 97-98 Amualmnh (AV),613,615
xenge,2O7 mple of,209
insantanon
Boaom-up nodding 478 - 479 British Gwitational (BG) sy$em of
speeds, 212-213 of,213,214
Nadonal Electric Code, 329 -r3O
wie,330-332
479,481
Amplitude, dedned,324 Aagula motion, 212-214 acleration, 214
trreler;rjon,2W-209,214
Bams, rond momots of m in, 180182 Bell shaped we, 588 Bionediel oginering, profesion of, 18 Bonds,616 Bmlm opemtiom for solid modeling
Gage (AWG), 329
330
apui,tors,333-334 defiftd,326 elecrriel porcr onsmption, 330
-
mlution of enginming problm, 84 Compsites, 518-519 Compoud intect, 600 Conpmive (ultinate) strmgth, 50,
82-
260-26r,5M,5W-508 material prcpeny of, 5O4,507-508 material slrction od, 50 mmment of fore of,'260 -26L
Compuational tmls 37 O - 459 elecronic spradsher, 372- 418
Es@\372-418 FOKTRAN,373 inuoduaion to, 371
Irronx MfrLAB,419-459 Computer numerielly ontrolled (CNC)
m6,
l)),4//
Concpmaliation, dcign prccss, 44 Conmte,512-514 Conditional statemms, MA|L-AB, 433-
Dependent vriables, 535, 562 Deprciation of value, 616 Dsimtive ruIs, 563 Dcigr prce, 42* 48, See ahoEng-
nedogdcign
elaniel md elatronic engiaeoing
476-4n
Ohm'slffi,328 parallel
ciroia, 332-333
qteroion lins, 468
photmision, 322-323
fillea,468-469 hiddm lins, 465-466 impome of in engineiag,462-
PwqomPtion,330 pwplans,323 rcidmtial
power distribution, 326,
oncptualizadon, 44 omuzinre, 47 definition of,42 dcign miablc,47 waluaion, 45
Not m Ssle (NTS)' 468
wpleof,47-48
onhogra.phic
prop€rty of, 505 marerial election m4 51
fwible olution rgi on, 47 - 48 objative function, 47-47 optimiation, 45-46
solid
t7,476-4n,486 &awngtfo\476-4n
them:l onductivity,
prGotation,46
rymbols for, 483-485, 485 - 486
profsion of, 16-17
436
el-se amrnnd" 434-436
lf qlmd"434-435
Conduction, 51, 294 -297, 505
Fouiot
Lm, 295-296
harnnsfq nd294-297 mtsial
505
problm defiaitioa ad mdeffanding, 43-44 rognizing need for a prcdud or s-
intset, engineing ethia
Conflict of
md, 112 Confl.ic rcolurion, dcign tr-rcrk,53 Comemtion of energr, 145,348-350
Comation
Cowtim
m,
145, 229 of of meclmi aI
346
347
-
Amerio Sciety of Hating Reeigaating, ud Aironditionng End-
Dm
(ASHRAX),72
moagment, mdhods of,72 lroimm lwel (MCL), 67 mitIm lwel goat (MCLG), 67 Contrarts, eqS""erirg ethic ad 112 Convation, 300-304 6f m mistane (coefi cient), 302 - 303
fored,301-302
fre
(natural), 301-302 tmnsfer oeftcients, 301-302 Convmiom, 134 -136, 291 -293, 293
hat
294,620-AL fomulx for t*.Femtue, A0-621 hat transfer md, 293-294 mrroment, mits of, L34 -136 tmpemture, mits of, 297 -293 Coordim sysreos len$h md, t54-
-
r55 Copper
ad is allgrc, 510-511
Cnyyi$s,56-57 Cosine rule, 159, 622 Cct analysis, re Amual wonh (AV);
Pmt
Futue wonh (FV);
(Pv)
Coulomb (C), SI
mit
of
worh
elmiql ru-
mt,320 Craton
tm
role
oi
52
Crcssctional ag, 165-166 Cmulatiw ft equen cy, 581 -582 Cmulatiw &equocy polygon, 581
Cmmt, reeElctric mmt md
related
Pmm
420 fiming 407- 411,450-451
Cwe
63
miablc,
5M,
506 -507
mwement of,222-223
562
dqimtive nrls,563 indepmdcrt ruiablc, 562 mte ofchange an4 562 Differential equatio n, 57 0 -57 2 Dimmion linc, dwings, 468 Dimmional homogmeity, 136-137 Dimmiom md units, 126 -151, 467-
469. See abo Dranng Fmdamoai dimmione Physiel lmrc md obsnatiomi Units dimmional homogeneity, 136137
dwtngn{467-469
ecginerirg omponou od grtm, L4t-143 findammtal dimeroiore, 12i7 -L28
nmsiel plutiom, 138-1 39 physiel lms ald obwiom,
143-
r46 signifi mt digits, 139-141 symbolic solutioro, 138-139
systm of uits, 128-133 ,o16ci"go 467_469
tolewang,467-469
qymbols for graphia, 486 49 -50, 503 Elatromotive fore (emf), 321 Elaoonic spmdshets, se Excel Elmmt by elmmt operatiom, MAf-
Drinking watu staanads, 65, 67 - 68 Dry frictional forc, 240
oM&,
MAILAB,4n,4n
48-49
mualwrti (AV),613,615
t7 9, 2(A -267, 268 -269 I
/y
forc ud,
264 -267, 268 -269 mechmiel work m4 268-269
moment of a fore md,264-267 squred, defined, 179
Dime
I0O -l 01, 462 - 497
National Smrdads Imritute
(ANSI),467
- 436
of, L45, 348-350
dastic,345-346
6st lm of thmodynamic, 348-350 gwiational potmtial, 343 Hoorer Dm, mginering maml, 367-369 mahmical, 341-369 potcndal,343-344
life<)'de ost, 616 ModGed Aqlemted Ccr Rewery Systm (MACRS),616 nominal interat nte, 603
&qn^,293-294,W-348
Pf€nt mnh (P'\r),
605
615
-606, 613 -
wwered"14l-369
wrknd"342-348 Energr efficiency mtio (EER), 360-361
Enghcring anal}6i$ 3e4, 43O,
609
613,61t-616
-
mualwonh (A\tr),615
simple intemu 599 EFective interct nte, 603-6M
dcision makiog based ot,613-616 eonomic fomulr, 609-513
Effiaency,355-361
Excel funcdom for, 384
oefficimt of pufome (COP), 360-36r dennd,355
360-
prentmrh @w),6i3-615 Enginreriog
361
symm,358-361
FWAC
futuercnh (FV'),615-616 interct-time &mr, 61 0-6 12 MAJI,AB finctiom for, 430
enerry efficiency mtio (EER),
powc plans, 355*356 Elasdc mge, defined, 238 Elauic mt md rela:ed pmeten, 128, 319 -340. See ako Ciuoia, Moton
altmting
mmt
Arcriro Vire
(rc), 323 -326
Gage
(AVG),329-
330
qpadms,333-334 ciroitr,,3'26,328-334 defined,321
direa
civil mginaning,476
elecnomotire forc (enf), 321 fun&mental dimeroion, s a, 128,
md, 100-101 dahed linc, 465-466
434
Ul -369. Sa
futue wonh (FV), 600-602,606609,615-616 inte6t, 599-600, 603- 6M
qtulinc,465-466
ommiqtion
-294,
Wienc.342-43
613
Powsm4 355-361
559
293
owrvation
@Epomd inter6t, 600 dcisiom bred on analysis,6l3-6L7 depraiation,616 effative interst rarc, 603-604 oghqirg malcis fomulas, 609-
intemal ombustion engines, 356 moton md pmps, 356-358
D:sate,
LAB,425-427 Elemos of a manix, 552 e 1 se smtercnt, MAII-{B,
abaHat naasfec Pmq \trork oreration of mahmiel 346 -347
Diret cpmion,
matrix algebn, 558-
Elmiel rai*ivity,
Energ, | 45'
Economia, 597-619
uitonmion, 134-136 Dhatmt (dc),323-326
Amsia
manrial propeny of, 5M,506-507 material selation md, 50
oL Jzu-Jzl voltage,321-323
Electriel md elecnonic engitrerhg, 16-
sh0owdiagm,598
dependent
Dwings,
D ereity, 50, 222 -223,
sls
- 467
seclllomlvi*,472-475 lins, 465 ol.id modeling 476 -482,493-497
bonds, 616
qlalu, 5 O -565
MATLAB,450-451
465-466
464
DiFermtial
Dmgoefficient, 167
Daylight wing time, 200
vim,
Diagonal maaix, 552-553
Es@|,407-4II Dched linc, onhogmphic dwings,
symbols for distdbuaon of,326-327
Economic fasmn of enginsing,
of a matrix, 557-559
Dzuschc Institut fir Nomung (DIN),
delred,
Cment directory MAIL{B, 420 Cweat dtrco ry wiadry, MAIIIB,
laden,468
Daign%iablc,47
Detenimt
Cantmirma,,67,72
reric ciroits 330-332
Ductile mterial behryior, 259 Dynmic ftiction, 240
synthcis,44-45
3n
ircmetic vis, 469-472
dsp
wce,43
reach mdprepantion,44
-232
*tgr,
4&
627
mot
(dc), 323-326
320-521
dimmion linc,468
KirchoFs
drmercioung 467-469
motors,334-337
ment lar, 324-325
eer,
27- 40. &e almSady
habir mginedng orgmiutions, involrement
with,36 gnduation plan, 37 helpfirl oroi&ntiom for, 37 -38 prepantion for, 27-40 smdying for, 31-35 tim, budgetilg of, 28-31 tmmition from high schol to ollege, 28 Enginering omponens md syrm,
t4t-r43
Enginering dcig!, 4l-8O, See aboDesign prm; Matcial dcion; Smdarrls md codc
adefot,57-67 onflic' rcolution,
ontamims,
53
metiods to manage,72
opyrights,56-57
dsignprm,42-48 mnomic Factos,48-49
INorx
528
Ea$aet'ng deign Qortinued) yaluting akmativs, 55 -56
Enginering roterials, 500-528,
See
ako
Matuial slrction; M*qial propenic ah,519-520 al'rmin"m,509
mterial *lcion, 49-51 patents,56-57 prcjc sheduling 53-54
sm@6rcr, )/-/z
mdmls
md seryie n.ks,56-57 Enginedng disciplins, 1 1, 12-14, 15-
miagmrtrshec,3T4
81-104
dcigr md 41-80 diriplinc of, 11, L2-14,15-21
Narional Smda& lrodrure (ANSD,509 aromic strumm of, 502
ompositc, 518-519
IF, 389
pre+nginuing
now( ), 384
products md
rerospae, 17-18
flua,519-521
regisoed profsional enginer (PE),
biomedial, 18 chemiel, 18 civil, 15-16 el€trical, 16-17 electronic, 16-17 enviromenal, 18-19
gEss,
2l
mrufactrring
5tc
126 electric cment md related
l1
194
rrc
md m-relatedpmffs, 128,218-235
tmpeffi
-relared
pmmete$, L28,282-318
prueto,
128,
Engineering graphics, 100 -101, 461 abo
Dwingi
-
Plonng5
Solid modeling
omuietionmd,
100-101
&awngs,462-497
'nprtane of,46L-4M solid nodeling 476 -482,493-497 symbols for, 483 - 485, 485 - 486
mels, 77 -80, 189 -L94, n 8 -27 r, 367 -369, 493 - 497,
Enginuing
526-528 Boeing
532
of stel plarc,
55r dcibel sle, 551 daisiom bcd on eonomic ualysis, 613-617 deflction of a bam,543-544 differentiat
elolu,
562
-5 65
77 omccial adrylne 493 -
497 Caterpillar 797 mining mck, 278281 cell phonc, hw they work, 7-80
atg ndpwet,367-369
enginering marcrials, 526 -528
forc,278-281 Hoover Dm,367-369
enghe, €oginffiing wel,526528 lengrh, 189-194 N* York City Wrer Tmel No, 3,
J*
189-794 solid modeling 493-497
apo nential madels, 5 46 -5 49 nequency dEtubunom, )du-)uz fundamental dimmiom ud, 127r28 integml elculu, 565-570 laminrr fluid velcit)z imide a pipe, 541 mear modeb, )5)-)J6, )4u mear spnn& >r)-)rO logarithmic models, 549-55 I
outd@r air quality slandaids, 68-70 Safe Drinking VaterAct, 65 Sufae Varer Tiernent Rule
(svTR),68 Enos, typs of in *pedmtal obsemtion,582 E*iq, L05-I22. See afu Code ofethio
dsignprm,45 -
32, 54L -546,
normal distdbution, 587 -594
372
role of, 52 418, 445 - 447.
prcbab{rty,5V-579
mlc K% oL J/J-Jla
@J1s.374-376,380-382 we fiaing 407-411 FilIomlffi4 380-382
sK6, )//-)/y stopping sight
disuna,
532, 541-543
temperatue distribution aaos
a
plain
wall,536-538 time, rcle of in, 197-198 Engineing profesi on, 2 -l 22 Accrediation Bord for Enginedng and Technologr
(ABE$,
A-n
262
@.S.),
47-48 Fibs (qptic) gl,s' 517-518 Fill ommd,380-382 Filet, dwings,468-469
Fihn rcistane (o#cimt), 302-303 Finishs, tm rcle o(,52 Fim lm of themodynmic,348-350
emn, 582 nte,210-211,224 volme,210-211
Fixed
Fls
M,224
Fluid friction, re Vrmsity Fluid matqials, 5 1 9-521 Foot (ft), U, S. Cutomaryuit oflength,
l)o
MAILAB,432
pmetm, 236-
z8l
ra,
mingover
a,244-2i6
bu& modulu of omprcibility, 263264 Caterpillar 797 miaiag t ,r"ls *Ctneerng |Wg, z/d-l6r omprshe (ultimate) strength, 261
d6^d,237 disvne, aarng at a, 2& -267 distane, acting ovo a 268-269 facmr of ufety @,S,), 262
fnaton,240-241 gaitanoml 237,242-244 See
ako
imFul*, 269-271 modulu of elasticity, 256 -259, 260 modulu of rigidity, 259-263 momertof a,?14-267 Nswton! lM, 241-244
fomulo,376-380 tunaons376-3W,387-389 imponane in ngindhg, 372-373 ioponing fila into MAfl-{B, 445447
insningells, olmm, ad rom,376
eeprepmtion for,27-40
logiel functiom, 387-389
ommon uaits of 8-12
mari: compuradom,
400 -
Hnke\lm,238,258-259
lia6
Cells: Ercel fuoaiom uithmetic operatioro, 376
slution o[,82-84
of*fety
&fuenheit fF), U.S. Cretomarymitof tmpemtue, 291-293 Fmd (F), boic uit for qpacitors, 334 Faible rclution region, dcign prre,
xtage racrion,2.69-271.
112
dsign altematim, 55-56
ErqL
5
-122
Evalvdot,45,55-56
nonlina
models,
LzO
112
tm
Factor
Fora ad forerelated
enginert med, 111-l12 enginsin& 106-109 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!, ll2-1I7, ode of, 107-111
mthematia h,532- 576
)4tJ-ttu
Exc€I, 1
lryel onaminmt goal
@ntram
Pste,383
Exponential functiorc, 385, 430
lwel ontamimt (MCL)
onfliaofinterct,
logiel,387-389
nigonomeaic 385
EPA, ree Environmqtal Protection Agqcy (EPA)
546-549,549-
togmmc, 16)
Xxmtiresrmic,87
(Ms-G),67
-57 6,
-.h ffowdiagm,598-599
onomio m4 597-619
r95-217 See
Enginering organiatiom, involvement with,36
ooling
time-rdated
mim
r97 -198, 530 -53r, 577-596,597-619
pamet6,
t52-
tmpemtm od
lroimm
Enginering papu, 85-86 Enginering prcblem, 82-84, 127 -128,
236-281
innoduction to, 12J length md length-relatd
st€I,511-512 water,520-527
pmetes,
specialiatioro of, ll, I3-L4 technologr progm, 2l -22 U.S. Buru of Iabor Satistia for I world PoPrl'iion' altr€ts of 6-8
-387,387-389 rc of, 3M
engin@irg arab{is qponential,3S5
relational opemton, 388 todsy ( ), 384
5-6
Air Act, 65, 68-69 drinking water mdads, 57-68 function of,65 indor air qualiry GACD, 70 hdmr air quality sandarls, /0-l2
rcod,514-515
-l5l
o6,
Clm
503-508
tianim,509-510
128,3t9-340 enerry md power, 341-36! fore md forcrelared llmetes,
497.
o4,
wie
Enviromental enginering prcfesion o4, 18-19 Enviromental Protation Agacy (EPA), 65,67-72
soli4 509-519
of labor Statisia for,
dimmiom md unis,
ad
l0
ofmatter, 502-503
prcpenic
\Feb sitc for, 12-13 Enginering findmenels, 124 -369
time
wel,526-
silion,5l
spaialiaion md, 13-14
128,
mgineing
plasic,5l5-516
nucla, 19-20 petloleu, 19
Buru
(water vapor), 520
lightweight neals, 509-5
phm
of Enginming
15
l5
mgn$im,510
19
materials,20-21 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, 2-3, 4-26 Ntional Sociery of Profesioual Engins (NSPE),3
@nqetg'5L2-514 opper md its allqre, 5 10-5 1 I dcign pr
Aqeditation Bqd for Enginming md Technolory (ABE$, M -24
ploning 389-399
eu6. lu)-1zz
Amerio
ukclwt,53-54 tmwork,51,52-53
omuietionmd
&6
pr6we,2U-255 sha modulu, 259 -260,263 spiae238-2r9
sn*,256 tmsile Sield) suength, 260 tjme, acl:ngqer,269-Z7l ants o[,237-238
work,268-269 Forad onvrction,
30
1-302
629
IupBx Fomat
Cellc md Font option, Exal, 401,403 f ormaL omds, Ml{fLAB, 423 Fornst Data Seie option, Exel, 397-
399
Foroulabr, Excel,373 Fomulro, elmonic qqtion of, 376Ex@I,376-380
MAruB,425
ls,
295-296
fprintf ommds,
MAILAB, 423-
424 (natunl) onvwtion, 301-302 F rqaencic, 20 L -203, 324 altemating (e) ffi nt nd, 324
Fc
ddn€d,201
andt(liot,294-297
Mrus
defr\d,293
Fouiq\ls,295-296
pre6m for,
inv
:&.emal thermal
nug,293-294 297 -)00,
sistane,
wrof,293-294
-304 Joule
pein& vatilation, md aironditioning ftIVAC) qretm, 7l -72,306,352 7
1
-72
(1,
work,
-156, 196 -198, 2t8 -221,
283-286,32fr-321 amout of a sbsane, 128,
elstriel
130
mot,
L28, 320 -321 enginming prcb.lems nd, 1 27 -128
length, 128, 152-156 lminou intemity, 128, 130 masr 128, 218-221 time, 128, 196-198
t)'ps of, 128 Fmdamenel uir, re Intmtional Sptem (SI) of mits of Enginering Pem (FE), l) Futue wonh (FI?), 600-602, 606-
Fm&rends
609,615-616 detemination of, 600-602 engineing ost malpis, 615-616
prent momt,
of 4 600-602 seria paymmt (muiry), of a 606-
Goenl Conferene onVeighs
Mm
nd
(CGMP), 128,130,320
Gla$,517-518 Global Positioning System (GPS, 157158 Grapb Prcpoty Editor, MlffL{B, 441
omuietion
see
Enginer-
irygraphio Gwitational fore, 237. qffi@.
242
-244
ztl
Nsrcdsls
of,242-244
Grl'n'ty, 2o7 -2o8, 222 -223
linw arclemtion
due to, 207 -2AB
sperfio222-223 Grek alphabet in mthemis, 533534,621
qrid omds,
MdlI-A3,440
Gmuped ftequency drsaibutiom, 580
Hadwo4 514-515 Haeprcity,51,505
mitof,269
mit
S
of ms,130,222
44r,443
qwc
Imp ontrol MAII.AB, 432whiLe qmmand,432-433 Lminou itrteffity, 128, 130
bm (fundamqal) dimeroion, tz8
uit
of,19
defined, 139
Imete6,
al:o
MsmdIwrelcedpmetm,
/ga:,Yol-
arervaion of,229-232 dewty,222-U3
Homwork presentation, 85-86
oondinre systm fo r, 154 -L55 fot (ft), U.S. Cwtonary mit o(, 156
fun&motal dimemion, c
,),
srsm
HydsJic qt'stm, 253-255
Hydrwuic
de6nod 245
Ideal gas lar, 289-290 IF function, Excel, 389
i f satmst, MAILAB, Irnponing Excel
- 206 297-300, 347 -308 2A5
Imulating materials, clothing (clo), 307-308 R-va[ue,297-3O0
sistala
m d, 297
Integal elculu, 565-570
-300
iatqnl nrla,569
role of iategmls, *,nFlc of, 565 -568 Integml rulo, 569 Intellctual property, 56
Inteml 599-600, 603-6M
ompou4
600 e$at of fteqrency of ompouding periods, 603
effectiYe rate,
603-6M
enginming mnomig m4 599-600, 603-604 nominal nre,603 nte ofbonds, 616 simple' 599 Intemal ombustion engins, efrciency
of,356
mmentsof,
157-160 meter (m), SI mit of, 155-156 New York City Vater Tmel No. 3, engineuing mmel, 189 -194 noninal vem rcmal sizes, 160-162 ndim s a mio of two lengths, f 63 mnd moment of area, 17-182 snain a a ntio of two lerytts, 163
Life
I)nq detaion,207-2O9
remge aceleration, 207 dre to gwity,207-208 insantaneou aclention, 207
Linru
equatiom, 452- 453, 538 -540, 560-561 MAII-AB, dutiom of simularou, 452-453
matrix algebm, olutiom of simultale-
ou,560-561 models, for, 538-540 system of, 539
inpulse, 269-271
Linru models, 535 -538, 538 -540 chmctqistia o{, 540 depodent ruiabla, 535 indep€ndfl t triebles, 536 linear equatiom for, 538-540 mear spmB )t)-)ro rcistivity vdus, 537 rystem of linru equtiom for, 539 ternpemtrc distribution uN a plaia
wall,536-538
Linar velmity, *enge
nte224
205
speed,
-206
205
a, 128,
218-221
kilogm,
152-156 Global Positioning Systm (GPS, 157-
Lina
frow
128,
volwe, L/2-177
- 435 md othc file into 434
MNTAB,45-U7
thermal
fin&mmul dimmioa, a a t)6
md aironditio"irg (FIVAC)
128,
2t8-235
ta,163-172,177-182
Huidiiy (wter wpor), 520 HVAC syrm, sseHaing votilaion,
a,
of, 130
ttold omm:nd, MAfLr{B, 443 Hmket lm,238,258-259 modulu of ehticity nd"258-259 springfore md 238 Horepowr (hp), U.S. Cwomary mis
r
Magnsim,510 Mmuectuing nginsing, profsion
L6gth md ldgth-relared See
433
forom4432
rza-rz)
ue
Histogm,580-581
logiel finctiom, Erel, 387-389 Ipeial operatoa MATI-IIB, 433 1oslos (x, y ) omd, MlffLAB,
mdela, SI
mrffi!
128,152-194.
Imtanmou velcity (sytd),
Gagepr*ue 251-252
Gmphiel
465-46
Independot role of, 52
mit of, 269,293-294 mer6r, mit of,293-294
Kinetic friction, 240
Lasr
Hidden linc, onhognphic dwings,
Inst
tm
- 472
La&re drtrings,468
themal omfonm4 306 mits ofpower of 352
vriabls, 536, 562 menu, Excel,376 Insmtanou rcceleradon, 207
.609 Gatherq,
360-
469
486
?61
ot Dower, ,)r,
tmpennre, 728,283-286
(EER),
385, 43o
Exel,385 Logic symbols for engineiag graphics,
,or _tq1
MOOn
152
M,{ILAB, 449
(kg), SI
205-206
MANAB,430
131
Kelvin (K), SI mit of temperoure, 130,
oerry eficiencT ratio
I4rithmic 6naio B,
SI
thmd
speed,
slar (magoitude) qmtity, 206 vaor (directional od mgniilde) qwtity,206
(CGMP), 128, 130
tiw, dwing,,
korcettrc 303
instanunou
of, 133
Invere of a muix, 561-562 lrcn, )l I
wiadchill factor, 304 -305
nd,
om4
Frj.qion fore,240-241
-128, 130,
of, 129-130
onvesion md, 293-294
Kilogm
In
28-
werydry
oeffi cimt of perfomc (COP), 360-361 efficimcl of, 359-36r
funciom Fudmental dimmiorc,
w
f
Gmenl Confeme on V'eigha md
elecnical sfstem, 202 natual,202 Frequency distributioff , 580-582 Fuels, hating value of, 312-314 Fmctiom, raeExcel finaioro; MAILAB
{
t3r,133 broe (fundumul) uia
onvecion,300-3M
air quality staoddds
campls of for
'ii ri:
Intenadonal System (SD of mits,
coefficiens, 301-302
imulating rntoials, 297-300 mdiation, 305-306 R-value,297-300
380,425
Fowier's
Hsr mnsfs,293-306
SI unit of, 130, 222
mmmentof,222 mommt of inmia, 220,224-227 mom6ntm"2Z7-229 periodic table of
chmiel dmmt6,
2t9-220 spai,frc gavity,222-723
speifrcvolmq222-223 spaificwight,223
Mm
moment of inenia, 22n,224-227 Material propeni*, J03-508 bulk modr:lu of omprsibiliry, 505 ompmion (uftimte) suength, 504,
507-508 dmity,504,5O6-507 electriol rcisiviry, 503
h*
qpacity, 505
mterial
mpls o[ 506-508
modulm of elmicity, 504, 506
moduluof rciliqe,504 modulu ofrigidity, 504, 506 modulu oftoughs,505
shs
modulm, 504, 506 snength+eweight mtio, 505 temile (yield) strength, 504, 507-5Og
thomal onduivity, 505 therual
cpmion,
wpbrpru,505
505, 508
Yisiq',505 Youagt modulu, 5M, 506 Matedal relection, 49 -51,259-263, 498 - 528, See a*o Fapjnaiag wtaials; Mcerial propertis britde behryior, 259 bulk modulu of ompmibilit',, 51, 505
ompreion strength, ompreive (ultimate) 26L,504
dmity,50,504
50
sumgth, 50,
INorx
630
Materlal x)
pm ad 49-51,499,501-
dcign 5n?
ofsfetf
(F.S.),
MAIL{B finctioro, 429 - 437 &ilt-in,429-497 mginering malys6, use for, 430
262
hat epacity, 51, 505 modulus of resilimc€, 50, 504 nodulu of elrticity, 50, 5M modulu of rigidity, 50, 259 -260, modulu of mughns, 50, 505
aigonometric, 430 eupomntial,430
5M
504 saength-toweight mio, 50 -51, 505 tmile (vield) snength, 250-262,
)w
themal onduaivitt', 51, 505 themal *pmion,51,505
npor prsue,
51, 505
visoity,5l,505 Yomg's modulus, 50, 5M Materials enginerin& profsion of, 2L
Mathemadc in engin
si\&
20-
531, 532 -
576,620-6'4
xafonalas 622-623 elslu,562-570 &padent wiablc,
535, 562
diFerential elculm, 562-565 differotial equetioro, 57 0 - 57 2
oginaing
problems,
*
models of,
53r,532-533 *ponential models, 546-549 fom;lx for,620-624 Grek alphabet in, 533-534,621 indepodent ruiabl*, 536, 562 integnl elcllu, 565 -570 lina equations, 538-540 linear models, 535-538, l0 loguithmic node.ls, 549 -5 5 1 marix algeb& 551-562 nonlina models, 5 4L -546, 546 -5 50 polynomial furctions, 541 -546 Rom maals in, 533-534
slopc,538-539 Vnbols in, 533-534
tmperature onrenion fomula,
a0-62r fomula.s, 620 uigonomoic formulm, 62L - 622
volme fomulas 624
MAII!{B, 419-459.
See
akoMfr'L},B
functiom
aithnetic opemr, 42L bxicida of,420-429
422, 425
wedning450-451 dspqrrmdg423,427 elment by elment opemtions, 425427
format @mmnds,423
f
ommmds, 423 - 424 imponingExcl and othafilc into,
445-447 eguations,
olutioro of simulta-
nsts,452-453 matrix omputatioro, 447 - 45O
mx:m opuaiorc,427-429 plodngr43S-445
nag6,424-425
lapamol,4i2-433
Modulus of rigidity, 50, 259 -263, 5M, 506 britde behryior, 259 omprcive (uhimate) snengh, 261 ductile behrior, 259 facor ofefety @.S), 262 matuial prcpeny of, 504, 506 matsial reladon od, 5O, 2@ -263
logiel operamn, 433 relational opmton, 434 ouditional memos, 433 - 436
M-file.463-437
Matri& de6oed" 400, 422, 552 Matrix algebr& 551-562 addition,553 bric defiaitions of, 551-553 dermimt of a manic 557-559
mmmtoffore
polynomial funcdonr, 541 -546
ral roos,545
sopping sight dime, 532, 541 -543 Normal disaibution, 587 -59 4 Nomal strain, 163 Nomal strs, 256 Not to Sqle (NTS), 468 now( ) fiuction, 384 NSPE, ssaNation l Society of Prcfssional Enginm Nucla enginwing prof€sion
of, 19-
20
Nuaiql
slutiom,
138
-139
Objective function, dcign
47
prm,47-
Optimiztion, daign
prc,
Oml
md prcentadom,
omuiedon
of,259-263
45-46
90
shs
Orgmim, t*m roleo4,52 Onhogmphic vim, dn liag, 4e - 4A
diagonal tutft,552-553 direct erparoion, 558 -559 elmmts of a marh, 552
modulm, 259 -260,263 temile (yield) snength, 260 -263 Modulu of toughns, 50, 505 Mole (nol), SI mit of nbsane momt,
invere ofa matrir, 561-562 equariore, solutiom of simula-
Moment, de6ne4 179
Pael
Momatof afore,264-267
fuaction, 383 Patents,56-57 Psiodic able of chemisl elmenn,219220 Pqodl207-203, 324 ahmadng (rc) mentmd 324 d€fned,201 oscillation, 202 Pmonality tnis of good mginen, 9-10 Paolm engineing prcfesion of, 19
linru
nou uing
550-551
muupu€uon,
))J-))o
multiplying a natrir bI mother, 554555 ular qmtitic, 552-553 -554 singula mtrix, 559 sqrc matrir, 552 subtraction,553
symmetdcffiix,556 mropos of a mix,556-557
uitmarix,553
Muix omputatiom, FreL 400 -
4A6
algebro 401
dmqs
of the marrix, 400 and Font option, 401,
Fomat C:lk
mrrh,
400
Marrix ompuatiom ,
imporane of in engnenng, 264 265
mememat
of,2-65-267
Momqt of inertia Ln -L82,
220, 224
227
te"
MI|II-{B, 427-
429,447-450
mplaof,447-450
invom4449
-
177-L82
w,22Q,224-227
ontiruou duty,337 duty c[de337 dciency of, 356-358 elrrtieJ,334-337
Mwing windw outside the deltop, MATLAB.42O Muhiplyiag a mrtrix by aothc marrix,
(Pa), SI
uit ofpm,246
Phag ofmacer, 502-503 Photcmision, 322-323
Phpiel lm od obr*adow L43-146 omrvation of mergr, 145 oroerwtion of
impome
of
m.
145
ir euginsing, 143-
145
limiatioro of, 147 Nwton's mnd lry of motion, 145
mnd
law of
thmodynmis, 144-
r1)
>>4->>>
o;oatiorc,427-429 Mauuity date, bonds, 616
516 Pamlel cireits, 332-333 Par value, bonds,
Pagte
intmittent duty, 337 slation of,334-336
403 sire of the
130
Momqwm,227-229 Motiwror, tem role of, 52 Motos,334-337,356-358
Plagidim' erEinedng srhie
Nmebox, Exel,373
Plam rm, approrfunatior of,
67
N*ional Elatric Code, 329 - 330 National Fire Prctecdon Asciation
Plaaic,515-516
MCLG),67
Ndonal Sciety of Profsionai E"gl-
Muim onumimtlwelgoal
(NFPA),62,329
Ma,582-586
ddned" 582 dsiatiom ftom, 583-584 gouped ftequocy distribution, 584)d)
mmtof,582-586 Mecbmiqlwork 268-X9 Memc,87
Menbr,373,420
w@,)/5
fonulu,425
lha
260
r6-$rain diagram, 257 Modulu of rsilio e, 50, 505
Mdimm @ntamimt ltrd (MCL),
{w
masial elation and, 50, 260 mmument of fore of,256-259, sfi
logrithmic,430
shru modulu, 650, 259-260, 263,
f print
50, 256-259, 260, 504,506 dasttc point,256-257 HmLe's lm, 258-259 material prcperty of, 5M, 506
45r-452
dectriel rcistivity, 49-50, 503 erginsing matsials, 500 -528
ra6c
Modulu of elasticity,
sfmbolic mathemadel
ductile behryior, 259
factor
wingrcrlspace, 424 soLve onmrnd,452-453
MAn-48.420
nen (NSPE), 3, 107, 120 -122 Nmon's lzm, 145, 241-244 6nt lry of notion, 241 gwitaaon, lm of, 242 -244 obrevadoamd, 145 mnd [s of motion, L45, 241-242 third lav of motion. 242 Nominal intemt mte. 603 Nominal vmu actual sizs, 160-162 Nonlinw modcls, 532, 54L-546, 546 ))l
Metabolic nte, 306-308 Mets (m), SI mit oflil$h, 130,155156
mling
M-fiIe,Mfrj/'3,463-437
deflertion of e 5 43 -5 44 sponotial models, 546-549 laminrr fuid velcity imide a pipe,
Mining engirsing, profcion of, 20 tem (MACPS),616
platc,546-549,549-
dcibelsqle,55l
Rwry
Sys-
541
loguithmic models, 549-551
I 12 16T, 17 0
1aa
plot (
)
onmd
MAILAB,
-
4l!-
MO Ploning with Xxel, t89-399, 4O7-4Ll
AddDaraom4393
Add Tmdllne optio$, 408-41
1
chat q'pes, relection o{,389-390
Chnt.wral4389-392
we
fining 407-411 Fomat Data Seie opdot, 397 -399 prcdue of,, 389-393 two sts of data with differot mgc, 393-399 XY (Sena) plots, 389
Plottingwith MAILAB, 438- 445,
45r
55r
bw,
Mixed cell referme, Exel, 381
ModifiedAclemed Colt
of steel
md
450
-
mfining,450-451 Gmph PrcparyEditor,44l
grrid ommds,440
Holdom4443 line pmpenies 439
loql-og (x,y)
@mtaa^d, ULA43
Iupsx pLot(
) @mmd,439-rg0
semilog (x,y) omnaa4 443
sfmbol prcpenic,439
tirle
(
'texr'
)
omd,
440
of
prrue,
U.S, Cts-
oi 355-356
323, 326, 327 , 330, 341-369 . alsoFnergr
See
wdad
330
587
od pmps, 356-358
p1nts,323,355-356 rsidential elecriql distribution,. 5Zt
3X'
uis of,351-31 mrkmd,350-355
)//-)2b
ald l12
r
a
ratio
oftrc
lengrhs, 153
mrg
Ran-kine CR), U.S.
Cutomrymitof
,4*;min&99 slide l4rcuts, 93-94 slide mrter,94 slide trmition, 96-97
Reinforced onoete, 513 Relraional opeoon, 388,
Relative
Prerngineti!& defiftd,
14 (P!7), @5 * 606, 613 - 615
dcmiqation of, 605-606
enginering os aulyris, 613 future mout, of a, 605
61
5
wria paymut (muiry),of u605606
Prenation, dsign proces, 45 ltsmtatiom, ge Crmmication
fum, 244 -255, PrM
See
a*oYapr
atulute25l-252
ell refmce,
Raso.iry,3?n'328,5n7 de6ned" 326
elatriql cirqia md,
rmm,25l-252 vapr,252-253
-300, 302 -303
Drinking
Warer
Selar qmdry, 206,
Act, 65 400, 422, 552,
553-554
Erel matrix omputatiom, 400 MAILAB mix omputatioos, 422 @trE dgeDra, >>2, >>5-))+ mr:ftipliccion of, 553-554 time, a a phpiel prcperty of, 206 Srundmommtof m,177-182
bem, mmpls
of in, 180-182
akoBn-
(EPA)
Amaim
National Standtrds Insdtute
terials
(ASIM,6l-62 FmFis & nomalistim
(ATNOR),63
599
British Stadrds Instituts (BSt), 63 C-e 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, 160-162
Deutschc Initut fiu Nomung @IN), 63
139
l4I
Silion,51 Simple
intact,
Sine
Slidc, PwuPoint pwntarioro, 93-97, 97-98, IOO mimation of, 97-98 imting ftom otha PmrPoint 6lc,
mcta,94 of, 131
Solid linc, onhogra.phic dwings, 465 Solid marerials, 509-5 19 Solid modeling, 476 -482,493-497 Bcing 777 omerciat airylane, engi-
nuingmd493-497 Bmlru
openriom for, 479, 481 bonom-up, 478-479
onputer nmaielly onrolled (CNC) mchines 47 imporane of in enginedng
42-
84. 138-138, 197-198. See aho Engineing Problem; Linear
muatiom malFis of,83 defining the prcblm, 82-83
mple ofpmtation
fur,84
ronoiel,
138-139 pametric fom' 83
simpli$ing the prcblm, 83 197-198
of, 82-83 symbolic, 138-139 time rcle of in, 197-198 steps
(EPL),65,6712 of U.S., 65-67
mmpls
(NFPA),62 D@Afor,58-60 orymiations for, 60 -62, 0-64 outdmr air quality, 68-70 Star
Un&mites laboruoris (lL), 62 United Sates, 60-62, 65-67 \ff'eb sitc for ogmizriorcof,63-64 Staristics, 577-596.
br
See
abo
Ptobabitty
graphc 580-581
bxic idq of,579 mulative freqwcy,
58
1
-582
daapoints,579
sftwrefor,476-477 top-dm,479-481 Solution of enginuing prcblm, 82-
steadyprre,
Envimmental Pmtation Agency
intemational, 62-64 National Fite Protection Asociaion
tnnsition,96-97
mit
drinkingmer, 67-68
ndor air qvality,T0-72 Intemtional Orgmiation for Sur
primitirc, 479-480
in mafiemcic, 533-
534 Rotational (angulgr) spds 2L2 -2L3 Rryhads, Ercel,373
Safe
m*vemqt of,2M-245
digits, 139-141 addition md rubtraction nrlec 171
478
Rsisos,330-331
Rom nmenls
fore acting owr m xe. 2M -255 gnge25L-252
,PueJttry,246-247
3?5, 32A
yalua,328,537
R-vaJae, 297
hydrcstario 245 hfdnulic systms, 253-255 imponane of in engineiag 245
Exc€l, 38t
Relatiw frequmcy, 578 -579 Relative huidiry, 520 Raidotial powa dimibudon, 326, 327
See
Agsncy
Asociation
Sofwood,,514-515
4r4
mdods,57-72. vircmmul Prctetion
Sandads
Signifiat
Slug, BI
atmospheric 249-250 buayarcy,247
mits of, 246
(PE),
Ex€I,388 MATLAB,434
visual optiom, 92 oncrete, 513
of
units
printing, 100
I)
Pmt
SI unis, saalntemational Systen (SI)
layoua,93-94
tmpentue, 291-293 Regisered prcfesional engins
tmplares,92-93
Shs$rsg256
94-96
F^eL375-376 MdII B,424-425
Spingforca 238-239 Sqwe matdx, 552 Smdard dwiarion, 5M -586 Sand*disigen I Ssige (SIS), 63
(ANSI),60 Amcim Sciety for Tbting md Ma-
multipliedon od division rule6'
i411
printingslidc, hmdous, md notc p€es, 100
Su
lstout,
bcic idm for 9l 94-96
?-63.
a&aModulu of rigidity matsial dation an4 50 mrroment of fore of,259-?.60, 263
582 Rmgs, mion of, 375-376,44-
inaning slida 6om otha PmetPoint
for,
ko.lect schedulhg 53 -54 Pmps, efficimcy of, 356 -358 Pyhagorro trigonomenic relationships, 621
Raldom
mimarion,9T-96
Pmt mrth
enginshg 5-6 Sha modulus, 50, 259 -260,
Radiarion, 30J-306
100
need
43
Prcgrs reports,87
Radim
PowqPoint pmtatioaxo 9l-1 00
prc, rognizirg
dcign
-59 4
raExcl
Spmdshees,
Swie,5-6,43
ethis
Speificwlure,222-223 Spaificrcighc 223 Sped sreVelcity
smilog (x,y) omd,MAfLAB,43
dcigrr pm, regnizing need br, 43 Profesiond respomibiliry, mgineting
367-369
198
vaifying raultr, 83 solve ommard, MAII-AB' 452- 453 Solm, tm mleof, 52 Spedficgwity,222-223 Spaific het, 310-312
Sqie ciruits, 330-332
5-6,43 enginering 5-6
HVAC applietions, 352, 358 -361 intemal ombustion engins, 356
files,
Smndary ell battery 321 Sctional vim, dwing 472-475
Prcducts,
mergrmd,ll-369 HovcDm, engincingmel,
rdon butom,98-99,
St:rime
nomal dimibution, 589-
staisnG mc,
d€fined, 341, 350-351 efficiency, 355-361
motors
Son4 defined 130
mdom,578 reladw fieqmcy, 578-579
Psq,
180
notes paga
defined,5578 disuibution, 587-594
nomal dimibution, outome,578
246
dwtdel,323
elatriel omumption,
ad
)/u-)/d we,
Prmr plarc,323, 355-356 efficiency
15
100 Prcbability, J77 -596. See ako b-Nc rds ot, bell sheped 588
Porotiometm,330
mit
Principlc and Pracrie of Enginering
&omPmPoint,
Polynomial fu nctiom, 5 4L -546 Population, satistiel data poinc, 579
tomary
distane squared, 179 inpome of in enginefug,177178 reeangle, 180
Itinting slida, hildoucs,
Polmm,515
Poudspasqweinch (ps},
cirde
Primitivem, 167 Itimitives, slid modeling Fsm,
charts,438-441 Plmbing md piping s)'mbols for mginering graphic' 486
ry
umtady (tramient) procsss, 197-
Pdmry ell battery, 321 of, 479-48O
631
6xed ercr, 582 ftequency disuibutions, 580
hiaogm,580-581
-582
mm,582*586
nmurc
of
mtral
tendency and
uri-
arion, 582-586
nomal distribution, 587 -59 4 population,579 probabilty nd,5V-579 mdom mors, 582 sandad deviadon, 584-586 stardad nonnal disnibution, 589592
ststmgric mos, 582
wime,584
Smbu,Prcl,373 Stady processs, 197-198
Ixorx
632
mterial slection
Steel,511-512
Stain o
a ratio
oftrc
lengths, 163 raio, 50-51, 505
Strength-to-weigfrt
Strcs, mauement of
loreof,256
m ooss
md 3l-35
Thpaidal
prepamtion,35
snrdy groups, 34 -35 d6e, budg€ting of, 28-31 Symbol prcpatic, MAI'L{B ploa,439 Symbolic mathematiol opentiom,
M TLAB,45r-452
Syrabolic solutions, 138-139 Symbols for e"ginaning gra.phia,
483-
485,485-486
nrle, 167, 170, 623 Trigonometric ftnctiorc, 385, 430, 62L622
50
Thermd omfort, 306-308 Thermal onductivity, sae Crnducrion
onvmiom
FLin& 33_34
o4,
See
hattru{s nd"293-294 mehmiel oerry, as 347-348 Thmal qpmion, 51, 3O9 - 310,
wmplc of,485-486 of, 483-485
logio 486 plumbing md pipiry, 486 Synmetric matdx, 556 Syntheis, dcign plm, 44-45 Systematic mon, 582
mpemrm-rdaed prcpaq ,00,302-304
of, 297
-
ffind lry of, 144-
l+) Thmosets,515 Time md timedated pmeters, 128, 195 -217, 269 -27 1, See abo Arcs)-
Tqmwork, 51, 52-53 ommon naita ofgwd 52-53 onfict rsolution, 53
emtion;
roles
in, 52
Tirhniel repors,
detailed, 87-90
absolute
287
288
-
anduclgon,294-297 onvatioa, 300-304 @nvesions, 29 1-293, 293 -294, 620-621 dtffsence,293-306 fuels, hating valm of, 312-314 fun&mmtal dimmion, r a. 128, 283-286 han uauGr,293-306
idel gs lm, 289-290 meswmmt of, 286-293 menbolic rate, 306-308 mdiatioa, 305-306 specific thermal
ha, 310-312 omfort, 306-308
thesml aag,,293-294
spmion, 309 -31 0 thennal mistance 297 - 30O, 302- 303 thermal
tursof,291-293 windchill factor, 3M-305 Tlnplats, PwsPoint pmu.tiom, 92-93
Tmsile (vield) strength, 50, 260-263,
504,507-508 mataial prcpeny of, 504,507-508
(UL), 62
Unit Eanix, 553 Units, 128-133, 134 -136, 1 55 -156, r 67, 17 4, 222, 237 -238, 246, 29t -29r, 295 -29 4, 520 -r27,
35r-354
ta"167 ql$m
of,
tlL-t32
of, 174 mte, 210-211
I/ats,520-521
'Watts (W), SI mits of
pw*,
351-352
enginming disciplinc, 1 2-13 orymiaiom of enginuing setd{ds ud coda, 63-64
eletrial,32-0-321 werydayreof, 133
fow237-238
Wei&u50-51,223
Gmeml Confaene on V'eights md Mrua (CGMP), 128, 130, 320
lrc222
matcial *laion m4 @fio?23
50-51
strength-to-weight mio, 50 -5
I
whi le ommend, M{ILAB, 432- 433
'\Findchill factor, 3M -305
\f@d,514-515
pwu,35l-354
ofin. 197-198
sprifrc,222-223
mis
Vebsits, 12-13,63-&
134-136
transfer, 293-294 Intemtional Systm (SI) of, f 28-131 length, 155-156
enginoing problm md dutionc
importane ofin enginerin& L72t73 pmeta of leagth, u a" L72-177
Volme flw
bass 130,320 British Gwitarional @G)
t/Z-tt7
eqnprc O\ | /O-t / / fonrias fot 175, 624
sfength
hm
mgular motion, 212-214 daylightwin& 200 role
Tinpemtue md tmperaturerelated panmeter, 128, 282-318, 620 - 62L
rtrmod)'nmic, 291 abslatererc,289-293 elibmtion of irotmen a for,
Angulr motionr Velcity
rgjeradon,2A7-2O9,214
dcigl tam, 51 infl uenc of perfomane of, 52 -53
DUA|aAOJ,
d&sell,L72
dedve4 132
Themoplasie,5l5 Tck chan for deigu process, 53-54 Tamrcrker, tro deof,S2
wrks,52L-323 Yolwe, 17 2 -177, 222 -223, 624. See almBuqncy
Undmitm labomoric
onmioq
321
powaplans,323
U.S, Cmomary mits, 132-133 Ultimate strogth, aa Comprsive
pmp€ny o[, 505, 50E matsial wlection m4 51 prcpeny of, 309310
fore (emf),
photmision, 322-323
505,
mtdid
Themodyumics,
electmmotive
sine nrle, 621
rcfrcimcof,508
pmatiom,
Yolege,32l-323
battsc,32l-322
Su)
Pythagora relatiomhipa, 621
508
q?
rxel-
M.{n-rB,€0
293-294
Visul optiom, PwsPoiot
cioe mle,622 formlllx fot 621-622
Themal rcistane, 297 - 300, 302 -304 onvation, 302-304 6f - rsistalce (coefficient), 302-303 imulating @terials, 297-300 R-valte, 297 -300, 302 - 303
elctriql,486 inponane
a4
of fore of, 260 -261
Thaml wtgr, 293-294,347-348. alr Hat ransfs
Strss-strain diagm, 257 Study habin, 28-31, 31-35 daily prepandon
mauemot
Worb 268 -269,
321
-326,
342
-
3
48,
flwof nafrc,203-205 forc actingovu, 269-271
prsue,246 systm of, 128-133
owtim
frequencis, 201-203 findmmal dimemiotr, d a, 128,
tmperuure, 291-293 U,S, Cutomary, 132-133
350 elctromotive
forc (mf),
msurmnt
of fore oe 268 -269
rclme
196-198
lina inpulse md, 269-271 mmmentof, 198-201
pmeten
involving lmgth
Umady
m4 205-
psiods,20l-203 sp€eds,
212-213
velxity,205-206 volme fw rare, 210-211
anc, 199-200,201
TimingPmPointprmtation i"& 99
rehw
509-510 title ('Eext') @m4M,{ILAB,440 Tide bu, Exel, 373 todey ( ) function, 384 Titanim,
Ton of reftigmtion (mling), U.S. tomq'mit ofPom' 352 Tml bunom, Exel,373 Topdom modeling 479 - 48 I Torqre, saeMomot of a fore
Cu-
TndmLs ad sqvia m"b, 56-57 Tmfrc flw, 203-205, 620
fomulr
for, 620 timerelared pameten of, 203 -205 Tmpore of a matrix, 556-557
of enugl
m4
qryrnd,342-348
174
(transimt) procesc, 197-
145,
348-
321
mshmel268-269
198
2LL
rcational (angular)
348-350,350-355
Vuu,preueo a,25I-252 Vapor prm, 5 1, 252 -253, 5O5 mataial prcperty ofi !0J matsial seletion u4 51
msuremot Vaiucg584
of
fore
of, 252
pmo4350-355 volrage
x,32L-326
\Forkbook, P-el,374
Vorlshea, Exel,
373, 374
d&ned.374
-253
naningr 3T4 tahn,373
Vrctorgmtity, 206, 4OA, 422"552 F"el madx omputationq 400
'Vorld populuion, a$cs of engineriag prcfesion by, 6-8
MAIL{B madx ompvp'dore,422 matsalgebn,552,553-554 time, a
a
phpiel
prcpeny of, 206
Ydcity,205-2o6,212-213 rcnge speed, 205 imtantanou sped 205-206
lnw,205-206 panrn€ter involving logth and time
u
a"
205-206,212-213
rotarional (anguk) Aeeds, 212-213
Ywsity, 5 l,
241 -241, 5 05 material property of, J05 marerial slction m4 51 masummt of fore of, 240
visou
(duid) friction, 240
-241
XY (Setter) plots, Exel, 389 17 chara, M.{IL{B, 438 - 441 Yrcld streng&, re Teuile strengch Young's modulu, 50, 257, 504, 506
CONVERSION EACTORS
Quantity
L.ogth
SI + 1
I
U.S. Customary
: I ft: in.
0.39370in.
cm
0.0328
ft
1 mm2 1 mm2
:
I
= 1.07F4 fe
cm2
0.L55
: 1 m2 :
= =
: 1 m3 :
1 m3
3.5315E-5
ff
2.402E-6ina
= II5.86IE-L2i{ = 24.025F.4lri.4 1.1586E-6 ft4
1m4:
2.40251F;6'L#
:
1in3
115.86 ft4
645.16 mr# 92,903 mmz
6.4516 crtf 929.03 cmz
:
mnf
ft' =
1
: 1 fii:
1in3
:
I t€ =
:
1 m4
0.3048 m
nl
35.315 fC
1mm4
30.48 cm
:0.0929
I in3
1mm4:
1 cm4
#
I
61024in3
(length)a
2.54 cm
1in2 = 6.4516V4m2
O.061024 in3
Second Moment ofArea
1cm4
1 in2
= 6.lO24E4i# : 3.5315E-B f€
1 mm3
1 cm3
: 7 f& :
#
1550
1mm3
7 cm3
ft
i* 10.76 f{
1 m2
Volume
I.07648-5
1 cm2
ft: I ir] : L f? : 1
1.558-3inz
= :
SI
in.:0.0254m
1m=39.3700in. 1m= 3.28ft Area
-+
Customary
I in. : 25.4 mm 1ft = 304.8 mm
mm = 0.03937 rn, mm : 0.00328 ft
: 1 cm :
1
U.S.
16387
28.317E6 mm3 16.387 cm3
28317
cm3
I.6387E-5 m3 0.028317 rr'f
1in4: 4l6.23lL3mr# I
f{ :
1
in4
L
fta
1 in4 L
:
8.63097F9 mnf 41.623 cma
:863110
:
cma
4L6.2318-9 ma
f{ = 8.631E-3
ma
CONVERSION FACTORS (Continued) Quantity Mass
SI -) U.S. Customanr lkg= 6A.5Z1E-3 slug lkg:2.20461b^
:
U.S.
Custonary
->
SI
I slug: 14.593kg 1lb. : 0.4536k9
slug/ff 1kg/m3 :0.062481b^lff
I slug/fii :5L5.7kglm3 1
lb./ ft'
Force
lN
I
lbu: 4.44t
Moment
1N.m:
Density
1
kg/m3
0.001938
= 224.8098-3Lb, 8.851in.lb
1 N.m :0.7376ft.|b
Pressure, Stress, Modulus of Elasticity,
Modulus ofRigidity
\F"rk, Energy Power
I Pa : 745.0377E-6 lb I in2 I Pa: 20.885E-3lblft2 1 kPa: 145.03778-6ksi
:
0.7375 ft'lb 1k\f/h :3.47214E38nr
1J
: : 1k\7 rW
l in.lb:
16.018 kg/m3
*
0.113 N.m
: 1.356 N.m I lblin2 : 6.8947E3 Pa ft.lb
1
Ilblff
:47.880Pa
l ksi: 1 1
6.8947831*a
ft.lb:
1.35581
Btu:
293.071E-6kVh
:
\7
0.7375 ft.lbA
1
ft.lb/s
3.4L2l4E3Brulh
1
Bru/h :293.07E4kV
1kV = 1.34lhp Temperature
:
"c=2"p(-32)
t
hP
:
'r :2'c(
1.3558
0.7457
+
3z)
kV
