16
ELECTRICAL MACHINES&
POWER SYSTEMS(Problems with Solutions)
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ELECTRICAL MACHINES&
POWER SYSTEMS(Problems with Solutions)
Dr. C.S. INDULKARB. Tech (Hons) (IIT Kharagpur), M.Sc. Tech. (Manchester), Ph.D. (Manchester)
Life Senior Member IEEE, F.I.E. (India)Associate of the Manchester College of Science & Technology
Former Professor & Head, Electrical Engineering Department (IIT Delhi)
S. CHAND & COMPANY LTD.(AN ISO 9001 : 2000 COMPANY)
RAM NAGAR, NEW DELHI – 110 055
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PREFACE
This book contains problems in Electrical Machines & Power Systems (Problems withsolutions). I have used these and other problems in the class - room for many years. In most of thesolutions I have deliberately avoided giving theoretical explanations, because an average studentshould know the theory well before attempting to solve any problem. However, in each chapter, Ihave provided a brief introduction related to the chapter so that the students are made aware of thecontents of the chapter before reading the problems and their solutions. The introduction related toeach chpapter contains objective type questions and their answers. The introductions contain briefnotes on the topics of the chapters and also include Indian Standards for testing and maintenance ofsubstation equipments, transformers, overhead lines, underground cables and materials. Theintroductions also contain the http://www addresses of the websites which include electrical machinesand power systems topics related to the chapter. After presenting the brief introductory notes on thetopic of each chapter, problems with detailed solutions are included in each chapter.
The book contains about 160 objective type questions and an equal number of solved problems.The contents of the book are also available on the author’s website: http://cindulkar.tropod.com/index.htm. The solutions to the problems have been developed on Excel spreadsheets. Therefore,educators can use the problems for examination or tutorial purposes, because it is easy to change theproblem data on the spreadsheet. Ready availability of the solutions for the new data is an advantageof using the spreadsheet.
To the student, who has this book or who is able to access these problems on the website, myadvice is to attempt a problem himself and compare the solution given in this book or the website. Ifthe answer is wrong, he should try to find, from his text-book, where he may have gone wrong andthen solve the problem a second time. If he is wrong again, then only he should refer to the solutiongiven in this book or the website. Thus used judiciously, I am sure; the book will be of great help tohim. Students preparing for University/Institute of Engineers (India) examinations or the LondonCity & Guilds and other professional examinations will find the book quite useful.
B3, Gokul Society AUTHORVasna RoadBaroda – 390 [email protected]
Disclaimer : While the authors of this book have made every effort to avoid any mistake or omission and have used their skill,expertise and knowledge to the best of their capacity to provide accurate and updated information. The author and S. Chand doesnot give any representation or warranty with respect to the accuracy or completeness of the contents of this publication and areselling this publication on the condition and understanding that they shall not be made liable in any manner whatsoever. S.Chandand the author expressly disclaim all and any liability/responsibility to any person, whether a purchaser or reader of this publicationor not, in respect of anything and everything forming part of the contents of this publication. S. Chand shall not be responsible forany errors, omissions or damages arising out of the use of the information contained in this publication.Further, the appearance of the personal name, location, place and incidence, if any; in the illustrations used herein is purelycoincidental and work of imagination. Thus the same should in no manner be termed as defamatory to any individual.
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CONTENTS SECTIONS PAGES
1. D. C. Machines 1 — 141.1 Introduction & objective questions with answers1.2 Problems with Solutions
2. Synchronous Machines 15 — 362.1 Introduction & objective questions with answers2.2 Problems with Solutions
3. Induction Motor 37 — 593.1 Introduction & objective questions with answers3.2 Problems with Solutions
4. Overhead Lines 60 — 884.1 Introduction & objective questions with answers4.2 Problems with Solutions
5. Underground Cables 89 — 1025.1 Introduction & objective questions with answers5.2 Problems with Solutions
6. Transformers 103 — 1136.1 Introduction & objective questions with answers6.2 Problems with Solutions
7. Circuit –breakers /Sub-stations 114 — 1267.1 Introduction & objective questions with answers7.2 Problems with Solutions
8. Power flows 127 — 1428.1 Introduction & objective questions with answers8.2 Problems with Solutions
9. Short-circuit calculations 143 — 1619.1 Introduction & objective questions with answers9.2 Problems with Solutions
10. Protection 162 — 17010.1 Introduction & objective questions with answers10.2 Problems with Solutions
11. Steady-state Stability 171 — 17811.1 Introduction & objective questions with answers11.2 Problems with Solutions
12. Transient Stability 179 — 19012.1 Introduction & objective questions with answers12.2 Problems with Solutions
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13. Over Voltages 191 — 19813.1 Introduction & objective questions with answers13.2 Problems with Solutions
14. Automatic Generation Control 199 — 20314.1 Introduction & objective questions with answers14.2 Problems with Solutions
15. Control of Voltage & Reactive Power 204 —21015.1 Introduction & objective questions with answers15.2 Problems with Solutions
16. Economic Operation 211 — 21716.1 Introduction & objective questions with answers16.2 Problems with Solutions
17. Per Unit System 218 — 22717.1 Introduction & objective questions with answers17.2 Problems with SolutionsIndex 228 — 229
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1.1 INTRODUCTIONElectricity, as is well known, is the most flexible form of energy that can be transported with
ease from one place to another for ultimate consumption by its user. In this scenario electricalmachines and transformers play a vital role whose importance cannot be overemphasized. Electricalmachines are energy converting links between electrical & mechanical networks. Therefore workingprinciples & characteristics of basic machines must be well understood by electrical engineers. Incase of DC machines, the following topics must be covered:
Provision of self and separate excitation. Ability of the same machine to work as a generator and a motor. Brief reference to the field of application of D.C. Machines.
1.1.1 Constructional features & types of D.C. MachinesDescribe by sketches.
1.1.2 WindingsSimple lap & wave windings, their characteristics, winding calculations.
1.1.3 E.m.f and Equations TorqueE.m.f. equation, open circuit characteristics, voltage build -up process for generator, failure
of shunt generators to build up, critical resistance.Torque equation; torque speed characteristics of shunt, series & compound wound motors.1.1.4 Commutation & Armature Reaction
Commutation: Retarded, accelerated and linear commutation. Methods of improvingcommutation.
Armature Reaction: shifting of magnetic neutral plane, effect of brush shift, calculation ofdemagnetizing & cross magnetizing ampere- turns, compensating winding, effect of armaturereaction on external characteristics.
Students should be aware of the conventional e.m.f. and torque expressions (E= Z n.p/A)volts and T = (1/2) Z I (p/A) N-m.1.1.5 Characteristics of Series, Shunt, and Compound Wound Generators & Motors
Load characteristics of shunt generators (external & internal), their pre-determination fromgiven open-circuit characteristics and armature resistance drops.1.1.6 Starting & Speed Control
Starting of D.C. motors, starters.Basic principles of speed control of a D.C. machine, flux and armature voltage control, speed
control of shunt, series & separately excited motors.1.1.7 Parallel Operation
Parallel operation of shunt and compound wound generators.
C H A P T E R 1D . C . M A C H I N E S
1
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2 Electrical Machines & Power Systems
1.1.8 Testing & EfficiencyLosses in D.C. machines, efficiency calculation from measurement of losses (Swinburne’s
method, Hopkinson Test, Field’s Test).1.1.9 Dynamics of D.C. Machines
This Excel spreadsheet module uses a separately excited D.C. motor. The armature inductanceis included in the equivalent circuit. The student can study the response of the motor armaturecurrent & speed with respect to time for a sudden change in load or applied voltage. The studentcan vary the parameters like damping (D), armature inductance (L), moment of inertia (J), armatureresistance Ra, applied voltage (V), and the load torque (TL). The student observes the effect ofparameter variations on the steady –state & transient response of the system. The response can beobtained either by the Laplace Transform method or by the Runge-Kutta routine.
This module uses a separately excited D.C. motor. The armature inductance is included inthe equivalent circuit. The student can study the response of the motor with a sudden change inload or applied voltage. The responses include the variation of armature current and speed withrespect to time. For example, the student varies the parameters like damping (D), armatureinductance (L), moment of inertia (J), armature resistance (Ra), applied voltage (V), and the loadtorque (TL).Data: Parameters:
Ra = 0.09724 , La = 10.4 mH, J = 0.068, K = 1.8, D = 0, TL = 0,V = 240 V, Ia(0) = 0, Wm (0) = 0, Max. time = 0.2 s.
The student instantly sees the effect of these parameters on the transient and steady stateresponse of the motor. The student can choose either the Laplace transform or the numericalRunge-Kutta algorithm to obtain the response.THEORY
(s)/V(s) = K/[La J.s2 + (La. D + Ra J)s + Ra D + K Kb]
eb = pdKdt
T = KiaElectrical power eb.ia=mech.power, d
Tdt
Therefore K=KbSudden Change in Applied Voltage, D = 0
K.V = 432, La.J = 0.0007, Ra.J = 0.0066, K2=3.24(s) = K.V/s[La.J.s2 +Ra.Js + K2] = K.Ea(s)[La.Js2 + Ra.Js + K2] ... (1)(s) = 432/s[.000707s2 + .006612s + 3.24]
F
0
50
100
150
00.05 0.1 0.15 0.2
Time
angular speed
Fig. 1.1(a). Angular speed Vs. Time
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D.C. Machines 3
Angular speed (t) is calculated by solving the above equation using Laplace inverse transform.The result is shown in Fig.1.1.9(a).
Armature CurrentIa(s)/ (s) =Js2/K, D = 0Ia(s)/(s) = J.s/k ... (2)
Product of w(s)/Ea(s) and Ia(s)/(s) gives (from equations (1) & (2) above)Ia(s)= (V/La)*1/s(s2 + Ra.s/La + K2/(La.J))
V/La = 0.0231Ra/La = 9.35
K2/(La*J) = 4581.4Ia(s) = 023077*1/s(s2 + 9.35s + 4581.448)
Ia (t) is obtained from the above equation and the response is given in Fig. 1.1.9(b).
0.000005
0.00001
0.000015
0.00002
0 0.08 0.15Time
Armature Current
Fig.1.1.9(b). Armature current Vs. Time
Transient BehaviourBehaviour under transient conditions is normally approached through the general theory,
typical conditions being sudden load or short-circuits on a generator, starting or sudden loadchange of a motor or a sudden change in voltage applied to a field winding.
Such calculations involve extensive analytical work. Some cases can however lead to a simplenumerical solution, as in Prob.1.2.12 below.Saturation
The theoretical characteristics may have to be significantly modified in practice owing tosaturation in the magnetic circuit, especially where the flux changes over a wide range as in aseries motor.
The basis for calculations involving the effects of saturation is normally the magnetizationcurve. The Prob. 1.2.11 below illustrates the procedure for a typical problem.
Other problems in which saturation plays an essential part include the self-excitation of ad.c. generator, effects of high armature reaction and generators with special voltage/currentcharacteristics, e.g., welding generators.Single - Phase Machines Of the many types of 1- phase commutator motors that have been devised and used in the past,the only two that have retained commercial importance are the 1-phase series motor and therepulsion motor. The 1- phase series motor is used on some railways for traction purposes and requires acompensating winding (series connected or short- circuited) to give a satisfactory performance. Much more common motor is the fractional -kW motor which can run successfully on eithera.c. or d.c. and is manufactured on a mass-production basis. Prob.1.2.13 below gives an exampleof a fractional-kW motor.
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4 Electrical Machines & Power Systems
OBJECTIVE QUESTIONS1. The speed of a d.c. series motor at no load is
(a) Zero(b) Infinity(c) 1500 r.p.m.Ans. (b)
2. A sinusoidal voltage of frequency 1 Hz is applied to the field of a d.c. generator.The armature voltage will be(a) 1 Hz square wave(b) 1 Hz sine wave(c) d.c. voltageAns. (b)
3. The function of the commutator in a d.c. machine is(a) to change a.c. to d.c.(b) to change alternating voltage to direct voltage(c) for easy control of voltageAns. (b)
4. The direction of rotation of a d.c. series motor can be reversed(a) by interchanging the supply terminals(b) by interchanging the field terminals(c) either (a) or (b).Ans. (b)
5. A d.c. shunt generator driven at normal speed in the normal direction fails to build uparmature voltage because(a) the armature resistance is high(b) the field current is not sufficiently high(c) there is no residual magnetism.Ans. (c)
6. If a d.c. motor is connected across an ac supply, the dc motor will(a) burn as the eddy currents in the field produce heat(b) run at its normal speed(c) run at a lower speed.Ans. (a)
7. What would happen if the field of dc shunt motors is opened ?(a) the speed of the motor will be reduced(b) it will run at its normal speed(c) the speed will be enormously high, and the motor will destroy itself.Ans. (c)
8. Which one of the following motors is used in a locomotive ?(a) Induction motor(b) dc series motor(c) Synchronous motor.Ans. (b)
9. The efficiency of a d.c. shunt generator has a maximum value when.(a) magnetic losses are equal to mechanical losses
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D.C. Machines 5
(b) armature copper loss is equal to the constant loss(c) field copper loss is equal to the constant loss.Ans. (b)
10. The dummy coil in the d.c. machine is used to(a) eliminate armature reaction(b) bring about mechanical balance of armature(c) eliminate reactance voltage.Ans. (b)
11. The sparking at the brushes in a d.c. generator is attributed to(a) quick reversal of current in the coil under commutation(b) armature reaction(c) reactance voltageAns. (c)
12. Interpoles in the armature of a d.c. generator are used to(a) neutralize the reactance voltage(b) neutralize the reactance voltage and the cross-magnetizing effect of the armaturereaction(c) neutralize the cross-magnetizing effect of the armature reaction.Ans. (b)
13. The mechanical power developed by the d.c. motor is maximum when(a) back e.m.f. = applied voltage(b) back e.m.f. = 0(c) back e.m.f. = 0.5 times the applied voltage.Ans. (c)
14. In series –parallel control method when two d.c. series motors are connected in series, thespeed of the set is(a) half of the speed of he motors when connected in parallel(b) ¼ th of the speed of the motors when connected in parallel(c) same as in parallel.Ans. (b)
15. The armature torque of the d.c. shunt motor is proportional to(a) armature current only(b) field flux only(c) armature current and flux both.Ans. (c)
16. The reversal of speed of motors used as electric braking is known as(a) regenerative braking(b) plugging(c) dynamic braking.Ans. (b)
17. The method used for braking in rolling mills is(a) plugging(b) regenerative braking(c) dynamic braking.Ans. (a)
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6 Electrical Machines & Power Systems
PROBLEMS WITH SOLUTIONS
Prob.1.2.1 D.C. Generator-Back e.m.f.When a d.c. generator is being driven at 1200 rpm, the generated e.m.f. is 125 V. What will
be the generated e.m.f. if (a) the field flux is decreased by 10% with the speed remaining unchangedand (b) if the speed is reduced to 1100 RPM ,the field flux remaining unchanged ?
Solution. (a)E1= 1251= 12= 0.9 × 1E2= E1× 2/1 = 0.9 × E1 = 112.5 Volts
Answer(b) N1= 1200 rpm
N2 = 1100 rpmE2 = E1× N2/N1 = 114.583 Volts Answer
Prob.1.2.2 D.C. Generator-generator & motor ActionA shunt machine has armature and field resistances of 0.04 and 100 respectively.
When connected to a 460 V d.c. supply and driven as a generator at 600 r.p.m, it delivers 50 kW.Calculate its speed when running as a motor and taking 50 kW from the same supply. Show thatthe direction of rotation of the machine as a generator and as a motor under these conditions isunchanged.Solution: P1, Generator power = 50000 W
V, Terminal Voltage = 460 VI, Generator line current = P1/V = 108.695 ARa = 0.04 Rf = 100 I, Motor line current = 108.695 AIf 2= Motor field current = V/Rf = Generator field current = 4.6 AIa1, Generator armature current = I + If 2= 113.295E1= Generator back e.m.f. = V + Ia1Ra = 464.531 VIa2= Motor armature current = I – If 2= 104.095 AE2, Motor back e.m.f. = V – Ia2 Ra=455.836Speed is proportional to back e.m.f.N1, Generator speed = 600 r.p.m.N2, Motor speed = N1(E2/E1) = 588.768 r.p.m.
The direction of field flux remains the same when operating as a motor or a generator.However, the direction of the armature current changes in the two modes of operation. Therefore,according to Fleming’s Left hand Rule for motor operation and the Right-Hand Rule for the generatoroperation, the direction of rotation is unchanged.Prob.1.2.3 D.C. Motor-Output Power & Torque
A 240 V d.c. motor has an armature resistance of 0.68 ohm and draws a a full-load currentof 2 4 A at a speed of 100 r.p.m. Calculate (a) the back e.m.f. (b) the output power developed, and(c) the torque developed.
Solution.V = 240 VIa = 24 ARa = 0.68 E, back e.m.f. = V-IaRa = 223.68 Volts Answer
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D.C. Machines 7
P, power developed = E × Ia = 5368.32 Watts AnswerN, Speed = 1000 RPM
T, Torque = 602
PN = 51.263 N-m Answer
Prob.1.2.4 D.C. Shunt motor-Effects of flux & torque on current and speedA shunt motor connected across a 440 V supply takes an armature current of 20 A and runs
at 500 r.p.m. The armature resistance is 0.6 . If the magnetic flux is reduced by 30 % and thetorque developed by the armature increases by 40 % ,what are the values of the armature currentand of the speed ?
Solution.V = 440Ia1 = 20Ra = 0.6E1 = V-Ia1× Ra = 428 Volts1 = 12 = 0.7 1E2 = New back e.m.f.E2 = E1× 2/1= E1× 0.7 = 299.6 Volts AnswerT1, Original torque = T1T2, New Torque = 1.4 T1Ia2 = New CurrentN1 = Original speed 500 r.p.m.E1, original back e.m.f. = E1T, Torque is proportional to E × Ia/N. Therefore,T2/T1= (E2/E1)(Ia2/Ia1)(N1/N2),orIa2/N2= (T2/T1)(Ia1/N1)(E1/E2) = 0.08 . . . (1)E2 = V-Ia2Ra.Therefore,Ia2 = (V-E2)/Ra = 234 AmpsE is proportional to and N.Therefore, N2 = N1(E2 / E1)(1/2) = N1((V – Ia2Ra)/E1)(1/2)
= 500((440 – Ia2 × 0.6)/428)(1/0.7)or,(440 – Ia2× 0.6)/N2 = 0.599 . . . (2)From equations (1) and (2),(440-Ia2 × 0.6)/Ia2 = 7.49or,Ia2 = 54.388 AFrom Eq .1,N2 = 679.851 rpm Answer
Prob.1.2.5 D.C. Generator-long Shunt CompoundA 4 pole 500V 25 kW long-shunt compound generator delivers full-load at the rated voltage.Calculate the emf generated, if the armature resistance is 0.03 , series field resistance is 0.04 ,and the shunt field resistance 200 . Contact drop per brush is 0.9 volts. Neglect armature reaction.
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Electrical Machines And PowerSystems (Problems With Solutions)
Publisher : SChand Publications ISBN : 9788121939010 Author : C. S. Indulkar
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