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C O N T E N T S

Introduction CHAPTER

1

vii

Mathematics Review PROBLEM S

1

SO LUTIO N S

3

R ECO M M EN D ED REFER ENCES

C H A P T E R

2

3

PROBLEM S

15

SO LUTIO N S

19

4

PROBLEM S

37

SOLUTIO N S

45

5

PROBLEM S

80

SO LU TIO N S

88

6

PROBLEM S

116

SO LUTIO N S

120

7

113

115

141

Per-Unit M ethods and Calculations PROBLEM S

143

SO LUTIO N S

149

R ECO M M EN D ED REFER EN CES

C H A P T E R

79

Balanced Three-P hase Circuits

R ECO M M EN D ED REFER EN CES

C H A P T E R

77

Single-Phase A C Circuits

R ECO M M EN D ED REFER EN CES

C H A P T E R

35

Basic Circuit Concepts and D C Circuit Analysis

R ECO M M EN D ED REFER EN CES

C H A P T E R

14

Electric and Magnetic Fields 15

R ECO M M EN D ED REFER ENCES

C H A P T E R

1

Transformers PROBLEM S

171

SO LUTIO N S

176

170

171

R ECO M M EN D ED REFER EN CES

196

143

37

iv

I

Contents

C H A P T E R

8

Transmission and Distribution Lines PROBLEM S

197

SOLUTIO N S

202

RECO M M EN D ED REFER EN CES

C H A P T E R

9

CHAPTER

10

11

PROBLEM S

218

SOLUTIO N S

222

PROBLEM S

252

SO LUTIO N S

255

R ECO M M EN D ED REFER EN CES

276

Power Flow Study

277

PROBLEM S

277

SO LUTIO N S

285

12

13

PROBLEM S

305

SO LUTIO N S

310

14

PROBLEM S

333

SO LUTIO N S

339

15

332

371

Power Quality and Power Electronics PROBLEM S

373

SO LU TIO N S

379

RECO M M EN D ED REFER EN CES

C H A P T E R

304

Generator Transient Behavior, Short-Circuit Study, and Power System Protection 333

RECO M M EN D ED REFER EN CES

C H A P T E R

251

Symmetrical Com ponents and Unbalance Factors 305

R ECO M M EN D ED REFER EN CES

C H A P T E R

217

250

Alternating-Current Machines

R ECO M M EN D ED REFER EN CES

C H A P T E R

216

Direct-Current Machines and Machine Basics

R ECO M M EN D ED REFER EN CES

C H A P T E R

197

393

Measurement, Reliability, and Lighting PROBLEM S

395

SO LUTIO N S

398

373

395

Contents

C H A P T E R

C H A P T E R

16

17

Codes, Standards, and Safety PROBLEM S

411

SO LUTIO N S

413

Engineering Econom ics PROBLEM S

419

SO LU TIO N S

423

R ECO M M EN D ED REFER ENCES

APPENDIX

Interest Tables

436

437

419

411

Introduction

vii

Introduction OUTLINE HOW TO USE THIS BOOK VII BECOMING A PROFESSIONAL ENGINEER VIII Education ■ Fundamentals of Engineering (FE/EIT) Exam a Experience a Professional Engineer Exam

ELECTRICAL ENGINEERING PROFESSIONAL ENGINEER EXAM

IX Examination Development ■ Examination Structure a Exam Dates a Exam

Procedure * Preparing for and Taking the Exam ■ Exam Day Preparations a What to Take to the Exam a Examination Scoring and Results

ACKNOWLEDGMENTS XIV

H O W TO U SE TH IS B O O K PE Power Electrical Engineering Problems & Solutions and its companion texts form a three-step approach to preparing for the Principles and Practice of Engineering (PE) exam: PE Power Electrical Engineering License Review Study Guide contains conceptual review of topics for the electrical and electronics subdiscipline of the PE exam, including key terms, equations, analytical methods, and reference data. In addition to reviewing concepts, the book provides solved examples to help you apply your understanding of equations and techniques as you proceed. Because it does not contain exam-style problems and solutions, the book can be brought into the open-book PE exam as one of your references. PE Power Electrical Engineering Problems & Solutions provides problems for you to solve to test your understanding of concepts and techniques. Ideally, you should solve these problems after completing your conceptual review. Then, compare your solutions to the detailed solutions provided, to get a sense of how well you have mastered the content and what topics you may want to review further. ■

Chapters 1-3 review some very fundamental material. While you are unlikely to see direct questions on these topics on the PE exam, you may need to use them in solving more involved problems. Use these chapters to confirm your understanding of foundation material. Chapters 4-17 focus on exam-specific topics. The questions are openended, rather than multiple-choice as on the PE exam. This question style is intended to give you a rigorous opportunity to test your mastery of concepts and analytical skills. Finally, we recommend that you pair this book with the PE Power Electrical & Electronics Engineering Sample Exam to optimize your preparation. The sample exam provides exam-style questions to give you practice in problem

viii

Introduction

solving. These practice problems test your mastery of exam topics and simulate what you will experience in the actual exam, including the tight time constraints. On the actual PE exam, you will have an average of six minutes per problem, so practicing your pace, as well as problem-solving skills, is important. Take the sample exam after you’re satisfied with your review of concepts and problem-solving techniques, to test your readiness for the real exam.

B E C O M IN G A P R O F E S S IO N A L EN G IN EER To achieve registration as a professional engineer there are four distinct steps: (1) education, (2) the Fundamentals of Engineering/Engineer-In-Training (FE/EIT) exam, (3) professional experience, and (4) the professional engineer (PE) exam, more formally known as the Principles and Practice of Engineering exam. These steps are described in the following sections.

Education The obvious appropriate education is a BS degree in electrical engineering from an ABET-accredited college or university. This is not an absolute requirement. Alternative, but less acceptable, education is a BS degree in something other than electrical engineering, or a degree from a nonaccredited institution, or four years of education but no degree.

Fundamentals of Engineering (FE/EIT) Exam Most people are required to take and pass this eight-hour multiple-choice examination. Different states call it by different names (Fundamentals of Engineering or Intern Engineer), but the exam is the same in all states. It is prepared and graded by the National Council of Examiners for Engineering and Surveying (NCEES). Review materials for this exam are found in other Kaplan AE books, such as Fundamentals of Engineering FE/EIT Exam Preparation.

Experience Typically, one must have several years of acceptable experience before being per­ mitted to take the Professional Engineer exam—the duration varies by state and you should check with your state licensing board for details. Both the length and character of the experience will be examined.

Professional Engineer Exam The second national exam is called Principles and Practice of Engineering by NCEES, but just about everyone else calls it the professional engineer exam. All states, plus Guam, the District of Columbia, and Puerto Rico, use the same NCEES exam.

Introduction

ix

ELECTRICAL EN G IN EERIN G P R O FE S SIO N A L EN G IN EER EXAM The reason for passing laws regulating the practice of electrical engineering is to protect the public from incompetent practitioners. Most states require engineers working on projects involving public safety to be registered, or to work under the supervision of a registered professional engineer. In addition, many private companies encourage or require engineers in their employ to pursue registration as a matter of professional development. Engineers in private practice who wish to consult or serve as expert witnesses typically also must be registered. There is no national registration law; registration is based on individual state laws and is administered by boards of registration in each of the states. You can find a list of contact information for and links to the various state boards of registration at the Kaplan AE Web site: www.kaplanengineering.com. This list also shows the exam registration deadline for each state.

Examination Development Initially the states wrote their own examinations, but beginning in 1966 the NCEES took over the task for some of the states. Now the NCEES exams are used by all states. This greatly eases the ability of an engineer to move from one state to another and achieve registration in the new state. The development of the engineering exams is the responsibility of the NCEES Committee on Examinations for Professional Engineers. The committee is composed of people from industry, consulting, and education, plus consultants and subject matter experts. The starting point for the exam is a task analysis survey, which NCEES does at roughly five- to ten-year intervals. People in industry, consulting, and education are surveyed to determine what electrical engineers do and what knowledge is needed. From this, NCEES develops what it calls a “matrix of knowledge” that forms the basis for the exam structure described in the next section. The actual exam questions are prepared by the NCEES committee members, subject matter experts, and other volunteers. All people participating must hold professional registration. Using workshop meetings and correspondence by mail, the questions are written and circulated for review. Although based on an understanding of engineering fundamentals, the problems require the application of practical professional judgment and insight.

Examination Structure The Electrical and Computer Engineering PE exam is an eight-hour exam consisting of 40 morning questions and 40 afternoon questions. All questions are multiplechoice format, each with four possible answer choices. As of April 2009, exam candidates must select one of three subdiscipline topics for their exam at the time they register with their state board: (1) Power, (2) Electrical and Electronics, or (3) Computer. Table 1 summarizes the topics covered in the Power examination and their relative emphasis as specified by NCEES. For more information about the topics and subtopics on the exam, visit the NCEES Web site at www.ncees.org.

Exam Dates NCEES prepares professional engineer exams for use on a Friday in April and October of each year. Some state boards administer the exam twice a year in their state, whereas others offer the exam once a year. The scheduled exam dates can be found on the NCEES Web site at www.ncees.org. People seeking to take a particular exam must apply to their state board several months in advance. Ta b le 1 PE Power Exam Topics

General Power Engineering (30%) Measurement and Instrumentation (7.5%) Special Applications (10%) Codes and Standards (12.5%)

Circuit Analysis (20%) Analysis (11%) Devices and Power Electronic Circuits (9%)

Rotating Machines and Electromagnetic Devices (20%) Rotating Machines (12.5%) Electromagnetic Devices (7.5%)

Transmission and Distribution (High, Medium, and Low Voltage) (30%) System Analysis (12.5%) Power System Performance (7.5%) Protection (10%)

Exam Procedure Before the morning four-hour session begins, proctors will pass out an exam book­ let, answer sheet, and mechanical pencil with eraser to each examinee. Candidates are not permitted to bring their own writing instruments or erasers into the exam. If you need an additional pencil during the exam, a proctor will supply one. Fill in the answer bubbles neatly and completely. Questions with two or more bubbles filled in will be marked as incorrect, so if you decide to change an answer, be sure to erase your original answer completely. The afternoon session will begin following a one-hour lunch break. In both the morning and afternoon sessions, if you finish more than 15 minutes early you may turn in your booklet and answer sheet and leave. In the last 15 minutes, however, you must remain to the end of the exam in order to ensure a quiet environment for those still working and an orderly collection of materials.

Preparing for and Taking the Exam Give yourself time to prepare for the exam in a calm and unhurried way. Many candidates like to begin several months before the actual exam. Target a number of hours per day or week that you will study, and reserve blocks of time for doing so. Creating a review schedule on a topic-by-topic basis is a good idea. Remember to allow time for both reviewing concepts and solving practice problems. In addition to review work that you do on your own, you may want to join a study group or take a review course. A group study environment might help you stay committed to a study plan and schedule. Group members can create additional practice problems for one another and share tips and tricks.

Introduction

xi

You may want to prioritize the time you spend reviewing specific topics accord­ ing to their relative weight on the exam, as identified by NCEES, or by your areas of relative strength and weakness. When you find a problem or topic during your study that needs more attention, make a note of it or look it up immediately. Don’t wait; you may forget to address something important. People familiar with the psychology of exam taking have several suggestions for people as they prepare to take an exam. 1. Exam taking involves, really, two skills. One is the skill of illustrating knowledge that you know. The other is the skill of exam taking. The first may be enhanced by a systematic review of the technical material. Exam-taking skills, on the other hand, may be improved by practice with similar problems presented in the exam format. 2. Since there is no deduction for guessing on the multiple-choice problems, answers should be given for all of them. Even when guessing, a logical approach is to attempt to first eliminate one or two of the four alternatives. If this can be done, the chance of selecting a correct answer obviously improves. 3. Problem statements may seem confusing at first glance. They may be designed to see if you can isolate critical data from extraneous information and make good typical assumptions. Read the problems carefully, make good assumptions, and select the best answer. 4. Plan ahead with a strategy for working through the exam. For example, you may want to take a two-pass approach. In the first pass, answer those problems that you find you can solve quickly and easily with reasonable certainty. Mark or list these problems so you don’t waste time looking at them again. Then, make a second pass to address those problems you find especially challenging or about which you feel uncertain. During the test, a busy hour or two away from a difficult problem might allow you a fresh insight into the solution. Even if you have answered every question before the exam session ends, do not walk out early. Use this time to go back and check your work, as well as revisit any problems about which you are still uncertain. 5. Read all four multiple-choice answers before making a selection. An answer in a multiple-choice question is sometimes a plausible decoy—not the best answer. 6. Read all four multiple-choice answers before making a selection. An answer in a multiple-choice question is sometimes a plausible decoy—not the best answer. 7. Do not change an answer unless you are absolutely certain you have made a mistake. Your first reaction is likely to be correct. 8. Do not sit next to a friend, a window, or other potential distractions.

Exam Day Preparations The exam day will be a stressful and tiring one. This will be no day to have unpleas­ ant surprises. For this reason we suggest that an advance visit be made to the examination site to determine the following: H

How much time should I allow for travel to the exam on that day? Plan to arrive about 15 minutes early. That way you will have ample time, but not too much time. Arriving too early, and mingling with others who also are anxious, will increase your anxiety and nervousness.

■ Where will I park? How does the exam site look? Will I have ample workspace? Where will I stack my reference materials? Will it be overly bright (sunglasses), cold (sweater), or noisy (earplugs)? Would a cushion make the chair more comfortable? □ Where are the drinking fountains and lavatory facilities? ■ What about food? Should I take something along for energy in the exam? A bag lunch during the break probably makes sense.

W hat to Take to the Exam The NCEES guidelines state that you may bring only the following reference materials and aids into the examination room for your personal use: H

Handbooks and textbooks, including the applicable design standards.

■

Bound reference materials, provided the materials remain bound during the entire examination. The NCEES defines “bound” as books or materials fas­ tened securely in their covers by fasteners that penetrate all papers. Examples are ring binders, spiral binders and notebooks, plastic snap binders, brads, screw posts, and so on. A battery-operated, silent, nonprinting, noncommunicating calculator from the NCEES list of approved calculators. For the most current list, see the NCEES Web site (www.ncees.org). You also need to determine whether or not your state permits preprogrammed calculators. Bring extra batteries for your calculator just in case; many people feel that bringing a second calculator is also a very good idea.

At one time, NCEES had a rule barring “review publications directed principally toward sample questions and their solutions” in the exam room. This set the stage for restricting some kinds of publications from the exam. State boards may adopt the NCEES guidelines, or adopt either more or less restrictive rules. Thus, an important step in preparing for the exam is to know what will—and will not—be permitted. We suggest that if possible you obtain a written copy of your state’s policy for the specific exam you will be taking. Occasionally there has been confusion at individual examination sites, so a copy of the exact applicable policy will not only allow you to carefully and correctly prepare your materials, but will also ensure that the exam proctors will allow all proper materials that you bring to the exam. As a general rule we recommend that you plan well in advance what books and materials you want to take to the exam. Then, use the same materials in your review that you will have in the exam. Being able to find the appropriate informa­

Introduction

xiij

tion quickly presupposes a lot of advance organization on your part, and familiar­ ity with your reference materials. Catalog your reference material by tabbing or marking important topics, equations, tables, and procedures. Keep everything well organized as you prepare these materials during the weeks before the exam, and be sure to insert new items in their correct places and catalog them properly.

License Review Books The review books you use to prepare for the exam are good choices to bring to the exam itself. After weeks or months of studying, you will be very familiar with their organization and content, so you’ll be able to quickly locate the material you want to reference during the exam. Keep in mind the caveat just discussed—some state boards will not permit you to bring in review books that consist largely of sample questions and answers.

Textbooks If you still have your university textbooks, they are the ones you should use in the exam, unless they are too out of date. To a great extent the books will be like old friends with familiar notation.

Bound Reference Materials The NCEES guidelines suggest that you can take any reference materials you wish, so long as you prepare them properly. You could, for example, prepare several vol­ umes of bound reference materials, with each volume intended to cover a particu­ lar category of problem. Maybe the most efficient way to use this book would be to cut it up and insert portions of it in your individually prepared bound materials. Use tabs so that specific material can be located quickly. If you do a careful and systematic review of electrical engineering, and prepare a lot of well-organized materials, you just may find that you are so well prepared that you will not have left anything of value at home.

Calculators Be sure you can perform needed calculations on your calculators prior to the test. You may need to work with exponentials and logarithms, trigonometries and their inverses, complex numbers (including polar-rectangular conversions, and com­ plex trigonometric and hyperbolic numbers), and perhaps some simple matrix operations. Calculators are usually kept in soft cases; if they are thrown together with books and other materials, buttons can be pressed and held unintentionally. Be sure you understand your calculator well enough to get out of unusual modes (such as graphics or programming) if you accidentally get into them. The exam is no time to find you have been inadvertently prevented from using your calculator.

Other Items In addition to the reference materials just mentioned, you should consider bringing the following to the exam: SI

Clock—You must have a time plan and a clock or wristwatch. Exam assignment paperwork—Take along the letter assigning you to the exam at the specified location. To prove you are the correct person, also bring something with your name and picture.

xiv

Introduction

Items suggested by advance visit—If you visit the exam site, you probably will discover an item or two that you need to add to your list. Clothes—Plan to wear comfortable clothes. You probably will do better if you are slightly cool. Box for everything—You need to be able to carry all your materials to the exam and have them conveniently organized at your side. Probably a card­ board box is the answer. Be sure to refer to your state boards and NCEES for any updates to exam requirements.

Examination Scoring and Results The questions are machine-scored by scanning. The answers sheets are checked for errors by computer. Marking two answers to a question, for example, will be detected and no credit will be given. Your state board will notify you whether you have passed or failed roughly three months after the exam. Candidates who do not pass the exam the first time may take it again. If you do not pass you will receive a report listing the percent­ ages of questions you answered correctly for each topic area. This information can help focus the review efforts of candidates who need to retake the exam. The PE exam is challenging, but analysis of previous pass rates shows that the majority of candidates do pass it the first time. By reviewing appropriate concepts and practicing with exam-style problems, you can be in that majority. Good luck!

A C K N O W LE D G M E N T S Howard Smolleck, PhD, PE provided the foundation for the PE Power Electrical Engineering License Review Study Guide. This problems and solutions book is meant to complement his work. The authors and publisher are grateful to many reviewers for providing extensive accuracy checking and recommendations on draft manuscript.

C

H

A

P

T

E

R

Mathematics Review PROBLEMS 1 SOLUTIONS 3 RECOMMENDED REFERENCES 14

PROBLEM S 1.1 The current in an electric circuit decreases exponentially with time as given by the equation: i(t) = e-a06t A Determine the value of t at which the current will be at 50 percent of its initial value. 1.2 The voltage across a capacitor decreases exponentially with time as given by the equation: v(t) = 10e~~aoo8t V Determine the value of t at which the voltage across the capacitor has reached 2 V. 1.3 The current in an electric circuit is given by the equation:

Where x, yt z and n are some circuit parameters. Determine the value of i for z = 20, x = 10, y = 50 and n = 0.2 using natural log. 1.4 Determine the exponential form of the complex number Z = 4 + /5. 1.5 Express the complex number Z = 5el0'927 in trigonometric or rectangular form.

1

2

Chapter 1 Mathematics Review

1.6

Perform the following arithmetic operations on the complex numbers Zj = 3 + i4 and Z2 = 8 + i 4. a-

Z - Z7 + Z2

b.

Z —Z;Z2

,

z =^

1.7 Determine the hyperbolic cosine value of the complex number Z = 3 + i4. 1.8 Determine the hyperbolic sine value of the complex number Zj 1.9

5.651 emns.

Determine the determinant of matrix B using the cofactor expansion method. [2 2 B }

1.10 Determine the determinant of matrix B using the cofactor expansion method. B

4

3 - 5

6

10

8

9

2

3

1.11 Determine the inverse of the matrix B: B—

2

2

2

3

1.12 Determine the inverse of the matrix A: 2 A= 3 4

3 - 4 -1 -7

-2 -6

1.13 Solve the following set of linear equations by substitution. 6X + 87 = 48 2X + 67 = 36 1.14 Solve the following set of linear equations by reduction. 5X + 157 = 45 10X4-207 = 50

Solutions

3

1.15 Using Cramer’s rale, solve the following set of linear equations: 9X —2Y —7Z = 30 —2 X + 5.57 —Z = 10 —I X —7 + 13Z = —10 1.16 Determine the global maximum and minimum of the function: f ( x ) = 5x6 —18x5 + 15x4 -1 0 1.17 Determine the first derivative of the following functions: a.

f ( x ) = l0 tfx f

b.

f(x ) =

c.

/ (*) = xex sin x

sin x sml[a + x)

1.18 Determine the integral of the following function:

f

x2 cos 2xdx

p

1.19 Evaluate f cos xdx

2

3

J2

1.20 Evaluate J* Ix* ~ xZ + ^\dx -i

SO LU TIO N S 1.1 Step 1: At 50 percent, the value of current is given as i = 0.5 A Therefore, we have 0.5 = e"006' A Step 2: Taking natural log on both sides of Equation (1.1), we have ln(0.5) = ln(e~006?) ln(0.5) = —0.06?, since \n{ex ) = x —0.693 = —0.06? *= 11.55

(1.1)

4

Chapter 1 Mathematics Review

1.2 Step 1: Substituting v = 2 V in the voltage equation, we have 2 = I0e~0-mt V Step 2: Taking natural log on both sides of Equation (1.2), we have ln(2) = ln(lO£ra008') ln(2) = In (10) + In (e~°'mt j , since In (xy) = In (x) + ln(_y) ln(2) = In (10) —0.008/, since ln(ex ) = x 0.693 = 2.303 —0.008/

1.3

Step 1: Taking natural log on both sides of the current equation, we have

In (j) = In (20) + In (10) - 0.2 x In (50) ln(z') = 2.996 + 2.303 - 0.782 = 4.517 Step 2: Because ln(ex) = x , we have i = e4517 = 91.561 A 1.4

Step 1: The modulus of the complex number Z is given as r = mod (Z) = ^ 4 2 + 52) = 6.403 Step 2: The argument of the complex number Z is given as 5

1

9 = arg(z) = tan_1 - = 51.34° = 0.896 rad .4 j Step 3: The exponential form of the complex number Z is given as Z = rei6 = 6.404e/0'896

(1.2)

Solutions

1.5

5

Step 1: The modulus of the complex number Z is given as r =5 and the argument of the complex number is given as 6 = 0.927 rad Step 2: The complex number Z in trigonometric or rectangular form is given as Z = r(cos^ + /sin6)) Z = 5 (cos (0.927) + i sin (0.927) Z = 3 + z'4

1.6 a). Addition of two complex numbers in rectangular or trigonometric form is given as Z —3 ~h z'4 -f- 8 -f" z"4 Z = (3 -f- 8) + / (4 + 4) Z = 11 + z'8 b). Multiplication of two complex numbers in rectangular or trigonometric form is given as Z = (3 + z'4)x(8 + z'4) Z = (3x8) + z'(4x8) + z'(3x4) + z'2 (4 x 4 ) Z = 24 + z'32 + z'12 —16, since z'2 = —1 Z = 8 + i44 c). Division of two complex numbers in rectangular or trigonometric form is performed by multiplying both the numerator and denominator with the conjugate of the denominator as shown: z = 3 ± j4 = (3 + (8 - M) = 40 + ^20 = Q; 8 + z'4 (8 + z'4)x(8-z4) 80

^ ^

1.7 The hyperbolic cosine of a complex number Z - a + ib in rectangular form is given as cosh (Z) = cosh (a + ib) = cosh (a) cos (b) + zsinh (a) sin (b) Therefore, for the given complex number Zp the hyperbolic cosine value is given as cosh (Zt ) = cosh (3 + z'4) = cosh (3) x cos (4) + zsinh (3) x sin (4) cosh (3 + z'4) = 10.068 x 0.6536 + *10.018 x (-0.757) = 6.58 + z'7.5836 i

Chapter 1 Mathematics Review

1.8 Step 1: Expressing the complex number Zx in rectangular from, we have Zj = rx{cosex +zsm 01) = 5.657(cos(O.785) + isin(O.785)) = 4 + i4 Step 2: The hyperbolic of the complex number Zx is given as sinh (Zj) = sinh (4 + i4) = sinh (4) cos (4) + i cosh (4) sin (4) sinh (4 + i4) = 27.29 x 0.6536 + *27.308 x (-0.7568) = 17.84 + *20.67 1.9 Step 1: The determinant of a 2 by 2 matrix in general using cofactor expansion method is aU

a \2

a 2l

a 22

a \ 1a 22

a \ 2 a 2\

Step 2: The determinant of the given matrix B is B 1.10

2

2

2

3

(2 x 3) —(2 x 2) = 2

Step 1: The determinant of a 3 by 3 matrix in general using cofactor expansion method is given as an

a12

ai3

^ 21

a22

a 23 =

a 3l

a 32

a 33

an

a22 a 32

a 23

al2

a 33

021

a 23

a 2i

a22

«31

a 32

+ a l3 «31

a 33

Step 2: The determinant of the given matrix B is

B

4

3

-i

6 9

10 2

8 3

10 2

8 3

6

8

9

3

+ (-5 )

6

10

9

2

|l?| = 4 x (10 x 3 —8 x 2) —3 x (6 x 3 —8 x 9) —5 x (6 x 2 —10 x 9) = 608 1.11

Step 1: The adjoint matrix of a 2 by 2 matrix is given as

Solutions

Step 2: The adjoint matrix of the given matrix B is given as 3 -2

^Adj

-2 2

Step 3: The determinant of the matrix B is given as \B\

2

2

2

3

(2x3) —(2 x 2 ) = 2

Step 4: The inverse of the matrix B is given as

B -1 _ ^Adj \S\

3

-2

-2

2

1.5 -1

-1 1

1.12 Step 1: The cofactor of element an of matrix A is given as

Step 2: The cofactor of element an of matrix A is given as =

10

Step 3: The cofactor of element a13 of matrix A is given as 3

-1

4

-7

17

Step 4: The cofactor of element a2l of matrix A is given as 3

-4

-7

-6

46

Similarly, determining the other cofactors, the cofactor matrix of A is given as

^cofac

-8

10

-1 7

46

4

26

-1 0

-8

-1 1

7

Chapter 1 Mathematics Review

Step 5: The adjoint matrix of A is given as AAdj = transpose (A,cofac

-8

46

10

10 -17

4 26

-8 11

Step 6: The determinant of matrix A is given as A = 2x

-1

-2

-7

-6

—3 x

-2

3 -4 )x + (' -6 4

3 4

-1 -7

|A| = 2 x (6 —14) —3 x (—18 + 8) —4 x (—21 + 4) = 82 Step 7: The inverse of the matrix A is given as

A*

1 > & 1

-8

46 4

10 -1 7

26

-1 0 -8 -11

82

|A|

-0.098

0.561

-0.122

0.122 -0.207

0.049 0.317

-0.098 -0.134

1.13 The two linear equations are given as 6 X + 87 —48

(1.13a)

2 X + 67 = 36

(1.13b)

Step 1: From Equation (1.13b), solving for variable X, we have 2 X = 36 —67 3 6 -6 7

(1.13c) Step 2: Substituting the value of X from Equation (1.13c) in Equation (1.13a), we have 6 x (18 —37) + 87 = 48 108 —187 + 87 = 48 —107 = 48 —108 Y=6 Step 3: Substituting the value of 7 in Equation (1.13c), we have

Solutions

9

1.14 The two linear equations are given as 5X + 1 5 7 = 45

(L14a)

10X + 207 = 50

(L14b)

Step 1: Multiply Equation (1.14a) with -2 resulting in Equation (1.14c): —10X —307 = —90

(1.14c)

Step 2: Adding Equation (1.14c) and Equation (1.14b) results in the elimination of variable X: -1 0 7 = -4 0 7=4 Step 3: Substituting the value of 7 in Equation (1.14a), we have 5X 4-15x4 = 45 5X = 45 —60 X = -3 1.15 The given set of linear equations can be expressed in matrix form as: Ax = B Where -2

9 A = -2 -7

5.5 -1

-7 -1 is known as the coefficient matrix. 13

is known as the unknown vector.

B

30 10 is known as the constant vector. -10

Step 1: The determinant of coefficient matrix A is given as A=9

5.5 -1

-2 -1 -(-2 ) 13 -7

-1 13

-7

-2

5.5

-7

-1

= 9 x 70.7 4- 2 x (-33) - 7 x 40.5 = 285

Chapter 1 Mathematics Review

Step 2: To determine the variable X, replace column 1 in the coefficient matrix A with the constant vector B as shown:

4 =

30

-2

10 -1 0

5.5 -1

-1 -1 13

The determinant of matrix A is given as U J = 30

5.5 -1

-1

/ _\ 10 - ( \- 2 )/ 13 -1 0

-1 13

+ (-7 )

10 -1 0

5.5 -1

30x70.5 + 2 x 1 2 0 - 7 x 4 5 = 2040

Step 3: The variable X is given as \Aj 2040 X = V4[ = ^ ^ = 7.158 A 285 Step 4: To determine the variable Y, replace column 2 in the coefficient matrix A with the constant vector B as shown: 9 A2 = - 2 -7

30 10

-■1 -1

-1 0

13

The determinant of matrix A 2 is given as \A2\ —1440 The variable Y is given as \A2 Y = Vn \A\

1440 285

5.053

Step 5: To determine the variable Z, replace column 3 in the coefficient matrix A with the constant vector B as shown: 9 a3 = - 2

-2 5.5

30 10

-7

-1

-1 0

The determinant of matrix A3 is given as Ia J = 990 The variable Z is given as

Solutions

l.16

11

Step 1: The first derivative of the function f(x) is given as , d f(x ) d{5xb —18x5 + 15*4 —10 1 / (x ) = - - - = - i -----------------------------' = 30.t - 90-V + 60x dx dx Step 2: The value of x at the extreme point i.e., either at maximum or minimum, or inflection point, is determined by equating the first derivative of the function to zero as: f (x ) =

= 30x5 - 90x4 + 60x3 = 30x3 (x2 - 3x + 2) = 0

Therefore, we have two equations: 30x3 = 0

(1.16a)

x2 - 3 x + 2 = 0

(1.16b)

Step 3: From Equation (1.16a), we have Xj = 0 And solving the quadratic Equation (1.16b), we have x2 = 1 and x3 = 2 Step 4: The second derivative of the function/(x) is given as ,5 n n J i r„( ^ d 2f (x) d f {x) d I30* - 90* + 60* ^

dx2

dx

f ”(x) = 150x4 - 360x3 + 180x2

dx (1 16c)

Step 5: Substituting x = x xin Equation (1.16c), we have

Therefore, x = xt represents the point where the function experiences an inflection point.

12Chapter

1 Mathematics Review

Step 6: Substituting x = x2 in Equation (1.16c), we have / " ( x 2) = -3 0 Because / > 2 ) < 0 , x2represents the point at which/(x) has a maximum value given as / ( x 2) = - 8 Step 7: Substituting x = x3in Equation (1.16c), we have f ( x 3) = 240 Because / (x3) > 0, x., represents the point at which fix) has a minimum value given as / ( x 3) = - 26 1.17

a). The first derivative of/(x) is given as i— \

/ i— \

4 / _7

i

/ (*) = —--------- = 10— ----- = 10—-— — = 10x —(x7 ' 4 dx dx dx 4v ' 5/

*\

dx

35 ! ■x

b). The first derivative of f(x) is given as ^ /

^ (* )=

. / xt/sinx sm(tf + x) —

. d sin (a + x) smx —

sin2 (a + x)

sin (a + x)cosx —sin (x) cos (a + x) —

sin2 (a + x)

sin (a) ~

sin2 (a + *)

c). The first derivative of f(x) is given as y — xex sin x

(1.17a)

Step 1: Taking natural log on both sides of Equation (1.17a), we have In (_y) = In (x) + In [ex j + In (sin x) In (_y) = In (x) + x + In (sin x)

(1.17b)

Solutions

13

Step 2: The first derivative of Equation (1.17b) is given as 1 dy y dx dy dx

1 x

sinx

,1d sm x ax

y

dy

xe smx

1 „ cosx + 1+ - — x smx

ex (sin x + x (sin x + cos x

-

1.18 Step 1: Let the function / (x) = x2 cos 2x be divided into two parts: g{x) = x2 h(x) = cos2x Step 2: The integral of the given function is determined using the integration by parts technique as follows: J* f (x)dx — j * g (x)h(x)dx = g (x) J* h (x)d x —J ^ g (X)J * h (x )d x dx x2 cos 2xdx = x 2J* cos 2xdx —J* 2x J* cos 2xdx sin2x 2

I

sin 2x 2

xcJFs isin n l2xdx xdx-

sin2x

cos2x

2xsin2x -----------a 2

' sin 2x I x cos 2x 2 1 2 1.19

1.20

fl'f /

sin 2xdx si

—cos 2x

sin 2x 4

C

7T

I

cos xdx = sin x

sin (?r)

sm

0-1

14

Chapter 1 Mathematics Review

RECOM M ENDED REFEREN CES The following Kaplan products may be helpful for those in need of additional mathematics review: College Algebra Exam File Linear Algebra Exam File ■

Probability and Statistics Exam File Differential Equations Exam File Calculus I Exam File Calculus II Exam File Calculus III Exam File

For more information, refer to www.kaplanaecengineering.com, and then search on these products.

C

H

A

P

T

E

R

Electric and Magnetic Fi©lds PROBLEMS 15 SOLUTIONS 19 RECOMMENDED REFERENCES 35

PROBLEM S 2.1

Determine the magnitude of the electric field at point A(3,2,5), given a positive charge of 1.602 x 10~19 C at point B (l,l,l). Assume the distance is measured in meters and the charge is located in free space.

2.2 A positive charge of 2 x 10~5 C is located at the origin (0,0,0). Determine the electric field at point A(4,4,4) with the distance measured in meters and the charge located in free space. 2.3 Two point charges, q{ and qv with charges 4 x 10 s C and -8 x 10~5 C, respectively, are located in free space at points A(4,3,-2), and B(-6,4,-4) in Cartesian coordinates. a. Determine the electric field E at point D(3,1,-2) b. Determine the force on 8 x 10~5 C charge at point D. 2.4 A line charge with current density Pi = 4z C/m is oriented along the z-axis. Determine the charge contained in a cylindrical tube of length 0.5 m centered on the z-axis between z = 0 and z = 0.5. 2.5

A circular disk of electric charge is characterized by a symmetric surface charge density that increases linearly with the radius R from zero to 10 C/m2 at R = 6 cm. Determine the total charge on the disk surface.

2.6 A spherical shell centered at the origin extends between the radius of 2 cm and 3 cm. If the volume charge density is given as 0.0003R C/m3, determine the total charge contained in the shell.

15

16

Chapter 2 Electric and Magnetic Fields

2.7 Determine the gradient of the following potential and evaluate the gradient at point A (l,-1,2). V = x 2y + xy2 + xz2

I

2.8 Determine the charge density of a region if the electric flux density of the region is given as A A A D = %x2y x+ 2zx y + xz2 z 2.9 Determine the divergence of the following electric field given in spherical a 7r ,7r coordinates at point A 4 2 a2 cosi

E

R2

a cosi R:

2.10 A positive charge of 30 (j C is located at the origin in free space. Determine the electric potential at a distance of 3 meters from the origin. 2.11

Determine the electric potential at point A with respect to point B, both located on the z-axis. Point A is at 4 meters from the origin on the positive z-axis, and point B is at 2 meters from the origin on the negative z-axis. The points A and B are in the electric field: 15 A V/m R3

E

~Tr

2.12 In a generator, there exists a loop of wire in free space, which approximates a square. The length of the sides of the square are 0.25 m. Determine the flux generated by the loop if it is rated to carry a current of 25 A. 2.13 Two infinitely long parallel wires are separated by a distance of 5 meters. The wires carry a current of 15 A in opposite directions. Determine the magnetic flux density at a point P in the plane of the wires between them and located 1 meter from one wire, and 4 meters from the other. 2.14 A proton moving at a speed of 2 x 106m/s in a magnetic field of magnetic flux density 3.5 T experiences a magnetic force of 2.5 x 10~13N. Determine the angle between the magnetic field and the proton’s velocity. 2.15 The force per unit length between two parallel wires of infinite length is measured to be 75 x 10~3N/m when the two wires carry equal current and repel each other. Determine the magnitude of the current if the two wires are separated by a distance of 0.15 meter.

i

2.16 Determine the curl of the vector field: -

A

.

. A

A

E = x2y x - (x3 + 4y 2 ) y+ l x 2yz2 z

Problems

17

2.17

A circular loop of 50 turns with a radius of 6 cm carries a current of 5 A. Determine the magnetic field intensity and flux density at the center of the loop.

2.18

A linear conductor of infinite length in free space placed along the z-axis carries a current of 10 A in the positive z-direction. Determine the magnetic flux density at a distance of 20 cm from the conductor in the x-y plane.

2.19 Determine the magnetic flux density in commercial iron if the associated magnetic field intensity is 100 A/m. 2.20 Determine the magnetic flux density given the vector magnetic potential as ( A A A 7r ir A — x 2.5 sin —z + y 6 cos — xz + z (sin 7r x + cos 7r y) ,60 , u 2.21

A magnetic circuit has a cross sectional area of 1.5 x 10~3m2, mean length of 0.25 m, and the magnetic material has a relative permeability of 50,000. Determine a. the reluctance of the magnetic circuit. b. the permeance of the magnetic circuit. c. the generated flux if the flux density is 0.5 T.

2.22 The magnetic circuit of Problem 2.21 has a winding of 200 turns. Determine the magnetomotive force (mmf) and the current required to produce a flux density of 0.5 T using the computed parameters from Problem 2,21. 2.23

The magnetic circuit of Problem 2.21 has a winding of 500 turns with a current of 1.5 A. Using the computed parameters from Problem 2.21, determine a. the magnetomotive force. b. the flux generated. c. the flux density.

2.24 A magnetic circuit consists of an air gap of 0.025 x 10~2 m length and a cross sectional area of 4.5 x 10^ m2. If the winding consists of 500 turns, determine the inductance of the winding with the assumption that the reluctance of the core is negligible. 2.25

A magnetic circuit with an air gap is shown in Figure 2.25. The cross sectional area of the magnetic core and the air gap is 12 x 10^ m2. The air gap length and the mean core length are 0.035 x 10~2m and 0.4 m respectively. If the winding consists of 300 turns carrying a current of 0.25 A, determine a. the generated flux if the relative permeability of the magnetic core is 65,000. b. the flux density. c. the inductance of the winding. d. the magnetomotive force.

Chapter 2 Electric and Magnetic Fields

i=0.25 A

J .

N = 300

T

g =0.035 x 10~2 m

F ig u re 2 .2 5

2.26 An inductor is constructed from an iron ring of mean radius 0.2 m and a core diameter of 0.04 m as shown in Figure 2.26. The relative permeability of the iron is given as 3000. There are 100 turns of current carrying conductor carrying a current of 2 A. Determine a. the magnetic field intensity in the magnetic core. b. the flux density in the core. c. the magnetic flux in the core. 0.04 m

F igu re 2 .2 6

2.27 Repeat Problem 2.26 with the ring modified by cutting a gap of 0.005 m at section A -A . 2.28

Consider the two-loop magnetic circuit shown in Figure 2.28. The winding consists of 300 turns carrying a current of 0.5 A. Each segment of the circuit is of length 0.1 m and a cross-sectional area of 1 x 10~3m2. If the magnetic core is nickel with a relative permeability of 600, determine the magnetic flux and magnetic flux density in section A-A’.

Solutions

19

0.2 m

Figu re 2 .2 8

2.29 For the two loop magnetic circuit given in Problem 2.28, determine a. the net reluctance of the magnetic circuit. b. the flux generated in section B-B ’. c. the magnetic field intensity in section B-B’. 2.30 Repeat Problem 2.28 with the two loop magnetic circuit modified by cutting a gap of 0.005 m at section A-A’.

SO L U T IO N S 2.1

Step 1: The distance RBAbetween points A and B is given as R1M = v (1 - 3 ) + (1“ 2) + ( 1 - 5 ) =4.582 m Step 2: The magnitude of the electric field at point A is given as p q 1.602x io -19 . _ , 1 A _11 V/m E a = ------------- - = ----------------------------------- - = 6.86x10 4tt£0 (Rba) 4 x t r x 8.85x 10~12 x (4.582)

2.2

Step 1: The distance between the origin and the point A is given as Roa = V(4 - °j + (4 - °) + (4 - °) = 6-928 m Step 2: The vector joining the origin and the point A in Cartesian coordinates is given as A

A

A

r0A = 4 a x + 4ay + 4az

20

Chapter 2

Electric and Magnetic Fields

Step 3: The unit vector in direction of rQAis given as A A A ax + ay + az 'OA OA

6.928

ROA

A A A 0.577 ax + ay + az

Step 4: The electric field at point A is given as A A A 0.577 x ax + ay + az

A

Et

rOA

4

( R 0A

A A A E a = 2.162xl03 x ax + ay + az

2.3

4 X 7TX

2x10“ 8.85x10 1 X6.9282

V/m

Step 1: The position vector rQAfrom the origin to point A is given as A rOA = 4 a x +

A A 3ay- 2 a z

The position vector rQBfrom the origin to point B is given as A A A r0B = - 6 a x + 4ay - 4 a z The position vector r

from the origin to point D is given as

roD = 3 a x+ a y- 2 a z Step 2: The distance between points D and A is given as OD

'O A

(3 - 4) + (1 - 3) + ( - 2 + 2) = 2.236 m

The distance between points D and B is given as v0D—r0B = -^(3 + 6) + (l —4) + (—2 + 4) = 9.695 m

Solutions

21

Step 3: The electric field E at point D due to the charges at A and B is given as /

= jJbNIL Wb Where N is the number of conductors I is the current flowing through the conductors L is the length of the conductor ju=jujur is the magnetic permeability Because the loop of wire is in free space, we have \x = fi0 = 47T x 10~7 H/m ^ = 4 x 7r x 10~7 x 1.0 x 25 x 0.25 = 7.85 tiWb

Solutions

2.13

Assuming that the two wires are in free space, the magnetic flux density at point P is given as f^O^A , 2 llTXr 2ttx,

B

rri

Where is the distance between point P and the wire A xBis the distance between point P and the wire B Substituting the given values, the magnetic flux density is given as 4x7r x 1CT7 x 15

B

2

2.14

25

X 7T X

1.0

4 x 7r x 1(T7 x 15 2x7r x 4.0

037.5 x 10

T

The force on a moving charge in a magnetic field is given as F = qvB sm 9N Where q is the charge of the proton (1.6022 x 10~19 C). v is the velocity of the proton. B is the magnetic flux density. 9 is the angle between the B field and velocity of the proton. Therefore, we have . , sm

2.5x10 -13 1.6022 x 10~19 x 2 x 106 x 3.5

F qvB

0.223

0 = sin-1 (0.223) = 12.88° or 167.12° 2.15 The force between two parallel conductors of infinite length is given as F = BIL

m ljl2 2pr

Where B is the magnetic flux density. 7j and I2 are the current flowing in the conductors. L is the length of the wire, r is the distance between the wires. Because both the wires carry the same current, we have / =

I

2n rF

for conductor length of 1 meter and in free space

I 2 x t t x 0 . 1 5 x 7 5 x 10 - 3

4 x 7 rx l0 - 7

237.17 A

26

Chapter 2 Electric and Magnetic Fields

2.16 Step 1: The curl of a vector field E is given as A

A

= Kbhb cos (- = 208 x 41.6 xcos(o° -(-5 3 .1 3 °)) + 2 0 8 x 41.6x co s(-120°

— f —173.13"

+208 x 41.6 x cos 120° -66.87 P34>= 3 x 208 x 41.6 x cos(53.13° I = 15.575 kW 5.7 Step 1: Because the load is delta connected, the phase and line voltage are equal. Therefore V.ph = VIT = 440 V LL Step 2: The magnitude of the phase current is given as 4 h~

T

25 = -y= —14.43 A V3 V3

Step 3: The magnitude of the load impedance is given as \%l | = = = 30.5 O 11 / Ph 14-43 Step 4: The power factor of the balanced delta connected system is given as 3 = —=-----------7.5x10 n ana/c = 0.3936 p fr = cos 9(\ — - 7 =^-------V3FLL/ L V3 x 4 4 0 x 2 5 Step 5: The resistance and reactance of the load are given as Rh = 1ZL| cos 9 = 30.5 x 0.3936 = 12 0 XL = |ZL| sin 9 = 30.5 x 0.9192 = 28.03 O 5.8

Step 1: Assuming positive phase sequence ABC, the three phases and line voltages are given as F l= F l = 1 2 0 Z 0 ° V Vbc = ^ Z - 120° = 120Z-120° V VCA = Vab Z120° = 120Z120° V

Solutions

127

Step 2: The three phase currents are given as

ab

ZL

120/ 0 ° 7.07 - y'7.07

12/45° A

Ibc = I ab / -120° = 1 2 / - 75° A Ica = I ab / 1 20° = 12/165° A Step 3: The three-phase complex power consumed by the load is given as * * ' A ' A ' A ' A A fai S*=Vat Lit + n c he +1 Vca ^ca \ ) \ ) \ J = 120/0° x 1 2 / - 45° + 1 2 0 / -120° x 12/75° +120/120° x 1 2 / - 165c Su = 4320/ - 45° = 3055 - y3055 VA

5.9 Step 1: For a delta connected load, the magnitude of the phase voltage is given as Vph, = VL = 208 V Step 2: The magnitude of the phase current is given as I ph

43.30 V3

25 A

V3

Step 3: The magnitude of the per phase impedance is given as Zph

ph

ph

208 25

8.32 0

The resistance per phase is given as Rvh = Zph cos9 = 8.32x0.8 = 6.65 O Because the power is consumed at a lagging power factor, the reactance component of the load is inductive and is given as X L = Zph sin0 = 8.32x0.6 = 5 O Therefore, the impedance per phase is given as Zph = * ph + jXiL — 6-65 + j5 O

Chapter 5 Balanced Three-Phase Circuits

Step 4: The three-phase power consumed by the load is given as p3, = V3Fl/ l cos0 = V3 x 208 x 43.03 x 0.8 = 12.479 kW Step 5: If the same impedances are connected in wye to the same source, the magnitude of the phase voltage is given as V h=~r = = 120 V p V3 V3 Step 6: The magnitude of the phase current is given as I = |Zph|

= — = 14.42 A 8.32

Because the load is wye connected, the line current is equal to the phase current. Step 7: The three-phase power consumed by the wye connected load is given as P3(p = 3 Fph/ ph cos0 — 3 x 120x 14.42x 0.8 = 4.15 kW Notes: The power factor is still 0.8 lagging, since the load impedances are same. The line current and the power absorbed with the load impedances connected in wye are one-third of that when the load is connected in delta. 5.10 Step 1: The three-phase reactive power is given as = VlO2 —(? = 8 kVAR

Step 2: The power factor angle is given as 0 = tan"

Qz

oo

128

= tan —1 — ,6,

53.13c

Step 3: The magnitude of the line-to-line voltage is given as 6x10-

3(j>

VT

■J3I,

COS

0

V 3xl5xcosf53.13

384.9 V

Solutions

129

Step 4: The magnitude of the per phase impedance is given as |Z I ph1I= -^ j - = ^j - = ^ ic^ ph

i L

= 14.81 ft

1 J

The resistance of the per phase impedance is given as i?ph = |Zph| cos0 = 14.81 x cos (53.13°) = 8.88 Q The reactance of the per phase impedance is given as Xph = |Zph| sin0 = 14.81 x sin f53.13°) = 11.84 Q Step 5: If the same impedances are connected in delta to the same source, the magnitude of the phase current is given as /

= ik - =

P |zP„S 14-81

= 25.98 A

The magnitude of the line current is given as / L = V 3 /ph = 7 3 x 2 5 .9 8 = 45 A Step 6: The three-phase power consumed is given as = S v hIh cos0 = V 3x 384.9 x 45 x cos(53.13°) = 18 kW

Notes: The power factor angle is still 53.13°, since the load impedances are the same. The line current and the power absorbed with the load impedances connected in delta are three times that of when the load is connected in wye.

130

Chapter 5 Balanced Three-Phase Circuits

5.11

Step 1: The delta connected load can be converted into equivalent wye connected load as shown in Figure 5.11a.

F igu re 5.11a

The equivalent wye connected impedance per phase is given as z .. = Z„.. = Z .. = an

bn

cn

Z .. ^

9

=

_

/1 2

J

3

= 3 -/4 S l

Step 2: The equivalent impedance per phase of the two wye connected impedances is given as ^ ph —

^ an^aV Zan + Z ..

5Z53.13° x 5 Z -53.13° 1 5Z53.130 + 5Z - 53.13c

— '

— 4 . 1 0 uI

Step 3: The phase voltage is given as A

a V

y 440Z0° = —pr = = 254Z0°V V3 v3

Step 4: The phase current is given as A

a V u 954/0° /„„ = - ^ = = 61.05/0” A ' Zpk 4.16 Step 5: The phase current through the load impedance per phase Zm is given as A

/

A

=/

Z

. .

5 /

_

1^°

------^ -----= 61.05Z00 x -------- ------ -------------- = 50.87Z - 53.13° A P Za n + Z an,, 5Z53.130 + 5 Z - 53.13° an

Solutions

131

The phase current through the load impedance per phase Za>n, is given as A

A _____

an_____

Z an a n + Za"n

---------- = 50.87/53.13° A = 61.05/0° x --------5/53.13° + 5 /-5 3 .1 3 °

Step 6: The magnitude of the line current drawn from the source is given as / L = /ph =61.05 A Step 7: Because the equivalent impedance per phase is purely resistive, the overall power factor is unity. Therefore, the three-phase power consumed by the loads is given as p34>= S v hl h cos 9 = \f3 x 440 x 61.05 x 1.0 = 46.526 kW 5.12 Step 1: The delta connected load is converted to an equivalent wye connected load as shown in Figure 5.12a. The per phase impedance of the equivalent wye connected load is given as an

Zj = 1+j1 n

a

A

a'

zT= 1+11 n Figure 5.12a

Step 2: A The phase current Ian is given as A

A vV„

r

__

r an

Z. I

— — __ V V

“ ~ Z- -

Su )

3 + _/4

= 24.02/-53.13° A

132

Chapter 5 Balanced Three-Phase Circuits

Step 3 : A The phase current / - • is given as

208

A

V .

A

I.. an

zr +z,,

zo°

l + j l + 3 + j8

1 2 .2 0 Z -66.03° A

Because I >. is the phase current of the equivalent wye connected load, a

the phase current A

I

A

of the original delta connected load is given as

I ,, in 0 = - ^ r Z - 30° = -

0 / __m °

Z - 30° = 7.04Z -96.03° A V3

Step 4: The line current drawn from the source is given as La = h n+ lA” = 24.02Z —53.13° + 12.20Z-66.03° = 36.0Z-57.46° A Step 5: The three-phase power consumed by the loads are given as /> = V3FAR/ Aa cose = S x 208 x 36 x cos f57.46°) = 6.98 kW

5.13 Step 1: The single phase equivalent of the three-phase balanced system on A-phase is as shown in Figure 5.13a. Z = 2+j2Q

Z T = 6+j8 Q

Figure 5.13a

Step 2: A The phase current I an drawn from the source is given as A

=

Z + ZT + ZL

254Z0C (2 + j 2) + (6 + j 8) + (8 + y'6)

11.23Z —45° A

Solutions

133

Step 3: A The phase voltage Van across the load is given as

Van =

- Im (Z + ZT) = 2 54/0” - 1 1.23Z - 45” x (8 + yiO) = 112.2Z - 8.09” V Step 4: The three-phase complex power consumed by the load is given as = 3 x 112.2Z - 8.09° x 11.23Z450 = 3780Z36.910 = 3022 + J2210 VA

5.14 Step 1: The wye equivalent impedance per phase of the load is given as 6 -J9

2 —j3 = 3.60Z —56.30° Q

The single phase equivalent of the three-phase balanced system with wye equivalent impedance of the load is as shown in Figure 5.14a.

F ig u re 5.14a

Step 2: A The A-phase line current I Aa is given as VAn ■Aa

Z -(- Zj -I- Z^

120Z0C 3 + j 4 + 6 + j 8 -j- 2 —j 3

8 .4 4 Z -39.28° A

Assuming positive phase sequence ABC, the other two line currents are given as hb =

^ - 120°

= 8.44Z -159.28° A

l Cc = I Aa Z120° = 8.44Z80.72° A

134

Chapter 5

Balanced Three-Phase Circuits

Step 3: A From the single phase equivalent circuit the phase voltage Van is given as Van = v \ - IAa (Z + ZT) = 120Z0° - 8.44Z - 39.28° x (3 + j 4 + 6 + j8) = 30.43Z - 95.48° V A

Because the phase voltage Van is across the equivalent wye impedance per phase of the delta connected load, assuming positive phase sequence the phase voltage Vab is given as Vab = yf3Vm Z30° = V3 x 30.43Z - 95.48° x Z30° = 52.70Z - 65.48° V The other two phase voltages based on the positive phase sequence are given as Vbc =Vab/ l - 120° = 30.43Z - 65.48° x Z -120° = 52.70Z -185.48° V Vca = Vab Z120° = 30.43Z - 65.48° x Z120° = 52.70Z54.52° V Step 4: The phase current of the load can be determined from the line currents. A

Based on the positive phase sequence, the phase current I * is given as A

t

a

Iab = ^ Z 30° = V3

R 4 4 /

__ 3 0

J 73

78°

Z30° = 4.87Z - 9.28“ A

The other two phase currents based on the positive phase sequence are given as be A

Tabh Z -120° = 4.87Z - 9.28° x Z -120° = 4.87Z -129.8° A A

Ica = l ca Z120° = 4.87Z - 9.28° x Z120° = 4.87Z110.2° A

Solutions

5.15

135

Step 1: The single phase equivalent of the three-phase balanced system is as shown in Figure 5.15a. Z = 2+}2Q

Zj. = 3+}4 Q

V = 254 Z 0°

M

F igu re 5.15a

Step 2: Because the source and the induction motor are wye connected, the line and the phase currents are equal. Therefore, the phase current of the induction motor is given as /p

3(j>

N

Va„ COS I

50x10 Z - c o s l (p f)

254x0.8

Z -c o s ^ 1(0.8) = 8 1 .9 Z - 36.86° A

The other two line currents based on the positive phase sequence are given as I*Bb = l Aa Z -120° = 81.9Z - 36.86° x Z -120° = 81.9Z -156.86° A l Cc - IAa Z120° = 81.9Z - 36.86° x Z120° = 81.9Z83.140 A Step 3: A The phase voltage V • • at the terminals of the source is given as VAn = Vm + ha

z t

= 254Z0° + 81.9Z - 36.86° x 5Z53.13° = 657Z10° V

The line voltage V An, Z30° = V3 x 657Z100 x Z30° = 1138Z400 V

136

Chapter 5 Balanced Three-Phase Circuits

The other two line voltages at the terminals of the source are given as Fd A , = FA Bv Z -120° - 1 138Z400 x Z -120° = 1138Z - 80° V BC / C A , = vA B[ Z120° = 1138Z40° xZ120° = 1138Z1600 V Step 4: The internal phase voltage of the source is given as A A A v'An j,= = V.J vAn.. + 1AaZ = 657Z100 + 81.9Z - 36.8° x 2.828Z450 = 888Z9.5° V A

The internal line voltage VAB of the source based on positive phase sequence is given as VAB = S VAn Z30° = V3 x 888Z9.50 x Z30° = 1538Z39.5° V The other two internal line voltages of the source are given as VBC = VA3Z - 120° = 1538Z39.50 x Z -120° = 1538Z - 80.5° V VCA = VAB Z120° = 1538Z39.50 x Z120° = 1538Z159.50 V Step 5: The three-phase complex or apparent power supplied by the source is given as ^ = 3 ^ - IAa

3x888Z9.5° x81.9Z36.8°

218.1Z46.3° =150.7 + y'157.75 kVA Therefore, the real and reactive power supplied by the source are given as D(p = 150.7

kW

Qh = 157.75 kVAR 5.16 Step 1: The output power of the synchronous motor is given as Pout = 7]Pm = 0.8 x 75 x 103 = 60 kW The horsepower output of the motor is given as HP = j k = 60 x 1Q3 = 80.42 hp 746 746

Solutions

137

Step 2: The magnitude of the line current drawn from the source is given as / —-

- i-- 75X l° 3 —-1 --- Pi" --— X2 3 JA lV V3Fl cos6> V 3 x 440 x 0.8

JLt

5.17

Step 1: The input power of IMa is given as p, = C

in

746 _ 50 X 746 ??„ 0.95

^ HP, X

r?0

^

fcw

Step 2: The input line current of IM is given as A

J

—=—— ------Z - cos"1(pfa) = IS Z - cos"1(0.8) = 64.4Z - 36.86° A V3Fl c o s 0& V3x 440x0.8 Step 3: The input power of IMbis given as ^ b = / ^ = :H g Lx746 = 75 x 746 = 65 g2 ^ %

%

0.85

Step 4: The input line current of IMb is given as Ih =

------ Z - cos-1(pf b) = V3Fl cos0„

V

'

^ 5'8 2 x l ° 3 Z - cos-1(0.6) = 143.94Z - 53.13° A V ’

7 3 x 4 4 0 x 0 .6

Step 5: The line current at the source is given as IL = I a+ Ib = 64.4Z - 36.86° + 143.94Z - 53.13° = 206.54Z - 48.12° A Step 6: The overall power factor at the source is given as p f = cos 6 = cos (cf) —Lp) = cos(0° - (—48.12°)] = 0.66 lagging (since the current is lagging) 5.18

Step 1: The real power consumed by loada is given as Pa = |»Sa|cos6&= 6 0 x 0 .8 = 48 kW The reactive power consumed by loada is given as Q&=

Isin0a = 60 x 0.6 = 36 kVAR

138

Chapter 5 Balanced Three-Phase Circuits

Therefore the apparent or complex power consumed by loadais given as 5a = / >.+ y a = 4 8 + 736 kVA Step 2: The real power consumed by loadb is given as Pb = |£b| cos 9h = 80 x 0.6 = 48 kW The reactive power consumed by loadb is given as Qh = |Sb| sin 9h = 60 x sin (—cos-1 (0.6)) = —64 kVAR Therefore the apparent or complex power consumed by loadb is given as Sh =Ph + jQ h = A % -j6A kVA Step 3: The total apparent or complex power supplied by the source is given as s u = S, + Sh = (48 + ;36) + (48 - j64) = 96 - J2S = 100/ -16.26° kVA Step 4: The line current at the source is given as * /

$3

h =

,

V

100 x l0 3Z - 16.26°

131.21Z16.260 A

V 3x440Z0c

Note: The current leads the voltage since the source is consuming reactive power. Step 5: The overall power factor at the source is given as p f = cos 9 = cos (—ip) = cos (o° —16.26° ] = 0.96 (lead) 5.19 Step 1: The power factor angle of the induction motor at full load is given as 9m = cos-1 (pfm) = cos-1 (0.85) = 31.78° Step 2: The reactive power consumed by the induction motor is given as Qm = Pm tan9m = 5 0 x 103 x ta n (31.78°) = 30.97 kVAR Step 3: The new overall power factor angle is given as ^new = COST1(/?/new) = cos_1 (0.96) = 16.26c

Solutions

139

Step 4: The new reactive power supplied by the source is given as 0new

= Pm tan ^new = 50 x 103 x tan (16.26°) = 14.58 kVAR

Note: The real power consumed by the induction motor remains the same. Step 5: Because the induction motor consumes the same amount of reactive power, the reactive power that needs to be supplied by the capacitor bank in parallel is given as Qo = Qmw ~ Q m = 14.58 - 30.97 = -16.39 kVAR Note: The negative sign indicates that the reactive power is supplied. Step 6: If the capacitor bank is connected in delta to supply Qc, the magnitude of the phase current / of the capacitor bank is given as S,. 3Kph

L

-1 6 .3 9 x l0 3 3x440

12.41 A

The capacitive reactance per phase when connected in delta is given as 440 Ph ________ L

-12.41

-35.45 O /,

The capacitance per phase required to produce Qcwhen connected in delta is given as C= -

1 2ixfXc

1 2 x 7 rx 6 0 x (—35.45)

74.81 fjFI

Step 7: If the capacitor bank is connected in wye to supply Qc, the magnitude of the phase current I of the capacitor bank is given as 16.39 xlO3 3KPh

3x

-21.50 A

440

The capacitive reactance per phase when connected in wye is given as 440 X

KPh L

,s , = -1 1 .8 1 0 -21.50

140

Chapter 5

Balanced Three-Phase Circuits

The capacitance per phase required to produce Qcwhen connected in wye is given as C=

1 2ttJX c

1 2 x 7 rx 6 0 x (—11.81)

224.5 /iFlcf)

5.20 Step 1: The power factor angle at which the induction motor is operating is given as 6m = cos 1(pfm) = cos 1(0.8) = 36.86° Step 2: The reactive power consumed by the induction motor (an induction motor always consumes reactive power) is given as Qm = Pmtan [ p f ) = — L2— / _ cos- ‘ (0.8) = 0.625Z - 36.86° = 0.5 - /0.374 pu VlP/i 2.0 X 0.8 V ' A

The current / 2 through the load L2is given as A = - ^ - Z c o s " ‘ (p /2) = — Zcos-' (0. 6) = 1.666Z53.13‘’ = 1 . 0 + /1.332pu VLPh 2.0 x 0.6 y ’

Step 4: A The current Is drawn from the source is given as Is = / 1+ / 2 = (0 .5 -y*0.374)+ (1.0 + 7*1.332) = 1.5 + 7*0.958 = 1.779/32.56° pu Step 5: A The source voltage Vs is given as Vs =VL+ i s ZT = 2 .0 /0 ° +1.779/32.56° x 1.033/53.13° =2.80/40.48° pu

Solutions

155

Step 6: The complex power supplied by the source is given as A

¥

A

s = v. Is = 2.80/40.48° x 1.779Z - 32.56° = 4.981Z7.920 = 4.933 + y0.686 pu V

J

The complex power supplied by the source and the source voltage actual values are given as S = S(pu)Sbase = (4.936 + y0.686) x 10 x 103 = 49.36 +

7 6 .8 6

kVA

Vs = Vs (pu)Vbase = 2.80Z40.48° x 220 - 616Z40.48° V 6.8 Because the base values are of the three-phase system, the factor >/3 is not used in calculations based on per-unit system. Step 1: The source voltage expressed in per-unit is given as Vs V,b ase

Vs(Pu)

10.5 11

0.954 pu

The magnitude of the complex power delivered by the source expressed in per-unit is given as 5000

S{pu) Stase

0.833 pu

6000

Step 2: A The magnitude of the current / drawn from the source is given as / c

S __ 0.833 Vs ~ 0.954

0.8731 pu

Step 3: The transmission line impedance expressed in per-unit is given as ZT{pu)

ZT ^ base \

K sef cbase

{( ^\21 (l 1 x 10 )

0.148 pu

6000 xlO 3 V

/

The voltage drop due to the transmission line is given as Vdrop = ISZT = 0.8731 x 0.148 = 0.1292 pu

156

Chapter 6 Per-Unit Methods and Calculations

6.9

Step 1: Because the change in base is only in the voltage, the per-unit impedances on the new base is given as m ew

v>base

r y o ld

ly n e w base

new

Z rp

___

---

old V,base

r jo ld

Aj'J'

^L — ^L

/ 440>'2 = (8.88 + y'2 2 .2 2 ) x 1 0 " 3 pu 660

(0.01 + 7 O.OI) x

440 660

\2

rrn ew \ base j

old V,base

r jn e w __ r jo ld

(0.02 + ;0.05)x

(0.04 + y'1.5) x

ty n e w

base

= (4.4 + y'4.4)xl0 ^3 pu

/ 440n2 660

(0.0177 + y0.66) pu

Step 2: The actual values of the impedances on the old base is given as Z = Z f ( p u ) Z ° !ai = Z f ( p i i )

(0.02 + y'0.05):

ZT = Z f { p u ) Z t L = Z f { p u )

(0.01 + yO.Ol): (

ZL = Z i d{ p u ) Z t L = Z i \ p u )

,

v? old NZ vh b ase -iold base

(440)

(440)'

0.387 + 70.387 O

5x10(440)

(0.04 + yl.5)

0.774 + j'1.936 O

5 xlO 3

5 x 10

1.548 + 758.08 O

Step 3: The actual values of the impedances on the new base is given as v2 rrnew (660) " base (8.88 + 7*22.22) x 10“ 0.774 + 71.936 Q Z = Z ™ { p u ) Z Z e = Z nr ( Pu) nnew 5xl0~ ^base (

zT=zr(pu)zfz =zr(pu)

'

new V,base

\2

anew

(4.4 +

7 4 .4 ) xlO-

base

zL=zripu)zze=zr{pu)

jr n e w base an ew ^ base

(0.0177 + 7O.66)

(660)2 5 xlO 3 (660)2 5 xlO 3

0.383 + 70.383 O

1.54 + 757.499 O

6.10 Step 1: Because the change in base is only in the kVA, the per-unit impedance on the new base is given as r^new _

an ew base

old

s,base

(0.05 + 7O.O2)

10x10 100 xlO3

5 + j2 pu

Solutions

157

Step 2: The actual impedance based on the old base is given as old

ryold .

Z f(p u )Z Z = Z r(p u )

old v> base old s base

= (0.05 + y0.02)x

11x10-

2

100 x io 3

60.5 + j 24.2 Q

Step 3: The actual impedance based on the new base is given as ( /jr n e w \2 base

zi = z r W

c = r W

rtnew ^ b a se

11x10 —(5 + j2)

10x10

60.5 + ;24.2 Q,

1—k o o

6.11 Because the base is changed in both voltage and kVA, the per-unit impedance per phase on the new base is given as 2 x2 rinew '14.4' Kbase mew __ y o ld base X 0.051 + y0.078 pu = (0.015 + 70.023) it — Zj T ~told rrnew 150 j I 11 base base

J

6.12 Step 1: The base values based on the low voltage side are given as V.base = 110 V and S. base= 2.5 kVA The leakage reactance in per-unit with respect to the low voltage side is given as v l, , X[ X\ 0.06 X L(pu) = - — = -------------------------------- ^ = —---- ^ - = 0.0124 pu f c j (n o ) Sbase 2.5 xlO3 Step 2: The turns ratio a of the transformer is given as N N

K V)

440

110 0.25

Step 3: The leakage reactance referred to the high voltage side is given as X ^ = ^ f = 5 ^ = 0.96 0 a1 .25 Step 4: The base values based on the high voltage side are given as v base = 440 V and Sh =2.5kVA base

158

Chapter 6

Per-Unit Methods and Calculations

The leakage reactance in per-unit with respect to the high voltage side is given as -2 /,_\ Xt XI ^ ( / W) = ^ = 7^

0.96 = — ^ — = 0.0124 pu K s e ) (440) Sbase 2.5 x 103

Note: The leakage reactance in per-unit is the same on both the low and high voltage sides of the transformer. 6.13 Step 1: The base values based on the secondary ratings of T1 are given as V.base = 66 kV and S.base =10kVA Step 2: The turns ratio a of the transformer T1 is given as JVj N2

Vx V2

6.6 66

0.1

The turns ratio a of the transformer T2 is given as -i'y -— N± = — ^ L—- 66 --- 15 Nh0 V2 V., 4.4 The base values referred to the secondary of T2 is given as — = ^ = 4.4 kV a2 15 Step 3: The load impedance in per-unit referred to the secondary of T2 is given as z l {pu ) =

~ ^=

~ 2 = - ----4 ~° 2 —0-207 pu K l) 4.4 xlO 3 Stm

1 0 x l0 3

Step 4: The load impedance ZL referred to the primary side of T2, or the secondary side of Tl, is given as ZL — (a2f ZL = 152 X 400 = 90 kO Step 5: The per-unit impedance of ZLreferred to the primary side of T2, or the secondary side of Tl, is given as „2 zl

/ \ Z l2 Z\ 90 x 103 \PU) = = ----- ~ r = ------------ r = °-207 Pu b ase

(V b a se

)

^ b a se

(66X103

10X103

Solutions

159

Step 6: The load impedance ZLreferred to the primary side of T1 Z^ ^ ( ^ f z 2 =0.12 x 9 0 x l0 3 = 0.9 kO Step 7: The base values referred to the primary of T1 is given as C

=

=0 .1 x 66 = 6.6 kV

Step 8: The per-unit impedance of ZLreferred to the primary side of T1 is given as 0.9 xlO3

Z l(p u )~ Z L ' (K lf

0.207 pu

(6.6 xlO3'2 10 xlO3

6.14 Step 1: The leakage reactance XLl of T1 in per-unit expressed in the new base is given as 2

X

new __ old LI ~ A L\

base iold base

/

base •rrnew base ,

0.05 x

\

/

440 U -5 x l0

v6.6,

0.00326 pu

Step 2: The base voltage on the secondary side of T1 is given as Tl Vtbase

__

z: 'vbase ax

1.5 xlO3 440 [4400

15 kV

Step 3: The transmission line impedance ZTin per-unit is given as 6 + y5

ZT (pu)

(1.33+ yl. 11)xlO"4 pu

15 xlO3

base anew base

5 x l0 3

Step 4: The leakage reactance XL2 of T2 in per-unit expressed in the new base is given as ynew L2

__ ~

S;base &old

\rold A

I,2

\

base /

y o ld base rrnew ' base

(

'i2

= 0.08 x

\9-0/

-x

4.2 x 10 15 x 10

\2

Step 5: The base voltage on the secondary side of T2 is given as T2 Vbase l

VT h lase

15 xlO3 750 V 4200 210

0.00348 pu

Chapter 6 Per-Unit Methods and Calculations

Step 6: The load impedance ZLin per-unit is given as ^ M

="

z=

= -T—

K l) rrnew

f =

M

= ° '0267 + i'0'0356 P“

5x10

base

Step 7: The impedance diagram is as shown in Figure 6.14a.

6.15 The new values in part B of the system are given as Vfase = 10 kV and S w =10 kVA Step 1: The new base impedance in part B of the system is given as K sef (lOxlO3 z L e = 1~ = J~ = -----------r SZ, 1 0 x 10

= 10

kn

Step 2: The transmission line impedance in per-unit on the new base is given as

Z rW = l r

i ^

= (a 2 + /o -2)xi